Table of Contents – Part 1 i Table of Contents – Part 1 1 | The Study of Life .............................................................................................................. 1 Review Questions ......................................................................................................................................................1 Critical Thinking Questions ........................................................................................................................................4 Test Prep for AP® Courses .........................................................................................................................................6 2 | The Chemical Foundation of Life ....................................................................................... 9 Review Questions ......................................................................................................................................................9 Critical Thinking Questions ......................................................................................................................................14 Test Prep for AP® Courses .......................................................................................................................................18 Science Practice Challenge Questions .....................................................................................................................20 3 | Biological Macromolecules ............................................................................................. 25 Review Questions ....................................................................................................................................................25 Critical Thinking Questions ......................................................................................................................................34 Test Prep for AP® Courses .......................................................................................................................................41 Science Practice Challenge Questions .....................................................................................................................48 4 | Cell Structure ................................................................................................................. 58 Review Questions ....................................................................................................................................................58 Critical Thinking Questions ......................................................................................................................................65 Test Prep for AP® Courses .......................................................................................................................................71 Science Practice Challenge Questions .....................................................................................................................76 5 | Structure and Function of Plasma Membranes ............................................................... 85 Review Questions ....................................................................................................................................................86 Critical Thinking Questions ......................................................................................................................................89 Test Prep for AP® Courses .......................................................................................................................................95 Science Practice Challenge Questions ...................................................................................................................100 Advanced Placement Biology Instructor’s Solution Manual Copyright 2018, Rice University. 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This ancillary resource is intended for instructor use only and may not be uploaded, redistributed, or reproduced without written approval by OpenStax. ii Table of Contents – Part 1 6 | Metabolism ................................................................................................................. 105 Review Questions ..................................................................................................................................................105 Critical Thinking Questions ....................................................................................................................................114 Test Prep for AP® Courses .....................................................................................................................................121 Science Practice Challenge Questions ...................................................................................................................125 7 | Cellular Respiration ...................................................................................................... 136 Review Questions ..................................................................................................................................................136 Critical Thinking Questions ....................................................................................................................................143 Test Prep for AP® Courses .....................................................................................................................................149 Science Practice Challenge Questions ...................................................................................................................155 8 | Photosynthesis............................................................................................................. 163 Review Questions ..................................................................................................................................................163 Critical Thinking Questions ....................................................................................................................................169 Test Prep for AP® Courses .....................................................................................................................................173 Science Practice Challenge Questions ...................................................................................................................179 9 | Cell Communication ..................................................................................................... 186 Review Questions ..................................................................................................................................................186 Critical Thinking Questions ....................................................................................................................................191 Test Prep for AP® Courses .....................................................................................................................................197 Science Practice Challenge Questions ...................................................................................................................201 10 | Cell Reproduction ....................................................................................................... 208 Review Questions ..................................................................................................................................................208 Critical Thinking Questions ....................................................................................................................................213 Advanced Placement Biology Instructor’s Solution Manual Table of Contents – Part 1 iii Test Prep for AP® Courses .....................................................................................................................................218 Science Practice Challenge Questions ...................................................................................................................225 11 | Meiosis and Sexual Reproduction ............................................................................... 229 Review Questions ..................................................................................................................................................229 Critical Thinking Questions ....................................................................................................................................232 Test Prep for AP® Courses .....................................................................................................................................237 Science Practice Challenge Questions ......................................................................................................................... 12 | Mendel's Experiments and Heredity ........................................................................... 240 Review Questions ..................................................................................................................................................240 Critical Thinking Questions ....................................................................................................................................246 Test Prep for AP® Courses .....................................................................................................................................249 Science Practice Challenge Questions ...................................................................................................................254 13| Modern Understandings of Inheritance ....................................................................... 265 Review Questions ..................................................................................................................................................265 Critical Thinking Questions ....................................................................................................................................268 Test Prep for AP® Courses .....................................................................................................................................270 Science Practice Challenge Questions ...................................................................................................................275 14 | DNA Structure and Function ....................................................................................... 280 Review Questions ..................................................................................................................................................280 Critical Thinking Questions ....................................................................................................................................291 Test Prep for AP® Courses .....................................................................................................................................303 Science Practice Challenge Questions ...................................................................................................................312 15 | Genes and Proteins .................................................................................................... 320 Review Questions ..................................................................................................................................................320 Critical Thinking Questions ....................................................................................................................................326 Advanced Placement Biology Instructor’s Solution Manual Copyright 2018, Rice University. 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This ancillary resource is intended for instructor use only and may not be uploaded, redistributed, or reproduced without written approval by OpenStax. iv Table of Contents – Part 1 Test Prep for AP® Courses .....................................................................................................................................332 Science Practice Challenge Questions ...................................................................................................................336 16 | Gene Regulation......................................................................................................... 346 Review Questions ..................................................................................................................................................346 Critical Thinking Questions ....................................................................................................................................352 Test Prep for AP® Courses .....................................................................................................................................359 Science Practice Challenge Questions ...................................................................................................................367 17 | Biotechnology and Genomics ..................................................................................... 369 Review Questions ..................................................................................................................................................369 Critical Thinking Questions ....................................................................................................................................376 Test Prep for AP® Courses .....................................................................................................................................381 Science Practice Challenge Questions ...................................................................................................................384 18 | Evolution and Origin of Species .................................................................................. 389 Review Questions ..................................................................................................................................................389 Critical Thinking Questions ....................................................................................................................................395 Test Prep for AP® Courses .....................................................................................................................................404 Science Practice Challenge Questions ...................................................................................................................423 19 | The Evolution of Populations ...................................................................................... 434 Review Questions ..................................................................................................................................................434 Critical Thinking Questions ....................................................................................................................................438 Test Prep for AP® Courses .....................................................................................................................................444 Science Practice Challenge Questions ...................................................................................................................452 20 | Evolutionary Relationships of Life on Earth ................................................................. 459 Review Questions ..................................................................................................................................................459 Critical Thinking Questions ....................................................................................................................................466 Advanced Placement Biology Instructor’s Solution Manual Table of Contents – Part 1 v Test Prep for AP® Courses .....................................................................................................................................469 Science Practice Challenge Questions ...................................................................................................................479 21 | Viruses ....................................................................................................................... 486 Review Questions ..................................................................................................................................................486 Critical Thinking Questions ....................................................................................................................................493 Test Prep for AP® Courses .....................................................................................................................................498 Science Practice Challenge Questions ...................................................................................................................504 22 | Prokaryotes: Bacteria and Archaea ............................................................................. 510 Review Questions ..................................................................................................................................................510 Critical Thinking Questions ....................................................................................................................................520 Test Prep for AP® Courses .....................................................................................................................................531 Science Practice Challenge Questions ...................................................................................................................542 23 | Plant Form and Physiology ......................................................................................... 553 Review Questions ..................................................................................................................................................553 Critical Thinking Questions ....................................................................................................................................570 Test Prep for AP® Courses .....................................................................................................................................582 Science Practice Challenge Questions ...................................................................................................................590 24 | The Animal Body: Basic Form and Function ................................................................. 602 Review Questions ..................................................................................................................................................602 Critical Thinking Questions ....................................................................................................................................607 Test Prep for AP® Courses .....................................................................................................................................612 25 | Animal Nutrition and the Digestive System ................................................................. 623 Review Questions ..................................................................................................................................................623 Critical Thinking Questions ....................................................................................................................................631 Test Prep for AP® Courses .....................................................................................................................................639 Science Practice Challenge Questions ...................................................................................................................650 Advanced Placement Biology Instructor’s Solution Manual Copyright 2018, Rice University. 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This ancillary resource is intended for instructor use only and may not be uploaded, redistributed, or reproduced without written approval by OpenStax. vi Table of Contents – Part 1 26 | The Nervous System ................................................................................................... 652 Review Questions ..................................................................................................................................................652 Critical Thinking Questions ....................................................................................................................................654 Test Prep for AP® Courses .....................................................................................................................................657 Science Practice Challenge Questions ...................................................................................................................668 27 | Sensory Systems......................................................................................................... 680 Review Questions ..................................................................................................................................................680 Critical Thinking Questions ....................................................................................................................................688 Science Practice Challenge Questions ...................................................................................................................696 28 | The Endocrine System ................................................................................................ 701 Review Questions ..................................................................................................................................................701 Critical Thinking Questions ....................................................................................................................................710 Test Prep for AP® Courses .....................................................................................................................................717 Science Practice Challenge Questions ...................................................................................................................725 29 | The Musculoskeletal System....................................................................................... 731 Review Questions ..................................................................................................................................................731 Critical Thinking Questions ....................................................................................................................................738 Science Practice Challenge Questions ...................................................................................................................743 30 | The Respiratory System .............................................................................................. 748 Review Questions ..................................................................................................................................................748 Critical Thinking Questions ....................................................................................................................................753 Test Prep for AP® Courses .....................................................................................................................................758 Science Practice Challenge Questions ...................................................................................................................764 31 | The Circulatory System ............................................................................................... 766 Review Questions ..................................................................................................................................................766 Critical Thinking Questions ....................................................................................................................................776 Advanced Placement Biology Instructor’s Solution Manual Table of Contents – Part 1 vii Test Prep for AP® Courses .....................................................................................................................................780 Science Practice Challenge Questions ...................................................................................................................785 32 | Osmotic Regulation and Excretion .............................................................................. 791 Review Questions ..................................................................................................................................................791 Critical Thinking Questions ....................................................................................................................................798 Test Prep for AP® Courses .....................................................................................................................................807 33 | The Immune System ................................................................................................... 823 Review Questions ..................................................................................................................................................823 Critical Thinking Questions ....................................................................................................................................831 Test Prep for AP® Courses .....................................................................................................................................838 Science Practice Challenge Questions ...................................................................................................................848 34 | Animal Reproduction and Development ..................................................................... 852 Review Questions ..................................................................................................................................................852 Critical Thinking Questions ....................................................................................................................................861 Test Prep for AP ® Courses ....................................................................................................................................869 Science Practice Challenge Questions ...................................................................................................................878 Advanced Placement Biology Instructor’s Solution Manual Copyright 2018, Rice University. All Rights Reserved. This ancillary resource is intended for instructor use only and may not be uploaded, redistributed, or reproduced without written approval by OpenStax. 1 | The Study of Life 1 1 | THE STUDY OF LIFE REVIEW QUESTIONS 1 What is a suggested and testable explanation for an event called? A Discovery B Hypothesis C Scientific method D Theory Solution 2 The solution is (B). A hypothesis is a testable explanation for an event. Which branch of science is NOT considered a natural science? A Astronomy B Biology C Computer science D Physics Solution 3 The solution is (C). Computer science is a practical and scientific approach to computation and its application. It is not considered a natural science. What is the name for the formal process through which scientific research is checked for originality, significance, and quality before being accepted into the scientific literature? A Publication B Public speaking C Peer review D Scientific method Solution 4 The solution is (C). Peer review is the process in which scientific papers are reviewed by experts in the scientist’s field to ensure that a scientist’s research is original, significant, logical, and thorough. What are (i) two topics that are likely to be studied by biologists and (ii) two areas of scientific study that would fall outside the realm of biology? A (i) diseases affecting humans, pollution affecting species’ habitat; (ii) calculating surface area of rectangular ground, functioning of planetary orbits B (i) calculating surface area of rectangular ground, functioning of planetary orbits; (ii) formation of metamorphic rocks, galaxy formation and evolution Advanced Placement Biology Instructor’s Solution Manual 2 1 | The Study of Life C (i) plant responses to external stimuli, functioning of planetary orbits; (ii) formation of metamorphic rocks, galaxy formation and evolution D (i) plant responses to external stimuli, study of the shape and motion of physical objects; (ii) formation of metamorphic rocks, galaxy formation and evolution Solution 5 The solution is (A). Diseases affecting humans and pollution affecting species are two topics that can be studied by biologists, while calculating the surface area of a rectangular plot of land and the function of planetary orbits are two topics that fall outside of the realm of biology. Which statement is an example of deductive reasoning? A Most swimming animals use fins; therefore, fins are an adaptation to swimming. B Mitochondria are inherited from the mother; therefore, maternally inherited traits are encoded by mitochondrial DNA. C Small animals lose more heat than larger animals. One would not expect to find wild mice in the North and South Poles. D Water conservation is a major requirement to survive in the desert. Large leaves increase loss of water by evaporation. Therefore, desert plants should have smaller leaves. Solution 6 The solution is (D). Water conservation is a major requirement to survive in the desert. On plants, broad, thin leaves lose more water to evaporation than smaller, thicker leaves. Therefore, desert plants should have smaller leaves. The answer best demonstrates how a general principle or law is used to forecast specific results. Why are viruses NOT considered living? A Viruses are not made of cells. B Viruses do not have genetic material. C Viruses have DNA and RNA. D Viruses are obligate parasites and require a host. Solution 7 The solution is (A). Viruses are not made up of cells. They are made up of DNA or RNA surrounded by a protein coat. Biologists characterize living organisms as being composed of cells. The presence of a membrane-enclosed nucleus is a characteristic of what? A Bacteria B Eukaryotic cells C All living organisms D Prokaryotic cells Solution The solution is (B). Unlike prokaryotic cells, eukaryotic cells contain a membranebound nucleus. Advanced Placement Biology Instructor’s Solution Manual 1 | The Study of Life 8 3 What is a group of individuals of the same species living in the same area called? A A community B An ecosystem C A family D A population Solution 9 The solution is (D). A population is a group of individuals that belong to the same species, live in a particular geographical area, and have the capability to interbreed. Which sequence represents the hierarchy of biological organization from the most inclusive to the least complex level? A Biosphere, ecosystem, community, population, organism B Organelle, tissue, biosphere, ecosystem, population C Organism, organ, tissue, organelle, molecule D Organism, community, biosphere, molecule, tissue, organ Solution 10 The solution is (A). The biosphere is the highest level, which is the most inclusive, and the organism is the lowest level and least inclusive in the given hierarchy. Where in a phylogenetic tree would you expect to find the organism that had evolved most recently? A At the base B At the nodes C At the branch tips D Within the branches Solution 11 The solution is (C). The "tips" of the tree branches represent the recently evolved taxa. What is a characteristic that is NOT present in all living things? A Homeostasis and regulation B Metabolism C Nucleus containing DNA D Reproduction Solution The solution is (C). A nucleus containing DNA is not present in all living organisms. Advanced Placement Biology Instructor’s Solution Manual 4 1 | The Study of Life CRITICAL THINKING QUESTIONS 12 Is mathematics a natural science? Explain your reasoning. A No, it is not a natural science because it is not used in the study of the natural world. B No, it is not a natural science. Mathematics focuses on understanding mathematical relations and calculations, which is useful in natural sciences but which is distinct. C Yes, it is a natural science. Mathematics deals with verifying the experimental data. D Yes, it is a natural science. It uses chemical and physical measurements. Solution 13 The solution is (B). No, it is not a natural science. Mathematics does not involve measuring aspects of the natural world. Although the scientific method is used by most of the sciences, it can also be applied to everyday situations. A situation is given below. Using the scientific method try to arrange the given steps in the correct order. Situation: 1. If the car doesn’t start the problem might be in the battery. 2. Car doesn’t start. 3. After changing the battery. Car starts working. 4. The car should start after charging the battery or changing the battery. 5. The car doesn’t start because the battery is dead. 6. The car doesn’t start even after charging the battery, the battery must have stopped working. A 1, 2, 3, 4, 5, 6 B 2, 1, 3, 4, 5, 6 C 2, 1, 5, 4, 6, 3 D 2, 1, 5, 6, 3, 4 Solution 14 The solution is (B). 2, 1, 5, 4, 6, 3 is the correct sequence of the situation given. Read questions 1 through 4. Question 1. Is macaroni and cheese tastier than broccoli soup? Question 2. Are hummingbirds attracted to the color red? Question 3. Is the moon made out of green cheese? Question 4. Is plagiarism dishonest? Which questions lend themselves to investigation using scientific methods? In other words, is the hypothesis falsifiable (can be proven false)? Advanced Placement Biology Instructor’s Solution Manual 1 | The Study of Life 5 A Questions 1 and 2 are subjective and cannot be disproven using scientific methods. Questions 3 and 4 can be tested using scientific methods. B Questions 3 and 4 are subjective and cannot be disproven using scientific methods. Questions 1 and 2 can be tested using scientific methods. C Questions 1 and 3 are subjective and cannot be disproven using scientific methods. Questions 2 and 4 can be tested using scientific methods. D Questions 1 and 4 are subjective and cannot be disproven using scientific methods. Questions 2 and 3 can be tested using scientific methods. Solution 15 The solution is (D). Questions 1 and 4 are subjective because taste, dishonesty, and plagiarism are subjective, whereas questions 2 and 3 can be tested using scientific methods. The hypotheses for questions 2 and 3 are falsifiable (can be disproven). Consider the levels of organization of the biological world and place each of these items in order from narrowest level of organization to most encompassing: skin cell, elephant, water molecule, planet Earth, tropical rain forest, hydrogen atom, wolf pack, liver. A Hydrogen atom, water molecule, skin cell, liver, elephant, wolf pack, tropical rain forest, planet Earth B Hydrogen atom, skin cell, water molecule, liver, elephant, wolf pack, tropical rain forest, planet Earth C Hydrogen atom, skin cell, water molecule, liver, wolf pack, elephant, tropical rain forest, planet Earth D Water molecule, hydrogen atom, skin cell, liver, elephant, wolf pack, tropical rain forest, planet Earth Solution 16 The solution is (A). The narrowest level of organization to the most-encompassing is: hydrogen atom, water molecule, skin cell, liver, elephant, wolf pack, tropical rainforest, planet Earth. What scientific evidence did Carl Woese use to determine that there should be a separate domain for Archaea? A A sequence of DNA B A sequence of rRNA C A sequence of mRNA D A sequence of tRNA Solution 17 The solution is (B). Woese used sequences of rRNA as evidence to determine that there should be a separate domain for Archaea. Both astronomy and astrology study the stars. Which one is considered a natural science? Explain your reasoning. Advanced Placement Biology Instructor’s Solution Manual 6 1 | The Study of Life A Astrology is a natural science because it indirectly influences human affairs and the natural world. B Astronomy is a natural science because it deals with observations and predictions of events in the sky, which are based on the laws of physics. C Astrology is a natural science because it deals with observations and predictions of events in the sky, and influences human affairs and the natural world. D Astrology is a natural science because it deals with the study of asteroids and comets, which is based on the laws of natural sciences. Solution The solution is (B). Astrology does involve observations of celestial bodies, but it is a way to predict human affairs based on astronomical data that do not rely on testable hypotheses. Astronomy is the science of observing, measuring, and predicting events in space based on past observations and the laws of physics. TEST PREP FOR AP® COURSES 18 Which structure is conserved in all living organisms and points to a common origin? A All living organisms have mitochondria that produce energy. B All living organisms store genetic material in DNA/RNA. C All living organisms use the energy from sunlight. D All living organisms have a nucleus. Solution 19 The solution is (B). This is true of all organisms and, therefore, logically suggests a common origin. Which statement is the strongest argument in favor of two organisms, A and B, being closely related evolutionarily? A A and B look alike. B A and B live in the same ecosystem. C A and B use the same metabolic pathways. D The DNA sequences of A and B are highly homologous. Solution 20 The solution is (D). All living organisms store genetic material in DNA/RNA so this is one of the strongest ways to probe for similarities. In the phylogenetic tree shown, which organism is most distantly related to organism 2? Advanced Placement Biology Instructor’s Solution Manual 1 | The Study of Life 7 A Organism 1 B Organism 3 C Organism 4 D Organism 5 Solution 21 The solution is (D). Organism 5 is most distantly related to organism 2 as it is evident in the phylogenetic tree that 5 diverged from the main branch much earlier than rest of the organisms. In the phylogenetic tree shown, which organism is the most recent common ancestor of organisms 1 and 3? A Organism A B Organism B C Organism C D Organism D Solution 22 The solution is (B). Organism B is the recent common ancestor of 1 and 3 because B is the base of 1 and 3. The French scientist Jacques Monod famously said, “Anything found to be true of E. coli must also be true of elephants.” How is this statement based on the notion that living organisms share a common ancestor? A E. coli is a eukaryote and shares similarities with most living organisms. B E. coli is a prokaryote. The various metabolic processes and core functions in E. coli share homology with higher organisms. C E. coli contains a nucleus and membrane-bound cell organelles that are shared by all living organisms. D E. coli is a prokaryote and reproduces through binary fission, which is common to most living organisms. Solution 23 The solution is (B). All organisms share core structures and functions; therefore, basic research on E. coli metabolic pathways, molecular biology, and other topics applies to higher organisms. Birds have been reclassified as reptiles. What is one line of evidence that has led to this reclassification? Advanced Placement Biology Instructor’s Solution Manual 8 1 | The Study of Life A Archeopteryx is the connecting link between birds and reptiles, which shows that birds and reptiles are related. B Birds have scales, so they have the same origin as reptiles. C Birds and reptiles have the same circulatory and excretory systems and both are egglaying animals. D Birds and reptiles have similar anatomical and morphological features. Solution The solution is (B). Birds have feathers, which are understood to be modified scales, as well as scales on their feet. They show some skeletal similarity, which supports their reclassification as reptiles. Advanced Placement Biology Instructor’s Solution Manual 2 | The Chemical Foundation of Life 9 2 | THE CHEMICAL FOUNDATION OF LIFE REVIEW QUESTIONS 1 What are atoms that vary in the number of neutrons found in their nuclei called? A Ions B Isotopes C Isobars D Neutral atoms Solution 2 The solution is (B). Isotopes are atoms that have the same number of protons but differ in the number of neutrons. Potassium has an atomic number of 19. What is its electron configuration? A Shells 1 and 2 are full, and shell 3 has nine electrons. B Shells 1, 2, and 3 are full, and shell 4 has three electrons. C Shells 1, 2, and 3 are full, and shell 4 has one electron. D Shells 1, 2, and 3 are full, and no other electrons are present. Solution 3 The solution is (C). Shells 1, 2, and 3 are full, as they hold a total of 18 electrons, and shell 4 has the remaining one electron and is unfilled. Which type of bond exemplifies a weak chemical bond? A Covalent bond B Hydrogen bond C Ionic bond D Nonpolar covalent bond Solution 4 The solution is (B). A hydrogen bond forms between hydrogen and an electronegative atom. It is weak because electrons are not shared as they are in a covalent bond nor are they transferred to form opposite, attracting charges, as they are in ionic bonds. Instead, hydrogen has a slight positive charge when covalently bonded to a more electronegative atom, because that atom draws the hydrogen’s electron away from it, causing it to become attracted to nearby electronegative atoms with unpaired electrons. Which statement is false? A Electrons are unequally shared in polar covalent bonds. B Electrons are equally shared in nonpolar covalent bonds. Advanced Placement Biology Instructor’s Solution Manual 10 2 | The Chemical Foundation of Life C Hydrogen bonds are weak bonds based on electrostatic forces. D Ionic bonds are generally stronger than covalent bonds. Solution 5 The solution is (D). Ionic bonds are generally stronger than covalent bonds. If xenon has an atomic number of 54 and a mass number of 108, how many neutrons does it have? A 27 B 54 C 100 D 108 Solution 6 The solution is (B). The number of neutrons can be calculated by subtracting the element’s atomic number from its mass number. When you subtract 54 from 108, the number of neutrons is 54. What forms ionic bonds? A Atoms that share electrons equally B Atoms that share electrons unequally C Ions with similar charges D Ions with opposite charges Solution 7 The solution is (D). Ions with similar charges will not likely come together to form ionic bonds; they must be oppositely charged. Based on the information provided, which of the following statements is correct? A In NH2 , the nitrogen atom acquires a partial positive charge and the hydrogen atoms acquire a partial negative charge. B In H2O, the hydrogen atoms acquire a partial negative charge, and the oxygen atom acquires a partial positive charge. Advanced Placement Biology Instructor’s Solution Manual 2 | The Chemical Foundation of Life 11 C In aqueous HCl, the hydrogen atom acquires a partial positive charge, and the chlorine atom acquires a partial negative charge. D In LiF, the lithium atom acquires a positive charge, and the fluorine atom acquires a negative charge. Solution 8 The solution is (C). In HCl(aq), the hydrogen atom acquires a partial positive charge, and the chlorine atom acquires a partial negative charge. Which statement is NOT true? A Water is polar. B Water can stabilize the temperature of nearby air. C Water is essential for life. D Water is the most abundant molecule in Earth’s atmosphere. Solution 9 The solution is (D). Water is the most abundant molecule in Earth’s atmosphere. Why do hydrogen and oxygen form polar covalent bonds within water molecules? A Hydrogen is more electronegative than oxygen, generating a partial negative charge near the hydrogen atoms. B Hydrogen is more electronegative than oxygen, generating a partial positive charge near the hydrogen atoms. C Oxygen is more electronegative than hydrogen, generating a partial negative charge near the oxygen atom. D Oxygen is more electronegative than hydrogen, generating a partial positive charge near the oxygen atom. Solution 10 The solution is (C). The oxygen atom nucleus is more attractive to the electrons of a hydrogen atom than the hydrogen nucleus is to the oxygen’s electrons. Therefore, the hydrogen atom acquires a partial positive charge while the oxygen atom acquires a partial negative charge. What happens to the pH of a solution when acids are added? A The pH of the solution decreases. B The pH of the solution increases. C The pH of the solution increases and then decreases. D The pH of the solution stays the same. Solution The solution is (A). An acid is a substance that increases the H ion concentration of the solution by dissociating its H atoms. Thus, the pH of the solution decreases on addition of a hydrogen atom. Advanced Placement Biology Instructor’s Solution Manual 12 11 2 | The Chemical Foundation of Life Which statement is true? A Acids and bases cannot mix together. B Acids and bases can neutralize each other. C Acids, not bases, can change the pH of a solution. D Acids donate hydroxide ions ( OH ); bases donate hydrogen ions ( H ). + Solution 12 The solution is (B). Acids add H ions in a solution, while bases add OH ions to a solution. These ions neutralize each other by forming water molecules. What is water’s property of adhesion? A The force that allows surface water molecules to escape and vaporize B The attraction between water molecules and other molecules C The intermolecular force between water molecules D The force that keeps particles dispersed in water Solution 13 The solution is (B). Adhesion is the tendency of different particles to cling to one another. It is sometimes a strong adhesive force, such as when water is exposed to charged surfaces. In a solution, what kind of molecule binds up excess hydrogen ions? A Acid B Base C Donator D Isotope Solution 14 The solution is (B). Bases bind excess H ions in solution, resulting in a neutral solution. For example, the addition of OH to H produces water. What is the maximum number of atoms or molecules a single carbon molecule can bond with? A 4 B 1 C 6 D 2 Solution The solution is (A). The carbon atom has four electrons in its valence shell. Therefore, bonding four more electrons will complete its octet, allowing it to attain a stable configuration. A carbon molecule can therefore bond with a maximum of four other atoms. Advanced Placement Biology Instructor’s Solution Manual 2 | The Chemical Foundation of Life 15 13 Which statement is true? A Molecules with the formulas CH3CH2OH and C3H6O2 could be structural isomers. B Molecules must have a single bond to be cis-trans isomers. C To be enantiomers, a molecule must have at least three different atoms or groups connected to a central carbon. D To be enantiomers, a molecule must have at least four different atoms or groups connected to a central carbon. Solution 16 The solution is (D). To be enantiomers, a molecule must have at least four different atoms or groups connected to a central carbon. What is NOT a functional group that can bond with carbon? A Carbonyl B Hydroxyl C Phosphate D Sodium Solution 17 The solution is (D). Sodium is not a functional group; it is an element. Thus, it does not bond with carbon as a functional group. Which functional group is NOT polar? A Carbonyl B Hydroxyl C Methyl D Sulfhydryl Solution 18 The solution is (C). Methyl groups are nonpolar because carbon and hydrogen have similar electronegativity. This means the covalent bond between carbon and hydrogen has electrons that are equally shared by the carbon and hydrogen. Therefore, it is a nonpolar molecule. What are enantiomers? A Hydrocarbons consisting of closed rings of carbon atoms B Isomers with similar bonding patterns differing in the placement of atoms along a double covalent bond C Molecules that share the same chemical bonds but are mirror images of each other D Molecules with the same chemical formula that differ in the placement of their chemical bonds Solution The solution is (C). Enantiomers are molecules that share the same chemical bonds but are mirror images of each other. Advanced Placement Biology Instructor’s Solution Manual 14 2 | The Chemical Foundation of Life CRITICAL THINKING QUESTIONS 19 What are the mass number and atomic number of carbon-12 and carbon-13, respectively? A The mass number and atomic numbers of carbon-12 are 12 and 6, while those of carbon-13 are 13 and 6. B The mass number and atomic numbers of carbon-12 are 12 and 6, while those of carbon-13 are 13 and 12. C The mass number and atomic numbers of carbon-12 are 12 and 6, while those of carbon-12 are 13 and 13. D The mass number and atomic numbers of carbon-12 are 12 and 12, while those of carbon-13 are 13 and 12. Solution 20 The solution is (A). The mass number and atomic number of carbon-12 are 12 and 6, while those of carbon-13 are 13 and 6. Use the figure to determine the mass and atomic numbers. Why are hydrogen bonds and van der Waals interactions necessary for cells? A Hydrogen bonds and van der Waals interactions form weak associations between molecules, providing the necessary shape and structure for DNA and proteins to function in the body. B Hydrogen bonds and van der Waals interactions form strong associations between molecules, providing the necessary shape and structure for DNA and proteins to function in the body. C Hydrogen bonds and van der Waals interactions form weak associations between different molecules, providing the necessary shape and structure for acids to function in the body. D Hydrogen bonds and van der Waals interactions form strong associations between the same molecules, providing the necessary shape and structure for acids to function in the body. Advanced Placement Biology Instructor’s Solution Manual 2 | The Chemical Foundation of Life Solution 21 15 The solution is (A). Hydrogen bonds and van der Waals interactions form weak associations between different molecules or within different regions of the same molecule. They provide the structure and shape necessary for proteins and DNA within cells so that they function properly. Using the figure, which two groups will form a strong ionic bond? A Group 1 and Group 17 B Group 1 and Group 14 C Group 14 and Group 18 D Group 1 and Group 18 Solution 22 The solution is (A). Group 1 and Group 17 will form a strong ionic bond. Group 1 has one valence electron, so to obtain a stable configuration, it will donate the electron readily. Group 17 has 7 electrons in its valence shell. In order to obtain a stable configuration, it will accept one electron readily. Why can some insects walk on water? A Insects can walk on water because of its high surface tension. B Insects can walk on water because it is a polar solvent. C Insects can walk on water because they are less dense than water. D Insects can walk on water because they are denser than water. Solution The solution is (A). Some insects can walk on water, although they are denser than water, because of the surface tension of water. Surface tension is a result of cohesion, or the attraction between water molecules at the surface of the body of water (the liquid-air/gas interface). Advanced Placement Biology Instructor’s Solution Manual 16 23 2 | The Chemical Foundation of Life Which statement describes how buffers help prevent drastic swings in pH? A Buffers absorb excess hydrogen and hydroxide ions to prevent increases or decreases in pH. An example is the bicarbonate system in the human body. B Buffers absorb excess hydrogen ions to prevent increases or decreases in pH. An example is the bicarbonate system in the human body. C Buffers absorb excess hydroxide ions to prevent increases or decreases in pH. An example is the bicarbonate system in the human body. D Buffers absorb excess hydrogen and hydroxide ions to prevent increases or decreases in pH. An example is carbonate system in the human body. Solution 24 The solution is (A). Buffers absorb the free hydrogen ions and hydroxide ions produced by chemical reactions. Because they can bond to these ions, they prevent increases or decreases in pH. An example of a buffer system is the bicarbonate system in the human body. This system is able to absorb hydrogen and hydroxide ions to prevent changes in pH and keep cells functioning properly. What are three examples of how the characteristics of water are important in maintaining life? A First, the lower density of water as a solid versus a liquid allows ice to float, forming an insulating surface layer for aquatic life. Second, the high specific heat capacity of water insulates aquatic life or bodily fluids from temperature changes. Third, the high heat of vaporization of water allows animals to cool themselves by sweating. B First, the higher density of water as a solid versus a liquid allows ice to float, forming an insulating surface layer for aquatic life. Second, the high specific heat capacity of water insulates aquatic life or bodily fluids from temperature changes. Third, the low heat of vaporization of water allows animals to cool themselves by sweating. C First, the lower density of water as a solid versus a liquid allows ice to float, forming an insulating surface layer for aquatic life. Second, the low specific heat capacity of water insulates aquatic life or bodily fluids from temperature changes. Third, the high heat of vaporization of water allows animals to cool themselves by sweating. D First, the lower density of water as a solid versus a liquid allows ice to float, forming an insulating surface layer for aquatic life. Second, the low specific heat capacity of water insulates aquatic life or bodily fluids from temperature changes. Third, the low heat of vaporization of water allows animals to cool themselves by sweating. Solution The solution is (A). The lower density of ice compared to liquid water allows it to float on water. In lakes and ponds, ice will form on the surface of water creating an insulating barrier that protects the animals and plant life in the pond from freezing. Water’s lower density in its solid form is due to the orientation of hydrogen bonds as it freezes: the water molecules are pushed farther apart compared to liquid water. Advanced Placement Biology Instructor’s Solution Manual 2 | The Chemical Foundation of Life 17 Water is used by warm-blooded animals to more evenly disperse heat in their bodies. Water has the highest specific heat capacity of any liquid, a property caused by hydrogen bonding between water molecules. In many living organisms, the evaporation of sweat allows organisms to cool to maintain homeostasis of body temperature. This is because water has a high heat of vaporization. As liquid water heats up, hydrogen bonding makes it difficult to separate the liquid water molecules from each other. Other examples include water’s solvent properties as well as water’s cohesive and adhesive properties. 25 What is the pH scale, and how does it relate to living systems? Give an example of how drastic pH changes are prevented in living systems. A The pH scale ranges from 0 to 14, where anything below 7 is acidic and above 7 is alkaline. The bicarbonate system in the human body buffers the blood. B The pH scale ranges from 0 to 14, where anything below 7 is alkaline and above 7 is acidic. The bicarbonate system in the human body buffers the blood. C The pH scale ranges from 0 to 7, where anything below 7 is acidic and above 7 is alkaline. Water in the human body buffers the blood. D pH scale ranges from 0 to 7, where anything below 4 is acidic and above 4 is alkaline. Water in the human body buffers the blood. Solution 26 The solution is (A). The pH scale ranges from zero to 14. Pure water has a pH of seven, which is known as neutral pH. The pH of human cells and blood is maintained at pH seven. Anything below pH seven is acidic and anything above seven is alkaline. Extreme pH is usually considered inhospitable for life. Buffers maintain pH homeostasis. For example, the bicarbonate system in the human body absorbs hydrogen and hydroxide ions to prevent changes in pH. What property of carbon makes it essential for organic life? A Carbon can form up to four covalent bonds, allowing it to form long chains. B Carbon can form more than four covalent bonds, allowing it to form long chains. C Carbon can form more than four covalent bonds, but it can only form short chains. D Carbon can form up to four covalent bonds, but it can only form short chains. Solution The solution is (A). Carbon is found in all living things because it can form up to four covalent bonds. These strong covalent bonds allow the formation of long chains that combine to form proteins and DNA. Advanced Placement Biology Instructor’s Solution Manual 18 27 2 | The Chemical Foundation of Life What property of carboxyl makes carboxyl-containing molecules hydrophilic? Which macromolecules contain carboxyl? A Carboxyl groups release H+, making its parent molecule hydrophilic. It is found in amino acids and fatty acids. B Carboxyl groups absorb H+, making its parent molecule hydrophilic. It is found in phospholipids and triglycerides. C Carboxyl groups release OH , making its parent molecule hydrophilic. It is found in phospholipids, phosphates, and triglycerides. D Carboxyl groups release OH , making its parent molecule hydrophilic. It is found in phospholipids and DNA. Solution 28 The solution is (A). The carboxyl group is found in amino acids, some amino acid side chains, and the fatty acids that form triglycerides and phospholipids. This carboxyl group ionizes to release hydrogen ions (H+) from the COOH group, resulting in the negatively charged COO – group, which contributes to the hydrophilic nature of whatever molecule it is found on. Compare and contrast saturated and unsaturated triglycerides. A Saturated triglycerides contain single bonds and are solid at room temperature, while unsaturated triglycerides contain double bonds and are liquid at room temperature. B Saturated triglycerides contain double bonds and are solid at room temperature, while unsaturated triglycerides contain single bonds and are liquids at room temperature. C Saturated triglycerides contain single bonds and are liquid at room temperature, while unsaturated triglycerides contain double bonds and are solid at room temperature. D Saturated triglycerides contain double bonds and are liquid at room temperature, while unsaturated triglycerides contain single bonds and are solid at room temperature. Solution The solution is (A). Saturated triglycerides contain no double bonds between carbon atoms; they are usually solid at room temperature. Unsaturated triglycerides contain at least one double bond between carbon atoms; they are usually liquid at room temperature. TEST PREP FOR AP® COURSES 29 What property of water makes it a good insulator within the bodies of endothermic (warm-blooded) animals? A Adhesion B Surface tension Advanced Placement Biology Instructor’s Solution Manual 2 | The Chemical Foundation of Life 19 C Heat of vaporization D Specific heat capacity Solution 30 The solution is (D). Specific heat is defined as the amount of heat one gram of a substance must absorb or lose to change its temperature by 1 °C Therefore, warmblooded animals use water to more evenly disperse heat in their bodies. It takes a large amount of energy to heat or cool water. The unique properties of water are important in biological processes. For the following three properties of water, define the property and give one example of how the property affects living organisms: Cohesion Adhesion High heat of vaporization A Cohesion is the attraction between water molecules, which helps create surface tension. Insects can walk on water because of cohesion. Adhesion is the attraction between water molecules and other molecules. Water moving up from the roots of plants to the leaves as a result of capillary action is because of adhesion. Heat of vaporization is the amount of energy required to convert liquid into gas. This property helps humans maintain homeostasis of body temperature by evaporation. B Cohesion is the attraction between water and other molecules, which helps create surface tension. Insects can walk on water because of cohesion. Adhesion is the attraction between water molecules. Water moving up from the roots of plants to the leaves as a result of capillary action is because of adhesion. Heat of vaporization is the amount of energy required to convert liquid into gas. This property helps humans maintain homeostasis of body temperature by evaporation. C Cohesion is the attraction between water molecules, which helps create surface tension. Insects can walk on water because of cohesion. Adhesion is the attraction between water molecules and other molecules. Water moving up from the roots of plants to the leaves as a result of capillary action is because of adhesion. Heat of vaporization is the amount of energy required to convert solid into gas. This property helps humans maintain homeostasis of body temperature by evaporation. Solution The solution is (A). Cohesion is the attraction between water molecules, which helps create surface tension. Insects can walk on water because of cohesion. Adhesion is the attraction between water molecules and other molecules. Water moving up from the roots of plants to the leaves as a result of capillary action is because of adhesion. Heat of vaporization is the amount of energy required to convert liquid into gas. This property helps humans maintain homeostasis of body temperature by evaporation. Advanced Placement Biology Instructor’s Solution Manual 20 2 | The Chemical Foundation of Life SCIENCE PRACTICE CHALLENGE QUESTIONS 2.1 Atoms, Isotopes, Ions, and Molecules: The Building Blocks 31 At a time when the theory of evolution was controversial (the year following the Scopes Monkey Trial), Macallum (Physiological Reviews, 2, 1926) made an observation that is still contested by some who do not see the pattern in the data in the table showing percentages (g solute/100 g solution) of major biologically important inorganic elements in a variety of sources. Source Na K Ca2 Mg2 Cl Ocean water 0.306 0.011 0.012 0.0037 0.55 Lobster 0.903 0.0337 0.0438 0.0156 1.547 Dog fish 0.5918 0.02739 0.01609 0.0146 0.9819 Sand shark 0.6173 0.0355 0.0184 0.0172 1.042 Cod 0.416 0.0395 0.0163 0.00589 0.6221 Pollock 0.4145 0.017497 0.01286 0.00608 0.5613 Frog 0.195 0.0233 0.00627 0.00155 0.2679 Dog lymph 0.3033 0.0201 0.0085 0.0023 0.4231 Blank blank blank blank blank Blood 0.302 0.0204 0.0094 0.0021 0.389 Lung 0.2956 0.02095 0.00839 0.0021 Testes 0.3023 0.01497 0.00842 0.001914 0.3737 Human 0.3425 Abdominal 0.2935 0.0164 0.0091 0.00184 0.3888 cavity A. Using a spreadsheet, or by sharing calculations with your classmates, construct a quantitative model of these data from these percentages as ratios of mass fractions relative to that of sodium, %X/%Na. Of course, you will not be asked to use a spreadsheet on the AP Biology Exam. However, you may be assessed on the ability to develop a quantitative model through the transformation of numerical data. The question that led Macallum to investigate the elemental composition of different species and compare these with the composition of seawater follows from the central organizing principle of biology: the theory of evolution. B. The elements in the table all occur in aqueous solution as ions. Cells expend energy to produce a charge gradient and an ion gradient that is necessary for basic cellular functions. Large differences in the concentrations of ions inside versus outside of the cell results in stresses that the cell must expend energy to relieve. Based on this constraint on the total number of ions, connect this refined model based on ratios of ion concentration rather than absolute ionic concentrations to the modern concept of shared ancestry. Advanced Placement Biology Instructor’s Solution Manual 2 | The Chemical Foundation of Life 21 Frequently, a follow-up question regarding scientific data on the AP Biology Exam will ask you to pose questions that are raised by the data. You will be awarded credit for scientific questions. These questions usually look for a cause-and-effect relationship, and are testable. C. Examine relative concentrations of potassium and magnesium ions in terrestrial and marine organisms. Pose a question that could be investigated to connect concentrations of these ions to adaptations to a change in the environment. Macallum noted the high potassium to sodium ratio relative to seawater, and made this claim about what the ratio implied about the oceans of early Earth: “At once it is suggested that as the cell is older than its media is [presently] the relative proportions of the inorganic elements in it are of more ancient origin than the relative proportions of the same amount of elements which prevail in the media, blood plasma and lymph or in the ocean and river water of today.” D. In your own words, summarize the argument that Macallum is using to justify this claim. Solution Sample answer: A. Source K Ca2 Mg2 Cl Ocean water 0.036 0.039 0.012 1.8 Lobster 0.037 0.048 0.017 1.7 Dog fish 0.046 0.027 0.025 1.7 Sand shark 0.057 0.030 0.028 1.7 Cod 0.095 0.039 0.014 1.5 Pollock 0.042 0.031 0.015 1.35 Frog 0.12 0.032 0.0080 1.4 Dog lymph 0.07 0.028 0.0076 1.4 blank blank Blood 0.067 0.031 0.0070 1.3 Lung 0.0715 0.028 0.0071 1.2 Testes 0.049 0.028 0.0063 1.2 Abdominal cavity 0.056 0.031 0.0062 1.3 Human blank blank Table of Ratios %ion/%Na B. The hypothesis that all animal life originated in the oceans seems to be supported by the observation that the ratios of ions in the tissues of terrestrial animals are similar to those found in ocean water. The absolute amounts of ions may vary from Advanced Placement Biology Instructor’s Solution Manual 22 2 | The Chemical Foundation of Life animal to animal, however, relatively similar ratios point to inherited cellular mechanisms that require similar ratios of ions. C. The lower concentrations of K and Mg2 in terrestrial animals compared to those found in ocean water can be attributed to adaptation to life in air rather than a reflection of concentrations present in ancient oceans. One question to pose would be why the concentrations of K and Mg2 changed over time in the oceans. Examination of ancient sediments, and studies of how the weathering of rock adds salts into the ocean may provide insights into the composition of ancient oceans. D. The proportion of ions in the blood and tissues of terrestrial animals are a reflection of the composition of salts in the marine environment existing at the time when animals moved to land. It is not tied to present day ocean composition. 2.2 Water 32 Approximately half the energy that flows through Earth’s biosphere is captured by phytoplankton, photosynthetic microscopic organisms in the surface waters of the oceans. Scientists think the growth of phytoplankton in the Atlantic Ocean is limited by the availability of nitrogen, whereas growth in the Pacific Ocean is limited by the availability of iron. The concentration of oxygen (O2) in the atmosphere of early Earth was low and, therefore, so was the concentration of dissolved oxygen in the early ocean. Because insoluble iron oxides (rust) do not form in the absence of oxygen, soluble iron ions Fe2+ were more available in the early ocean than at present. Nitrogen (N2), while always abundant in the atmosphere, was not biologically available until the evolution of molybdenum-based enzymes that allow nitrogen fixation. The graphs (Anbar and Knoll, Science, 297, 2002) show models of concentrations of two trace elements, iron (Fe) and molybdenum (Mo), in ocean waters (first), as well as the concentration of various elements dissolved in seawater versus within cells (second). The model describes the change over time of these elements from early Earth (>1.85 billion years ago, Gya) to a modern era (<1.25 Gya) and a period of transition between these. Surface waters of the oceans lie to the left of the vertical double line while deep ocean water lies to the right. Modern concentrations of dissolved iron and molybdenum (relative to dissolved carbon) are shown. Advanced Placement Biology Instructor’s Solution Manual 2 | The Chemical Foundation of Life 23 A. The principal chemical processes of life today have been conserved through evolution from early Earth conditions. Using this fact, justify the selection of these data shown in the graphs in the construction of a model of ocean photosynthetic productivity. Iron and molybdenum are two of 30 elements that are required by the chemical processes supporting life on Earth. Concentrations of these two and 15 other elements are shown in the second graph. Of these elements, the three most abundant in cells are also found in seawater in approximately the same concentrations. By increasing the mass of phytoplankton in the ocean, we may be able to compensate for the increasing concentration of carbon produced by the combustion of gas, oil, and coal. Advanced Placement Biology Instructor’s Solution Manual 24 2 | The Chemical Foundation of Life B. Select, with justification, the element or elements that, if added in large amounts to the ocean, could boost the growth of phytoplankton. C. Before implementing a large-scale geo-engineering effort to avert the effects of climate change due to carbon pollution, we must test the legitimacy of this solution. Describe a plan for collecting data that could be used to evaluate the effect of enrichment on phytoplankton productivity. Solution Sample answer: A. The first graph shows the concentration of rate-limiting elements, including iron and molybdenum, available for photosynthesis. At the surface molybdenum is abundant. Iron is less available, perhaps because it reacts with oxygen in the atmosphere to form insoluble iron oxide that sinks. B. The second graph shows that iron, manganese, and cobalt are also abundant in the ocean than in the cells. This means cells will have a hard time obtaining enough of these elements, and adding more would increase productivity. Manganese is an essential component of the reaction center of photosystem II where it is involved in the splitting of water molecules. C. In controlled laboratory conditions, supplemental metal could be added to phytoplankton grown and the productivity could be measured. It is important to stress that the oceans are vast areas with different local conditions. The unintended effects of overgrowth of phytoplankton must also be addressed. Eutrophication, or overgrowth of algae leading to dead zones in lakes and coastal areas, occurs when excess nitrogen is present in the water. Advanced Placement Biology Instructor’s Solution Manual 3 | Biological Macromolecules 25 3 | BIOLOGICAL MACROMOLECULES REVIEW QUESTIONS 1 Dehydration synthesis leads to the formation of what? A Monomers B Polymers C Carbohydrates only D Water only Solution 2 The solution is (B). Polymers are formed through dehydration synthesis, during which monomers combine to release a water molecule. What is removed during the formation of nucleic acid polymers? A Carbon B Hydroxyl groups C Phosphates D Amino acids Solution 3 The solution is (C). Unlike proteins, carbohydrates, and lipids, the molecule that is released in the formation of nucleic acid polymers is pyrophosphate (two phosphate groups bound together). During the breakdown of polymers, which reaction takes place? A Condensation B Covalent bond C Dehydration D Hydrolysis Solution 4 The solution is (D). Hydrolysis is a reaction in which a water molecule is used during the breakdown of another compound or a polymer. Energy is released as a result of which chemical reaction? A Condensation B Dehydration synthesis C Hydrolysis D Dissolution Solution The solution is (C). Hydrolysis reactions typically release energy that can be used to power cellular processes. Advanced Placement Biology Instructor’s Solution Manual 26 5 3 | Biological Macromolecules In the metabolism of the cell, why is hydrolysis used? A Hydrolysis is used to break down polymers. B Hydrolysis is used to form linkages in DNA. C Hydrolysis is used to produce proteins. D Hydrolysis is used to synthesize new macromolecules. Solution 6 The solution is (A). Polymers can be broken down by hydrolysis, or the addition of water. Plant cell walls contain what in abundance? A Cellulose B Glycogen C Lactose D Starch Solution 7 The solution is (A). The cell walls of plants are mostly made of cellulose, which provides structural support to the cell. What makes up the outer layer of some insects? A Carbohydrate B Protein C RNA D Triglyceride Solution 8 The solution is (A). Arthropods like insects have an outer skeleton, called the exoskeleton, which protects their internal body parts. The exoskeleton is mainly made up of chitin, a nitrogen-containing polysaccharide. What is an example of a monosaccharide? A Cellulose B Fructose C Lactose D Sucrose Solution 9 The solution is (B). Fructose is a simple ketose monosaccharide found in many plants. Cellulose and starch are examples of — A disaccharides B lipids Advanced Placement Biology Instructor’s Solution Manual 3 | Biological Macromolecules 27 C monosaccharides D polysaccharides Solution 10 The solution is (D). A long chain of monosaccharides linked by glycosidic bonds is known as a polysaccharide, which can contain the same or different monosaccharides. Starch is the stored form of sugars in plants and is made up of a mixture of amylose and amylopectin, while cellulose is made up of glucose monomers that are linked by -1-4 glycosidic bonds. What type of bond joins the molecules in the disaccharide lactose, and what molecule is joined with glucose to form lactose? A A glycosidic bond between glucose and lactose B A glycosidic bond between glucose and galactose C A hydrogen bond between glucose and sucrose D A hydrogen bond between glucose and fructose Solution 11 The solution is (B). Lactose is a disaccharide consisting of the monomers glucose and galactose, which are joined together by a glycosidic bond. What is structurally different about cellulose when compared to starch? A An extra hydrogen atom is left on the monomer. B -1,4 glycosidic linkages are used. C -1,6 glycosidic linkages are used. D An extra hydroxyl group is removed during synthesis. Solution 12 The solution is (B). Starch is made up of glucose monomers that are joined by -1-4 or -1-6 glycosidic bonds. Cellulose is made up of glucose monomers that are linked by -1-4 glycosidic bonds only. What are classified as lipids? A Disaccharides and cellulose B Essential amino acids C mRNA and DNA D Oils and waxes Solution 13 The solution is (D). Lipids are a group of naturally occurring molecules that include oils, waxes, and other hydrophobic substances. What is cholesterol specifically classified as? A A lipid B A phospholipid Advanced Placement Biology Instructor’s Solution Manual 28 3 | Biological Macromolecules C A steroid D A wax Solution 14 The solution is (C). Cholesterol is a steroid having four linked carbon rings with a short tail. It is mainly synthesized in the liver and is the precursor to many steroid hormones. Which fat serves as an animal’s major form of energy storage? A Cholesterol B Glycerol C Phospholipid D Triglycerides Solution 15 The solution is (D). Triglycerides, stored in adipose tissue, are a major form of energy storage in animals. Fat cells are designed for continuous synthesis and breakdown of triglycerides in animals. Which hormones are made from cholesterol? A Estradiol and testosterone B Insulin and growth hormone C Progesterone and glucagon D Prolactin and thyroid hormone Solution 16 The solution is (A). Cholesterol is the precursor to many steroid hormones such as testosterone and estradiol, which are secreted by the gonads and adrenal endocrine glands. Which characteristic is NOT true for saturated fats? A They are solid at room temperature. B They have single bonds within the carbon chain. C They contain mostly hydrogen and carbon atoms. D They tend to dissolve in water easily. Solution The solution is (D). Saturated fats are solid at room temperature, and have single bonds between carbon and hydrogen atoms. However, they are nonpolar molecules. Water is a polar solvent, so it is not true that saturated fats do not dissolve in water. Advanced Placement Biology Instructor’s Solution Manual 3 | Biological Macromolecules 17 29 Which fat has the least number of hydrogen atoms? A Trans fat B Saturated fat C Unsaturated fat D Wax Solution 18 The solution is (C). Unlike saturated fats, in which extensive hydrogenation leads to close packing of molecules and results in a solid state, unsaturated fats contain the least number of hydrogen atoms, which results in multiple bonds between carbon atoms and loose packing of molecules. Of what are phospholipids important components? A The double bond in hydrocarbon chains B The plasma membrane of animal cells C The ring structure of steroids D The waxy covering on leaves Solution 19 The solution is (B). Phospholipids are major constituents of the plasma membrane, the outermost layer of animal cells. Phospholipids are responsible for the dynamic nature of the plasma membrane. What is diacylglycerol 3-phosphate? A A phospholipid B A phosphatidylcholine C A phosphatidylserine D A phosphatidate Solution 20 The solution is (D). A phosphate group alone attached to a diglycerol does not qualify as a phospholipid; it is phosphatidate (diacylglycerol 3-phosphate), the precursor of phospholipids. What is the basic structure of a steroid? A Four fused hydrocarbon rings B Glycerol with three fatty acid chains C Two fatty-acid chains and a phosphate group D Two six-carbon rings Solution The solution is (A). All steroids have four linked carbon rings and some of them have a short tail. Cholesterol and many hormones such as estrogen and cortisol are steroids. Advanced Placement Biology Instructor’s Solution Manual 30 21 3 | Biological Macromolecules Besides its use in hormone production, for what does the body use cholesterol? A mRNA transport B Production of bile salts C Water reabsorption in the kidney D Wax production Solution 22 The solution is (B). Cholesterol is also the precursor of bile salts, which help in the emulsification of fats and their subsequent digestion. Where is cholesterol found in cell membranes? A Attached to the inner side of the membrane B Attached to the outer side of the membrane C Floating in the phospholipid tail layer D Penetrating both lipid layers Solution 23 The solution is (D). Cholesterol is a component of the plasma membrane of animal cells and is found within the phospholipid bilayer. The plasma membrane of each cell is made up of a continuous double layer of phospholipids, interwoven with cholesterol and proteins. Which type of body cell would have a higher amount of cholesterol in its membrane? A A cartilage cell B A liver cell C A red blood cell D A spleen cell Solution 24 The solution is (C). Blood cells tend to travel all along the blood vessels of the body and therefore require flexibility to travel. The cholesterol embedded in their membrane provides them with the required flexibility. What is a function of proteins in cells? A Energy storage B Gene storage and access C Membrane fluidity D Structure Solution The solution is (D). Proteins provide structure and support for cells. They also fulfill many other functions. For example, they allow the body to move (e.g., actin), catalyze reactions as enzymes, act as cellular messengers and receptors, and provide defense against pathogens as antibodies. Advanced Placement Biology Instructor’s Solution Manual 3 | Biological Macromolecules 25 31 What type of protein facilitates or accelerates chemical reactions? A An enzyme B A hormone C A membrane transport protein D A tRNA molecule Solution 26 What type of amino acids would you expect to find on the surface of proteins that must interact closely with water? Solution 27 The solution is (A). Enzymes accelerate, or catalyze, chemical reactions. Almost all metabolic processes in the cell need enzymes to occur at rates fast enough to sustain life. Enzymes increase the rate of a reaction by lowering its activation energy. Amino acids that contain acidic, basic, or polar side groups are most likely to be found on the surface of water-soluble proteins. Proteins with these charged ide groups can interact more readily with water than if the protein had a neutral net charge. What are the monomers that make up proteins called? A Amino acids B Chaperones C Disaccharides D Nucleotides Solution 28 The solution is (A). Amino acids are organic compounds (monomers) that combine to form proteins. There are 20 amino acids, 10 of which are referred to as essential amino acids in humans because the body cannot synthesize them and they must be supplied by the diet. Where is the linkage made that combines two amino acids? A Between the R group of one amino acid and the R group of the second B Between the carboxyl group of one amino acid and the amino group of the other C Between the six-carbon of both amino acids D Between the nitrogen atoms of the amino groups in the amino acids Solution The solution is (B). A peptide bond is formed by dehydration between the amino group of one amino acid and the carboxyl group of a second amino group. Each amino acid has the same fundamental structure, which consists of a central carbon atom, also known as the alpha carbon, bonded to an amino group (NH2), a carboxyl group (COOH), and a hydrogen atom. The fourth position bound to the central carbon is occupied by a side chain that distinguishes the amino acids from each other. Advanced Placement Biology Instructor’s Solution Manual 32 29 3 | Biological Macromolecules The -helix and the -pleated sheet are part of which protein structure? A The primary structure B The secondary structure C The tertiary structure D The quaternary structure Solution 30 The solution is (B). The local folding of the polypeptide in some regions gives rise to the secondary structure of the protein. The -helix and -pleated sheets are the secondary structure of proteins that forms because of hydrogen bonding between carbonyl and amino groups in the peptide backbone. Which structural level of proteins is most often associated with their biological function? A The primary structure B The secondary structure C The tertiary structure D The quaternary structure Solution 31 The solution is (C). Tertiary structure is generally stabilized by nonlocal interactions, most commonly by the formation of a hydrophobic core, but also through salt bridges, hydrogen bonds, disulfide bonds, and even posttranslational modifications. Therefore, the tertiary structure is what controls the basic function of the protein. Some proteins acquire their biological functions only in their quaternary structures, for example, antibodies. What may cause a protein to denature? A Changes in pH B High temperatures C The addition of some chemicals D All of the above Solution 32 The solution is (D). If the protein is subject to changes in temperature, pH, or exposure to chemicals, the protein structure may change, losing its shape without losing its primary sequence, in a process known as denaturation. What is a protein’s chaperone? A A chemical that assists the protein in its enzymatic functions B A second protein that completes the quaternary structure Advanced Placement Biology Instructor’s Solution Manual 3 | Biological Macromolecules 33 C A chemical that helps the protein fold properly D A chemical that functions as a cofactor for the protein Solution 33 The solution is (C). Proteins receive assistance in the folding process from protein helpers known as chaperones that associate with the target protein during the folding process. They prevent aggregation of polypeptides, and then disassociate themselves from the protein once the target protein is folded. What are the building blocks of nucleic acids? A Nitrogenous bases B Nucleotides C Peptides D Sugars Solution 34 The solution is (B). DNA and RNA are nucleic acids made up of monomers known as nucleotides. Each nucleotide is made up of three components: a nitrogenous base, a pentose sugar, and a phosphate group. What may a nucleotide of DNA contain? A Ribose, uracil, and a phosphate group B Deoxyribose, uracil, and a phosphate group C Deoxyribose, thymine, and a phosphate group D Ribose, thymine, and a phosphate group Solution 35 The solution is (C). Each nucleotide consists of a pentose sugar (deoxyribose in DNA and ribose in RNA), a nitrogenous base (adenine, cytosine, guanine, or thymine), and a phosphate group. What is DNA’s structure described as? A A step ladder B A double helix C A tertiary protein-like structure D A barber pole Solution The solution is (B). DNA has a double-helical structure with the two strands running in opposite directions (antiparallel), connected by hydrogen bonds and complementary to each other. Advanced Placement Biology Instructor’s Solution Manual 34 36 3 | Biological Macromolecules What is found in RNA that is NOT in DNA? A Deoxyribose and adenine B Fructose and thymine C Glucose and quinine D Ribose and uracil Solution 37 The solution is (D). The nucleotide molecule RNA consists of ribose sugar. Also, in RNA, uracil replaces thymine and pairs with adenine (U-A). What is the smallest type of RNA? A mRNA B miRNA C rRNA D tRNA Solution 38 The solution is (B). MicroRNAs (miRNAs) are the smallest RNA molecules and their role involves the regulation of gene expression. Where is the largest amount of DNA found in a eukaryotic cell? A Attached to the inner layer of the cell membrane B In the nucleus C In the cytoplasm D On ribosomes Solution The solution is (B). DNA in the form of nucleosomes is found inside the nucleus. CRITICAL THINKING QUESTIONS 39 The word hydrolysis is defined as the lysis of water. How does this apply to polymers? A Polymers break apart by separating water into hydrogen and a hydroxyl group that are added to the monomers. B Polymers are synthesized by using the energy released by the breaking of water molecules into hydrogen and a hydroxyl group. C Polymers are separated into monomers, producing energy and water molecules. D Polymers are hydrolyzed into monomers using water in a process called dehydration synthesis. Solution The solution is (A). In the process of hydrolysis, water is separated into a hydrogen atom and a hydroxyl group. These are added to the separated monomers to complete their structures. Advanced Placement Biology Instructor’s Solution Manual 3 | Biological Macromolecules 40 35 What role do electrons play in dehydration synthesis and hydrolysis? A Electrons are added to OH and H ions in dehydration synthesis. They are removed from OH and H in hydrolysis. B Electrons are transferred from OH and H ions to monomers in dehydration synthesis. They are taken up by the H and OH ions from monomers in hydrolysis. C Electrons are removed from OH and H in dehydration synthesis. They are added to OH and H in hydrolysis. D Electrons are transferred from monomers to H and OH ions in hydrolysis. They are transferred from OH and H to monomers in dehydration synthesis. Solution 41 The solution is (A). Electrons are added to the hydroxyl group and the hydrogen ion during dehydration synthesis to constitute water. During hydrolysis, they are removed from the OH and H of water to create ions to reconstitute monomers. Which bodily process would most likely be hindered by a lack of water in the body? A Digestion B Protein synthesis C DNA copying D Breathing Solution 42 The solution is (A). Digestion would be negatively affected by an inability to perform hydrolysis. Digestion requires hydrolysis to break larger polymers within food molecules into monomers. Why is it impossible for humans to digest food that contains cellulose? A There is no energy available in fiber. B An inactive form of cellulase in the human digestive tract renders it undigested and removes it as waste. C The acidic environment in the human stomach makes it impossible to break the bonds in cellulose. D Human digestive enzymes cannot break down the -1,4 glycosidic linkage in cellulose, which requires a special enzyme absent in humans. Solution The solution is (D). Human digestive enzymes cannot break down the -1-4 linkage. It requires a special enzyme, cellulase, which is secreted by bacteria and protists in the digestive tract of herbivores. Advanced Placement Biology Instructor’s Solution Manual 36 43 3 | Biological Macromolecules Which statement describes some of the differences between cellulose and starch? A Cellulose is unbranched and starch is branched. Both molecules are found in animals. B Starch is unbranched and cellulose is branched. Both molecules are found in plants. C Starch is branched and cellulose is unbranched. Both molecules are found in plants. D Cellulose is branched and starch is unbranched. Both molecules are found in animals. Solution 44 The solution is (C). Starch contains branches and cellulose does not. Both molecules are made by plants but neither is made by animals. Which statement best describes the production of sucrose, maltose, and lactose? A Glucose and fructose combine to form sucrose. Glucose and galactose combine to form lactose. Two glucose monomers combine to form maltose. B Glucose and fructose combine to form sucrose. Glucose and galactose combine to form maltose. Two glucose combine to form lactose. C Two glucose combine to form lactose. Glucose and galactose combine to form sucrose. Glucose and fructose combine to form maltose. D Two galactose combine to form sucrose. Fructose and glucose combine to form lactose. Two glucose combine to form maltose. Solution 45 The solution is (A). Glucose and fructose are combined to form sucrose. Glucose and galactose are combined to form lactose. Two glucose monomers are combined to form maltose. What are the four classes of lipids? Provide a common example for each one. A Lipids like margarine, waxes like the coating on feathers, phospholipids like cell membrane constituents, steroids like cholesterol B Lipids like phosphatidylserine, waxes like phosphatidic acid, phospholipids like oleic acid, steroids like epinephrine C Lipids like phosphatidic acid, waxes like margarine, phospholipids like phosphatidylcholine, steroids like testosterone D Lipids like cholesterol, waxes like the coating on feathers, phospholipids like phosphatidylserine, steroids like margarine Solution 46 The solution is (A). Margarine is a fat with higher trans fatty acid content, wax is present as a feather-coating material, cell membranes are made of phospholipids, and cholesterol is a steroid. What are three functions that lipids serve in plants and/or animals? A Lipids serve in the storage of energy, as a structural component of hormones, and as signaling molecules. B Lipids serve in the storage of energy, as carriers for the transport of proteins across the membrane, and as signaling molecules. Advanced Placement Biology Instructor’s Solution Manual 3 | Biological Macromolecules 37 C Lipids serve in the breakdown of stored energy molecules, as signaling molecules, and as structural components of hormones. D Lipids serve in the breakdown of stored energy molecules, as signaling molecules, and as channels for protein transport. Solution 47 The solution is (A). Lipids serve as a source of long-term energy storage, and as a structural component of some hormones in animals. Hormones are signaling molecules. A class of lipids, the phospholipids, is a major constituent of cell membranes in both plants and animals. How are trans fats created? Why have they been banned from some restaurants? A Trans fat is produced by the hydrogenation of oil that makes it more saturated and isomerized. It increases LDL in the body. B The dehydrogenation of oil forms trans fat, which contains single bonds in its structure. It increases HDL in the body. C Trans fat is produced by the dehydrogenation of oils, which makes it unsaturated. It increases LDL in the body. D The hydrogenation of oil makes the trans fat, which contains double bonds in its structure. It decreases the HDL in the body. Solution 48 The solution is (A). Trans fats resemble saturated fats in their chemical and physical characteristics, and they increase LDL. They are produced by bubbling hydrogen gas through unsaturated lipids under pressure, adding hydrogen atoms to the fatty acids in the lipids. How do phospholipids contribute to cell membrane structure? A Phospholipids orient their heads toward the polar molecules and tails in the interior of the membrane, forming a bilayer. B Phospholipids orient their tails toward the polar molecules of water solutions and their heads toward in the interior of the membrane, forming a bilayer. C Phospholipids orient their heads toward the nonpolar molecules and their tails toward the interior of the membrane, forming a bilayer. D Phospholipids orient their tails toward the polar molecules and their heads toward the nonpolar side of the membrane, forming a bilayer. Solution The solution is (A). The hydrophilic head orients toward polar molecules such as water or the cytoplasm of the cell, whereas the hydrophobic tails of the molecules orient toward other nonpolar molecules. This forms the middle of the membrane, with heads on the outermost and innermost surfaces of the cell membrane. Advanced Placement Biology Instructor’s Solution Manual 38 49 3 | Biological Macromolecules What type of compound functions in hormone production, contributes to membrane flexibility, and is the starting molecule for bile salts? A All steroid molecules help in the mentioned functions. B Cholesterol, which is a lipid and also a steroid, functions here. C Glycogen, which is a multi-branched polysaccharide of glucose, is the compound. D Phosphatidylcholine, which is a phospholipid with a choline head group, serves these functions. Solution 50 The solution is (B). Cholesterol is a steroid and is involved in hormone production, membrane flexibility, and production of bile salts. What part of the cell membrane gives flexibility to the structure? A Carbohydrates B Cytoskeleton filaments C Lipids D Proteins Solution 51 The solution is (C). Lipids add flexibility to the membrane, allowing it to bend and twist as necessary. How do the differences in amino acid sequences lead to different protein functions? A Different amino acids produce different proteins because of the secondary bonds they form. B Differences in amino acids lead to the recycling of proteins, which produces other functional proteins. C Different amino acids cause rearrangements of amino acids to produce a functional protein. D Differences in amino acids cause post-translational modification of the protein, which reassembles to produce a functional protein. Solution 52 The solution is (A). Differences in amino acid sequences result in different configurations of the finished protein. This allows different protein shapes to bind with different chemicals, giving each protein its function. What causes the changes in protein structure through its three or four levels of structure? A The primary chain forms secondary - helix and - pleated sheets that fold onto each other forming the tertiary structure. B The primary structure undergoes alternative splicing to form secondary structures that fold on other protein chains to form tertiary structures. Advanced Placement Biology Instructor’s Solution Manual 3 | Biological Macromolecules 39 C The primary structure forms secondary - helix and - pleated sheets. These further undergo phosphorylation and acetylation to form the tertiary structure. D The primary structure undergoes alternative splicing to form a secondary structure, and then disulfide bonds give way to tertiary structures. Solution 53 The solution is (A). The primary structure is based on the bonds between individual amino acids while the secondary structure is based on the formation of alpha and beta pleated sheets. The tertiary structure describes the folding of the secondary structure. What structural level of proteins is functional? Why? A The secondary structure is functional as it attains its two-dimensional shape, which has the necessary bonds. B The tertiary structure is functional, as it possesses the geometric shape showing the necessary loops and bends. C The tertiary structure is functional as it has the non-covalent and covalent bonds along with the subunits attached at the right places, which help it function properly. D The quaternary structure is functional, as it has the essential set of subunits. Solution 54 The solution is (B). The folding pattern is what creates the shape of the protein, which determines its use as well as, in the case of enzymes, what substrates it can bind to. How does a chaperone work with proteins? A Chaperones assist proteins in folding. B Chaperones cause the aggregation of polypeptides. C Chaperones associate with proteins once the target protein is folded. D Chaperones escort proteins during translation. Solution 55 The solution is (A). Chaperones (or chaperonins) associate with the target protein during the folding process. They act by preventing aggregation of polypeptides that make up the complete protein structure, and they disassociate from the protein once the target protein is folded. What is a difference between DNA and RNA? A DNA is made from nucleotides, while RNA is not. B DNA contains deoxyribose and thymine, while RNA contains ribose and uracil. C DNA contains adenine, while RNA contains guanine. D DNA is double stranded, while RNA may be double stranded in animals. Advanced Placement Biology Instructor’s Solution Manual 40 3 | Biological Macromolecules Solution 56 The solution is (B). Both DNA and RNA are made of nucleotides containing a sugar (deoxyribose in DNA, ribose in RNA), a nitrogenous base (DNA has thymine, RNA has uracil). Both also have adenine, cytosine, guanine, and a phosphate group. Which molecule carries information in a form that is inherited from one generation to another? A DNA B mRNA C Proteins D tRNA Solution 57 The solution is (A). Hereditary information is stored in the sequence of nucleotides found in DNA. What are the four types and functions of RNA? A mRNA is a single-stranded transcript of DNA. rRNA is found in ribosomes. tRNA transfers specific amino acids to a growing protein strand. miRNA regulates the expression of mRNA strands. B mRNA is a single-stranded transcript of rRNA. rRNA is translated in ribosomes to make proteins. tRNA transfers specific amino acids to a growing protein strand. miRNA regulates the expression of the mRNA strand. C mRNA regulates the expression of the miRNA strand. rRNA is found in ribosomes. tRNA transfers specific amino acids to a growing protein strand. miRNA is a singlestranded transcript of DNA. D mRNA is a single-stranded transcript of DNA. rRNA transfers specific amino acids to a growing protein strand. tRNA is found in ribosomes. miRNA regulates the expression of the mRNA strand. Solution The solution is (A). Messenger RNA (mRNA) is a single-stranded copy of the sequencing in DNA. It leaves the nucleus and attaches to ribosomes for protein synthesis. Ribosomal RNA (rRNA) plus protein make up the ribosomes, which are used as points of protein synthesis. Transfer RNA (tRNA) is a single-stranded RNA that carries or transfers specific amino acids to the growing protein chain. MicroRNA (miRNA) is a single-stranded RNA that has been folded back on itself. This is trimmed into a short strand of about 20 nucleotide pairs. One of the strands is degraded, while the other binds onto protein. The miRNA-protein complex can attach to mRNA with complementary sequences and functions to regulate the expression of that mRNA strand. Advanced Placement Biology Instructor’s Solution Manual 3 | Biological Macromolecules 41 TEST PREP FOR AP® COURSES 58 Urey and Miller constructed an experiment to illustrate the early atmosphere of Earth and possible development of organic molecules in the absence of living cells. Which assumption did Urey and Miller make regarding conditions on Earth? A Electric sparks occurred to catalyze the reaction. B The composition of the gases in the atmosphere C There was sufficient oxygen for creating life. D It produced water-soluble organic molecules. Solution 59 The solution is (B). They assumed the composition of gases in the atmosphere included methane, ammonia, and hydrogen gases. Urey and Miller proposed that a series of reactions occurred that ultimately resulted in amino acid formation. What is true based on their theory? A Hydrogen, methane, water, and ammonia combined to create amino acids. B Hydrogen, methane, and oxygen combined to create macromolecules. C Nitrogenous bases combined to form monomers, then RNA. D Periodic elements combined to create molecules, then DNA. Solution 60 The solution is (A). Their experiment resulted in the spontaneous formation of amino acids, which form from hydrogen, nitrogen, and other compounds. How does Urey and Miller's model support the claim that simple precursors present on early Earth could have assembled into the complex molecules necessary for life? A The simple molecules assembled to form amino acids and nucleic acids. B The organic molecules assembled to form large complexes, such as water and methane. C The inorganic molecules assembled to form amino acids and nucleic acids. D The inorganic molecules assembled to form large complexes, such as water and methane. Solution 61 The solution is (A). Miller and Urey’s experiment resulted in the production of organic molecules, such as amino acids and nucleic acids, from the building blocks of water, methane, ammonia, and hydrogen gas. This was done in the absence of living organisms, but could be used for the development of them. Which statement most accurately describes the importance of the condensation stage during Urey and Miller’s experiment? A Condensed water enabled the formation of monomers. B Condensation and evaporation simulated lightning storms. Advanced Placement Biology Instructor’s Solution Manual 42 3 | Biological Macromolecules C Condensation and evaporation simulated the water cycle. D Condensed water enabled the formation of polymers. Solution 62 The solution is (C). Condensation and evaporation are main components of the water cycle. According to the findings of the Urey and Miller experiment, the primitive atmosphere consisted of water in the form of steam, methane, ammonia, and hydrogen gases. If there was so much hydrogen gas in the early atmosphere, why is there so little now? A Hydrogen gas is so light with a molecular weight of 1 that the excess diffused into space over time and is now absent from the atmosphere. B Hydrogen combined with ammonia to make ammonium. C It was all used up in the production of organic molecules. D The excess hydrogen gas was dissolved in the early oceans. Solution 63 The solution is (A). Hydrogen gas is so light (molecular weight of 1) that the excess diffused into space over time and is now absent from the atmosphere. Could the primitive atmosphere illustrated by the Urey and Miller experiment be reproduced on today’s Earth? Why or why not? A The primitive atmosphere cannot be created due to the oxidizing atmosphere and lack of hydrogen. B The primitive atmosphere can be created because the atmosphere is reducing and Earth has sufficient hydrogen to reproduce the conditions. C The primitive atmosphere cannot be created due to the presence of abundant water and hydrogen in the atmosphere. D The primitive atmosphere can be created because the atmosphere is oxidizing and has less hydrogen. Solution 64 The solution is (A). The atmosphere reproduced in the Urey and Miller experiment could not exist on present-day Earth. Most of the hydrogen gas has been used or diffused into space. The presence of large amounts of oxygen has created an oxidizing atmosphere that would break down any organic molecules that might be produced. What is structurally different between starch and cellulose that gives them different physical properties? A Cellulose is formed by - 1-4 glycosidic linkages and crosslinks, making it rigid, while starch has - 1-4 and - 1-6 glycosidic linkages without the tight crosslinks of cellulose. B Cellulose has rigid - 1-4 glycosidic linkages, while starch has less rigid - 1-4 glycosidic linkages. Advanced Placement Biology Instructor’s Solution Manual 3 | Biological Macromolecules 43 C Cellulose has amylose and amylopectin, making it more rigid than starch. D Starch has amylose and amylopectin that make it more rigid than cellulose. Solution 65 The solution is (B). Starch is made up of glucose monomers that are joined by - 1-4 or - 1-6 glycosidic bonds. Cellulose is made up of glucose monomers that are linked by - 1-4 glycosidic bonds. Complex polymers are built from combinations of smaller monomers. What type of reaction is illustrated in the figure, and what is the product of the reaction? A A synthesis reaction producing glucose B A hydrolysis reaction producing fructose C A condensation reaction producing lactose D A dehydration reaction producing water Solution The solution is (D). Two six-carbon rings with hydroxyl groups are shown. The hydroxyl group on one is highlighted red, and the hydrogen of a hydroxyl group of the other is highlighted red. An arrow points to two five carbon rings connected by an oxygen. The molecule is sucrose and is formed by the condensation of glucose and fructose. Advanced Placement Biology Instructor’s Solution Manual 44 66 3 | Biological Macromolecules The fatty acids of triglycerides are classified as saturated, unsaturated, or trans fat. What about the structure of these compounds gives them their physical characteristics? A Saturated fats and trans fats contain the greatest possible number of hydrogen atoms, while unsaturated fats do not. B Saturated and unsaturated fats have stable configurations, while trans fats are transient. C Unsaturated fats and trans fats have some double-bonded carbon atoms, while saturated fats do not. D Unsaturated and trans fats are the same; fatty acids are only found on opposite sides of a trans fat. Solution 67 The solution is (C). Unsaturated fats and trans fats have some double-bonded carbon atoms, while saturated fats do not. Carbohydrates serve various functions in different animals. Arthropods, like insects, crustaceans, and others, have an outer layer, called the exoskeleton, which protects their internal body parts. This exoskeleton is made mostly of chitin. Chitin is also a major component of the cell walls of fungi, the kingdom that includes molds and mushrooms. Chitin is a polysaccharide. What is the major difference between chitin and other types of polysaccharides? A Chitin is a nitrogen-containing polysaccharide, with repeating units of N-acetyl- -Dglucosamine, a modified sugar. B Chitin is similar to amylase, but with sulfur linkages between the monomers. C Chitin is similar to inulin, a polysaccharide with fructose plus additional glucose monomers. D Chitin contains phosphate groups that give it a stiffness not found in other polysaccharides. Solution 68 The solution is (A). Chitin is a nitrogen-containing polysaccharide, with repeating units of N-acetyl- -D-glucosamine, a modified sugar. Which categories of amino acids would you expect to find on the surface of a soluble protein, and which would you expect to find in the interior? A Nonpolar and charged amino acids will be present on the surface and polar in the interior of the membrane, whereas nonpolar will be found in the membraneembedded proteins. B Nonpolar and uncharged proteins will be found on the surface with nonpolar in the interior, while only nonpolar will be found in the embedded proteins. Advanced Placement Biology Instructor’s Solution Manual 3 | Biological Macromolecules 45 C Polar and charged amino acids will be found on the surface, whereas nonpolar will be found in the interior. D Polar and charged amino acids will be found on the surface of a membrane protein, whereas nonpolar will be found in the interior. The membrane protein will be polar and hydrophobic. Solution 69 The solution is (C). Polar amino acids—such as proline, asparagine, and glutamine— would be found at the surface of a soluble protein, while nonpolar amino acids— such as leucine, methionine, and glycine—would be oriented toward the interior. You have been identifying the sequence of a segment of a protein. The sequence to date is: leucine-methionine-tyrosine-alanine-glutamine-lysine-glutamate. You insert arginine between the leucine and methionine. What effect would this insertion have on the segment? A Arginine is a negatively charged amino acid and could attach to the glutamate at the end of the segment. B Inserting arginine places a positively charged amino acid in a portion that is nonpolar, creating the possibility of a hydrogen bond in this area. C There would be no effect other than an additional amino acid. D The arginine could attach to the lysine and bend the protein chain at this point. Solution 70 The solution is (B). Inserting arginine places a positively charged amino acid in a portion that is nonpolar, creating the possibility of a hydrogen bond in this area. What happens if even one amino acid is substituted for another in a polypeptide? Provide a specific example. A The change will definitely not be sufficient to have any effect on the function and structure of the protein. B The amino acid may not show any significant effect on the protein structure and function, or it may have a significant effect, as in the case of hemoglobin in individuals with sickle cell trait. C These changes would increase the possibility of having extra bends and loops in the proteins, as seen in Leber congenital disease. D These changes would modify the structures of proteins, making them nonfunctional. Solution The solution is (B). There are two possibilities when one amino acid is substituted for another: there may be no effect on the protein if the amino acid’s position was not critical to the tertiary structure of the protein, or it may cause an extra bond to be made or not made that would significantly alter the functional structure. A classic example of a single amino acid change that significantly modifies the function of a protein is in the case of sickle cell hemoglobin in which a valine replaces a glutamic acid. Advanced Placement Biology Instructor’s Solution Manual 46 3 | Biological Macromolecules 71 HIV is an RNA virus that affects CD4 cells, also known as T helper cells, in the human body. Which mechanism is most likely responsible for the fast rate at which HIV can spread? A Recombination B Mutation C Reassortment D Replication errors Solution 72 The solution is (B). The high rate of mutations allows HIV to develop resistance to antiviral drugs. Furthermore, as the virus mutates, it is not recognized any longer by existing antibodies and is not tagged for destruction. For many years, scientist believed that proteins were the source of heritable information. There are many thousands of different proteins in a cell, and they mediate the cell’s metabolism, producing the traits and characteristics of a species. Researchers working with DNA viruses proved that it is DNA that stores and passes on genes. They worked with viruses that have an outer coat of protein and a DNA strand inside. How did scientists prove that it was DNA, not protein, that is the primary source of heritable information? A The DNA and protein of the virus were tagged with different isotopes and exposed to the host cell whereas only the DNA was transferred to the host. B The DNA was tagged with an isotope, which was retained in the virus, proving it to be the genetic material. C The viral protein was tagged with an isotope, and the host cell was infected by it. This protein was transferred to the host. D The viral DNA, when sequenced, was found to be present in the host cell, proving it to be the hereditary material instead of protein. Solution The solution is (A). Researchers tagged the DNA and protein coat of the virus with different isotopes. Then, they exposed host cells to the virus and determined the Advanced Placement Biology Instructor’s Solution Manual 3 | Biological Macromolecules 47 tagged proteins did not enter the cell, but the tagged viral DNA did. The virus was replicated in the host cells, showing that the DNA was responsible for the passing of genetic information. 73 The genetic code is based on each amino acid being coded for by a distinctive series of three nucleic acid bases called a codon. The following is a short segment of DNA, using the slash symbol ( / ) to separate the codons for easy viewing: ATC/GTT/GAA/CTG/TAG/GAT/AAA A change has occurred in the segment resulting in the following: ATC/GTT/GTA/CTG/TAG/GAT/AAA What kind of change has occurred? A A substitution of T for A, resulting in a coding change for the third codon B An addition of C for G, lengthening the strand and changing every codon past the addition C A deletion of an A, resulting in a shortening and changing every codon past the deletion D No change has occurred; the same base was replaced with the same base. Solution 74 The solution is (A). In the third codon, A has been substituted by T, which may or may not change the amino acid. A change in DNA on a chromosome affects all proteins made from that gene for the life of the cell. A change in the RNA involved in protein production is short lived. What is the difference between the effects of the changes in the two types of nucleic acids? A DNA is the genetic material that is passed from parent cells to daughter cells and to future generations. B DNA would not affect the individuals as the proteins made are finally altered and modified. RNA would cause harm to the person as the RNA encoded by the DNA and is not altered. C DNA is the genetic material and is transferred from one generation to another, making use of repair mechanisms for every mutation. The RNA does not use a repair mechanism. D DNA, when mutated, makes use of the repair mechanisms and can be repaired, whereas RNA is not repaired and is transferred in generations. Solution The solution is (A). The DNA in the chromosome will last the life of the cell, and possibly beyond, as shown in the DNA harvested from archeological research. Any protein made from the information on the chromosome will reflect the change in DNA. The effect is permanent and will possibly cause the expression of a genetic disease. If the change happened in a gamete, the change may be transmitted to an Advanced Placement Biology Instructor’s Solution Manual 48 3 | Biological Macromolecules offspring. Any change in the type of RNA lasts only as long as that strand of RNA is intact. All RNA is eventually degraded and replaced with new strands, depending on the needs of the cell. Any change in the RNA will only have an effect until it disappears with the degradation of that RNA. RNA changes alone are not permanent. SCIENCE PRACTICE CHALLENGE QUESTIONS 3.1 Synthesis of Biological Macromolecules 75 The capture of radiant energy through the conversion of carbon dioxide and water into carbohydrates is the engine that drives life on Earth. Ribose, C5H10O5, and hexose, C6H12O6, form stable 5- and 6-carbon rings. The numbering of the carbons on these rings is important in organizing the description of the role these molecules play in biological energy transfer and information storage and retrieval. Glycolysis is a sequence of chemical reactions that converts glucose to two 3-carbon compounds called pyruvic acid. A. Create visual representations to show how when bonds in the glucose molecules are broken between carbon number 1 and the oxygen atom and between carbons 3 and 4, two molecules of pyruvic acid are produced. Several enzymes in the cell are involved in converting glucose to pyruvic acid. These enzymes are proteins whose amino acid sequences provide these functions. This protein structure is information that was inherited from the cell’s parent and is stored in deoxyribonucleic acid (DNA). The “deoxyribo” component of that name is a shorthand for 2-deoxyribose. B. Create a visual representation of 2-deoxyribose, 5-phosphate by replacing the OH at carbon 2 with a hydrogen atom and replacing the OH at carbon 5 with a hydrogen phosphate ion, HPO32– , whose structure is shown in problem AP3.2. Use your representation to show that both phosphorylation (the addition of a phosphate ion) at Advanced Placement Biology Instructor’s Solution Manual 3 | Biological Macromolecules 49 carbon 5 and removal of the hydroxide at carbon 2 produce water molecules in an aqueous solution where hydrogen ions are abundant. DNA is a polymer formed from a chain with repeated 2-deoxyribose, 5-phosphate molecules. C. Create a visual representation of three 2-deoxyribose, 5-phosphate molecules forming a chain in which an oxygen atom in the phosphate that is attached to the 5-carbon replaces the OH on the 3-carbon of the next ribose sugar. Solution Sample answer: A. Drawing should show that by breaking the molecule of glucose between carbon 1 and the oxygen atom and between carbons 3 and 4 two molecules of 3 carbon atoms are produced: one molecule will contain carbons 1, 2, and 3. The second molecule will contain carbons 4, 5, and 6. Hydrogen atoms are lost in oxidation reactions. B. A molecule of deoxyribose phosphate is shown. When this molecule forms, an OH group is lost from position two and position five. Each OH group can react with a hydrogen ion to form water. C. Advanced Placement Biology Instructor’s Solution Manual 50 3 | Biological Macromolecules 3.2 Carbohydrates 76 Cells are bounded by membranes composed of phospholipids. A phospholipid consists of a pair of fatty acids that may or may not have carbon-carbon double bonds, fused at the carboxylic acid with a three-carbon glycerol that is terminated by a phosphate, as shown in the figure below. Most cell membranes comprise two phospholipid layers with the hydrophilic phosphate ends of each molecule in the outer and inner surfaces. The hydrophobic chains of carbon atoms extend into the space between these two surfaces. The exchange of matter between the interior of the cell and the environment is mediated by this membrane with selective permeability. A. Pose questions that identify The important characteristics of this lipid bilayer structure The molecules that must be acquired from the environment and eliminated from the cell Relationships between the structures of these molecules and the structure of the bilayer Advanced Placement Biology Instructor’s Solution Manual 3 | Biological Macromolecules 51 Because the plasma cell membrane has both hydrophilic and hydrophobic properties, few types of molecules possess structures that allow them to pass between the interior of the cell and the environment through passive diffusion. The fluidity of the membrane affects passive transport, and the incorporation of other molecules in the membrane, in particular cholesterols, has a strong effect on its fluidity. Fluidity is also affected by temperature. Measurements of the speed of movement of oxygen molecules, O2, through three types of membranes were made (Widomska et al., Biochimica et Biophysica Acta, 1,768, 2007) and compared to the speed of movement of O2 through water. These measurements were carried out at four different temperatures. One type of membrane was obtained from the cells in the eyeball of a calf (lens lipid). Synthetic membranes composed of palmitic acid with cholesterol (POPC/CHOL) and without cholesterol (POPC) were also used. The results from these experiments are shown in the table. Material 15 °C 25 °C 35 °C 45 °C Lens lipids 15 cm/s 30 cm/ 65 cm/s 110 cm/s POPC/CHOL 15 cm/s 30 cm/s 65 cm/s 110 cm/s POPC 55 cm/s 100 cm/s 155 cm/s 280 cm/s Water 45 cm/s 55 cm/s 65 cm/s 75 cm/s B. Represent these data graphically. The axes should be labeled, and different symbols should be used to plot data for each material. C. Analyze the data by comparing transport of oxygen through the biological membrane, water, and the synthetic membranes. Consider both membrane composition and temperature in your analysis. The plasma membrane separates the interior and the exterior of the cell. A potential to do work is established by defining regions inside and outside the cell with different concentrations of key molecules and net charges. In addition to the membrane defining the cell boundary, eukaryotic cells have internal membranes. D. Explain how internal membranes significantly increase the functional capacity of the cells of eukaryotes relative to those of prokaryotes. Solution Sample answer: A. Phospholipids orient themselves in solution forming a bilayer that allows stable interaction with the polar inside and outside of the cell while separating the two environments. Nutrients must be acquired from the environment and waste must be eliminated. Ions must move across the membrane to maintain homeostasis. Certain gases, such as oxygen and carbon dioxide, must also enter or leave the cell. Gases and can cross freely the membrane. The membrane is also permeable to small hydrophobic molecules, but is impermeable to ions and polar molecules. However, small polar molecules can enter or leave the cell through transporters. Ions can enter or leave Advanced Placement Biology Instructor’s Solution Manual 52 3 | Biological Macromolecules through channels. Large polar molecules can be packaged into lipid vesicles that fuse with the membrane, allowing the molecules to move across. B. By convention, the independent variable is plotted on the x-axis and the dependent is plotted on the y-axis. In this graph, temperature is the independent variable and rate of diffusion is the dependent variable. Each medium should be represented by a separate line. The graph should have a title, the axes should be labeled and show units. A legend showing the symbol for each medium should be included. C. The speeds and slopes for the plots representing the lens membrane and POPC/CHOL are similar. The speed of diffusion of oxygen in water is the least affected by change in temperature as should be expected. Both the speed of diffusion in and the slope are greater in POPC. D. The increase in membrane surface increases the potential to do work by allowing different types of concentration gradients to be established inside the same cell. 3.3 Lipids 77 Proteins are polymers whose subcomponents are amino acids connected by peptide bonds. The carboxylic acid carbon, O=C–OH , of one amino acid can form a bond with the amine, NH2 , of another amino acid. In the formation of this peptide bond, the amine replaces the OH to form O=C–NH2 . The other product of this reaction is water, H2O . Advanced Placement Biology Instructor’s Solution Manual 3 | Biological Macromolecules 53 Amino acids can be synthesized in the laboratory from simpler molecules of ammonia (NH3), water (H2O), methane (CH4), and hydrogen (H2) if energy is provided by processes that simulate lightning strikes or volcanic eruptions (Miller, Science, 117, 1953; Johnson et al., Science, 322, 2008). A. The synthesis of amino acids in solutions under laboratory conditions consistent with early Earth was a step toward an explanation of how life began. Pose a question that should have been asked but was not until 2014 (Parker et al., Angewandte Chemie, 53, 2014), when these solutions that had been stored in a refrigerator were analyzed. The diversity and complexity of life begins in the variety of sequences of the 20 common amino acids. B. Apply mathematical reasoning to explain the source of biocomplexity by calculating the possible variations in a polymer composed of just three amino acids. Polarity in a bond between atoms occurs when electrons are distributed unequally. Polarity in a molecule also is caused by charge asymmetry. Life on Earth has evolved within a framework of water, H2O, one of the most polar molecules. The polarities of the amino acids that compose a protein determine the properties of the polymer. The electric polarity of an amino acid in an aqueous solution depends on the pH of the solution. Here are three forms of the general structure of an amino acid. C. Qualitatively predict the relationship between solution pH and the form of the amino acid for three solutions of pH: pH < 7, pH = 7, and pH > 7. Advanced Placement Biology Instructor’s Solution Manual 54 3 | Biological Macromolecules The properties of proteins are determined by interactions among the amino acids in the peptide-bonded chain. The protein subcomponents, especially amino R (variable) groups, can interact with very strong charge-charge forces, with attractive forces between groups of atoms with opposite polarities and with repulsive forces between groups of atoms with the same or no polarity. Attractive polar forces often arise between molecules through interactions between oxygen and hydrogen atoms or between nitrogen and hydrogen atoms. D. Consider particular orientations of pairs of three different amino acids. Predict the relative strength of attractive interaction of all pairs; rank them and provide your reasoning. In an amino acid, the atoms attached to the -carbon are called the R group. Interactions between R groups of a polypeptide give three-dimensional structure to the one-dimensional, linear sequence of amino acids in a polypeptide. E. Construct an explanation for the effect of R-group interactions on the properties of a polymer with drawings showing molecular orientations with stronger and weaker polar forces between R groups on asparagine and threonine and between asparagine and alanine. Advanced Placement Biology Instructor’s Solution Manual 3 | Biological Macromolecules Solution 55 Sample answer: A. Can amino acids form in the absence of volcanic eruptions and lightning (rare events) if common inorganic molecules can act as catalysts and facilitate synthesis of amino acids from ammonia, methane, water, and hydrogen? B. The number of possible variations if of the 20 possible amino acids form a peptide is 20 20 20 8,000 because each amino acid added independently from the pool of 20 amino acids. C. At acidic pH when pH < 7, excess H+ ions will bind to amino acids. The prevalent form will be the structure on the right with an ionized carboxyl group and NH2 bound to the alpha carbon. At neutral pH, pH=7, the center structure which is electrically neutral is prevalent. At basic pH, pH > 7, the structure on the left, with a positively charged amine group NH3 and a carboxyl group, is prevalent. D. From the strongest to the weakest interactions: Threonine has a carboxyl group facing an amine on asparagine. Positive and negative charges form an ionic bond. Serine and threonine will attract because of hydrogen bonding between the hydrogen in the hydroxyl group of threonine and the oxygen in the carbonyl group of serine. Last the serine and asparagine will attract each other as they are both polar molecules. E. The amide R group of asparagine will interact more strongly with the hydroxyl of threonine because both R groups are polar. Alanine which has a hydrophobic R group–CH2 will not interact strongly with either asparagine or alanine side chains. The asparagine amide group will form hydrogen bonding with the carboxyl group on the alpha carbon of either threonine or asparagine. 3.4 Proteins 78 The nucleobase part of deoxyribonucleic acid encodes information in each component in the sequence making up the polymer. There are five nucleobases that are commonly represented by only a single letter: A (adenine), C (cytosine), G (guanine), T (thymine), and U (uracil). These molecules form a bond with the 1-carbon of deoxyribose. In this Advanced Placement Biology Instructor’s Solution Manual 56 3 | Biological Macromolecules problem, we need to look at the molecules in slightly more detail so that you can develop the ability to explain why DNA, and sometimes RNA, is the primary source of heritable information. Edwin Chargaff and his team isolated nucleobases from salmon sperm and determined the fraction of each (Chargaff et al., Journal of Biological Chemistry, 192, 1951). Experiments in which the fraction of all four nucleobases was determined are shown. Also shown are averages of two standard deviations and the sum of total fractions for each experiment. Precision is calculated with each average. Shown below are the chemical structures of these four nucleobases. In these structures, the nitrogen that attaches to the 2-deoxyribose, 5-phosphate polymer is indicated as N*. The partial charges of particular atoms are indicated with δ+ and δ−. A. Analyze Chargaff’s data in terms of the partial charges on these molecules to show how molecular interactions affect the function of these molecules in the storage and retrieval of biological information. Experiment Adenine Guanine Cytosine Thymine Total 5 0.28 0.20 0.21 0.27 0.96 6 0.30 0.22 0.20 0.29 1.01 7 0.27 0.18 0.19 0.25 0.89 8 0.28 0.21 0.20 0.27 0.94 11 0.29 0.18 0.20 0.27 0.94 12 0.28 0.21 0.19 0.26 0.94 13 0.30 0.21 0.20 0.30 1.01 Average 0.29 ± 0.02 0.20 ± 0.03 0.20 ± 0.01 0.27 ± 0.02 0.96 ± 0.08 Advanced Placement Biology Instructor’s Solution Manual 3 | Biological Macromolecules 57 The interactions between nucleobase molecules are strong enough to produce the association of pairs observed in Chargaff’s data. However, these pairs are bonded by much weaker hydrogen bonds, which are chemical bonds within the molecules. Demonstrating understanding of the replication of DNA requires the ability to explain how the two polymer strands of the double helix interact and grow. To retrieve information from DNA, the strands must be separated. The proteins that perform that task interact with the polymer without forming new chemical bonds. In their paper (Watson and Crick, Nature, 3, 1953) announcing the structure of the polymer considered in this problem, Watson and Crick stated, “It has not escaped our notice that the specific pairing we have postulated immediately suggests a possible copying mechanism for the genetic material.” Eschenmoser and Lowenthal (Chemical Society Reviews, 21, 1992) asked why the 5-carbon sugar ribose is used in DNA when the 6-carbon sugar glucose is so common in biological systems. To answer the question, they synthesized polymeric chains of artificial DNA using glucose. They discovered that the strength of the interaction between pairs of nucleobases increased in the DNA with glucose. Paired strands of hexose-based polymers were more stable. The AP Biology Curriculum Framework (College Board, 2012) states, “The double-stranded structure of DNA provides a simple and elegant solution for the transmission of heritable information to the next generation; by using each strand as a template, existing information can be preserved and duplicated with high fidelity within the replication process. However, the process of replication is imperfect.” B. Based on the findings of Eschenmoser and Lowenthal, why didn’t DNA evolve to use glucose rather than hexose? What does this have to do with the idea that “replication is imperfect” in DNA? Thoroughly explain your answers. Solution Sample answer: A. Adenine and thymine are present in a similar percentage. Cytosine and guanine are also present in a similar percentage. The results are reproducible within small error margins. The molecular interactions between the single two pairs are stabilized by opposite partial charges. B. The interactions between the nitrogenous bases should be weak enough for the two strands of DNA to separate during replication and transcription and specific enough that adenine pairs only with thymine and cytosine with guanine. More stable interactions would require too much energy to open the double helix. Weaker interactions also mean that mismatches between nitrogenous bases are possible, making replication an imperfect process that may lead to mutations if errors are not repaired. Advanced Placement Biology Instructor’s Solution Manual 58 4 | Cell Structure 4 | CELL STRUCTURE REVIEW QUESTIONS 1 When viewing a specimen through a light microscope, what is a method that scientists use to make it easier to see individual components of cells? A A beam of electrons B High temperatures C Radioactive isotopes D Special stains Solution 2 The solution is (D). Special stains are used to enhance visualization of the cell or certain cellular components under a microscope. Cells may also be stained to highlight metabolic processes or to differentiate between live and dead cells in a sample. What is the basic unit of life? A Cell B Organism C Organ D Tissue Solution 3 The solution is (A). The cell is the basic unit of life. All organisms consist of one or more cells. Which statement is part of the cell theory? A All living organisms are made of cells. B All cells contain DNA that they pass on to daughter cells. C All cells depend on their surroundings to provide energy. D All cells have a nucleus. Solution 4 The solution is (A). The cell theory given by Schleiden and Schwann states the cell is the basic unit of life and all the living cells are composed of cells. The third tenet of the cell theory is that every cell originates from another cell. There is no spontaneous generation. What could most effectively be visualized with a scanning electron microscope? A Cells swimming in a drop of pond water B Details of structures inside cells Advanced Placement Biology Instructor’s Solution Manual 4 | Cell Structure 59 C A three-dimensional view of the surface of a membrane D The movement of molecules inside the cell Solution 5 The solution is (C). The SEM uses the electron beam to view the surface of a dried and fixed specimen. The molecules in the specimen get excited and release secondary electrons, which are captured by a detector. Who was the first to clearly identify and name individual cells? A Anton van Leeuwenhoek B Matthias Schleiden C Robert Hooke D Theodor Schwann Solution 6 The solution is (C). Robert Hooke coined the term cell when he was viewing cork tissue through a lens and saw some box-like structures. Which observation contributed to the cell theory? A Animal and plant cells have nuclei and organelles. B Nonliving material cannot give rise to living organisms. C Prokaryotic and eukaryotic cells are surrounded by a plasma membrane. D Viruses replicate. Solution 7 The solution is (B). The theory states cells reproduce and give rise to other cells, which implies nonliving materials cannot give rise to living organisms; that means that there is no spontaneous generation. To obtain some materials and remove waste, what process is used by prokaryotes? A Cell division B Diffusion C Flagellar motion D Ribosomes Solution 8 The solution is (B). The changes in the concentration help the movement of the particles and waste across the plasma membrane. This process is known as diffusion, where molecules move from high to low concentration. When bacteria lack fimbriae, what are they less likely to do? A Adhere to cell surfaces B Retain the ability to divide C Swim through bodily fluids D Synthesize proteins Advanced Placement Biology Instructor’s Solution Manual 60 4 | Cell Structure Solution 9 The solution is (A). Fimbriae, the “attachment pili,” are used by bacteria to adhere to one another or to surfaces. They can contribute to motility by causing twitching. What is a difference between prokaryotic and eukaryotic cells? A Both cells have a nucleus, but prokaryotic cells lack cytoplasm. B Both cells have cytoplasm, but prokaryotic cells lack a nucleus. C Both cells have DNA, but prokaryotic cells lack a cell membrane. D Both cells have a cell membrane, but prokaryotic cells lack DNA. Solution 10 The solution is (B). Prokaryotic cells lack a well-defined nucleus where DNA is surrounded by a nuclear membrane. Instead, DNA is folded into a compact structure called the nucleoid. Eukaryotic cells contain complex organelles that carry out their chemical reactions. Prokaryotes lack many of these complex organelles, although they have a variety of unique structures of their own. However, most prokaryotic cells can exchange nutrients with the outside environment faster than most eukaryotic cells. Why is this so? A Most prokaryotic cells are smaller and have a higher surface-to-volume ratio than eukaryotic cells. B Most prokaryotic cells are larger and have a higher surface-to-volume ratio than eukaryotic cells. C Most prokaryotic cells are smaller and have a lower surface-to-volume ratio than eukaryotic cells. D Prokaryotic cells are larger and have a lower surface-to-volume ratio than eukaryotic cells. Solution 11 The solution is (A). Due to their small size (0.1 to 5 µm in diameter), ions and organic molecules that enter prokaryotic cells can quickly diffuse to other parts of the cell. Likewise, waste products produced within a prokaryotic cell can also quickly diffuse out. By contrast, eukaryotic cells have developed different structural adaptations to enhance intracellular transport. What is surrounded by one phospholipid bilayer? A Lysosomes B Ribosomes C Nucleolus D Nucleus Solution The solution is (D). The nucleus is surrounded by a single phospholipid bilayer, the nuclear membrane. Protein-lined channels called nuclear pores allow traffic of molecules such as RNA through the nuclear membrane. Advanced Placement Biology Instructor’s Solution Manual 4 | Cell Structure 12 61 Peroxisomes got their name because hydrogen peroxide is — A a cofactor for the organelles’ enzymes B incorporated into their membranes C produced during their oxidation reactions D used in their detoxification reactions Solution 13 The solution is (C). Certain enzymes found in peroxisomes use molecular oxygen to remove hydrogen from specific organic substrates, producing hydrogen peroxide. Because of the production of hydrogen peroxide, the organelle has been named peroxisome. In plant cells, what carries out the function of the lysosomes? A Nuclei B Peroxisomes C Ribosomes D Vacuole Solution 14 The solution is (D). The vacuole plays a role similar to lysosomes and fulfills many functions. It contains digestive enzymes, isolates harmful materials, stores waste products and secondary compounds such as pigments in petals, latex, and alkaloids that deter predators. It helps maintain the pressure within a cell by changing its content of water, contributes to the balance of pH, exports products, and stores proteins for seed germination. What is found in both eukaryotic and prokaryotic cells? A Mitochondrion B Nucleus C Ribosomes D Centrosomes Solution 15 The solution is (C). Prokaryotes and eukaryotes both require proteins for their survival; thus, both contain ribosomes, the protein-synthesizing machinery found in both prokaryotic and eukaryotic cells. Which structure is NOT found in prokaryotic cells? A Plasma membrane B Chloroplast C Nucleoid D Ribosome Advanced Placement Biology Instructor’s Solution Manual 62 4 | Cell Structure Solution 16 The solution is (B). Chloroplasts and other organelles are found only in the eukaryotic cells. Chloroplasts and mitochondria originated in engulfed bacteria that were not digested by the host; they lost their autonomy and became symbionts. Where would you find DNA, the genetic material, in an animal cell? A In the centriole B Only in the mitochondria C In the mitochondria and the nucleus Solution 17 The solution is (C). DNA is present in the nucleus as well as mitochondrion. Mitochondria contain their own DNA and ribosomes. Most DNA is located in the nucleus. What is most likely to have the greatest concentration of smooth endoplasmic reticulum (SER)? A A cell that secretes enzymes B A cell that destroys pathogens C A cell that makes steroid hormones D A cell that engages in photosynthesis Solution 18 The solution is (C). Steroid-secreting cells are characterized by abundant SER. For example, cells in the gonads and the adrenal gland have an extensive network of SER because they synthesize cholesterol as a precursor for steroid hormones or take up this substrate from plasma lipoproteins. Many of the enzymes for sterol and steroid synthesis are localized in the SER. Which sequence correctly lists in order the steps involved in the incorporation of a protein within a cell membrane? A Synthesis of the protein on the ribosome; modification in the Golgi apparatus; packaging in the endoplasmic reticulum; modification in the vesicle B Synthesis of the protein on the lysosome; modification in the Golgi; packaging in the vesicle; distribution in the endoplasmic reticulum C Synthesis of the protein on the ribosome; modification in the endoplasmic reticulum; tagging in the Golgi; distribution via the vesicle D Synthesis of the protein on the lysosome; packaging in the vesicle; distribution via the Golgi; modification in the endoplasmic reticulum Solution The solution is (C). The protein is synthesized by ribosomes and modified by the endoplasmic reticulum. The sorting, tagging, packaging, and distribution of lipids and proteins takes place in the Golgi apparatus, which is then distributed by vesicles. Advanced Placement Biology Instructor’s Solution Manual 4 | Cell Structure 19 63 What is NOT a component of the endomembrane system? A Endoplasmic reticulum B Golgi apparatus C Lysosome D Mitochondrion Solution 20 The solution is (D). The endomembrane system does not include the membranes of mitochondria or chloroplasts because these organelles have separate DNA. These were actually engulfed bacteria that evolved into symbionts. What has the ability to disassemble and reform quickly? A Intermediate filaments and microtubules B Microfilaments and intermediate filaments C Microfilaments and microtubules D Only intermediate filaments Solution 21 The solution is (C). Microfilaments, such as actin, and microtubules found in flagellum have the capacity to polarize and depolarize quickly to perform their functions. What does NOT play a role in intracellular movement? A Intermediate filaments and microtubules B Microfilaments and intermediate filaments C Microfilaments and microtubules D Only intermediate filaments Solution 22 The solution is (D). Recall that intermediate filaments provide the cell its shape and help in fastening organelles in place. Which components of the cytoskeleton are responsible for the contraction of muscles? A Intermediate filaments B Microfilaments C Microtubules Solution The solution is (B). Microfilaments are made up of actin filaments, which serve as a track for movement of a motor protein called myosin. When actin and myosin slide past each other, the muscles contract. Advanced Placement Biology Instructor’s Solution Manual 64 23 4 | Cell Structure What type of junctions prevents the movement of chemicals between two adjacent animal cells? A Desmosomes B Gap junctions C Plasmodesmata D Tight junctions Solution 24 The solution is (D). Tight junctions composed of occludins and claudins are essential for forming an impermeable barrier to fluids. They are typically found in cells of the epithelial tissues that line exposed internal and external surfaces in animal bodies. For example, cells in the intestinal mucosa are joined by tight junctions to prevent leakage of intestinal content directly into the underlying tissues. Gap junctions are formed by — A gaps in the cell wall of plants B protein complexes that form channels between cells C tight, rivet-like regions in the membranes of adjacent cells D a tight knitting of membranes Solution 25 The solution is (B). Gap junctions are made of protein complexes that make way for the movement of the substances across the cells. For example, cardiac muscle cells are connected by gap junctions that allow rapid transfer of the electric signal for synchronized muscle contraction. Some animal cells produce an extensive extracellular matrix. You would expect their ribosomes to synthesize large amounts of which proteins? A Actin B Collagen C Myosin D Tubulin Solution 26 The solution is (B). Collagen is interwoven with proteoglycans, and collectively these molecules form the extracellular matrix. Actin, myosin, and tubulin are found inside the cells and are part of the cytoskeleton. Which molecule is typically found in the extracellular matrix? A Nucleic acids such as DNA B Peptidoglycans C Cellulose D Proteoglycans Advanced Placement Biology Instructor’s Solution Manual 4 | Cell Structure Solution 65 The solution is (D). Proteoglycans are a major component of the extracellular matrix. Nucleic acids are found inside the cell, and cellulose forms the cell wall of plant cells. CRITICAL THINKING QUESTIONS 27 Which element of the cell theory has practical applications in health care because it promotes the use of sterilization and disinfection? A All cells come from preexisting cells. B All living organisms are composed of one or more cells. C A cell is the basic unit of life. D A nucleus and organelles are found in prokaryotic cells. Solution 28 The solution is (A). All cells come from preexisting cells. If all microorganisms are removed or killed, then the environment, medical devices, or other items are sterile or, at least, have fewer potential pathogens. What are the advantages and disadvantages of light microscopes? What are the advantages and disadvantages of electron microscopes? A Advantage: In light microscopes, the light beam does not kill the cell. Electron microscopes are helpful in viewing intricate details of a specimen and have high resolution. Disadvantage: Light microscopes have low resolving power. Electron microscopes are costly and require killing the specimen. B Advantage: Light microscopes have high resolution. Electron microscopes are helpful in viewing surface details of a specimen. Disadvantage: Light microscopes kill the cell. Electron microscopes are costly and low resolution. C Advantage: Light microscopes have high resolution. Electron microscopes are helpful in viewing surface details of a specimen. Disadvantage: Light microscopes can be used only in the presence of light and are costly. Electron microscopes use a short wavelength of electrons and hence have lower magnification. D Advantage: Light microscopes have high magnification. Electron microscopes are helpful in viewing surface details of a specimen. Disadvantage: Light microscopes can be used only in the presence of light and have lower resolution. Electron microscopes can be used only for viewing ultra-thin specimens. Solution The solution is (A). The advantages of light microscopes are that they are easily obtained, and the light beam does not kill the cells. However, typical light microscopes are somewhat limited in the amount of detail they can reveal. Electron microscopes are ideal because you can view intricate details; however, they are bulky, costly, and preparation for the microscopic examination kills the specimen. Advanced Placement Biology Instructor’s Solution Manual 66 29 4 | Cell Structure Mitochondria are observed in plant cells that contain chloroplasts. Why do you find mitochondria in photosynthetic tissue? A Mitochondria are not needed but are an evolutionary relic. B Mitochondria and chloroplasts work together to use light energy to make sugars. C Mitochondria participate in the Calvin cycle/light-independent reactions of photosynthesis. D Mitochondria are required to break down sugars and other materials for energy. Solution 30 The solution is (D). The cell needs energy constantly. When light is available, the cell can produce energy directly from photosynthesis; but, in the dark, it generates energy as ATP mainly through respiration, which involves the activity of mitochondria. In what situation(s) would the use of a light microscope be ideal? Why? A A light microscope is used to view the details of the surface of a cell, as it cannot be viewed in detail by the transmission microscope. B A light microscope allows visualization of small cells that have been stained. C A standard light microscope is used to view living organisms with little contrast to distinguish them from the background, which would be harder to see with the electron microscope. D A light microscope reveals the internal structures of a cell, which cannot be viewed by transmission electron microscopy. Solution 31 The solution is (B). Light microscopes are advantageous for viewing living organisms, but since individual cells are generally transparent, their components are not distinguishable unless they are colored with special stains. Staining, however, usually kills the cells. The major role of the cell wall in bacteria is protecting the cell against changes in osmotic pressure, pressure caused by different solute concentrations in the environment. Bacterial cells swell but do not burst in low solute concentrations. What happens to bacterial cells if a compound that interferes with the synthesis of the cell wall is added to an environment with low solute concentrations? A Bacterial cells will shrink due to the lack of cell wall material. B Bacterial cells will shrink in size. C Bacterial cells may burst due to the influx of water. D Bacterial cells remain normal; they have alternative pathways to synthesize cell walls. Solution The solution is (C). The cells will burst because of the movement of water inside the cell because there is no cell wall protection against bursting. Important antibiotics such as penicillin kill bacteria by interfering with the cell wall synthesis. Advanced Placement Biology Instructor’s Solution Manual 4 | Cell Structure 32 67 There is a lower limit to cell size. What determines how small a cell can be? A The cell should be large enough to escape detection. B The cell should be able to accommodate all the structures and metabolic activities necessary to survival. C The size of the cell should be large enough to reproduce itself. D The cell should be large enough to adapt to the changing environmental conditions. Solution 33 The solution is (B). A cell must be large enough to accommodate all the structures and metabolic activities that are required to sustain life. A smaller size also facilitates predation and reduces the ability to bind to surfaces. Which statement is a possible explanation for the presence of a rigid cell wall in plants? A Plants remain exposed to changes in temperature and thus require rigid cell walls to protect themselves. B Plants are subjected to variations in osmotic pressure, and a cell wall helps them against bursting or shrinking. C Plant cells have a rigid cell wall to protect themselves from grazing animals. D Plant cells have a rigid cell wall to prevent the influx of waste material. Solution 34 The solution is (B). Plant cells are subject to extreme changes in osmotic pressures. The rigid cell wall protects tissues against excessive shrinking or bursting of cells. Bacteria do not have organelles, yet the same reactions that take place on the mitochondria inner membrane, the phosphorylation of ADP to ATP, and chloroplasts, photosynthesis, take place in bacteria. Where do these reactions take place? A These reactions take place in the nucleoid of the bacteria. B These reactions occur in the cytoplasm present in the bacteria. C These reactions occur on the plasma membrane of bacteria. D These reactions take place in the mesosomes. Solution The solution is (C). ATP formation and photosynthesis occur on the plasma membrane. In some photosynthetic bacteria such as the cyanobacteria, photosynthesis occurs on loose internal membranes. Advanced Placement Biology Instructor’s Solution Manual 68 35 4 | Cell Structure What are the structural and functional similarities and differences between mitochondria and chloroplasts? A Similarities: double membrane, intermembrane space, ATP production, contain DNA. Differences: Mitochondria have inner folds called cristae; chloroplast contains accessory pigments in thylakoids, which form grana and stroma. B Similarities: DNA, intermembrane space, ATP production, and chlorophyll. Differences: Mitochondria have a matrix and inner folds called cristae; chloroplast contains accessory pigments in thylakoids, which form grana and a stroma. C Similarities: double membrane and ATP production. Differences: Mitochondria have intermembrane space and inner folds called cristae; chloroplast contains accessory pigments in thylakoids, which form grana and a stroma. D Similarities: double membrane and ATP production. Differences: Mitochondria have intermembrane space, inner folds called cristae, ATP synthase for ATP synthesis, and DNA; chloroplast contains accessory pigments in thylakoids, which form grana and a stroma. Solution 36 The solution is (A). Both are enveloped in a double membrane, have an intermembrane space, and make ATP. Mitochondria and chloroplasts have DNA, and mitochondria have inner folds called cristae and a matrix, while chloroplasts have chlorophyll and accessory pigments in the thylakoids that form stacks (grana) and a stroma. Is the nuclear membrane part of the endomembrane system? Why or why not? A The nuclear membrane is not a part of the endomembrane system, as the endoplasmic reticulum is a separate organelle of the cell. B The nuclear membrane is considered a part of the endomembrane system, as it is continuous with the Golgi body. C The nuclear membrane is part of the endomembrane system, as it is continuous with the rough endoplasmic reticulum. D The nuclear membrane is not considered a part of the endomembrane system, as the nucleus is a separate organelle. Solution 37 The solution is (C). Because the external surface of the nuclear membrane is continuous with the rough endoplasmic reticulum, which is part of the endomembrane system, it is part of the endomembrane system. What happens to the proteins that are synthesized on free ribosomes in the cytoplasm? Do they go through the Golgi apparatus? A These proteins move through the Golgi apparatus and enter in the nucleus. B These proteins go through the Golgi apparatus and remain in the cytosol. Advanced Placement Biology Instructor’s Solution Manual 4 | Cell Structure 69 C The proteins do not go through the Golgi apparatus and move into the nucleus for processing. D The proteins do not go through the Golgi apparatus and remain free in the cytosol. Solution 38 The solution is (D). Most of the proteins synthesized on free ribosomes stay in the cytosol. They do not go through the Golgi apparatus to be tagged for export or integration in membranes. What are the similarities and differences between the structures of centrioles and flagella? A Centrioles and flagella are made of microtubules but show different arrangements. B Centrioles are made of microtubules, but flagella are made of microfilaments, and both show the same arrangement. C Centrioles and flagella are made of microfilaments. Centrioles have a 9 + 2 arrangement. D Centrioles are made of microtubules, flagella are made of microfilaments, and both have different structures. Solution 39 The solution is (A). Centrioles and flagella are made of microtubules. Centrioles have two rings of nine microtubule “triplets” arranged at right angles to one another. Flagella, in turn, show 9 + 2 arrangement. Inhibitors of microtubule assembly, vinblastine for example, are used for cancer chemotherapy. How does an inhibitor of microtubule assembly affect cancerous cells? A The inhibitors restrict the separation of chromosomes by the mitotic spindle. B The inhibition of microtubules interferes with the synthesis of proteins. C The inhibitors bind the microtubule to the nuclear membrane. D The inhibitors interfere with energy production. Solution 40 The solution is (A). Cancerous cells divide rapidly. By inhibiting microtubules’ movement, cell division stops because the chromosomes cannot be separated by the centrioles. How do cilia and flagella differ? A Cilia are made of microfilaments and flagella of microtubules. B Cilia are helpful in the process of engulfing food. Flagella are involved in the movement of the organism. C Cilia are short and found in large numbers on the cell surface, whereas flagella are long and fewer in number. D Cilia are found in prokaryotic cells and flagella in eukaryotic cells. Solution The solution is (C). Cilia and flagella are alike in that they are made up of microtubules. Cilia are short, hair-like structures that exist in large numbers and Advanced Placement Biology Instructor’s Solution Manual 70 4 | Cell Structure usually cover the entire surface of the plasma membrane. In contrast, flagella are long, hair-like structures; when flagella are present, a cell has just one or two. 41 In which human tissues would you find desmosomes? A Bone cells and cartilage cells B Muscle cells and skin cells C Nerve cells and muscle cells D Secretory cells and muscle cells Solution 42 The solution is (B). Muscle and skin cells undergo most of the mechanical stress, thus show the presence of desmosomes. If there is a mutation in the gene for collagen, such as the one involved in Ehlers-Danlos syndrome, and the individual produces defective collagen, how would it affect coagulation? A The syndrome affects the clotting factors and platelet aggregation. B The syndrome leads to hypercoagulation of blood. C Coagulation is not affected because collagen is not required for coagulation. D The syndrome occurs due to the breakdown of platelets. Solution 43 The solution is (A). The Ehlers-Danlos syndrome affects the connective tissues and indirectly the coagulation process as well. It manifests in type I to type X, which show the malfunctioning of different clotting factors and platelet aggregation. This leads to impaired coagulation, causing patients to bruise and bleed easily. The research on this disease is ongoing. How does the structure of a plasmodesma differ from that of a gap junction? A Gap junctions are essential for transportation in animal cells, and plasmodesmata are essential for the movement of substances in plant cells. B Gap junctions are found to provide attachment in animal cells, and plasmodesmata are essential for the attachment of plant cells. C Plasmodesmata are essential for communication between animal cells, and gap junctions are necessary for attachment of cells in plant cells. D Plasmodesmata help in transportation, and gap junctions help in attachment in plant cells. Solution The solution is (A). A plasmodesma is a channel between the cell walls of two adjacent plant cells. Plasmodesmata allow materials to pass from the cytoplasm of one plant cell to the cytoplasm of an adjacent cell. A gap junction is a protein-lined pore that allows water and small molecules to pass between adjacent animal cells. Advanced Placement Biology Instructor’s Solution Manual 4 | Cell Structure 71 TEST PREP FOR AP® COURSES 44 Which organism appears first in the fossil record? A Archaea B Fish C Protists D Plants Solution 45 The solution is (A). The earliest fossils reported so far are layered structures called stromatolites from Greenland dating from 3.7 billion years ago and hydrothermal vent structures from Canada dating from 3.7 to 4.2 billion years ago. Both structures are of microbial origin. Why is it challenging to study bacterial fossils and determine if the fossils are members of the domain Archaea rather than Bacteria? A Bacteria lack rigid structures and thus do not form fossils. B Bacteria have rigid structures, but their fossil impression is scarce. C Fossils of bacteria are rarely found because bacteria were not abundant in the past. D Fossils of bacteria changes over time due to the presence of new bacteria living on them. Solution 46 The solution is (A). Bacteria do not easily form fossils because they lack rigid structures such as shells or skeletons. It is hard to identify the biochemical features that differentiate archaea from bacteria in fossils. Pictured are two cells along with their radius. What does cell (b) likely have when compared to cell (a)? A Smaller surface area and larger volume B Larger surface area and smaller volume C Smaller surface area-to-volume ratio D Larger surface area-to-volume ratio Advanced Placement Biology Instructor’s Solution Manual 72 4 | Cell Structure Solution 47 The solution is (C). Cell (b) will have the lowest ratio and would not exchange materials as rapidly as cell (a). Consider the shapes. The diameter of the sphere is equal to 1 mm and the side of the cube is also equal to 1 mm. What is the ratio of the surface-to-volume ratios for the sphere and the cube? A 3:1 B 4:1 C 1:1 D 2:1 Solution The solution is (C). The surface area-to-volume ratio of circle to square will be the same. Remember that the diameter (D), not the radius (r), is equal to 1 mm. Therefore, r D / 2 . For the sphere: area 4 r 2 4 D / 2 mm2 D2 mm2 mm2 2 volume 4 / 3 r 3 4 / 3 D / 2 mm3 4 / 3 D / 2 4 / 3 D3 / 8 mm3 1 / 6 mm3 3 3 area/volume / 1 / 6 mm1 6 mm1 For the cube: area 6 s2 6 1 6 mm2 volume s3 1 mm3 area/volume 6 mm1 48 Consider the shapes. The diameter of the sphere is equal to 1 mm and the side of the cube is also equal to 1 mm. What is true regarding the surface-to-volume ratios of the cube and the sphere? A The sphere will have a higher surface area than the cube. Advanced Placement Biology Instructor’s Solution Manual 4 | Cell Structure 73 B The sphere will have a higher volume than the cube. C The sphere will have a higher surface area-to-volume ratio than the cube. D Their surface area-to-volume ratios will be equal. Solution The solution is (D). Remember that the diameter (D), not the radius (r), is equal to 1 mm. Therefore, r D / 2 . For the sphere: area 4 r 2 4 D / 2 mm2 D2 mm2 mm2 2 volume 4 / 3 r 3 4 / 3 D / 2 mm3 4 / 3 D / 2 4 / 3 D3 / 8 mm3 1 / 6 mm3 3 3 area/volume / 1 / 6 mm1 6 mm1 For the cube: area 6 s2 6 1 6 mm2 volume s3 1 mm3 area/volume 6 mm1 49 What is the major consideration in setting the lower limit of cell size? A The cell must be large enough to fight the pathogens. B The cell must be large enough to attach to a substrate. C The lower limit should be small enough for the cell to move in the fluid efficiently. D The cell size must be small as to fit all the processes and structures to support life. Solution 50 The solution is (D). The cells must be large enough to include all of the necessary structures and materials, including DNA, ribosomes, enzymes, and other cellular structures, determining the lower end of the scale. It should be able to support life. Which structure has the same general structure in the domains Archaea, Bacteria, and Eukarya, pointing to a common origin? A Centriole B Cytoplasmic membrane C Golgi apparatus D Nucleus Solution The solution is (B). Cytoplasmic membrane is found in all the three domains. The centriole, Golgi apparatus, and the nucleus are found only in the domain Eukarya. Archaea and Bacteria are prokaryotes. Advanced Placement Biology Instructor’s Solution Manual 74 51 4 | Cell Structure Why does the structure of the cytoplasmic membrane point to a common ancestor? A The presence of a cytoplasmic membrane in every organism does not point to a common ancestry. B The similar arrangement of phospholipids and proteins points to common ancestry. C The lipid nature of the membrane makes it the most primitive trait. D The similar effect of temperature on the membrane makes it the ancestral trait. Solution 52 The solution is (B). The similar arrangement of macromolecules, phospholipids, and embedded proteins in the lipid bilayer makes it a primitive trait. Archaea, Bacteria, and Eukarya display a closely related general architecture of the plasma membrane. Which organelles would be present in high numbers in the leg muscles of a marathon runner? A Centrioles B Chloroplasts C Mitochondria D Peroxisome Solution 53 The solution is (C). Mitochondria produce the required energy in the form of ATP to run a marathon. The higher number of mitochondria is needed to sustain a prolonged activity. Macrophages ingest and digest many pathogens. Which organelle plays a major role in the activity of macrophages? A Chloroplast B Lysosome C Nucleus D Peroxisome Solution 54 The solution is (B). Lysosomes contain the hydrolytic enzymes that digest the ingested pathogen. The pathogens are engulfed in a vesicle called a phagosome that fuses with a lysosome in the cytoplasm forming, a phagolysosome where digestion takes place. This process is called phagocytosis. You are looking at a sample under a light microscope and observe a new type of cell. You come to the conclusion that it is a bacterium and not a eukaryotic cell. What would you observe to come to this conclusion? A The cell has a cell wall. B The cell has a flagellum. Advanced Placement Biology Instructor’s Solution Manual 4 | Cell Structure 75 C The cell does not have a nucleus. D The cell is small. Solution 55 The solution is (C). Only a bacterial cell would lack a nucleus. Plant and fungal cells, and some protists have cell walls. Flagella are found in both prokaryotic and eukaryotic cells. Bacteria are generally smaller than eukaryotic cells, but there are notable exceptions. The bacterium Thiomargarita namibiensis can be seen with the naked eye. Thiomargarita namibiensis is a large single-cell organism, which can reach lengths of 700 μm. The cell is classified as a bacterium. What is the main argument to justify the classification? A This organism shows simple diffusion for the uptake of nutrients and is thus classified as a bacterium. B This organism does not show presence of any cell organelles and is thus classified as a bacterium. C This organism appears pearl-like and exists in long chains and is thus classified as a bacterium. D This organism demonstrates characteristics of gram-negative bacteria and is thus classified as a bacterium. Solution 56 The solution is (B). Like the other prokaryotes, Thiomargarita namibiensis does not possess a nucleus or any other organelles thus is classified as a bacterium. Radioactive amino acids are fed to a cell in culture for a short amount of time. This is called a pulse. You follow the appearance of radioactive proteins in the cell compartments. In which organelles and in what order does radioactivity appear? A Endoplasmic reticulum - lysosomes - Golgi body - vesicle - extracellular region B Endoplasmic reticulum - vesicles - Golgi body - vesicles - extracellular region C Golgi Body - vesicles - endoplasmic reticulum - vesicles - extracellular region D Nucleus - endoplasmic reticulum - Golgi body - vesicle - extracellular region Solution 57 The solution is (B). Radioactive proteins will first appear associated with the rough endoplasmic reticulum where proteins are synthesized on ribosomes; then in vesicles in transit to the Golgi apparatus; in the Golgi apparatus itself where they are modified for export; and finally, in vesicles associated with exocytosis. The extracellular matrix interacts with which cellular structure? A Cytoskeleton B Lysosome Advanced Placement Biology Instructor’s Solution Manual 76 4 | Cell Structure C Nucleus D Smooth endoplasmic reticulum Solution 58 The solution is (A). The cytoskeleton, which provides the cell with rigidity and shape, is made up of microfilaments and intermediate filaments that interact with the extracellular matrix. Which structure or structures allow bacteria to move about? A Fimbriae only B Flagella only C Flagella and fimbriae D Plasmid and capsule Solution 59 The solution is (C). Flagella, made of flagellin, and fimbriae, made of pilin, are both involved in the movement of the bacteria in the fluid or on surfaces such as rocks, or the mucous membranes lining body cavities. Cells lining the intestine absorb a lot of nutrients. How did those cells adapt to their function? A Cells use cilia to move nutrients to their surface. B Cells grow much larger than adjacent cells to increase intake. C Cells are flat and thin to absorb more nutrients. D Membrane folds called microvilli increase the surface area. Solution The solution is (D). The microvilli are the cellular membrane protrusions of the intestinal membrane, which increase the surface area for maximum absorption. SCIENCE PRACTICE CHALLENGE QUESTIONS 4.3 Eukaryotic Cells 60 Describe structural and functional similarities between mitochondria and chloroplasts that provide evidence of common ancestry. Solution Both organelles are enclosed by a pair of phospholipid membranes with folds that define inner and outer regions. Both have their own DNA and ribosomes. These similarities are unlikely to have arisen independently and therefore provide evidence of a common ancestor. Core conserved properties include information storage using DNA, electron transport chains in both respiration and photosynthesis, ATP-based energy storage, and inner membrane separation, creating potential gradient due to a proton gradient associated with membrane-bound proteins facilitating active transport. These similarities are unlikely to have arisen independently and therefore provide evidence of a common ancestor. Advanced Placement Biology Instructor’s Solution Manual 4 | Cell Structure 61 Explain how the structural and functional differences between mitochondria and chloroplasts provide evidence of adaptations among common ancestral organisms. Solution 62 77 In mitochondria, organic compounds and oxygen are converted to carbon dioxide and water. In the chloroplast, carbon dioxide and water are converted to organic compounds and oxygen. This provides evidence of a common ancestor from which these two organelles diverged. Electrochemically, the function of the mitochondria and chloroplasts are complementary. In the mitochondria, carbon in organic carbon compounds is oxidized by oxidative phosphorylation to carbon dioxide with the synthesis of ATP from ADP. This oxidation is coupled to the reduction of molecular oxygen to the oxygen in water. In the chloroplast, the carbon of carbon dioxide is reduced in the synthesis of organic carbon molecules while the oxygen in water is reduced to molecular oxygen. Electrochemically, the structure of the mitochondria and chloroplasts are complementary. In the chloroplast, in the space contained by the inner membrane, referred to as the thylakoid lumen, the concentration of protons is higher than in the space bounded by the outer and inner membranes. The proton gradient is established by a light-activated sequence of reduction-oxidation reactions. This proton gradient is coupled to the synthesis of ATP from ADP by the thylakoid membrane-bound enzyme ATP synthase. In contrast, the mitochondrion in the space bounded by the inner membrane, the matrix, has a lower concentration of protons. This concentration gradient is produced by the active transport associated with a sequence of reduction-oxidation reactions among membrane-bound proteins. This gradient is coupled to the synthesis of ATP from ADP by the membrane-bound enzyme ATP synthase. These differences arise through of the interaction of light with chromophores. The explanation of these complementary structures and functions is provided by the theory of endosymbiosis, in which both organelles diverge from a photosynthetic prokaryotic ancestor. Examine the differences and similarities in the structural features of animal and plant cells. Justify the claim that both animals and plants have common ancestors based on your observations. Solution Students could enumerate these structural similarities: Nucleus with nuclear membrane Smooth and rough endoplasmic reticulum Plasma membrane Nucleolus Golgi apparatus Mitochondria Peroxisome Advanced Placement Biology Instructor’s Solution Manual 78 4 | Cell Structure Cytoskeleton Vacuole Students could enumerate these differences: Chloroplasts Lysosomes Plastids Cell walls So many similarities in structure lead to the conclusion that it is unreasonable that all of these structures arose independently, and so there must be a common ancestor. 63 What conserved core processes are common to both animals and plants? Construct an explanation of the differences based on the selective advantages provided in different environments. Solution Students could enumerate the following common core processes: Information storage and retrieval from a nucleus separated by a nuclear membrane Free-energy transduction by mitochondria Coordinated synthetic processes distributed throughout the cell interior Post-synthesis processing of proteins at the Golgi apparatus Structural integrity and transport provided by a cytoskeleton Peroxisome for recycling waste Plasma membrane regulation of transport and signaling into and out of the cell Students could enumerate the following differences in processes: Photosynthesis within chloroplasts Cell rigidity provided by a cell wall and a central vacuole Students’ explanations of the divergence revealed by process differences should include Selective advantage in an environment with high-intensity radiation is provided by specialized chromophores Selective advantage in an environment with stresses associated with intense mechanical forces and/or variations in concentrations of water in the environment is provided by the external mechanical support provided by a cell wall Advanced Placement Biology Instructor’s Solution Manual 4 | Cell Structure 79 Selective advantage where mobility is limited is provided by free-energy procurement through interaction with radiation Selective advantage where free-energy resources are distributed is provided through mobility that arises with deformable cells 4.6 Connections Between Cells and Cellular Activities 64 Louis Sullivan described architectural design as “form follows function.” For example, a window is designed to add light to a space without heat transport. A door is designed to allow access to a space. Windows and doors have different functions and therefore take different forms. Biological systems are not designed, but selected from random trials by interaction with the environment. Apply Sullivan’s principle to explain the relationship of function and form for each pair of cellular structures below. A. Plasma membrane and endoplasmic reticulum B. Mitochondrion and chloroplast C. Rough endoplasmic reticulum and smooth endoplasmic reticulum D. Flagella and cilia E. Muscle cells and secretory cells Solution A. The plasma membrane forms the outer boundary of the cell and regulates the transport of materials to and from the environment. One function of the endoplasmic reticulum is to provide a surface on which subcellular structures manage protein and lipid production for processes inside or outside of the cell. Therefore, the plasma membrane has a relatively small area-to-volume ratio and few folds that increase surface area, while the ER is highly convoluted. B. The function of mitochondria is the synthesis of ATP from ADP using oxygen and organic carbon compounds. The function of chloroplasts is the synthesis of ATP from ADP and organic carbon compounds using radiant energy, carbon dioxide, and water. Therefore, mitochondria have no systems for the capture of light energy, while chloroplasts do. C. The function of the rough ER is the synthesis of proteins within ribosomes, while the function of the smooth ER is the synthesis of materials other than protein. Therefore, the surface of the rough ER is studded with ribosomes, while the surface of the smooth ER is not. D. The function of flagella is motility. The function of the cilia is primarily the redistribution of materials on the cell surface. Therefore, the structure of the flagella is elongated and directional, while cilia are shorter and distributed over the surface. E. The function of muscle cells is movement that is achieved by metabolically active changes in cell shape. The function of secretory cells such as those of the liver, pancreas, or glands is the production of protein-rich materials. Therefore, muscle cells are dense with rough ER, while secretory cells are dense with smooth ER. Advanced Placement Biology Instructor’s Solution Manual 80 4 | Cell Structure 65 Complex multicellular organisms share nutrients and resources, and their cells communicate with each other. A society may encourage cooperation among individuals while discouraging selfish behavior to increase the overall success of the social system, sometimes at the expense of the individual. Scientific questions are testable and often attempt to reveal a mechanism responsible for a phenomenon. Pose three questions that can be used to examine the ways in which a social system regulates itself. Be prepared to share these in small group discussions with your classmates about the similarities between these regulatory strategies and the analogous roles of plasmodesmata and gap junctions in cell communication. Solution Sample answers: How are material resources distributed among individuals in a society? Students might be drawn to contrast the exchange of time for money and the exchange of money for goods with withdrawal from a shared cache of goods. The former is analogous to active transport among cells through structures such as plasmodesmata in plants and gap junctions in animals, while the latter is analogous to diffusion over a concentration or pressure gradient. How does a society exert control on individuals? Students might be drawn to contrast the regulation of individual behaviors through a delegated judicial system with an authoritarian system. The former is analogous to the coordinated exchange among cells of molecules that regulate the behaviors of others and is consistent with many relatively small and possibly specialized cells. The latter is analogous to regulatory control predominantly within the cell in competition with other cells for resources. How is information shared within a society? Students might be drawn to contrast simple cell adhesion and clustering with diffusion of materials through the extracellular space with active transport along specialized structures such as plasmodesmata or gap junctions. These are analogous to the isolation (hermit-like), low-density populations with very low-speed communication, and high-density populations that are tightly connected. How are resources allocated for shared infrastructure? Students might be drawn to contrast societies by rates of taxation. Higher rates of taxation produce a more structured support for resource sharing analogous to greater investment of free energy in cytoskeleton. Advanced Placement Biology Instructor’s Solution Manual 4 | Cell Structure 81 66 Plasmodesmata in vascular plants and gap junctions in animals are examples of specialized features of cells. Mechanisms by which transport occurs between cells evolved independently within several eukaryotic clades. Explain, in terms of cellular cooperation, the selective advantages provided by such structures. Solution 67 How are individuals within a society differentiated? Students might be drawn to contrast educational systems. In some societies, individual roles are differentiated at an early age, and these roles are relatively unchanging. In other societies, differentiation is deferred and/or roles change over time, requiring retraining. These would be analogous to cell differentiation and the ability of some cells under particular environmental conditions to redifferentiate. In some societies, the roles of individuals are highly specialized, while in others, roles are very similar. This would be analogous to the transition to complex multicellular organisms. Student explanations of the selective advantages provided should include spatial and temporal coordination: organisms with tissue and/or subcellular structures that are specialized functions and event timing that are integrated are more efficient than organisms without these characteristics. Coordination is achieved through cellcell signaling. Mammalian red blood cells have no nuclei, must originate in other tissue systems, are relatively long lived, are small with shapes that actively respond to their environment, and are metabolic anaerobes. Other vertebrates have red blood cells that are usually nucleated and are often relatively large, aerobic, self-replicating, and short lived. To connect these facts to biology, you need to ask questions. The questions that you pose will depend on the path your class is taking through the curriculum. Begin by summarizing what you know: What are the functions of a eukaryotic cell nucleus? What is the approximate average size of a human red blood cell? What is the range of blood vessel diameters in adult humans? What is the range of red blood cell size in vertebrates? What is the average lifetime of a human red blood cell? How can you show how cell production is stimulated using examples from particular systems? How is cell death controlled? What biochemical cycles are associated with anaerobic and aerobic respiration, and what are the important differences between these? What process is involved in the transport of oxygen and carbon dioxide into and out of red blood cells? Advanced Placement Biology Instructor’s Solution Manual 82 4 | Cell Structure What behaviors and dynamic homeostatic processes might be associated with the properties of red blood cells in mammalian and nonmammalian organisms? What do you know about the evolutionary divergences among vertebrates? Your summary has revealed some similarities and differences among vertebrate erythrocyte and circulatory system structures. Scientific questions are testable. They can be addressed by making observations and measurements and analyzing the resulting data. A. Pose three scientific questions that arise from your summaries of what you know about erythrocytes and capillary size. B. For each question you pose, predict what you believe would be the answer and provide reasoning for your prediction. C. Describe an approach you think can be used to obtain data to test your prediction. D. In the production of mammalian red blood cells, erythrocytes that have not yet matured and are still synthesizing heme proteins are surrounded by a macrophage. Predict the role of the macrophage in the maturation of a red blood cell. Solution Sample answers: What are the functions of a eukaryotic cell nucleus? The function of the nucleus is storage of genetic information and transcription of that information into RNA. What is the approximate, average size of a human red blood cell? between 5 and 10 micrometers What is the range of blood vessel diameters in adult humans? between 3 centimeters and 5 micrometers What is the range of red blood cell size in vertebrates? between 1 and 20 micrometers If only Chapters 3 and 4 have been considered, then the first two bullets are appropriate. Within these, and with a bit of help from Google, students should be able to locate these answers. A and B. The questions that students are able to pose will also depend on the instructional sequence. What are selection pressures? Fitness will be determined by oxygen availability and temperature variation, as well as other factors that are integrated with the homeostatic strategy of the organism. What are the trade-offs of a coordinated interaction among tissue systems versus an autonomous cell production? Cell division does not require the Advanced Placement Biology Instructor’s Solution Manual 4 | Cell Structure 83 integration of tissue systems. So rather than investing in energy costs of maintenance and communication, self-reproductive red blood cells would be short lived and large in number, requiring a different type of energy investment. Are cells that transport oxygen while not consuming oxygen more efficient? Efficiency could be measured in terms of energy provided through aerobic respiration at the target cell per energy required to deliver the resource. The denominator would include the cost of production for the anucleate strategy and the cost of aerobic respiration for the nucleated strategy. Are the structures of capillaries different in mammalian and reptilian or avian organisms? Thermal regulation and pulmonary structures vary, as well as heme proteins, and the sizes and densities of capillaries are consistent with these factors. How is thermoregulation affected by the cellular structure of the red blood cell? Anucleate and nucleate strategies are coordinated with thermal regulation, which is coordinated with temperature variation and behaviors. How are production and apoptosis of red blood cells different in mammalian and nonmammalian vertebrates? Mammals generally have red blood cells without nuclei or mitochondria that are synthesized in bone marrow. Reptiles and amphibians have nucleated cells with mitochondria, and multiple organs are reported as the site of synthesis, for example liver and kidney, depending on maturity. Birds have both nuclei and mitochondria. How are low oxygen states in tissues communicated to process that control erythrocyte production? Hypoxia detection is at the level of the tissues that secrete erythropoietin. It is a well conserved hormone in all vertebrates. Erythropoietin induces cell division in tissues that produce red blood cells, such as bone marrow. What do we know about the phylogeny of this signaling system? Vertebrates differ in whether their red blood cells contain mitochondria and a nucleus. Mammal red blood cells lack a nucleus and mitochondria in their mature form, but the red blood cells of fish, reptiles, and birds contain mitochondria and a nonfunctional nucleus that cannot divide. We could expect that the ancestors of systems without mitochondria emerged in a period with low atmospheric oxygen concentrations and that those with mitochondria emerged where oxygen levels were higher. What do we know about the phylogeny of mammalian and nonmammalian oxygen transport? Mollusks, insects, and other invertebrates have an open circulatory system, though there is a pump. The structure of the pump varies with the number of chambers and their connection to pulmonary organs. So the conserved features of oxygen transport are connected among vertebrates and invertebrates. C. Data that connect the organism to its environment, including behaviors, and to its ancestors have been pursued for each of the questions posed above. Advanced Placement Biology Instructor’s Solution Manual 84 4 | Cell Structure D. The surrounding macrophage forms an erythropoietic niche that increases availability of iron and, when the nucleus has been isolated, cleans up the debris through phagocytosis. 68 Mitochondria have DNA that encode proteins related to the structures and functions of the organelles. The replication appears to occur continuously; however, many questions about control of replication rate and segregation during mitosis are yet unanswered. Many diseases are caused by mitochondrial dysfunction. Mitophagy, as the name suggests, leads to the destruction of mitochondria. Predict whether or not cellular control mechanisms involving the regulation of mitochondrial DNA by the nucleus exist. Make use of what you know about selection and homeostasis as they apply to both the organism and to the organelle. Solution Sample answer: Although mitochondria possess DNA and protein-synthesizing machinery, they are not autonomous. Because mitochondria provide energy for metabolism through oxidative phosphorylation, the control of the number of mitochondria in the cell is essential to the maintenance of homeostasis. Therefore, the cell must control both the biogenesis and the destruction of mitochondria. The fact that mitochondria cannot replicate outside of the cell also implies that, ultimately, DNA replication is under the control of the nucleus. It is possible to predict that mtDNA replication is likely under the control of nuclear-encoded proteins, for example, mitochondrial transcription factors. Mitophagy is a process by which the cell regulates organelle number in response to developmental or physiological signals. This process offers an evolutionary advantage by allowing the cell to destroy defective or superfluous mitochondria and recycle their content. Advanced Placement Biology Instructor’s Solution Manual 5 | Structure and Function of Plasma Membranes 85 5 | STRUCTURE AND FUNCTION OF PLASMA MEMBRANES REVIEW QUESTIONS 1 Which plasma membrane component can be either found on its surface or embedded in the membrane structure A Carbohydrates B Cholesterol C Glycolipid D Protein Solution 2 The solution is (D). Proteins can be found either on the surface or embedded in the membrane structure. In addition to a plasma membrane, a eukaryotic cell has organelles, such as mitochondria, that also have membranes. In which way would these membranes differ? A The proportion of phosphate within the phospholipids will vary. B Only certain membranes contain phospholipids. C Only certain membranes are selectively permeable. D The proportions of proteins, lipids, and carbohydrates will vary. Solution 3 The solution is (D). The membranes differ in proportions of proteins, lipids, and carbohydrates, depending on the organelle or the cell type. For example, the myelinated membrane of neurons is rich in lipids. The membranes of mitochondria contain a high percentage of proteins. Which characteristic of a phospholipid increases the fluidity of the membrane? A Cholesterol B Its head C Saturated fatty acid tail D Unsaturated fatty acid tail Solution The solution is (D). Unsaturated fatty acids contain some double bonds between adjacent carbon atoms; a double bond results in a bend in the string of carbons, resulting in fluidity. Cells adapt to variations in temperature by changing the ratio of saturated to unsaturated phospholipids in the membranes. Advanced Placement Biology Instructor’s Solution Manual 86 4 5 | Structure and Function of Plasma Membranes How would an organism maintain membrane fluidity in an environment where temperatures fluctuate from very high to very low? A Greater proportion of unsaturated phospholipids in the membranes B Greater proportion of saturated phospholipids in the membranes C Greater proportion of carbohydrates in the membranes D Greater proportion of proteins in the membranes Solution 5 The solution is (A). The unsaturated fatty acids have “kinks” in their tails, which increases fluidity in the membrane at low temperatures. The membranes with saturated fatty acid tails in their phospholipids will “freeze,” or solidify. According to the fluid mosaic model of the plasma cell membrane, what is the location of carbohydrates in the cell membranes? A Carbohydrates are in contact with the aqueous fluid both inside and outside the cell. B Carbohydrates are present only on the interior surface of a membrane. C Carbohydrates are present only on the exterior surface of a membrane. D Carbohydrates span only the interior of a membrane. Solution 6 The solution is (C). Carbohydrates are generally attached to proteins on the outer surface of the membrane and form glycoproteins, which play an important role in cell-cell recognition. Carbohydrates can also bind to lipids, forming glycolipids. What do double bonds in phospholipid fatty acid tails contribute to? A The fluidity of membranes B The hydrophobic nature of membranes C The hydrophilic nature of membranes D The prevention of high temperatures from increasing the fluidity of membranes Solution 7 The solution is (A). Double bonds in phospholipid fatty acid tails increase the fluidity of a membrane because they create a bend in the molecules, preventing the tight packing of the fatty acid tails and the ensuing rigidity. What is the principal force driving movement in diffusion? A Concentration gradient B Membrane surface area C Particle size D Temperature Solution The solution is (A). Diffusion is the net movement of a substance from a region of high concentration to a region of low concentration. The principal driving force in diffusion is this concentration gradient. Advanced Placement Biology Instructor’s Solution Manual 5 | Structure and Function of Plasma Membranes 8 87 Which of the following is an example of passive transport across a membrane? A The movement of H+ into a thylakoid disc during photosynthesis B The uptake of glucose in the intestine C The uptake of mineral ions into root hair cells of plants D The movement of water from the descending loop of a nephron into the interstitium Solution 9 The solution is (D). Water crosses the collecting duct epithelium by osmosis, which is passive transport. Protein channels called aquaporin facilitate the movement of water. Hydrogen ions use the energy of photosynthesis to cross into the thylakoids against their concentration gradient. They flow out into the stroma through the ATP synthase along their gradient. The uptake of glucose by the intestine epithelium and mineral ions into roots are examples of active transport. Water moves via osmosis across plasma cell membranes in which direction? A From an area with a high concentration of other solutes to a lower one B From an area with a high concentration of water to one of lower concentration C From an area with a low concentration of water to one of higher concentration D Throughout the cytoplasm Solution 10 The solution is (B). Osmosis is the spontaneous net movement of water from a region of higher concentration to lower concentration of water. For example, red blood cells suspended in distilled water will swell and burst because water is flowing inside the cells. What problem is faced by organisms that live in fresh water? A They will have higher concentrations of body solutes. B Without compensating mechanisms, their bodies tend to take in too much water. C They have no way of controlling their tonicity. D Their bodies tend to lose too much water to their environment. Solution 11 The solution is (B). The fish are hypertonic to the fresh water environment they live in; thus, too much water diffuses into their body. Compensating mechanisms include excreting diluted urine and taking up salt through their gills. Which of the following questions can be asked about organisms that live in fresh water? A Will their bodies take in too much water? B Can they control their tonicity? C Can they survive in salt water? D Will their bodies lose too much water to their environment? Advanced Placement Biology Instructor’s Solution Manual 88 5 | Structure and Function of Plasma Membranes Solution 12 The solution is (A). Organisms contain salts and solutes that make their internal fluid hypertonic to freshwater. Losing water to their environment is not the challenge they face. They must limit the intake of water, and they must develop mechanisms to get rid of excess water. Why must active transport of molecules across plasma membranes function continuously? A Diffusion cannot occur in certain cells. B Diffusion is constantly moving solutes in opposite directions. C Facilitated diffusion works in the same direction as active transport. D Not all membranes are amphiphilic. Solution 13 The solution is (B). Diffusion allows movement down the concentration gradient to maintain equilibrium. Active transport must continuously pump solutes against the concentration gradient to counteract diffusion. How does the sodium-potassium pump make the interior of the cell negatively charged? A By expelling anions B By pulling in anions C By expelling more cations than it takes in D By taking in and expelling an equal number of cations Solution 14 The solution is (C). The sodium-potassium pump transports three sodium ions outside the cell and takes two potassium ions into the cell, making the interior of the cell less positively charged, therefore, more negatively charged. What is the difference between primary and secondary active transport? A Primary active transport is indirectly dependent on ATP, while secondary active transport is directly dependent on ATP. B Primary active transport is directly dependent on ATP, while secondary active transport is indirectly dependent on ATP. C Primary active transport does not require ATP, while secondary active transport is indirectly dependent on ATP. D Primary active transport is indirectly dependent on ATP, while secondary active transport does not require ATP. Solution The solution is (B). Primary active transport is directly dependent on ATP and transports compounds against their concentration gradient, whereas secondary active transport is indirectly dependent on ATP. For example, glucose and sodium ions are part of a co-transport where a primary Na+/K+ active transport across the membrane of epithelial cells powers the transport of glucose from the lumen of the intestine into the cell against a concentration gradient. Advanced Placement Biology Instructor’s Solution Manual 5 | Structure and Function of Plasma Membranes 15 89 What happens to the membrane of a vesicle after exocytosis? A It leaves the cell. B It is disassembled by the cell. C It fuses with and becomes part of the plasma membrane. D It is used again in another exocytosis event. Solution 16 The solution is (C). It fuses with and becomes part of the plasma membrane as the contents of the vesicle are secreted out of the cell. In what important way does receptor-mediated endocytosis differ from phagocytosis? A It transports only small amounts of fluid. B It does not involve the pinching off of the membrane. C It brings in only a specifically targeted substance. D It brings substances into the cell, while phagocytosis removes substances. Solution The solution is (C). Receptor-mediated endocytosis brings in only targeted substances whereas phagocytosis is not specific. CRITICAL THINKING QUESTIONS 17 Why do phospholipids tend to spontaneously orient themselves into something resembling a membrane such as the lipid-bilayer sphere, single-layer lipid sphere, and lipid-bilayer sheet? A Phospholipids are amphipathic molecules. The polar head faces toward water, and the nonpolar fatty acid tails face toward other fatty acid tails. B Phospholipids are lipophilic molecules. The polar head faces toward water, and the nonpolar fatty acid tails face toward other fatty acid tails. C Phospholipids are amphipathic molecules. The nonpolar head faces toward other fatty acid tails, and the polar fatty acid tails face toward water. D Phospholipids are hydrophilic molecules. The polar head faces toward water, and the nonpolar fatty acid tails face toward other fatty acid tails. Solution 18 The solution is (A). The hydrophobic, nonpolar regions must align with each other for the structure to have minimal potential energy and, consequently, higher stability. The fatty acid tails of the phospholipids cannot mix with water, but the phosphate head of the molecule can. Thus, the head orients to water, and the tail to other lipids. The fluid mosaic model described the plasma membrane as a mosaic of components. Why is it advantageous for the plasma membrane to be fluid in nature? A Fluidity allows greater flexibility to the cell and the motion of membrane components required for transport. Advanced Placement Biology Instructor’s Solution Manual 90 5 | Structure and Function of Plasma Membranes B Fluidity helps only in transport of some materials, but does not contribute to the flexibility. C Fluidity helps in maintaining the pH of the intracellular fluid, and helps in maintaining the physiological pH of the cell. D Fluidity helps provide mechanical strength to the plasma membrane. Solution The solution is (A). The fluid characteristic of the plasma membrane allows greater flexibility to the cell than it would if the membrane were rigid. It also allows the motion of membrane components required for some types of membrane transport. Advanced Placement Biology Instructor’s Solution Manual 5 | Structure and Function of Plasma Membranes 19 91 List four components of a plasma membrane and explain their function. A Phospholipids: form the bilayer; carbohydrates: help in adhesion; cholesterol: provides flexibility; integral proteins: form transporters; peripheral proteins: part of the cell’s recognition sites. B Phospholipids: form the bilayer; carbohydrates: help in adhesion; cholesterol: forms transporters; integral proteins: provide flexibility; peripheral proteins: part of the cell’s recognition sites. C Phospholipids: form the bilayer; carbohydrates: part of the cell’s recognition sites; cholesterol: provides flexibility to the membrane; integral proteins: form transporters; intermediate filaments: help in adhesion. D Phospholipids: form the bilayer; carbohydrates: function as adhesion; cholesterol: provides flexibility to the membrane; integral proteins: form transporters; intermediate filaments: part of the cell’s recognition sites. Solution The solution is (A). Phospholipids are the main fabric of the membrane. The hydrophilic heads are in contact with aqueous fluid both inside and outside of the cell. The hydrophobic tails face each other forming a lipid bilayer. Carbohydrates are found on the exterior surface of cells and are bound either to proteins (forming glycoproteins) or to lipids (forming glycolipids) and cell-cell interactions. Cholesterol is attached between phospholipids and between the two phospholipid layers. It tends to dampen the effects of temperature on the membrane. Integral proteins are embedded in the membrane structure. They include transporters, channels, enzymes, and so on. Peripheral proteins are found on the exterior and interior surfaces of membranes. They include enzymes, transporters, signaling receptors, and so on. 20 Which explanation identifies how the following affect the rate of diffusion: molecular size, temperature, solution density, and the distance that must be traveled? A Larger molecules move faster than lighter molecules. Temperature affects the molecular movement. Density is directly proportional to the molecular movement. Greater distance slows the diffusion. B Larger molecules move more slowly than lighter molecules. Increasing or decreasing temperature increases or decreases the energy in the medium, affecting molecular movement. Density is inversely proportional to molecular movement. Greater distance slows the diffusion. Advanced Placement Biology Instructor’s Solution Manual 92 5 | Structure and Function of Plasma Membranes C Larger molecules move more slowly than lighter molecules. Temperature does not affect the rate of diffusion. Density is inversely proportional to molecular movement. Greater distance speeds up the diffusion. D Larger molecules move more slowly than lighter molecules. Increasing or decreasing temperature increases or decreases the energy in the medium, affecting molecular movement. Density is inversely proportional to the molecular movement. Greater distance speeds up the diffusion. Solution 21 The solution is (B). Larger molecules move more slowly than lighter ones. It takes more energy in the medium to move them along. Increasing or decreasing temperature increases or decreases the energy in the medium, affecting molecular movement. The denser or more viscous a solution is, the harder it is for molecules to move through it, causing diffusion to slow down due to friction. Living cells require a steady supply of nutrients and a steady rate of waste removal. If the distance these substances need to travel is too great, diffusion cannot move nutrients and waste materials efficiently enough to sustain life. Both of the regular intravenous solutions administered in medicine, normal saline and lactated Ringer’s solution, are isotonic. Why is this important? A Isotonic solutions maintain equilibrium and avoid the exchange of materials to or from the blood. B Isotonic solutions disrupt equilibrium and allow for better exchange of materials in the blood cells. C Isotonic solutions increase the pH of the blood and allow for better absorption of saline in the blood cells. D Isotonic solutions decrease the pH of the blood and avoid the exchange of materials to or from the blood cells. Solution 22 The solution is (A). Injection of isotonic solutions ensures that there will be no perturbation of the osmotic balance and no water taken from tissues or added to them from the blood. If a doctor injected a patient with what was labeled as an isotonic saline solution, but then the patient died and an autopsy revealed that many red blood cells had burst, would it be true that the injected solution was really isotonic. Why or why not? A False, the solution was hypertonic. B False, the solution was either hypotonic or hypertonic. C False, the solution was hypotonic. D True, the solution was isotonic. Solution The solution is (C). The solution would have been hypotonic. Hypotonic solutions have a lower salt concentration compared to the red blood cells; therefore, the cells have low water potential compared to the saline solution. The water moves from Advanced Placement Biology Instructor’s Solution Manual 5 | Structure and Function of Plasma Membranes 93 the region of high water potential to low water potential, causing the red blood cells to swell and burst. 23 How does the sodium-potassium pump contribute to the net negative charge of the interior of the cell? A The sodium-potassium pump forces out three (positive) Na+ ions for every two (positive) K+ ions it pumps in; thus, the cell loses a net positive charge of one at every cycle of the pump. B The sodium-potassium pump expels three K+ for every two Na+ inside the cells, creating a net positive charge outside the cell and a net negative charge inside the cell. C The sodium-potassium pump helps the development of a negative charge inside the cell by making the membrane more permeable to negatively charged proteins. D The sodium-potassium pump helps in the development of a negative charge inside the cell by making the membrane impermeable to positively charged ions. Solution 24 The solution is (A). The sodium/potassium pump maintains a negative potential across the cell membrane by expelling three positive ions (Na+) for every two positive ions (K+) pumped into the cell. The net result is that there is a deficit in positive charges. Na+ ions are constantly expelled to balance osmotic pressure between the cell and its environment; otherwise, the extracellular fluid would become hypotonic and water would flow into the cells, causing them to swell and burst. Potassium is a necessary nutrient to maintain the function of our cells. What would happen to a person who is deficient in potassium? A The excess sodium disrupts the membrane components. B The excess sodium increases action potential generation. C The cells would not be able to get rid of extra sodium. D The cells would not be able to bring in sodium. Solution 25 The solution is (C). Cells typically have a high concentration of potassium in the cytoplasm and a high concentration of sodium outside of the cells. A deficiency of potassium would prevent cells from getting rid of sodium using the sodiumpotassium pump. Which statement describes processes of receptor-mediated endocytosis, exocytosis, and the changes in the membrane organization involved with each? A Receptor-mediated endocytosis involves the binding of a ligand to its receptor, resulting in the formation of a clathrin-coated vesicle that enters the cell. In exocytosis, waste material is enveloped in a vesicle that fuses with the interior of the plasma membrane via attachment proteins. Advanced Placement Biology Instructor’s Solution Manual 94 5 | Structure and Function of Plasma Membranes B In receptor-mediated endocytosis, waste material is enveloped in a membrane that fuses with the interior of the plasma membrane via attachment proteins. Exocytosis involves the opsonization of the receptor and its ligand in clathrin-coated vesicles. C In receptor-mediated endocytosis, waste material is enveloped in a membrane that fuses with the interior of the plasma membrane via attachment proteins. Exocytosis involves the opsonization of the receptor and its ligand in caveolae-coated vesicles. D Receptor-mediated endocytosis involves the opsonization of the receptor and its ligand in clathrin-coated vesicles that enter the cell. In exocytosis, waste material is enveloped in a membrane that fuses with the exterior of the plasma membrane via attachment proteins. Solution 26 The solution is (A). In receptor-mediated endocytosis, clathrin is attached to the cytoplasmic side of the plasma membrane. The process involves the inward budding of plasma membrane vesicles containing proteins with receptor sites specific to the molecules being absorbed. After the binding of a ligand to the receptor, a signal is sent through the membrane, leading to membrane coating and formation of a membrane invagination. The receptor and its ligand are then opsonized in clathrin-coated vesicles. Exocytosis is the process by which cells expel the contents of secretory vesicles out of the cell membrane into the extracellular space. Waste material is enveloped in a membrane, which fuses with the interior of the plasma membrane via attachment proteins. This fusion opens the membranous envelope on the exterior of the cell, and the waste material is expelled into the extracellular space. Describe the process of potocytosis. How does it differ from pinocytosis? A Potocytosis is a form of receptor-mediated endocytosis where molecules are transported via caveolae-coated vesicles. Pinocytosis is a form of exocytosis used for excreting excess water. B Potocytosis is a form of exocytosis where molecules are transported via clathrincoated vesicles. Pinocytosis is a form of receptor-mediated endocytosis used for excreting excess water. C Potocytosis is a form of receptor-mediated endocytosis where molecules are transported via caveolae-coated vesicles. Pinocytosis is a mode of endocytosis used for the absorption of extracellular fluid. D Potocytosis is a form of receptor-mediated endocytosis used for the absorption of water. Solution The solution is (C). Potocytosis is a type of receptor-mediated endocytosis in which small molecules are transported across the plasma membrane of a cell. The molecules are transported by caveolae rather than clathrin-coated vesicles. Pinocytosis is a mode of endocytosis in which small particles are brought into the cell. It is used primarily for the absorption of extracellular fluids. Advanced Placement Biology Instructor’s Solution Manual 5 | Structure and Function of Plasma Membranes 95 TEST PREP FOR AP® COURSES 27 One type of mutation in the CFTR protein prevents the transport of chloride ions through the channel. What is most likely to be observed in the lungs of patients with this mutation? A Dehydrated epithelial cells B Dehydrated mucus C Mucus with excess water D Mucus with a high electrolyte concentration Solution 28 The solution is (B). When the normal movement of chloride ions is prevented, water that follows by osmosis does not move to produce freely flowing mucus. The result is cells that line the airways of the lungs secrete mucus that is thick and sticky. Arsenic poisoning disrupts ATP production by inhibiting several of the enzymes in the oxidative phosphorylation pathway. Some of the symptoms of arsenic poisoning are similar to cystic fibrosis (CF; a hereditary disorder that results in difficulty breathing and frequent lung infections). What impact may arsenic poisoning have on components of the plasma membrane and transport that would result in CF-like symptoms? A Arsenic poisoning disrupts ATP production, leading to a decreased transport of Cl ions by epithelial cells. This leads to decreased electrolyte concentration in the mucus and retention of water into the cells. The mucus becomes dehydrated, as in CF. B Arsenic poisoning disrupts the Na+/ Cl pump, leading to decreased transport of Cl– ions out of the epithelial cells. This increases the electrolyte concentration in the mucus and movement of water out of the cells. The mucus becomes hydrated, as in CF. C Arsenic poisoning affects the oxidative phosphorylation pathway, leading to a decreased transport of Na+ ions out of the epithelial cells. This leads to increased electrolyte concentration in the mucus and movement of water into the cells. The mucus becomes dehydrated, as in CF. D Arsenic poisoning disrupts the binding sites for Cl ions, leading to a decreased transport of Cl– ions outside the epithelial cells. This leads to decreased electrolyte concentration in the mucus and movement of water out of the cells. The mucus becomes hydrated, as in CF. Solution The solution is (A). The CFTR protein requires ATP to transport Cl ions out of epithelial cells to the covering mucus. Arsenic poisoning disrupts ATP production; therefore, Cl ions will not be transported out of epithelial cells, total electrolyte concentration in the mucus will not increase, and water will not move out of epithelial cells to mucus via osmosis. The mucus will become dehydrated, as seen in CF. Advanced Placement Biology Instructor’s Solution Manual 96 29 5 | Structure and Function of Plasma Membranes In individuals with a normally functioning CFTR protein, which substances are transported via active transport? A Cl B Mucus C Na+ D Water Solution 30 The solution is (A). The CFTR protein transports Cl ions out of epithelial cells and into mucus in the intercellular space using ATP hydrolysis. Water follows by osmosis, maintaining a fluid mucus in the passageways of the body, e.g., in the lungs and digestive tract. The sodium-potassium (Na+/K+) pump functions like an anti-porter transporting Na+ and K+ icons across membranes using ATP. This protein spans the membrane with intracellular and extracellular domains. It has a binding site for Na+, K+, and ATP. An experiment was conducted to determine the locations of these binding sites. Artificial cells were created and incubated in buffers containing ATP, ouabain (or g-strophanthin, a cardiac glycoside), Na+, and K+ in varying combinations inside and outside of the cells, as indicated in the given table. The transport of Na+ and K+ was measured to determine the activity of the Na+/K+ pump. ATP present inside Experiment cells? ATP present outside cells? Ouabain present inside cells? Ouabain present outside Was Na+ cells? transported? Was K+ transported? 1 Yes Yes No No Yes Yes 2 Yes No No No Yes Yes 3 No Yes No No No No 4 No No No No No No 5 Yes No Yes Yes No No 6 Yes No Yes No No Yes Yes No 7 Yes No No Yes Which of the following conclusions is supported by the data? A Ouabain does not disrupt ATP binding to the Na+/K+ pump. B ATP is required for the transport of Na+ and not for transport of K+. Advanced Placement Biology Instructor’s Solution Manual 5 | Structure and Function of Plasma Membranes 97 C The ATP binding site of the Na+/K+ pump is located on the intracellular domain of the pump. D The ATP binding site of the Na+/K+ pump is located on the extracellular domain of the pump. Solution 31 The solution is (C). The ATP binding site of the Na+/K+ pump is located on the intracellular domain of the pump. If ATP is not present extracellularly but present inside the cell, the exchange takes place. If there is no ATP inside the cell, there is no exchange. If ATP is present outside the cell but not inside, the exchange does not take place whether ouabain is present or absent. There is no evidence presented that ATP cannot bind to the pump in the presence of ouabain. Paramecia are unicellular protists that have contractile vacuoles to remove excess intracellular water. In an experimental investigation, Paramecia were placed in salt solutions of increasing osmolarity. The rate at which a Paramecium’s contractile vacuole contracted to pump out excess water was determined and plotted against the osmolarity of the solutions, as shown in the graph. Which of the following is the correct explanation for the data? A At higher osmolarity, lower rates of contraction are required because more salt diffuses into the Paramecium. B In an isosmotic salt solution, there is no diffusion of water into or out of the Paramecium, so the contraction rate is zero. C The contraction rate increases as the osmolarity decreases because the amount of water entering the Paramecium by osmosis increases. D The contractile vacuole is less efficient in solutions of high osmolarity because of the reduced amount of ATP produced from cellular respiration. Solution The solution is (C). The contraction rate increases as the osmolarity of the solution decreases because water will diffuse into the paramecium’s cytoplasm. The increased contraction rate pumps the incoming water out of the cell. Advanced Placement Biology Instructor’s Solution Manual 98 32 5 | Structure and Function of Plasma Membranes What is most likely to happen if Paramecia are moved from a hypertonic solution to solutions of decreasing osmolarity? A The rate of contraction would increase with decreasing osmolarity because more water diffuses into the Paramecium. B The rate of contraction would decrease with decreasing osmolarity because more water diffuses into the Paramecium. C The rate of contraction would increase with decreasing osmolarity because more salt diffuses into the Paramecium. D The rate of contraction would decrease with decreasing osmolarity because more salt diffuses into the Paramecium. Solution 33 The solution is (A). The rate of contraction would increase with decreasing osmolarity because more water diffuses into the Paramecium. Describe the Na+/K+ pump, labeling the binding sites for Na+, K+, and ATP. Explain how the data indicate the location of the binding sites for Na+ and K+ on the pump. Based on the data in the table, which statement describes the location of the binding sites for Na+ and K+ on the pump? ATP present inside Experiment cells? ATP present outside cells? Ouabain present inside cells? Ouabain present outside Was Na+ cells? transported? Was K+ transported? 1 Yes Yes No No Yes Yes 2 Yes No No No Yes Yes 3 No Yes No No No No 4 No No No No No No 5 Yes No Yes Yes No No 6 Yes No Yes No No Yes 7 Yes No No Yes Yes No + A The binding of Na occurs on the outer surface of the cell, as its transportation remains unaffected by the presence of ouabain. The binding of K+ occurs on the inner surface of the cell, as its transportation is blocked when ouabain is present inside the cell. B The binding of K+ occurs on the outer surface of the cell, and its transportation is blocked when ouabain is present outside the cell. The binding of Na+ occurs on the inner surface of the cell, and its transportation is blocked by the presence of ouabain. Advanced Placement Biology Instructor’s Solution Manual 5 | Structure and Function of Plasma Membranes 99 C The binding of K+ occurs on the outer surface of the cell, and the binding of Na+ occurs on the inner surface of the cell because they are not transported when ATP is absent. D The binding of Na+ occurs on the outer surface of the cell, and the binding of K+ occurs on the inner surface of the cell because they are not transported when ATP is absent. Solution 34 The solution is (B). K+ binds to the outer surface of the cell while Na+ binds to the inner surface of the cell. The addition of ouabain to the outside of the cells inhibits transport of K+ only, while ouabain inside the cell inhibits Na+ transport. An experiment was set up to determine the movement of molecules through a dialysistubing bag into water. A dialysis-tubing bag containing 5% lactose and 5% fructose was placed in a beaker of distilled water, as illustrated in the given figure. After four hours, fructose is detected in the distilled water outside of the dialysis-tubing bag, but lactose is not. What conclusions can be made about the movement of molecules in this experiment? A Fructose, being a monosaccharide, diffused through the dialysis bag into the distilled water. However, lactose, being a disaccharide, could not diffuse through the dialysis bag. B Fructose was homogenized by lactose, allowing the fructose to diffuse through the dialysis bag and into the distilled water. Lactose is not homogenized, so it could not pass through the dialysis bag. C Fructose and lactose are oppositely charged and separated out due to the force of repulsion. D Fructose diffused because of the pore specificity of the semipermeable membrane, not because of its concentration gradient. Solution The solution is (A). Fructose is a monosaccharide, so it will be able to cross the membrane of the bag. Lactose is a disaccharide and is unable to cross the membrane. Therefore, fructose is observed in the distilled water outside of the dialysis-tubing bag, but lactose is not. Advanced Placement Biology Instructor’s Solution Manual 100 35 5 | Structure and Function of Plasma Membranes This is the 3D structure of clathrin, with an individual subunit highlighted in blue. What is a functional reason for this structure? A Clathrin stabilizes the section of the membrane forming the vesicle. B Clathrin is heat resistant. C The interlocking clathrin subunits provide a rigid cell structure. D Clathrin fuses with the trans Golgi apparatus. Solution 36 The solution is (A). Clathrin stabilizes the membrane-forming lattice of the vesicle. Which of the following statements appropriately describe the role of clathrin in neutrophils based on your understanding of phagocytosis? A Clathrin tethers the antigen to the cytoskeleton. B Clathrin opposes phagocytosis. C Clathrin marks the antigen on the invading cell for phagocytosis by neutrophils. D Clathrin stabilizes the inward facing surface of the plasma membrane, which engulfs the antigen. Solution 37 The solution is (D). Clathrin is involved in the engulfing of the antigen once the neutrophils bind to the target. Mast cells produce signals that activate inflammation. Neutrophils are phagocytic white blood cells. Monocytes are the largest type of white blood cell that engulf pathogens. Based on the information provided in the chapter, which cell types produce endosomes? A Monocytes and mast cells B Neutrophils, monocytes, and mast cells C Neutrophils and mast cells D Neutrophils and monocytes Solution The solution is (D). Neutrophils and monocytes produce endosomes for degrading the pathogens. Mast cells are not phagocytic. SCIENCE PRACTICE CHALLENGE QUESTIONS 5.2 Passive Transport 38 Membrane fluidity is influenced by the number of C-C double bonds (unsaturation) in the hydrocarbon tails of the lipids composing cell membranes. Fluidity is also dependent on temperature. The transit of materials through the cell membrane is controlled by fluidity. To maintain homeostasis, all organisms, including the simple bacterium E. coli, must sense the temperature of the environment and adapt to changes. Advanced Placement Biology Instructor’s Solution Manual 5 | Structure and Function of Plasma Membranes 101 Samples of E. coli were grown at four different temperatures, and then researchers determined the fatty acid composition of their plasma membranes. The data are shown in the given table. Fatty Acid 10 °C 20 °C 30 °C 40 °C Myristic 17% 14% 14% 16% Palmitic 18% 25% 29% 48% Palmitoleic 26% 24% 23% 9% Oleic 38% 34% 30% 12% Ratio (U/S) Fatty acid compositions of the plasma membrane of E. coli were incubated at the temperatures shown. Myristic and palmitic acid are saturated, while palmitoleic and oleic acids each have one C-C double bond. Blank Blank Blank Blank A. Analyze the data to calculate the ratio of the fraction of unsaturated (U) to the fraction of saturated (S) fatty acids in the plasma membrane, and complete the table. B. Graph the ratio U/S versus growth temperature. C. Explain the response of E. coli to the temperature of the environment. D. We know that the temperature of the environment is sensed by E. coli through the temperature-dependent confirmation of enzymes that convert a single bond in the lipid tail to a double bond, and vice versa. Explain how the discovery of a mutant strain of E. coli could lead to this insight. Solution Teaching Tip—This question connects concepts drawn from Big Ideas 2, 4, and 1. Advanced Placement Biology Instructor’s Solution Manual 102 5 | Structure and Function of Plasma Membranes A. The ratio is calculated here as the sum of the percent unsaturated fatty acid divided by the sum of the percent saturated fatty acid. Fatty acid 10 °C 20 °C 30 °C 40 °C Myristic 17% 14% 14% 16% Palmitic 18% 25% 29% 48% Palmitoleic 26% 24% 23% 9% Oleic 38% 34% 30% 12% Ratio (U/S) 1.8 1.5 1.2 0.32 B. C. The organism has a composition of the plasma membrane in which unsaturation increases with decreasing temperature. Homeostasis is maintained by response to the environment. D. A mutant was discovered in which the enzyme that converts a single bond to a double is defective. This organism cannot synthesize unsaturated fatty acids; therefore, it is not adapted for life at lower temperatures and dies. 39 Aquaporins, which allow for the movement of water across a cell membrane, are gated. Both low and high pH within a plant cell can cause alterations of the membrane-spanning protein. Describe the advantage of this feedback mechanism. Predict how conditions of flooding or drought could activate this mechanism. Solution To maintain the pressure within a plant cell, water concentration must be regulated. When water is transported into the cell, the pH drops, and a negative feedback loop changes the configuration of the membrane-spanning peptides; the gate closes. When water concentrations with the cell are low, the pH is high, and the gate opens. In flooding or drought conditions, the regulation cannot simply be reversed. Instead, drought stress and flooding have been shown to induce (drought) and suppress (flooding) the synthesis of proteins that provide the channel. Advanced Placement Biology Instructor’s Solution Manual 5 | Structure and Function of Plasma Membranes 103 5.3 Active Transport 40 Rice plants grown in high-salt environments can actively transport sodium ions into the vacuole by the antiporter movement of protons out of the vacuole. In a study aimed at the development of salt-tolerant rye, researchers produced several varieties of transgenic rye. Measurements of height and stem diameter for the transgenic varieties (TG1–TG4) are compared with the wild-type varieties WT1 and WT2. Shown in the table are the means and standard deviations from measurements of a very large sample size. Variety Height (cm) Stem Thickness (cm) WT1 9.667 ± 0.333 1.975 ± 0.095 WT2 11.867 ± 0.376 2.238 ± 0.204 TG1 15.420 ± 1.146 2.723 ± 0.261 TG2 15.600 ± 0.909 2.903 ± 0.323 TG3 14.925 ± 0.767 2.633 ± 0.073 TG4 16.100 ± 0.682 3.160 ± 0.169 A. Analyze the data. Are the heights and stem thicknesses in the transgenic plants significantly different than in the wild-type plant? Justify your claim with evidence. B. Are the heights and stem thicknesses among the transgenic plants significantly different? Justify your claim with evidence. C. Plants from which these data were taken were grown in 10 mM NaCl solutions. Pose one question that researchers can investigate by growing the same varieties in a series of lower salinity conditions. D. The Na+/H+ antiporter is an active transport system. Briefly explain negative feedback regulation of the movement of sodium into the vacuole of rye cells. Solution A. In each transgenic variety, the height of the plant is significantly different from the wild-type varieties. A comparison of the intervals of two measurements is not significant if the intervals overlap. (More sensitive tests of significance are out of scope on the AP Biology Exam.) Stem thickness is more complex. TG1 and WT2 nearly overlap, so stem thickness might not be significantly different for TG1. In each of the other cases, one may be confident that the transgenic varieties have thicker stems. B. Overlap of the measurement intervals occurs for each, so they are not different. C. Since the goal is the development of salt-tolerant rye plants, the question underlying this project is probably, “Does introduction of the genes that express the antiporter movement of sodium and hydrogen increase yield?” But a preliminary question that can lead to a selection among the transgenic varieties might be, “Do height or stem thickness depend on the salt concentration in the environment?” Advanced Placement Biology Instructor’s Solution Manual 104 5 | Structure and Function of Plasma Membranes D. An antiporter is bidirectional. As sodium diffuses into the vacuole, hydrogen ions diffuse out along their concentration gradient. Hydrogen ions are pumped into the vacuole by and ATPase-driven process. As the pH of cytoplasm falls, the antiport process halts. Advanced Placement Biology Instructor’s Solution Manual 6 | Metabolism 105 6 | METABOLISM REVIEW QUESTIONS 1 Energy can be taken in as glucose, and then has to be converted to a form that can be easily used to perform work in cells. What is the name of the latter molecule? A Anabolic molecule B Cholesterol C Electrolyte D Adenosine triphosphate Solution 2 The solution is (D). Sugars are stored in the body, usually as starch or glycogen. Glucose forms the basic unit of these sugars, which serve as a source of energy. ATP contains potential for a quick burst of energy that can be harnessed to perform cellular work. When cellular respiration occurs, what is the primary molecule used to store the energy that is released? A AMP B ATP C mRNA D Phosphate Solution 3 The solution is (B). ATP is considered the energy currency of cells. It contains the potential for a quick burst of energy that can be harnessed to perform cellular work. DNA replication involves unwinding two strands of parent DNA, copying each strand to synthesize complementary strands, and releasing the resulting two semiconserved strands of DNA. Which description is accurate for this process? A This process is anabolic. B This process is catabolic. C This process is both anabolic and catabolic. D This process is metabolic, but it is neither anabolic nor catabolic. Solution The solution is (A). This is an anabolic process. An anabolic process is one in which simpler molecules combine to form complex molecules with input of energy. Synthesis of new DNA strands from nucleic acid building blocks is an anabolic process. Advanced Placement Biology Instructor’s Solution Manual 106 4 6 | Metabolism Which process is a catabolic process? A Digestion of sucrose B Dissolving sugar in water C DNA replication D RNA translation Solution 5 The solution is (A). Digestion of sucrose is a catabolic process because sucrose, being a disaccharide, is a complex molecule. Its digestion produces the simpler compounds glucose and fructose with the release of energy. What food molecule used by animals for energy and obtained from plants is most directly related to the use of sun energy? A Glucose B Protein C Triglycerides D tRNA Solution The solution is (A). Glucose is synthesized by plants when light energy is converted to chemical energy through the process of photosynthesis. Glucose is then passed on to the food chain and is then used as a part of the animal cell’s cellular respiration reactions, in which it is first broken down in the process of glycolysis to make ATP, the chemical energy the cell uses to perform work. The image below shows how plants absorb solar energy and use it to drive the process of photosynthesis, thereby producing glucose and oxygen as by-products which are then used for the animal cell’s cellular respiration needs. Advanced Placement Biology Instructor’s Solution Manual 6 | Metabolism 6 107 What reaction will release the largest amount of energy to help power another reaction? A AMP to ATP B ATP to ADP C DNA to proteins D Glucose to starch Solution 7 The solution is (B). The hydrolysis of ATP is the reaction yielding the largest amount of energy to help another power reaction. ATP hydrolysis occurs when a molecule of ATP (the cell’s energy currency) reacts with a molecule of water, leading to the transformation of ATP into ADP (adenosine triphosphate) and inorganic phosphate (Pi). Several reactions in the cell are coupled with this hydrolysis reaction, thus leading to a large number of reactions being powered by this reaction. Consider a pendulum swinging. Which type(s) of energy is/are associated with the pendulum in the following instances? 1. The moment at which it completes one cycle, just before it begins to fall back toward the other end 2. The moment that it is in the middle between the two ends 3. Just before it reaches the end of one cycle, before step 1 A 1. Potential and kinetic 2. Potential and kinetic 3. Kinetic B 1. Potential 2. Potential and kinetic 3. Potential and kinetic C 1. Potential 2. Kinetic 3. Potential and kinetic D 1. Potential and kinetic 2. Kinetic 3. Kinetic Solution 8 The solution is (C). The energy possessed by the pendulum after completion of one cycle would be potential since it is at some elevation and energy is in stored form. When the pendulum is in between the two ends, then it is in motion and thus possesses kinetic energy. Which of the following best describes energy? A The transfer of genetic information B The ability to assemble a large number of functional catalysts Advanced Placement Biology Instructor’s Solution Manual 108 6 | Metabolism C The ability to store solar output D The ability to do work Solution 9 The solution is (D). The ability to do work or cause some change is called energy. What is the ultimate source of energy on this planet? A Glucose B Plants C Metabolic pathways D The sun Solution 10 The solution is (D). The main source of energy for this planet is the sun. Solar energy includes light, radio waves, and X-rays, all of which provide energy on Earth. Which molecule is likely to have the most potential energy? A ATP B ADP C Glucose D Sucrose Solution The solution is (D). Sucrose is a disaccharide molecule made of glucose and fructose. The stored potential energy in it would be the greatest. 11 Which of the following is the best way to judge the relative activation energies between two given chemical reactions? A Compare the G values between the two reactions. B Compare their reaction rates. C Compare their ideal environmental conditions. D Compare the spontaneity between the two reactions. Solution 12 The solution is (B). The activation energy of a particular reaction determines the rate at which it will proceed. The higher the activation energy, the slower the chemical reaction will be. Thus, relative activation energies between two chemical reactions can be judged by comparing their reaction rates. Which term in the Gibbs free energy equation denotes enthalpy? A G B H C S D T Advanced Placement Biology Instructor’s Solution Manual 6 | Metabolism Solution 13 109 The solution is (B). Total energy change in a system is called enthalpy and is denoted by H . Which chemical reaction is more likely to occur? A Dehydration synthesis B Endergonic C Endothermic D Exergonic Solution 14 The solution is (D). Exergonic reactions can occur spontaneously, and free energy is released from the reaction. Which comparison or contrast between endergonic and exergonic reactions is false? A Both endergonic and exergonic reactions require a small amount of energy to overcome an activation barrier. B Endergonic reactions have a positive G, and exergonic reactions have a negative G. C Endergonic reactions consume energy, and exergonic reactions release energy. D Endergonic reactions take place slowly, and exergonic reactions take place quickly. Solution 15 The solution is (A). Both endergonic and exergonic reactions require a small amount of energy to overcome an activation barrier. Is each system high or low entropy? 1. Perfume the instant after it is sprayed into the air 2. An unmaintained 1950s car compared with a brand new car 3. A living cell compared with a dead cell A 1. Low 2. High 3. Low B 1. Low 2. High 3. High C 1. High 2. Low 3. High D 1. High 2. Low 3. Low Advanced Placement Biology Instructor’s Solution Manual 110 6 | Metabolism Solution 16 The solution is (A). The entropy in case (1) would be low, since the disorder of the perfume is low immediately after it is sprayed; (2) would be high because, over time, the disorder in a 1950s car would increase; and (3) would be low because a living cell would have less disorder in comparison to dead cell. What counteracts entropy? A Energy release B Endergonic reactions C Input of energy D Time Solution 17 The solution is (C). Input of energy lowers the entropy of the system and decreases the disorder. Thus, input of energy counteracts high entropy and makes the system stable. What is the best example of the first law of thermodynamics? A A body getting warmer after exercise B A piece of fruit spoiling in the fridge C A power plant burning coal and producing electricity D An exothermic chemical reaction Solution 18 The solution is (C). A power plant burning coal and producing electricity is an example of the first law of thermodynamics. Energy is transformed from one form to another. What is the difference between the first and second laws of thermodynamics? A The first law involves creating energy, while the second law involves expending it. B The first law involves expending energy, while the second involves creating it. C The first law involves conserving energy, while the second law involves the inability to recapture energy. D The first law discusses creating energy, while the second law discusses the energy requirement for reactions. Solution 19 The solution is (C). The first law of thermodynamics states that energy can neither be created nor destroyed. It is just transformed from one state to other. The second law of thermodynamics states that every energy transfer involves some loss of energy in an unusable form, such as heat energy, resulting in a more disordered system. What best describes the effect of inputting energy into a living system? A It decreases entropy within the system. B It fuels catabolic reactions. Advanced Placement Biology Instructor’s Solution Manual 6 | Metabolism 111 C It causes enthalpy. D The energy is used to produce carbohydrates. Solution 20 The solution is (A). This is the best answer because inputting energy into a living system will allow cells to perform a variety of functions that decrease entropy, such as moving substances against their concentration gradient, repairing damage to cells and molecules, and storing energy as ATP. Why is ATP considered the energy currency of the cell? A It accepts energy from chemical reactions. B It holds energy at the site of release from substrates. C t is a protein. D It can transport energy to locations within the cell. Solution 21 The solution is (A). This is the best answer because inputting energy into a living system will allow cells to perform a variety of functions that decrease entropy, such as moving substances against their concentration gradient, repairing damage to cells and molecules, and storing energy as ATP. What is ATP made from? A Adenosine + high-energy electrons B ADP + pyrophosphate C AMP + ADP D The conversion of guanine to adenosine Solution 22 The solution is (B). An ATP molecule is made from the addition of pyrophosphate to an ADP molecule, which contains two phosphate groups. What is true about the energy released by the hydrolysis of ATP? A It is equal to 57 kJ/mol . B The cell harnesses it as heat energy to perform work. C It is primarily stored between the alpha and beta phosphates. D It provides energy to coupled reactions. Solution 23 The solution is (D). When ATP is used in a reaction, it transfers its third phosphate with energy to the chemical reaction. It phosphorylates another molecule and makes the energy available. Also, ATP hydrolysis performs work in cells by energy coupling. What part of ATP is broken to release energy for use in chemical reactions? A The adenosine molecule Advanced Placement Biology Instructor’s Solution Manual 112 6 | Metabolism B The bond between the first and second phosphates C The bond between the first phosphate and the adenosine molecule D The bond between the second and third phosphates Solution The solution is (D). The terminal phosphate bond of ATP is broken upon hydrolysis of ATP to ADP. This phosphate molecule, also referred to as the phosphate, is linked to the other two phosphates of ATP via high-energy bonds called phosphoanhydride bonds. Advanced Placement Biology Instructor’s Solution Manual 6 | Metabolism 24 113 What does an allosteric inhibitor do? A Binds to an enzyme away from the active site and changes the conformation of the active site, increasing its affinity for substrate binding. B Binds to an active site and blocks it from binding substrate. C Binds to an enzyme away from the active site and changes the conformation of the active site, decreasing its affinity for the substrate. D Binds directly to the active site and mimics the substrate. Solution 25 The solution is (C). An allosteric inhibitor binds to an enzyme away from the active site and changes the conformation of the active site, decreasing its affinity for the substrate. An allosteric inhibitor is a molecule that binds to an enzyme, thus inhibiting the enzyme’s activity. An allosteric inhibitor may be a competitive or noncompetitive inhibitor. What happens if an enzyme is NOT functioning in a chemical reaction within a living organism that needs it? A The reaction stops. B The reaction proceeds, but much more slowly. C The reaction proceeds faster without the interference. D There is no change in the reaction rate. Solution 26 The solution is (B). The reaction proceeds, but much more slowly. If the enzyme needed for the reaction is not be available, then the reaction will occur slowly, potentially so slowly that an organism may die before the reaction is complete. Which of the following is NOT true about enzymes? A They increase the G of reactions. B They are usually made of amino acids. C They lower the activation energy of chemical reactions. D Each one is specific to the particular substrate, or substrates, to which it binds. Solution 27 The solution is (A). They increase G of reactions. Enzymes have no effect on G of a reaction, that is, they do not change it even if a reaction is exergonic or endergonic. Which analogy best describes the induced-fit model of enzyme-substrate binding? A A hug between two people B A key fitting into a lock C A square peg fitting through the square hole and a round peg fitting through the round hole of a children’s toy D The fitting together of two jigsaw puzzle pieces Advanced Placement Biology Instructor’s Solution Manual 114 6 | Metabolism Solution 28 The solution is (B). The enzyme changes its configuration to bind to the transition state of the substrate, which validates the lock and key analogy. What is the function of enzymes? A To increase the G of reactions B To increase the H of reactions C To lower the entropy of the chemicals in the reaction D To lower the activation energy of a reaction Solution The solution is (D). Enzymes are proteins that accelerate chemical reactions by lowering their activation energies. As such, enzymes are catalysts of chemical reactions, causing an increase in the rate at which reactions occur. CRITICAL THINKING QUESTIONS 29 What is the connection between anabolic and catabolic chemical reactions in a metabolic pathway? A Catabolic reactions produce energy and simpler compounds, whereas anabolic reactions involve the use of energy to make more complex compounds. B Catabolic reactions produce energy and complex compounds are formed, whereas in anabolic reactions, free energy is used by complex compounds to make simpler molecules. C Catabolic reactions use energy and gives simpler compounds, whereas in anabolic reactions, energy is produced and simpler compounds are used to make complex molecules. D Catabolic reactions produce energy and water molecules, whereas in anabolic reactions, this free energy is used by simpler compounds to make only proteins and nucleic acids. Solution The solution is (B). Catabolic reactions give out energy and simpler compounds, whereas in anabolic reactions, free energy is used by simpler compounds to make complex molecules. Anabolic reactions depend on energy to complete their reactions. This energy is obtained as a result of catabolic reactions, so each one leads to the next, with the energy coming from one to run the next one. A catabolic reaction occurs when a complex compound breaks down, giving simpler molecules with the release of energy. The energy released is used to power the anabolic reactions, where simpler molecules use the energy to form complex molecules. Therefore, one reaction leads to another. Advanced Placement Biology Instructor’s Solution Manual 6 | Metabolism 30 115 Does physical exercise involve anabolic processes, catabolic processes, or both? Give evidence for your answer. A Physical exercise involves both catabolic and anabolic processes. Glucose is broken down into simpler compounds during physical activity. The simpler compounds are then used to provide energy to the muscles for contraction by the anabolic pathway. B Physical exercise is just a catabolic process. Glucose is broken down into simpler compounds during physical activity, and the simpler compounds are then used to provide energy to the muscles for contraction. C Physical activity involves only anabolic processes. Glucose is broken down into simpler compounds during physical activity, and the simpler compounds are then used to provide energy to the muscles for contraction by anabolic pathways. D Physical exercise involves both anabolic and catabolic processes. Cellulose is broken down into simpler compounds during physical activity. The simpler compounds are then used to provide energy to the muscles for contraction by anabolic pathways. Solution 31 The solution is (A). Physical exercise involves both catabolic and anabolic reactions. Glucose must be broken down in metabolic pathways, which are both catabolic and anabolic, and used to supply energy to muscles for contraction. Glucose is broken down to provide energy to the body. This is a catabolic process, and the simpler molecules use that energy to form complex compounds. This is an anabolic pathway. How do chemical reactions play a role in energy transfer? A Energy from the breakdown of glucose and other molecules in animals is released as ATP, which transfers energy to other reactions. B Energy from the breakdown of glucose and other molecules in animals is released in the form of NADP, which transfers energy to other reactions. C Energy is released in the form of glucose from the breakdown of ATP molecules. These ATP molecules transfer energy from one reaction to other. D Energy is released in the form of water from the breakdown of glucose. These molecules transfer energy from one reaction to other. Solution 32 The solution is (A). Energy must be stored in a cell in the form of a bond between two molecules; it cannot be stored in the cell as free energy. The energy is attached to a chemical such as ATP, which transfers the energy from one reaction to another and from one area of a cell to another safely. Energy never exists in the living cells as free energy. These ATP molecules are obtained from the metabolism of glucose. What are two cellular functions that require energy? A Phagocytosis helps amoebae take up nutrients, and pseudopodia help the amoebae move. B Phagocytosis allows amoebae to move, and pseudopodia help in the uptake of nutrients. Advanced Placement Biology Instructor’s Solution Manual 116 6 | Metabolism C Phagocytosis helps amoebae take up nutrients, and cilia help amoebae move. D Phagocytosis helps amoebae in cell division, and pseudopodia help amoebae move. Solution 33 The solution is (A). Phagocytosis helps amoebae in uptake of nutrients, and pseudopodia help the amoebae move. Amoebae take up some nutrients through phagocytosis and combine the resulting vacuoles with lysosomes to digest materials, as humans do. Amoebae move by constricting actin and myosin within the cell, just as human muscles move. The protrusion in cytoplasm that helps in movement is called pseudopodia. What is the conversion of energy that takes place when the sluice of a dam is opened? A Potential energy stored in the water held by the dam will convert to kinetic energy when the water falls through the opening of the sluice. B Kinetic energy stored in the water held by the dam will convert to potential energy when the water falls through the opening of the sluice. C Potential energy stored in the water held by the dam will convert to electrical energy when the water falls through the opening of the sluice. D Hydrothermal energy stored in the water held by the dam will convert to kinetic energy when the water falls through the opening of the sluice. Solution 34 The solution is (A). The potential energy of the water held in the reservoir behind the dam converts to kinetic energy as it falls through the opening of the sluice. Potential energy is referred to the energy stored in a system due to its position. Therefore, water held in a dam possesses potential energy, whereas kinetic energy is referred to as energy possessed by the motion of the object. When the sluice of the dam is opened, the potential energy is converted to kinetic energy. What is the difference between a spontaneous reaction and one that occurs instantaneously? A A spontaneous reaction is one that releases free energy and moves to a more stable state. Instantaneous reactions occur rapidly with a sudden release of energy. B A spontaneous reaction is one that uses free energy and moves to a more stable state. Instantaneous reactions occur rapidly with a sudden release of energy. C A spontaneous reaction is one that releases free energy and moves to a more stable state. Instantaneous reactions occur rapidly within a system by the uptake of energy. D A spontaneous reaction is one that occurs rapidly with the sudden release of energy. Instantaneous reactions release free energy and move to a more stable state. Solution The solution is (A). A spontaneous reaction occurs without the input of energy and decreases the free energy of a system, making the system more stable and releasing that energy to do work. An instantaneous reaction occurs very rapidly with a sudden release of energy. A spontaneous reaction does not require energy to carry out the reaction. It releases free energy that is used by the system to move to a stable state. Advanced Placement Biology Instructor’s Solution Manual 6 | Metabolism 117 The difference lies in the duration of reactions: spontaneous reactions can take seconds to years, while instantaneous reactions occur right away. 35 Which option describes the position of the transition state on a vertical energy scale, from low to high, relative to the position of the reactants and products, for both endergonic and exergonic reactions? A The transition state of the reaction exists at a lower energy level than the reactants. Activation energy is always positive, regardless of whether the reaction is exergonic or endergonic. B The transition state of the reaction exists at a higher energy level than the reactants. Activation energy is always positive, regardless of whether the reaction is exergonic or endergonic. C The transition state of the reaction exists at a lower energy level than the reactants. Activation energy is always negative, regardless of whether the reaction is exergonic or endergonic. D The transition state of the reaction exists at an intermediate energy level than that of the reactants. Activation energy is always positive, regardless of whether the reaction is exergonic or endergonic. Solution 36 The solution is (B). The transition state of the reaction exists at a higher energy state than the reactants and, thus, activation energy is always positive regardless of whether the reaction is an exergonic or endergonic reaction. Imagine an elaborate ant farm with tunnels and passageways through the sand where ants live in a large community. Now imagine that an earthquake shook the ground and demolished the ant farm. In which scenario, before or after the earthquake, was the ant farm system in a state of higher or lower entropy? Why? A The ant farm is in the state of higher entropy after the earthquake, and energy must be spent to bring the system to low entropy. B The ant farm is in the state of lower entropy after the earthquake, and energy must be spent to bring the system to high entropy. C The ant farm is in the state of higher entropy before the earthquake, and energy is given out of the system after the earthquake. D The ant farm is in the state of lower entropy before the earthquake, and energy is given out of the system after the earthquake. Solution The solution is (A). The ant farm is in a state of higher entropy (disorder) after the earthquake. The tunnels have been destroyed, and energy must be spent to rebuild them. Entropy is defined as the measure of randomness or disorder. Therefore, after the earthquake, the randomness or disorder increases, leading to higher entropy. Advanced Placement Biology Instructor’s Solution Manual 118 6 | Metabolism Energy should be given to the system to rebuild the tunnels and passageways to lower the entropy. 37 Energy transfers take place constantly in everyday activities. Think of two scenarios: cooking on a stove and driving. How does the second law of thermodynamics apply to these scenarios? A Heat is lost into the room while cooking and into the metal of the engine during gasoline combustion. B Heat gained while cooking helps to make the food, and heat released due to gasoline combustion helps the car accelerate. C The energy given to the system remains constant during cooking, and more energy is added to the car engine when the gasoline combusts. D The energy given to the system for cooking helps to make food, and energy in the car engine remains conserved when gasoline combustion takes place. Solution 38 The solution is (A). In both examples, there is an input of energy that results in work being done—cooking and moving the car—and loss of heat as a result. The heat loss travels into the room during cooking and into the metal of the engine during gasoline combustion. Energy must be continuously put into the systems to maintain the activities. Run out of natural gas or propane, and the cooking stops. Run out of gasoline, and the car stops. The second law of thermodynamics states that no energy transfer is efficient, and some form of energy in the form of heat is released. While cooking, heat is lost and travels into the room. The engine metal heats up during gasoline combustion. If propane or gas runs out, then both of the activities would stop. What does it mean for a system to be at a higher level of entropy? How can it be reduced? A Higher level of entropy refers to a higher state of disorder in the system, and it can be reduced by input of energy to lower the entropy. B Higher level of entropy refers to a higher state of symmetry in the system, and it can be reduced by the release of energy to lower the entropy. C Higher level of entropy refers to low disorder in the system, and it can be reduced by input of energy to increase the entropy. D Higher level of entropy refers to a higher state of disorder in the system, and it can be reduced by providing a catalyst to lower the entropy. Solution The solution is (A). The system has lost energy and is in a higher state of disorder. Energy or work must be invested into the system to restore order and lower entropy. When a system is in a state of disorder, then it possesses high entropy. The entropy can be reduced by input of energy, which reduces the randomness and lowers the entropy of the system. Advanced Placement Biology Instructor’s Solution Manual 6 | Metabolism 39 119 When the air temperature drops and rain turns to snow, which law of thermodynamics is exhibited? A First law of thermodynamics B Second law of thermodynamics C Third law of thermodynamics D Zeroth law of thermodynamics Solution 40 The solution is (A). This is an example of the first law of thermodynamics. Energy is being transferred as the atmosphere gets colder and the water in rain solidifies into snow. The first law of thermodynamics states that energy can neither be created nor destroyed. It is transferred from one state to another. Energy is being transferred as the atmosphere gets colder and the rainwater solidifies into snow. How does ATP supply energy to chemical reactions? A ATP dissociates, and the energy released by breaking a phosphate bond within ATP is used for phosphorylation of another molecule. ATP hydrolysis also provides energy to power coupling reactions. B ATP uses energy to power exergonic reactions by hydrolysis of ATP molecules. The free energy released as a result of ATP breakdown is used to carry out metabolism of products. C ATP uses energy to power endergonic reactions by dehydration of ATP molecules. The free energy released as a result of ATP breakdown is used to carry out metabolism of products. D ATP uses the energy released from coupling reactions, and that energy is used to power the endergonic and exergonic reactions. Solution 41 The solution is (A). ATP transfers its third phosphate with its energy to the chemical in the reaction, phosphorylating that chemical and making the energy available. When ATP is used in a reaction, it transfers its third phosphate with energy to the chemical reaction. It phosphorylates another molecule and makes the energy available. Also, ATP hydrolysis performs work in cells by energy coupling. Is the EA for ATP hydrolysis relatively low or high? Explain your reasoning. A EA for ATP hydrolysis is low because considerable energy is released. B EA for ATP hydrolysis is high because considerable energy is released. C EA for ATP hydrolysis is intermediate because considerable energy is released. D EA for ATP hydrolysis is high because a low amount of energy is released. Solution The solution is (B). The EA of ATP hydrolysis is high due to the high energy level of the compound and the ultimate energy release. Advanced Placement Biology Instructor’s Solution Manual 120 42 6 | Metabolism What is phosphorylation as it occurs in chemical reactions? A The attachment of a phosphate to another molecule to facilitate a chemical reaction B The uptake of a phosphorous molecule by an ATP molecule to power chemical reactions C The release of a third phosphorous molecule of ATP during hydrolysis D The breakdown of a pyrophosphate molecule which gives phosphate ions Solution 43 The solution is (A). Phosphorylation is the attachment of a high-energy phosphate from ATP to another molecule to facilitate a chemical reaction. If a chemical reaction could occur without an enzyme, why is it important to have one? A Enzymes are important because they give the desired products only from the reaction. B Enzymes are important because the products are obtained consistently with time. C Enzymes are important because they do not disturb the concentration of the products. D Enzymes are important because energy remains conserved and no loss of energy occurs. Solution 44 The solution is (A). Chemical reactions can occur without an enzyme catalyzing them, but not at a rate consistent with life. Enzymes speed up reactions that allow chemicals to be available for an organism’s growth and maintenance. However, enzymes are also very specific to their host molecules. This ensures that only reactions whose products are needed will be catalyzed. How does enzyme feedback inhibition benefit a cell? A Feedback inhibition benefits the cell by blocking the production of the products by changing the configuration of enzymes. This will prevent the cells from becoming toxic. B Feedback inhibition benefits the cell by blocking the production of the reactants by changing the configuration of enzymes. This will prevent the cells from becoming toxic. C Feedback inhibition benefits the cell by blocking the production of the products by changing the configuration of reactants. This will prevent the cells from becoming toxic. D Feedback inhibition benefits the cell by blocking the production of the products by reducing the reactants. This will prevent the cells from becoming toxic. Solution The solution is (A). Feedback inhibition allows the cell to maintain a relatively constant level of a product without having to degrade the enzyme. The products bind to the allosteric site of the enzyme and change its configuration. The reaction eventually stops. This way, overproduction of the product does not occur, which would be toxic. Advanced Placement Biology Instructor’s Solution Manual 6 | Metabolism 45 121 What type of reaction allows chemicals to be available for an organism’s growth and maintenance in a timely manner? A Enzymatically facilitated reactions B Redox reactions C Catabolic reactions D Hydrolysis of ATP Solution The solution is (A). Enzymatically facilitated reactions occur at a speed that brings reactants together to provide the biochemical molecules needed for life. The reactions would occur without the enzymes, but not fast enough to supply the products needed by cells. Enzymes catalyze a reaction and accelerate the rate of chemical reaction. Therefore, the products are made in timely manner and can be available on time for an organism’s growth and maintenance. TEST PREP FOR AP® COURSES 46 Cell metabolism is a complex process that uses many types of chemicals in a variety of processes. Which statement is true? A A loss of free nucleotides would result in cancer. B A loss of assorted carbohydrates would result in mitosis. C A loss of triglycerides would result in cell death. D A loss of enzymes would result in cell death. Solution 47 The solution is (D). A lack of enzymes would result in an extreme slowdown of metabolic pathways because reactions would only be able to occur spontaneously. Since reactions occurring spontaneously do not necessarily occur quickly, the cell would die as necessary processes slowed or halted. Which pair of descriptors of chemical reactions go together? A Anabolic and exergonic B Exergonic and dehydration synthesis C Endergonic and catabolic D Hydrolysis and exergonic Solution 48 The solution is (D). A hydrolysis reaction involves addition of water to break bonds, thus liberating energy. What is the underlying principle that supports the idea that all living organisms share the same core processes and features? A All organisms must harvest energy from their environment and convert it to ATP to carry out cellular functions. B Plants produce their own energy and pass it on to animals. Advanced Placement Biology Instructor’s Solution Manual 122 6 | Metabolism C Herbivores, carnivores, and omnivores coexist for the survival of all. D Glucose is the primary source of energy for all cellular functions. Solution 49 The solution is (A). All organisms must harvest energy from their environment and convert it to ATP to carry out cellular functions. It is a requirement of life that organisms metabolize energy. It has been accepted that life on Earth started out as single-celled, simple organisms, which then evolved into complex organisms. How did evolution proceed to produce such a wide variety of living organisms from a simple ancestor? A Prokaryotes produced the fungi, then the protists, which then branched to plants and animals. B Protists evolved first, then the prokaryotes, which branched into the fungi, plants, and animals. C Prokaryotes produced the protists, which branched into the fungi, plants, and animals. D Prokaryotes produced the protists, then the fungi, which branched into plants and animals. Solution 50 The solution is (C). Prokaryotes produced the protists, which branched into the fungi, plants, and animals. Single-celled organisms that are the prokaryotes appeared first on Earth. They produced multicellular organisms called protists. The protists further branched into fungi, plants, and animals. Plants make glucose through a pathway called photosynthesis. The amount of energy captured from light can be expressed as the number of energy-containing molecules used to make one molecule of glucose. Which option best states the number of each molecule needed? A 54 molecules of ATP and 18 molecules of NADPH B 18 molecules of ATP and 12 molecules of NADPH Advanced Placement Biology Instructor’s Solution Manual 6 | Metabolism 123 C 24 molecules of ATP and 18 molecules of NADPH D 12 molecules of ATP and 18 molecules of NADPH Solution 51 The solution is (B). The amount of energy needed to make one molecule of glucose from 6 molecules of carbon dioxide is 18 molecules of ATP and 12 molecules of NADPH, or a total of 54 molecules of ATP. What is an anabolic pathway? What is an example of an anabolic pathway used by cells in their metabolism? A Anabolic pathways involve the breakdown of nutrient molecules into usable forms. An example is the harvesting of amino acids from dietary proteins. B Anabolic pathways involve the breakdown of nutrient molecules into usable forms. An example is the use of glycogen by the liver to maintain blood glucose levels. C Anabolic pathways build new molecules out of the products of catabolic pathways. An example is the separation of fatty acids from triglycerides to satisfy energy needs. D Anabolic pathways build new molecules out of the products of catabolic pathways. An example is the linkage of nucleotides to form a molecule of mRNA. Solution 52 The solution is (D). Anabolic pathways build new molecules out of the products of catabolic pathways. An example is the linkage of nucleotides to form a molecule of mRNA. If glucose is broken down through aerobic respiration, a number of ATP molecules can be made from the energy extracted. How many molecules of ATP are possible? A 2 to 4 B 36 to 38 C 10 to 12 D 24 to 30 Solution 53 The solution is (B). When glucose is broken down through aerobic respiration, 36– 38 molecules of ATP can be produced. Two molecules are obtained from the Krebs cycle, and 34 molecules of ATP are produced from the electron transport chain. Plants must have adequate resources to complete their functions. If they do not have what they need, there are changes in the organism’s metabolism. What happens to the metabolism of a plant that does not have adequate sunlight? A Photosynthesis slows and less glucose is produced for energy use. B The plant switches to anaerobic metabolism. C The plant goes into a dormant state until the sunlight returns. D The plant flowers quickly to reproduce while it can. Advanced Placement Biology Instructor’s Solution Manual 124 6 | Metabolism Solution 54 The solution is (A). Photosynthesis slows and less glucose is produced for energy use. Sunlight is an extremely important requirement for the plants to carry out the photosynthesis reaction and manufacture their food. When the sunlight is not adequate, the rate of photosynthesis decreases and, as a result, less glucose is produced for energy use. Water deficiency is arguably the easiest deficiency to detect in plants. This is because plants that lack water will wilt since water within the plant’s cells helps to support the plant’s weight. Plant cells become water deficient because their cells use the water for metabolic processes. What happens to the metabolism of a plant that does NOT have adequate water? A Photosynthesis is inhibited, less glucose is produced, and water used by the cells is not replaced. B The plant increases its breakdown of glucose to create more water at the end of the process. C The plant will stop photosynthesizing for long periods of time until it has enough water to do so. D The cell will bring in more carbon dioxide to compensate for the lack of water, allowing glucose synthesis to continue. Solution 55 The solution is (A). Photosynthesis is inhibited, less glucose is produced, and water that leaves the cells is not replaced, causing wilting of the plant. When adequate water is not present, photosynthesis will decrease or stop altogether. This is because water is an important reactant in the photosynthesis reaction. As a result, the product of the photosynthesis reaction, glucose, will decrease. Wilting of plants will occur as a result of diminished water in the cells. Enzymes facilitate chemical reactions that result in changes to a substrate. How does the induced-fit model of enzymes and substrates explain their function? A Both enzymes and substrates undergo dynamic changes, inducing the transitions state of the substrate. B The enzyme induces a change in the substrate but is not changed itself during the reaction. C The substrates attach to the enzyme, and the chemical reaction proceeds. D The enzyme changes shape to fit the substrate, causing the transition state to occur. Solution The solution is (A). Both enzymes and substrates undergo dynamic changes, inducing the transition state of the substrate. When the enzyme and substrate come together, their interaction causes a mild shift in the enzyme’s structure. This confirms an ideal binding arrangement between the enzyme and the transition state of the substrate. Advanced Placement Biology Instructor’s Solution Manual 6 | Metabolism 56 125 Enzyme inhibitors play an important part in the control of enzyme functions, allowing them to continue, or inhibiting them for a period of time. Which inhibitor affects the initial rate, but does NOT affect the maximal rate? A Allosteric B Competitive C Noncompetitive D Uncompetitive Solution The solution is (B). The effect of competitive inhibition is to increase the substrate concentration required to achieve a given reaction speed. So, the rate of the reaction will increase. SCIENCE PRACTICE CHALLENGE QUESTIONS 6.1 Energy and Metabolism 57 Activation energy is required for a reaction to proceed, and it is lower if the reaction is catalyzed. Sucrose (table sugar) is a disaccharide. When we eat sucrose, it is converted to carbon dioxide and water, as with other carbohydrates. 1. Identify whether the breakdown of sucrose is endergonic or exergonic. Explain the reasoning for your identification. 2. Based on your identification, explain whether cubes of sugar can be stored in a sugar bowl, by creating a diagram similar to Figure 6.10. 3. If table sugar is placed in a spoon held over a high flame, the sugar is charred and becomes a blackened mixture composed primarily of carbon. Create a visual representation that includes a chemical equation to explain the role of the flame in this process. 4. In terms of your answers to questions 1‒3, predict whether sugar cubes in a bowl placed in a dish of water can be stored on a table, and justify your prediction. 5. [Extension] The energy of activation of a chemical reaction can be determined by measurement of the effect of temperature on reaction rate. The natural logarithm of the reaction rate constant is a linear function of the inverse of the temperature in Kelvin degrees. The negative of the slope of that graph is the energy of activation divided by the universal ideal gas constant, R 8.314 J/Kmol . Using the following data (R. Wolfenden and Yang Yean, Journal of the American Chemical Society, 2008 Jun 18; 130(24): 7,548–7,549), evaluate the energy of activation of the following reaction: Advanced Placement Biology Instructor’s Solution Manual 126 6 | Metabolism sucrose → fructose + glucose Temperature (K) ln(rate) 440 3.8 4.5 5 6 423 403 388 A. Construct a graph of ln(rate) versus 1/T(K) and determine the energy of activation for the uncatalyzed reaction. B. Based on the data, explain the importance of enzymes for time scales characteristic of living systems on Earth—that is to say, life as we know it. The time scale required for half of the molecules of initial sucrose to remain can be estimated. The relationship between the half-life and the activation energy is: t1 2 0.69 10EA 2.3RT At a temperature of 300 K, approximately room temperature, RT is equal to 2,494 J/mole. Solution Sample answer: 1. The breakdown of sucrose to fructose and glucose is an exergonic process. One could calculate the free-energy change and find that it was negative. One may also note that this is an important process in digestion in which higher free energy is degraded to lower states, which are then converted to even lower free-energy states; respiration is a cascade. While there are chemical reactions within respiration that are endergonic, they must be coupled to exergonic reactions. One may misconceive that a process such as the degradation of sucrose, which in living systems is catalyzed by an enzyme, is endergonic and coupled to the exergonic configuration changes in the catalytic protein. However, the protein configuration is restored, returning to its own initial free-energy state, cyclically. So, while the catalyst changes the kinetics (the rate) of the reaction, it does not drive the reaction forward. 2. This means that if the sugar cube can be isolated from biological systems that could catalyze the conversion, the sugar cube would be stable. The diagram in Figure 6.10 would then represent the free energy of the sucrose sugar cube at the left and the glucose and fructose products at the right. Separating these is a free-energy barrier. 3. If we place the sugar cube on a teaspoon and expose it to a flame, the energy provided by the flame will allow the system to move to another, lower freeenergy state consisting of a mixture of carbon, carbon monoxide, carbon dioxide, and water. Advanced Placement Biology Instructor’s Solution Manual 6 | Metabolism 127 4. If we place the sugar cube in a glass of water and allow microbial life to participate, then the sugar becomes a nutrient. The catalytic power of the proteins expressed by the microbes will allow the system to move to the lower state. 5. A graph can be constructed from the data according to the model suggested. -3,00 0,0022 -3,50 -4,00 0,0023 0,0024 0,0025 0,0026 1/T(K) y = -6 846,00x + 11,77 -4,50 -5,00 ln(rate) -5,50 -6,00 A. Multiplying the slope by 8.314 J/molK gives an energy of activation of 57 kJ/mole. B. We can conclude from the data that this is a very large energy of activation for transformation and that the time scale characteristic of life as we know it would not be consistent with respiration conducted by uncatalyzed processes. 58 Physical exercise involves both anabolic and catabolic processes. For each process, explain an expected outcome and describe an example of a specific exercise that can lead to the expected outcome. Solution Sample answer: Physical exercise can lead to weight loss through the respiration of energy stored in tissues. Exercise can also lead to the synthesis of new tissue and likely weight gain. The former process is catabolic, and the latter is anabolic. 6.2 Potential, Kinetic, Free, and Activation Energy 59 Explanations in science are often constructed by analogy. Explanations of the behavior of a poorly understood phenomenon can often be constructed by analogy to a phenomenon that is well understood. For each of the following cellular functions that require free energy, describe a parallel human activity and identify a source of free energy for that activity. For example, the synthesis of proteins can be expected to proceed as an assembly of a small set of subcomponents, just as the construction of a building is accomplished by gathering and joining materials. It is consistent with our analogy to expect that there must be a free-energy resource that is consumed in the synthesis of proteins, just as hydrocarbon fuels are a source of energy for the construction of a building. Advanced Placement Biology Instructor’s Solution Manual 128 Solution 6 | Metabolism Students might propose one of the following analogies to guide the understanding of the behavior of a biological system: Storing and retrieving genetic information: We expect the storage of biological information to be analogous to the storage of cultural data, such as texts, in which information is contained in the arrangement of subcomponents, such as letters, numbers, notes, or colors. Though the subcomponents of the information may be few, the retrieval of cultural data can lead to nearly infinite forms of expression, and the process of data retrieval is complex and elegant. Likewise, arrangements of a very small number of nucleotides can be translated into the diversity of life. Sending chemical messages: We expect the coordination of subsystems within an organism to be accomplished by messaging. We expect the subcomponents of the messages to be simple and highly redundant. We expect the sequence and timing of messages to be as important as the information provided by the message. All of these features are characteristic of human communication systems and of the cell-cell signaling that coordinates cell behavior and integrates tissue systems. Digesting food: Harvesting, preparing, cooking, and delivering food involves multiple specialized actors in coordinated action. We expect metabolism to be an analogous process of energy harvesting and delivery. The output from each step has coevolved with its input. Some processes, such as hunting and gathering, have relatively low efficiency, but are less demanding in terms of specialization and coordination. In some processes, highly organized physical structures are required by the system. By analogy, glycolysis has a lower efficiency than the Krebs cycle, which involves the complex structure of the mitochondria. Regulating transport: Transport of materials can involve containerized passage through a system of locks in which free energy is consumed. This is analogous to the conformational changes of a protein spanning a cell membrane. At the other end, less complex materials can simply leak through a barrier. And between these are the porins whose channels are like streams that may be dammed. Advanced Placement Biology Instructor’s Solution Manual 6 | Metabolism 129 6.3 The Laws of Thermodynamics 60 Each process in the figure shows examples of endergonic and exergonic processes. 1. Identify each process as endergonic or exergonic and provide reasoning for your identification that includes your definition of the system. 2. For each process, does entropy increase or decrease? Explain your reasoning in terms of changes in the amount of order within the system. 3. For each process, is there an input of energy? Explain your reasoning in terms of (a) the source of the energy input into the system and (b) the interaction between the system and its environment that provides that input of energy. Solution There are many answers to these questions, and the answers depend on how one defines the system. The photo shows a compost pile. (1) If we consider the contents bound by the wooden box to be the system and the process to be the decomposition of the compost, then the system is exergonic. Although the process of photosynthesis is endergonic through coupling to radiant energy, the respiration processes that dominate the behavior of the system are all leading to the decomposition of complex organic structures to produce much less ordered products—carbon dioxide and water. (2) The entropy of the system increases as order decreases. (3) If the compost pile is immersed in cold air for a sufficient period of time, respiration might cease. Gardeners do work on the system to keep that from happening during the winter. They frequently stir the compost pile to maintain conditions that support microbial life. If the compost pile becomes too warm, it can also create unstable conditions. So, without energy inputs, the compost pile may not be self-sustaining. Even without the work done on the system by the gardener, there is an energy exchange through heating between the compost pile and its surroundings. The photo shows chicks hatching. (1) and (2) Considering the chick to be the system, the system is strongly exergonic. The new tissue that is rapidly growing is at a much Advanced Placement Biology Instructor’s Solution Manual 130 6 | Metabolism higher free energy than the nutrient materials within the shell. Considering the shell to be the system, the process is clearly endergonic; the shell must be broken by that beak, and the highly symmetric shell is more ordered than the shell fragments. Considering the chick and the shell to be the system, the system is exergonic; we know that the process will advance as free energy of the shell and its nutrients are coupled to the free energy of the growing chick. (3) However, without heating, the system may not continue to function, so changes in S are not sufficient. The photo shows sand art. (1) and (2) If we consider the ordered ridges relative to the planar surface to be the system, then it is highly ordered, and, as we are told in the caption, this order is in the process of degrading. In the transformation from the ordered to the disordered state, the products have a lower free energy than the reactants. The system is exergonic even if a barrier prevented wind, water, or other environmental disruptions the system. The barrier would prevent kinetic and not thermodynamic effects, and the system would ultimately degrade, though very slowly. (3) The possibility of spontaneous disorder, even in the sealed system, shows that no energy input is needed to drive the process. The photo shows spheres rolling down a hill. Considering a sphere to be the system, (1) and (2,) we see that the process is spontaneous only if (3) there is an input from the gravitational work done on the sphere. So, for that system, the process of rolling is endergonic. Considering a sphere and the Earth to be the system, we see that the process is a spontaneous rearrangement of parts within the system; the center-ofmass of the sphere-Earth system spontaneously decays from the slightly high freeenergy state of the sphere at the top of the hill. But (3) there is no work done on the system after the sphere has been placed at the top of the hill. 61 Energy transfers occur constantly in daily activities. Think of two scenarios: cooking on a stove and driving a car. For each scenario, describe the system and explain how the second law of thermodynamics applies to the system in terms of energy input and change in entropy. Solution Sample answer: Cooking on a stove involves the transfer of energy as heating from combustion of fuel to pan and then from the pan to contents of the pan. Considering the pan and then the contents of the pan to be the system, energy is added to the system from the surroundings through heating. In this process, the transport of heat energy creates atomic-scale disorder as the speeds of atomic increase and the correlations of their motions decrease. Driving a car is very different. Considering the car to be the system, the entropy of the system decreases as the speed and direction of the system become more ordered. The second law is consistent with this behavior since the entropy of the surroundings increases though the decomposition of high-energy fuels to low free-energy exhaust. In fact, we have become aware that the entropy production of driving is a central challenge of our era. Advanced Placement Biology Instructor’s Solution Manual 6 | Metabolism 62 131 Consider a simple process that illustrates the change in entropy when energy is transferred. Take a block of ice as a system with a temperature of 0 °C. This is water as a solid, so it has a high structural order. This means that the molecules are in a fixed position. As a result, the entropy of the system is low. 1. Take a block of ice as a system with a temperature of 0 °C. This is water as a solid, so it has a high structural order. This means that the molecules are in a fixed position. As a result, the entropy of the system is low. 2. Allow the ice to melt at room temperature. Describe changes in the motion and interactions of water molecules before and after melting. Explain where the energy came from whose transfer produced melting. Predict the effect of the energy transfer on the entropy on the system, and justify your prediction. 3. Heat the water until the temperature reaches the boiling point. Explain what happens to the entropy of the system when the water is heated. 4. Continue to heat the water at the constant temperature of the boiling point. Describe changes in the motion and interactions of water molecules before and after boiling. Predict the effect of the energy transfer on the entropy of the system, and justify your prediction. 5. [Extension/Connection] Molecules of water have simple responses to heating: The molecules move faster and interact less strongly with other neighboring molecules. Consider the primary producers of an aquatic ecosystem in summer. Describe the source of energy transfer to the system of photosynthetic plants and algae. Predict changes in the system in response. Explain what happens to the entropy of this trophic level when energy transfer occurs. Now consider the primary producers and their aqueous environment as the system. Explain what happens to the entropy of this system composed of photosynthetic organisms and their abiotic environment. 6. Predict the change in entropy of the system when both autotrophs and their abiotic environment are considered. Justify your prediction. Predict the signs of the entropy changes in both biotic and abiotic components of this system. Predict the relative magnitudes of these entropy changes, and justify your prediction. Solution Sample answer: 1. N/A 2. Heating causes the molecules to become less aligned and slightly decrease the volume in which they are constrained to move; liquid water has a slightly higher density that water ice. However, as melting occurs, there is no change in average speed since melting occurs at constant temperature. The more disordered system has a higher entropy and the increase is paid through transfer of energy from the surrounding environment. Advanced Placement Biology Instructor’s Solution Manual 132 6 | Metabolism 3. As the system continues to gain heat energy from the surroundings, the speeds of the molecules and disorder increase, so the entropy of the system increases. 4. When the boiling temperature is reached, further energy transport through heating causes a second change of phase and the system is vaporized. Increasing disorder and increasing volume increase the entropy of the system. Collisions with other molecules broaden the distribution of kinetic energies of molecules within the system, but the average temperature remains unchanged until the system is entirely vaporized. 5. Energy is transferred to the photosynthetic system primarily as sunlight. The response of the system is the synthesis of materials, reproduction, and growth. Order increases, and so does entropy. If we include the abiotic environment, water, and atmospheric gases, it must be that the entropy decreases. This is a requirement of the second law of thermodynamics. 6. The magnitude of the entropy increase in the abiotic environment must be larger than the magnitude of the entropy decrease in the biotic environment. Otherwise, the behavior of the system overall violates the second law. Some students may suggest that since it is a law and laws are made to be broken, this provides insufficient evidence. Remind them that in a system, “law” has a different meaning: laws are never broken, by definition. However, scientists are always trying to find ways to break scientific laws and disprove theories. When this occurs, a new theory that increases our understanding of the world results. 6.4 ATP: Adenosine Triphosphate 63 The sodium-potassium pump is an example of free-energy coupling. The free energy derived from exergonic ATP hydrolysis is used to pump sodium and potassium ions across the cell membrane. The hydrolysis of one ATP molecule releases 7.3 kcal/mol of free energy G –7.3 kcal/mol. If it takes 2.1 kcal/mol of free energy to move one Na+ across the membrane G 2.1 kcal/mol , how many sodium ions could be moved by the hydrolysis of one ATP molecule? Show your calculations to provide reasoning for your answer. Solution Sample answer: Assume that only these two processes need to be considered. 1 ATP 7.3kcal 1Na+ 3.5Na+ 1 ATP 21kcal The factors display the reasoning. If 7.3 kcal are available for each ATP and 2.1 kcal are required to transport 1 sodium ion, the number of sodium ions transported per ATP is the ratio. 64 Is the EA for ATP hydrolysis in cells likely relatively low or high compared to the EA for the combustion of gasoline in an internal combustion engine? Advanced Placement Biology Instructor’s Solution Manual 6 | Metabolism 133 1. Explain your reasoning in terms of the relative stabilities of ATP and gasoline compared to air in which no catalysts are present. 2. Describe how the role of the enzyme ATPase in the hydrolysis of ATP in a cell differs from a spark in the cylinder of an internal combustion engine. 3. Describe a strategy for collecting data that can be used to measure the energies of activation (EA) of each of these two processes with instruments that can measure concentrations of reactions produced in each system. Solution Sample answer: The spark of the internal combustion engine is similar to a catalyst in that the free energy of the original state is recovered through the generation of electrical energy by the rotational motion produced by the spark. However, in the absence of a spark, gasoline can spontaneously combust at temperatures above its ignition temperature. The activation energy can be estimated as comparable with RT, where R is the universal gas constant and T is the ignition temperature, roughly 5 kJ/mole. The activation energy of ATP hydrolysis is usually quoted as 57 kJ/mole. Rather than look up the numbers, it would be useful to consider the likelihood that ATP will spontaneously become ADP. Metabolic and cellular integrity would suffer since these processes are regulated by the availability of ATP. 6.5 Enzymes 65 Vitamin B12 is a coenzyme involved in a wide variety of cellular processes. Synthesis of vitamin B12 occurs only in bacteria; in animals, these bacteria populate anaerobic environments in the gut. Consequently, vegan diets in developing nations and diets common to developing nations provide no source of B12. Researchers (Ghosh et al. http://dx.doi.org/10.3389/fnut.2016.00001) found that rats whose diets contained limited (L) and no (N) B12 displayed symptoms that were not observed in the control group (C) whose diet included B12 and was otherwise identical. Chemical analysis of adipocytokines in the plasma after feeding periods of 4 and 12 weeks are shown in the following table. Adipocytokines Tissue of Origin Feeding duration (weeks) C L N Leptin (pg/L) 4 5.7 ± 0.21 5.8 ± 0.25 6.1 ± 0.25 Adipose 12 5.8 ± 0.39 6.5 ± 0.36 9.9 ± 0.68 MCP-1 (mg/L) 4 43.0 ± 1.18 44.4 ± 1.95 46.9 ± 2.08 Monocytes 12 43.2 ± 2.47 45.3 ± 3.02 49.5 ± 1.27 IL-6 (mg/L) 4 150 ± 3.2 154 ± 4.5 184 ± 8.0 Monocytes 12 151 ± 6.7 176 ± 11.0 185 ± 8.2 The sample size for these data is small: n = 6 within each group. Also shown in the table are cells in which these cytokine messages originate. Adipose cells store fats. Monocytes are white blood cells of the immune system. Over the 12 weeks of feeding, the weights of all three groups were equivalent, while the percent of body fat increased relative to the Advanced Placement Biology Instructor’s Solution Manual 134 6 | Metabolism control for the rats fed a diet of limited and no B12: 40 percent (N) and 20 percent (L), respectively. A. Identify which adipocytokines show significant increases relative to the control group, after only 4 weeks of treatment. Justify your identification. B. Identify which adipocytokines show only significant increases relative to the control group, after 12 weeks of treatment. Justify your identification. C. Identify which adipocytokines show significant increases relative to the control group, after 4 weeks of treatment but no further increase after 12 weeks. Justify your identification. Adipocytokines are chemical messengers that regulate metabolism and blood vessel production and dilation. High concentrations of adipocytokines are commonly found among individuals with abnormal autoimmune response. Monocyte chemoattractant protein 1 (MCP-1) is involved in the trafficking or guiding of monocytes to damaged tissue, as in a wound. In mice, leptin receptors of cells in the hypothalamus suppress hunger. Interleukin (IL-6) is released to initiate and then regulate inflammation in response to an infection. (The mice in this study were not infected or wounded.) D. Construct an explanation, with reasoning based on the evidence provided by these data, for the observed variations in adipocytokines. Many noncommunicable diseases are associated with abnormal autoimmune responses, and the number of diseases that involve abnormal autoimmune response is increasing. Many autoimmune diseases, such as diabetes and heart disease, occur in developed nations at a much higher frequency than in developing nations. E. Evaluate, based on these data concerning the effect of restrictions on the availability of B12, the following question: Does the increased lack of exposure to pathogens in developed nations lead to reduced or abnormal immune response? Solution Sample answer: A. Leptin and IL-6 are likely to be significantly different than the control group for the N diet. The intervals of the measured values of these cytokines do not overlap with those of the control measurements. With such a small sample size, the conclusion of significance would be very similar to that obtained with a t-test. And application of a t-test is out of scope for the AP Biology Exam. B. After 12 weeks of treatment, both leptin and IL-6 are significantly different than the control. Justification for this claim is the same as given in (A). C. The increase in concentrations of leptin continues to rise during the longer treatment period. However, the concentration of IL-6 stops increasing. Again, the claim is based on a comparison of the intervals of measured values. D. During the treatment, the percent body fat of the animals whose diets are deficient in B12 increases relative to the control. The constancy of MCP-1 is consistent with the absence of wounding that would evoke an inflammatory Advanced Placement Biology Instructor’s Solution Manual 6 | Metabolism 135 response. The increase in leptin is consistent with the increase in adipose tissue. The interleukin increase indicates that the physiological state of the animals deficient in B12 have responded as if the system is infected. Adipose cells or their secretion products are provoking an immune response from the host. E. Students may need to reread the problem narrative to obtain information about the source of B12. Populations in developed nations may be less exposed to bacteria. Consequently, there may be a diminished supply of B12, or other dietary supplements provided by a higher and more diverse population of gut bacteria. The evaluation of this question should connect the role and origin of B12 with the example of autoimmune diseases and obesity in developed nations. Good scientific questions can be pursued experimentally, taking into account all of the available data and are congruent with other good questions. This question is a good scientific question and is referred to as the hygiene hypothesis. 66 Using an example, explain how enzyme feedback inhibition regulates a cellular process. Solution Students might consider stability. A system under negative feedback remains at the set point. They might consider efficiency. A system under negative feedback uses resources on demand rather than consume free energy to produce redundant, unused subcomponents. Students might focus on the enzyme part of this question and respond that, without catalysis, the system must be greatly simplified or operate on a time scale that is inconsistent with life. Students might focus on the specific part of the question and point out that the investment in greater information content allows more precise communication. Advanced Placement Biology Instructor’s Solution Manual 136 7 | Cellular Respiration 7 | CELLULAR RESPIRATION REVIEW QUESTIONS 1 What is the most important energy currency used by cells? A ATP B ADP C AMP D Adenosine Solution 2 The solution is (A). ATP has high-energy bonds between its phosphate molecules, allowing energy to be stored by cells within ATP molecules. What happens when a chemical is reduced during a reaction? A The compound is reduced to a simpler form. B An electron is added to the chemical. C A hydrogen atom is removed from the substrate. D It acts as a catabolic reaction. Solution 3 The solution is (B). Addition of an electron to a substrate causes its reduction, and the donor is said to be oxidized. Which molecules are oxidizing agents? A FAD+ and NAD+ B FADH2 and NADH C FAD and FADH2 D NAD+ and NADH Solution 4 The solution is (A). FAD+ and NAD+ have a tendency to accept 2H+ and 2e− to become reduced and form FADH2 and NADH2, respectively. Which reaction releases energy? A AMP + phosphate → ADP + H2O B ADP + phosphate → ATP + H2O C ATP + H2O → ADP + phosphate D AMP + H2O → ATP + phosphate Solution The solution is (C). Hydrolysis of ATP releases energy and frees an inorganic phosphate. This energy is used for various chemical reactions. Advanced Placement Biology Instructor’s Solution Manual 7 | Cellular Respiration 5 137 During the second half of glycolysis, what occurs? A ATP is used up. B Fructose is split in two. C ATP is produced. D Glucose becomes fructose. Solution 6 The solution is (C). During the pay-off phase of glycolysis, substrate level phosphorylation takes place twice, creating a net of 2 ATP molecules per glucose. GLUTs are integral membrane proteins that assist in the facilitated diffusion of glucose into and out of cells. What reaction in glycolysis prevents glucose from being transported back out of the cell? A Hexokinase dephosphorylates glucose using ATP, creating a glucose molecule that can’t cross the hydrophilic portion of the plasma membrane. B Hexokinase phosphorylates glucose using ADP, creating a glucose molecule that can’t cross the hydrophobic interior of the plasma membrane. C Hexokinase dephosphorylates glucose using ADP, creating a glucose molecule that can’t cross the hydrophilic portion of the plasma membrane. D Hexokinase phosphorylates glucose using ATP, creating a glucose molecule that can’t cross the hydrophobic interior of the plasma membrane. Solution 7 The solution is (D). Hexokinase phosphorylates glucose using ATP. The negative charge of the phosphate prevents the passage of phosphorylated glucose through the hydrophobic interior of the plasma membrane. How many ATP molecules are used and produced per molecule of glucose during glycolysis? A The first half of glycolysis uses 2 ATPs, and the second half of glycolysis produces 4 ATPs. B The first half of glycolysis produces 2 ATPs, and the second half of glycolysis uses 4 ATPs. C The first half of glycolysis uses 4 ATPs, and the second half of glycolysis produces 2 ATPs. D The first half of glycolysis produces 4 ATPs, and the second half of glycolysis uses 2 ATPs. Solution The solution is (A). The enzymes hexokinase and PFK utilize 2 ATPs each in the first half of the cycle. The second half produces 4 ATPs by substrate-level phosphorylation. This occurs twice. Advanced Placement Biology Instructor’s Solution Manual 138 8 7 | Cellular Respiration What is removed from pyruvate during its conversion into an acetyl group? A Oxygen B ATP C Vitamin B D Carbon dioxide Solution 9 The solution is (D). During the oxidation of pyruvate, a molecule of carbon dioxide along with NADH is produced. What do the electrons added to NAD+ do in aerobic respiration? A They become part of a fermentation pathway. B They go to another pathway for ATP production. C They energize the acetyl group in the citric acid cycle. D They are converted to NADP. Solution 10 The solution is (B). NADH undergoes oxidative phosphorylation and enters the electron transport chain to produce ATP. GTP, which can be converted to ATP, is produced during which reaction of the citric acid cycle? A Isocitrate into α-ketoglutarate B Succinyl-CoA into succinate C Fumarate into malate D Malate into oxaloacetate Solution 11 The solution is (B). During the citric acid cycle, two oxidative decarboxylation reactions occur, producing 2 CO2 and 2 NADH. In addition, two dehydrogenation reactions producing 1 NADH and 1 FADH2 and 1 substrate-level phosphorylation occur. GTP is produced during the fifth reaction of the citric acid cycle. This reaction is catalyzed by the enzyme succinyl-CoA synthetase. How many NADH molecules are produced on each turn of the citric acid cycle? A One B Two C Three D Four Solution 12 The solution is (C). The citric acid cycle produces 3 NADH with the help of 2 oxidative decarboxylations and 1 dehydrogenation. What compound receives electrons from NADH? Advanced Placement Biology Instructor’s Solution Manual 7 | Cellular Respiration 139 A FMN B Ubiquinone C Cytochrome c1 D Oxygen Solution The solution is (A). Complex I (NADH dehydrogenase) contains an FMN subunit that immediately accepts electrons from NADH and passes them to CoQ. Advanced Placement Biology Instructor’s Solution Manual 140 13 7 | Cellular Respiration Chemiosmosis involves the movement of what? Where does it occur? A Electrons across the cell membrane B Hydrogen atoms across a mitochondrial membrane C Hydrogen ions across a mitochondrial membrane D Glucose through the cell membrane Solution 14 The solution is (C). Hydrogen ions produced during the electron transport chain are pumped across the inner mitochondrial membrane. These ions help maintain the correct pH gradient for the efficient production of ATP. What is the function of an electron in the electron transport chain? A To dephosphorylate ATP, producing ADP B To power active transport pumps C To reduce heme in complex III D To oxidize oxygen Solution 15 The solution is (B). As electrons move across the electron transport chain, hydrogen ions are pumped across the inner membrane of the mitochondria. What would be the outcome if hydrogen ions were able to diffuse through the mitochondrial membrane into the mitochondria without the need for integral membrane proteins? A ATP would not be produced. B Pyruvate would not be produced. C Citric acid would not be produced. D Carbon dioxide would not be produced. Solution 16 The solution is (A). ATP is produced with the help of ATP synthase embedded as integral proteins in the inner mitochondrial membrane. These structures allow a sequential passage of H+ ions into the matrix, generating ATP. Which fermentation method can occur in animal skeletal muscles? A Lactic acid fermentation B Alcohol fermentation C Mixed acid fermentation D Propionic fermentation Solution The solution is (A). The animal muscle cells contain lactate dehydrogenase, the enzyme responsible for the formation of lactic acid. Advanced Placement Biology Instructor’s Solution Manual 7 | Cellular Respiration 17 141 Which molecules are produced in glycolysis and used in fermentation? A Acetyl-CoA and NADH B Lactate, ATP, and CO2 C Glucose, ATP, and NAD+ D Pyruvate and NADH Solution 18 The solution is (D). Each glucose molecule produces 2 pyruvate and 2 NADH molecules. These are further used in the fermentation process to form either lactate or ethanol. What are the products of alcohol fermentation? A Methane and NADH B Lactic acid and FAD+ C Ethanol and NAD+ D Pyruvic acid and NADH Solution 19 The solution is (C). Ethanol and NAD+ are produced. Pyruvate first forms the acetaldehyde that further utilizes NADH to form ethanol and NAD +. In the first step of glycolysis, what is glucose transformed into? A Glucose-6-phosphate B Fructose-1,6-bisphosphate C Dihydroxyacetone phosphate D Phosphoenolpyruvate Solution 20 The solution is (A). Conversion of glucose into glucose-6-phosphate is the first step of glycolysis. What is beta-oxidation? A The main process used to break down glucose B The main process used to assemble glucose C The main process used to break down fatty acids D The main process used to remove amino groups from amino acids Solution 21 The solution is (C). Beta-oxidation involves breaking down fatty acids using two carbon fragments from the carboxyl end of fatty acids. It occurs in the mitochondria, and the - carbon of fatty acids is oxidized. Which statement about catabolic pathways is false? A Carbohydrates can feed into oxidative phosphorylation. B Glycerol can be broken down into glucose and feed into glycolysis. Advanced Placement Biology Instructor’s Solution Manual 142 7 | Cellular Respiration C Amino acids can feed into pyruvate oxidation. D Fatty acids can feed into the citric acid cycle. Solution 22 The solution is (A). As carbohydrates eventually transform into glucose, they can be fed into glycolysis. Oxidative phosphorylation is only performed by the electron transport chain by oxidizing NADH. What impact, if any, do high levels of ADP have on glycolysis? A They increase the activity of enzymes involved with glycolysis. B The high levels decrease the activity of enzymes involved with glycolysis. C They have no effect on the activity of any enzymes involved with glycolysis. D The high levels slow down all pathways involved with glycolysis. Solution 23 The solution is (A). ADP is an activator and a positive regulator of certain enzymes like phosphofructokinase-1. The control of which enzyme exerts the greatest control of glycolysis? A Hexokinase B Phosphofructokinase C Glucose-6-phosphatase D Aldolase Solution 24 The solution is (A). Hexokinase is responsible for phosphorylating glucose into glucose-6-phosphate which traps and prevents its passage through the plasma membrane. Therefore, hexokinase is needed for glycolysis to start. What does NOT occur as ATP concentration increases relative to ADP? A Decreased activity of phosphofructokinase B Increased activity of pyruvate kinase C Decreased activity of isocitrate dehydrogenase D Slowdown of the electron transport chain Solution The solution is (B). In high levels of ATP, the pyruvate kinase activity tends to decrease because it is a negative regulator/inhibitor of pyruvate kinase. Acetyl-CoA also shows this inhibition. Advanced Placement Biology Instructor’s Solution Manual 7 | Cellular Respiration 143 CRITICAL THINKING QUESTIONS 25 Why is it beneficial for cells to use ATP rather than directly use the energy stored in the bonds of carbohydrates to power cellular reactions? What are the greatest drawbacks to harnessing energy from the bonds of several different compounds? A ATP is readily available in the form of a single unit that provides a consistent, appropriate amount of energy. The cell would need to tailor each reaction to each energy source if it harvested energy from different compounds. B ATP energy cannot activate the ROS-dependent stress response, whereas food molecules are responsible for activating ROS. C ATP is low in energy, but food molecules possess higher levels of energy that cells can use. D ATP is readily available to cells, unlike compounds that have to first be phosphorylated to release their energy. Solution 26 The solution is (A). ATP provides the cell with a way to handle energy in an efficient manner. The molecule can be charged, stored, and used as needed. Moreover, the energy from hydrolyzing ATP is delivered in a consistent amount. Harvesting energy from the bonds of several different compounds would result in energy deliveries of different quantities. What role does NAD+ play in redox reactions? A NAD+, an oxidizing agent, can accept electrons and protons from organic molecules and get reduced to NADH. B NAD+, a reducing agent, can donate its electrons and protons to organic molecules. C NAD+, an oxidizing agent, can accept electrons from organic molecules and get reduced to NADH2. D NAD+, a reducing agent, can donate its electrons and protons to inorganic molecules. Solution 27 The solution is (A). NAD+ is the oxidized form of the molecule. It is an oxidizing agent; NAD+ can accept electrons from organic molecules. After NAD+ accepts two electrons and a proton, it becomes NADH, which is the reduced form of the molecule. NAD+ has a tendency to accept electrons and protons to become reduced to NADH. This reduced form can further act as a reducing agent. Therefore, NAD can play an oxidizing as well as a reducing role in redox reactions. In the following general reaction, how are electrons transferred? Explain the role of each species. Remember that R represents a hydrocarbon molecule and RH represents the same molecule with a particular hydrogen identified. RH + NAD+ → NADH + R Advanced Placement Biology Instructor’s Solution Manual 144 7 | Cellular Respiration A RH acts as a reducing agent and donates its electrons to the oxidizing agent NAD+, forming NADH and R. B NAD+, the oxidizing agent, donates its electrons to the reducing agent RH, forming R and NADH. C RH acts as an oxidizing agent and donates electrons to the reducing agent NAD +, producing NADH and R. D NAD+, the reducing agent, accepts electrons from the oxidizing agent RH, producing NADH and R. Solution 28 The solution is (B). NAD+ is the oxidizing agent, which can accept electrons from RH, the reducing agent. This causes NAD+ to be reduced to NADH and RH to be oxidized to R. RH, the reducing agent, has electrons to donate, and the oxidizing agent, NAD +, has a tendency to accept electrons, which in turn produces NADH and R as products. Nearly all organisms on Earth carry out some form of glycolysis. How does this fact support or not support the assertion that glycolysis is one of the oldest metabolic pathways? A To be present in so many different organisms, glycolysis was probably present in a common ancestor rather than evolving many separate times. B Glycolysis is present in nearly all organisms because it is an advanced and recently evolved pathway that has been widely used because it is so beneficial. C Glycolysis is absent in a few higher organisms. This contradicts the assertion that it is one of the oldest metabolic pathways. D Glycolysis is present in some organisms and absent in others, which may or may not support the assertion that it is one of the oldest metabolic pathways. Solution 29 The solution is (A). If glycolysis evolved relatively late, it likely would not be as universal in organisms as it is. It probably evolved in very primitive organisms and persisted, with the addition of other pathways of carbohydrate metabolism that evolved later. Red blood cells (RBCs) do not perform aerobic respiration, but they do perform glycolysis. Why do all cells need an energy source, and what would happen if glycolysis were blocked in a red blood cell? A Cells require energy to perform certain basic functions. Blocking glycolysis in RBCs causes imbalance in the membrane potential, leading to cell death. B Cells need energy to perform cell division. Blocking glycolysis in RBCs interrupts the process of mitosis, leading to nondisjunction. C Cells maintain the influx and efflux of organic substances using energy. Blocking glycolysis stops the binding of CO2 to the RBCs, causing cell death. D Cells require energy to recognize attacking pathogens. Blocked glycolysis inhibits the process of recognition, causing invasion of the RBCs by a pathogen. Advanced Placement Biology Instructor’s Solution Manual 7 | Cellular Respiration Solution 30 145 The solution is (A). All cells must consume energy to carry out basic functions such as pumping ions across membranes. A red blood cell would lose its membrane potential if glycolysis were blocked, and it would eventually die. What is the primary difference between a circular pathway and a linear pathway? A The reactant and the product are the same in a circular pathway, but different in a linear pathway. B The circular pathway components get exhausted, whereas those of the linear pathway do not and are continually regenerated. C Circular pathways are not suited for amphibolic pathways, whereas linear pathways are. D Circular pathways contain a single chemical reaction that is repeated, while linear pathways have multiple events. Solution 31 The solution is (D). In a circular pathway, the final product of the reaction is also the initial reactant. The pathway is self-perpetuating, as long as any of the intermediates of the pathway are supplied. Circular pathways are able to accommodate multiple entry and exit points, thus being particularly well suited for amphibolic pathways. In a linear pathway, one trip through the pathway completes the pathway, and a second trip would be an independent event. Cellular respiration breaks down glucose and releases carbon dioxide and water. Which step in the oxidation of pyruvate produces carbon dioxide? A Removal of a carboxyl group from pyruvate releases carbon dioxide. The pyruvate dehydrogenase complex comes into play. B Removal of an acetyl group from pyruvate releases carbon dioxide. The pyruvate decarboxylase complex comes into play. C Removal of a carbonyl group from pyruvate releases carbon dioxide. The pyruvate dehydrogenase complex comes into play. D Removal of an acetyl group from pyruvate releases carbon dioxide. The pyruvate dehydrogenase complex comes into play. Solution 32 The solution is (D). Two molecules of pyruvate are produced by glycolysis. A carboxyl group is removed from pyruvate, releasing a molecule of carbon dioxide. This step proceeds twice because there are two pyruvate molecules produced at the end of glycolysis. A carboxyl group is released with the help of pyruvate dehydrogenase, releasing carbon dioxide. What three steps are included in the breakdown of pyruvate? A Pyruvate dehydrogenase removes a carboxyl group from pyruvate, producing carbon dioxide. Dihydrolipoyl transacetylase oxidizes a hydroxyethyl group to an acetyl group, producing NADH. Finally, an enzyme-bound acetyl group is transferred to CoA, producing a molecule of acetyl-CoA. Advanced Placement Biology Instructor’s Solution Manual 146 7 | Cellular Respiration B Pyruvate dehydrogenase oxidizes a hydroxyethyl group to an acetyl group, producing NADH. It further removes a carboxyl group from pyruvate, producing carbon dioxide. Finally, dihydrolipoyl transacetylase transfers an enzyme-bound acetyl group to CoA, forming an acetyl-CoA molecule. C Pyruvate dehydrogenase transfers enzyme-bound acetyl group to CoA, forming an acetyl CoA molecule. It then oxidizes a hydroxyethyl group to an acetyl group, producing NADH. Dihydrolipoyl transacetylase removes a carboxyl group from pyruvate, producing carbon dioxide. D Pyruvate dehydrogenase removes a carboxyl group from pyruvate, producing carbon dioxide. Dihydrolipoyl dehydrogenase transfers enzyme-bound acetyl groups to CoA, forming an acetyl-CoA molecule. Finally, a hydroxyethyl group is oxidized to an acetyl group, producing NADH. Solution 33 The solution is (A). Step 1. A carboxyl group is removed from pyruvate, releasing a molecule of carbon dioxide. The result is a two-carbon hydroxyethyl group bound to the enzyme pyruvate dehydrogenase. Step 2. The hydroxyethyl group is oxidized to an acetyl group, and the electrons are picked up by NAD+, forming NADH. Step three: The enzyme-bound acetyl group is transferred to CoA, producing a molecule of acetyl-CoA. How do the roles of ubiquinone and cytochrome c differ from the other components of the electron transport chain? A CoQ and cytochrome c are mobile electron carriers, while NADH dehydrogenase and succinate dehydrogenase are bound to the inner mitochondrial membrane. B CoQ and cytochrome covalently bind electrons, while NADH dehydrogenase and succinate dehydrogenase are bound to the inner mitochondrial membrane. C CoQ and cytochrome c are bound to the inner mitochondrial membrane, while NADH dehydrogenase and succinate dehydrogenase are mobile electron carriers. D CoQ and cytochrome c covalently bind electrons, while NADH dehydrogenase and succinate dehydrogenase are mobile electron carriers. Solution 34 The solution is (A). CoQ and cytochrome c are transport molecules and referred to as small electron carriers. Ubiquinone (CoQ) is the only component of the electron transport chain that is not a protein. Also, the other components of the electron transport chain are large complexes anchored to the inner mitochondrial membrane. What accounts for the different number of ATP molecules that are formed through cellular respiration? A Transport of NADH from cytosol to mitochondria is an active process that decreases the number of ATP produced. B The ATPs produced are utilized in the anaplerotic reactions that are used for the replenishment of the intermediates. Advanced Placement Biology Instructor’s Solution Manual 7 | Cellular Respiration 147 C Most of the ATPs produced are rapidly used for the phosphorylation of certain compounds found in plants. D A large number of ATP molecules are used in the detoxification of xenobiotic compounds produced during cellular respiration. Solution 35 The solution is (A). Few tissues except muscle produce the maximum possible amount of ATP from nutrients. The intermediates are used to produce needed amino acids, fatty acids, cholesterol, and sugars for nucleic acids. When NADH is transported from the cytoplasm to the mitochondria, an active transport mechanism is used, which decreases the amount of ATP that can be made. The electron transport chain differs in composition between species, so different organisms will make different amounts of ATP from their electron transport chains. Which statement best describes complex IV in the electron transport chain? A Complex IV consists of an oxygen molecule held between the cytochrome and copper ions. The electrons flowing finally reach the oxygen, producing water. B Complex IV contains a molecule of flavin mononucleotide and iron-sulfur clusters. The electrons from NADH are transported here to coenzyme Q. C Complex IV contains cytochrome b, c, and Fe-S. Here, the proton motive Q cycle takes place. D Complex IV contains a membrane-bound enzyme that accepts electrons from FADH2 to make FAD. This electron is then transferred to ubiquinone. Solution 36 The solution is (A). The cytochromes in complex IV hold an oxygen molecule very tightly between the iron and copper ions until the oxygen is completely reduced. The reduced oxygen then picks up two hydrogen ions from the surrounding medium to make water. What is the primary difference between fermentation and anaerobic respiration? A Fermentation uses only glycolysis, and its final electron acceptor is an organic molecule, whereas anaerobic respiration uses glycolysis, TCA, and the ETC, but finally gives electrons to an inorganic molecule. B Fermentation uses glycolysis, TCA, and ETC, but finally gives electrons to an inorganic molecule, whereas anaerobic respiration uses only glycolysis, and its final electron acceptor is an organic molecule. C Fermentation uses glycolysis, and its final electron acceptor is an inorganic molecule, whereas anaerobic respiration uses glycolysis, TCA, and ETC, but finally gives electrons to an organic molecule. D Fermentation uses glycolysis, TCA, and ETC, but finally gives electrons to an organic molecule, whereas anaerobic respiration uses only glycolysis, and its final electron acceptor is an inorganic molecule. Advanced Placement Biology Instructor’s Solution Manual 148 7 | Cellular Respiration Solution 37 The solution is (A). Fermentation uses glycolysis only. Anaerobic respiration uses all three parts of cellular respiration, including the parts in the mitochondria like the citric acid cycle and electron transport; it also uses a different final electron acceptor instead of oxygen gas. What type of cellular respiration is represented in the following equation, and why? CO2 + H2 + NADH → CH4 + H2O + NAD+ A Anaerobic respiration, because the final electron acceptor is inorganic B Aerobic respiration, because oxygen is the final electron acceptor C Anaerobic respiration, because NADH donates its electrons to a methane molecule D Aerobic respiration, because water is being produced as a product Solution 38 The solution is (A). NAD+ is regenerated from NADH using an inorganic molecule (carbon dioxide) as a final electron acceptor. Would you describe metabolic pathways as inherently wasteful or inherently economical, and why? A Metabolic pathways are economical due to feedback inhibition. Also, intermediates from one pathway can be utilized by other pathways. B Metabolic pathways are wasteful because they perform uncoordinated catabolic and anabolic reactions that waste some of the energy that is stored. C Metabolic pathways are economical due to the presence of anaplerotic reactions that replenish the intermediates. D Metabolic pathways are wasteful because most of the energy produced is utilized in maintaining the reduced environment of the cytosol. Solution 39 The solution is (A). They are very economical. The substrates, intermediates, and products move between pathways and do so in response to finely tuned feedback inhibition loops that keep metabolism balanced overall. Intermediates in one pathway may occur in another, and they can move from one pathway to another fluidly in response to the needs of the cell. What lipids are connected to glucose catabolism pathways, and how are they connected? A Cholesterol and triglycerides can be converted to glycerol-6-phosphate that continues through glycolysis. B Glucagon and glycogen can be converted to 3-phosphoglyceraldehyde, which is an intermediate of glycolysis. C Chylomicrons and fatty acids get converted to 1,3-bisphosphoglycerate that continues in glycolysis, forming pyruvate. D Sphingolipids and triglycerides form glucagon that can be fed into glycolysis. Advanced Placement Biology Instructor’s Solution Manual 7 | Cellular Respiration Solution 40 149 The solution is (A). Triglycerides can be both made and broken down through parts of the glucose catabolism pathways. Glycerol can be phosphorylated to glycerol-3phosphate, which continues through glycolysis. How does citrate from the citric acid cycle affect glycolysis? A Citrate and ATP are negative regulators of phosphofructokinase-1. B Citrate and ATP are negative regulators of hexokinase. C Citrate and ATP are positive regulators of phosphofructokinase-1. D Citrate and ATP are positive regulators of hexokinase. Solution 41 The solution is (A). Citrate from the citric acid cycle indicates that an alternative energy source is available and thus inhibits phosphofructokinase-1. ATP is another negative regulator of PFK-1. Why might negative feedback mechanisms be more common than positive feedback mechanisms in living cells? A Negative feedback mechanisms maintain homeostasis, whereas positive feedback drives the system away from equilibrium. B Positive feedback mechanisms maintain a balanced amount of substances, whereas negative feedback restricts them. C Negative feedback turns the system off, making it deficient in certain substances. Positive feedback balances out these deficits. D Positive feedback brings substance amounts back to equilibrium, while negative feedback produces excess amounts of the substance. Solution The solution is (A). Negative feedback mechanisms control the process by turning the system off at the appropriate step. Positive feedback accelerates the processes, taking the system away from equilibrium. TEST PREP FOR AP® COURSES 42 The table shows the amount of oxygen consumed (third column) by different animals (first column) at different temperatures. This type of apparatus measures the change in volume of air to detect the removal of oxygen. However, organisms produce carbon dioxide as they take in oxygen. To provide accurate measurements, what would you need to add to the setup? Advanced Placement Biology Instructor’s Solution Manual 150 7 | Cellular Respiration A A substance that removes carbon dioxide gas B A plant that will add oxygen to allow an animal to breathe C A glucose reserve D A substance that adds carbon dioxide gas Solution The solution is (A). A substance to remove carbon dioxide should be added, as the respirometer measures the change in the gas volume. The animals produce carbon dioxide that, if not removed, would bias the results. 43 According to the data, the crickets at 25 °C have greater oxygen consumption per gram of tissue than do the crickets at 10 °C. This trend in oxygen consumption is the opposite of that in mice. The difference in trends in oxygen consumption among crickets and mice is due to what? A Their difference in size B Their mode of nutrition C Their difference in metabolic heat production D Their mode of ATP production Solution 44 The solution is (C). Crickets are ectotherms, meaning that their metabolic rate is generally higher at higher ambient temperatures. This is why the crickets at 25 °C have a higher metabolic rate, as evidenced by higher oxygen consumption, than crickets at a lower temperature. Where in a cell does glycolysis take place in both prokaryotes and eukaryotes? A The cytosol Advanced Placement Biology Instructor’s Solution Manual 7 | Cellular Respiration 151 B The mitochondria C The plasma membrane D The nucleus Solution 45 The solution is (A). Glycolysis occurs in the cytosol in eukaryotes as well as in prokaryotes. A new species of obligate anaerobe, a bacterium, has been found that lives in hot, acidic conditions. While other pathways may also be present, which metabolic pathway is the most likely to be present in this species? A Aerobic respiration B The citric acid cycle C Oxidative phosphorylation D Glycolysis Solution 46 The solution is (D). Glycolysis is a metabolic pathway that is common to almost all organisms. What evidence provides the strongest support that glycolysis is an older and more conserved pathway than the citric acid cycle? A Glycolysis is the primitive pathway, found in all three domains. It also occurs in anaerobic conditions and in the cytosol. B This pathway occurs in the cytosol, is found in all animals and plants, and does not require oxygen. C Glycolysis takes place in anaerobic conditions, can metabolize cholesterol and fatty acids, and occurs even in methanogens. D This pathway only occurs in the mitochondria. It is highly flexible because it is found in almost all organisms. Solution 47 The solution is (A). Glycolysis is found in all three domains, while the citric acid cycle is not. This suggests that glycolysis is the more primitive pathway. What is Structure X in the graphic? A The inner mitochondrial membrane Advanced Placement Biology Instructor’s Solution Manual 152 7 | Cellular Respiration B The mitochondrial matrix C A eukaryotic plasma membrane D The cytosol Solution 48 The solution is (A). The phospholipid bilayer membrane is the inner mitochondrial membrane. The embedded structures depict the complexes of the electron transport chain. What would be the most direct result of blocking Structure Z in the graphic? A Cytochrome c would not pass electrons from complex III to complex IV. B Ubiquinone would not pass electrons from complex III to complex IV. C NADH would not be converted to NAD+, and the electron transport chain would stop. D No protons would be pumped across the membrane. Solution The solution is (A). Structure Z is cytochrome c. If it gets blocked, then electrons wouldn’t be able to pass from complex III to complex IV. Advanced Placement Biology Instructor’s Solution Manual 7 | Cellular Respiration 49 153 Where do the electrons moving along the membrane in the figure come from, and where do the electrons end up? A The electrons are released by NADH and FADH2 and finally accepted by oxygen to form water. B The electrons are given off by water and finally accepted by NAD + and FAD+ to produce the energy currencies NADH and FADH2. C The electrons are emitted by ubiquinone and are, in turn, transferred from complex I to complex II. Water finally accepts the electrons. D The electrons are given out by NADH and FADH2 and are, in turn, finally accepted by H2O. Solution 50 The solution is (A). The electrons come from NADH and FADH2, which are produced during glycolysis and the citric acid cycle. The electrons from NADH and FADH join with H+ and O2 to form water. Glucose catabolism pathways are sequential and lead to the production of energy. What is the correct order of the pathways for the breakdown of a molecule of glucose as shown in the formula? C6H12O6 + O2 → CO2 + H2O + energy A Oxidative phosphorylation → citric acid cycle → oxidation of pyruvate → glycolysis B The oxidation of pyruvate → citric acid cycle → glycolysis → oxidative phosphorylation C Glycolysis → oxidation of pyruvate → citric acid cycle → oxidative phosphorylation D Citric acid cycle → glycolysis → oxidative phosphorylation → oxidation of pyruvate Solution The solution is (C). Glycolysis produces pyruvate, and then this pyruvate undergoes oxidation to produce CO2 and acetyl-CoA, the latter of which enters the citric acid cycle to produce more ATP, NADH, and FADH. NADH and FADH2 donate electrons to the electron transport chain to produce ATP and water molecules through oxidative phosphorylation. Advanced Placement Biology Instructor’s Solution Manual 154 51 7 | Cellular Respiration Which statement most directly supports the claim that different species of organisms use different metabolic strategies to meet their energy requirements for growth, reproduction, and homeostasis? A During cold periods, pond-dwelling animals can increase the number of unsaturated fatty acids in their cell membranes, while some plants make antifreeze proteins to prevent ice crystal formation in their tissues. B Bacteria lack introns, while many eukaryotic genes contain many of these intervening sequences. C Carnivores have more teeth that are specialized for ripping food, while herbivores have more teeth specialized for grinding food. D Plants generally use starch molecules for storage, while animals use glycogen and fats for storage. Solution 52 The solution is (D). Having different mechanisms of storing energy (starch molecules in plants, glycogen in animals) is a way to strategize the conservation and use of energy. According to their requirements, these organisms can then use the stored energy in a sustainable way. Which statement best describes how the citric acid cycle relates to glycolysis, oxidative phosphorylation, and chemiosmosis? A Glycolysis produces pyruvate, which is converted to acetyl-CoA and enters the citric acid cycle. This cycle produces NADH and FADH2, which donate electrons to the electron transport chain to pump protons and produce ATP through chemiosmosis. Production of ATP using an electron transport chain and chemiosmosis is called oxidative phosphorylation. B The citric acid produces pyruvate, which converts to glucose to enter glycolysis. This pathway produces NADH and FADH2, which enter oxidative phosphorylation to produce ATP through chemiosmosis. C Citric acid produces NADH and FADH2, which undergo oxidative phosphorylation. This produces ATP by pumping protons through chemiosmosis. The ATP produced is utilized in large amounts in the process of glycolysis. D Glycolysis produces pyruvate, which directly enters the citric acid cycle. This cycle produces the energy currency that undergoes the electron transport chain to produce water and ATP. Solution The solution is (A). Pyruvate is produced by glycolysis during the catabolism of glucose. Pyruvate is converted to acetyl-CoA, which enters the citric acid cycle and is oxidized to CO2. The electrons extracted are picked up by NAD+ and FAD+, and then carried by NADH and FADH2 to the electron transport chain of oxidative phosphorylation. NADH and FADH2 transfer their electrons to the electron transporters embedded in the inner mitochondrial membrane. As the electrons move through the electron transport chain, the energy produced is used to pump protons Advanced Placement Biology Instructor’s Solution Manual 7 | Cellular Respiration 155 from the mitochondrial matrix to the intermembrane space. These protons then diffuse back through the inner mitochondrial membrane through ATP synthase during chemiosmosis. This facilitates the addition of a phosphate to ADP, forming ATP. SCIENCE PRACTICE CHALLENGE QUESTIONS 7.1 Energy in Living Systems 53 Combustion of carbohydrates, like in a fireplace, is a reduction-oxidation reaction in which the carbon atom is oxidized and the oxygen atom is reduced, producing water and carbon dioxide. Oxidative phosphorylation and glycolysis are also reduction-oxidation reactions that produce the same products. Explain the differences and similarities among these abiotic and biotic processes in terms of the changes in entropy and heat that contribute to the free energy extracted from chemical bonds, the spontaneity of each, and the role of catalysis. Solution This question is very open ended and could be used as the basis for small group discussion. Sample answers: A fire generates almost entirely heat and a large increase in entropy, so the resulting free energy is negligible. Oxidative phosphorylation produces very little heat, so the bond energy is captured as free energy. Glycolysis is less efficient, and the heat produced is useful for thermoregulation. Both the abiotic and biotic processes are exergonic. The development of elegant catalysis and structural features for metabolism and its regulation is very different from the unregulated and random fire. But all of these processes have activation energies, including the spark. 7.3 Oxidation of Pyruvate and the Citric Acid Cycle 54 A. [Extension] Living systems require free energy to carry out cellular functions and employ various strategies to capture, use, and store free energy. Explain the advantage that the higher energy efficiency per kilogram of the Krebs cycle provides to you compared to a metabolism based on glycolysis alone. Your explanation should make use of all the following facts: G for glycolysis is 135 kJ/mole of glucose. G for aerobic respiration is 2,880 kJ/mole glucose. The basal metabolic rate of mammals is often represented as 300 kJ/day m0.75 . The molar mass of glucose is 180 g/mole. B. Explain the bioenergetic difference between aerobic and anaerobic respiration in terms of the difference between free-energy production and power. Your explanation should make use of all the following facts: Advanced Placement Biology Instructor’s Solution Manual 156 7 | Cellular Respiration Power is the rate of free-energy production. Cancer cells derive most of their free energy from glycolysis. Enzymes of the citric acid (Krebs) cycle form coordinate complexes on the cytoskeleton within the mitochondria. C. The life cycle of the human parasite Trypanosoma brucei is divided between the body of the tsetse fly and the human blood stream. The parasite causes “sleeping sickness” in Sub-Saharan Africa. Within the human bloodstream, the parasite depends on glycolysis, with enzymes compartmentalized in a membrane-bound organelle called the glycosome. In the insect host, the parasite utilizes glycolysis as well as substrate-level and oxidative phosphorylation. Explain the advantage of a life cycle in the human host that employs anaerobic respiration with a rate of free-energy production that is enhanced by compartmentalization in the glycosome and a life cycle in the insect host that is aerobic. D. Predict the advantages of a biological system that uses both glycolysis and oxidative phosphorylation. Your prediction should make use of all the following facts: Solution Signaling can be used to detect low-oxygen environments and to regulate response. Some cells, such as muscle and blood cells, must function in both low- and highoxygen environments. Glycolysis is reversible. The citric acid cycle is not reversible. Thermoregulation is needed for homeostasis. It is very hard for students to learn to construct explanations. They need to remember that it must be testable and account for all of the available facts. Identifying the facts that must be included in an explanation helps them to develop the habit of organizing the evidence. Sample answer: A. Basal metabolic rate is calculated by the student to be between 4,000 and 7,000 kJ/day. Dividing that by the free energy conversion for glucose shows that about 40 moles of glucose, or 7.6 kg or 17 lb, would need to be consumed per day. In contrast, the same calculation gives about 1 lb/ day for aerobic respiration. G 300 kJ/kg 0.75day m0.75 5640 kJ/day assuming 50 kg day 5640 kJ 1 mole glucose 180 g 41.8 mole glucose 7.5 kg glucose day 135 kJ mole glucose That’s about 1,500 teaspoons per day, or about 2 teaspoons per waking minute. One measure of efficiency might be the ability to work other than feed one’s self. Advanced Placement Biology Instructor’s Solution Manual 7 | Cellular Respiration 157 B. Thermodynamics is really thermostatics. Often the rate matters more. Cancer cells are growing rapidly, so we know that the rate of energy consumption is high. However, they are also primarily using glycolysis for this production. I read that there are effects of mitochondrial damage and restricted blood flow (and so hypoxic). The student should puzzle over this and read about it—there is a wellknown concept called the Warburg model that looks at the effect that a switch to anaerobic respiration has on cancer cells. On the other hand, we know that the Krebs cycle is highly coordinated and not diffusion limited. The need for structural support is often discussed, and there is work that shows the attachment of clusters of enzymes in a coordinated manner to the cytoskeleton. However, the idea of a metabolon is of growing interest. The glycolytic complex metabolon is now discussed. C. A connection to compartmentalization can be made with Trypanosoma. And to make it all as ambiguous as biology is, there is an order of organisms that have a specialized organelle for glycolysis. The claim is made that Trypanosoma brucei (sleeping sickness) has a metabolic rate 50 times higher than is common for glycolysis. This compartmentalization compensates for the lower output of anaerobic respiration. The advantage of anaerobic respiration in the human host is that venal blood is depleted of oxygen. The insect host has no similar inhomogeneous oxygen supply. Also, the fly’s hemolymph is rich in proline, and the degradation of this amino acid can be used as input to the Krebs cycle. D. The selective advantage is provided through a more flexible response to changes in the environment. The facts here point to the ability to switch to anaerobic respiration under oxygen depletion, to store fat through gluconeogenesis, and to use low efficiency metabolism to generate heating. 7.4 Oxidative Phosphorylation 55 Dinitrophenol (DNP) was used in the manufacture of munitions in World War I. In the 1930s, it was used as a weight-loss drug. Use in the United States cannot be regulated by the U.S. Food and Drug Administration (FDA) because DNP is considered a dietary supplement. Attempts to ban the drug in the United Kingdom following the death of four users in 2015 failed in Parliament. DNP is a small molecule that is soluble in the mitochondrial inner membrane. The hydroxyl group reversibly dissociates a proton. Advanced Placement Biology Instructor’s Solution Manual 158 7 | Cellular Respiration A. Predict the effect of DNP on the electrochemical gradient across the inner mitochondrial membrane. B. Explain how DNP can be used to reduce weight. C. The effects of DNP can be reversed by administering glucose. However, treatment with a combination of glucose and 2-deoxyglucose, which is an inhibitor of glycolysis, does not reverse the effects of DNP. Explain, in terms of the products of glycolysis, why this reversal of the effects of DNP was unexpected. (Hint—It might be useful to review the reactants and products of glycolysis.) D. Obesity correlates with an epidemic of other health issues, such as elevated blood pressure, heart disease, and diabetes type II. A slow-release form of DNP (CRMP) is patented. With slow-release technology, a drug can be delivered in small doses over time from a pill whose matrix limits solubility. A simple but nonscientific question that can be raised is: Will a slow-release drug retard progress toward behavioral changes that can reduce the magnitude of this epidemic? Scientific questions can be pursued by testing the outcomes predicted by possible answers. Refine this question for discussion in small groups. Be prepared to justify the merits of your question. Solution Sample answer: A. Because DNP makes the inner mitochondrial membrane “leaky” to protons, the electrochemical gradient across the membrane is disrupted. B. DNP produces weight loss by partial starvation. The hydroxyl group of DNP is protonated on the intermembrane space side of the inner membrane where hydrogen ion concentrations are high. Diffusing through the membrane, the proton is dissociated in the matrix where the hydrogen ion concentrations are low. The net effect is the transport of a proton through the membrane. Molecules that have this effect are called protonophores or proton translocators. The production of ATP and NADH in the electron transport chain is driven by the electrochemical gradient produced by the relatively low concentration of protons in the matrix. The transport of protons to the matrix lowers the electrochemical potential. The ETS is said to be decoupled. C. Flooding with glucose increases glycolysis. The fact that treatment with 2deoxyglucose, which inhibits glycolysis, along with glucose supports this implication. However, for each glucose converted to two pyruvates, two hydrogen ions are released. The increase in pH shifts the equilibrium in the equation shown further to the left, which would increase the leakage of protons. The work that first reported this finding (Nakamura et al., 1989, Cardiovasc. Res. Apr., 23(4):286–94) suggested increased levels of ATP in the intermembrane space might be responsible. In the 1990s, discoveries of ATP-dependent potassium channels began to be reported. Transport of potassium into the matrix would repair the electrochemical potential difference, reversing the effects of DNP. Claims are not made on the Internet that, when using DNP, you should also take a dose of insulin. Advanced Placement Biology Instructor’s Solution Manual 7 | Cellular Respiration 159 D. Useful questions will produce predictions that can be tested. The following might be among those that students pose: 56 What data are available concerning the long-term success rates of dieting or exercise as diabetes treatment? If CRMP is patented, will it be regulated as a prescription drug and not a dietary supplement? Will mitochondria in all cells, particularly muscle and nerve tissue, be affected, or can only tissues that have a prominent role in homeostasis, such as liver or white adipose tissue, be targeted? (Inventors of CRMP use an ether derivative of DNP that allows it to selectively target the liver.) As shown in the figure, cyanide inhibits the electron transport chain by competing with O2 molecules for the cytochrome c oxidase heme group. Carbon monoxide (CO) has a similar effect. Both cyanide and carbon monoxide cause poisoning in victims of smoke inhalation. A. Predict the effects of these poisons on the following properties of mitochondria just after exposure: the pH of the intermembrane space, the concentration of NADH, and the rate of production of ATP in the matrix. Justify your predictions. B. Rotenone is a poison that blocks the transfer of electrons from Complex I of the electron transport chain to ubiquinone. Methylene blue is a molecule with many uses involving its reduction-oxidation properties. Recent studies show the effectiveness of methylene blue in increasing the body’s metabolic rate and as a treatment for Alzheimer’s patients. The oxidized form of methylene blue is reduced by NADH, and its reduced form is oxidized by O2. Explain the use of methylene blue as an antidote for rotenone poisoning. Advanced Placement Biology Instructor’s Solution Manual 160 7 | Cellular Respiration Solution Sample answer: A. When the cytochrome is blocked, electrons cannot be transferred to O 2. The concentration of molecular oxygen increases. The transfer of electrons to cytochrome is also blocked, and so the NADH concentration increases. Hydrogen ions cannot be transported over the membrane so the pH of the intermembrane space drops. B. Methylene blue is currently a hot supplement available on the web; athletes and students want improved performance. Measurements of respiration rate indicate that the effects are real. Since methylene blue is able to act as a redox surrogate for Complex II and III, the effect of rotenone is counteracted by the transfer of electrons from NADH to methylene blue to cytochrome. 7.5 Metabolism without Oxygen 57 E. coli are enteric (gut-dwelling) facultative anaerobic bacteria. (Facultative anaerobes can grow with or without free oxygen. Obligatory anaerobes grow only in the absence of free oxygen.) Researchers planned to grow cultures of E. coli under a range of conditions to model the transition from strictly anaerobic to aerobic respiration. The oxygen content of atmospheres at constant total pressure will be controlled by volumes of nitrogen and oxygen gases. Ratios of volume, r VO2 / VN2 between 0 and 0.25 of shaken growth flasks, can be measured in terms of optical density, which is the percent of transmission of light through a sample of the growing E. coli culture. A rule of thumb is that the range of strict anaerobes is when r 0.01, and the boundary for aerobic respiration is when r = 0.05. A large number of flasks that can be constantly shaken at fixed temperature, and from which samples can be taken without atmospheric contamination, are available for this study. These results of the experiment will be used to infer growth rates of E. coli along the entire 7.5-m length of the average human intestine (small intestine and large intestine), where the oxygen content varies from atmospheric to anaerobic conditions. The retention time of food in the small intestine, whose average length is 2.5 m, is approximately 4 h. The retention time of food over the entire length of the intestine is between 24 and 72 h. A. Describe and apply a mathematical model that can be used to represent the variation of oxygen environments of a bacterium that is being transported with the food along the length of the intestine. B. Design the experimental sampling times in terms of growth intervals of interest in this study: (i) the time when the bacteria are passing the small-large intestine boundary; (ii) the time when the bacteria reach the end of the large intestine; and (iii) the time when the bacteria reach facultative anaerobic conditions, r 0.05. Advanced Placement Biology Instructor’s Solution Manual 7 | Cellular Respiration 161 C. Sketch a graph that predicts the distribution of aerobic, facultative anaerobic, and obligatory anaerobic bacteria along the length of the entire intestine based on these parameters. Keep in mind that anaerobes have a lower respiration rate. Solution Sample answer: A. The oxygen ratio must go from 0.25 to 0 over a distance of 7.5 m. To make progress, begin by assuming the simplest form (you might also use 0.25e − ax since the ratio cannot be negative) rO2 = 0.25 – 0.03x, where x is the distance along the intestine 0 x 7.5 m and rO2 is the oxygen ratio. To answer Part B, we also need a model that involves the dependence of rO2 on time. The speed of food through the intestine can be used to obtain the rO 2(t) relationship. The range of total time is very broad (and that is good because it introduces conditional modeling) 24 h t 72 h. So, the speed is between 0.3 and 1 m/h. Then since x speed time , we have a second relationship 0.25 – 0.01t rO2 t 0.25 – 0.03t , where t is measured in hours. B. The times of interest are (i) passing the small-large intestine boundary, x = 2m, r = 0.19, and 2 h t 6 h ; (ii) reached the end of the large intestine, 24 h t 72 h; and (iii) the boundary to facultative anaerobic conditions, r 0.05; 6 h t 20 h. C. A reasonable sketch could look like the following: 7.6 Connections of Carbohydrate, Protein, and Lipid Metabolic Pathways 58 White snakeroot is a plant that contains chemicals that deactivate the enzyme lactate dehydrogenase. Humans who consume milk from cows or goats that eat white snakeroot can become ill. Symptoms of milk poisoning include vomiting, abdominal pain, and tremors, which become worse after exercise. Beyond childhood, most people do not express the enzyme lactase that catalyzes the breakdown of lactose into glucose and galactose. Consumption of milk can produce symptoms similar to those of milk poisoning. After a period of consumption of dairy foods, though, prebiotic adaptation (changes in the Advanced Placement Biology Instructor’s Solution Manual 162 7 | Cellular Respiration microbes in the intestine) imparts lactose tolerance. Since dairy foods are a valuable source of calcium, proteins, and vitamin D, considerable research has been conducted to characterize adaptation. Explain the similarities and differences between the effect of milk poisoning by white snakeroot and lactose intolerance, and the possibility of prebiotic adaptation for each. Solution Sample answer: As described in the text, in the absence of an active form of lactate dehydrogenase in the liver, lactic acid cannot be converted to pyruvate. Lactose, another component of dairy food, cannot be metabolized without the enzyme lactase. The human population recently originated in Europe, where dairy food resources produced selection favoring the production of lactase by adults. Populations recently originating in Africa and Asia do not express lactase as adults. However, consumption of dairy products leads to changes in gut bacteria. Such a prebiotic adaptation could not protect Midwestern residents of the 19th century against milk poisoning. Both gut bacteria and their human host are susceptible to disruption by white snakeroot. Advanced Placement Biology Instructor’s Solution Manual 8 | Photosynthesis 163 8 | PHOTOSYNTHESIS REVIEW QUESTIONS 1 Which component is NOT used by both plants and cyanobacteria to carry out photosynthesis? A Carbon dioxide B Chlorophyll C Chloroplasts D Water Solution 2 The solution is (C). The chloroplast is not a component used by both plants and cyanobacteria. Photosynthesis occurs in chloroplasts of eukaryotes, while in prokaryotes like cyanobacteria, photosynthesis occurs within the membrane and cytoplasm. Why are chemoautotrophs NOT considered the same as photoautotrophs if they both extract energy and make sugars? A Chemoautotrophs use wavelengths of light not available to photoautotrophs. B Chemoautotrophs extract energy from inorganic chemical compounds. C Photoautotrophs prefer the blue side of the visible light spectrum. D Photoautotrophs make glucose, while chemoautotrophs make galactose. Solution 3 The solution is (B). Chemoautotrophs extract energy from inorganic chemical compounds. Chemoautotrophs are organisms that use inorganic molecules, rather than sunlight, as energy sources to synthesize their own food. They are mostly bacteria and archaea that live in hostile sea environments where they are the primary producers. In which compartment of the plant cell do the light-independent reactions of photosynthesis take place? A Mesophyll B Outer membrane C Stroma D Thylakoid Solution The solution is (C). The light-independent reactions take place in the stroma of the chloroplast because ATP and NADPH are produced on the stroma side of thylakoid. Also, Calvin cycle enzymes are found in the stroma, so light-independent reaction takes place in stroma. Advanced Placement Biology Instructor’s Solution Manual 164 4 8 | Photosynthesis What is a part of grana? A The Calvin cycle B The inner membrane C Stroma D Thylakoids Solution 5 The solution is (D). The stack of the tube-like strands connecting thylakoids is called granum. What are two major products of photosynthesis? A Chlorophyll and oxygen B Oxygen and carbon dioxide C Sugars/carbohydrates and oxygen D Sugars/carbohydrates and carbon dioxide Solution 6 The solution is (C). The photosynthesis reaction involves the reactants carbon dioxide and water. The products of the reaction are oxygen, carbohydrates, and mainly glucose. What is the primary energy source for cells? A Glucose B Starch C Sucrose D Triglycerides Solution 7 The solution is (A). Glucose is a ready source of energy since its carbon atoms are easily oxidized to form carbon dioxide, releasing energy in the process. It can be metabolized via aerobic (cellular respiration) and anaerobic (fermentation) means, whereby the cell uses the by-products of these reactions to produce ATP, the cell’s energy currency. Which portion of the electromagnetic radiation originating from the sun is harmful to living tissues? A Blue B Green C Infrared D Ultraviolet Solution The solution is (D). Ultraviolet waves are high-energy waves that penetrate tissues and damage cells and DNA. The resulting damage to DNA from ultraviolet radiation can lead to the formation of thymidine dimers, a mutation that can potentially lead Advanced Placement Biology Instructor’s Solution Manual 8 | Photosynthesis 165 to cancer if many are accumulated within a cell and are not repaired through the cell’s DNA repair mechanisms. 8 The amount of energy in a wave can be measured using what trait? A Color intensity B Distance from trough to crest C The amount of sugar produced D Wavelength Solution 9 The solution is (D). The amount of energy is the distance measured by two successive points in the wave that are characterized by same phase, called the wavelength. What portion of the electromagnetic radiation emitted by the sun has the least energy? A Gamma B Infrared C Radio D X-rays Solution 10 The solution is (C). Radio waves possess the least energy in the electromagnetic spectrum and have long wavelengths. What is the function of carotenoids in photosynthesis? A They supplement chlorophyll absorption. B They are visible in the fall during leaf color changes. C They absorb excess energy and dissipate it as heat. D They limit chlorophyll absorption. Solution 11 The solution is (D). Carotenoids absorb excess energy and dissipate it as heat. In photosynthesis, carotenoids serve as an efficient molecule for the disposal of excess energy. Excess energy can damage the leaf; therefore, carotenoids, which reside in thylakoid membranes, absorb excess energy and dissipate that energy as heat. Which structure is NOT a component of a photosystem? A Antenna molecule B ATP synthase C Primary electron acceptor D Reaction center Solution The solution is (B). ATP synthase is not a component of photosystem. It is present in thylakoid membranes of chloroplasts that create ATP during photosynthesis. Advanced Placement Biology Instructor’s Solution Manual 166 12 8 | Photosynthesis Which complex is NOT involved in producing the electrochemical gradient for ATP synthesis? A Antenna complex B Cytochrome complex C Photosystem I D Photosystem II Solution 13 The solution is (C). Photosystem I is not involved in the establishment of conditions for ATP synthesis because it captures energy to reduce NADP+ into NADPH. What can be calculated from a wavelength measurement of light? A A specific portion of the visible spectrum B Color intensity C The amount of energy of a wave of light D The distance from trough to crest of the wave Solution 14 The solution is (C). The amount of energy of the wave can be calculated by its wavelength. It is the distance between two consecutive points of a wave. Which molecule must enter the Calvin cycle continually for the light-independent reactions to take place? A CO2 B RuBisCO C RuBP D 3-PGA Solution 15 The solution is (A). CO2 is required to be fixed and converted into organic form (glyceraldehyde 3 phosphate to glucose) continually for use by plants. Stage 1 of carbon fixation initiates in the presence of CO2, which is used to convert inorganic carbon to an organic compound. Which order of molecular conversions is correct for the Calvin cycle? A RuBP + G3P → 3-PGA → sugar B RuBisCO → CO2 → RuBP → G3P C RuBP + CO2 → [RuBisCO]3-PGA → G3P D CO2 → 3-PGA → RuBP → G3P Solution The solution is (C). RuBP and CO2 are needed with the enzyme RuBisCO to produce 3-PGA molecules. ATP and NADPH are used to convert 3-PGA molecules into G3P. Advanced Placement Biology Instructor’s Solution Manual 8 | Photosynthesis 16 167 Which statement correctly describes carbon fixation? A The conversion of CO2 into an organic compound B The use of RuBisCO to form 3-PGA C The production of carbohydrate molecules from G3P D The use of ATP and NADPH to reduce CO2 Solution 17 The solution is (A). Carbon fixation is the term used to describe the conversion of inorganic carbon into organic form. The conversion of CO2 (inorganic form) into an organic compound (3-PGA) in stage 1 of the Calvin cycle is an example of carbon fixation. Which substance catalyzes carbon fixation? A 3-PGA B NADPH C RuBisCO D RuBP Solution 18 The solution is (C). RuBisCO is an enzyme that is used to catalyze the reaction between RuBP molecules and CO2 to produce an organic molecule called G3P. This comprises stage 1 of the Calvin cycle and is called carbon fixation. Which pathway is used by both plants and animals? A Carbon fixation B Cellular respiration C Photosystem II D Photosynthesis Solution 19 The solution is (B). Both plants and animals perform cellular respiration. The difference is that animals get their food by eating it, while plants make their own food by photosynthesis. Which organism is a heterotroph? A Cyanobacterium B Intestinal bacterium C Kelp D Pond algae Solution The solution is (B). Heterotrophs are essentially organisms that cannot “make” their own food, and thus must obtain their nutrients through the ingestion or absorption of organic sources of carbon. Of the available answer choices, intestinal bacteria such as E. coli are heterotrophic. Advanced Placement Biology Instructor’s Solution Manual 168 20 8 | Photosynthesis What is the role of ribulose 1,5, bisphosphate carboxylase/oxygenase, abbreviated RuBisCO, in photosynthesis? A It catalyzes the reaction between CO2 and ribulose bisphosphate (RuBP). B It catalyzes the reaction that produces glyceraldehyde3-phosphate (G3P). C It catalyzes the reaction that regenerates RuBP. D It catalyzes the reaction utilizing ATP and NADPH. Solution 21 The solution is (A). RuBisCO catalyzes the reaction between CO2 and ribulose bisphosphate (RuBP). What is the product of the Calvin cycle? A Glucose B Glyceraldehyde-3-phosphate C Phosphoglycerate (PGA) D Sucrose Solution The solution is (B). In addition to glyceraldehyde-3-phosphate, three molecules of adenosine diphosphate (ADP) and 2 molecules of NADP are produced. However, the latter two products are regenerated and are reused in the light-dependent reactions of photosynthesis. CRITICAL THINKING QUESTIONS 22 What are the roles of ATP and NADPH in photosynthesis? A ATP and NADPH are forms of chemical energy produced from the light-dependent reactions to be used in the light-independent reactions that produce sugars. B ATP and NADPH are forms of chemical energy produced from the light-independent reactions to be used in the light-dependent reactions that produce sugars. C ATP and NADPH are forms of chemical energy produced from the light-dependent reactions to be used in the light-independent reactions that produce proteins. D ATP and NADPH are forms of chemical energy produced from the light-dependent reactions to be used in the light-independent reactions that use sugars as reactants. Solution 23 The solution is (A). ATP and NADPH are forms of chemical energy produced from the light-dependent reactions to be used in the light-independent reactions that produce sugars. NADPH and ATP are produced during the light-dependent reactions. The energy captured during photosynthesis is used to convert ADP to ATP and to reduce NADP to NADPH. The NADPH and ATP thus produced enter the lightindependent reactions. What is the overall outcome of the light reactions in photosynthesis? Advanced Placement Biology Instructor’s Solution Manual 8 | Photosynthesis 169 A NADPH and ATP molecules are produced during the light reactions and are used to power the light-independent reactions. B NADPH and ATP molecules are produced during the light reactions and are used to power the light-dependent reactions. C Sugar and ATP are produced during the light reactions and are used to power the light-independent reactions. D Carbon dioxide and NADPH are produced during the light reactions and are used to power the light-dependent reactions. Solution 24 The solution is (A). NADPH and ATP molecules are produced during the light reactions and are used to power the light-independent reactions. NADPH and ATP molecules are the end products of the light reactions. The free energy stored in ATP and NADPH is used to power light-independent reactions. PSII captures energy to create protein gradients to make ATP, while PSI captures energy to reduce NADP + to NADPH. How does the equation relate to both photosynthesis and cellular respiration? A Photosynthesis uses energy to build carbohydrates, while cellular respiration metabolizes carbohydrates. B Photosynthesis uses energy to metabolize carbohydrates, while cellular respiration builds carbohydrates. C Photosynthesis and cellular respiration both use carbon dioxide and water to produce carbohydrates. D Photosynthesis and cellular respiration both metabolize carbohydrates to produce carbon dioxide and water. Solution The solution is (A). Photosynthesis uses energy to build carbohydrates, while cellular respiration metabolizes carbohydrates. Energy is needed to fuel the production of carbohydrates and for the anabolic reactions of metabolism. Photosynthesis absorbs light energy to build carbohydrates in chloroplasts, and aerobic cellular respiration releases energy by using oxygen to metabolize carbohydrates in mitochondria and cytoplasm. In photosynthesis, plants absorb sunlight to build carbohydrates from Advanced Placement Biology Instructor’s Solution Manual 170 8 | Photosynthesis carbon dioxide and water. In cellular respiration, cells metabolize carbohydrates to release energy. This released energy is used by cells for cellular activities. 25 How is the energy from the sun transported within chloroplasts? A When photons strike photosystem (PS) II, pigments pass the light energy to chlorophyll a molecules that excite an electron, which is then passed to the electron transport chain. The cytochrome complex transfers protons across the thylakoid membrane and transfers electrons from PS-II to PS-I. The products of the lightdependent reaction are used to power the Calvin cycle to produce glucose. B When photons strike PS I, pigments pass the light energy to chlorophyll, molecules that excite electrons, which is then passed to the electron transport chain. The cytochrome complex then transfers protons across the thylakoid membrane and transfers electrons from PS-II to PS-I. The products of the light-dependent reaction are used to power the Calvin cycle to produce glucose. C When photons strike PS II, pigments pass the light energy to chlorophyll molecules that in turn excite electrons, which are then passed to the electron transport chain. The cytochrome complex transfers protons across the thylakoid membrane and transfers electrons from PS-I to PS-II. The products of the light-dependent reaction are used to power the Calvin cycle to produce glucose. D When photons strike PS II, pigments pass the light energy to chlorophyll molecules that excite electrons, which are then passed to the electron transport chain. The cytochrome complex transfers protons across the thylakoid membrane and transfers electrons from PS II to PS I. The products of the light-independent reaction are used to power the Calvin cycle to produce glucose. Solution 26 The solution is (A). This is the most accurate description of the chemical processes involved with photosynthesis. When a photon strikes PS II, pigments pass the light energy to chlorophyll a molecules that excite an electron, which are then passed to the electron transport chain. The cytochrome complex transfers protons across the thylakoid membrane and transfers electrons from PS II to PS I. The products of the light-dependent reactions are used to power the Calvin cycle to produce glucose. Explain why X-rays and ultraviolet light wavelengths are dangerous to living tissues. A UV and X-rays are high-energy waves that penetrate the tissues and damage cells. B UV and X-rays are low-energy waves that penetrate the tissues and damage cells. C UV and X-rays cannot penetrate tissues and damage the cells. D UV and X-rays can penetrate tissues and thus do not damage the cells. Solution The solution is (A). UV and X-rays are high-energy waves that penetrate the tissues and damage cells. X-rays and UV light are higher in energy than the wavelengths used for photosynthesis. These high-energy waves can penetrate tissues and damage cells and DNA, explaining why both X-rays and UV rays can be harmful to living organisms. Advanced Placement Biology Instructor’s Solution Manual 8 | Photosynthesis 171 Ultraviolet rays and X-rays prove detrimental for plants’ cell function. They are highenergy waves that penetrate the tissues and cause damage to the cells and DNA of plants. 27 If a plant were to be exposed to only red light, would photosynthesis be possible? A Photosynthesis would not take place. B The rate of photosynthesis would increase sharply. C The rate of photosynthesis would decrease drastically. D The rate of photosynthesis would decrease and then increase. Solution 28 The solution is (B). When a plant is exposed to both red and far-red light, then a phenomenon called the Emerson effect takes place. There is a sharp increase in the rate of photosynthesis after chloroplasts are exposed to wavelengths of 670 nm (farred light for PS II) and 700 nm (red light for PS I). The Emerson effect means that when a plant is exposed to both red light and far-red light, then there is a sharp increase in the rate of photosynthesis. Which statement describes the electron transfer pathway from photosystem II to photosystem I in the light-dependent reactions? A After splitting water in PS II, high-energy electrons are delivered through the chloroplast electron transport chain to PS I. B After splitting water in PS I, high-energy electrons are delivered though the chloroplast electron transport chain to PS II. C After the photosynthesis reaction, the released products like glucose help in the transfer of electrons from PS II to PS I. D After the completion of the light-dependent reactions, the electrons are transferred from PS II to PS I. Solution 29 The solution is (A). After splitting water in PS II, high-energy electrons are delivered though the chloroplast electron transport chain to PS I. The excited electron removed from the photosystem must then be replaced. In PS II, the electron comes from the splitting of water, which releases oxygen as a waste product. In PS I, the electron comes from the chloroplast electron transport chain. The reaction center of PS II (called P680) delivers its high-energy electrons, one at a time, to a primary electron acceptor and through the electron transport chain to PS I. Photolysis of water releases high-energy electrons and oxygen as a waste product. These highenergy electrons are sent to PS I through the chloroplast electron transport chain that reduces NADP+ to NADPH. What will happen to a plant leaf that loses CO2 too quickly? A There will be no effect on the rate of photosynthesis. B Photosynthesis will slow down or possibly stop. Advanced Placement Biology Instructor’s Solution Manual 172 8 | Photosynthesis C Photosynthesis will increase exponentially. D Photosynthesis will decrease and then increase. Solution 30 The solution is (B). Photosynthesis will slow down or possibly stop. Loss of gases, especially of CO2, will affect photosynthesis in the leaf, and Calvin cycle will slow down and eventually stop because of inadequate carbon to fix in the system. A continuous supply of atmospheric CO2 is thus required. Carbon in the form of CO2 must be taken from the atmosphere and attached to an existing organic molecule in the Calvin cycle. Therefore, the carbon is bound to the molecule. The products of the cycle only occur because of the added carbon. What are the products of the Calvin cycle, and what is regenerated? A The product of the Calvin cycle is glyceraldehyde-3 phosphate, and RuBP is regenerated. B The product of the Calvin cycle is glyceraldehyde-3 phosphate, and RuBisCO is regenerated. C The product of the Calvin cycle is a 3-PGA molecule, and glyceraldehyde-3 phosphate is regenerated. D The product of the Calvin cycle is glyceraldehyde-3 phosphate, and oxygen is regenerated. Solution The solution is (A). The product of the Calvin cycle is glyceraldehyde-3 phosphate, and RuBP is regenerated. The carbon cycle is the cyclical use of carbon in organic molecules for energy pathways and its release into the atmosphere after these pathways have been completed. The carbon is then fixed from the atmosphere by photosynthetic organisms to an organic molecule, and the cycle continues. The Calvin cycle produces glyceraldehyde-3 phosphate. The cycle must be repeated three times to yield the product. RuBP is generated from the G3P molecules that remain in the cycle to fix more CO2. 31 How do desert plants prevent water loss from the heat, which would compromise photosynthesis? A By using CAM photosynthesis and by closing stomatal pores during the night B By using CAM photosynthesis and by opening of stomatal pores during the night C By using CAM photosynthesis and by keeping stomatal pores closed at all times D By bypassing CAM photosynthesis and by keeping stomatal pores closed at night Solution The solution is (B). Desert plants use CAM photosynthesis to reduce water loss. In CAM photosynthesis, desert plants open their stomata at night to collect carbon dioxide from the air. At night, temperatures are cooler, preventing a large amount of water loss by transpiration. The collected carbon dioxide is then stored within the leaves as malate. This allows the plants to close their stomata during the day, Advanced Placement Biology Instructor’s Solution Manual 8 | Photosynthesis 173 reducing water loss by evapotranspiration while still having a supply of carbon dioxide to use for photosynthesis. 32 Why are carnivores, such as lions, dependent on photosynthesis to survive? A Because the prey of lions are generally herbivores, which depend on heterotrophs B Because the prey of lions are generally smaller carnivorous animals, which depend on nonphotosynthetic organisms C Because the prey of lions are generally herbivores, which depend on autotrophs D Because the prey of lions are generally omnivores, which depend only on autotrophs Solution 33 The solution is (A). The prey of lions are generally herbivores, which depend on heterotrophs. These herbivores rely on the plants and other photosynthetic organisms for food. Therefore, without photosynthetic organisms, the carnivore’s prey would not be present. Lions generally prey on herbivorous animals, which are dependent on photosynthetic organisms. Why does it take three turns of the Calvin cycle to produce G3P, the initial product of photosynthesis? A To fix enough carbon to export one G3P molecule B To fix enough oxygen to export one G3P molecule C To produce RuBisCO as an end product D To produce ATP and NADPH for fixation of G3P Solution The solution is (A). In stage 3 of the Calvin cycle, RuBP, the molecule that starts the cycle, is regenerated so that the cycle can continue. All of the extra G3P is used in this regeneration process. Because the G3P exported from the chloroplast has three carbon atoms, it takes three turns of the Calvin cycle to fix enough carbon to export one G3P. G3P is produced after three turns of Calvin cycle because the carbon dioxide molecule is incorporated one at a time, so the cycle must be completed three times to produce a single three-carbon (G3P) molecule. TEST PREP FOR AP® COURSES 34 Photosynthesis and cellular respiration are found throughout the eukaryotic world. They are complementary to each other because they each use products of the other process. What do the two pathways share? A Chloroplasts and mitochondria B Photosystems I and II C The cytochrome complex D Thylakoids Advanced Placement Biology Instructor’s Solution Manual 174 8 | Photosynthesis Solution 35 The solution is (C). Photosynthesis and cellular respiration both require the cytochrome complex for electron transport, but the roles in each process differ. What evidence exists that the evolution of photosynthesis and cellular respiration support the concept that there is a common ancestry for all organisms? A All organisms perform cellular respiration using oxygen and glucose, which are produced by photosynthesis. B All organisms perform cellular respiration using carbon dioxide and glucose, which are produced by photosynthesis. C All organisms perform cellular respiration using oxygen and lipids, which are produced by photosynthesis. D All organisms perform cellular respiration using carbon dioxide and lipids, which are produced by photosynthesis. Solution 36 The solution is (A). All organisms perform cellular respiration using oxygen and glucose, which are produced by photosynthesis. The principal energy source used throughout the planet is glucose or some form of carbohydrate. The ultimate source of this chemical is photosynthesis, which is found in every organism that produces glucose. The process of photosynthesis produces oxygen as a product that is used by all organisms in cellular respiration. These two pathways form the foundation of metabolism in nearly all of the organisms in existence. Primitive anaerobic prokaryotic microorganisms are thought to be the earliest organisms developed, with others following once photosynthesis allowed oxygen to exist in the atmosphere. Cellular respiration then became possible and built on some of the chemical reactions that already existed in photosynthesis. What are the correct labels for the indicated parts of a chloroplast? A A. stroma, B. outer membrane, C. granum, D. thylakoid, E. inner membrane B A. outer membrane, B. stroma, C. granum, D. thylakoid, E. inner membrane C A. outer membrane, B. stroma, C. granum, D. inner membrane, E. thylakoid D A. stroma, B. outer membrane, C. inner membrane, D. granum, E. thylakoid Advanced Placement Biology Instructor’s Solution Manual 8 | Photosynthesis Solution 37 175 The solution is (B). The correct labels are A. outer membrane, B. stroma, C. granum, D. thylakoid, E. inner membrane. What cellular features and processes are similar in both photosynthesis and cellular respiration? A Both processes are contained in organelles with single membranes, and both use a version of the cytochrome complex. B Both processes are contained in organelles with double membranes, and neither uses a version of the cytochrome complex. C Both processes are contained in organelles with double membranes and use a version of the cytochrome complex. D Both processes are contained in organelles with single membranes, and neither uses a version of the cytochrome complex. Solution 38 The solution is (C). Both processes are contained in organelles with double membranes (chloroplasts and mitochondria), and both use a version of the cytochrome complex in their respective electron transport chains. Why do the light-dependent reactions of photosynthesis take place in the thylakoid? A Photosystem I is anchored to the membrane, but not photosystem II. B The cytochrome complex requires a membrane for chemiosmosis to occur. C The light-dependent reactions depend on the presence of carbon dioxide. D Light energy is absorbed by the thylakoid membrane. Solution 39 The solution is (B). The cytochrome complex requires a membrane for chemiosmosis to occur. The cytochrome complex in the thylakoid membrane of chloroplasts facilitates ATP generation by the movement of hydrogen ions across the thylakoid membrane. This ATP is further used to power the Calvin cycle. Metabolic pathways both produce and use energy to perform their reactions. How does the Calvin cycle help to harness, store, and use energy in its pathway? A The Calvin cycle harnesses energy in the form of 6 ATP and 6 NADPH that are used to produce fructose-3-phosphate (F3P) molecules. These store the energy captured from photosynthesis. The cycle uses this energy to regenerate RuBP. B The Calvin cycle harnesses energy in the form of 6 ATP and 6 NADPH that are used to produce glyceraldehyde-3-phosphate (GA3P) molecules. These store the energy captured from photosynthesis. The cycle uses this energy to regenerate RuBP. Advanced Placement Biology Instructor’s Solution Manual 176 8 | Photosynthesis C The Calvin cycle harnesses energy in the form of 3 ATP and 3 NADPH that are used to produce glyceraldehyde-3-phosphate (GA3P) molecules. These store the energy captured from photosynthesis. The cycle uses this energy to regenerate the RuBP. D The Calvin cycle harnesses energy in the form of 6 ATP and 3 NADPH that are used to produce glyceraldehyde-3-phosphate (GA3P) molecules. These store energy captured from photosynthesis. The cycle uses this energy to regenerate RuBP. Solution 40 The solution is (B). The Calvin cycle harnesses energy in the form of the 6 ATP and 6 NADPH that are used to produce glyceraldehyse-3-phosphate (GA3P) molecules. These store the energy captured from photosynthesis. The cycle uses this energy to regenerate the RuBP that starts the cycle. Based on the diagram, what would most likely cause a plant to run out of NADP? A Missing the ATP synthase enzyme B Exposure to light C Lack of water preventing H+ and NADP+ from forming NADPH D Not enough CO2 Solution 41 The solution is (D). As you can see in the diagram, CO2 is the only listed option that is directly involved in the Calvin cycle, which produces NADP+. As temperatures increase, gases such as CO2 diffuse faster. As a result, plant leaves will lose CO2 at a faster rate than normal. If the amount of light impacting on the leaf and the amount of water available is adequate, predict how this loss of gas will affect photosynthesis in the leaf. A Loss of gases, mainly CO2 , will not affect photosynthesis in the leaf, as adequate amounts of water and light are still present which will let the Calvin cycle run smoothly. Advanced Placement Biology Instructor’s Solution Manual 8 | Photosynthesis 177 B Loss of gases, mainly CO2 , will affect photosynthesis in the leaf, as the Calvin cycle will become faster to compensate for the loss. C Loss of gases, mainly CO2 , will not affect photosynthesis in the leaf, as stored reservoirs of CO2 in the leaf can be utilized in such times. D Loss of gases, mainly CO2 , will affect photosynthesis in the leaf, as the Calvin cycle will slow down and possibly stop because of inadequate carbon to fix in the system. Solution 42 The solution is (D). Loss of gases, especially of CO2 , will affect photosynthesis in the leaf as the Calvin cycle will slow down and eventually stop because of inadequate carbon to fix in the system. A continuous supply of atmospheric CO2 is thus required. How do the cytochrome-complex components involved in photosynthesis contribute to the electron transport chain? A Photosystem I excites the electron as it moves down the electron transport chain into photosystem II. B Plastoquinone and plastocyanine perform redox reactions that allow the electron to move down the electron transport chain into photosystem I. C ATP synthase de-excites the electron as it moves down the electron transport chain into photosystem I. D RuBisCO excites the electron as it moves down the electron transport chain into photosystem II. Solution The solution is (B). Plastoquinone and plastocyanine perform redox reactions that allow the electron to move down the electron transport chain into photosystem I. The cytochrome complexes perform oxidation-reduction reactions that, in essence, pass on the electron to the next stop on the ETC. 43 How do membranes in chloroplasts contribute to the organelles’ essential functions? A The inner membrane contains the chemicals needed for the Calvin cycle and components of the light-dependent reactions. The thylakoid membrane contains photosystems I and II, as well as the enzyme NAD+ reductase. B The inner membrane contains only the chemicals needed for the Calvin cycle. The thylakoid membrane contains components of the light-dependent reactions, photosystems I and II, and the enzyme NAD+ reductase. C The inner membrane contains components of the light-dependent reactions as well as photosystems I and II. The thylakoid membrane contains the chemicals needed for the Calvin cycle and the enzyme NAD+ reductase. D The inner membrane contains the chemicals needed for the Calvin cycle, components of the light-dependent reactions, and photosystems I and II. The thylakoid membrane contains the enzyme NAD+ reductase. Advanced Placement Biology Instructor’s Solution Manual 178 8 | Photosynthesis Solution 44 The solution is (A). The inner membrane contains the chemicals needed for the Calvin cycle and components of the light-dependent reactions. The thylakoid membrane contains photosystems I and II, as well as the enzyme NAD + reductase. The inner membrane contains the chemicals needed for the Calvin cycle and provides a home for the components of the light-dependent reactions. Photosystem I and II are housed within the thylakoid membrane, as are the enzymes NAD + reductase, which catalyze the reduction of NADP and ATP synthase, which plays an important role in ATP production through the chemiosmosis process across the membrane. If the absorption spectrum of photosynthetic pigments was restricted to the green portion of the spectrum, which pigment or pigments would be affected the least? A Carotenoids B Chlorophyll a C Chlorophyll b D Chlorophyll c Solution 45 The solution is (A). Carotenoids would not be affected if the absorption spectrum of photosynthetic pigments were restricted to the green portion of the spectrum, because carotenoids absorb violet-blue-green light. How can the passage of energy from light until it is captured in the primary electron acceptor be described? A Chlorophyll molecules in the photosystems are excited and pass the energy to the primary electron acceptor where the energy is used to excite electrons from the splitting of water. B Chlorophyll a molecules in the photosystems are excited and pass the energy to the primary electron acceptor where the energy is used to excite electrons from the splitting of water. C Chlorophyll b molecules in the photosystems are excited and pass the energy to the primary electron acceptor where the energy is used to excite electrons from the splitting of water. D Chlorophyll molecules in the photosystems absorb light and get excited in the primary electron acceptor from where the energy is used to excite electrons from the splitting of water. Solution The solution is (A). The chlorophyll molecules in the photosystem are excited and pass the energy to the primary electron acceptor where the energy is used to excite electrons from the split of water. Excitation occurs in chlorophyll a molecules and then energy is passed to the primary electron acceptor where the energy is used for exciting electrons from the splitting of water. Advanced Placement Biology Instructor’s Solution Manual 8 | Photosynthesis 179 SCIENCE PRACTICE CHALLENGE QUESTIONS 8.2 The Light-Dependent Reaction of Photosynthesis 46 On a hot, dry day, plants close their stomata to conserve water. Explain the connection between the oxidation of water in photosystem II of the light-dependent reactions and the synthesis of glyceraldehyde-3-phosphate (G3PA) in the light-independent reactions. Predict the effect of closed stomata on the synthesis of G3PA, and justify the prediction. Solution 47 The emergence of photosynthetic organisms is recorded in layers of sedimentary rock known as a banded iron formation. Dark-colored and iron-rich bands composed of hematite (Fe2O3) and magnetite (Fe3O4) only a few millimeters thick alternate with lightcolored and iron-poor shale or chert. Hematite and magnetite can form precipitates from water that have high concentrations of dissolved oxygen. Shale and chert can form under conditions that have high concentrations of carbonates (CO3−2). These banded iron formations appeared 3.7 billion years ago (and became less common 1.8 billion years ago). Justify the claim that these sedimentary rock formations reveal early Earth conditions. Solution 48 Sample answer: Feedback is used to maintain water within cells by closing stomata. When water cannot be lost by vaporization, the redox couple between water and NADP+ is broken: Electrons cannot be delivered to reduce NADP+. The lightindependent reactions then must stop since they are regulated by the supply of NADPH. The product of the light-independent reactions is G3PA. Sample answer: These systems indicate formation from waters that fluctuated over time in the dominant dissolved oxygen-containing chemical species. Based on these, it may be reasoned that the atmosphere was oscillating between carbon dioxide rich and oxygen rich. Because oxygen is a product of photosynthesis that consumes carbon dioxide, the formations can reveal the emergence on early Earth of photosynthetic organisms. When a high-oxygen atmosphere eventually became stable, the banded iron formation ceased. The diagram summarizes the light reactions of photosynthesis. Advanced Placement Biology Instructor’s Solution Manual 180 8 | Photosynthesis The diagram shows light-dependent reactions of photosynthesis, including the reaction centers, electron transport chains, and the overall reactions within each of these. The free energy per electron is shown for the oxidation-reduction reactions. The free change of the captured radiant energy is shown. 2NADP 2H 2H2O 3ADP 3Pi O2 4H 2NADPH 3ATP A. In the overall mass balance equation for the light reactions shown above, identify the source of electrons for the synthesis of NADPH. B. Calculate the number of electrons transferred in this reaction. C. Using the free energies per electron displayed, calculate the free-energy change of the light-dependent reactions. D. Given that the free-energy change for the hydrolysis of ATP is 31.5 kJ/mole and the free-energy change for the formation of NADPH from NADP+ is 18 kJ/mole, calculate the total production of free energy for the light reactions. E. Using this definition of energy efficiency, calculate the efficiency of the light reaction of photosynthesis: energy efficiency = free energy produced/energy input. Solution Sample answer: A. The source of the electrons in the light-dependent reactions of photosynthesis is the oxygen atom in water. Advanced Placement Biology Instructor’s Solution Manual 8 | Photosynthesis 181 B. Each oxygen atom loses two electrons, so four electrons are transferred in this reaction. C. The free energy per electron of the conversion of NADP+ to NADPH is identified in the chart as 29 kJ/electron. The free energy per electron of the conversion of water to molecular oxygen is 77 kJ/electron. The change in free energy is 77 – 29 106 kJ/electron. There are four electrons transferred, so the overall free energy change is 420 kJ. D. There are three ATP produced from three ADP with a free energy change of 94.5 kJ. There are two NADPH produced from two NADP+ for a total of 36 kJ. Then the free energy captured in chemical bonds of ATP and NADPH is 131 kJ. E. The total free-energy change of 420 kJ 131 kJ 550 kJ . You often see an estimate of 20 percent efficiency. The radiant-energy input according to the diagram is 2,800 kJ and 550 / 2800 0.20 . The true photosynthetic efficiency is more difficult to determine; the number of photons captured is only approximately eight. The free energies that are available from tables are for pH of 7 and standard temperature and pressure. 49 Algae can be used for food and fuel. To maximize profit from algae production under artificial light, researchers proposed an experiment to determine the dependence of the efficiency of the process used to grow the algae on light intensity (“brightness”) that will be purchased from the electric company. The algae will be grown on a flat sheet that will be continuously washed with dissolved carbon dioxide and nutrients. Light-emitting diodes (LEDs) will be used to illuminate the growth sheet. Photodiodes placed above and below the sheet will be used to detect light transmitted through and reflected from the algal mat. The intensity of light can be varied, and the algae can be removed, filtered, and dried. The amount of stored energy in the algal mats can be determined by calorimetry. Advanced Placement Biology Instructor’s Solution Manual 182 8 | Photosynthesis A. Identify a useful definition of efficiency for this study and justify your choice. B. Frequencies of light emitted by the LEDs will not be variables but must be specified for the construction of the apparatus. Identify the frequencies of light that should be used in the experiment, and justify your choice. C. Evaluate the claim that the experiment is based on the assumption that there is an upper limit on the intensity of light used to support growth of algae. Predict a possible effect on algal growth if light with too great an intensity is used, and justify the prediction. D. Design an experiment by describing a procedure that can be used to determine the relationship between light intensity and efficiency. Solution Sample answer: A. The product is biomass, and the component of the cost that is considered here is the price paid for the brightness of light used to grow the algae. So, efficiency could be defined as the value of the biomass divided by the cost of electrical energy. It could be defined as the energy contained in the algae divided by the energy used to produce light. It could be defined as the profit from the sale of the biomass (or its derivatives) divided by the cost of the power. B. LEDs with a distribution of wavelengths matched to the absorbance maxima for chlorophyll a should be chosen. C. Light with too great an intensity can damage tissue, denature enzymes, etc. Light produces heat, and thermoregulation is required for homeostasis. D. Seed the mat with a fixed volume at fixed concentration of algae. At fixed frequency, irradiate algae for a fixed period of time. Remove the algal mat and measure biomass. Repeat varying light intensity. Then repeat this sequence at fixed light intensity, varying pulse frequency. 50 The classical theory of evolution is based on a gradual transformation, the accumulation of many random mutations that are selected. The biological evidence for evolution is overwhelming, particularly when one considers what has not changed: core conserved characteristics. A. Describe three conserved characteristics common to both chloroplasts and mitochondria. Some hypotheses that have been proposed to account for biological diversity are saltatory, involving sudden changes, rather than gradualist. In defense of the classical gradualist theory of evolution, nearly all biologists in the late 1960s rejected the theory of endosymbiosis as presented by Lynn Margulis in 1967. Advanced Placement Biology Instructor’s Solution Manual 8 | Photosynthesis 183 B. Suppose that you want to disprove the theory of endosymbiosis. Explain how the following evidence could disprove the theory: i. A “transitional species” with cellular features that are intermediate cells with and without mitochondria ii. A “transitional organelle” with some features, such as compartmentalized metabolic processes, but not other features, such as DNA Explain how the following evidence supports the theory of endosymbiosis: iii. Bacteria live within your intestines, but you still have a separate identity. iv. No one has directly observed the fusion of two organisms in which a single organism results. Solution Sample answer: A. The following could be identified: DNA and DNA processing Inner/outer membrane ATP- and NADH-based energetics Chemiosmosis including ETS and multiple analogous proteins B. i. A transitional species would have intermediate structural and functional features. The existence of other gradual variations of this transitional species would mean that the simplest solution would be gradualist. ii. Similarly, a transitional organelle—for example, something like a glycosome or a metabolon—that, with a molecular phylogeny such as protein homologues, would disprove endosymbiosis with the accumulation of sufficient evidence. iii. Endosymbiosis is not the same as symbiosis. In this context, the organism has systems with integrated reproduction and growth in which critical, obligatory functions are provided by one or the other. iv. Evidence of evolution is often obtained by observation of the necessary consequences of an event without observation of the event. So, absence of direct evidence is not evidence of absence. 8.3 Using Light Energy to Make Organic Molecules 51 Discovering the carbon-fixation reactions (or light-independent reactions) of photosynthesis earned Melvin Calvin a Nobel Prize in 1961. The isolation and identification of the products of algae exposed to 14C revealed the path of carbon in photosynthesis. 14C was fed to the algal culture in the form of a bicarbonate ion (HCO3−). Advanced Placement Biology Instructor’s Solution Manual 184 8 | Photosynthesis To agitate the culture, air, which contains CO2, was bubbled through the system, so there were two sources of carbon. Since Calvin’s experiment, research has focused on the way carbon from a solution containing bicarbonate ions is absorbed by algae. In aqueous solutions, the bicarbonate anion (HCO3−) is in equilibrium with dissolved CO2, as shown in the equation. H HCO3 H2O CO2 In a later experiment, Larsson and Axelsson (1999) used acetazolamide (AZ), a carbonate anhydrase inhibitor, to inhibit enzymes that convert bicarbonate into carbon dioxide. They also used disulfonate (DIDS), an inhibitor of the transport of anions, such as the bicarbonate ion, through the plasma membrane. A. Pose a scientific question that can be pursued with AZ and DIDS in terms of the path of carbon in photosynthesis. B. The plasma membrane is permeable to the nonpolar, uncharged carbon dioxide molecule. However, the concentration of carbon dioxide in solution can be very small. Explain how the enzyme carbonate anhydrase can increase the availability of carbon dioxide to the cell. C. Larsson Sand Axelsson conducted experiments in which the growth medium was fixed at two different pH levels and determined the effects of AZ and DIDS on the rate of photosynthesis by measuring oxygen concentrations at various times. The results are shown in the side-by-side graphs. The arrows indicate the time points during which HCO3−, AZ, and DIDS were added to each system. In which system, A or B, is there a strong reliance on the bicarbonate ion as the mechanism of carbon uptake by the cell? Justify your answer using the data. D. If both systems are dosed with the same concentrations of bicarbonate ion, in which system, A or B, is the pH higher? Justify your answer using the data and the bicarbonatecarbon dioxide equilibrium equation. Advanced Placement Biology Instructor’s Solution Manual 8 | Photosynthesis 185 Solution Sample answer: A. If AZ and DIDS are introduced into the light-independent reactions of photosynthesis in the middle of the reaction, how would the introduction of these inhibitors impact the amount of carbon produced? B. The enzyme resides on the exterior surface of the cell and releases carbon dioxide that is nonpolar and penetrates the membrane. The charged bicarbonate ion does not passively enter the cell. C. The data show that as bicarbonate is added, the equilibrium concentration of carbon dioxide increases and photosynthesis (detected by the production of oxygen) increases. However, some of that increase is due to the action of the anhydrase enzyme and the relative amounts of each pathway (entry as CO2 and entry as HCO3−), which are indicated by the slopes of the curve before and after the addition of AZ. In system A, the addition of AZ decreased the rate by a factor of almost 10; thus, under the conditions for system A, most C entered the cell as CO 2—produced by the surface-bound enzyme. After the addition of AZ, system A gets CO2 from that dissolved in solution. DIDS has no effect, so in system A there is almost no entry of bicarbonate into the cell. In system B, carbonate is the predominate source of carbon. After the addition of DIDS, photosynthesis shuts down. D. The results described in (C) tell us that system A is at a higher pH where, according to the equilibrium equation, the concentration of dissolved carbon dioxides is high. System B is at a higher pH where the concentration of dissolved carbon dioxide is small. Advanced Placement Biology Instructor’s Solution Manual 186 9 | Cell Communication 9 | CELL COMMUNICATION REVIEW QUESTIONS 1 Which property prevents the ligands of cell-surface receptors from entering the cell? A The molecules bind to the extracellular domain. B The molecules are hydrophilic and cannot penetrate the hydrophobic interior of the plasma membrane. C The molecules are attached to transport proteins that deliver them through the bloodstream to target cells. D The ligands are able to penetrate the membrane, directly influencing gene expression upon receptor binding. Solution 2 The solution is (B). The molecules are hydrophilic and cannot penetrate the hydrophobic interior of the plasma membrane. Hydrophobic ligands can easily cross the lipid bilayer, bringing about changes in gene expression. Hydrophilic ligand molecules cannot cross the plasma membrane and bind the cell-surface receptor. The secretion of hormones by the pituitary gland is an example of which type of signaling? A Autocrine signaling B Direct signaling across gap junctions C Endocrine signaling D Paracrine signaling Solution 3 The solution is (C). The secretion of hormones by the pituitary gland is an example of endocrine signaling helping in long-distance signaling. Why are ion channels necessary to transport ions into or out of a cell? A Ions are too large to diffuse through the membrane. B Ions are charged particles and cannot diffuse through the hydrophobic interior of the membrane. C Ions bind to hydrophobic molecules within the ion channels. D Ions bind to carrier proteins in the bloodstream, which must be removed before transport into the cell. Solution The solution is (B). Ions are charged particles and cannot diffuse through the hydrophobic interior of the membrane. Ions are charged particles and therefore cannot diffuse through the membrane and require ion channels for their transport. Advanced Placement Biology Instructor’s Solution Manual 9 | Cell Communication 4 187 Why are endocrine signals transmitted more slowly than paracrine signals? A The ligands are transported through the bloodstream and travel greater distances. B The target and signaling cells are close together. C The ligands are degraded rapidly. D The ligands do not bind to carrier proteins during transport. Solution 5 The solution is (A). The ligands are transported through the bloodstream and travel greater distances. In endocrine signaling, the ligands are transported at long distances. Aldosterone is a steroid hormone that regulates reabsorption of sodium ions in the kidney tubular cells. What is the probable mechanism of action of aldosterone? A It binds gated ion channels and causes a flow of ions in the cell. B It binds cell surface receptors and activates synthesis of cAMP. C It binds to cell surface receptors and activates a phosphorylation cascade. D It binds to an intracellular receptor and activates gene transcription. Solution 6 The solution is (D). It binds an intracellular receptor and activates gene transcription. Aldosterone is a steroid hormone that binds to intracellular receptors and promotes gene expression. The gas nitric oxide has been identified as a signaling molecule. Which mechanism of action would you expect from a gaseous molecule? A It binds to a G-protein-linked receptor. B It binds to a receptor tyrosine kinase. C It binds to a gated ion channel. D It binds to an intracellular receptor. Solution 7 The solution is (D). It binds to an intracellular receptor. Nitric oxide is a gas that can easily diffuse through the cell membrane. It binds to the intracellular receptors and brings about effector functions. How do DAG and IP3 originate? A They are formed by phosphorylation of cAMP. B They are ligands expressed by signaling cells. C They are hormones that diffuse through the plasma membrane to stimulate protein production. D They are the cleavage products of the inositol phospholipid, PIP2. Advanced Placement Biology Instructor’s Solution Manual 188 9 | Cell Communication Solution 8 The solution is (D). They are the cleavage products of the inositol phospholipid, PIP2. The enzyme phospholipase C cleaves PIP2 to form diacylglycerol (DAG) and inositol triphosphate (IP3). What property enables the residues of the amino acids serine, threonine, and tyrosine to be phosphorylated? A They are polar. B They are nonpolar. C They contain a hydroxyl group. D They occur more frequently in the amino acid sequence of signaling proteins. Solution 9 The solution is (C). They contain a hydroxyl group. Serine, threonine, and tyrosine are hydroxyl-group-containing amino acids, which enables them to be easily phosphorylated. Dopamine is a neurotransmitter in the brain that causes long-term responses in neurons and binds to a G-protein-linked receptor. Which chemical would you expect to increase in concentration after dopamine binds its receptor? A ATP B cAMP C Calcium ions D Sodium ions Solution 10 The solution is (B). When dopamine binds to the G-protein-coupled receptor, ATP is hydrolyzed by adenylyl cyclase to produce cAMP, which acts as a second messenger. The hormone insulin binds to a receptor tyrosine kinase on the surface of target cells. Which step takes place before phosphorylation of tyrosine residues? A A tyrosine kinase enzyme must be activated. B GDP is exchanged for GTP. C The receptor forms a dimer. D The insulin molecule is internalized in the cytoplasm. Solution 11 The solution is (C). The receptor forms a dimer in response to ligand binding, which activates the tyrosine kinase activity of the receptor, leading to autophosphorylation of the tyrosine residues on the receptor. What is the function of a phosphatase? A A phosphatase removes phosphorylated amino acids from proteins. B A phosphatase removes the phosphate group from phosphorylated amino acid residues in a protein. Advanced Placement Biology Instructor’s Solution Manual 9 | Cell Communication 189 C A phosphatase phosphorylates serine, threonine, and tyrosine residues. D A phosphatase degrades second messengers in the cell. Solution 12 The solution is (B). A phosphatase removes the phosphate group from phosphorylated amino acid residues in a protein. The phosphate group from a phosphorylated amino acid is removed by a phosphatase that helps in altering the activity of a protein. How does NF-κB induce gene expression? A A small, hydrophobic ligand binds to NF-κB, activating it. B NF-κB is phosphorylated and is then free to enter the nucleus to bind DNA. C NF-κB is a kinase that phosphorylates a transcription factor that binds DNA and promotes protein production. D Phosphorylation of the inhibitor IκB dissociates the complex between it and NF-κB, allowing NF-κB to enter the nucleus and stimulate transcription. Solution 13 The solution is (D). Phosphorylation of the inhibitor Iκ-B dissociates the complex between it and NF-κB, allowing NF-κB to enter the nucleus and stimulate transcription. When Iκ-B is bound to NF-κB, this complex cannot enter the nucleus. When Iκ-B is phosphorylated by PKC, it dissociates from the complex, releasing NF-κB. NF-κB acts as a transcription factor that enters the nucleus and initiates transcription. Apoptosis can occur in a cell under what conditions? A When a cell is infected by a virus B When a cell is damaged C When a cell is no longer needed D All of the above Solution 14 The solution is (D). Apoptosis occurs when a cell is no longer needed, is damaged, or is infected by a virus. Cancer cells that continue to divide when defective often show changes in what cellular function? A Apoptosis B Their mechanism of glycolysis C The mechanism of protein biosynthesis D Replication of DNA Solution The solution is (A). Cancer cells show defects in apoptosis. Although they are defective, they do not undergo cell suicide. Advanced Placement Biology Instructor’s Solution Manual 190 15 9 | Cell Communication Epinephrine mediates the fight-or-flight response of the body. One of the effects is to increase the amount of glucose available to muscles. What does the signaling pathway triggered by epinephrine cause to occur in liver cells? A Activation of metabolism B Cell division C Inhibition of glucose metabolism by liver cells D Synthesis of enzymes Solution 16 The solution is (A). Binding of epinephrine leads to increase in metabolic rate. Epinephrine signaling results in breakdown of glycogen and an increase in concentration of glucose. Which type of molecule acts as a signaling molecule in yeasts? A Autoinducer B Mating factor C Second messenger D Steroid Solution 17 The solution is (B). When a haploid yeast cell is ready to mate, it releases a signaling molecule called mating factor, which binds to the receptors present on the nearby yeast cells. Binding of mating factor arrests the normal cell cycle of the nearby cell and the cell enters the signaling cascade that includes protein kinases and GTPbinding proteins that are similar to G-proteins. When is quorum sensing triggered to begin? A A sufficient density of bacteria is present. B Bacteria release growth hormones. C Bacterial protein expression is switched on. D Treatment with antibiotics occurs. Solution 18 The solution is (A). A sufficient density of bacteria is present. When high cell density is present, more autoinducers are present, which bind to receptors, leading to enhanced gene expression and more production of autoinducers. Yeast-releasing mating factor can be classified as which type of signal? A Autocrine B Endocrine C Paracrine D Gap junction Advanced Placement Biology Instructor’s Solution Manual 9 | Cell Communication Solution 19 191 The solution is (C). Paracrine signaling helps communicate with nearby cells. Yeast cells that are present in close proximity communicate through paracrine signaling. The bioluminescent bacteria Vibrio fischeri produce luminescence only if the population reaches a certain density. What is the advantage of an autoinducer? A An autoinducer allows the producer to act independently of the presence of other cells. B An autoinducer does not diffuse away from the cell. C An autoinducer allows a positive feedback loop, which increases the response in proportion to the population size. D An autoinducer presents no advantage for the cell. Solution The solution is (C). An autoinducer allows a positive feedback loop, which increases the response in proportion to the population size. Production of an autoinducer allows a positive feedback loop for production of more autoinducer, which is directly proportional to cell density. CRITICAL THINKING QUESTIONS 20 What is the difference between intracellular signaling and intercellular signaling? A Intracellular signaling occurs between cells of two different species. Intercellular signaling occurs between two cells of the same species. B Intracellular signaling occurs between two cells of same species. Intercellular signaling occurs between cells of two different species. C Intracellular signaling occurs within a cell. Intercellular signaling occurs between cells. D Intracellular signaling occurs between cells. Intercellular signaling occurs within cells. Solution The solution is (C). Intracellular signaling occurs within a cell. Intercellular signaling occurs between the cells. Signaling between two cells is called intercellular signaling. Communication that occurs within a cell is called intracellular signaling. Advanced Placement Biology Instructor’s Solution Manual 192 21 9 | Cell Communication What are the differences between internal receptors and cell-surface receptors? A Internal receptors bind to ligands that are hydrophobic, and the ligand-receptor complex directly enters the nucleus, initiating transcription and translation. Cellsurface receptors bind to hydrophilic ligands and initiate a signaling cascade that indirectly influences the making of a functional protein. B Internal receptors bind to ligands that are hydrophilic, and ligand-receptor complex directly enters the nucleus, initiating transcription and translation. Cell-surface receptors bind to hydrophobic ligands and initiate a signaling cascade that indirectly influences the making of a functional protein. C Internal receptors bind to ligands that are hydrophobic and initiate the signaling cascade that indirectly influences the making of a functional protein. Cell-surface receptors bind to hydrophilic ligands, and a ligand-receptor complex directly enters the nucleus, initiating transcription and translation. D Internal receptors are integral membrane proteins that bind to hydrophobic ligands, initiating a signaling cascade, which indirectly influences the making of a functional protein. Cell-surface receptors bind to hydrophilic ligands, and the ligand-receptor complex directly enters the nucleus, initiating transcription and translation. Solution 22 The solution is (A). Their ligands enter the cell to bind the receptor. The complex formed by the internal receptor and the ligand then enters the nucleus, directly affecting protein production by binding to the chromosomal DNA and initiating the transcription of mRNA that codes for proteins. Cell-surface receptors, however, are embedded in the plasma membrane; their ligands do not enter the cell. Binding of the ligand to the cell-surface receptor initiates a cell-signaling cascade and does not directly influence the making of proteins, though it may involve the activation of intracellular proteins. Internal receptors bind to hydrophobic molecules. Receptorligand complexes directly influence the synthesis of proteins by binding to DNA in the nucleus. Cell-surface receptors bind to hydrophilic ligands, which cannot cross the plasma membrane, and lead to a signaling cascade, indirectly influencing the synthesis of functional proteins. Cells grown in the laboratory are mixed with a dye molecule that is unable to pass through the plasma membrane. If a ligand is added to the cells, the dye is observed entering the cells. What type of receptor did the ligand bind to on the cell surface? A G-protein-linked R receptor B Ligand-gated ion channel C Voltage-gated ion channel D Receptor tyrosine kinase Solution The solution is (B). An ion channel-linked receptor opened a pore in the membrane, which allowed the ionic dye to move into the cell. Binding of the ligand to the ligandAdvanced Placement Biology Instructor’s Solution Manual 9 | Cell Communication 193 gated ion channel receptor will lead to conformational changes in the receptor, leading to formation of a pore in the membrane that allows the dye to move in along with the ligand. 23 The same second messengers are used in many different cells, but the response to second messengers is different in each cell. How is this possible? A Different cells produce the same receptor, which bind to the same ligands, but have different responses in each cell type. B Cells produce variants of a particular receptor for a particular ligand through alternative splicing, resulting in different responses in each cell. C Cells contain different genes, which produce different receptors that bind to same ligand, activating different responses in each cell. D Cells produce different receptors that bind to the same ligand or the same receptor that binds to the same ligand with different signaling components, activating different responses in each cell. Solution 24 The solution is (D). Different cells produce different proteins, including cell-surface receptors and signaling pathway components. Therefore, they respond to the same ligands differently because the second messengers activate different pathways. Signal integration can also change the end result of signaling. Different receptors in different cells bind to same ligand and produce different responses. What would happen if the intracellular domain of a cell-surface receptor was switched with the domain from another receptor? A It would activate the pathway normally triggered by the receptor that contributed the intracellular domain. B It would activate the same pathway even after the intracellular domain is changed with the domain from another receptor. C The receptor will be mutated and become nonfunctional, not activating any pathway. D The receptor will become mutated and lead to continuous cell signaling, even in the absence of a ligand. Solution 25 The solution is (A). The binding of the ligand to the extracellular domain would activate the pathway normally activated by the receptor donating the intracellular domain. The intracellular domain determines the response. How would a chemical that blocks the binding of EGF to the EGFR interfere with the replication of cancerous cells that overexpress EGFR? A It will activate the EGFR pathway. B It will block the EGFR pathway. C It will have no effect, and the EGFR pathway will continue normally. D It will lead to overexpression of the EGFR pathway. Advanced Placement Biology Instructor’s Solution Manual 194 9 | Cell Communication Solution 26 The solution is (B). The chemical would interfere with the activation of the EGFR pathway. A chemical that blocks the binding EGF to the EGFR would block the EGFR pathway. How does the extracellular matrix control the growth of cells? A Contact of receptors with the extracellular matrix maintains equilibrium of the cell and provides optimal pH for the growth of the cells. B Contact of the receptor with the extracellular matrix helps maintain concentration gradients across membranes, resulting in the flow of ions. C The extracellular matrix provides nutrients for the cell. D The extracellular matrix connects the cell to the external environment and ensures correct positioning of the cell to prevent metastasis. Solution 27 The solution is (D). Receptors on the cell surface must be in contact with the extracellular matrix in order to receive positive signals that allow the cell to live. If the receptors are not activated by binding, the cell will undergo apoptosis. This ensures that cells are in the correct place in the body and helps prevent invasive cell growth, such as occurs in metastasis of cancer cells. Which options gives an example for each one of the following effects of a cell signal: on protein expression, cellular metabolism, and cell division? A Protein expression: binding of epinephrine (adrenaline) to a G-protein-linked receptor; cellular metabolism: the MAP-kinase cascade; cell division: promoted by the binding of the EGF to its receptor tyrosine kinase B Protein expression: the MAP-kinase cascade; cellular metabolism: binding of epinephrine (adrenaline) to a G-protein-linked receptor; cell division: promoted by the binding of the EGF to its receptor tyrosine kinase C Protein expression: binding of the EGF to its receptor tyrosine kinase; cellular metabolism: the MAP-kinase cascade; cell division: FAS-RAS signaling D Protein expression: RAS signaling; cellular metabolism: binding of the EGF to its receptor tyrosine kinase promotes an increase; cell division: binding of epinephrine (adrenaline) to a G-protein-linked receptor Solution The solution is (B). Protein expression: The MAP-kinase cascade is initiated by the binding of a signal to a receptor tyrosine kinase. In the fight-or-flight response caused by the binding of epinephrine (adrenaline) to a G-protein-linked receptor, whose signaling pathway activates the enzymes involved in gluconeogenesis, glycogen breaks down. Binding of the epidermal growth factor to its receptor tyrosine kinase will start a signaling cascade such as the RAS-RAF-MEK-ERK pathway, which leads to cell proliferation. The MAPK pathway leads to protein expression. Epinephrine results in increased metabolic rate in cases of fight-or-flight. Epidermal growth factor promotes cell division and proliferation. Advanced Placement Biology Instructor’s Solution Manual 9 | Cell Communication 28 195 The mitogen-activated protein (MAP) kinase cascade triggered by RTKs results in cell division. What are a few possible scenarios of abnormalities in the MAPK pathway leading to uncontrolled cell proliferation? A Gain of function mutation in RAS protein, mutation in Iκ-B, loss of function mutation in genes for MAPK kinase pathway, regulated phosphorylation cascade B Loss of function mutation in RAS protein and gain of function mutation in RAF protein, Iκ -B permanently bound to NF-κB, regulated phosphorylation cascade C RAS protein unable to hydrolyze its bound GTP, loss of function mutation in Iκ-B, gain of function mutation in genes for MAPK kinase pathway, unregulated phosphorylation cascade D Unregulated phosphorylation cascade, loss of function mutation in RAS and RAF protein, mutation in genes for MAPK kinase pathway, regulated phosphorylation cascade Solution 29 The solution is (C). If the RAS protein is locked in the on position because it cannot hydrolyze GTP, proliferation is not stopped. The Iκ-B molecule cannot bind to NF-κB and prevent its activity. Overexpression of the genes for the MAPK kinase pathway components can lead to overgrowth. Every step of the phosphorylation cascade must be tightly regulated to avoid uncontrolled cell proliferation. If the RAS protein is unable to hydrolyze its bound GTP, it will remain permanently active, loss of function mutation in Iκ-B will not allow Iκ-B to bind to NF-κB, rendering it permanently active, gain of function mutation in genes for MAPK kinase pathway will lead to overexpression of the pathway, unregulated phosphorylation cascade. All these situations will lead to uncontrolled cell division. What characteristics make yeast a good model for learning about signaling in humans? A Yeasts are prokaryotes. They have a short life cycle, are easy to grow, and share similarities with humans in certain regulating mechanisms. B Yeasts are eukaryotes. They have a short life cycle, are easy to grow, and share similarities with humans in certain regulating mechanisms. C Yeasts are single-celled organisms. They have a short life cycle, are easy to grow, and share similarities with humans in certain regulating mechanisms. D Yeasts are single-celled organisms. They have a complex life cycle like that of humans and share similarities in regulating mechanisms. Solution The solution is (C). Yeasts are eukaryotes and have many of the same systems that humans do. However, they are single celled, so they are easy to grow, have a short generation time, and are much simpler than humans. Advanced Placement Biology Instructor’s Solution Manual 196 30 9 | Cell Communication Why is signaling in multicellular organisms more complicated than signaling in singlecelled organisms? A Multicellular organisms coordinate between distantly located cells; single-celled organisms communicate only with nearby cells. B Multicellular organisms involve receptors for signaling; single-celled organisms communicate by fusion of plasma membrane with the nearby cells. C Multicellular organisms require more time for signal transduction than single-celled organisms because they show compartmentalization. D Multicellular organisms require more time for signal transduction than single-celled organisms because they lack compartmentalization. Solution 31 Biofilms are a prominent danger in infectious disease treatment today because it is difficult to find drugs that can penetrate the biofilm. What characteristics would a drug have if it aimed to prevent bacteria from forming biofilms in the first place? Explain your answer. Solution 32 The solution is (A). Multicellular organisms must coordinate many different events in different cell types, which may be very distant from each other. Single-celled organisms are only concerned with their immediate environment and the presence of other cells in the area. Multicellular organisms take more time in signal transduction than single-celled organisms because they coordinate between cells that are distantly located. A drug aimed at preventing biofilm formation would block the receptors of autoinhibitors. By blocking auto-inhibitors, quorum sensing would not occur in bacterial populations, and these colonies would not turn on genes that form biofilms. These drugs may cause bacterial populations to increase because the bacteria would continue to divide even after they reach “quorum”. Therefore, antibacterial drugs that inhibit bacterial reproduction would have to be used in addition to an autoinhibitor blocker. Supports the hypothesis that signaling pathways appeared early in evolution and are well conserved using the yeast mating factor as an example. A Signaling in yeast uses the RTK pathway and is evolutionarily conserved, like epinephrine signaling in humans. B Signaling in yeast uses G-protein-coupled receptors for signaling and is evolutionarily conserved, like epinephrine signaling in humans. C Signaling in yeast uses an endocrine pathway and is evolutionarily conserved, like epinephrine signaling in humans. D Mating factor in yeast uses an autocrine signaling pathway and is evolutionarily conserved, like epinephrine signaling in humans. Advanced Placement Biology Instructor’s Solution Manual 9 | Cell Communication Solution 197 The solution is (B). Signaling pathways have been identified in many bacteria, protists, and fungi. The mating factor signaling pathway in yeast uses a G-proteinlinked receptor, a G-protein that activates a MAPK kinase cascade. Yeast produces mating factor, which binds to the receptors present on the surrounding yeast cells, and initiates a cell-signaling cascade including protein kinases and GTP-binding proteins, which are similar to G-proteins. G-proteins are evolutionarily conserved since they are used in various signaling mechanism in humans. TEST PREP FOR AP® COURSES 33 Upon ingestion of bacteria, white blood cells release a chemical messenger into the blood stream that causes the synthesis of inflammation response proteins by liver cells. What is this is an example of? Advanced Placement Biology Instructor’s Solution Manual 198 9 | Cell Communication A Autocrine signaling B Endocrine signaling C Paracrine signaling D Synaptic signaling Solution 34 The solution is (B). The chemical messenger is a hormone secreted by the endocrine system. Molecules do not flow between the endothelial cells in the brain capillaries. The membranes of the cells must be joined by what? A Gap junctions B Ligand-gated channels C Synapses D Tight junctions Solution 35 The solution is (D). Tight junctions prevent the flow of molecules between adjacent cells. What are the possible benefits of having autocrine signaling? A Autocrine signaling helps to communicate with distantly located cells. B Autocrine signaling connects nearby cells. C Autocrine signaling helps to amplify the signal by inducing more signaling production from the cell itself. D Autocrine signaling is specific only for the cell that produced it. Solution 36 The solution is (C). Autocrine signals may induce a cell to produce more signal molecules by inducing proliferation and increasing the signal—for example, helper T. Autocrine signals coordinate responses by groups of identical cells. Cancer cells can use autocrine signals to escape the normal cell proliferation control and multiply abnormally. Autocrine signaling involves the action of the hormone on the same cell from which it was produced and helps in increasing the signal for a better response. If a chemical is an inhibitor of the enzyme adenylyl cyclase, which step in the G-protein signaling pathway would be blocked? A Activation of gene transcription B Exchange of GTP for GDP C Ligand-bound receptor activation of G-protein D Synthesis of cAMP Solution The solution is (D). Adenylyl cyclase hydrolyzes ATP to ADP to synthesize cAMP. Therefore, a chemical that inhibits the activity of adenylyl cyclase will affect the formation of cAMP. Advanced Placement Biology Instructor’s Solution Manual 9 | Cell Communication 37 199 Thyroid hormone is a lipid-soluble signal molecule that crosses the membrane of all cells. Why would a cell fail to respond to the thyroid hormone? A The MAPK cascade leading to cell activation is defective in the target cells. B The DNA sequence it binds to underwent a mutation. C There is no intracellular receptor for thyroid hormone in the cell. D The second messenger does not recognize the signal from the receptor. Solution The solution is (C). There is no intracellular receptor for thyroid hormone in the cell. The thyroid hormone triggers its signaling pathway only in the presence of its intracellular receptor; it would fail if this receptor were not found. 38 The poison from the krait snake’s bungarotoxin binds irreversibly to acetylcholine receptors, interfering with acetylcholine binding at the synapse. What is the effect of bungarotoxin binding on the post synaptic cell? A cAMP production is inhibited. B Bungarotoxin G-proteins are not activated. C Ion movement in the cell is inhibited. D Phosphorylation cascade is inhibited. Solution 39 The solution is (C). Ion movement in the cell is inhibited. Bungarotoxin binds irreversibly to acetylcholine receptors, interfering with acetylcholine binding at the synapse. Therefore, it will directly inhibit the passage of ions across the postsynaptic nerve and the signal is not transmitted. In autoimmune lymphoproliferative syndrome (ALPS), lymphocytes that multiplied during an infection persist in the body and damage tissue. The syndrome is caused by a mutation in the FAS gene, which encodes a cell-surface receptor. Which signaling pathway does the receptor initiate? A Activated metabolism B Apoptosis C Cell division D Cell differentiation Solution The solution is (B). The lymphocytes persist because they do not undergo apoptosis; therefore, it is the pathway affected. Advanced Placement Biology Instructor’s Solution Manual 200 40 9 | Cell Communication Place the following events in their sequential order: 1. Protein kinase A is activated. 2. Glycogen breaks down. 3. Epinephrine binds to G-protein-linked receptor. 4. G-protein activates adenylyl cyclase. 5. GTP is exchanged for GDP on the G-protein. 6. ATP is converted into cAMP. A 1, 3, 5, 4, 6, 2 B 3, 5, 4, 1, 6, 2 C 3, 4, 5, 1, 6, 2 D 3, 5, 4, 6, 1, 2 Solution 41 The solution is (D). Epinephrine binds to G-protein-linked receptor, resulting in the activation of G-protein by the exchange of GDP with GTP. Activated G-protein stimulates protein kinase A and leads to glycogen breakdown. The RAS protein is a G-protein connected with the response to RTKs that initiates the MAPK kinase cascade when GDP is released and GTP uploaded. Mutations in the RAS protein that interfere with its GTPase activity are common in cancer. What is the connection between the inability of RAS to hydrolyze GTP and uncontrolled cell proliferation? A RAS, when bound to GTP, becomes permanently inactive even in the presence of the ligand and no longer regulates cell division. B RAS, when bound to GTP, becomes permanently active even in the absence of the ligand and no longer regulates cell division. C RAS, when bound to GTP, forms a dimer after binding to the ligand and causes uncontrolled division, but it remains inactive when the ligand is absent. D RAS, when bound to GTP, does not form a dimer after binding to the ligand but stimulates downstream signaling to occur and causes uncontrolled cell division. Solution The solution is (B). RAS is the first major step in the MAPK kinase cascade. If it is always active, the transcription signals keep promoting cell division. When RAS is unable to hydrolyze its bound GTP, it remains permanently active even in the absence of the ligand, and cell undergoes unregulated division. Advanced Placement Biology Instructor’s Solution Manual 9 | Cell Communication 42 201 Common medications called - blockers bind to G-protein-linked receptors in heart muscles, blocking adrenaline. They are prescribed to patients with high blood pressure. Can you formulate a hypothesis on their mechanism of action? A Adrenaline has a stimulatory effect on heart rate and blood pressure. - blockers are antagonistic to adrenaline and produce an inhibitory effect. B Adrenaline has both a stimulatory and an inhibitory effect on heart rate and blood pressure. - blockers bind to G-protein and stimulate the inhibitory effect of adrenaline. C Adrenaline has an inhibitory effect on heart rate and blood pressure. - blockers have a synergistic effect along with adrenaline, producing an inhibitory effect. D Adrenaline has both a stimulatory and an inhibitory effect on heart rate and blood pressure. - blockers bind to G-protein and intervene with the inhibitory effect of adrenaline. Solution The solution is (A). Adrenaline promotes rapid heartbeat, increasing cardiac output and blood pressure. Adrenaline increases heart rate. - blockers are provided to patients with high blood pressure because they bind to G-protein and produce an inhibitory effect, decreasing the blood pressure. SCIENCE PRACTICE CHALLENGE QUESTIONS 9.1 Signaling Molecules and Cellular Receptors 43 The figure shows a series of states for typical G-protein signal transduction. Use this representation to describe the following stages in this signaling process: A. Between A and B B. Between B and C C. Between C and D D. Between D and E E. Between E and A Advanced Placement Biology Instructor’s Solution Manual 202 9 | Cell Communication Solution Teaching Tip: Prior Knowledge: Chapters 3‒7 Sample answer: A. Between A and B, the signal ligand attaches to the protein and induces a conformational change in the subunit. B. Between B and C, the GDP dissociates from the protein. C. Between C and D, the high concentration of GTP within the cell causes the replacement of GDP with GTP. D. Between D and E, the subunit activated by GTP dissociates from the protein. E. Between E and A, GTP dephosphorylates, returning the subunit to its original configuration which then re-associates with protein. 9.2 Propagation of the Signal 44 Tyrosine kinase receptors are pairs of proteins that span the plasma membrane. On the extracellular side of the membrane, one or more sites are present that bind to signaling ligands such as insulin or growth factors. On the intracellular side, the ends of peptide chains on each protein phosphorylate the other member of the pair, providing active docking sites that initiate cellular responses. The signal is switched off by dissociation of the ligand. For each ligand-receptor system, the equilibrium constant, k, controls the distribution of receptor-bound and unbound ligands. In systems with large values of k, a site is likely to be occupied, even at low concentrations of ligand. When k is small, the likelihood of binding is low, even when the concentration of ligand is high. To initiate a new stimulus response cycle for the receptor, the ligand must dissociate. Larger values of k mean that the receptor is more likely to be occupied and thus unavailable to bind another ligand. Some ligand-binding systems have multiple binding sites. For example, hemoglobin binds four oxygen molecules, whereas myoglobin has only a single binding site. When multiple binding sites are present, the presence of an already-bound ligand can cooperatively affect the binding of other ligands on the same protein. For hemoglobin, the binding is positively cooperative. The affinity of oxygen for heme increases as the number of bound oxygen molecules increases. Advanced Placement Biology Instructor’s Solution Manual 9 | Cell Communication 203 A. Describe the features in the graph for hemoglobin that demonstrate positive cooperativity. B. The insulin receptor (IR) is a tyrosine kinase receptor that has two sites to which insulin can attach. IR is negatively cooperative. In the graph in (A), the dependence of the bound fraction on available insulin is similar to the curve for k 1 with negative cooperativity. Describe the features of this curve in the graph that demonstrate negative cooperativity. C. When viewed from above the cell surface, the representation shows receptors with one and two bound insulin molecules. Explain the negative cooperation for this receptor based on the free energy of conformational changes in the receptor-peptide chains. D. E. Explain the advantages in terms of selection of two-site binding with negative cooperation relative to one-site binding. F. Three binding curves with negative cooperativity and different values of k are shown on the graph. Describe conditions in which there is an advantage in having a low value of k with negative cooperativity. Solution Sample answer: A. Positive cooperativity is shown by the slope of the curve. The slope rises and then falls as the sites become saturated. B. Negative cooperativity is shown by the slope of the curve. The slope falls immediately. C. Free-energy expenditure is required for changes in protein configuration (binding of the ligand). The second configuration requires more free energy because the protein is now constrained by the attached insulin. Advanced Placement Biology Instructor’s Solution Manual 204 9 | Cell Communication D. Two binding sites can increase the range of response of the ligand-receptor system. Very high concentrations of ligand can be managed when necessary and low concentrations can be accommodated with less free-energy expenditure. The affinity for the second ligand will also be lower so that dissociation will be faster and the receptor will increase the frequency of stimulation-response. E. As described for (D), if k is large, then the frequency of response-cycle completion will be smaller. The message sent by the next ligand is wasted unless that receptor has been restored to the original configuration. A smaller value of k shifts the receptor toward a condition to receive the next signal. 45 Organisms, including plants, have evolved chemical signaling pathways to direct physiological responses to environmental changes. Stomata are pores, typically on the underside of leaves, that regulate CO2, O2, and H2O exchange between plants and the external environment. This interaction controls photosynthetic rate and transpiration rate. The opening and closing of stomata are controlled by specialized guard cells that surround the stomatal pore. The osmotic state within the guard cells determines their turgor; when the guard cells are flaccid, stomata close. Turgor in the guard cells is regulated by the active transport of several ions, including K+ and H+, across the plasma membrane. Several environmental factors can cause stomatal closing: water deficit, darkness, microbes, ozone, and sulfur dioxide and other pollutants. Intracellular carbon dioxide concentration and light can trigger stomata to open. The system is regulated by a phytohormone (plant hormone) called abscisic acid (ABA) and the amino acid precursor of the synthesis of a second phytohormone called ethylene (ACC). The second messengers NO and Ca2 in the signal response to changes in the concentrations of these hormones activate transcription factors that affect ion transport across guard cell membranes. High CO2 levels and light also alter phytohormone concentrations. A. Explain why plants must regulate the opening and closing of stomata. Explain how this response relates to the capture of free energy for cellular processes. B. Construct an explanation in terms of the water potential, Y, for the efflux (outward flow) of H+ during water stress (drought). C. Consider a scenario involving environmental factors, such as water stress and daylight, which have opposing effects on the opening and closing of stomata; stomata would be signaled to close under drought conditions and to open during photosynthesis. Pose two scientific questions regarding the response of the system, one involving the phytohormones ABA and ACC, and the second involving the concentration of second messengers. D. The data shown in the figure were obtained by treating rockcress (Arabidopsis) with doses of ABA, ACC, and ABA plus ACC. Using the terms and and or, describe the expected and unexpected responses of the system just after 10 min and around 45 min, as displayed by these data. Advanced Placement Biology Instructor’s Solution Manual 9 | Cell Communication 205 E. Researchers are investigating the interactions among multiple signaling pathways, a phenomenon referred to as “crosstalk.” The same second messengers, NO and Ca2+, are used in many different signaling pathways. Construct an explanation by analogy to other phenomena in which combining a small set of events (for example, 0 and 1 in a computer, the musical scale, or the R, G, and B components of a color) can lead to a vast assortment of outcomes. Solution Sample answer: A. The rate of conversion of carbon dioxide and water to oxygen and C3 or C4 sugars determines the rate of free-energy production. B. As ions are pumped through the plasma membrane, the water potential within the cell increases and the cell swells. C. Possible answers: Researchers in the field are asking these questions: If these hormones lead to the same cellular response, then isn’t this redundant? If these hormones lead to opposing cellular responses, then isn’t this something that should have been selected against? Are the hormones present in a Y/N manner, or are their “quorum-like” evaluations of their equilibrium concentrations that are operating to allow an integrated signal—if there are n and m discrete concentration levels for each, then the information content for a pair of interacting hormones increases as the number n m. Is there time dependence in the response of the system to each receptor that produces cycles that contain further information? This seems to be the direction that this work is taking. Advanced Placement Biology Instructor’s Solution Manual 206 9 | Cell Communication Does the secondary messenger have a target that is near and the lifetime of the messenger is short? Does each signaling pathway operate independently or are there interactions that lead to additional information? This seems to be the direction that the field is going, crosstalk among pathways. Teaching Tip: Like all “pose a question,” this is going to require guidance from the instructor if good questions are to be produced. By visiting small group discussions and making clear that the questions have value and can be either good or bad (not all questions are good questions), students can be encouraged to think scientifically. D. Clearly the combination of the two produces a surprising effect: just after 10 min, the signal to close the stomata is sent by both ABA and ACC and, as expected, a signal ABA and ACC amplifies the effect. However, around 45 min, the stomata close if the signal is ABA or ACC. But ABA and ACC open the stomata. E. With just 3 colors, 27 combinations can be made. If there are 16 bits (possible values of 0 or 1 in the representation of a number on a computer), 65,536 integer values can be expressed. Thinking of each bit as receptor state (ligand present or not present) demonstrates the potential range of cell responses. 9.3 Response to the Signal 46 Construct a graphical representation of information as a function of time during the transduction of a signal along a signaling pathway. B. Annotate your representation for a specific signaling system, such as the effect of epinephrine on the free energy released from glucose. Solution Advanced Placement Biology Instructor’s Solution Manual 9 | Cell Communication 207 9.4 Signaling in Single-Celled Organisms 47 Bacteria and fungi produce several extracellular chemicals, including antibiotics that affect other organisms in the environment. Antibiotics are also produced industrially in large bacteria-containing fermentation tanks. However, antibiotics that have been used by humans to control microbes are now found at subinhibitory concentrations in the environment. Low levels of antibiotics in the environment are mutagenic for bacteria and promote the development of antibiotic resistance. Bacteria produce chemical signals that detect population density and regulate gene expression, a phenomenon called quorum sensing. Density is signaled by the extracellular concentration of small amino acid derivatives. To combat antibiotic resistance, an emerging strategy for the control of bacterial disease is quorum quenching. A. Describe the advantage of antibiotics to the organisms that produce them. B. Based on the name of the emerging strategy for controlling bacterial infections, describe a possible mechanism by which bacteria determine their population density. Justify the claim that quorum quenching may provide a more sustainable approach to disease control than the use of antibiotics. Solution Sample answer: A. Antibiotics are extracellular chemical signals. They are used as communication within a species (there are also examples of communication between species called eavesdropping) and to provide a competitive advantage for resources. B. The success of genes that produce resistance leads to the long-term selection of resistant bacterial strains. Even if antibiotic use were better regulated, subinhibitory concentrations are found to induce mutations, and even with greater restraint, the long-term effect is the lost effectiveness of these materials in the treatment of disease associated with bacterial infection. If a molecule controlled behavior rather than killed bacteria, a selective pressure would less likely drive evolution of bacteria. The quorum signal would be turned off by a “quorum quencher,” such as an enzyme that modifies the molecule that is sensed in the detection of a quorum. A. Use your graph to describe trends in the amount of information rather than the actual magnitude. In sketching your graph, consider how the shape of the curve would change during these events: i. Extracellular first messenger ii. Receptor binding and conformational changes iii. Release of second messengers iv. Cellular responses v. Halt signal and degrade intermediates Advanced Placement Biology Instructor’s Solution Manual 208 10 | Cell Reproduction 10 | CELL REPRODUCTION REVIEW QUESTIONS 1 A diploid cell has how many times the number of chromosomes as a haploid cell? A Four times B Half C One-fourth D Twice Solution 2 The solution is (D). In a diploid cell, the haploid number doubles when an offspring receives one set from the father and another set from the mother. The first level of DNA organization in a eukaryotic cell is maintained by which molecule? A Cohesin B Condensin C Chromatin D Histone Solution 3 The solution is (D). The histones are the alkaline proteins present in the eukaryotic cells that the DNA strands are wound around, thereby packaging the DNA strand into structural units known as nucleosomes. What inherited feature, in specific combinations, determines an organism’s traits? A Cell membranes B Genes C Proteins D RNA Solution 4 The solution is (B). Each gene is a factor that is inherited from the parents by offspring and possesses alleles of a single trait. What are identical copies of chromatin held together by cohesin at the centromere called? A Histones B Nucleosomes C Chromatin D Sister chromatids Solution The solution is (D). Sister chromatids are held together at the kinetochore region by cohesin, which assists them in cell division. Advanced Placement Biology Instructor’s Solution Manual 10 | Cell Reproduction 5 209 Chromosomes are duplicated during what stage of the cell cycle? A G1 phase B Prophase C Prometaphase D S phase Solution 6 The solution is (D). In the S phase, DNA replication and centrosome duplication occur. The DNA replication results in the formation of identical pairs of DNA, known as sister chromatids. Which event does NOT occur during some stages of interphase? A DNA duplication B Increase in cell size C Organelle duplication D Separation of sister chromatids Solution 7 The solution is (D). Interphase includes G0, G1, S, and G2 phases, which in the end produces a chromosome made up of sister chromatids. It is in metaphase of the mitotic phase that separation of sister chromatids occurs. Attachment of the mitotic spindle fibers to the kinetochores is a characteristic of which stage of mitosis? A Anaphase B Prophase C Prometaphase D Metaphase Solution 8 The solution is (C). During prometaphase, the individual chromosomes bind to the mitotic spindle at the kinetochore and arrange themselves at the metaphase plate. The fusing of Golgi vesicles at the metaphase plate of dividing plant cells forms what structure? A Actin ring B Cell plate C Cleavage furrow D Mitotic spindle Solution The solution is (B). In the plant cells during cytokinesis, the Golgi vesicles, which are transported to form the phragmoplast at metaphase plate, fuse to form a structure called a cell plate. Advanced Placement Biology Instructor’s Solution Manual 210 9 10 | Cell Reproduction What would be the outcome of blocking S phase of interphase? A The cell would enter karyokinesis. B DNA replication would not occur. C Centrosomes would be duplicated. D The cytoskeleton would be dismantled. Solution 10 The solution is (B). S phase is responsible for the replication or synthesis of DNA, which, if blocked, would not copy DNA into sister chromatids. At which cell cycle checkpoint does external forces have the greatest influence? A G1 checkpoint B G2 checkpoint C M checkpoint D G0 checkpoint Solution 11 The solution is (A). The G1 checkpoint is influenced as it checks for the presence of essential nutrients and enzymes. The growth hormones and factors are essential to inform the cell about the local conditions. If the M checkpoint is NOT cleared, what stage of mitosis will be blocked? A Prophase B Prometaphase C Metaphase D Anaphase Solution 12 The solution is (D). The M checkpoint occurs just before the anaphase. It checks if all the sister chromatids are properly attached to the kinetochore because the cells will undergo separation in the anaphase. Which protein is a positive regulator that phosphorylates other proteins when activated? A p53 B Retinoblastoma protein (Rb) C Cyclin D Cyclin-dependent kinase (Cdk) Solution The solution is (D). Cdk’s are protein kinases that, with the help of cyclins, have the ability to phosphorylate other proteins when activated. Advanced Placement Biology Instructor’s Solution Manual 10 | Cell Reproduction 13 211 Which negative regulatory molecule can trigger apoptosis if vital cell cycle events do NOT occur? A p53 B p21 C Retinoblastoma protein (Rb) D Cyclin-dependent kinase (Cdk) Solution 14 The solution is (A). p53 is the major negative regulatory molecule since it has the capacity to put the cell on halt or induce apoptosis during any DNA damage or if the damage cannot be repaired. What is the main prerequisite for clearance at the G2 checkpoint? A The cell has reached a sufficient size. B The cell has an adequate stockpile of nucleotides. C An accurate and complete DNA replication has occurred. D Proper attachment of mitotic spindle fibers to kinetochores has occurred. Solution 15 The solution is (C). The G2 checkpoint follows S phase, and correctly duplicated chromosomes will allow cell cycle progression at the G2 checkpoint. What do you call changes to the order of nucleotides in a segment of DNA that codes for a protein? A Proto-oncogenes B Tumor suppressor genes C Gene mutations D Negative regulators Solution 16 The solution is (C). Gene mutations are responsible for altering the sequence of nucleotides. The shape and function of protein produced by these genes are likely altered. Human papillomavirus can cause cervical cancer. The virus encodes E6, a protein that binds p53. Based on this fact and what you know about p53, what effect do you think E6 binding has on p53 activity? A E6 activates p53. B E6 protects p53 from degradation. C E6 mutates p53. D E6 binding marks p53 for degradation. Solution The solution is (D). Binding of E6 to p53 leads to the degradation of tumor suppressor protein, which in turn encourages the development of cancerous growth. Advanced Placement Biology Instructor’s Solution Manual 212 17 10 | Cell Reproduction What is a gene that codes for a positive cell cycle regulator called? A Kinase inhibitor B Oncogenes C Proto-oncogenes D Tumor suppressor genes Solution 18 The solution is (C). Proto-oncogenes are proteins that help in cell cycle progression. If they get mutated, the cell progression may get stopped. Which molecule is a Cdk inhibitor or is controlled by p53? A Anti-kinase B Cyclin C p21 D Rb Solution 19 The solution is (C). If DNA damage is found in replication, p53 becomes active and triggers synthesis of Cdk inhibitors, also known as p21. Which eukaryotic cell cycle events are missing in binary fission? A Cell growth B DNA duplication C Karyokinesis D Cytokinesis Solution 20 The solution is (C). Prokaryotic cells do not possess nuclei, and, hence, they do not undergo karyokinesis. Which statement about binary fission is false? A In both prokaryotic and eukaryotic cells, the outcome of cell reproduction is a pair of daughter cells, which are genetically identical to the parent cell. B Karyokinesis is unnecessary in prokaryotes because there is no nucleus. C Replication of the prokaryotic chromosome begins at the origin of replication and continues in both directions at once. D The mitotic spindle draws the duplicated chromosomes to the opposite ends of the cell, followed by formation of a septum and two daughter cells. Solution The solution is (D). There is no need for a mitotic spindle in prokaryotic cells because they do not undergo karyokinesis. Advanced Placement Biology Instructor’s Solution Manual 10 | Cell Reproduction 21 213 The formation of what structure, which will eventually form the new cell walls of the daughter cells, is directed by FtsZ? A Contractile ring B Cell plate C Cytoskeleton D Septum Solution The solution is (D). The FtsZ triggers the accumulation of proteins at the center of the cell to promote cell wall synthesis for division. This wall is known as the septum. CRITICAL THINKING QUESTIONS 22 How would you compare and contrast a human somatic cell to a human gamete? A Somatic cells have 46 chromosomes and are diploid, whereas gametes have half as many chromosomes as found in somatic cells. B Somatic cells have 23 chromosomes and are diploid, whereas gametes have half as many chromosomes as are present in somatic cells. C Somatic cells have 46 chromosomes and are haploid, whereas gametes have 23 chromosomes and are diploid. D Somatic cells have 46 chromosomes with one sex chromosome. In gametes, 23 chromosomes are present with two sex chromosomes. Solution 23 The solution is (A). Human somatic cells have 46 chromosomes: 22 pairs and 2 sex chromosomes. This is the 2n, or diploid, condition. Human gametes have 23 chromosomes, one each of 23 unique chromosomes, one of which is a sex chromosome. This is the n, or haploid, condition. The somatic cells have 46 chromosomes with 22 autosomes and 2 sex chromosomes. They are considered diploid. But in gametes, 23 chromosomes are found with only one sex chromosome; they are haploid. Eukaryotic chromosomes are thousands of times longer than a typical cell. How can chromosomes fit inside a eukaryotic nucleus? A The genetic material remains distributed in the nucleus, mitochondria, and chloroplast. B The genome is present in a looped structure; thus, it fits the size of the nucleus. C The DNA remains coiled around proteins to form nucleosomes. D The genetic material remains bound to the nuclear envelope, forming invaginations. Solution The solution is (C). The DNA double helix is wrapped around histone proteins to form structures called nucleosomes. Nucleosomes and the linker DNA in between them are coiled into a 30-nm fiber. During cell division, chromatin is further condensed by packing proteins. Advanced Placement Biology Instructor’s Solution Manual 214 24 10 | Cell Reproduction Briefly describe the events that occur in each phase of interphase. A G1: assessment for DNA damage; S: duplication of genetic material; G2: duplication and dismantling organelles B G1: duplication of organelles; S: duplication of DNA; G2: assessment of DNA damage C G1: synthesis of DNA; S: synthesis of organelle genetic material; G2: assessment of DNA damage D G1: preparation for DNA synthesis; S: assessment of DNA damage; M: division of cell Solution 25 The solution is (A). During G1, the cell increases in size, the genomic DNA is assessed for damage, and the cell stockpiles energy reserves as well as the components to synthesize DNA. During the S phase, the chromosomes, centrosomes, and centrioles (animal cells) duplicate. During the G2 phase, the cell recovers from the S phase, continues to grow, duplicates some organelles, and dismantles other organelles. Chemotherapy drugs, such as vincristine and colchicine, disrupt mitosis by binding to tubulin (the subunit of microtubules) and interfering with microtubule assembly and disassembly. Exactly what mitotic structure do these drugs target, and what effect would that have on cell division? A The drugs bind tubulin and inhibit the binding of spindle to the chromosome. This can arrest the cell cycle. B The drugs bind the tubulin, which leads to an error in the chromosome separation. This could lead to apoptosis. C The drugs bind the tubulin, thereby inhibiting their division in S phase. This inhibits cell division. D The drugs bind the spindle fiber and hinder the separation of chromatins. This promotes the division spontaneously. Solution 26 The solution is (A). The mitotic spindle is formed of microtubules. Microtubules are polymers of the protein tubulin; therefore, it is the mitotic spindle that is disrupted by these drugs. Without a functional mitotic spindle, the chromosomes will not be sorted or separated during mitosis. The cell will arrest in mitosis and die. Why might a cell that has just completed cytokinesis enter the G0 phase instead of the G1 phase? A Some cells are physiologically inhibited from undergoing any division and remain in the G0 phase to provide assistance to their neighboring cells. B Some cells reproduce only under certain conditions and, until then, they remain in the G0 phase. C Suspected DNA damage can lead the cell to undergo the G0 phase. D The lack of important components of cell division makes cells stay in the G0 phase. Advanced Placement Biology Instructor’s Solution Manual 10 | Cell Reproduction Solution 27 215 The solution is (B). Many cells temporarily enter G0 until they reach maturity. Some cells are only triggered to enter G1 when the organism needs to increase that particular cell type. Some cells only reproduce following an injury to the tissue. Some cells never divide once they reach maturity. Which general conditions must be met at each of the three main cell cycle checkpoints? A G1 checkpoint: assessment of DNA damage; G2: assessment of new DNA; M checkpoint: segregation of sister chromatids in anaphase B G1 checkpoint: energy reserves for S phase; G2 checkpoint: assessment of new DNA; M checkpoint: attachment of spindle to kinetochore C G1 checkpoint: assessment of DNA damage; G2 checkpoint: energy reserves for duplication; M checkpoint: attachment of spindle to kinetochore D G1 checkpoint: energy reserves for S phase; S checkpoint: synthesis of DNA, G2 checkpoint: assessment of new DNA Solution 28 The solution is (B). The G1 checkpoint monitors adequate cell growth, the state of the genomic DNA, adequate stores of energy, and materials for S phase. At the G2 checkpoint, DNA is checked to ensure that all chromosomes were duplicated and there are no mistakes in newly synthesized DNA. What is the role of the positive cell cycle regulators compared to that of the negative regulators? A Positive regulators promote the cell cycle, but negative regulators block the cell cycle. B Positive regulators block the cell division in cancerous cells, but negative regulators promote in such cells. C Positive regulators promote the cell cycle, but negative regulators arrest the cell cycle until certain events have occurred. D Positive regulators show positive feedback mechanisms, but negative regulators show negative feedback in the cell cycle. Solution 29 The solution is (C). Positive cell regulators, such as cyclin and Cdk, perform tasks that advance the cell cycle to the next stage. Negative regulators such as Rb, p53, and p21 block the progression of the cell cycle until certain events have occurred. What occurs at the M checkpoint? What would happen if the M checkpoint failed? A The M checkpoint checks for proper attachment of sister chromatids, and, if it fails, then cells may undergo nondisjunction of chromosomes. B The M checkpoint checks if the DNA is damaged and promotes its repair. If it fails, then the daughters end up with damaged DNA. Advanced Placement Biology Instructor’s Solution Manual 216 10 | Cell Reproduction C The M checkpoint ensures the proper duplication of DNA, and, if it fails, the cells may undergo nondisjunction of chromosomes. D The M checkpoint ensures that all the components required for cell division are available, and, if it fails, the cell cycle will be inhibited. Solution The solution is (A). The M checkpoint determines whether all the sister chromatids are correctly attached to the spindle microtubules. Normally, the cycle will not proceed until the kinetochores of each pair of sister chromatids are attached to at least two spindle fibers arising from opposite poles of the cell. This ensures that the sister chromatids separate during anaphase and that each daughter cell contains the proper amount of DNA. If the M checkpoint fails, then it is possible that sister chromatids might not separate during mitosis. If the spindle microtubules only attach to one sister chromatid and the cell cycle is not halted by the M checkpoint, then both sister chromatids will be pulled to one pole during anaphase (nondisjunction). The result will be one daughter cell with too many chromosomes and one daughter cell with too few chromosomes (aneuploidy). 30 Which regulatory mechanisms might be lost in a cell producing faulty p53? A Assessment of damaged DNA, recruiting repair enzymes, and binding of spindle to kinetochore B Quality of DNA, triggering apoptosis, and recruiting repair enzymes C Quality of DNA, binding of spindle to kinetochore, and assessment of DNA repair D Triggering apoptosis, recruiting repair enzymes, and proper binding of spindle to kinetochore Solution 31 The solution is (B). Regulatory mechanisms that might be lost include monitoring of the quality of the genomic DNA, recruiting of repair enzymes, and the triggering of apoptosis. DNA damage and quality are detected by p53. It triggers apoptosis in case of damage, and it recruits enzymes. p53 can trigger apoptosis if certain cell cycle events fail. How does this regulatory outcome benefit a multicellular organism? A The apoptosis helps in controlling the consumption of energy by the extra cells. B The apoptosis inhibits the production of faulty proteins, which could be produced due to the DNA damage. C The process of apoptosis stops the invasion of viruses in the other cells. D The cells are killed due to the production of reactive oxygen species produced, which could harm the organism. Solution The solution is (B). If a cell has damaged DNA, the likelihood of producing faulty proteins is higher. The daughter cells of such a damaged parent cell would also produce faulty proteins, which might eventually become cancerous. If p53 Advanced Placement Biology Instructor’s Solution Manual 10 | Cell Reproduction 217 recognizes this damage and triggers the cell to self-destruct, the damaged DNA is degraded and recycled. No further harm comes to the organism. A healthy cell is triggered to divide instead. Apoptosis is a vital mechanism, as p53 checks for the DNA damage in the cell, which could produce faulty proteins when expressed. This could prove fatal for a multicellular organism. 32 Which processes do eukaryotic cell division and binary fission have in common? A DNA duplication, division of cell organelles, division of the cytoplasmic contents B DNA duplication, segregation of duplicated chromosomes, and division of the cytoplasmic contents C Formation of a septum, DNA duplication, division of the cytoplasmic contents D Segregation of duplicated chromosomes, formation of a septum, division of cell organelles Solution 33 The solution is (B). The common processes of eukaryotic cell division and binary fission are DNA duplication, segregation of duplicated chromosomes, and division of the cytoplasmic contents. The formation of what structure, that will eventually form the new cell walls of the daughter cells, is directed by FtsZ? A Contractile ring B Cell plate C Cytoskeleton D Septum Solution The solution is (D). The FtsZ triggers the accumulation of proteins at the center of the cell to promote cell wall synthesis for division. This wall is known as the septum. TEST PREP FOR AP® COURSES 34 Which statement cannot be inferred from the karyotype shown? Advanced Placement Biology Instructor’s Solution Manual 218 10 | Cell Reproduction A The cell contains DNA. B The cell contains 46 chromosomes. C The cell is diploid. D The cell is prokaryotic. Solution 35 The solution is (D). The cell is prokaryotic. Prokaryotes have a single circular chromosome, not multiple chromosomes. How can DNA, which in humans measures approximately 2 m, fit inside a human cell that is about 10 μm? How does the organization of the genetic material in eukaryotes differ from that of prokaryotes? A The DNA is found wrapped around histones to form nucleosomes, which further compact and ultimately form linear chromosomes. The prokaryotic genome is found as a loop and in eukaryotes as a double-stranded linear structure. B The DNA is wrapped around the nucleosomes to show a compact structure. The eukaryotes show a loop structure, and prokaryotes show a double-stranded linear genome. C The genetic material shows ringed heterochromatin structure. The prokaryotes show multiple loops, and eukaryotes show a condensed chromatin. D The genetic material is wrapped around histones. The prokaryotes have a condensed structure in nucleoids, but eukaryotes have a double-stranded linear structure. Solution The solution is (D). DNA is compacted to fit in a cell. In the first level of compaction, short stretches of the DNA double helix wrap around a core of histone proteins, forming a nucleosome that is linked by linker DNA. The next level of compaction occurs as the nucleosomes and linker DNA are coiled into a 30-nm chromatin fiber. In the third level of packing, a variety of fibrous proteins are used to pack the chromatin. Advanced Placement Biology Instructor’s Solution Manual 10 | Cell Reproduction 219 In eukaryotes, the genome consists of several double-stranded linear DNA molecules forming chromosomes. In prokaryotes, the genome is composed of a single doublestranded DNA molecule in the form of a loop or circle. 36 Which statement about structure 1 on the karyotype is NOT true? A Structure 1 consists of homologous chromosomes. B The two parts of structure 1 will have genes in different loci. C The two parts of structure 1 originate from different parents. D The two parts of structure 1 will have slightly different sequences of nucleotides. Solution 37 The solution is (B). The two parts of structure 1 will have genes in different loci. Homologous chromosome will have genes in the same loci. However, the loci may have different variations of the genes. Based on the karyotype provided, the nondisjunction of which chromosome causes Down syndrome? Advanced Placement Biology Instructor’s Solution Manual 220 10 | Cell Reproduction A Chromosome 21 B Chromosome 22 C X chromosome D Y chromosome Solution 38 The solution is (A). The figure depicts three copies of chromosome 21, while the other chromosomes have two chromatids. This condition is found in Down syndrome and occurs during anaphase I when one pair of homologous chromosomes fails to separate. What is the sequence of mitotic cell cycle for one pair of chromosomes that is undergoing normal mitotic division? A Anaphase - metaphase - prophase - cytokinesis B Anaphase - prophase - metaphase - cytokinesis C Prophase - anaphase - metaphase - cytokinesis D Prophase - metaphase - anaphase - cytokinesis Advanced Placement Biology Instructor’s Solution Manual 10 | Cell Reproduction Solution 221 The solution is (D). The figure shows the sequence prophase – metaphase – anaphase – cytokinesis. There is a cell labeled prophase, with two chromosomes. During prophase, spindle fibers emerge from centrosomes and the nuclear envelope breaks down. A cell labeled metaphase shows the two chromosomes are lined up at the center of the cell. A cell labeled anaphase shows one chromosome and a sister chromatid pulled to one pole of the cell and just a sister chromatid pulled to the other pole. Finally, in cytokinesis, two cells are depicted; one has a full chromosome and a sister chromatid, the other has only a single sister chromatid. 39 In a study on cell division, researchers culture synchronously dividing human cells with thymidine, which causes the cells to arrest at the G1 boundary. The cells are then placed in medium lacking thymidine, which releases the block, and the cells begin to divide again. Starting with Sample A and ending with Sample D, the DNA content of the cells is measured at different times after thymidine is removed. Results for four samples (A–D) are shown in the graph. Which sample presents the expected results for cells in S phase? A Sample A B Sample B C Sample C D Sample D Solution The solution is (B). Sample B shows relative content of DNA that has been doubled. This doubling of the DNA content happens only in S phase, which immediately follows G1. Advanced Placement Biology Instructor’s Solution Manual 222 40 10 | Cell Reproduction In a study on cell division, researchers culture synchronously dividing human cells with thymidine. This causes the cells to arrest at the G1 boundary. The cells are then placed in medium lacking thymidine, which releases the block, and the cells begin to divide again. Starting with Sample A and ending with Sample D, the DNA content of the cells is measured at different times after thymidine is removed. Results for four samples (A-D) are shown in the graph. Explain what is happening in terms of the cell cycle and DNA content in sample B. A All the contents of the cell have been doubled. B The DNA content of the cell has doubled. C Two cells have been fused. D The cells are showing the semiconservative mechanism of cell division Solution 41 The solution is (B). Sample B occurred after the S phase of the cell cycle, which causes the DNA content of the cell to double after all DNA is copied. Li-Fraumeni syndrome (LFS1) is a rare hereditary disorder that leads to a predisposition to cancer. This hereditary disorder is linked to mutations in the tumor-suppressor gene encoding the transcription factor p53. p53 acts at the G1 checkpoint. If damaged DNA is detected, p53 halts the cell cycle. As p53 levels rise, the production of p21 is triggered. p21 enforces the halt in the cell cycle. A variant of Li-Fraumeni, called LFS2, is thought to occur due to a mutation of the CHK2 gene, which is also a tumor-suppressor gene. CHK2 regulates the action of p53. Which of the following cascades is most likely to occur in a normal cell that does not contain the LFS mutation? A 1. Cell cycle progression 2. p53 3. p21 4. CHK2 B 1. p53 2. p21 Advanced Placement Biology Instructor’s Solution Manual 10 | Cell Reproduction 223 3. CHK2 4. Cell cycle progression C 1. p21 2. p53 3. CHK2 4. Cell cycle progression D 1. CHK2 2. p53 3. p21 4. Cell cycle progression Solution The solution is (D). CHK2 is a kinase that checks for the DNA damage and phosphorylates the p53 to produce p21 (inhibitor of Cdk’s). After the repair process, the cell cycle progresses. Advanced Placement Biology Instructor’s Solution Manual 224 42 10 | Cell Reproduction The insulin growth factor (IGF-1) promotes cell proliferation as shown in the diagram. The expression of which protein in the diagram is controlled through negative feedback? A Active Cdk4 B Cyclin D1 C Cyclin D1/Cdk4 complex D IGF-1 Solution 43 The solution is (C). The Cdk4 bound to cyclin becomes active, and this is responsible for inhibiting the cyclin D1-Cdk4 complex. Why are p53, p21, and CHK2 considered tumor-suppressor genes, NOT proto-oncogenes? Give an example of a proto-oncogene. A p53, p21, and CHK2 suppress the proteins that regulate the cell cycle, whereas protooncogenes, like phosphorylated Rb, help in cell cycle progression. B p53, p21, and CHK2 are negative cell cycle regulators, whereas Cdk’s are protooncogenes, which could cause cancer when mutated. C p53, p21, and CHK2 suppress the proteins that regulate the cell cycle, whereas Rb is considered a proto-oncogene because it is the most primitive. D The three proteins help stop the formation of tumors, whereas Cdk’s are called protooncogenes because they are the most primitive of all. Solution The solution is (A). A proto-oncogene is a segment of DNA that codes for one of the positive cell cycle regulators. If such a gene becomes mutated, producing a hyperactivated protein product, it is considered an oncogene. A tumor-suppressor gene is Advanced Placement Biology Instructor’s Solution Manual 10 | Cell Reproduction 225 a segment of DNA that codes for one of the negative cell cycle regulators. If such a gene becomes mutated such that the protein product becomes less active, the cell cycle will run unchecked. A single oncogene can initiate abnormal cell division; however, tumor suppressors lose their effectiveness only when both copies of the gene are damaged. The Cdk gene is considered a proto-oncogene. The three proteins are tumor suppressor proteins; they regulate the cell cycle and any mutation in these will cause tumor. The Cdk gene, if it undergoes mutation, would lead to cancer and is called a proto-oncogene as a result. SCIENCE PRACTICE CHALLENGE QUESTIONS 10.2 The Cell Cycle 44 Many biological processes are synchronized with the 24-h rotational period of Earth. Circadian (24-h) periodicity is common across phyla. One of these processes is the cell cycle. The currently accepted explanation is that the low-oxygen atmosphere of early Earth had no ozone layer to filter out the solar ultraviolet radiation that damages DNA. Completing the S phase of the cell cycle at night provided a selective advantage. The internal clock controlling the cell cycle and the circadian clock became synchronized. Research has demonstrated that changes in one clock, either the circadian clock or the cell cycle clock, disrupt timing in the other. The question was, which clock controls the other? Researchers have found that the circadian clock, which can be observed by fluorescent markers on proteins that carry the circadian signal, can be disrupted by changes in light, nutrition, or exposure to the steroid dexamethasone. Nutrition can also disrupt the cell cycle clock. Rat fibroblasts (cells constantly undergoing mitosis) were cultured on medium containing different levels of fetal bovine serum (FBS) with and without the addition of dexamethasone. Confluence is a phenomenon that occurs in tissue culture when the surface of the growth medium becomes covered with cells and the cells stop dividing. The circadian and cell cycle periods were measured. Run FBS Dexamethasone Confluence Circadian Period (h) Cell Cycle Period (h) A 0% No No 24 ± 0.5 24 ± 0.5 B 10% No No 21.9 ± 1.1 21.3 ± 1.3 C 15% No No 19.4 ± 0.5 18.6 ± 0.6 D 10% Yes No 24.2 ± 0.5 20.1 ± 0.94 E 20%* Yes No 21.25 ± 0.36 19.5 ± 0.42 F 20% Yes No 29 ± 1.05 16.05 ± 0.48 G 10% Yes Yes 24 ± 0.5 n/a *Subsets of samples with 20% FBS and dexamethasone were clustered around two means for each measured period. Advanced Placement Biology Instructor’s Solution Manual 226 10 | Cell Reproduction A. Based on these data, describe the connections between the circadian period and the cell cycle period for each of the experimental conditions. B. Based on these data, justify the claim that, in cells that are actively dividing, the circadian period is set by the cell cycle period rather than the reverse. Solution Sample answer: A. The periods of both clocks when dexamethasone is absent are initially identical. Growth on FBS is known to accelerate the cell cycle. The circadian period is equal to the cell cycle period, implying causation. Higher concentrations of the nutrient are consistent with this. When the circadian rhythm is disrupted by dexamethasone, the cell cycle period is unaffected, though the circadian period is lengthened (runs B and D). This is consistent with only one (E) of the measurements made at higher FBS levels. Finally, when the cell cycle stops, the circadian period returns to its original value. B. Conclusion: The circadian clock is always expressed, but during periods of rapid cell division, the circadian period becomes synchronized with the cell cycle period; although at much shorter cell cycle periods, the circadian period becomes erratic. Teaching note—This is a very surprising result in this field. Studies suggest a correlation between disruption of the circadian clock, for example a shift to latenight working hours, and cancer. This would involve entrainment of the cell cycle by the circadian cycle, and the opposite is shown by this work to be the case. 45 Cells in different tissues of a fully developed human show significant variations in the length of time that they remain in the G0 phase of the cell cycle: muscle (lifetime), nerve (lifetime), adipose (years), liver (years), erythrocyte (months), bone osteoclasts (weeks), leukocyte (days), and epidermal (hours). For each of these types of tissues, propose a reason based on internal and external factors and function that might account for the differences among their longevities. Solution 46 Sample answer: Muscle: constrained by available space in a developed tissue Nerve: constrained by function because it is the intercellular connections that imbue function Adipose, liver, erythrocyte: Replacement rate is proportional to stresses. Leukocyte: constrained by the demands of phagocytosis Epidermal: hours; includes skin, stomach, and lung—all tissues in contact with a hostile chemical environment Describe the essential components and results of mitosis and the activities that occur during interphase to prepare the cell for mitosis. Solution Sample answer: During G1, the cell increases in size, the genomic DNA is assessed for damage, and the cell stockpiles energy reserves as well as the components to Advanced Placement Biology Instructor’s Solution Manual 10 | Cell Reproduction 227 synthesize DNA. During the S phase, the chromosomes, the centrosomes, and the centrioles (animal cells) duplicate. During the G2 phase, the cell recovers from the S phase, continues to grow, duplicates some organelles, and dismantles other organelles. 10.4 Cancer and the Cell Cycle 47 Cancer comprises many diseases with a common cause: uncontrolled cell growth. Cancer is a complex response to a host of environmental mutagens as well as the accumulation of random mutations. Since the “war on cancer” began in 1971, the death rate due to cancer has changed very little despite the discovery of several tumor-suppressor genes, including p53. A. Briefly describe the multiple functions of p53, including the role of p53 in apoptosis. B. A principle of biology is that “form follows function.” The protein p53, which has multiple functions, is named for its molecular mass—approximately 53 kDa. This is not a large polymer by comparison with other proteins; for example, ATP synthase, which has only one function, has a molecular mass of approximately 550 kDa. Based on analogies to processes involved in cellular signaling, create a model(s) to explain how so many functions can be supported by a single, relatively simple structure. C. Mutational signatures of p53 are shown in the figure (G.P. Pfeifer et al., Nature, 21(48), 2002) for the three types of cancer with the highest death rates in the United States: lung (~225,000 deaths in 2016), breast (246,000), and colorectal (381,000). These data can be obtained by sequencing the gene that encodes p53. Approximately 85 percent of lung cancers occur in smokers. Based on these data, calculate how many deaths due to lung cancer among nonsmokers were reported in 2016. How much does smoking increase the likelihood of death due to lung cancer? D. As shown under each graph in the figure, particular transversions (replacement of a pyrimidine by a purine of vice versa) or transitions (replacement of a purine or pyrimidine by the alternative purine or pyrimidine) are features of specific mutational signatures. Based on these data, identify the transversion or transition that seems to be induced by cigarette smoke. Advanced Placement Biology Instructor’s Solution Manual 228 10 | Cell Reproduction E. Using your answer to (B), predict possible mechanisms—that is, transversion or transition—for the different mutational signatures among lung cancers of smokers and those of other cancers, and for the very similar mutational signatures of lung cancers of nonsmokers and of breast and colorectal cancers. The partitioning of function along the length of the protein can lead to functional and nonfunctional segments. It is believed that the transversions due to smoking are caused by polyaromatic hydrocarbons. The hotspots for these mutations lie in the segment that binds to DNA. The transition hotspots are in segments that regulate apoptosis. Solution Sample answer: A. As described in the text, p53 detects DNA damage, stops the cell cycle at G1/S, activates DNA repair, and initiates apoptosis. B. p53 achieves multifunctionality with several segments of amino acid sequence responsible for different functions. Segments activate transcription factors that lead to the expression of genes, much as a signal receptor can produce messengers that lead to expression. These segments activate expression of enzymes that achieve the other functions. In addition, a large segment binds to DNA to manage checks of integrity and initiate repair. C. If 85 percent of the 225,000 lung cancer deaths occur among smokers, then 0.85 225,000 deaths can be attributed to death due to lung cancer triggered by smoking. Nonsmokers may also have lung cancer, and the ratio is 0.85/0.15, which is a factor of between 5 and 6. Second-hand smoke is sometimes suggested as the cause of the disease in nonsmokers, although other mutagens such as asbestos fibers are thought to be causes. D. The population of smokers with lung disease shows a significant increase in the G T transversion. Guanine is a pyrimidine, and thymine is a purine. This replacement creates kinks that are much less likely to be silent mutations than are the transitions. Cancer is often modeled as the result of an accumulation of mutations, and the dominant mutations among the nonsmoker lung, breast, and colon-rectal cancers are transitions. E. The partitioning of function along the length of the protein can lead to functional and nonfunctional segments. It is believed that the transversions due to smoking are caused by polyaromatic hydrocarbons. The hotspots for these mutations lie in the segment that binds to DNA. The transition hotspots are in segments that regulate apoptosis. Advanced Placement Biology Instructor’s Solution Manual 11 | Meiosis and Sexual Reproduction 229 11 | MEIOSIS AND SEXUAL REPRODUCTION REVIEW QUESTIONS 1 How many and what type of daughter cells does meiosis produce? A Four haploid B Four diploid C Two haploid D Two diploid Solution 2 The solution is (A). In meiosis, there are two rounds of nuclear division—meiosis I and meiosis II. This results in four nuclei and four daughter cells, each with half the number of chromosomes from the parents. What structure is most important in forming the tetrads? A Centromere B Chiasmata C Kinetochore D Synaptonemal complex Solution 3 The solution is (D). The synaptonemal complex is a protein lattice that forms between homologous chromosomes and is a key part in forming the tetrad and maintaining the synapsis between the strands. It disassembles at the end of prophase I. At which stage of meiosis are sister chromatids separated from each other? A Anaphase I B Anaphase II C Prophase I D Prophase II Solution 4 The solution is (B). The sister chromatids are pulled apart by the kinetochore microtubules and move toward opposite poles during anaphase II. At metaphase I, homologous chromosomes are connected only at what structures? A Chiasmata B Kinetochores C Microtubules D Recombination nodules Advanced Placement Biology Instructor’s Solution Manual 230 11 | Meiosis and Sexual Reproduction Solution 5 The solution is (A). Chiasmata—sites where crossing over has occurred—are formed during prophase I and link the homologous chromosomes until they are separated in anaphase 1. What phase of mitotic interphase is missing from meiotic interkinesis? A G0 phase B G1 phase C G2 phase D S phase Solution 6 The solution is (D). The S phase, or synthesis phase, is where the DNA of the chromosomes is replicated. It is not present in interkinesis, which is a brief interlude between meiosis I and II in some species. What part of meiosis is most similar to mitosis? A Reduction division B Interkinesis C Meiosis I D Meiosis II Solution 7 The solution is (D). Mitosis and meiosis II are similar processes since the number of chromosomes is conserved in both and no genetic variation is introduced. What is NOT true during crossing over? A Chiasmata are formed. B Nonsister chromatids exchange genetic material. C Recombination nodules mediate cross-over events. D Spindle microtubules guide the movement of chromosomal material. Solution 8 The solution is (D). Spindle microtubules are not directly involved with crossing over. During which phase does the second round of genetic variation occur during meiosis? A Anaphase I B Metaphase I C Prophase II D Genetic variation only occurs during prophase I. Solution The solution is (B). The random assortment of homologous chromosomes at the metaphase plate is the second mechanism that introduces variation into the gametes or spores. Advanced Placement Biology Instructor’s Solution Manual 11 | Meiosis and Sexual Reproduction 9 231 Which type of life cycle has both a haploid and a diploid multicellular stage? A Alternation of generations B Asexual C Diploid-dominant D Haploid-dominant Solution 10 The solution is (A). Species with a life cycle that alternates alternate between haploid and diploid multicellular stages. What is a source of variation in asexual reproduction? A Crossing over of chromosomes B Mutation of DNA C Random assortment of chromosomes D There is no variation in asexual reproduction. Solution 11 The solution is (B). A mutation is a change that occurs in the nucleotides of the DNA. It is a source of genetic variation, often the only source, in organisms that reproduce asexually. What is a likely evolutionary advantage of sexual reproduction over asexual reproduction? A Sexual reproduction involves fewer steps. B Sexual reproduction results in variation in the offspring. C Sexual reproduction is more metabolically efficient. D Sexual reproduction uses up fewer resources in a given environment. Solution 12 The solution is (B). Sexual reproduction increases genetic variability by shuffling combinations of genes and chromosomes. This can provide an evolutionary advantage. What is a disadvantage of sexual reproduction over asexual forms of reproduction? A Half the population is capable of carrying offspring. B Identical offspring are not produced. C Adaptation to rapidly changing environments is more difficult. D Mutation rates are slower. Solution The solution is (A). This is one disadvantage of sexual reproduction—that only half of the population (the females) can carry offspring. Reproduction tends to be slower. Advanced Placement Biology Instructor’s Solution Manual 232 13 11 | Meiosis and Sexual Reproduction Fungi typically display which type of life cycle? A Alternation of generations B Asexual C Diploid-dominant D Haploid-dominant Solution 14 The solution is (D). Fungi have haploid-dominant life cycles. The haploid multicellular stage produces specialized haploid cells that fuse to form a diploid zygote, which immediately undergoes meiosis to produce haploid cells. What is a haploid cell produced in a diploid-dominant organism by meiosis called? A Gamete B Gametophyte C Spore D Sporophyte Solution The solution is (A). Gametes are unicellular haploid cells produced in diploiddominant organisms. They fuse with other gametes during fertilization to produce diploid zygotes. CRITICAL THINKING QUESTIONS 15 What happens to tetrads after they form? A Prophase I of meiosis forms the tetrads. They line up at the midway point between the two poles of the cell to form the metaphase plate. There is an equal chance of a microtubule fiber encountering a maternally or a paternally inherited chromosome. Orientation of each tetrad is independent of the orientation of other tetrads. B Prophase II of meiosis forms the tetrads. They line up at the midway point between the two poles of the cell to form the metaphase plate. There is an equal chance of microtubule fiber encountering maternally or paternally inherited chromosome. Orientation of each tetrad is independent of the orientation of other tetrads. C Prophase I of mitosis forms the tetrads. They line up at the midway between the two poles of the cell to form the metaphase plate. There is an equal chance of a microtubule fiber encountering a maternally or a paternally inherited chromosome. Orientation of each tetrad is independent of the orientation of other tetrads. D Prophase I of meiosis forms the tetrads. They line up at the midway between the two poles of the cell to form the metaphase plate. There is a chance of microtubule fiber encountering a maternally inherited chromosome. Orientation of each tetrad is independent of the orientation of other tetrads. Solution The solution is (A). The tetrads line up at the midway point between the two poles of the cell to form the metaphase plate. There is an equal chance of a microtubule fiber Advanced Placement Biology Instructor’s Solution Manual 11 | Meiosis and Sexual Reproduction 233 encountering a maternally or a paternally inherited chromosome. Orientation of each tetrad is independent of the orientation of other tetrads. In prophase I of meiosis, the homologous chromosomes form the tetrads. In metaphase I, these pairs line up at the midway point between the two poles of the cell to form the metaphase plate. Because there is an equal chance that a microtubule fiber will encounter a maternally or paternally inherited chromosome, the arrangement of the tetrads at the metaphase plate is random. Any maternally inherited chromosome may face either pole. Any paternally inherited chromosome may also face either pole. The orientation of each tetrad is independent of the orientation of the other tetrads. When the microtubules pull the tetrads apart, the sister chromatids remain attached to each other. 16 What distinguishes metaphase I from metaphase II? A Metaphase I occurs when chromosomes appear in homologous pairs on the spindle. Metaphase II has a single line of chromosomes on the spindle. A pair of chromosomes is pulled apart and migrates toward a pole in anaphase I, while in anaphase II, sister chromatids separate. Telophase I reconstitutes the nucleus and loosens the chromosomes, while telophase II mimics telophase I. B Prophase I condenses the chromosomes and eliminates the nuclear membrane. The microtubules arrange in a spindle. Prophase II mimics prophase I. Metaphase I occurs when chromosomes appear in homologous pairs on the spindle. Metaphase II has a single line of chromosomes on the spindle. Pairs of chromosomes are pulled apart and migrate toward the poles during anaphase I, while in anaphase II, sister chromatids separate. Telophase I reconstitutes the nucleus and condenses the chromosomes, while telophase II mimics telophase I. C Prophase I condenses the chromosomes and adds a nuclear membrane. The microtubules arrange in a spindle. Prophase II mimics prophase I. Metaphase I occurs when chromosomes appear in homologous pairs on the spindle. Metaphase II has a single line of chromosomes on the spindle. A pair of chromosomes is pulled apart and migrates toward the poles in anaphase I, while in anaphase II, sister chromatids separate. Telophase I reconstitutes the nucleus and loosens the chromosomes, while telophase II mimics telophase I. D Prophase I condenses the chromosomes and eliminates the nuclear membrane. The microtubules arrange in a spindle. Prophase II mimics prophase I. Metaphase I occurs when chromosomes appear in homologous pairs on the spindle. During metaphase II, the chromosomes line up in a double line across the spindle. Each pair of chromosomes is pulled apart and migrates toward the poles in anaphase I, while in anaphase II, sister chromatids separate. Telophase I reconstitutes the nucleus and loosens the chromosomes, while telophase II mimics telophase I. Solution The solution is (A). The difference in metaphase between meiosis I and II reflects how meiosis I involves the separation of homologous chromosomes and meiosis II Advanced Placement Biology Instructor’s Solution Manual 234 11 | Meiosis and Sexual Reproduction represents the splitting of sister chromatids. This causes the differences in alignment between metaphase I and II and the subsequent differences in later phases. 17 Though the stages of meiosis have the same names as the stages of mitosis, they exhibit fundamental differences. What are the main differences between the two processes? A Meiosis differs from mitosis in that the number of chromosomes is halved and genetic variation is introduced in meiosis, but not in mitosis. B Meiosis differs from mitosis in that the number of chromosomes is halved and genetic variation is reduced in meiosis, but not in mitosis. C Metaphase and telophase portions of meiosis and mitosis are the same. Meiosis and mitosis are also the same, except for the number of chromosomes. Anaphase I and anaphase II are different. D Prophase and telophase portions of meiosis and mitosis are the same. Meiosis II and mitosis are also the same and have the same number of chromosomes. Anaphase I and anaphase II are different. Solution The solution is (A). Meiosis cycles through two rounds of cell division, while mitosis happens once. Meiosis I differs from mitosis in that the number of chromosomes is halved, and crossing over (in prophase I) and independent assortment (in metaphase I) adds genetic variation. Meiosis II is essentially the same as mitosis; the number of chromosomes does not change, and no more genetic variation is introduced. However, the end result is four haploid cells instead of the two diploid cells that are produced in mitosis. Also, an S phase does not precede meiosis II as it does in mitosis. In meiosis, the number of chromosomes is halved in order to make haploid gametes, rather than the diploid body cells produced during mitosis. Genetic variation is introduced during meiosis through crossing over. In mitosis, the resulting cells are genetic clones. 18 How does the orientation of homologous chromosomes during metaphase I of meiosis contribute to greater variation in gametes? A The random alignment of homologous chromosomes at the metaphase plate ensures the random destination of the chromosomes in the daughter cells. B Because homologous chromosomes dissociate from the spindle fibers during metaphase I, they move randomly to the daughter cells. C The homologous chromosomes are paired tightly during metaphase I and undergo crossover as the synaptonemal complex forms a lattice around them. D Recombination of maternal and paternal chromosomes occurs in metaphase I, because the homologous chromosomes are not connected at their centromeres. Advanced Placement Biology Instructor’s Solution Manual 11 | Meiosis and Sexual Reproduction Solution 19 235 The solution is (A). The pairs of homologous chromosomes that line up in metaphase I are randomly oriented: maternal or paternal chromosomes may be on either side of the cell’s equator and encounter a microtubule that will pull them randomly to either daughter cell. So, the gametes made from meiosis contain a mixture of chromosomes originally derived from both parents, but randomly assorted. How does the Red Queen’s catchphrase, “It takes all the running you can do to stay in the same place,” describe coevolution between competing species? A When a sexually reproducing species and an asexually reproducing species compete for the same resources, they both “run [evolve] in the same place” because the increased genetic variation in the sexually reproducing species balances the loss in energy it uses to find and attract mates. B When one species gains an advantage with a favorable variation, selection increases on another species with which it competes. This species must also develop an advantage or it will be outcompeted. The two species “run [evolve] to stay in the same place.” C When one species develops a mutation that decreases its ability to survive, a competing species will become better able to survive even though it has not changed in any way. In effect, this species “runs [evolves] to stay in the same place.” D When two asexually reproducing species encounter rapid environmental change, the species that is also able to reproduce sexually will outcompete the other. This way it can “run [evolve] to stay in the same place.” Solution The solution is (B). When one species gains an advantage with a favorable variation, selection increases on another species with which it competes. This species must also develop an advantage or it will be outcompeted. The two species “run [evolve] to stay in the same place.” Each tiny advantage gained by favorable variation gives a species an edge over close competitors, predators, parasites, or even prey. The only method that will allow a coevolving species to maintain its own share of the resources is to also continually improve its fitness. As one species gains an advantage, this increases selection on the other species; it must also develop an advantage or it will be outcompeted. The net effect is like running to stay in the same place. The evolution of two competing species is tightly connected. If one species has a mutation that enables it to take advantage of its competitor in any way, the other species will survive only if it also evolves a way to take the same or another advantage. Neither one “wins”; each must “run in place” just to survive. The variations that result from meiosis enable coevolution to happen, often quickly. 20 Which three processes lead to variation among offspring that have the same two parents? A Genetic recombination, fertilization, meiosis B Crossing over, random chromosome assortment, genetic recombination Advanced Placement Biology Instructor’s Solution Manual 236 11 | Meiosis and Sexual Reproduction C Meiosis, crossing over, genetic recombination D Fertilization, crossing over, random chromosome assortment Solution 21 The solution is (D). Random (or independent) assortment mixes the chromosomes derived from each parent. Crossing over creates a genetic combination that did not exist prior to the exchange. Fertilization occurs randomly between an egg and a large number of available sperm. All three produce genetic variation. Compare the three main types of life cycles in multicellular organisms and give an example of an organism that employs each. A In a diploid dominant cycle, the multicellular diploid stage is present, as in humans. Haploid-dominant life cycles have a multicellular haploid stage, as in fungi. In alternation of generations, haploid-dominant and diploid-dominant stages alternate, as in plants. B In a diploid-dominant cycle, the unicellular diploid stage is present, as in humans. In a haploid-dominant life cycle, a unicellular haploid stage is present, as in fungi. In alternation of generations, haploid-dominant and diploid-dominant stages alternate, as in plants. C In a diploid-dominant cycle, a multicellular haploid stage is present, as in humans. In a haploid-dominant life cycle, a multicellular diploid stage is present, as in fungi. In alternation of generations, haploid-dominant and diploid-dominant stages alternate, as in plants. D In a diploid-dominant cycle, a multicellular diploid stage is present, as in algae. In a haploid-dominant life cycle, a multicellular haploid stage is present, as in plants. In alternation of generations, haploid-dominant and diploid-dominant stages alternate, as in fungi. Solution The solution is (A). In a diploid dominant cycle, the multicellular diploid stage is present, as in humans. Haploid-dominant life cycles have a multicellular haploid stage, as in fungi. In alternation of generations, haploid-dominant and diploiddominant stages alternate, as in plants. There are three main categories of life cycles in multicellular organisms: (1) diploid-dominant, in which the multicellular diploid stage is the most obvious life stage; humans are an example; (2) haploid-dominant, in which the multicellular haploid stage is the most obvious life stage; examples include fungi and some algae; and (3) alternation of generations, in which the two stages are apparent to different degrees depending on the group, as with plants and some algae. In diploid-dominant life cycles, the multicellular diploid stage is present such as with most animals, including humans; haploid-dominant, in which the multicellular haploid stage is the most obvious life stage, such as with all fungi and some algae; and alternation of generations, in which the two stages are apparent to different degrees depending on the group, as with plants and some algae. Advanced Placement Biology Instructor’s Solution Manual 11 | Meiosis and Sexual Reproduction 237 TEST PREP FOR AP® COURSES 22 Reproductive cells in most species are different from the cells that make up the rest of the organism. What are the body cells called, and how are they different from the reproductive cells? A Body cells are called gametes and have half the number of chromosomes found in reproductive cells. B Body cells are called somatic cells and have the same number of chromosomes as reproductive cells. C Body cells are called somatic cells and have double the number of chromosomes found in reproductive cells. D Body cells are called gametes and have double the number of chromosomes found in reproductive cells. Solution 23 The solution is (C). Body cells are called somatic cells and have double the number of chromosomes found in reproductive cells. When the haploid reproductive cells merge during fertilization, the diploid number of chromosomes is restored, and the cells produced by subsequent mitotic divisions are somatic. Spores are structures produced by some plants and all fungi. What is true about them? A Spores are haploid reproductive cells that can produce haploid organisms through mitosis. B Spores are haploid precursors to gametes that give rise to gametes when environmental conditions are favorable. C Spores are haploid reproductive cells that can produce diploid cells without fertilization. D Spores are haploid cells formed only during asexual reproduction and so are not formed by meiosis. Solution 24 The solution is (A). Spores are haploid reproductive cells that can form multicellular haploid organisms after rounds of mitosis. They can also fuse with other spores to produce diploid zygotes. In prophase I, the homologous chromosomes are paired up and linked together. What binds the chromosomes together and maintains their alignment? A Cohesin proteins B Tetrads C The centromere D Synaptonemal complex Advanced Placement Biology Instructor’s Solution Manual 238 11 | Meiosis and Sexual Reproduction Solution 25 The solution is (D). The synaptonemal complex is a protein lattice that forms between homologous chromosomes during prophase I when the chromosomes undergo synapsis. One of the ways that sexual reproduction enhances the diversity of offspring from the same parents is through a process called crossing over. What entities does this occur between during prophase I? A Sister chromatids B Tetrads C Nonhomologous chromosomes D Nonsister chromatids of homologous chromosomes Solution 26 The solution is (D). Exchange of genetic material between nonsister chromatids of homologous chromosomes is called crossing over. There are three sources of genetic variation in sexual reproduction. Which one is NOT considered random? A All are random. B Crossing over C Egg and sperm fertilization D Tetrad alignment on the meiotic spindle Solution 27 The solution is (A). Fertilization, crossing over during prophase 1, and tetrad alignment during metaphase I are all random events. Which one of the three types of life cycles of sexually reproducing organisms does NOT have a multicellular haploid stage? A Alternation of generations B Diploid-dominant C Haploid-dominant D They all have a multicellular haploid stage in their life cycles. Solution 28 The solution is (B). In diploid-dominant organisms, haploid cells are produced, and fertilization occurs when the male and female gametes fuse. There is no multicellular haploid life stage. How are spores produced in haploid-dominant and alternation of generation life cycles? A By gametophytes B By germ cells C Through mitosis D Through meiosis Advanced Placement Biology Instructor’s Solution Manual 11 | Meiosis and Sexual Reproduction Solution 29 239 The solution is (D). During sexual reproduction, specialized haploid cells form a diploid zygote. The zygote undergoes meiosis to form four haploid cells called spores. What is one thing that is true of haploid-dominant life cycles but NOT of alternation of generation life cycles? A Meiosis B (+) and (−) mating types C Spores D A free-living haploid stage Solution The solution is (B). Haploid-dominant organisms, such as fungi and some algae, spend the dominant part of their life cycles as multicellular haploids. During sexual reproduction, specialized haploid cells from two individuals—designated the (+) and (−) mating types—join to form a diploid zygote. The zygote immediately undergoes meiosis to form four haploid cells, called spores. The spores undergo mitosis to form the multicellular haploid individual. Organisms that alternate generations, like plants, have both haploid and diploid multicellular life stages. In some organisms, like ferns, both stages are free living. In others, only one stage is free living. Meiosis occurs in both life cycles, so both exhibit genetic variation through crossing over, chromosome assortment, and fertilization. Advanced Placement Biology Instructor’s Solution Manual 240 12 | Mendel's Experiments and Heredity 12 | MENDEL'S EXPERIMENTS AND HEREDITY REVIEW QUESTIONS 1 Mendel performed hybridizations by transferring pollen to the female ova from what part of the male plant? A Anther B Pistil C Stigma D Seed Solution 2 The solution is (A). Mendel performed hybridization by transferring the pollen from the anther of the male plant. What is one of the seven characteristics that Mendel observed in pea plants? A Flower size B Leaf shape C Seed texture D Stem color Solution 3 The solution is (C). One of the seven contrasting characteristics used by Mendel in pea plants was seed texture. Imagine you are performing a cross involving garden pea plants. What F1 offspring would you expect if you cross true-breeding parents with green seeds and yellow seeds? Yellow seed color is dominant over green. A 100 percent yellow-green seeds B 100 percent yellow seeds C 50 percent yellow, 50 percent green seeds D 25 percent green, 75 percent yellow seeds Solution The solution is (B). A cross between true breeding plants with green and yellows seeds would result in 100 percent yellow seeds because yellow is dominant over green color. Advanced Placement Biology Instructor’s Solution Manual 12 | Mendel's Experiments and Heredity 4 241 Consider a cross to investigate the pea pod texture trait, involving constricted or inflated pods. Mendel found that the traits behave according to a dominant/recessive pattern in which inflated pods were dominant. If you performed this cross and obtained 650 inflated-pod plants in the F2 generation bred from true-breeding stock, approximately how many constricted-pod plants would you expect to have? A 600 B 165 C 217 D 468 Solution The solution is (C). The phenotypic ratio of a monohybrid cross is 3 : 1. The predicted number of constricted pods in the F2 generation will be 217. 650 217 3 5 The observable traits expressed by an organism are described as its — A alleles B genotype C phenotype D zygote Solution 6 The solution is (C). The observable traits of an organism are referred as its phenotype. A recessive trait will be observed in individuals that are what for that trait? A Diploid B Heterozygous C Homozygous or heterozygous D Homozygous Solution 7 The solution is (D). Homozygous individuals will express the recessive trait. If black and white true-breeding mice are mated and the result is all gray offspring, what inheritance pattern would this be indicative of? A Codominance B Dominance C Incomplete dominance D Multiple alleles Advanced Placement Biology Instructor’s Solution Manual 242 12 | Mendel's Experiments and Heredity Solution 8 The solution is (C). The phenotype produced is a combination of the dominant and recessive phenotypes. The offspring produced has an intermediate phenotype between black and white and hence show incomplete dominance. The ABO blood groups in humans are controlled by the IA, IB, and I alleles. The IA allele encodes the A blood group antigen, IB encodes B, and I encodes O. Both A and B are dominant to O. If a heterozygous blood type A parent (iAi) and a heterozygous blood type B parent (iBi) mate, one-quarter of their offspring will have AB blood type (IAIB) in which both antigens are expressed equally. Therefore, the ABO blood groups are an example of — A codominance and incomplete dominance B incomplete dominance only C multiple alleles and incomplete dominance D multiple alleles and codominance Solution 9 The solution is (D). ABO blood groups in humans are expressed from the IA, IB, and I alleles and therefore are examples of multiple alleles. Since both the alleles IA and IB express simultaneously resulting in AB blood group, the ABO blood group is also an example of codominance. In a mating between two individuals that are heterozygous for a recessive lethal allele that is expressed in utero, what genotypic ratio—homozygous dominant : heterozygous : homozygous recessive—would you expect to observe in the offspring? A 1:2:1 B 3:1:1 C 1:2:0 D 0:2:1 Solution 10 The solution is (C). The genotypic ratio 1 : 2 : 0 will be obtained after mating between two heterozygous individuals for a recessive lethal allele since the offspring with the homozygous recessive allele will not survive in utero. The forked line and probability methods make use of what probability rule? A Monohybrid rule B Product rule C Sum rule D Test cross Solution The solution is (B). The values along each forked pathway can be multiplied because each gene assorts independently. Advanced Placement Biology Instructor’s Solution Manual 12 | Mendel's Experiments and Heredity 11 243 In pea plants, smooth seeds (S) are dominant to wrinkled seeds (s). The Punnett square shows a genetic cross of two plants that are heterozygous for the seed shape trait. What is the missing genotype? A SS B Ss C sS D ss Solution 12 The solution is (D). The ss genotype is obtained when the s allele of one parent combines with s allele of another parent. If the inheritance of two traits fully obeys Mendelian laws of inheritance, where may you assume that the genes are located? A On any autosomal chromosome or chromosomes B On Y chromosomes C On the same chromosome D On separate chromosomes Solution 13 The solution is (D). If the inheritance of two traits fully follows the Mendelian laws of inheritance, the genes assort independently, so they are located on separate chromosomes. How many different offspring genotypes are expected in a trihybrid cross between parents heterozygous for all three traits? How many phenotypes are expected if the traits behave in a dominant and recessive pattern? A 64 genotypes; 16 phenotypes B 16 genotypes; 64 phenotypes C 8 genotypes; 27 phenotypes D 27 genotypes; 8 phenotypes Solution The solution is (D). There are 27 genotypes and 8 phenotypes expected in a trihybrid cross. Review Table 12.5: General Rules for Multihybrid Crosses. Advanced Placement Biology Instructor’s Solution Manual 244 14 12 | Mendel's Experiments and Heredity Four-o’clock flowers may be red, pink, or white. In the crossing of true-breeding red and true-breeding white plants, all the offspring are pink. What is the correct genotype of the offspring if the red parent has genotype RR and the white parent has genotype rr? Use a Punnett square. A RR and Rr B Rr and rr C Rr only D RR only Solution 15 The solution is (C). Rr is the only possible genotype of the offspring obtained from a cross of true-breeding red and true-breeding white plants. Which cellular process underlies Mendel’s law of independent assortment? A Chromosomes align randomly during meiosis. B Chromosomes can exchange genetic material during crossover. C Gametes contain half the number of chromosomes of somatic cells. D Daughter cells are genetically identical to parent cells after mitosis. Solution 16 The solution is (A). The law of independent assortment states that allele pairs separate independently at the time of gamete formation. This means that traits are transmitted to offspring independently of one another. While studying meiosis, you observe that gametes receive one copy of each pair of homologous chromosomes and one copy of the sex chromosomes. This observation is the physical explanation of Mendel’s law of — A dominance B independent assortment C random distribution of traits D segregation Solution 17 The solution is (D). The law of segregation states that the two copies of each hereditary factor segregate during the production of gametes. The offspring acquires one factor from each parent as a result of segregation. In some primroses, the petal color blue is dominant. A cross between a true-breed blue primrose and a white primrose yields progeny with white petals. A second gene at another locus prevented the expression of the dominant coat color. What is this an example of? A Codominance B Hemizygosity Advanced Placement Biology Instructor’s Solution Manual 12 | Mendel's Experiments and Heredity 245 C Incomplete dominance D Epistasis Solution 18 The solution is (D). A cross between a true-breed blue primrose and a white primrose which yields a progeny with white petals is an example of epistasis since the white color allele masks or interferes with the expression of the blue color allele. Purple flowers (P) are dominant over red flowers (p), and long pollen grains are dominant over round pollen grains. When purple flowers and long pollen grain plants were crossed with plants with white flowers and round pollen grains, all the F1 plants showed purple flowers and long pollen grains. The F1 plants were crossed and the results are in the table. What conclusions about the physical relationship between the traits can be drawn from the experiment? A The traits are probably linked. B The traits follow the law of independent assortment. C The traits are located on different chromosomes. D There was epistasis. Solution 19 The solution is (A). The probability of having purple flowers and long pollen grains is higher. Therefore, it is likely that the genes for purple flower color and long pollen grains are in close proximity, and the probability for them to be inherited together is higher. When the expression of one gene pair masks or modifies the expression of another, what do the genes show? A Codominance B Epistasis C Incomplete dominance D Partial linkage Solution The solution is (B). The antagonistic interaction between two genes, such that one gene masks or interferes with the expression of another gene, is called epistasis. Advanced Placement Biology Instructor’s Solution Manual 246 12 | Mendel's Experiments and Heredity CRITICAL THINKING QUESTIONS 20 Why is the garden pea an excellent system for studying inheritance? A The garden pea has flowers that close tightly to promote cross-fertilization. B The garden pea has flowers that close tightly to prevent cross-fertilization. C The garden pea does not mature in one season and is a perennial plant. D Male and female reproductive parts attain maturity at different times, promoting selffertilization. Solution 21 The solution is (B). The garden pea has flowers that close tightly during selfpollination. This helped to prevent accidental or unintentional fertilizations that could have diminished the accuracy of Mendel’s data. Garden peas naturally adopt ways to promote self-fertilization and prevent cross-fertilization. Self-fertilization leads to true breeding lines that avoid the appearance of unexpected traits in offspring that might occur if the plants were not true breeding. How would you perform a reciprocal cross to test stem height in the garden pea? A First cross is performed by transferring the pollen of a heterozygous tall plant to the stigma of a true-breeding dwarf plant. Second cross is performed by transferring the pollen of a heterozygous dwarf plant to the stigma of a true-breeding tall plant. B First cross is performed by transferring the pollen of a true-breeding tall plant to the stigma of a true-breeding dwarf plant. Second cross is performed by transferring the pollen of a true-breeding dwarf plant to the stigma of a true-breeding tall plant. C First cross is performed by transferring the pollen of a true-breeding tall plant to the stigma of a heterozygous dwarf plant. Second cross is performed by transferring the pollen of a heterozygous dwarf plant to the stigma of a true-breeding tall plant. D First cross is performed by transferring the pollen of a heterozygous tall plant to the stigma of a heterozygous dwarf plant. Second cross is performed by transferring the pollen of a heterozygous tall plant to the stigma of a heterozygous dwarf plant. Solution 22 The solution is (B). Two sets of P parents would be used. In the first cross, pollen would be transferred from the anther of a true-breeding tall plant to the stigma of a true-breeding dwarf plant. In the second cross, pollen would be transferred from the anther of a true-breeding dwarf plant to the stigma of a true-breeding tall plant. For each cross, F1 and F2 offspring would be analyzed to determine whether offspring traits were affected according to which parent donated each trait. Flower position in pea plants is determined by a gene with axial and terminal alleles. Given that axial is dominant to terminal, what are the possible F1 and F2 genotypes and phenotypes from a cross involving parents that are homozygous for each trait? Express genotypes with conventional genetic abbreviations. Advanced Placement Biology Instructor’s Solution Manual 12 | Mendel's Experiments and Heredity 247 A F1: all AA-axial; F2: AA-axial and aa-terminal B F1: all aa-terminal; F2: AA-axial and Aa-terminal C F1: AA-axial and Aa-terminal; F2: all AA-axial D F1: all Aa-axial; F2: AA-axial, Aa-axial, and aa-terminal Solution 23 The solution is (D). Because axial is dominant, the gene would be designated A. F1 would be all heterozygous Aa with an axial phenotype. F2 would have possible genotypes of AA, Aa, and aa; these would correspond to axial, axial, and terminal phenotypes, respectively. F1 genotypes will all be hybrids (Aa) of both parents. Therefore, all the flowers in F1 will be axial. F2 genotypes will segregate in the ratio of 1AA : 2Aa : 1aa. Therefore, the F2 generation will contain both axial and terminal flower in the ratio 3 : 1. Use a Punnett square to predict the offspring in a cross between a dwarf pea plant (homozygous recessive) and a tall pea plant (heterozygous). What is the phenotypic ratio of the offspring? A 1 tall : 1 dwarf B 1 tall : 2 dwarf C 3 tall : 1 dwarf D 1 dwarf : 4 tall Solution 24 The solution is (A). The Punnett square would be 2 × 2 and will have t and t along the top, and T and t along the left side. Clockwise from the top left, the genotypes listed within the boxes will be Tt, Tt, tt, and tt. The phenotypic ratio will be 1 tall : 1 dwarf after crossing a homozygous recessive dwarf pea plant and a heterozygous tall pea plant. Can a human male be a carrier of red-green color blindness? A Yes, males can be the carriers of red-green color blindness since color blindness is autosomal dominant. B No, males cannot be the carriers of red-green color blindness since color blindness is X-linked. C No, males cannot be the carriers of red-green color blindness since color blindness is Y-linked. D Yes, males can be the carriers of red-green color blindness since color blindness is autosomal recessive. Solution The solution is (B). No, males either have colored vision or are color blind. They cannot be carriers because an individual needs two X chromosomes to be a carrier. This is an example of hemizygosity. Advanced Placement Biology Instructor’s Solution Manual 248 25 12 | Mendel's Experiments and Heredity What are the genotypes and genotypic proportions of a cross between AABBCc and Aabbcc parents? Use the probability method for your calculations. A Possible genotypes are AABBcc, AaBbCc, and AaBbcc, and the ratio is 1 : 2 : 1. B Possible genotypes are AABbcc, AaBbCc, and AaBbcc, and the ratio is 1 : 3 : 1. C Possible genotypes are AABbCc, AABbcc, AaBbCc, and AaBbcc, and the ratio is 1 : 1 : 1 : 1. D Possible genotypes are AABbcc, AaBbCC, and AaBbcc, and the ratio is 1 : 1 : 1. Solution 26 The solution is (C). Considering each gene separately, the cross at A will produce offspring of which half are AA and half are Aa; B will produce all Bb; C will produce 1 1 1 half Cc and half cc. Proportions then are 1 , or AABbCc; continuing 4 2 2 1 1 1 for the other possibilities yields AABbcc, AaBbCc, and AaBbcc. The 4 4 4 proportions therefore are 1 : 1 : 1 : 1. How does the segregation of traits result in different combinations of gametes at the end of meiosis? A The chromosomes randomly align during anaphase I at the equator. Separation of bivalent chromosomes occurs during metaphase I of meiosis I. Similarly, separation of sister chromatids occurs at metaphase II of meiosis II. At the end of meiosis II, four different gametic combinations are produced, each containing a haploid set of chromosomes. B The chromosomes randomly align during metaphase I at the equator, and separation of homologous chromosomes occurs during anaphase I. Similarly, separation of sister chromatids occurs at anaphase II of meiosis II. At the end of meiosis II, four different gametic combinations are produced, each containing a haploid set of chromosomes. C The chromosomes randomly align during prophase I at the equator, and separation of sister chromatids occurs during metaphase I of meiosis I. Similarly, separation of bivalent chromosomes occurs at metaphase II of meiosis II. At the end of meiosis II, four different gametic combinations are produced, each containing a diploid set of chromosomes. D The chromosomes randomly align during prophase I at the equator, and separation of bivalent chromosomes occurs during anaphase I of meiosis I. Similarly, separation of homologous chromosomes occurs at metaphase II of meiosis II. At the end of meiosis II, four different gametic combinations are produced, each containing a diploid set of chromosomes. Solution The solution is (B). The chromosomes randomly align during metaphase I at the equator, and separation of homologous chromosomes occurs during anaphase I. Similarly, separation of sister chromatids occurs at anaphase II of meiosis II. At the end of meiosis II, four different gametic combinations are produced, each containing Advanced Placement Biology Instructor’s Solution Manual 12 | Mendel's Experiments and Heredity 249 a haploid set of chromosomes. Four different combinations of gametes are produced: RT, Rt, rT, and rt. Separation of bivalent chromosomes occurs at anaphase I of meiosis I, resulting in two daughter cells. Separation of sister chromatids occurs at anaphase II of meiosis II, resulting in four different gametic combinations each containing a haploid set of chromosomes. 27 In Section 12.3 Laws of Inheritance, an example of epistasis was given for summer squash. Cross white WwYy heterozygotes to demonstrate the phenotypic ratio of 12 white: 3 yellow : 1 green that was given in the text. A Twelve offspring are white because the W gene is epistatic to the Y gene. Three offspring are yellow, because w is not epistatic. The green offspring is obtained when the recessive form of both genes (wwyy) is present. B Twelve offspring are white because W gene is hypostatic to Y gene. Three offspring are yellow because Y is epistatic to w. The green offspring is obtained when the dominant form of both the genes (WWYY) is present. C Twelve offspring are white because W gene is dominant. Three offspring are yellow because Y is dominant and w is recessive. The green offspring is obtained when the recessive form of both the genes (wwyy) is present, showing codominance. D Twelve offspring are white because W is epistatic to Y gene. Three offspring are yellow because Y is hypostatic to w. The green offspring is obtained when the recessive form of both the genes (wwyy) are present, showing codominance. Solution The solution is (C). The cross can be represented as a 4 × 4 Punnett square, with the following gametes for each parent: WY, Wy, wY, and wy. For all 12 of the offspring that express a dominant W gene, the offspring will be white. The three offspring that are homozygous recessive for w but express a dominant Y gene will be yellow. The remaining wwyy offspring will be green. TEST PREP FOR AP® COURSES 28 The trait for widow’s peak can be considered a monoallelic dominant trait in humans. If a man with a widow’s peak and a woman with a straight hairline have a child together, what is the probability that the child will inherit the widow’s peak if the father’s mother had a straight hairline? A 0.25 B 0.50 C 0.75 D 1 Solution The solution is (B). The father is heterozygous dominant for the trait (Ww), and the mother is homozygous recessive (ww). The offspring produced from such parents would be 50 percent Ww (widow’s peak) and 50 percent ww (straight hairline). The probability of having a child with the dominant trait is 50 percent, or 0.50. Advanced Placement Biology Instructor’s Solution Manual 250 12 | Mendel's Experiments and Heredity Advanced Placement Biology Instructor’s Solution Manual 12 | Mendel's Experiments and Heredity 29 251 Don’t like brussels sprouts? Blame your genes. The chemical PTC (phenylthiocarbamide), which is nearly identical to a compound found in the cabbage family, tastes very bitter to some people. Others cannot detect a taste. The ability to taste PTC is incompletely dominant and is controlled by a gene on chromosome 7. A woman who finds brussels sprouts mildly distasteful—in other words, who can taste PTC weakly—has a child with a man who hates brussels sprouts—in other words, who can taste PTC strongly. What is the probability that their son likes brussels sprouts—in other words, cannot taste PTC? A 0 B 0.25 C 0.50 D 1 Solution 30 The solution is (A). The mother is a weak taster for PTC test as she does not like brussels sprouts, which means she is heterozygous for the trait (Tt). The father is a strong taster (TT) for PTC taste, so he hates brussels sprouts. Therefore, the probability of their child (Tt-weak taster) liking brussels sprouts will be 0. Tay-Sachs disease is an autosomal recessive disorder that causes severe problems in neurons. Children who receive two copies of the gene rarely live beyond the age of five. There is no cure for the disease. During a genetic screening, a couple is told that both partners carry the recessive gene. What kind of issue must the couple confront? A Scientific B Financial C Ethical D Educational Solution 31 The solution is (C). Parents will have to consider if it is right to undergo prenatal screening and terminate a pregnancy, which may be unethical. A couple has three daughters. What is the probability that the next child they have will be a daughter? A 0% B 25% C 50% D 100% Solution The solution is (C). The probability that the couple’s next child will be a daughter will be 50 percent because there are equal chances of having a son or a daughter. Advanced Placement Biology Instructor’s Solution Manual 252 32 12 | Mendel's Experiments and Heredity What is the probability that a couple will have three daughters? A 1 2 B 1 3 C 1 6 D 1 8 Solution 33 1 . Therefore, the probability 2 1 1 1 1 that a couple will have three daughters is . 2 2 2 8 The solution is (D). The probability of having a girl is Petunias can be blue, red, or violet. When a blue flower is crossed with a red flower, all the resulting flowers are violet. When a violet flower is crossed with a red flower, about half of the flowers are violet and half are red. How do you characterize the color trait? A Complete dominance B Codominance C Incomplete dominance D Sex linked Solution 34 The solution is (C). A cross between a blue (BB) flower and a red (RR) flower produces all violet (BR) flowers as a result of incomplete dominance. Both alleles when present produce intermediate inheritance, in which one allele for the trait is not completely expressed over its paired allele. Petunias can be blue, red, or violet. When a blue flower is crossed with a red flower, all the resulting flowers are violet. Two violet petunias are crossed. What is the most probable result of the cross? A 75 percent blue and 25 percent red B 50 percent blue and 50 percent red C 75 percent red and 25 percent blue D 25 percent blue, 50 percent violet, and 25 percent red Solution The solution is (D). This is a case of incomplete dominance where one allele for the trait is not completely expressed over its paired allele, resulting in an intermediate phenotype (violet). The cross between two violet petunia flowers Advanced Placement Biology Instructor’s Solution Manual 12 | Mendel's Experiments and Heredity 253 (BR) will produce 25 percent blue flowers (BB), 75 percent violet flowers, and 25 percent red flowers (RR). 35 Fruit flies (Drosophila melanogaster) with a wild-type phenotype have gray bodies and red eyes. Certain mutations can cause changes to these traits. Mutant flies may have a black body and/or cinnabar eyes. To study the genetics of these traits, a researcher crossed a true-breeding wild-type male fly with a true-breeding female fly with a black body and cinnabar eyes. All of the F1 progeny displayed a wild-type phenotype. Which statement is correct about the traits observed? A Gray body and cinnabar eyes are dominant. B Eye color is sex linked. C Body color is sex linked. D Gray body and red eyes are dominant. Solution 36 The solution is (D). A cross between wild-type males (GGRR) and mutant females (ggrr) produce all wild-type progenies (GgRr), showing that gray body is dominant over black body, and red eyes are dominant over cinnabar eyes. Female flies from the F1 generation were crossed with true-breeding male flies with black bodies and cinnabar eyes. The table represents the predicted outcome and the data obtained from the cross. Which assumption led to the predicted numbers? A The traits assort independently. B The traits are located on the X chromosome. C The traits are on the same chromosome. D The female flies were homozygous for wild-type alleles. Solution 37 The solution is (A). All the phenotypes are observed in equal numbers (1 : 1 : 1 : 1). Therefore, the traits follow the law of independent assortment. Cats can be black, yellow, or calico (black and yellow patches). Coat color is carried on the X chromosome. What type of inheritance is color coat in cats? A Codominance B Incomplete dominance Advanced Placement Biology Instructor’s Solution Manual 254 12 | Mendel's Experiments and Heredity C Codominance, sex linked D Incomplete dominance, sex linked Solution 38 The solution is (C). Coat color is governed by two alleles, black (B) and yellow (Y). When both alleles are present, they exhibit codominance. The development of colored patches requires two X chromosomes, and hence under normal conditions, the calico phenotype is observed in females and not in males. Cats can be black, yellow, or calico (black and yellow patches). Coat color is carried on the X chromosome. A yellow cat is crossed with a black cat. Assume that the offspring are both male and female. What are the phenotypes of the offspring and in what proportions? A All the cats are yellow. B All the cats are black. C All the cats are calico. D There is not enough information to answer the question. Solution The solution is (D). The question does not state the sex of either parent cat. Information is missing. SCIENCE PRACTICE CHALLENGE QUESTIONS 12.2 Characteristics and Traits 39 The gene SLC24A5 encodes an antiporter membrane protein that exchanges sodium for calcium (R. Ginger et al., JBC, 2007). This process has a role in the synthesis of the melanosomes that cause skin pigmentation. A mutation in this gene affecting a single amino acid occurs in humans. The homozygous mutant gene is found in 99 percent of humans with European origins. Both the wild type and mutant display codominance. A. Representing the wild-type form of the gene as +/+ and the mutant form of the gene as m/m for two homozygous parents, construct a Punnett square for this cross using the first grid below. Annotate your representation to identify the phenotypes with high (H), intermediate (I), and low (L) melanosome production. Use the second grid to represent an F2 generation from the offspring of the first cross. Use annotation to show the phenotype. Advanced Placement Biology Instructor’s Solution Manual 12 | Mendel's Experiments and Heredity 255 F1 m + m Blank + blank blank blank Blank Blank Blank Blank Blank Blank blank F2 B. Draw sister chromatids at anaphase II for both parents in the F1 generation, and annotate your drawing to identify each genotype of the gametes using the cells of the Punnett square. C. Explain which of Mendel's laws is violated by codominance. D. Suppose that these data were available to evaluate the claim that the wild-type and mutant forms of SLC24A5 are codominant: F2 blank blank Phenotype Observed Expected H 1,206 I 2,238 blank blank L 1,124 Complete the table. Explain the values expected in terms of the genotype of the offspring. blank E. Using a 2 statistic at the 95 percent confidence level, evaluate the claim that the wildtype and mutant forms of SCLO24A5 are codominant. The definition of the statistic 2 C Oi Ei 2 Ei , where is the chi-square test statistic, c is the significant level of the test (we will use 0.05), O is the observed value for variable i, and E is the expected value for variable i. The chi-square statistic table is provided in the AP Biology Exam. p 1 2 3 0.05 3.84 5.99 7.82 4 5 6 9.49 11.07 12.59 Degrees of Freedom Advanced Placement Biology Instructor’s Solution Manual 7 8 14.07 15.51 256 12 | Mendel's Experiments and Heredity Solution Sample answer: A. F1 m m + +/m I +/m I + +/m I +/m I F2 m m m/m + L +/m I + +/m I +/+ H B. The drawing shows a pair of chromatids for each parent on which there only wildtype on one pair and only mutant on the other. C. Mendel thought that genes were either dominant or recessive. He did not consider the idea that both traits can be dominant at once. D. Heterozygotes express the both alleles. E. Phenotype Observed Expected (o − e)2/e H 1,206 1,142 3.59 I 2,238 2,284 0.93 L 1,124 1,142 0.28 The sum of the last column is 4.80, which is less than 5.99 (using the table of critical values provided at the Exam), so the claim of codominance cannot be rejected based on these data. 40 Adrenoleukodystrophy (ALD) is a genetic disorder in which lipids with very high molecular weights are not metabolized and accumulate within cells. Accumulation of these fats in the brain damages the myelin that surrounds nerves. This progressive disease has two causes: an autosomal recessive allele, which causes neonatal ALD, and a mutation in the ABCD1 gene located on the X chromosome. A controversial treatment is the use of Lorenzo’s oil, which is expensive; despite this treatment, neurological degradation persists in many patients. Gene therapy as a potential treatment is currently in trials but is also very costly. An infant patient exhibits symptoms of neonatal ALD, which are difficult to distinguish from the X-linked form of the disease. The infant’s physician consults electronic health records to construct a pedigree showing family members who also presented symptoms similar to ALD. The pedigree is shown in this diagram. The infant patient is circled. Symbols for males (o) and females (m) are filled when symptoms are present. Advanced Placement Biology Instructor’s Solution Manual 12 | Mendel's Experiments and Heredity 257 A. Using the pedigree, explain which form of ALD (neonatal or X linked) is present in the infant. B. Sharing of digital records among health providers is one method proposed to improve the quality and reduce the cost of health care in the United States. The privacy of electronic health records is a concern. Pose three questions that must be addressed in developing policies that balance the costs of treatments and diagnoses, patient quality of life, and risks to individual privacy. Solution Sample answer: A. The disease is not X linked. It might be neonatal. The great aunt disproves sex linked. Autosomal recessive: X linked: Advanced Placement Biology Instructor’s Solution Manual 258 12 | Mendel's Experiments and Heredity B. Questions might arise from these types of considerations: 41 The U.S. Department of Health and Human Services cited 380 breaches of healthcare records involving 500 or more patients in 2011. Are these records secure? Who is responsible for their security? Insurance companies have traditionally charged rates that are fixed by preexisting conditions. Access to care could be affected. Should rates be adjusted to genome as they are now according to certain habits? Potential employers may have an interest in these records. Can employers gain access? Will the precedent of mandatory drug testing be broadened? These records might have prosecutorial value that affects the rights of both the accused and the accuser. My DNA is admissible for forensic purposes, so will it also be used to infer guilt or innocence of intent or “natural tendencies.” In some scenarios (the movie Gattaca, for example) the meaning of “health records” could evolve with unintended consequences if the whole genome is recorded. In the most efficient societies, your role is matched to your abilities—can these be evaluated without my participation? Two genes, A and B, are located adjacent to each other (linked) on the same chromosome. In the original cross (P0), one parent is homozygous dominant for both traits (AB), whereas the other parent is recessive (ab). Characteristic Alleles Chromosome Seed color Yellow (I)/green (i) 1 Seed coat and flowers Colored (A)/white (a) 1 Mature pods Smooth (V)/wrinkled (v) 4 Flower stalk From leaf axils (Fa)/umbellate at top of plant (fa) 4 Height >1m (Le)/−0.5 m (le) 4 Unripe pods Green (Gp)/yellow (gp) 5 Mature seeds Smooth (R)/wrinkled (r) 7 A. Describe the distribution of genotypes and phenotypes in F1. B. Describe the distribution of genotypes and phenotypes when F1 is crossed with the ab parent. C. Describe the distribution of genotypes and phenotypes when F1 is crossed with the AB parent. D. Explain the observed non-Mendelian results in terms of the violation of the laws governing Mendelian genetics. Advanced Placement Biology Instructor’s Solution Manual 12 | Mendel's Experiments and Heredity Solution 259 Sample answer: A. AB × ab where AB and ab can be treated as a single gene results in only AaBb. B. The possible gametes from AaBb when treated as a single gene are AB and ab. So, the result of xab is AaBb and aabb. The phenotype is 1 : 1 with equal probabilities doubly dominant and doubly recessive. C. The possible gametes from AaBb when treated as a single gene are AB and ab. So, the result of xAB is AABB and AaBb. All offspring are dominant in both genes. D. Mendelian genetics assumes that the genes assort independently. If they cannot crossover—because they are adjacent—then the outcomes are as described and non-Mendelian. 12.3 Laws of Inheritance 42 Gregor Mendel’s 1865 paper described experiments on the inheritance of seven characteristics of Pisum sativum, shown in the first column in the table below. Many years later, based on his reported outcomes and analysis of the inheritance of a single characteristic, Mendel developed the concepts of genes, their alleles, and dominance. These concepts are defined in the second column of the table, using conventional symbols for the dominant allele for each characteristic. Even later, the location of each of these genes on one of the seven chromosomes in P. sativum was determined, as shown in the third column. A. Before the acceptance of what Mendel called “factors” as the discrete units of inheritance, the accepted model was that the traits of progeny were “blended” traits of the parents. Evaluate the evidence provided by Mendel’s experiments in disproving the blending theory of inheritance. B. Mendel published experimental data and analysis for two experiments involving the inheritance of more than a single characteristic. He examined two-character inheritance of seed shape and seed color. He also reported three-character inheritance of seed shape, seed color, and flower color. Evaluate the evidence provided by the multiple-character experiments. Identify which of the following laws of inheritance depend upon these multiple-character experiments for support: i. During gamete formation, the alleles for each gene segregate from each other so that each gamete carries only one allele for each gene. ii. Genes for different traits can segregate independently during the formation of gametes. iii. Some alleles are dominant, whereas others are recessive. An organism with at least one dominant allele will display the effect of the dominant allele. iv. All three laws can be inferred from the single-character experiments. C. As shown in the table, some chromosomes contain the gene for more than one of the seven characteristics Mendel studied, for example, seed color and flowers. The table Advanced Placement Biology Instructor’s Solution Manual 260 12 | Mendel's Experiments and Heredity below shows, with filled cells above the dashed diagonal line, the combinations of characteristics for which Mendel reported results. In the cells below the dotted diagonal line, identify with an X each cell where deviations from the law or laws identified in part B might be expected. D. Explain the reasons for the expected deviations for those combinations of characteristics identified in part C. E. In one of the experiments reported by Mendel, deviations from the law identified in part B might be expected. Explain how the outcomes of this experiment were consistent with Mendel’s laws. Solution Sample answer: A. Characters are always in one of two forms for these experiments. There is no progeny with intermediate characters. B. The correct choice is ii. Independent assortment requires that genes are not correlated. This can only be tested by considering the distribution of outcomes from more than one gene. C. D. Deviations are expected if the genes are on the same chromosome. E. Outcomes from the two-character experiment involving seed shape and color showed independent assortment, although they are on the same chromosome. Recombination through crossing over restores the independence. Advanced Placement Biology Instructor’s Solution Manual 12 | Mendel's Experiments and Heredity 43 261 A dihybrid cross involves two traits. A cross of parental types AaBb and AaBb can be represented with a Punnett square: This representation clearly organizes all of the possible genotypes and reveals the 9 : 3 : 3 : 1 distribution of phenotypes and a 4 × 4 grid of 16 cells. Expressed as a fraction of the 16 possible genotypes of the offspring, the phenotypic ratio describes the probability of each phenotype among the offspring: 3 (AA, Aa, aA) × 3 (BB, bB, Bb)/16 = 9/16 3 (AA, Aa, aA) × 1 (bb) /16 = 3/16 1 (aa) × 3 (BB, bB, Bb) = 3/16 1 (aa) × 1 (bb) = 1/16 A. Using the probability method, calculate the likelihood of these phenotypes from each dihybrid cross: Recessive in the gene with alleles A and a from the cross AaBb × aabb Dominant in both genes from the cross AaBb × aabb Recessive in both genes from the cross AaBb × aabb Recessive in either gene from the cross AaBb × aabb A Punnett square representation of a trihybrid cross, such as the self-cross of AaBbCc, is more cumbersome because there are eight columns and rows (2 × 2 × 2 ways to choose parental genotypes) and 64 cells. A less tedious representation is to calculate the number of each type of genotype in the offspring directly by counting the unique permutations of the letters representing the alleles. For example, the probability of the cross AaBbCc × AaBbCc is 3 (AA, Aa, aA) × 3 (BB, Bb, bB) × 3 (CC, Cc, cC)/64 = 27/64. Advanced Placement Biology Instructor’s Solution Manual 262 12 | Mendel's Experiments and Heredity B. Using the probability method, calculate the likelihood of these phenotypes from each trihybrid cross: Recessive in all traits from the cross AaBbCc × aabbcc Recessive in the gene with alleles C and c and dominant in the other two traits from the cross AaBbCc × AaBbCc Dominant in the gene with alleles A and a and recessive in the other two traits from the cross AaBbcc × AaBbCc C. The probability method is an easy way to calculate the likelihood of each particular phenotype, but it doesn’t simultaneously display the probability of all possible phenotypes. The forked line representation described in the text allows the entire phenotypic distribution to be displayed. Using the forked line method, calculate the probabilities in a cross between AABBCc and Aabbcc parents: Solution All traits are recessive: aabbcc. Traits are dominant at each loci, A? B? C? Traits are dominant at two genes and recessive at the third. Traits are dominant at one gene and recessive at the other two. Sample answer: A. Recessive in the gene with alleles A and a from the cross AaBb × aabb: 1 (aa) × 2 (bb or Bb) /4 = 1/2 Dominant in both genes from the cross AaBb × aabb: 1 (Aa) × 0 (bb) /4 + 1 (Bb) × 0 (aa) /4 = 0 Recessive in both genes from the cross AaBb × aabb: 1 (aa) × 1 (bb) /4 = 1/4 Recessive in either gene from the cross AaBb × aabb: [1 (aa) × 2 (Bb or bb) + 1 (Aa) × 2 (bb)]/4 = 1 Recessive in all traits from the cross AaBbCc × aabbcc: 1 (aa) × 1 (bb) × 1 (cc) /8 = 1/8 Recessive in the gene with alleles C and c and dominant in the other two traits from the cross AaBbCc × AaBbCc: 1 (cc) × 3 (AA, Aa, aA) ו 3 (BB, Bb, bB) = 9/64 Dominant in the gene with allele A and a and recessive in the other two traits from the cross AaBbcc × AaBbCc: 1 (bb) × 1 (cc) × 3 (AA, Aa, aA) /64 = 3/64 B. Advanced Placement Biology Instructor’s Solution Manual 12 | Mendel's Experiments and Heredity 263 C. We count the ways of particular phenotypes Having all recessive: aabbcc—just one way Having all dominant at each: A? B? C?—3 ways for each, such as AA, Aa, and aA—27 ways Having dominant at two genes and recessive at the third: A?B?cc, A?bbC?, and aaB?C? is 9 + 9 + 9 = 27 Having dominant at one gene and recessive at the other two: A?bbcc, aaB?cc, and aabbC? is 3 + 3 + 3 = 9 The probability of each phenotype is equal to the number of ways of making it from the genotypes divided by the total number of genotypes: All dominant: 27/64 = 0.42 Dominant A and recessive at B and C: 3/64 = 0.047 Dominant B and recessive at A and C: 3/64 = 0.047 Dominant C and recessive at A and B: 3/64 = 0.047 Dominant A and B and recessive at C: 9/64 = 0.14 Dominant A and C and recessive at B: 9/64 = 0.14 Dominant B and C and recessive at A: 9/64 = 0.14 All recessive: 1/64 = 0.0156 The sum of the probabilities of each outcome is equal to 1. 44 Construct a representation showing the connection between the process of meiosis and the transmission of six possible phenotypes from parents to F2 offspring. The phenotypes are labeled A, a, B, b and C, c. Expression of each phenotype is controlled by a separate Mendelian gene. Your representation should show the proportion of every possible combination of phenotypes (e.g., ABC, AbC, etc.) that will be present in the F2 offspring. Advanced Placement Biology Instructor’s Solution Manual 264 12 | Mendel's Experiments and Heredity Solution Advanced Placement Biology Instructor’s Solution Manual 13 | Modern Understandings of Inheritance PAGE \* MERGEFORMAT 277275 13| MODERN UNDERSTANDINGS OF INHERITANCE REVIEW QUESTIONS 1 When comparing humans (or in Drosophila), are X-linked recessive traits observed more frequently in males, in similar numbers between males and females, more frequently in females, or is the frequency different depending on the trait? A In more males than females B In more females than males C In males and females equally D In different distributions depending on the trait Solution 2 The solution is (A). Affected males result from mothers who are affected (homozygous) or who are carriers (heterozygous) for the X-linked recessive trait. Males possess only one X chromosome; therefore, males will always have the affected trait if obtained from a carrier or affected mother. Which recombination frequency corresponds to perfect linkage and violates the law of independent assortment? A 0 B 0.25 C 0.5 D 0.75 Solution 3 The solution is (A). The recombination frequency ranges from 0 percent (linked genes) to 50 percent (unlinked genes). Zero represents genes in close proximity to each other and segregated together, thus violating Mendel’s law of independent assortment. Which recombination frequency corresponds to independent assortment and the absence of linkage? A 0 B 0.25 C 0.5 D 0.75 l Advanced Placement Biology Instructor’s Solution Manu 266 13 | Modern Understandings of Inheritance Solution 4 The solution is (C). The 0.50 recombination frequency corresponds to the absence of linkage. Genes undergoing recombination assort independently of each other into one of two gamete cells. Based on the diagram, which statement is true? A Recombination of the body color and red/cinnabar eye alleles will occur more frequently than recombination of the alleles for wing length and aristae length. B Recombination of the body color and aristae length alleles will occur more frequently than recombination of red/brown eye alleles and the aristae length alleles. C Recombination of the gray/black body color and long/short aristae alleles will not occur. D Recombination of the red/brown eye and long/short aristae alleles will occur more frequently than recombination of the alleles for wing length and body color. Solution The solution is (D). Recombination of the red/brown eye and long/short aristae alleles will occur more frequently than recombination of the alleles for wing length and body color because of the greater map distance between the red/brown eye and long/short aristae. 5 Which code describes position 12 on the long arm of chromosome 13? A 13p12 B 13q12 C 12p13 D 12q13 Advanced Placement Biology Instructor’s Solution Manu 13 | Modern Understandings of Inheritance Solution PAGE \* MERGEFORMAT 277275 The solution is (B). 13q12 code describes position 12 on the long arm of chromosome 13. l Advanced Placement Biology Instructor’s Solution Manu 268 6 13 | Modern Understandings of Inheritance Assume a pericentric inversion occurred in one of two homologs prior to meiosis. The other homolog remains normal. During meiosis, what structure, if any, would these homologs assume in order to pair accurately along their lengths? A V formation B Cruciform C Loop D A pairing would not be possible. Solution The solution is (C). A loop is formed during a crossover between a chromosome with pericentric inversion and its normal homologous chromosome. This results in the formation of chromatids with deletions or entire segments missing. CRITICAL THINKING QUESTIONS 7 Which statement best describes the Chromosomal Theory of Inheritance? A The theory was proposed by Charles Darwin. It describes the units of inheritance between parents and offspring, as well as the processes by which those units control offspring development. B The theory was proposed by Boveri-Sutton. It describes linkage, recombination, and crossing over, and states that Mendelian genes have specific loci on chromosomes, which undergo segregation and independent assortment. C The theory was proposed by Charles Darwin. It states the Mendelian genes have two alternate forms and undergo independent assortment. It helped increase the understanding of linkage and recombination. D The theory was proposed by Boveri-Sutton. It describes the units of inheritance between parents and offspring as well as the processes by which those units control offspring development. Solution 8 The solution is (B). The theory was proposed by Boveri-Sutton. It describes linkage, recombination, and crossing over and states the Mendelian genes have specific loci on chromosomes, which undergo segregation and independent assortment. In a test cross for two characteristics (dihybrid cross), can the predicted frequency of recombinant offspring be 60 percent? Why or why not? A No, the predicted frequency of recombinant offspring ranges from 0 percent (for linked traits) to 50 percent (for unlinked traits) because of both parental and nonparental cases. B Yes, the predicted frequency of recombinant offspring can be 60 percent if genes are located very far from each other. Advanced Placement Biology Instructor’s Solution Manu 13 | Modern Understandings of Inheritance PAGE \* MERGEFORMAT 277275 C Yes, the predicted frequency can be 60 percent if crossing over occurs during every meiotic event. D No, the predicted frequency can never be 60 percent due to the presence of mutations such as deletions. Solution 9 The correct solution is (A). No. The predicted frequency of recombinant offspring ranges from 0 percent (for linked traits) to 50 percent (for unlinked traits). Which statement best describes how nondisjunction can result in an aneuploid zygote? A Nondisjunction only occurs when homologous chromosomes do not separate during meiosis I, resulting in the formation of gametes containing n + 1 and n − 1 chromosomes. B Nondisjunction only occurs when sister chromatids do not separate in meiosis II, resulting in the formation of gametes containing n + 1 and n − 1 chromosomes. C Nondisjunction is the failure of homologous chromosomes to separate during meiosis I or the failure of sister chromatids to separate during meiosis II, leading to the formation of n + 1/n − 1/n chromosomes. D Nondisjunction occurs when the sister chromatids fail to separate during mitosis II, resulting in the formation of gametes containing n + 1 and n − 1/n chromosomes. l Advanced Placement Biology Instructor’s Solution Manu 270 13 | Modern Understandings of Inheritance Solution 10 The solution is (C). Nondisjunction is the failure of homologous chromosomes to separate during meiosis I or the failure of sister chromatids to separate during meiosis II, leading to the formation of n + 1/n − 1/n chromosomes. Which answer correctly identifies the various chromosomal aberrations and their worstcase negative consequences? A Nondisjunction - aneuploid gametes; duplication - physical and mental abnormalities; deletion - lethal to a diploid organism; inversion - chromosomal breaks in gene; and translocations - effects depend on how positions of genes are altered B Nondisjunction - physical and mental abnormalities; inversion - genetic imbalance; duplication - aneuploid gametes; translocations - chromosomal breaks in the gene; and deletion - effects depend on how positions of genes are altered C Deletion - aneuploid gametes; translocations - physical and mental abnormalities; duplication - effects depend on positions of genes; nondisjunction - causes genetic imbalance lethal to a diploid organism; and aneuploidy - leads to various syndromes D Nondisjunction - chromosomal breaks in gene; duplication - physical and mental abnormalities; deletion - genetic imbalance lethal to a diploid organism; inversion aneuploid gametes; and translocations - effects depend on positions of genes Solution The solution is (A). Nondisjunction - aneuploid gametes; duplication - physical and mental abnormalities; deletion - lethal to a diploid organism; inversion chromosomal breaks in gene; translocations - effects depend on how positions of genes are altered. TEST PREP FOR AP® COURSES 11 The following figure represents a Drosophila linkage map for genes A–E. The numbers between the gene loci are the relative map units between each gene. Based on the linkage map, which two genes are most likely to segregate together? A Genes A and B B Genes B and C C Genes C and D D Genes D and E Solution The solution is (C). Genes C and D are at minimum distance from each other and, therefore, are most likely to segregate together. Advanced Placement Biology Instructor’s Solution Manu 13 | Modern Understandings of Inheritance 12 PAGE \* MERGEFORMAT 277275 A test cross was made between true-breeding EEWW flies and eeww flies. The resulting F1 generation was then crossed with eeww flies. One hundred offspring in the F2 generation were examined, and it was discovered that the E and W genes were not linked. Which is the correct genotype of the F2 offspring if the genes were linked and if the genes were NOT linked? A Linked: 50 percent EEWW and 50 percent eeww; not linked: 25 percent EeWw, 25 percent, Eeww 25 percent eeWw, and 25 percent eeww B Linked: 25 percent Eeww, 50 percent eeWw; not linked: parental genotypes EeWw and eeww C Linked: genotypes EeWw and eeww, and recombinant genotypes Eeww and eeWw in the F2 generation are nearly the same irrespective of their linkage D Linked: mostly with parental genotypes, Eeww and eeWw; unlinked: 25 percent EeWw and eeww with 75 percent Eeww and eeWw Solution 13 The solution is (A). The predicted genotypes are (parental): EEWW, eeww; (recombinants): Eeww, eeWw. If the genes are not linked, each genotype should be found in 25 progeny. If the genes were linked, there would be more progeny with the parental genotypes than the recombinant genotypes. A cross was made with true-breeding AABB flies and true-breeding aabb flies. The resulting F1 generation then was crossed with true-breeding aabb flies. Based on the linkage map, which F2 generation genotype ratio is most likely to be observed? A Number observed: AaBb (46), Aabb (4), aaBb (4), Aabb (46) B Number observed: AaBb (4), Aabb (46), aaBb (46), Aabb (4) C Number observed: AaBb (25), Aabb (25), aaBb (25), Aabb (25) D Number observed: AaBb (50), Aabb (0), aaBb (0), Aabb (50) Solution The solution is (C). This is a case of test cross. Therefore, the ratio obtained will be 1 : 1 : 1 : 1 (25 : 25 : 25 : 25). l Advanced Placement Biology Instructor’s Solution Manu 272 14 13 | Modern Understandings of Inheritance Which symptom is most likely associated with the disorder shown in the karyotype? A Lethality B Infertility C Heart and bleeding defects D Short stature and stunted growth Solution 15 The solution is (B). The karyotype represents the 47 XXY condition, known as Klinefelter syndrome. This syndrome leads to infertility. Which option describes the disorder shown in the karyotype and the social, ethical, or medical issue related to the disorder? Advanced Placement Biology Instructor’s Solution Manu 13 | Modern Understandings of Inheritance PAGE \* MERGEFORMAT 277275 A Down syndrome (47 XY +21) causes intellectual disability, vision problems, congenital heart disease, and susceptibility to cancer. Healthcare providers often do not discuss the positive aspects of raising a child with Down syndrome and often provide out-ofdate information. B Klinefelter syndrome (47 XXY) causes intellectual disability, vision problems, congenital heart disease, and susceptibility to cancer. Arguments often are made against abortion of an affected fetus. C Klinefelter syndrome (47 XXY) causes sterility and reduced testosterone production. Arguments often are made against informing insurance companies about a diagnosis of this disease. D Down syndrome (47 XY +21) causes sterility and lower testosterone production. Arguments often are made against informing insurance companies about a diagnosis of this disease. Solution 16 The solution is (C). The karyotype shows Klinefelter syndrome, with 47 XXY. Medical issues related to the disorder include sterility, and lower testosterone production. An ethical issue concerned with the disorder is whether information about genetic diagnosis should be given to insurance companies. Which gene order is the most likely outcome of an inversion mutation in the chromosome shown? A RSTUV B RRSTUV C RSUV D RTSUV Solution The solution is (D). RTSUV is most likely the outcome of an inversion mutation. Inversion leads to a 180-degree rotation of the chromosomal segment, leading to abnormal gene order. l Advanced Placement Biology Instructor’s Solution Manu 274 17 13 | Modern Understandings of Inheritance With the help of the diagram given, which statement most appropriately describes nondisjunction and its genetic consequences? A Nondisjunction occurs when a homologous pair is unable to separate during meiosis I, resulting in the formation of gametes containing n + 1 and n − 1 chromosomes. This is called aneuploidy. B Nondisjunction occurs due to the inability of sister chromatids to separate during meiosis II, resulting in the formation of gametes containing n + 1 and n − 1 chromosomes. This results in heart and bleeding defects. C Nondisjunction is the failure of homologous chromosomes to separate during meiosis I or failure of sister chromatids to separate during meiosis II. This results in aneuploid gametes. D Nondisjunction occurs when a pair of homologous chromosomes fails to segregate during meiosis II resulting in the formation of gametes containing n + 1, n − 1, or n numbers of chromosomes. This results in abnormal growth patterns. Solution The solution is (C). Nondisjunction occurs when a pair of homologous chromosomes fails to separate during meiosis I or sister chromatids fail to separate during meiosis II. Nondisjunction leads to individuals containing aneuploid gametes. Advanced Placement Biology Instructor’s Solution Manu 13 | Modern Understandings of Inheritance 18 PAGE \* MERGEFORMAT 277275 Klinefelter syndrome is — A more severe than Down syndrome due to gene deletions in Klinefelter syndrome B more severe than Down syndrome due to trisomy in Klinefelter syndrome C less severe than Down syndrome due to monosomy in Down syndrome D less severe than Down syndrome due to X-inactivation in this disorder Solution The solution is (D). Klinefelter syndrome is less severe as compared to Down syndrome. In Klinefelter syndrome, the extra copy of the X chromosome is inactivated. In Down syndrome, an extra copy of the chromosome 21 means individuals have three copies of each chromosome 21 gene instead of two, making it difficult for cells to properly control how much protein is made. Producing too much or too little protein can have serious consequences. SCIENCE PRACTICE CHALLENGE QUESTIONS 13.1 Chromosomal Theory and Genetic Linkages 19 Drosophila that are true breeding for the traits straight wings (S) and red eyes (R) are crossed with flies that are true breeding for curved wings (s) and brown eyes (r). A test cross then is made between the offspring and the true-breeding ssrr flies. A. Use the symbols S, s, R, and r to construct a representation of the parental genotypes in the test cross. B. If these genes are located on different chromosomes, use a Punnett square to construct a representation of the offspring of the test cross. C. Predict the distribution of genotypes and phenotypes resulting from the test cross. D. As it happens, these genes are both on chromosome II as shown below. Use the symbols S, s, R, and r to construct a representation of the parental and recombinant genotypes in the test cross. E. Suppose that 500 flies are produced in the test cross. Apply mathematical methods to calculate the expected number of recombinant offspring using the linear map units (LMU) shown in the diagram. l Advanced Placement Biology Instructor’s Solution Manu 276 13 | Modern Understandings of Inheritance Solution Sample answer: A. The F1 generation is SSRR × ssrr to give SsRr for one parent in the test cross. The other is ssrr. B. sr Description SR SsRr Long wing, red eye Sr Ssrr Long wing, brown eye sR ssRr Short wing, red eye blank sr ssrr Short wing, brown eye C. The genotypes and phenotypes will be 1 : 1 : 1 : 1. D. The parentals are SsRr and ssrr. The recombinants are ssRr and Ssrr. E. The distance in linkage map units is 29, so the recombinant frequency is 0.29. So, 145 recombinant offspring are expected. 20 Studies like the one described in question AP12.1 were carried out by Morgan and Sturtevant beginning in 1911. The discovery of linkage was made by Bateson and Punnett in 1900. They crossed a true-breeding purple (P) plant with long seeds (L) with a truebreeding red (r) plant with round seeds (l). They then performed a self-cross between the F1 generation. They obtained the F2 data shown in the table. Phenotype Genotype(s) Observed Purple, long 4,831 Purple, round 390 Red, long 393 Red, round 1,338 Expected Advanced Placement Biology Instructor’s Solution Manu 13 | Modern Understandings of Inheritance Phenotype PAGE \* MERGEFORMAT 277275 Genotype(s) Observed Total Expected 6,952 A. Use the symbols P, p, L, and l to construct a representation of the F2 genotypes and complete the second column in the table. B. Complete the fourth column table above by recording values of the predicted numbers of plants with each genotype. C. Apply a 2 test at the 95 percent confidence level to evaluate the claim that these data confirm linkage. The definition of the statistic 2 (o e)2 e and this table are provided at the AP Biology Exam. P 1 2 3 4 5 6 7 8 0.05 3.84 5.99 7.82 9.49 11.07 12.59 14.07 15.51 0.01 6.64 9.32 11.34 13.28 15.09 16.81 Degrees of Freedom 18.48 20.09 D. At first Bateson and Punnett did not see that these genes are located on the same chromosome and proceed to measure the linkage distance between them, taking the first step toward creating a gene map. Justify a selection of data and the procedure from which data could be collected that would have provided the necessary evidence to confirm linkage and recombination. Solution Sample answer: A. and B. Phenotype Genotype(s) Observed Expected Purple, long PpLl, PpLL, PPLl, PPLL 4,831 3,910.5 Purple, round Ppll and PPll 390 1,303.5 Red, long ppLl and ppLL 393 1,303.5 Red, round ppll 1,338 434.5 Total 6,952 C. The result is clearly larger than the critical value of 7.82 (3 classes). (4,831 3,910.5)2 (390 1,303.5)2 (393 1,303.5)2 (1,338 434.5)2 3,910.5 1,303.5 1,303.5 434.5 217 640 636 1,879 2 D. They needed to do a test cross of the F1 generation with the homozygous recessive. Then they could have determined the fraction of the offspring with l Advanced Placement Biology Instructor’s Solution Manu 278 13 | Modern Understandings of Inheritance parental (PpLl and ppll) and recombinant (ppLl and Ppll) genotypes. From this they might have made the leap that young Sturtevant made. 21 Review the observations that provided researchers with evidence in support of the Chromosomal Theory of Inheritance. A. Evaluate the dependence of these observations on improvements in a critical technology during the period from 1850 to 1940. Identify this technology and describe how this technology allowed scientists to make the connection between chromosomes and genes. (As a hint, the name chromosome is taken from the Greek word chroma, which means “colored” or “stained.”) B. Mendel’s laws of inheritance are explained by the chromosomal theory. Use these observations to justify: Solution The law of segregation The law of independent assortment Sample answer: A. The evidence is microscopic. All of the observations identified in the text can only be made with the microscope: 1. During meiosis, homologous chromosome pairs migrate as discrete structures that are independent of other chromosome pairs. 2. The sorting of chromosomes from each homologous pair into pre-gametes appears to be random. 3. Each parent synthesizes gametes that contain only half of their chromosomal complement. 4. Even though male and female gametes (sperm and egg) differ in size and morphology, they have the same number of chromosomes, suggesting equal genetic contributions from each parent. 5. The gametic chromosomes combine during fertilization to produce offspring with the same chromosome number as their parents. B. Points (1) and (3) explain the law of segregation. Points (2) and (1) explain independent assortment. 22 Errors in the transmission of genetic information to future generations are essential. Otherwise, organisms could not evolve over time. Some errors in the synthesis of new DNA during S phase in either meiosis or mitosis are not repaired. These errors usually involve single nucleotides. Errors that occur during prophase I of meiosis that are not corrected can involve the exchange of sequences between homologous chromosomes (duplications) or even nonhomologous chromosomes (translocations). Duplications usually are retained, and the organism remains viable without a change in phenotype. Translocations are usually lethal or significantly alter phenotype. In eukaryotes, Advanced Placement Biology Instructor’s Solution Manu 13 | Modern Understandings of Inheritance PAGE \* MERGEFORMAT 277275 duplications and the shuffling of parental genes through recombination are important sources of variation. Construct an explanation of the role of duplication as a source of raw material for future mutations and selection and contrast this type of variation with recombination. Solution 23 Sample answer: Duplication of a gene allows a second copy to undergo mutation independently of the original gene. As the two diverge, new functions result which provides a gradual (but not so gradual as recombination) means of change. If only recombination was the source of variation, then pairs of genes that are very close together would remain parental and never vary. Bacteria and Archaea reproduce asexually, and genetic material is in a closed loop. In both domains, genetic material is transferred horizontally, and polyploidy is common. Polyploidy is common in plants and occurs in invertebrate animals but is less common in vertebrates. In all domains, multiple copies of genes (gene duplication) are common. Based on this information, compare and contrast the mechanisms that provide genetic variation in the three domains: Bacteria, Archaea, and Eukarya. Solution Sample answer: Multiple copies of the genome—like duplication—provide raw material for the development of an alternative set of proteins. The rapidity of reproduction and development in unicellular organisms combined with multiple copies of the genome increases the frequency of presentation of new genotypes for selection. Horizontal gene transfer further increases the shuffling of information— even originating with other species. l Advanced Placement Biology Instructor’s Solution Manu l Ошибка! Неизвестный аргумент ключа. o 14 | DNA Structure and Functi 14 | DNA STRUCTURE AND FUNCTION REVIEW QUESTIONS 1 Who was the first person to isolate the material that came to be known as nucleic acids? A Frederick Griffith B Friedrich Miescher C James Watson D Oswald Avery Solution 2 The solution is (B). DNA was first isolated from white blood cells by Friedrich Miescher. What is bacterial transformation? A The transformation of a bacterium occurs during replication. B It is the transformation of a bacterium into a pathogenic form. C Transformation of bacteria involves changes in its chromosome. D Transformation is a process in which external DNA is taken up by a cell, thereby changing morphology and physiology. Solution 3 The solution is (D). Transformation is a process in which external DNA is taken up by a cell, thereby changing morphology and physiology. What type of nucleic acid material is analyzed the most frequently in forensics cases? A Cytoplasmic rRNA B Mitochondrial DNA C Nuclear chromosomal DNA D Nuclear mRNA Solution 4 The solution is (C). Forensics looks at the nuclear genetic material. The experiments by Hershey and Chase helped confirm that DNA was the hereditary material on the basis of the finding of what? A Radioactive phages were found in the pellet. B Radioactive cells were found in the supernatant. C Radioactive sulfur was found inside the cell. D Radioactive phosphorus was found in the cell. ******************************************************** 14 | DNA Structure and Function Solution 5 PAGE \* MERGEFORMAError! Unknown switch argument.*** The solution is (D). Radioactive phosphorous was found in the heavier particles that settled as pellets. The heavier bacterial cells settled down and formed pellets. If DNA of a particular species was analyzed and it was found that it contains 27 percent A, what would be the percentage of T? A 23% B 27% C 30% D 54% Solution 6 The solution is (B). Because A binds to T, there is the same proportion of A and T in each DNA molecule. If the sequence of the 5′ to 3′ strand is AATGCTAC, then the complementary sequence has which sequence? A 3′-AATGCTAC-5′ B 3′-CATCGTAA-5′ C 3′-TTACGATG-5′ D 3′-GTAGCATT-5′ Solution 7 The solution is (C). A binds to T and C binds to G in DNA molecules. The DNA double helix does NOT have what? A Antiparallel configuration B Complementary base pairing C Major and minor grooves D Uracil Solution 8 The solution is (D). Uracil is a nucleotide base that is present in RNA and not in DNA. What is a purine? A A double-ring structure with a six-membered ring fused to a five-membered ring B A single six-membered ring C A six-membered ring D Three phosphates covalently bonded by phosphodiester bonds Solution 9 The solution is (A). A double-ring structure with a six-membered ring fused to a fivemembered ring. What is the name of the method developed by Fred Sanger to sequence DNA? A Dideoxy chain termination ******************************************************** l Ошибка! Неизвестный аргумент ключа. o B Double helix determination ******************************************************** 14 | DNA Structure and Functi 14 | DNA Structure and Function PAGE \* MERGEFORMAError! Unknown switch argument.*** C Polymerase chain reaction D Polymer gel electrophoresis Solution 10 The solution is (A). The Dideoxy Chain Termination method was developed by Fred Sanger. It is a sequencing method based on the use of chain terminators. What happens when a dideoxynucleotide is added to a developing DNA strand? A The chain extends to the end of the DNA strand. B The DNA stand is duplicated. C The chain is not extended any further. D The last codon is repeated. Solution 11 The solution is (C). If a ddNTP is added to a growing a DNA strand, the chain is not extended any further because the free 3′OH group needed to add another nucleotide is not available. In eukaryotes, what is DNA wrapped around? A Histones B Polymerase C Single-stranded binding proteins D Sliding clamp Solution 12 The solution is (A). In eukaryotes, the DNA is wrapped around proteins known as histones to form structures called nucleosomes. Which enzyme is only found in prokaryotic organisms? A DNA gyrase B Helicase C Ligase D Telomerase Solution 13 The solution is (A). DNA gyrase helps to maintain the supercoiled structure in prokaryotes. Where is uracil found? A Chromosomal DNA B Helicase C Mitochondrial DNA D mRNA ******************************************************** l Ошибка! Неизвестный аргумент ключа. o Solution 14 14 | DNA Structure and Functi The solution is (D). Uracil is a nitrogenous base found in mRNA molecules. Its complementary base pair is adenine. What prevents the further development of a DNA strand in Sanger sequencing? A The addition of DNA reductase B The addition of dideoxynucleotides C The elimination of DNA polymerase D The addition of uracil Solution 15 The solution is (B). If a ddNTP is added to a growing DNA strand, the chain is not extended any further because the free 3′OH group needed to add another nucleotide is not available. ddNTP lack the 3′OH group on the five-carbon sugar. What is NOT one of the proteins involved during the formation of the replication fork? A Helicase B Ligase C Origin of replication D Single-stranded binding proteins Solution 16 The solution is (C). The origin of replication is the point at which the DNA unwinds. In which direction does DNA replication take place? A 5′ to 3′ B 3′ to 5′ C 5′ D 3′ Solution 17 The solution is (A). DNA polymerase adds nucleotides from 5′ to 3′ direction. Meselson and Stahl’s experiments proved that DNA replicates by which mode? A Conservative B Converse C Dispersive D Semiconservative Solution The solution is (D). The semiconservative mode of replication suggested that each of the two parental DNA strands act as a template for new DNA to be synthesized. After replication, each double-stranded DNA includes one parental or “old” strand and one “new” strand. ******************************************************** 14 | DNA Structure and Function PAGE \* MERGEFORMAError! Unknown switch argument.*** ******************************************************** l Ошибка! Неизвестный аргумент ключа. o 18 14 | DNA Structure and Functi Which set of results was found in Meselson and Stahl’s experiments? A The original chromosome was kept intact and a duplicate was made. B The original chromosome was split and half went to each duplicate. C The original chromosome was mixed with new material and each duplicate strand contained both old and new. D The original chromosome was used as a template for two new chromosomes and discarded. Solution 19 The solution is (B). The semiconservative method suggested that each of the two parental DNA strands act as a template for new DNA to be synthesized. After replication, each double-stranded DNA includes one parental or “old” strand and one “new” strand. It was found in Meselson and Stahl’s experiment. Which enzyme initiates the splitting of the double DNA strand during replication? A DNA gyrase B Helicase C Ligase D Telomerase Solution 20 The solution is (B). Helicase opens the DNA helix by breaking hydrogen bonds between the nitrogenous bases ahead of the replication fork. Which enzyme is most directly responsible for the main process of producing a new DNA strand? A DNA pol I B DNA pol II C DNA pol III D DNA pol I, DNA pol II, and DNA pol III Solution 21 The solution is (C). DNA polymerase III is the main enzyme in DNA replication that adds nucleotides in 5’ to 3’ direction. Which portion of a chromosome contains Okazaki fragments? A Helicase B Lagging strand C Leading strand D Primer ******************************************************** 14 | DNA Structure and Function Solution PAGE \* MERGEFORMAError! Unknown switch argument.*** The solution is (B). The replication of the strand that occurs in a direction away from the replication fork is known as lagging strand and contains Okazaki fragments. ******************************************************** l Ошибка! Неизвестный аргумент ключа. o 22 14 | DNA Structure and Functi What does the enzyme primase synthesize? A DNA primer B Okazaki fragments C Phosphodiester linkage D RNA primer Solution 23 The solution is (D). Primase synthesizes RNA primers to initiate synthesis by DNA polymerase, which can add nucleotides only in the 5′ to 3′ direction. The ends of the linear chromosomes are maintained by what? A DNA polymerase B Helicase C Primase D Telomerase Solution 24 The solution is (D). Telomerase maintains the linear end of the chromosomes. It consists of a catalytic part and a built-in RNA template. What is the difference in the rate of replication of nucleotides between prokaryotes and eukaryotes? A Eukaryotes are 50 times slower. B Eukaryotes are 20 times faster. C Prokaryotes are 100 times slower. D Prokaryotes are 10 times faster. Solution 25 The solution is (D). The advantage in prokaryotes is that RNA and protein synthesis occurs much more quickly than the eukaryotes. It is 10 times faster than eukaryotes. What are Autonomously Replicating Sequences (ARS)? A Areas of prokaryotic chromosomes that initiate copying B Portions of prokaryotic chromosomes that can be transferred from one organism to another C Areas of eukaryotic chromosomes that are equivalent to the origin of replication in E. coli D Portions of eukaryotic chromosomes that replicate independent of the parent chromosome Solution The solution is (C). Areas of eukaryotic chromosomes that are equivalent to the origin of replication in E. coli. ******************************************************** 14 | DNA Structure and Function 26 PAGE \* MERGEFORMAError! Unknown switch argument.*** What type of body cell does NOT exhibit telomerase activity? A Adult stem cells B Embryonic cells C Germ cells D Liver cells Solution 27 The solution is (D). Liver cells do not exhibit telomerase activity. Adult, embryonic, and germ cells do. During proofreading, which enzyme reads the DNA? A DNA polymerase B Helicase C Topoisomerase D Primase Solution 28 The solution is (A). DNA polymerase edits DNA by proofreading every newly added base. If a prokaryotic cell is replicating nucleotides at a rate of 100 per second, how fast would a eukaryotic cell be replicating nucleotides? A 1,000 per second B 100 per second C 10 per second D 1 per second Solution 29 The solution is (C). Rate of replication in prokaryotes is 10 times than eukaryotes. If replication in prokaryotes is 100 per second, then in eukaryotes it will be 10 per second. Which type of point mutation would have no effect on gene expression? A Frameshift B Missense C Nonsense D Silent Solution The solution is (D). Silent mutation is the one in which a nucleotide is substituted, but there is no effect on the protein sequence or gene expression. ******************************************************** l Ошибка! Неизвестный аргумент ключа. o 30 14 | DNA Structure and Functi Which type of point mutation would result in the substitution of a stop codon for an amino acid? A Frameshift B Missense C Nonsense D Silent Solution 31 The solution is (C). Nonsense mutation introduces a stop codon in place of an amino acid. A woman has developed skin cancer and she is pregnant. She is worried that her child will be born with the cancer she has while carrying the baby. Should she be worried? A Yes, the cancer can spread to the baby. B No, the mutations causing the cancer are in somatic cells, not reproductive germ cells. C Yes, the mutations can be passed on to the child through the placenta. D No, UV light only affects adult, somatic cells. Solution 32 The solution is (B). The mutation causing the cancer does not occur in the germ cells of the woman, but in the somatic cells. Therefore, it will not affect the baby. What is the initial mechanism for repairing nucleotide errors in DNA? A DNA polymerase proofreading B Mismatch repair C Nucleotide excision repair D Thymine dimers Solution 33 The solution is (A). DNA polymerase is an efficient enzyme, but can make mistakes while adding nucleotides during replication. It edits the DNA by proofreading every newly added base. Nucleotide excision repair often is employed when UV exposure causes the formation of what? A Phosphodiester bonds B Purine conjugates C Pyrimidine dimers D Tetrad disassembly ******************************************************** 14 | DNA Structure and Function Solution PAGE \* MERGEFORMAError! Unknown switch argument.*** The solution is (C). Thymine and cytosine are pyrimidine dimers. On long exposure of UV rays, thymine dimers are formed, which puts people at higher risk of developing skin cancer. Nucleotide excision repair is a DNA repair mechanism that excises the thymine dimers in normal individuals. CRITICAL THINKING QUESTIONS 34 Explain Griffith’s transformation experiments. What did he conclude from them? A Two strains of S. pneumoniae were used for the experiment. Griffith injected a mouse with heat-inactivated S strain (pathogenic) and R strain (nonpathogenic). The mouse died and S strain was recovered from the dead mouse. He concluded that external DNA is taken up by a cell that changed morphology and physiology. B Two strains of Vibrio cholerae were used for the experiment. Griffith injected a mouse with heat-inactivated S strain (pathogenic) and R strain (nonpathogenic). The mouse died and S strain was recovered from the dead mouse. He concluded that external DNA is taken up by a cell that changed morphology and physiology. C Two strains of S. pneumoniae were used for the experiment. Griffith injected a mouse with heat-inactivated S strain (pathogenic) and R strain (nonpathogenic). The mouse died and R strain was recovered from the dead mouse. He concluded that external DNA is taken up by a cell that changed morphology and physiology. D Two strains of S. pneumoniae were used for the experiment. Griffith injected a mouse with heat-inactivated S strain (pathogenic) and R strain (nonpathogenic). The mouse died and S strain was recovered from the dead mouse. He concluded that mutation occurred in the DNA of the cell that changed morphology and physiology. Solution 35 The solution is (A). Two strains of S. pneumoniae were used in Griffith’s transformation experiments. The R strain is nonpathogenic. The S strain is pathogenic and causes death. When Griffith injected a mouse with the heatinactivated S strain and a live R strain, the mouse died. The S strain was recovered from the dead mouse. Thus, Griffith concluded that something had passed from the heat-killed S strain to the R strain, transforming the R strain into S strain in the process. Which answer best explains why radioactive sulfur and phosphorus were used to label bacteriophages in the Hershey and Chase experiments? A Protein was labeled with radioactive sulfur and DNA was labeled with radioactive phosphorus. Phosphorus is found in DNA, so it will be tagged by radioactive phosphorus. B Protein was labeled with radioactive phosphorus and DNA was labeled with radioactive sulfur. Phosphorus is found in DNA, so it will be tagged by radioactive phosphorus. ******************************************************** l Ошибка! Неизвестный аргумент ключа. o 14 | DNA Structure and Functi C Protein was labeled with radioactive sulfur and DNA was labeled with radioactive phosphorus. Phosphorus is found in DNA, so DNA will be tagged by radioactive sulfur. D Protein was labeled with radioactive phosphorus and DNA was labeled with radioactive sulfur. Phosphorus is found in DNA, so DNA will be tagged by radioactive sulfur. Solution 36 The solution is (A). Hershey and Chase labeled one batch of phage with radioactive sulfur, 35S, to label the protein coat. Another batch of phage was labeled with radioactive phosphorus, 32P. Because phosphorus is found in DNA, but not protein, the DNA and not the protein would be tagged with radioactive phosphorus. How can Chargaff’s rules be used to identify different species? A The amount of adenine, thymine, guanine, and cytosine varies from species to species and is not found in equal quantities. They do not vary between individuals of the same species and can be used to identify different species. B The amount of adenine, thymine, guanine, and cytosine varies from species to species and is found in equal quantities. They do not vary between individuals of the same species and can be used to identify different species. C The amount of adenine and thymine is equal to guanine and cytosine and is found in equal quantities. They do not vary between individuals of the same species and can be used to identify different species. D The amount of adenine, thymine, guanine, and cytosine varies from species to species and is not found in equal quantities. They vary between individuals of the same species and can be used to identify different species. Solution 37 In the Avery, Macleod, and McCarty experiments, what conclusion would the scientists have drawn if the use of proteases prevented the transformation of R strain bacteria? Solution 38 The solution is (A). The content of DNA is different indifferent species and the amounts of adenine, thymine, guanine, and cytosine are found in different quantities. Therefore, the amounts of adenine, thymine, guanine, and cytosine are consistent for a species and can be used to identify that species. The conclusion would be that proteins are the heritable material in cells instead of nucleic acids. Describe the structure and complementary base pairing of DNA. A DNA is made up of two strands that are twisted around each other to form a helix. Adenine pairs up with thymine and cytosine pairs with guanine. The two strands are antiparallel in nature; that is, the 3′ end of one strand faces the 5′ end of the other strand. Sugar, phosphate, and nitrogenous bases contribute to the DNA structure. ******************************************************** 14 | DNA Structure and Function PAGE \* MERGEFORMAError! Unknown switch argument.*** B DNA is made up of two strands that are twisted around each other to form a helix. Adenine pairs up with cytosine and thymine pairs with guanine. The two strands are antiparallel in nature; that is, the 3′ end of one strand faces the 5′ end of the other strand. Sugar, phosphate, and nitrogenous bases contribute to the DNA structure. C DNA is made up of two strands that are twisted around each other to form a helix. Adenine pairs up with thymine and cytosine pairs with guanine. The two strands are parallel in nature; that is, the 3′ end of one strand faces the 3′ end of the other strand. Sugar, phosphate, and nitrogenous bases contribute to the DNA structure. D DNA is made up of two strands that are twisted around each other to form a helix. Adenine pairs up with thymine and cytosine pairs with guanine. The two strands are antiparallel in nature; that is, the 3′ end of one strand faces the 5′ end of the other strand. Only sugar contributes to the DNA structure. Solution 39 The solution is (A). DNA is made up of two strands that are twisted around each other to form a right-handed helix. Base pairing takes place between a purine and pyrimidine; namely, A pairs with T and G pairs with C. Adenine and thymine are complementary base pairs, and cytosine and guanine are also complementary base pairs. The base pairs are stabilized by hydrogen bonds; adenine and thymine form two hydrogen bonds and cytosine and guanine form three hydrogen bonds. The two strands are anti-parallel in nature; that is, the 3′ end of one strand faces the 5′ end of the other strand. The sugar and phosphate of the nucleotides form the backbone of the structure, whereas the nitrogenous bases are stacked inside. Which answer provides a brief summary of the Sanger sequencing method? A Frederick Sanger’s sequencing is a chain termination method that is used to generate DNA fragments that terminate at different points using dye-labeled dideoxynucleotides. DNA is separated by electrophoresis on the basis of size. The DNA sequence can be read out on an electropherogram generated by a laser scanner. B Frederick Sanger’s sequencing is a chain elongation method that is used to generate DNA fragments that elongate at different points using dye-labeled dideoxynucleotides. DNA is separated by electrophoresis on the basis of size. The DNA sequence can be read out on an electropherogram generated by a laser scanner. C Frederick Sanger’s sequencing is a chain termination method that is used to generate DNA fragments that terminate at different points using dye-labeled dideoxynucleotides. DNA is joined together by electrophoresis on the basis of size. The DNA sequence can be read out on an electropherogram generated by a laser scanner. D Frederick Sanger’s sequencing is a chain termination method that is used to generate DNA fragments that terminate at different points using dye-labeled dideoxynucleotides. DNA is separated by electrophoresis on the basis of size. The DNA sequence can be read out on an electropherogram generated by a magnetic scanner. ******************************************************** l Ошибка! Неизвестный аргумент ключа. o Solution 40 14 | DNA Structure and Functi The solution is (A). In Frederick Sanger's dideoxy chain termination method, dyelabeled dideoxynucleotides are used to generate DNA fragments that terminate at different points. The DNA is separated by capillary electrophoresis on the basis of size, and from the order of fragments formed, the DNA sequence can be read. The DNA sequence readout is shown on an electropherogram that is generated by a laser scanner. Compare and contrast the similarities and differences between eukaryotic and prokaryotic DNA. A Eukaryotes have a single, circular chromosome, while prokaryotes have multiple, linear chromosomes. Prokaryotes pack their chromosomes by super coiling, managed by DNA gyrase. Eukaryote chromosomes are wrapped around histone proteins that create heterochromatin and euchromatin, which is not present in prokaryotes. B Prokaryotes have a single, circular chromosome, while eukaryotes have multiple, linear chromosomes. Prokaryotes pack their chromosomes by super coiling, managed by DNA gyrase. Eukaryote chromosomes are wrapped around histone proteins that create heterochromatin and euchromatin, which is not present in prokaryotes. C Prokaryotes have a single, circular chromosome, while eukaryotes have multiple, linear chromosomes. Eukaryotes pack their chromosomes by super coiling, managed by DNA gyrase. Prokaryote chromosomes are wrapped around histone proteins that create heterochromatin and euchromatin, which is not present in eukaryotes. D Prokaryotes have a single, circular chromosome, while eukaryotes have multiple, linear chromosomes. Prokaryotes pack their chromosomes by super coiling, managed by DNA gyrase. Eukaryote chromosomes are wrapped around histone proteins that create heterochromatin and euchromatin, which is present in prokaryotes. Solution 41 The solution is (B). Prokaryotes have a single, circular chromosome, while eukaryotes have multiple, linear chromosomes. Prokaryotes pack their chromosomes into the cell using super coiling, managed DNA gyrase. Eukaryote chromosomes are wrapped around histone proteins that create heterochromatin and euchromatin that is not found in prokaryotic cells. DNA replication is bidirectional and discontinuous. How can you explain your understanding of those concepts? A DNA polymerase reads the template strand in the 3′ to 5′ direction and adds nucleotides only in the 5′ to 3' direction. The leading strand is synthesized in the direction of the replication fork. Replication on the lagging strand occurs in the direction away from the replication fork in short stretches of DNA called Okazaki fragments. B DNA polymerase reads the template strand in the 5′ to 3′ direction and adds nucleotides only in the 5′ to 3′ direction. The leading strand is synthesized in the ******************************************************** 14 | DNA Structure and Function PAGE \* MERGEFORMAError! Unknown switch argument.*** direction of the replication fork. Replication on the lagging strand occurs in the direction away from the replication fork in short stretches of DNA called Okazaki fragments. C DNA polymerase reads the template strand in the 3′ to 5′ direction and adds nucleotides only in the 5′ to 3′ direction. The leading strand is synthesized in the direction away from the replication fork. Replication on the lagging strand occurs in the direction of the replication fork in short stretches of DNA called Okazaki fragments. D DNA polymerase reads the template strand in the 5′ to 3′ direction and adds nucleotides only in the 3′ to 5′ direction. The leading strand is synthesized in the direction of the replication fork. Replication on the lagging strand occurs in the direction away from the replication fork in long stretches of DNA called Okazaki fragments. Solution 42 The solution is (A). DNA polymerase can add nucleotides only in the 5’ to 3’ direction. DNA polymerase recognizes the 3’OH end as its landing site; thus, polymerase “reads” the template strand in the 3’ to 5’ direction and builds the new DNA complementary DNA polymer in the 5’ to 3’ direction. One strand—called the leading strand—is synthesized continuously in the direction of the replication fork (the direction in which helicase is separating the two strands), with polymerase adding new nucleotides one-by-one. However, replication of the other strand— called the lagging strand—occurs in a direction away from the replication fork, in short stretches of DNA known as Okazaki fragments. How did the scientific community learn that DNA replication takes place in a semiconservative fashion? A Meselson and Stahl experimented with E. coli. DNA grown in 15N was heavier than DNA grown in 14N. When DNA in 15N was switched to 14N media, DNA sedimented halfway between the 15N and 14N levels after one round of cell division, indicating 50 percent presence of 14N. This supports the semiconservative replication model. B Meselson and Stahl experimented with S. pneumonia. DNA grown in 15N was heavier than DNA grown in 14N. When DNA in 15N was switched to 14N media, DNA sedimented halfway between the 15N and 14N levels after one round of cell division, indicating 50 percent presence of 14N. This supports the semiconservative replication model. C Meselson and Stahl experimented with E. coli. DNA grown in 14N was heavier than DNA grown in 15N. When DNA in 15N was switched to 14N media, DNA sedimented halfway between the 15N and 14N levels after one round of cell division, indicating 50 percent presence of 14N. This supports the semiconservative replication model. D Meselson and Stahl experimented with S. pneumonia. DNA grown in 15N was heavier than DNA grown in 14N. When DNA in 15N was switched to 14N media, DNA sedimented halfway between the 15N and 14N levels after one round of cell division, indicating complete presence of 14N. This supports the semiconservative replication model. ******************************************************** l Ошибка! Неизвестный аргумент ключа. o Solution 43 14 | DNA Structure and Functi The solution is (A). Meselson and Stahl experimented with E. coli grown first in heavy nitrogen (15N) then in 14N. DNA grown in 15N is heavier than DNA grown in 14N, and sediments to a lower level in cesium chloride solution in an ultracentrifuge. When DNA grown in 15N is switched to media containing 14N, after one round of cell division the DNA sediments halfway between the 15N and 14N levels, indicating that it now contains 50 percent 14N. In subsequent cell divisions, an increasing amount of DNA contains 14N only. These data support the semiconservative replication model. Why is half of DNA replicated in a discontinuous fashion? A Replication of the lagging strand occurs in the direction away from the replication fork in short stretches of DNA, since access to the DNA is always from the 5′ end. This results in pieces of DNA being replicated in a discontinuous fashion. B Replication of the leading strand occurs in the direction away from the replication fork in short stretches of DNA, since access to the DNA is always from the 5′ end. This results in pieces of DNA being replicated in a discontinuous fashion. C Replication of the lagging strand occurs in the direction of the replication fork in short stretches of DNA, since access to the DNA is always from the 5′ end. This results in pieces of DNA being replicated in a discontinuous fashion. D Replication of the lagging strand occurs in the direction away from the replication fork in short stretches of DNA, since access to the DNA is always from the ′' end. This results in pieces of DNA being replicated in a discontinuous fashion. Solution 44 The solution is (A). Since access to the DNA strand is always from the 5′ end, the replication of one strand, called the lagging strand, occurs in a direction away from the replication fork, in short stretches of DNA. This results in pieces of DNA being replicated in a discontinuous fashion. These pieces will be joined into a single strand of DNA. Explain the events taking place at the replication fork. If the gene for helicase is mutated, what part of replication will be affected? A Helicase separates the DNA strands at the origin of replication. Topoisomerase breaks and reforms DNA’s phosphate backbone ahead of the replication fork, thereby relieving the pressure. Single-stranded binding proteins prevent reforming of DNA. Primase synthesizes RNA primer which is used by DNA polymerase to form a daughter strand. If helicase is mutated, the DNA strands will not be separated at the beginning of replication. B Helicase joins the DNA strands together at the origin of replication. Topoisomerase breaks and reforms DNA’s phosphate backbone after the replication fork, thereby relieving the pressure. Single-stranded binding proteins prevent reforming of DNA. Primase synthesizes RNA primer which is used by DNA polymerase to form a daughter ******************************************************** 14 | DNA Structure and Function PAGE \* MERGEFORMAError! Unknown switch argument.*** strand. If helicase is mutated, the DNA strands will not be joined together at the beginning of replication. ******************************************************** l Ошибка! Неизвестный аргумент ключа. o 14 | DNA Structure and Functi C Helicase separates the DNA strands at the origin of replication. Topoisomerase breaks and reforms DNA’s sugar backbone ahead of the replication fork, thereby increasing the pressure. Single-stranded binding proteins prevent reforming of DNA. Primase synthesizes DNA primer which is used by DNA polymerase to form a daughter strand. If helicase is mutated, the DNA strands will be separated at the beginning of replication. D Helicase separates the DNA strands at the origin of replication. Topoisomerase breaks and reforms DNA’s sugar backbone ahead of the replication fork, thereby relieving the pressure. Single-stranded binding proteins prevent reforming of DNA. Primase synthesizes DNA primer which is used by RNA polymerase to form a parent strand. If helicase is mutated, the DNA strands will be separated at the beginning of replication. Solution The solution is (A). A replication fork is formed when helicase separates the DNA strands at the origin of replication. The DNA tends to become more highly coiled ahead of the replication fork. Topoisomerase breaks and reforms DNA’s phosphate backbone ahead of the replication fork, thereby relieving the pressure that results from this supercoiling. Single-strand binding proteins bind to the single-stranded DNA to prevent the helix from re-forming. Primase synthesizes an RNA primer. DNA polymerase III uses this primer to synthesize the daughter DNA strand. If helicase is mutated and cannot function, the DNA strands will not be separated at the beginning of replication. 45 What are Okazaki fragments, and how they are formed? A Okazaki fragments are short stretches of DNA on the lagging strand, which is synthesized in the direction away from the replication fork. B Okazaki fragments are long stretches of DNA on the lagging strand, which is synthesized in the direction of the replication fork. C Okazaki fragments are long stretches of DNA on the leading strand, which is synthesized in the direction away from the replication fork. D Okazaki fragments are short stretches of DNA on the leading strand, which is synthesized in the direction of the replication fork. Solution 46 The solution is (A). Replication of the lagging strand in DNA replication occurs in a direction away from the replication fork in short stretches of DNA known as Okazaki fragments. Compare and contrast the roles of DNA polymerase I and DNA ligase in DNA replication. A DNA polymerase I removes the RNA primers from the developing copy of DNA. DNA ligase seals the ends of the new segment, especially the Okazaki fragments. B DNA polymerase I adds the RNA primers to the already developing copy of DNA. DNA ligase separates the ends of the new segment, especially the Okazaki fragments. ******************************************************** 14 | DNA Structure and Function PAGE \* MERGEFORMAError! Unknown switch argument.*** C DNA polymerase I seals the ends of the new segment, especially the Okazaki fragments. DNA ligase removes the RNA primers from the developing copy of DNA. D DNA polymerase I removes the enzyme primase from the developing copy of DNA. DNA ligase seals the ends of the old segment, especially the Okazaki fragments. Solution 47 The solution is (A). DNA polymerase I has exonuclease activity that removes RNA primers from the developing copy while DNA ligase seals the ends of the new segment, especially the Okazaki fragments. If the rate of replication in a particular prokaryote is 900 nucleotides per second, how long would it take to make two copies of a 1.2 million base pair genome? A 22.2 min B 44.4 min C 45.4 min D 54.4 min Solution 48 The solution is (B). The rate of replication equals 900 nucleotides per second. For 1.2 million base pairs, the time taken for replication would be 1.2 million 900 22.2 min (1,333.3 s). To make two copies, the time taken will be 22.2 2 44.4 min. How do the linear chromosomes in eukaryotes ensure that their ends are replicated completely? A The ends of the linear chromosomes are maintained by the activity of the telomerase enzyme. B The ends of the linear chromosomes are maintained by the formation of a replication fork. C The ends of the linear chromosomes are maintained by the continuous joining of Okazaki fragments. D The ends of the linear chromosomes are maintained by the action of the polymerase enzyme. Solution The solution is (A). Telomerase enzyme replicates the ends of chromosomes by attaching to the 3′ chromosomal end of the DNA strand. When the 3′ end is elongated, then DNA polymerase adds complementary nucleotides to the end of chromosomes. ******************************************************** l Ошибка! Неизвестный аргумент ключа. o 49 14 | DNA Structure and Functi What is the best way to compare and contrast prokaryotic and eukaryotic DNA replication? A A prokaryotic organism’s rate of replication is 10 times faster than that of eukaryotes. Prokaryotes have a single origin of replication and use five types of polymerases, while eukaryotes have multiple sites of origin and use 14 polymerases. Telomerase is absent in prokaryotes. DNA pol I is the primer remover in prokaryotes, while in eukaryotes it is RNase H. DNA pol III performs strand elongation in prokaryotes and pol δ and pol ε do the same in eukaryotes. B A prokaryotic organism’s rate of replication is 10 times slower than that of eukaryotes. Prokaryotes have a single origin of replication and use five types of polymerases, while eukaryotes have multiple sites of origin and use 14 polymerases. Telomerase is absent in eukaryotes. DNA pol I is the primer remover in prokaryotes, while in eukaryotes it is RNase H. DNA pol III performs strand elongation in prokaryotes and pol δ and pol ε do the same in eukaryotes. C A prokaryotic organism’s rate of replication is 10 times faster than that of eukaryotes. Prokaryotes have five origins of replication and use a single type of polymerase, while eukaryotes have a single site of origin and use 14 polymerases. Telomerase is absent in prokaryotes. DNA pol I is the primer remover in prokaryotes, while in eukaryotes it is RNase H. DNA pol III performs strand elongation in prokaryotes and pol δ and pol ε do the same in eukaryotes. D A prokaryotic organism’s rate of replication is 10 times slower than that of eukaryotes. Prokaryotes have a single origin of replication and use five types of polymerases, while eukaryotes have multiple sites of origin and use 14 polymerases. Telomerase is absent in prokaryotes. DNA pol I is the primer remover in eukaryotes, while in prokaryotes it is RNase H. DNA pol III performs strand elongation in prokaryotes and pol δ and pol ε do the same in eukaryotes. Solution The solution is (A). Prokaryotic organisms have a single origin of replication, while eukaryotic ones have multiple sites. The rate of replication of prokaryotic cells is approximately 10 times that of eukaryotes. There are five types of DNA polymerases used by prokaryotes and 14 in eukaryotes. Telomerase functions in eukaryotic cells, but not prokaryotic one. The RNA primer remover in prokaryotic organisms is DNA pol I, but RNase H in eukaryotic cells. Strand elongation is performed by DNA pol III in prokaryotes and by Pol δ, pol ε in eukaryotic organisms. ******************************************************** 14 | DNA Structure and Function 50 PAGE \* MERGEFORMAError! Unknown switch argument.*** What would be the consequence of a mutation in a mismatch repair enzyme? How would this affect the function of a gene? A Mismatch repair corrects the errors after the replication is completed by excising the incorrectly added nucleotide and adding the correct base. Any mutation in a mismatch repair enzyme would lead to more permanent damage. B Mismatch repair corrects the errors during the replication by excising the incorrectly added nucleotide and adding the correct base. Any mutation in the mismatch repair enzyme would lead to more permanent damage. C Mismatch repair corrects the errors after the replication is completed by excising the added nucleotides and adding more bases. Any mutation in the mismatch repair enzyme would lead to more permanent damage. D Mismatch repair corrects the errors after the replication is completed by excising the incorrectly added nucleotide and adding the correct base. Any mutation in the mismatch repair enzyme would lead to more temporary damage. Solution 51 The solution is (A). Some errors are not corrected during replication, but instead are corrected after replication is completed; this type of repair is known as mismatch repair. The enzymes recognize the incorrectly added nucleotide and excise it; this then is replaced by the correct base. If this remains uncorrected, it may lead to more permanent damage. A mutation has occurred in the DNA and in the mRNA for a gene. Which one would have a more significant effect on gene expression? Why? A Both will result in the production of defective proteins. The DNA mutation, if not corrected, is permanent, while the mRNA mutation will affect only proteins made from that mRNA strand. Production of defective protein ceases when the mRNA strand deteriorates. B Both will result in the production of defective proteins. The DNA mutation, if not corrected, is permanent, while the mRNA mutation will not affect proteins made from that mRNA strand. Production of defective protein continues when the mRNA strand deteriorates. C Only DNA will result in the production of defective proteins. The DNA mutation, if not corrected, is permanent. Production of defective protein ceases when the DNA strand deteriorates. D Only mRNA will result in the production of defective proteins. The mRNA mutation will affect only proteins made from that mRNA strand. Production of defective protein ceases when the mRNA strand deteriorates. Solution The solution is (A). Both will result in a defective protein produced from the gene information. The DNA mutation, if not corrected, is permanent; the mRNA mutation will affect only proteins made from that RNA strand. When the mRNA strand deteriorates, the production of the defective protein ceases. ******************************************************** l Ошибка! Неизвестный аргумент ключа. o 52 14 | DNA Structure and Functi What are the effects of point mutations on a DNA strand? A Mutations can cause a single change in an amino acid. A nonsense mutation can stop the replication or reading of that strand. Insertion or deletion mutations can cause a frame shift. This can result in nonfunctional proteins. B Mutations can cause a single change in amino acid. A missense mutation can stop the replication or reading of that strand. Insertion or deletion mutations can cause a frame shift. This can result in nonfunctional proteins. C Mutations can cause a single change in amino acid. A nonsense mutation can stop the replication or reading of that strand. Substitution mutations can cause a frame shift. This can result in nonfunctional proteins. D Mutations can cause a single change in amino acid. A nonsense mutation can stop the replication or reading of that strand. Insertion or deletion mutations can cause a frame shift. This can result in functional proteins. Solution 53 The solution is (A). If one base is replaced by another base, but the coding for an amino acid is not changed, there is no effect on the DNA strand and a silent mutation has occurred. A missense mutation happens when a point mutation causes a change in a single amino acid. A nonsense mutation causes a stop message to be read, and the replication or reading of that strand is stopped at that point. Insertion or deletion mutations cause a frame shift from that point on and a non-functional protein will result. What is the significance of mutations in tRNA and rRNA? A Mutations in tRNA and rRNA would lead to the production of defective proteins or no protein production. B Mutations in tRNA and rRNA would lead to changes in the semiconservative mode of replication of DNA. C Mutations in tRNA and rRNA would lead to production of a DNA strand with a mutated single strand and normal other strand. D Mutations in tRNA and rRNA would lead to skin cancer in patients of Xeroderma pigmentosa. Solution The solution is (A). A mutation in a single type of tRNA will affect the transfer of one amino acid. The result of this will be a decrease in proteins that require this amino acid, as it will not be brought to the assembly points of proteins in adequate amounts. A mutation in rRNA will affect the assembly of all proteins and lead to significant deficiencies in these molecules. ******************************************************** 14 | DNA Structure and Function PAGE \* MERGEFORMAError! Unknown switch argument.*** TEST PREP FOR AP® COURSES 54 What would Chase and Hershey have concluded if the supernatant contained radioactive labeled-phosphorus? A DNA was the primary source of heritable information. B RNA was the primary source of heritable information. C Protein was the primary source of heritable information. D Phages were the primary source of heritable information. Solution 55 The solution is (C). Protein would have been identified as the primary source of heritable information. Which piece of evidence supports that the material Miescher discovered was DNA? A The precipitate contained sulfur. B The precipitate contained oxygen. C The precipitate contained phosphorus. D The precipitate contained protein. Solution 56 The solution is (C). The precipitate contained phosphorus, which is abundant in DNA. How are forensic scientists able to use DNA analysis to identify individuals? A Comparison of DNA from a known source or individual with analysis of the sequence of an unknown sample of DNA allows scientists to find out if both of them are similar or not. B DNA from the unknown sample is sequenced and analyzed. The result of the analysis then is matched with any random population. The matching individual then helps in forensics. C Comparison of DNA from a known source or individual with analysis of the sequence of bases in strands of an unknown sample of RNA allows scientists to find out if both of them are similar or not. D Comparison of DNA from a known source or individual with analysis of the sugars and phosphates in strands of an unknown sample of DNA allows scientists to find out if both of them are similar or not. Solution The solution is (A). Analysis of the sequence of bases in strands of DNA and their comparison to DNA from a known source or individual allows scientists to state that they are the same, or very similar, thus identifying the unknown source of the sample of DNA. ******************************************************** l Ошибка! Неизвестный аргумент ключа. o 57 14 | DNA Structure and Functi What were the contributions of Francis Crick, James Watson, and Rosalind Franklin to the discovery of the structure of DNA? A Franklin used X-ray diffraction methods to demonstrate the helical nature of DNA, while Watson and Crick formulated the double-stranded structural model of DNA. B Franklin, Watson, and Crick first employed the technique of X-ray diffraction to understand the storage of DNA. Since it did not work out, Watson and Crick then ran experiments to ascertain the DNA structure. C Watson and Crick used X-ray diffraction methods to demonstrate the helical nature of DNA, while Franklin formulated the double-stranded structural model of DNA. D Watson and Crick used X-ray diffraction methods to demonstrate the helical nature of DNA, while Franklin formulated the double-stranded structural model of DNA. Solution 58 The solution is (A). Franklin was using X-ray diffraction methods to understand the structure of DNA. Watson and Crick were able to piece together the puzzle of the DNA molecule on the basis of Franklin's data. What do RNA and DNA have in common? A Both contain four different nucleotides. B Both are usually double-stranded molecules. C Both contain adenine and uracil. D Both contain ribose. Solution 59 The solution is (A). Both contain four different nucleotides. What would be a good application of plasmid transformation? A To make copies of DNA B To isolate a change in a single nucleotide C To separate DNA fragments D To sequence DNA Solution 60 The solution is (C). DNA fragments move through gel based on their negative charge. How do the components of DNA fit together? A DNA is composed of nucleotides, consisting of a five-carbon sugar, a phosphate, and a nitrogenous base. DNA is a double-helical structure in which complementary base pairing occurs. Adenine pairs with thymine and guanine pairs with cytosine. Adenine and thymine form two hydrogen bonds and cytosine and guanine form three hydrogen bonds. The two individual strands of DNA are held together by covalent bonds between the phosphate of one nucleotide and sugar of the next. The two strands run antiparallel to each other. ******************************************************** 14 | DNA Structure and Function PAGE \* MERGEFORMAError! Unknown switch argument.*** B DNA is composed of nucleotides, consisting of a five-carbon sugar, a phosphate, and a nitrogenous base. DNA is a double-helical structure in which complementary base pairing occurs. Adenine pairs with cytosine and guanine pairs with thymine. Adenine and cytosine form two hydrogen bonds and guanine and thymine form three hydrogen bonds. The two individual strands of DNA are held together by covalent bonds between the phosphate of one nucleotide and sugar of the next. The two strands run antiparallel to each other. C DNA is composed of nucleotides, consisting of a five-carbon sugar, a phosphate, and a nitrogenous base. DNA is a double-helical structure in which complementary base pairing occurs. Adenine pairs with cytosine and guanine pairs with thymine. Adenine and cytosine form three hydrogen bonds and guanine and thymine form two hydrogen bonds. The two individual strands of DNA are held together by covalent bonds between the phosphate of one nucleotide and sugar of the next. The two strands run antiparallel to each other. D DNA is composed of nucleotides, consisting of a five-carbon sugar, a phosphate, and a nitrogenous base. DNA is a double-helical structure in which complementary base pairing occurs. Adenine pairs with cytosine and guanine pairs with thymine. Adenine and cytosine form three hydrogen bonds and guanine and thymine form two hydrogen bonds. The two individual strands of DNA are held together by covalent bonds between the phosphate of one nucleotide and sugar of the next. The two strands run parallel to each other. Solution 61 The solution is (A). DNA is made up of two strands that are twisted around each other to form a right-handed helix. Base pairing takes place between a purine and pyrimidine; namely, A pairs with T and G pairs with C. Adenine and thymine are complementary base pairs, and cytosine and guanine are also complementary base pairs. The base pairs are stabilized by hydrogen bonds; adenine and thymine form two hydrogen bonds, and cytosine and guanine form three hydrogen bonds. The two strands are anti-parallel in nature; that is, the 3′ end of one strand faces the 5′ end of the other strand. The sugar and phosphate of the nucleotides form the backbone of the structure, whereas the nitrogenous bases are stacked inside. What is the best way to describe the Sanger DNA sequencing method used for the human genome sequencing project? A A DNA sample is denatured by heating and then put into four tubes. A primer, DNA polymerase, and all four nucleotides are added. Limited quantities of one of the four dideoxynucleotides (ddNTPs) are added to each tube respectively. Each one of them carries a specific fluorescent label. Chain elongation continues until a fluorescent ddNTP is added to the growing chain, after which chain termination occurs. Gel electrophoresis is performed and the length of each base is detected by laser scanners with wavelengths specific to the four different ddNTPs. B A DNA sample is denatured by heating and then put into four tubes. A primer, RNA polymerase, and all four nucleotides are added. Limited quantities of one of the four ******************************************************** l Ошибка! Неизвестный аргумент ключа. o 14 | DNA Structure and Functi dideoxynucleotides (ddNTPs) are added to each tube respectively. Each one of them carries a specific fluorescent label. Chain elongation continues until a fluorescent ddNTP is added to the growing chain, after which chain termination occurs. Gel electrophoresis is performed and the length of each base is detected by laser scanners with wavelengths specific to the four different ddNTPs. C A DNA sample is denatured by heating and then put into four tubes. A primer, DNA polymerase, and all four nucleotides are added. Limited quantities of one of the four dideoxynucleotides (ddNTPs) are added to each tube respectively. Each one of them carries a specific fluorescent label. Chain elongation continues until a fluorescent ddNTP is removed from the growing chain, after which chain termination occurs. Gel electrophoresis is performed and the length of each base is detected by laser scanners with wavelengths specific to the four different ddNTPs. D A DNA sample is denatured by heating and then put into four tubes. A primer, DNA polymerase, and all four nucleotides are added. Limited quantities of one of the four deoxynucleotides (dNTPs) are added to each tube respectively. Each one of them carries a specific fluorescent label. Chain elongation continues until a fluorescent dNTP is added the growing chain, after which chain termination occurs. Gel electrophoresis is performed and the length of each base is detected by laser scanners with wavelengths specific to the four different dNTPs. Solution 62 The solution is (B). The DNA sample to be sequenced is denatured or separated into two strands by heating it to high temperatures. The DNA is divided into four tubes in which a primer, DNA polymerase, and all four nucleotides (A, T, G, and C) are added. In addition to each of the four tubes, limited quantities of one of the four dideoxynucleotides are added to each tube respectively. The tubes are labeled as A, T, G, and C according to the ddNTP added. For detection purposes, each of the four dideoxynucleotides carries a different fluorescent label. Chain elongation continues until a fluorescent dideoxy nucleotide is incorporated, after which no further elongation takes place. After the reaction is over, electrophoresis is performed. Even a difference in length of a single base can be detected. The sequence is read from a laser scanner. What process is illustrated in the figure? ******************************************************** 14 | DNA Structure and Function PAGE \* MERGEFORMAError! Unknown switch argument.*** ******************************************************** l Ошибка! Неизвестный аргумент ключа. o 14 | DNA Structure and Functi A Transcription B Mutation C Excision D Translation Solution 63 The solution is (A). RNA is being made from DNA. How does the model of DNA replication illustrate the function of topoisomerase? A Topoisomerase relieves the pressure that results from supercoiling by breaking and reforming DNA’s phosphate backbone ahead of the replication fork. B Topoisomerase increases the pressure to increase supercoiling by breaking and reforming DNA’s phosphate backbone ahead of the replication fork. C Topoisomerase relieves the pressure that results from supercoiling by breaking and reforming DNA’s nucleotide base pairs ahead of the replication fork. D Topoisomerase relieves the pressure that results from separation of DNA strands by breaking and reforming DNA’s phosphate backbone ahead of the replication fork. Solution 64 The solution is (A). The DNA tends to become more highly coiled ahead of the replication fork. Topoisomerase breaks and reforms DNA’s phosphate backbone ahead of the replication fork, thereby relieving the pressure that results from this supercoiling. Flamingos have genotypes for white feathers yet often appear with pink feathers within the same population. What is most likely affecting the phenotype of some flamingos, causing their feathers to turn pink in an isolated population? A Weather variations B Dietary changes C DNA mutations D Translation failure Solution 65 The solution is (A). Weather variations would affect the entire isolated population. What can be the result of DNA failing to undergo repair after too much UV exposure? A Second-degree burns B A malignant melanoma C A breakdown of deep layers of the skin D A sunburn Solution The solution is (B). UV light exposure can cause melanoma. ******************************************************** 14 | DNA Structure and Function PAGE \* MERGEFORMAError! Unknown switch argument.*** ******************************************************** l Ошибка! Неизвестный аргумент ключа. o 66 14 | DNA Structure and Functi What type of change can occur in the DNA of a chromosome that is termed a chromosomal mutation? A Substitution B Translocation C Missense D Transversion Solution 67 The solution is (B). Sometimes a piece of DNA from one chromosome may get translocated to another chromosome or to another region of the same chromosome; this chromosomal mutation is known as translocation. Why are patients with Xeroderma pigmentosa more prone to cancer than the rest of the population? A Xeroderma pigmentosa patients cannot employ the nucleotide excision repair mechanism. When these patients are exposed to UV light, thymine dimers are formed and they are not able to repair this defect. These dimers distort the structure of DNA and cause them to have a high risk of contracting skin cancer. B Xeroderma pigmentosa patients can employ the nucleotide excision repair mechanism. When these patients are exposed to UV light, the thymine dimers are formed and they are able to repair this defect. These dimers do not distort the structure of DNA and they have moderate risk of contracting skin cancer. C Xeroderma pigmentosa patients cannot employ the nucleotide excision repair mechanism. When these patients are exposed to UV light, the adjacent adenine forms dimers and they are not able to repair this defect. These dimers distort the structure of DNA and they have high risk of contracting skin cancer. D Xeroderma pigmentosa patients cannot employ the nucleotide excision repair mechanism. When these patients are exposed to UV light, the adjacent thymine cannot form thymine dimers and they are not able to repair this defect. The nonformation of dimers distorts the structure of DNA and they have high risk of contracting skin cancer. Solution The solution is (D). Affected individuals have skin that is highly sensitive to UV rays from the sun. When individuals are exposed to UV, pyrimidine dimers are formed; people with Xeroderma pigmentosa are not able to repair the damage. These are not repaired because of a defect in the nucleotide excision repair enzymes, whereas in normal individuals, the thymine dimers are excised and the defect is corrected. The thymine dimers distort the structure of the DNA double helix, and this may cause problems during DNA replication. People with Xeroderma pigmentosa may have a higher risk of contracting skin cancer than those who do not have the condition. ******************************************************** 14 | DNA Structure and Function 68 PAGE \* MERGEFORMAError! Unknown switch argument.*** You are looking at two fragments of DNA. Both have the sequence CATTCTG on one strand and GTAAGAC on the other. One of the fragments is exposed to UV light, the other is not. What will happen to the fragments and how might these mutations be repaired? A The fragment exposed to UV light contains thymine dimers. Thymines lying adjacent to each other can form thymine dimers when exposed to UV light. They can be repaired by nucleotide excision. B The fragment exposed to UV light contains adenine dimers. Adenines lying adjacent to each other can form dimers when exposed to UV light. They can be repaired by nucleotide excision. C The fragment exposed to UV light contains thymine dimers. Thymines lying parallel to each other can form thymine dimers when exposed to UV light. They can be repaired by nucleotide excision. D The fragment exposed to UV light contains thymine dimers. Thymines lying adjacent to each other can form thymine dimers when exposed to UV light. They can be synthesized by nucleotide excision. Solution 69 The solution is (A). Nucleotide excision repairs thymine dimers. When exposed to UV, thymines lying adjacent to each other can form thymine dimers. In normal cells, they are excised and replaced. How can mutations increase variation within a population? A Substitution mutations may cause a different amino acid to be placed at a specific location, causing small changes in the protein. Frameshift mutations usually cause multiple amino acid changes, increasing chances that a new protein will form, leading to radically different characteristics in the offspring. B Substitution mutations may cause multiple amino acid changes, increasing chances that a new protein will form, leading to radically different characteristics in the offspring. Frameshift mutations may cause a different amino acid to be placed at a specific location, causing small changes in a protein. C Substitution mutations may cause a different amino acid to be placed at a specific location, resulting in major changes to the protein and leading to radically different characteristics in the offspring. Frameshift mutations cause multiple amino acid differences in a protein, leading to small changes in the protein. D Substitution mutations result in a different amino acid being placed at a specific position in a protein, causing small changes. Silent mutations could result in new characteristics possessed by an offspring when a stop codon is substituted for an amino acid. Solution The solution is (A). Substitution mutations can cause a different amino acid to be placed at a specific location in a protein that could alter its characteristics and ******************************************************** l Ошибка! Неизвестный аргумент ключа. o 14 | DNA Structure and Functi change the population over time. It might result in more resistance to bacterial infections or even to direct sunlight. Frameshift mutations could, conceivably, make an entirely different protein, giving the resulting offspring a new characteristic not possessed previously. SCIENCE PRACTICE CHALLENGE QUESTIONS 14.1 Historical Basis of Modern Understanding 70 The proof that DNA, not protein, is the carrier of genetic information involved a number of historical experiments, including transformation or horizontal gene transfer (HGT), which is the uptake and expression of extracellular DNA. A. As described in the figure, transformation or HGT was first reported by Griffith in 1928 in an experiment in which the following occurred: 1. Heat-treated, pathogenic bacteria recovered their pathogenicity when incubated with nonpathogenic bacteria. 2. Plasmids were transferred to nonpathogenic bacteria from pathogenic bacteria through conjugation. 3. Nonpathogenic bacteria acquired pathogenicity when incubated in a broth containing heat-treated, pathogenic bacteria. 4. Polysaccharide cell capsules from pathogenic bacteria were transferred to nonpathogenic bacteria. B. Griffith’s experiment, however, left undetermined the identity of the cellular component that encoded genetic information. The identity of DNA as the carrier of genetic information was resolved through the experiments by Martha Chase and Alfred Hershey because they observed the following: 1. Injections with a serum containing chemically isolated polysaccharides and nonpathogenic bacteria were not lethal. ******************************************************** 14 | DNA Structure and Function PAGE \* MERGEFORMAError! Unknown switch argument.*** 2. Pathogenic bacterial DNA that was radioactively labeled using a phosphorus isotope was not present in mice that died. 3. Bacteriophages from a bacterial culture grown in a nutrient-containing medium and radioactively labeled using a sulfur isotope transferred the label to bacteria incubated in an unlabeled nutrient-containing medium. 4. Bacteriophages from a bacterial culture grown in a nutrient-containing medium and radioactively labeled using a sulfur isotope did not transfer the label to bacteria incubated in an unlabeled nutrient-containing medium. C. Transformation and transduction increase variation within populations of bacteria and archaebacteria by the following: 1. Transferring DNA among different species 2. Transferring free DNA across the cell membrane without energy expenditure 3. Transferring DNA between different strains of the same species of bacteria 4. Phagocytosis of bacteriophages The evolution of antibiotic resistance via HGT poses a challenge to medical technology. On the other hand, transformation often is assayed by incorporating an antibiotic-resistance gene in the plasmid to be transferred into the host organism. In natural environments, bacterial and archaebacterial cells become competent (able to transport DNA through the cytoplasmic membrane) in response to stress such as UV radiation, high population density, or heat shock. Such conditions are often difficult to model in the laboratory, where competence can be induced by high concentrations of divalent cations, Ca 2+ or Mg2+, or electrical shock. In either setting, extracellular DNA can be transported into the cell, and (to a good approximation) uptake is proportional to the concentration of extracellular DNA. D. Identify a factor that might affect transformation or HGT. Then, design a plan to evaluate the dependence of transformational efficiency (defined as the number of transformations per gram of extracellular DNA) of plasmids that transfer antibiotic resistance to a particular strain of Escherichia coli that is not resistant on that factor. Solution Sample answer: A. (3) B. (4) C. (1) D. The factor is identified. It might be any environmental factor, such as temperature, salinity, or pH. The method of control could be specified, such as heat bath or buffer, but the target of assessment would be the identification of the factor and the need to vary it measurably while fixing values of other factors. One of these fixed values should be population density which, based on lab experiences, the ******************************************************** l Ошибка! Неизвестный аргумент ключа. o 14 | DNA Structure and Functi student should be able to identify as optical density. Assay could be fluorescent if the student had prior experience with this method or the use of antibiotic resistance. Cells that had been transformed would then be counted in terms of the number of viable or fluorescent colonies. A test of the assumption that uptake is proportional to extracellular DNA should be included in the design with variation in DNA concentration at fixed values of other parameters. 71 Prior to the work of Hershey and Chase, scientists thought that inheritance involved “nucleoproteins.” The amount of information to be transmitted between generations did not seem consistent with the chemical simplicity of the few nucleotides found in polymers of deoxyribonucleic acids in comparison to the diversity of protein polymers. Briefly explain: Solution The relationship between the structure of polymeric DNA and the information stored The relationship between the interactions between base pairs on complementary strands of the double helix and Chargaff’s observation on the relative abundance of nucleotides in DNA The meaning of the statement from the Nature publication on the structure of DNA by Watson and Crick: “It has not escaped our notice that the specific pairing we have postulated immediately suggests a possible copying mechanism for the genetic material.” Sample answer: Although the number of amino acids and the 3-D shape increases the complexity of proteins, a very long, linear sequence of four bases stores sufficient information. The pairing is the AT and CG forms that are supported by the structure with a pair of strands, each with corresponding bases. The authors glibly point to the creation of a whole field, genomics, based on the weakness of the hydrogen bonding between the complementary strands. 14.2 DNA Structure and Sequencing 72 In 1977, Fred Sanger developed a method to determine the order of nucleotides in a strand of DNA. Sanger won a Nobel Prize for his work, and his method of sequencing based on dideoxy chain termination (see figure) has been foundational to the rapid development of more modern, rapid, and inexpensive methods of sequencing. The challenge of the $1,000 in one-day sequencing of the human genome was achieved in 2016 by next-generation sequencing (NGS), a “catch-all” term describing several sequencing methods. ******************************************************** 14 | DNA Structure and Function PAGE \* MERGEFORMAError! Unknown switch argument.*** A. Using the diagrams for reference, explain the effect of the addition of dideoxynucleotides on chain growth of the DNA strand that is copied during sequencing in terms of the structures of dideoxyribose and deoxyribose. B. Suppose that a single strand to be sequenced is 5′CGAGTACG3′. In the presence of each of the four deoxynucleotides and the dideoxynucleotide ddCTP, describe the strands that would be formed from this template. Include in your description an annotation indicating the 3′ and 5′ ends of the fragments resulting from the procedure. C. Next-generation sequencing makes termination technology very rapid and relatively inexpensive. All babies born in the United States are currently screened by statemandated tests for several genetic conditions. The number of conditions tested ranges from 29 (GA and KS) to 59 (IL and MS). It is proposed that whole-genome sequencing should be mandatory for all newborns. The Genetic Information Nondiscrimination Act (2008) prevents health insurers from denying coverage or increasing costs of premiums based on genetic information. It also prohibits employers from making use of these data for hiring, firing, or promotion. The act passed in the House with a vote of 420 to 3, although it was lobbied against by organizations representing business (human resources, health insurance, and manufacturers), including the U.S. Chamber of Commerce. The act does not cover life, long-term care, or disability insurance. Pose three questions that are relevant to the use of whole-genome data. Solution Sample answer: A. Without the OH at the 3’ position a Pi group cannot bond and extend the sugar backbone. ******************************************************** l Ошибка! Неизвестный аргумент ключа. o 14 | DNA Structure and Functi B. Chain termination would occur at each G leading to the fragments 3′CTCATDC5′, 3′CATGC5′, and 3′C5′. C. Questions could address ownership (ownership is a difficult question—is the DNA on a licked postage stamp still owned?—similar to questions of permissible searches), the relative benefits to society and perhaps to the individual of genomic information could include reduced cost of insurance, reduced cost of medical treatment, increased efficiency and effectiveness of medical treatment, the possible creation of caste systems. A really good scientific question might be the extent to which possibly identifiable, distinct phenotypes such as intelligence, athleticism, beauty, artistic talent, etc., can be associated with a gene or an array of genes, as opposed to environmental factors or behaviors. A search will show that the omission of life insurance from GINA has led to some discussion and some possible concerns. The student might question the reason that life insurance or disability insurance was omitted. Questions might address whether or not courts have heard claims based on GINA—there have been some and they are easily found. Questions about data security will be asked. 14.3 Basics of DNA Replication 73 Our understanding of the mechanisms of DNA replication is important to research on cancer and aging. Additionally, the molecular basis of Mendelian genetics was established. A. The mechanism of DNA replication was investigated by Meselson and Stahl. The diagram from their 1958 paper summarizes their findings. Describe how this representation illustrates the manner in which DNA is copied for transmission between generations. ******************************************************** 14 | DNA Structure and Function PAGE \* MERGEFORMAError! Unknown switch argument.*** B. During the synthesis of new strands of DNA from the parent strands, DNA polymerase can only add nucleotides at the terminal 3’ of a growing strand. Using the diagram, describe the similarities and differences between the DNA replication of both strands. C. Shown at the left end of the upper parent strand is the six-base repeat sequence TTAGGG. In humans, this is the repeated, telomeric sequence that is attached to the telomere. The RNA primer in humans spans 10 base pairs, unlike in the drawing where it spans only three. In somatic cells, an enzyme called telomerase no longer functions. Explain the function of telomerase in the development of stem cells and cancer cells, and the inhibition of telomerase in programmed cell death or apoptosis. Solution Sample answer: A. The semiconservative nature of replication is illustrated by this very effective representation in which the newly synthesized strands are unshaded. The student ******************************************************** l Ошибка! Неизвестный аргумент ключа. o 14 | DNA Structure and Functi might be expected to describe the use of centrifugation to separate light and heavy isotopes of nitrogen. B. The description should include the breaking of hydrogen bonds by helicase, the attachment of DNA polymerase to the leading strand, the creation of a template for synthesis on the lagging strand with RNA primer and the synthesis of fragments as required by the 3’ to 5’ direction of addition. Names of specific enzymes are not expected to be memorized and this item illustrates how they would be provided but that the information provided can be used. C. Telomerase as assessable content is out of scope although it is described in the text. What would be appropriate for assessment on the AP Biology Exam would be the recognition that the RNA primer might not have a sufficient number of bases remaining on the lagging strand as DNA pol approaches the telomere. This results in a segment of the lagging side not being copied. Without telomerase to repair the lost telomeric sequence, the number of repeats is reduced until eventually the cell cannot divide, leading to cell death. Cancer cells activate telomerase and so avoid cell death. 14.4 DNA Replication in Prokaryotes 74 The mitochondria of eukaryote cells contain their own circular DNA (mtDNA), consistent with their origin according to the theory of endosymbiosis. The mitochondrial genome is highly conserved in Eukarya. In humans, the 50 to 100 mitochondria in each of the cells in most tissues have 5 to 10 copies of the genome. Each has 37 genes that primarily encode proteins of the electron transport chain. Point mutations in which a single nucleotide is incorrectly placed is not repaired because the error-checking provided by DNA polymerase is not present in the mitochondria. The mutation rate for mtDNA is approximately 100 times higher than the mutation rate for nuclear DNA. The simultaneous existence of multiple alleles in each cell is likely, a condition called heteroplasmy. In mammals, sperm mitochondria are destroyed prior to fertilization. A. Explain how point mutations in mtDNA can result in a loss of function in critical cellular components such as cytochrome c yet not be lethal to the cell. B. Oocyte mitochondria are randomly segregated during meiosis, resulting in variation in the frequency of mtDNA mutations in offspring relative to the parent. Explain how a loss of function does not accumulate, lowering the metabolic performance from generation to generation. As described in the Evolution Connection in this chapter of the text, a fossil fingertip found in a Siberian cave revealed an evolutionary link between Neanderthals and Denisovans. Fossils from 28 individuals were located in the “pit of bones,” Sima de los Huesos, in Spain, thousands of miles from the Siberian cave. In 2013, mtDNA from a femur of one of these individuals was compared with mtDNA of Denisovans, Neanderthals, and ******************************************************** 14 | DNA Structure and Function PAGE \* MERGEFORMAError! Unknown switch argument.*** modern humans. It was found that the Sima fossil shared many more alleles with Denisovans than with either Neanderthals or modern humans. In 2016, the same group of scientists who sequenced the mtDNA from the femur of one of the Sima fossils partially sequenced the DNA from that fossil, showing a clear connection to Neanderthals. C. Analyze these data to draw alternative conclusions regarding the relatedness of the three fossils and support each with evidence. D. Design a plan to differentiate or resolve these alternative conclusions. Solution Sample answer: A. Multiple copies of mtDNA allow expression even though some copies code for proteins that do not function. B. As mutations accumulate the cell become less efficient. When a cell line has less than a threshold number of functional mitochondrial, the cell dies and that cell line is deleted. This applies to both somatic cells and to gametes. C. The mtDNA evidence indicates that the Sima people are more closely related to the Denosivans. However, the DNA evidence indicates the opposite conclusion. D. The recovery of both mtDNA and nuclear DNA was a major feat. However, suppose that the mtDNA reveals a maternal ancestor of different lineage. Because of the very high mutation rate of mtDNA large variations between family lines occur while an ancestral line within a family remains very similar. mtDNA from other bones at Sima might reveal other maternal lines and this would be consistent with the DNA result, implying a common maternal ancestor related to Denosivans. Evidence that would confirm this would be provided by a third, as yet undiscovered or unstudied, population whose mtDNA could be sequenced. ******************************************************** **************************336*************************** 15 | GENES AND PROTEINS REVIEW QUESTIONS 1 What is the flow of information for protein synthesis according to the central dogma? A DNA to mRNA to protein B DNA to mRNA to tRNA to protein C DNA to protein to mRNA to protein D mRNA to DNA to mRNA to protein Solution 2 The solution is (A). DNA carries genetic information that is copied onto an mRNA template to make a particular protein. The DNA of virus A is inserted into the protein coat of virus B. The combination virus is used to infect E. coli. The virus particles produced by the infection are analyzed for DNA and protein. What results would you expect? A DNA and protein from B B DNA and protein from A C DNA from A and protein from B D DNA from B and protein from A Solution 3 The solution is (B). DNA is the genetic material, not protein. So when virus A is infected in the protein coat of virus B, the new virus produced will have DNA and protein of virus A. Protein is not the genetic material and cannot be inherited. The AUC and AUA codons in mRNA both specify isoleucine. What feature of the genetic code explains this? A Complementarity B Degeneracy C Nonsense codons D Universality Solution The solution is (B). The genetic code is a triplet code, with each DNA or RNA codon consisting of three nucleotides that encode one amino acid. It is degenerate, as most amino acids can be specified by more than one codon. ******************************************************** **************************************************335*** 4 How many nucleotides are in 12 mRNA codons? A 12 B 24 C 36 D 48 Solution 5 The solution is (C). Twelve mRNA codons would have 36 nucleotides, since each codon consists of three nucleotides. Which molecule does NOT contain genetic information? A DNA B mRNA C Protein D RNA Solution 6 The solution is (C). Proteins are made up of amino acids and do not contain genetic information. Which molecule in the central dogma can be compared to a disposable photocopy of a book kept on reserve in the library? A DNA B mRNA C Protein D tRNA Solution 7 The solution is (B). Messenger RNA is like a disposable photocopy of a book kept on reserve in the library. It carries a copy of the instructions from the nucleus to other parts of the cell and usually has a short half-life. Which subunit of the E. coli polymerase confers specificity to transcription? A B C D Solution The solution is (D). The subunit is involved in transcription initiation. It confers transcriptional specificity so that the polymerase begins to synthesize mRNA from an appropriate initiation site. ******************************************************** **************************336*************************** 8 Why are the 10 and 35 regions of prokaryotic promoters called consensus sequences? A They are identical in all bacterial species. B They are similar in all bacterial species. C They exist in all organisms. D They have the same function in all organisms. Solution 9 The solution is (B). At the 10 and 35 regions upstream of the initiation site, there are two promoter consensus sequences, or regions, that are similar across all promoters and across various bacterial species. The sequence that signals the end of transcription is called the — A promoter B stop codon C TATA box D terminator Solution 10 The solution is (D). Transcription continues until RNA polymerase reaches a stop or terminator sequence at the end of a gene. If the protein is missing, will a prokaryotic gene be terminated? A It depends on the gene. B No, the rho protein is essential. C Transcription termination is not required. D Yes, the rho protein is not involved in transcription. Solution 11 The solution is (A). It depends on the sequence of the gene as to whether the transcription termination is rho dependent or independent. If the rho protein is missing, then termination can occur if the sequence consists of short strings of adenines. Which feature of promoters can be found in both prokaryotes and eukaryotes? A GC box B Octamer box C TATA box D Solution 10 and 35 sequences The solution is (C). The TATA box is a DNA sequence found in the promoter regions of both prokaryotes and eukaryotes. ******************************************************** **************************************************335*** 12 At what stage in the transcription of a eukaryotic gene would TFII factors be active? A Elongation B Initiation C Processing D Termination Solution 13 The solution is (B). The initiation of transcription in eukaryotes requires TFII. It binds to the promoter region and then recruits RNA polymerase II. Which polymerase is responsible for the synthesis of 5S rRNA? A Polymerase I B Polymerase II C Polymerase III D Ribonuclease I Solution 14 The solution is (C). Polymerase III transcribes a variety of structural RNAs that includes the 5S pre-rRNA, transfer pre-RNAs (pre-tRNAs), and small nuclear pre-RNAs. What transcripts will be most affected by low levels of -amanitin? A 18S and 28S rRNAs B 5S rRNAs and tRNAs C Other small nuclear RNAs D Pre-mRNAs Solution 15 The solution is (D). Pre-mRNAs are transcribed by RNA polymerase II and are extremely sensitive to low levels of -amanitin. Which feature distinguishes eukaryotic transcription from bacterial transcription? A Eukaryotic transcription does not start at a consensus sequence. B Eukaryotic transcription does not require an initiation complex. C Eukaryotic transcription and translation do not take place at the same time. D Eukaryotic transcription does not require a termination sequence. Solution 16 The solution is (C). In prokaryotes, both transcription and translation takes place in the cytoplasm, while in eukaryotes, the transcription takes place in the nucleus and translation occurs in the cytoplasm. A poly-A sequence is added at the — A 5′ end of a transcript in the nucleus B 3′ end of a transcript in the nucleus ******************************************************** **************************336*************************** C 5′ end of a transcript in the cytoplasm D 3′ end of a transcript in the cytoplasm Solution 17 The solution is (B). At the 3′ end of the transcript, a poly-A sequence is added. Polymerase A adds a string of approximately 200 A residues. Which pre-mRNA processing step is important for initiating translation? A Adding a poly-A tail B RNA editing C Splicing D Adding the 7-methylguanosine cap Solution 18 The solution is (D). Capping occurs at the 5′ end while pre-mRNA synthesis is still going on. A 7-methylguanosine cap is added by a phosphate linkage at the 5′ end of the growing transcript. Where are the RNA components of ribosomes synthesized? A Cytoplasm B Endoplasmic reticulum C Nucleus D Nucleolus Solution 19 The solution is (D). The nucleolus in the eukaryotic cell is a condensed region where ribosomes are formed. The ribosomal subunits are synthesized in the nucleolus, and then exported to the cytoplasm before use. What processing step enhances the stability of pre-tRNAs and pre-rRNAs? A Cleavage B Methylation C Nucleotide modification D Splicing Solution 20 The solution is (B). Methylation is the addition of CH3 moiety for the stability of pretRNAs and pre-rRNAs. What are introns? A DNA sequences to which polymerases bind B Processed mRNA C Translated DNA sequences in a gene D Untranslated DNA sequences in a gene ******************************************************** **************************************************335*** Solution 21 The solution is (D). Untranslated DNA sequences that are transcribed in pre-mRNAs and consist of noncoding or intervening sequences are called introns. Until introns are removed, translation does not occur. What is often the first amino acid added to a polypeptide chain? A Adenine B Leucine C Methionine D Thymine Solution 22 The solution is (C). AUG on an mRNA, where translation begins, always specifies methionine. In any given species, there are at least how many types of aminoacyl tRNA synthetases? A 20 B 40 C 100 D 200 Solution 23 The solution is (A). Each tRNA molecule is linked to its correct amino acid by a group of enzymes called aminoacyl tRNA synthetases. At least one type of aminoacyl tRNA synthetase exists for each amino acid. In prokaryotic cells, ribosomes are found in/on the — A cytoplasm B mitochondrion C nucleus D endoplasmic reticulum Solution 24 The solution is (A). Prokaryotic cells do not contain organelles. The ribosomes lie in the cytoplasm of prokaryotes. Peptide bond synthesis in prokaryotic translation is catalyzed by — A a ribosomal protein B a cytoplasmic protein C mRNA itself D ribosomal RNA Solution The solution is (D). Peptidyl transferase is an RNA-based enzyme that is integrated into the 50S ribosomal subunit and catalyzes the formation of peptide bonds. ******************************************************** **************************336*************************** 25 What would happen if the 5′-methyl guanosine was NOT added to an mRNA? A The transcript would be degraded when the mRNA moves out of the nucleus to the cytoplasm. B The mRNA molecule would be stabilized and start the process of translation within the nucleus of the cell. C The mRNA molecule would move out of the nucleus and create more copies of the mRNA molecule. D The mRNA molecule would not be able to add the poly-A tail onto its strand at the 5′ end. Solution 26 The solution is (A). Without 5′ capping, the pre-mRNA transcript would be degraded and initiation of translation would be compromised. Which option is associated with the docking of mRNA on a ribosome in eukaryotic cells? A Kozak’s rules B Poly-A sequence C Shine-Dalgarno sequence D TATA box Solution The solution is (A). A Kozak consensus sequence is found in eukaryotic mRNA; the following consensus sequence must appear around the AUG: 5′-GCC(purine)CCAUGG-3′ CRITICAL THINKING QUESTIONS 27 If mRNA is complementary to the DNA template strand and the DNA template stand is complementary to the DNA non-template strand, why are the base sequences of mRNA and the DNA non-template strand not identical? Could they ever be? A No, they cannot be identical because the T nucleotide in DNA is replaced by the U nucleotide in RNA, and AUG is the start codon. B No, they cannot be identical because the T nucleotide in RNA is replaced by the U nucleotide in DNA. C They can be identical if methylation of the U nucleotide in RNA occurs, yielding a T nucleotide. D They can be identical if demethylation of the U nucleotide in RNA occurs, yielding a T nucleotide. Solution The solution is (A). DNA is different from RNA in that T nucleotides in DNA are replaced by U nucleotides in RNA. The start codon, AUG, includes a U nucleotide. Therefore, they could never be identical in base sequence. ******************************************************** **************************************************335*** 28 Imagine if there were 200 commonly occurring amino acids instead of 20. Given what you know about the genetic code, what would be the shortest possible codon length? A 4 B 5 C 2 D 3 Solution 29 The solution is (A). For 200 commonly occurring amino acids, codons consisting of four types of nucleotides would have to be at least four nucleotides long, because 44 = 256. There would be much less degeneracy in this case. What part of the central dogma is NOT always followed in viruses? A The flow of information in HIV is from RNA to DNA, then back to RNA to protein. Influenza viruses never go through DNA. B The flow of information is from protein to RNA in HIV, while the influenza virus converts DNA into RNA. C The flow of information is similar, but nucleic acids are synthesized as a result of translation in HIV and influenza viruses. D The flow of information is from RNA to protein. This protein is used to synthesize the DNA of the viruses in HIV and influenza. Solution 30 Suppose a gene has the sequence ATGCGTTATCGGGAGTAG. A point mutation changes the gene to read ATGCGTTATGGGGAGTAG. How would the polypeptide product of this gene change? Solution 31 The solution is (A). The flow of information goes from RNA to DNA back to RNA to protein in HIV. Other viruses such as the influenza virus never go through DNA. It would change from MRYRE to MRYGE. Explain the initiation of transcription in prokaryotes. Include all proteins involved. A In prokaryotes, the polymerase comprises five polypeptide subunits, two of which are identical. Four of these subunits, denoted , , , and , compose the polymerase core enzyme. The fifth subunit, , is involved only in the initiation of transcription. The polymerase, which comprises all five subunits, is called the holoenzyme. ******************************************************** **************************336*************************** B In prokaryotes, the polymerase comprises four polypeptide subunits, two of which are identical. These subunits, denoted , , , and ', compose the polymerase core enzyme. There is a fifth subunit that is involved in translation initiation. The polymerase, which comprises all four subunits, is called the holoenzyme. C In prokaryotes, the polymerase comprises five polypeptide subunits, two of which are identical. Four of these subunits, denoted , , , and , compose the polymerase holoenzyme. The fifth subunit, , is involved only in transcription initiation. The polymerase, which comprises all five subunits, is called the core enzyme. D In prokaryotes, the polymerase comprises five polypeptide subunits, two of which are identical. Four of these subunits, denoted , , , and , compose the polymerase core enzyme. The fifth subunit, , is involved only in termination. The polymerase, which comprises all five subunits, is called the holoenzyme. Solution 32 The solution is (A). Refer to Figure 15.7 and add the additional subunits of the holoenzymes that are not in the figure. How would you describe the difference between rho-dependent and -independent termination of transcription in prokaryotes? A Rho-dependent termination is controlled by the rho protein, and the polymerase stalls near the end of the gene at a run of G nucleotides on the DNA template. In rhoindependent termination, when the polymerase encounters a region rich in C-G nucleotides, the mRNA folds into a hairpin loop that causes the polymerase to stall. B Rho-independent termination is controlled by the rho protein, and the polymerase stalls near the end of the gene at a run of G nucleotides on the DNA template. In rhodependent termination, when the polymerase encounters a region rich in C–G nucleotides, the mRNA folds into a hairpin loop that causes the polymerase to stall. C Rho-dependent termination is controlled by the rho protein, and the polymerase begins near the end of the gene at a run of G nucleotides on the DNA template. In rhoindependent termination, when the polymerase encounters a region rich in C–G nucleotides, the mRNA creates a hairpin loop that causes the polymerase to stall. D Rho-dependent termination is controlled by the rho protein, and the polymerase stalls near the end of the gene at a run of G nucleotides on the DNA template. In rhoindependent termination, when the polymerase encounters a region rich in A–T nucleotides, the mRNA creates a hairpin loop that causes the polymerase to stall. Solution The solution is (A). Rho-dependent termination is controlled by the rho protein, which tracks along behind the polymerase on the growing mRNA chain. Near the end of the gene, the polymerase stalls at a run of G nucleotides on the DNA template. The rho protein collides with the polymerase and releases mRNA from the transcription bubble. Rho-independent termination is controlled by specific sequences in the DNA template strand. As the polymerase nears the end of the gene being transcribed, it encounters a region rich in C–G nucleotides. This creates an ******************************************************** **************************************************335*** mRNA hairpin that causes the polymerase to stall right as it begins to transcribe a region rich in A–T nucleotides. Because A–U bonds are less thermostable, the core enzyme breaks away. 33 What is the main structure that differentiates rho-dependent and -independent termination in prokaryotes? A Rho-independent termination involves the formation of a hairpin. B Rho-dependent termination involves the formation of a hairpin. C Rho-dependent termination stalls when the polymerase begins to transcribe a region rich in A–T nucleotides. D Rho-independent termination stalls when the polymerase begins to transcribe a region rich in G nucleotides. Solution 34 The solution is (A). Refer to Figures 15.7 and 15.8. Add the termination step, which may involve either the rho protein or a hairpin. Which step in the transcription of eukaryotic RNA differs the most from its prokaryotic counterpart? A The initiation step in eukaryotes requires an initiation complex with enhancers and transcription factors. Also, the separation of the DNA strand is different, as histones are involved. B The initiation step in prokaryotes requires an initiation complex with enhancers and transcription factors. Also, the separation of the DNA strand is different, as histones are involved. C The elongation step in eukaryotes requires an initiation complex with enhancers and transcription factors. Also, the separation of the DNA strand is different, as histones are involved. D The initiation step in eukaryotes requires an initiation complex with enhancers and transcription factors. Also, the separation of the DNA strand is different, as histones are not involved. Solution 35 The solution is (A). The initiation step in eukaryotes requires an initiation complex with enhancers and transcription factors. The separation of the DNA strand is also different as histones are involved. Would you be able to determine which RNA polymerase you isolated from a eukaryotic cell without analyzing its products? A No, they have the same -amanitin sensitivity in all products. B No, quantitative analysis of products is done to determine the type of polymerase. C Yes, they can be determined as they differ in -amanitin sensitivity. D Yes, they can be determined by the number of molecules that bind to the DNA. ******************************************************** **************************336*************************** Solution 36 The solution is (C). Yes, RNA polymerases differ in their sensitivity to -amanitin. Predict how alternative splicing may lead to an economy of genes. Do you need a different gene for every protein that the cell can produce? A Alternative splicing can lead to the synthesis of several polypeptides from a single gene. B Alternative splicing can lead to the synthesis of several forms of mRNA from a single gene. C Alternative splicing can lead to the synthesis of several forms of codons from a set of genes. D Alternative splicing can lead to the synthesis of several forms of ribosomes from a set of genes. Solution 37 The solution is (A). It is possible to synthesize several polypeptides from a single gene by alternative splicing. What is the major challenge in the production of RNA in eukaryotes compared to prokaryotes? A Exporting the mRNA across the nuclear membrane B Importing the mRNA across the nuclear membrane C Keeping the mRNA inside the nuclear membrane D Translating the mRNA into proteins within seconds Solution 38 The solution is (A). The major challenge is exporting the mRNA across the nuclear membrane. What would happen if the 5’ methyl guanosine was not added to an mRNA? A The transcript would degrade when the mRNA moves out of the nucleus to the cytoplasm. B The mRNA molecule would stabilize and start the process of translation within the nucleus of the cell. C The mRNA molecule would move out of the nucleus and create more copies of the mRNA molecule. D The mRNA molecule would not be able to add the poly-A tail on its strand at the 5’ end Solution 39 The solution is (A). Without the 5’ capping, the pre-mRNA transcript would degrade and initiation of translation would be compromised. How should the following DNA sequence (non-template strand) be transcribed and translated? 5′-ATGGCCGGTTATTAAGCA-3′ ******************************************************** **************************************************335*** A The mRNA would be 5′-AUGGCCGGUUAUUAAGCA-3′ and the protein will be MAGY. B The mRNA would be 3′-AUGGCCGGUUAUUAAGCA-5′ and the protein will be MAGY. C The mRNA would be 5′-ATGGCCGGTTATTAAGCA-3′ and the protein will be MAGY. D The mRNA would be 5′-AUGGCCGGUUAUUAAGCA-3′ and the protein will be MACY. Solution The solution is (A). The mRNA would be 5′-AUGGCCGGUUAUUAAGCA-3′. The protein would be MAGY. Even though there are six codons, the fifth codon corresponds to a stop, so the sixth codon would not be translated ******************************************************** **************************336*************************** 40 The RNA world hypothesis proposes that the first complex molecule was RNA, and it preceded protein formation. Which major function of ribosomal RNA supports this hypothesis? A rRNA has catalytic properties in the large subunit, and it assembles proteins. B rRNA is a protein molecule that helps in the synthesis of other proteins. C rRNA is essential for the transcription process. D rRNA plays a major role in posttranslational processes. Solution 41 The solution is (A). The rRNA in the large subunit has catalytic properties. It synthesizes the peptide bond. The main idea is that the molecule that assembles proteins cannot itself be a protein. In other words, a protein could not have synthesized the first protein. RNA subunits can self-assemble. A tRNA is chemically modified so that the bound amino acid is different than the one specified by its anticodon. Which codon in the mRNA would the tRNA recognize: the one specified by its anticodon or the one that matches the modified amino acid it carries? A The anticodon will match the codon in mRNA. B The anticodon will match with the modified amino acid it carries. C The anticodon will lose the specificity for the tRNA molecule. D The enzyme aminoacyl tRNA synthetase would lose control over the amino acid. Solution The solution is (A). The anticodon matches the codon in mRNA. TEST PREP FOR AP® COURSES 42 What characteristic of the genetic code points to a common ancestry for all organisms? A The code is degenerate. B The code contains 64 codons. C The genetic code is almost universal. D The code contains stop codons. Solution 43 The solution is (C). The code is universal. What process transfers heritable material to the next generation? A Replication B Splicing C Transcription D Translation Solution The solution is (A). Replication is the process by which two identical DNA molecules are produced from one DNA molecule. ******************************************************** **************************************************335*** 44 When comparing transcription of heritable information in prokaryotes and eukaryotes, which events are the same? A The transcription by polymerase, the recognition of a consensus sequence in the promoter, and the termination by a hairpin loop are conserved. B The translation by polymerase, the recognition of a consensus sequence in the promoter, and the termination by a hairpin loop are conserved. C The transcription by polymerase, the recognition of a highly variable sequence in the promoter, and the termination by a hairpin loop are conserved. D The transcription by polymerase, the recognition of a consensus sequence in the promoter, and the elongation by a hairpin loop are conserved. Solution 45 The solution is (A). Transcription by a polymerase, recognition of a consensus sequence in the promoter, and termination by a hairpin loop are conserved. Which cell structure does NOT contain heritable information? A Chloroplast B Cytoplasmic membrane C Mitochondria D Nucleus Solution 46 The solution is (B). The cytoplasmic membrane separates the interior of the cell from the outside environment. How does the enzyme reverse transcriptase violate the central dogma of molecular biology in HIV? A The enzyme reverse transcriptase reverse-transcribes the RNA in the genome of HIV into DNA. B The enzyme reverse transcriptase translates the RNA of the HIV into protein and then back into DNA. C The enzyme reverse transcriptase transcribes the DNA straight into the protein molecules. D The enzyme reverse transcriptase transcribes DNA into RNA, then again into DNA. There is no protein synthesis. Solution The solution is (A). The genome of HIV is made of two molecules of RNA, which are reverse transcribed to DNA in the host. The enzyme reverse transcriptase is virally encoded and packaged in the virus particle. ******************************************************** **************************336*************************** 47 Radioactive deoxythymidine triphosphate is supplied to the protist Euglena. After an interval of time, the cells are homogenized, and different fractions are analyzed for radioactivity content in large nucleic acid molecules. Which fraction will not be labeled? A Nucleus B Mitochondrion C Chloroplast D Plasma membrane Solution 48 The solution is (D). The plasma membrane does not contain DNA. You sequence a gene of interest and isolate the matching mRNA. You find that the mRNA is considerably shorter than the DNA sequence. Why is that? A There was an experimental mistake. The mRNA should be the same length as the gene. B The mRNA should be longer than the DNA sequence because the promoter is also transcribed. C The processed mRNA is shorter because introns were removed. D The mRNA is shorter because the signal sequence to cross the nuclear membrane was removed. Solution 49 The solution is (C). Introns are noncoding regions of an RNA transcript, which are spliced out before the RNA is translated into protein. Splicing decreases the length of mRNA in comparison to DNA molecule. A mutation in the promoter region of the gene for beta-globin can cause betathalassemia, a hereditary condition that causes anemia. Why would mutations in the promoter region lead to low levels of hemoglobin? A The globin chains produced are too long to form functional hemoglobin. B The globin chains are too short to form functional hemoglobin. C Fewer globin chains are synthesized because less mRNA is transcribed. D Globin chains do not fold properly and are nonfunctional. Solution 50 The solution is (D). Protein folding is determined by the amino acid sequence. Mutations in the promoter region do not affect the amino acid sequence of the protein. You are given three mRNA sequences: 1. 5′-UCG-GCA-AAU-UUA-GUU-3 2. 5′-UCU-GCA-AAU-UUA-GUU-3′ 3. 5′-UCU-GCA-AAU-UAA-GUU-3′ ******************************************************** **************************************************335*** Using the table, write the peptide encoded by each of the mRNA sequences. Codon on mRNA Amino Acid GCA alanine AAG lysine GUU valine AAU asparagine UGC cysteine UCG serine UCU serine UUA leucine UAA stop A (i) Serine-alanine-asparagine-leucine-valine; (ii) serine-alanine-asparagine-leucinevaline; (iii) serine-alanine-asparagine(-stop) B (i) Serine-phenylalanine-asparagine-leucine-valine; (ii) serine-alanine-asparagineleucine-valine; (iii) serine-alanine-asparagine(-stop) C (i) Serine-alanine-asparagine-leucine-valine; (ii) serine-alanine-asparagine(-stop); (iii) serine-alanine-asparagine-leucine-valine D (i) Serine-alanine-asparagine-leucine-valine; (ii) serine-arginine-asparagine-leucinevaline; (iii) serine-alanine-asparagine(-stop) Solution 51 The solution is (A). The three mRNA sequences are (i) serine-alanine-asparagineleucine-valine; (ii) serine-alanine-asparagine-leucine-valine; (iii) serine-alanineasparagine(-stop). Refer to the table. Codon on mRNA Amino Acid GCA alanine AAG lysine GUU valine AAU asparagine UGC cysteine UCG serine UCU serine UUA leucine UAA stop ******************************************************** **************************336*************************** You are given three mRNA sequences: 1. 5′-UCG-GCA-AAU-UUA-GUU-3 2. 5′-UCU-GCA-AAU-UUA-GUU-3′ 3. 5′-UCU-GCA-AAU-UAA-GUU-3′ Using the peptide encoded by each sequence, compare the three peptides obtained. How are peptides 2 and 3 different from 1? What would the consequence be for the cell in each case? A There is a silent mutation in peptide 2, and peptide 3 has a stop codon due to a mutation. B There is a silent mutation in peptide 3, and peptide 2 has a stop codon due to a mutation. C There is a different amino acid in peptide 2, and peptide 3 has a stop codon due to a mutation. D There isn’t a mutation in peptide 2, and peptide 3 has a stop codon due to a mutation. Solution The solution is (A). In the case of peptide 2, there is a silent mutation. Although the base has been substituted, the amino acid is the same because the code is degenerate. In peptide 3, the mutation introduced a stop codon. Translation stops at the asparagine. SCIENCE PRACTICE CHALLENGE QUESTIONS 15.1 The Genetic Code 52 Gamow (1954) proposed that the structure of DNA deduced by Watson and Crick (1953) could be interpreted as a way of forming roughly 20 “words”" of the common amino acids from the four “letters” A, T, C, and G that represent DNA nucleotides. Crick and coworkers (1961) used a method developed by Benzer to induce mutations in the DNA of a virus by the insertion of a single nucleotide. The mutant could not infect the bacterium Escherichia coli and neither could viruses with a second insertion of a second DNA nucleotide. However, a third nucleotide insertion restored the ability of the virus to infect the bacterium. In 1961, Nirenberg and Matthaei conducted a series of experiments to better understand the flow of genetic information from gene to protein. They discovered that in solutions containing the contents of ruptured E. coli bacterial cells from which DNA had been removed, polymers containing only one repeating amino acid, phenylalanine, would be synthesized if synthetic mRNA composed of only the single nucleotide, uracil (U), was added to the solution in which phenylalanine was also present. In solutions containing mRNA with only adenine (A) or cytosine (C) and the amino acids lysine or proline, polymers containing only these amino acids would be synthesized. The researchers found ******************************************************** **************************************************335*** that when ribosomes were removed by filtration, these polymers did not form. Nirenberg and Leder (1964) extended this work to include other nucleotides. A. Summarize the conclusions regarding the encoding and decoding of heritable information supported by these studies. Explain how these studies provided evidence to support the triplet code. Khorana (1960) developed a technique for synthesizing RNA composed of predictable distributions of repeated pairs or triplets of nucleotides. He found, for example, that RNA synthesized when A and U were present in relative concentrations of 4:1, respectively, will produce RNA sequences with these distributions determined by their relative probabilities: AAU:AAA, AUA:AAA, and UAA:AAA; 0.82 0.2 / 0.83 1 / 4 [calculated as follows: (i) 4/5 of the bases are A, so the likelihood of selecting A is 0.8; (ii) the selection is repeated to determine the second letter of the three-letter codon; (iii) the likelihood of selecting a U is 1 in 5; (iv) the probability of selecting the set AUU is the product; (v) similarly, the probability of AAA is (4/5)3; and (vi) the ratio of these probabilities is their relative likelihood]: AUU:AAA, UUA:AAA, and UAU:AAA; 0.8 0.22 / 0.83 1 / 16; and UUU:AAA; 0.23 / 0.83 1 / 64 . B. Based on Khorana’s findings, calculate the relative distributions of the following ratios of concentrations of RNA triplet sequences from mixtures in which the relative concentrations of guanine and cytosine, G:C, are 5:1. Ratio Relative Probabilities GGC:GGG GCG:GGG CGG:GGG GCC:GGG CGC:GGG CCG:GGG CCC:GGG C. Based on the work of Nirenberg, Matthaei, Leder, and Khorana, the following table was constructed (taken from Khorana's Nobel Prize address): ******************************************************** **************************336*************************** A solution containing the amino acids shown in the table and equal concentrations of the two nucleotides C and G is prepared. Predict the proteins that can be synthesized from this mixture in terms of each possible codon and their relative concentrations in terms of their amino acid repeat sequences. D. Describe the effects of the codons UAA, UAG, and UGA on protein synthesis. Solution Sample answer: A. Gamow (a physicist) proposed that a sequence of three nucleotide letters could be used to make a single amino acid word. Crick and colleagues found that the length of the word was always three letters. Nirenberg, Matthaei, and Leder showed that DNA could be synthesized in the absence of a living cell and confirmed Gamow's idea. B. Ratio Relative Probabilities GGC:GGG GCG:GGG CGG:GGG 5 / 6 2 1 / 6 / 5 / 6 3 1 / 5 0.20 GCC:GGG CGC:GGG CCG:GGG 1 / 6 2 5 / 6 / 5 / 6 3 1 / 25 0.04 CCC:GGG 1 / 6 / 5 / 6 3 3 1 / 125 0.08 C. pro (CCC), ser (CCG), gly (CGG), arg (CGC), gly (GGG), gly (GGC), ala (GCC), and gly (GCG) So, concentrations of alanine, proline, arginine, and serine are equal and each in concentrations that are one-quarter of the concentration of glycine. The specific names of the amino acids are not in scope. D. These are the stop codons and do not encode amino acids. ******************************************************** **************************************************335*** 15.2 Prokaryotic Transcription 53 The yeast life cycle is usually dominated by haploid cells, each with a single set of unpaired chromosomes. The cell propagates asexually, and the genetic material is replicated through mitosis. Cell division occurs every 2–4 h, leading to 60–100 generations in a single day. Yeast also reproduces sexually, particularly under adverse environmental conditions. When two haploid cells—with DNA containing complementary mating-type alleles—conjugate, a diploid zygote results. The diploid zygote can then complete the sexual segment of the life cycle through meiosis. After meiosis, four haploid spores are produced, which can germinate. Researchers can grow yeast easily on nutrient-containing plates. Because both asexual and sexual reproduction is rapid, yeast has become an important organism for the experimental investigation of mutagenesis and evolution among eukaryotes. Environmental factors, such as chemicals or radiation, induce mutations. High-energy UVc radiation of less than 1 min in duration will result in many mutated yeast cells. UV-c can be used to mutate a strain of yeast in which the synthesis of adenine is blocked. This mutation is observable because the ade-2 mutant has a red color when cultured on nutrient-containing plates. Exposure to UV-c also can result in additional mutations. In particular, one mutant, ade-7, changes the color of the ade-2 mutant to white. A. You have a UV-c lamp, culture plates, and growth chambers at 23 °C and 37 °C. You also have available known haploid strains that are (ade-2,+,+), where + denotes the wild type. Design a plan to determine the rate of UV-c-induced mutations in nutrientcontaining plates inoculated with yeast. Earth's ozone layer removes high-energy ultraviolet radiation, UV-c, from the solar radiation received at the surface. Lower-energy ultraviolet radiation, UV-b, strikes Earth’s surface. Damage to DNA induced by ultraviolet radiation occurs with the formation of bonds between an adjacent pair of pyrimidine nucleotides, thymine and cytosine, on the same strand of DNA. A repair enzyme, photolyase, which is activated by visible light, is present in plants and most animals, but not in humans. In characterizing the relationship between environmental mutagens and cell damage, a useful assumption is often made and referred to as the linear hypothesis. This assumption states that the extent of damage is proportional to the amount of radiation received. Mutation rates for a strain (preac) that does not produce photolyase and a wild-type (+) strain were studied. Cultures of the two strains of yeast were diluted, and nutrientcontaining plates were inoculated in triplicate at 23 °C. The plates were exposed to bright sunlight for varying time intervals. After exposure, the plates were incubated in the dark at 23 °C. After incubation between 1 and 8 h, data shown in the table were collected by counting the density of living cells relative to the control, and averaging these among replicates. B. Using the data table, graph the average survival fraction, relative to the wild-type control. Predict the number of mutations in a sample of 1,000 cells of the preac type that are exposed to bright sunlight for 15 s. ******************************************************** **************************336*************************** Incubation Time (h) 10-s Exposure 20-s Exposure 30-s Exposure 40-s Exposure 50-s Exposure 1 0.83 0.58 0.33 0.17 0.08 2 1.00 0.43 0.17 0.09 0.04 3 0.92 0.38 0.12 0.03 0.01 4 0.75 0.35 0.08 0.01 0.00 5 0.99 0.49 0.11 0.01 0.00 6 0.81 0.42 0.12 0.01 0.00 7 0.80 0.32 0.09 0.01 0.00 8 1.05 0.59 0.11 0.01 0.00 Mean 0.89 0.45 0.14 0.04 0.02 Standard 0.11 0.10 0.08 0.06 0.03 Deviation Yeast can also be used to study sexual reproduction, a somewhat puzzling phenomenon. Cloning of cells through mitosis is molecularly much less complex than meiosis, consumes less energy, and is less risky. Two alternative explanations for the evolution of sexual reproduction are popular. In one model, through assortment of genes, meiosis leads to an increase in the frequency of beneficial mutations. In the second model, detrimental mutations are purged from a population through sex. Studies using yeast (Gray and Goddard, Evol. Biol., 2012 and McDonald et al., Nature, 2012) have provided a mechanism to study these models. The graph shows the fitness (defined as the log of the ratio of the number of cells in successive generations) of yeast as a function of number of mitotic reproductions in yeast grown in low- and high-stress environments, and with and without alternating induction of sexual reproduction. ******************************************************** **************************************************335*** C. Based on these data, evaluate the merits of the alternative theories of the adaptive advantage provided by sexual reproduction. Solution Sample answer: A. The main point should be that population measurements at multiple times after inoculation should be made with replicas at each time. Also, low to high dilutions will allow less uncertainty in the counts. Measurements of population density could count colonies or use a hemocytometer. It is essential that measurements of number are made. Triplets for each dilution are exposed to UV-c light for intervals such as 0, 10, 20, 30, 40, and 50 s. After exposure, each is incubated at 23° in the dark. Students should choose low temperature so yeasts will avoid sexual reproduction. At intervals of time (guided by the statement that cloning occurs every few hours) such as 1, 2, 3, 4, 5 h, and so on, the plates are removed from the incubator and counts are made. In this first experiment, we are counting the number of cells that are red. Mutations in the ade-2 will leave the cell white. Because we have a control and will be forming a fraction relative to the control Nred, exposed and mutated/Nred, no exposure we count either red or non-red. Students who have done this or a similar lab may know that a mutation in ade-4 will counteract the effect of ade-7. They might then go on to propose a test to confirm that the mutation was at ade-2. In this case, after the plates have been used to obtain population density, a test cross should be made with (ade-2, +, ade-7) and perhaps also with (+, ade-4, +). This can be done by streaking a colony of non-red cells on a line down the middle of the plate. Then a test strain should be streaked ******************************************************** **************************336*************************** onto the plate in a series of lines perpendicular to the first streak. Where the streaks nearly intersect, there will be a region of sexually reproduced cells after they are shifted to incubation at the higher temperature. Observation of incubated cells is made over the next two days. If the products of sexual reproduction of the (ade-2, +, ade-7) test are red, we know that the ade-2 allele was mutated. If the products of sexual reproduction of the (+, ade-4, +) test are red, we know that the ade-7 allele was mutated. To determine the mutation rate, the ratio to the control is formed. This ratio should be approximately constant among samples taken at different times during incubation after exposure. Taking an average of this ratio (and standard deviation) and plotting that average against the period of exposure yields an approximately straight line (or so we assume) whose slope is the mutation rate in units of sec 1 . B. This task is going to be daunting for students with weak backgrounds in math, or who are unfamiliar with graphing programs. Excel produced the graph shown below. Students should also have labels on the axes. C. These results show that under low-stress conditions, there is no additional gain in fitness due to sexual reproduction. However, under stress, there is a dramatic gain in fitness. ******************************************************** **************************************************335*** 15.3 Eukaryotic Transcription 54 A. Describe the storage and retrieval of genetic information with the following model. Use the list to fill in the blanks with the letter corresponding to the correct term. A. amino acid F. translation B. tRNA G. protein C. DNA H. RNA polymerase D. transcription I. rRNA E. mRNA Within the cytoplasm, __ is synthesized from __ bound to __ in a sequence that corresponds to information provided by __. This process is called __. blank blank blank blank blank Within the nucleus, information originating in __ is encoded as a sequence of bases in __, which is synthesized by the enzyme __. This process is called __. blank blank blank blank B. During development, cell differentiation occurs, and the expression of genes is permanently switched off. Using the model in (A), explain where information flow is most effectively blocked. C. A chemical message is received by the cell regulating the timing of events controlled by gene expression. Using the model in (A), explain where information flow is most effectively managed. Solution Sample answer: A. Within the cytoplasm G is synthesized from A bound to B in a sequence that corresponds to the information provided by E. This process is called F. Within the nucleus, information originating in C is encoded as a sequence of bases in I that is synthesized by the enzyme H. This process is called D. B. Permanently blocking gene expression is accomplished by the irreversible modification of DNA, stopping transcription. This is often accomplished by methylation, a concept that is out-of-scope for the AP Biology Exam. C. Posttranscriptional regulation is most efficient since the feedback loop has signals that vary in time, and the already transcribed information is available quickly. 55 Structure and function in biology result from both the presence of genetic information and the expression of that information. Some genes are continually expressed, whereas the expression of most genes is regulated, commonly at the level of transcription. At the initiation of transcription, the TATA-binding protein (TBP) provides access to the DNA strand to be transcribed. The 5′TATAAA3’ sequence called the TATA box is found in prokaryotes, archaebacteria, and eukaryotes. Even among eukarya, when the TATA box is not present among eukaryotes, the initiation of transcription involves TBP. Scientists attribute this common characteristic to the relative thermostability of the A-T interaction. Hydrogen bonds hold the two strands of the DNA double helix together. This type of bond ******************************************************** **************************336*************************** has the smallest interaction energy of all intermolecular forces; as temperature increases, these bonds are broken. A. Explain the advantage, in terms of the energy required, which is provided by an AT-rich region in the sequence where transcription is initiated. B. The fact that the TATA box or the associated TBP are common to all domains provides evidence of common ancestry among all life. Pose a scientific question that would need to be addressed by a valid alternative explanation of this fact. C. A whole-genome survey of prokaryotes (Zheng and Wu, BMC Bioinformatics, 2010) showed that the relative amounts of guanine and cytosine in DNA poorly predicted the temperature range conditions that are suitable for an organism. Refine the question posed in (B), taking this result into account. Solution Sample answer: A. Because A and T interact with only two hydrogen bonds (rather than three in the C-G interaction), the energy required to open the helix is smaller in the AT-rich region. B. Is the difference between the A-T and G-C interactions large enough that the TATA box would be the outcome for any DNA structure? C. How can we know if an initiation region with CG-rich regions could not transcribe DNA? This result shows that the difference between CG-rich and AT-rich regions is small enough not to provide a robust prediction of temperature range. So the difference should not be expected to be sufficient to allow multiple evolutionary lines to all adopt the TATA box solution. There must be a common ancestor. 15.5 Ribosomes and Protein Synthesis 56 Only a fraction of DNA encodes proteins. The noncoding portion of a gene is referred to as the intron. The intron fraction depends upon the gene. Introns are rare in prokaryotic and mitochondrial DNA; in human nuclear DNA, this fraction is about 95 percent. The intron is transcribed into mRNA, but this noncoding mRNA is edited out before translation of the coding portion, or exon, of a gene. The edited exon segments are then spliced together by a spliceosome, a very large and complex collection of RNAs and proteins. Although introns do not encode proteins, they have functions. In particular, they amplify the expression of the exon, although the mechanism is unknown. When introns are very long, which is common among mammalian genes with roles in development, they can significantly extend the time required to complete transcription. Analysis of genes common to different plant and animal species shows many shared intronic positions and base sequences, although in some organisms, such as yeast, many introns have been deleted. Because introns do not encode proteins, mutations can remain silent and accumulate. ******************************************************** **************************************************335*** A. Introns are ancestral remnants that are replicated because they do not disadvantage the organism. Consider the claim that introns are “junk DNA.” Evaluate the claim with supporting evidence. B. Introns may be retained during transcription. Explain how the retention of a transcribed intron between two transcribed exons within a gene could do the following: Solution Block expression of one polypeptide sequence. Increase expression of a polypeptide. Alter the polypeptide expressed. Sample answer: A. The fact that such a significant burden is imposed and the risk (such as splicing errors) is so great imply that there has to be an advantage in maintaining the noncoding sequences. Reasonable arguments can be made that the introns are just artifacts and have been for the four decades since the discovery of the intron. The assessment here is of the student’s ability to provide supporting evidence. This evidence is summarized in the stem. B. Consider a linear sequence with exons on either side of a retained intron. If the intron contains a stop code the second exon will not translate to protein. If both exons encode the same protein, expression will be increased. If the intron has a length that is not a multiple of three, it can cause the mRNA to be translated with nonfunctional proteins at either or both of the exons. ******************************************************** **************************360************************ 16 | GENE REGULATION REVIEW QUESTIONS 1 Control of gene expression in eukaryotic cells occurs at which level(s)? A Only the transcriptional level B Epigenetic and transcriptional levels C Epigenetic, transcriptional, and translational levels D Epigenetic, transcriptional, translational, and post-translational levels Solution 2 The solution is (D). Eukaryotes exhibit a tightly regulated mechanism of gene expression. Control of gene expression in eukaryotic cells occurs at the epigenetic, transcriptional, translational, and post-translational levels. What do Figures X and Y in the figure illustrate? A Transcription and translation in a eukaryotic cell (Figure X) and a prokaryotic cell (Figure Y) B Transcription and translation in a prokaryotic cell (Figure X) and a eukaryotic cell (Figure Y) ******************************************************** ***********************************************359*** C Transcription in a eukaryotic cell (Figure X) and translation in a prokaryotic cell (Figure Y) D Transcription in a prokaryotic cell (Figure X) and translation in a eukaryotic cell (Figure Y) Solution 3 The solution is (B). Figure X depicts coupled transcription and translation in prokaryotes. Eukaryotes show compartmentalization within their cells; therefore, transcription occurs inside the nucleus and translation occurs in the cytosol. Figure Y shows transcription and translation in eukaryotes. If glucose is absent but lactose is present, the lac operon will be — A activated B repressed C partially activated D mutated Solution 4 The solution is (A). Lactose acts as an inducer of the lac operon. When glucose is present, catabolite repression occurs and the lactose operon is repressed. What would happen if the operator sequence of the lac operon contained a mutation that prevented the repressor protein from binding the operator? A In the presence of lactose, the lac operon will not be transcribed. B In the absence of lactose, the lac operon will be transcribed. C The cAMP-CAP complex will not increase RNA synthesis. D The RNA polymerase will not bind the promoter. Solution 5 The solution is (B). If the repressor cannot bind to the operator, the lac operon will be transcribed when lactose is absent, since the RNA polymerase can bind to the promoter and initiate transcription. What would happen if the operator sequence of the trp operon contained a mutation that prevented the repressor protein from binding to the operator? A In the absence of tryptophan, the genes trpA–E will not be transcribed. B In the absence of tryptophan, only genes trpE and trpD will be transcribed. C In the presence of tryptophan, the genes trpA–E will be transcribed. D In the presence of tryptophan, the trpE gene will not be transcribed. Solution The solution is (C). If the operator sequence of the trp operon contained a mutation that prevented the repressor protein from binding to the operator, constitutive expression of the trp operon would occur regardless of the presence or absence of the tryptophan in the medium. ******************************************************** **************************360************************ 6 What are epigenetic modifications? A Addition of reversible changes to histone proteins and DNA B Removal of nucleosomes from the DNA C Addition of more nucleosomes to the DNA D Mutation of the DNA sequence Solution 7 The solution is (A). Epigenetic modifications do not change the sequence of the DNA and only bring about reversible changes in histone, proteins, and DNA. Which statement about epigenetic regulation is false? A Histone protein charge becomes more positive when acetyl groups are added. B DNA molecules are modified within CpG islands. C Methylation of DNA and histones causes nucleosomes to pack tightly together. D Histone acetylation results in the loose packing of nucleosomes. Solution 8 The solution is (A). Acetylation has the effect of changing the overall charge of the histone tail from neutral to negative. Acetylation disrupts the association between nucleosomes and DNA, leading to weaker binding of the nucleosomal components. What is true of epigenetic changes? A They only allow gene expression. B They allow movement of histones. C They change the DNA sequence. D They are always heritable. Solution 9 The solution is (B). Epigenetic modifications result in the movement of histones to open or close a chromosomal region. Open chromosomal regions can be transcribed, while closed chromosomal regions cannot be transcribed. The binding of what is required for transcription start? A A protein B DNA polymerase C RNA polymerase D A transcription factor Solution The solution is (C). The binding of RNA polymerase is required for transcription initiation. ******************************************************** ***********************************************359*** 10 What would be the outcome of a mutation that prevented DNA-binding proteins from being produced? A Decreased transcription because transcription factors would not bind to transcriptionbinding sites B Decreased transcription because enhancers would not be able to bind to transcription factors C Increased transcription because repressors would not be able to bind to promoter regions D Increased transcription because RNA polymerase would be able to increase binding to promoter regions Solution 11 The solution is (B). Enhancer regions are the binding sites for the transcription factors. When a DNA-binding protein binds to DNA, the shape of the DNA changes. This shape change allows activators bound to the enhancer regions to interact with transcription factors bound to the promoter region and the RNA polymerase. What will result from the binding of a transcription factor to an enhancer region? A Decreased transcription of an adjacent gene B Increased transcription of a distant gene C Alteration of the translation of an adjacent gene D Initiation of the recruitment of RNA polymerase Solution 12 The solution is (B). Enhancers are the DNA sequences that influence the rate of transcription by up-regulating the gene expression. What is involved in post-transcriptional control? A Control of RNA splicing B Ubiquitination C Proteolytic cleavage D Phosphorylation Solution 13 The solution is (A). Post-transcriptional control includes the control of RNA splicing after transcription. Gene A is thought to be associated with color blindness. The protein corresponding to gene A is isolated. Analysis of the protein recovered shows there are actually two different proteins that differ in molecular weight that correspond to gene A. What is one reason that two proteins may correspond to the gene? ******************************************************** **************************360************************ A One protein had a 5’ cap and a poly-A tail in its mRNA, and the other protein did not. B One protein had a 5’ UTR and a 3’ UTR in its RNA, and the other protein did not. C The gene was alternatively spliced. D The gene produced mRNA molecules with differing stability. Solution 14 The solution is (C). The alternative splicing of any gene can lead to the formation of proteins varying in their molecular weights. Binding of an RNA-binding protein will change the stability of the RNA molecule in what way? A Increase B Decrease C Neither increase nor decrease D Either increase or decrease Solution 15 The solution is (D). Binding of an RNA-binding protein (RBP) will either increase or decrease the stability of the RNA molecule depending on the specific RBP that binds. A mutation in the 5’UTR that prevents any proteins from binding to the region will — A increase or decrease the stability of the RNA molecule B prevent translation of the RNA molecule C prevent splicing of the RNA molecule D increase or decrease the length of the poly-A tail Solution 16 The solution is (A). The binding of RBP’s to the 5′UTR can increase or decrease the stability of an RNA molecule, depending on the specific RBP that binds. Any mutation in the 5′UTR can increase or decrease the stability of the RNA molecule. What can post-translational modifications of proteins affect? A mRNA splicing B 5’ capping C 3’ polyadenylation D Chemical modifications Solution The solution is (D). Chemical modifications occur post-translationally and can affect protein function. ******************************************************** ***********************************************359*** 17 A mutation is found in eIF-2, which impairs the initiation of translation. The mutation could affect all but one of the following functions of eIF-2. Which function would NOT be affected? A The mutation prevents eIF-2 from binding to RNA. B The mutation prevents eIF-2 from being phosphorylated. C The mutation prevents eIF-2 from binding to GTP. D The mutation prevents eIF-2 from binding to the 40S ribosomal subunit. Solution 18 The solution is (B). When eIF-2 is wildtype, it does not usually undergo phosphorylation and translation occurs. Mutation can lead to phosphorylation of eIF-2. Phosphorylated eIF-2 undergoes a conformational change and cannot bind to GTP. Therefore, the initiation complex cannot form properly and translation cannot occur. What does the addition of an ubiquitin group to a protein do? A Increases the stability of the protein B Decreases translation of the protein C Increases translation of the protein D Marks the protein for degradation Solution 19 The solution is (D). The addition of an ubiquitin group to a protein marks the protein for degradation. What are cancer-causing genes called? A Transformation genes B Tumor suppressor genes C Oncogenes D Proto-oncogenes Solution 20 The solution is (C). Cancer-causing genes are called oncogenes. Targeted therapies are used in patients with a certain gene expression pattern. A targeted therapy that prevents the activation of the estrogen receptor in breast cancer would be beneficial to what type of patient? A Patients who express the EGFR receptor in normal cells B Patients with a mutation that inactivates the estrogen receptor C Patients with over-expression of ER alpha in their tumor cells D Patients with over-expression of VEGF, which helps in tumor angiogenesis Solution The solution is (C). A targeted therapy can prove to be beneficial for patients showing over-expression of estrogen receptors. ******************************************************** **************************360************************ 21 In a new cancer treatment, a cold virus is genetically modified so that it binds to, enters, and is replicated in cells, causing them to burst. The modified cold virus cannot replicate when wildtype p53 protein is present in the cell. How does this treatment treat cancer without harming healthy cells? A The modified virus only infects and enters cancer cells. B The modified virus replicates in normal and cancer cells. C The modified virus only infects and enters normal cells. D The modified virus replicates only in cancer cells. Solution 22 The solution is (D). The treatment can treat cancer when the modified virus replicates only in cancer cells. A drug designed to switch silenced genes back on in cancer cells would result in what? A Methylation of DNA and deacetylation of histones B Methylation of DNA and acetylation of histones C Deacetylation of DNA and methylation of histones D Acetylation of DNA and demethylation of histones Solution 23 The solution is (A). Preventing methylation of DNA and acetylation of histones can switch on the silenced genes. What are positive cell-cycle regulators that can cause cancer when mutated called? A Transformation genes B Tumor suppressor genes C Oncogenes D Mutated genes Solution The solution is (C). Positive cell-cycle regulators that can cause cancer when mutated are called oncogenes. CRITICAL THINKING QUESTIONS 24 What best distinguishes prokaryotic and eukaryotic cells? A Prokaryotes possess a nucleus whereas eukaryotes do not, but eukaryotes show greater compartmentalization that allows for greater regulation of gene expression. B Eukaryotic cells contain a nucleus whereas prokaryotes do not, and eukaryotes show greater compartmentalization that allows for greater regulation of gene expression. ******************************************************** ***********************************************359*** C Prokaryotic cells are less complex and perform highly regulated gene expression, whereas eukaryotes perform less-regulated gene expression. D Eukaryotic cells are more complex and perform less-regulated gene expression, whereas prokaryotic cells perform highly regulated gene expression. Solution 25 The solution is (B). Eukaryotic cells contain a nucleus whereas prokaryotes do not, allowing greater regulation of gene expression in eukaryotes. Which statement is correct regarding the distinction between prokaryotic and eukaryotic gene expression? A Prokaryotes regulate gene expression at the level of transcription, whereas eukaryotes regulate at multiple levels including epigenetic, transcriptional, and translational. B Prokaryotes regulate gene expression at the level of translation, whereas eukaryotes regulate at the level of transcription to manipulate protein levels. C Prokaryotes regulate gene expression with the help of repressors and activators, whereas eukaryotes regulate expression by degrading mRNA transcripts, thereby controlling protein levels. D Prokaryotes control protein levels using epigenetic modifications, whereas eukaryotes control protein levels by regulating the rate of transcription and translation. Solution 26 All the cells of one organisms share the genome. However, during development, some cells develop into skin cells while others develop into muscle cells. How can the same genetic instructions result in two different cell types in the same organism? Thoroughly explain your answer. Solution 27 The solution is (A). Because prokaryotes perform transcription and translation at the same time, they regulate gene expression at the transcription level whereas eukaryotes regulate gene expression at multiple levels. Different genetic programs are turned on or off when cells differentiate into different cell types (e.g. skin cells, muscle cells, etc.) As a result, cells express genes needed for the tissue in which they are located. Which statement describes prokaryotic transcription of the lac operon? A When lactose and glucose are present in the medium, transcription of the lac operon is induced. B When lactose is present but glucose is absent, the lac operon is repressed. C Lactose acts as an inducer of the lac operon when glucose is absent. D Lactose acts as an inducer of the lac operon when glucose is present. Solution The solution is (C). Environmental stimuli can increase or induce transcription in prokaryotic cells. In this example, lactose in the environment will induce the ******************************************************** **************************360************************ transcription of the lac operon, but only if glucose is not available in the environment. 28 The lac operon consists of regulatory regions such as the promoter as well as the structural genes lacZ, lacY, and lacA, which code for proteins involved in lactose metabolism. What would be the outcome of a mutation in one of the structural genes of the lac operon? A Mutation in structural genes will stop transcription. B Mutated lacY will produce an abnormal galactosidase protein. C Mutated lacA will produce a protein that will transfer an acetyl group to galactosidase. D Transcription will continue but lactose will not be metabolized properly. Solution 29 In some diseases, alteration to epigenetic modifications turns off genes that are normally expressed. Hypothetically, how could you reverse this process to turn these genes back on? Solution 30 The solution is (D). A mutation in one of the structural genes of the lac operon will not prevent transcription of the operon. However, depending on the type of mutation in the gene, an abnormal protein may be produced, which could prevent metabolism of lactose. Sample answer: To turn these genes back on, you would have to reverse the epigenetic modifications. For example, if the genes are turned off due to methylation, removing the methyl groups should turn the genes back on. Flowering Locus C (FLC) is a gene that is responsible for flowering in certain plants. FLC is expressed in new seedlings, which prevents flowering. Upon exposure to cold temperatures, FLC expression decreases and the plant flowers. FLC is regulated through epigenetic modifications. What type of epigenetic modifications is present in new seedlings and after cold exposure? A In new seedlings, histone acetylations are present; upon cold exposure, methylation occurs. B In new seedlings, histone deacetylations are present; upon cold exposure, methylation occurs. C In new seedlings, histone methylations are present; upon cold exposure, acetylation occurs. D In new seedlings, histone methylations are present; upon cold exposure, deacetylation occurs. ******************************************************** ***********************************************359*** Solution 31 The solution is (A). Methylation of DNA causes nucleosomes to pack tightly together. Transcription factors cannot bind the DNA, and genes are not expressed. Histone acetylation results in loose packing of nucleosomes. Transcription factors can bind the DNA, and genes are expressed. Since the FLC gene is expressed in new seedlings, the corresponding DNA likely has histone acetylation. Since gene expression of the FLC gene decreases upon cold exposure, the corresponding DNA is likely methylated in response to cold temperatures. A mutation within the promoter region can alter gene transcription. How can this happen? A Mutated promoters decrease the rate of transcription by altering the binding site for the transcription factor. B Mutated promoters increase the rate of transcription by altering the binding site for the transcription factor. C Mutated promoters alter the binding site for transcription factors to increase or decrease the rate of transcription. D Mutated promoters alter the binding site for transcription factors and thereby cease transcription of the adjacent gene. Solution 32 The solution is (C). A mutation in the promoter region can change the binding site for a transcription factor that normally binds to increase transcription. The mutation could either decrease the ability of the transcription factor to bind, thereby decreasing transcription, or it can increase the ability of the transcription factor to bind, thus increasing transcription. What could happen if a cell had too much of an activating transcription factor present? A The transcription rate would increase, altering cell function. B The transcription rate would decrease, inhibiting cell functions. C The transcription rate decreases due to clogging of the transcription factors. D The transcription rate increases due to clogging of the transcription factors. Solution 33 The solution is (A). If too much of an activating transcription factor were present, then transcription would be increased in the cell. This could lead to dramatic alterations in cell function. The wnt transcription pathway is responsible for key changes during animal development. The transcription pathway shown in the figure uses arrows to represent activation and perpendicular symbols to represent repression of wnt gene products. Based on the pathway, how would blocking wnt gene expression affect the production of Bar-1? ******************************************************** **************************360************************ Solution 34 Sample answer: Bar-1 production would decrease. If wnt production is blocked, then MIG-1, LIN-17 are not produced, which results in MIG-5 not being produced either. When MIG-5 is not produced, PRY-1 is produced because there is no inhibition from MIG-5. Once PRY-1 is produced, it will repress BAR-1 production. How can RBPs prevent miRNAs from degrading an RNA molecule? A RBPs can bind first to the RNA, thus preventing the binding of miRNA, which degrades RNA. B RBPs bind the miRNA, thereby protecting the mRNA from degradation. C RBPs methylate miRNA to inhibit its function and thus stop mRNA degradation. D RBPs direct miRNA degradation with the help of a DICER protein complex. Solution 35 The solution is (A). RNA-binding proteins (RBP) bind to the RNA and can either increase or decrease the stability of the RNA. If they increase the stability of the RNA molecule, the RNA will remain intact in the cell for a longer period of time than normal. Since both RBPs and miRNAs bind to the RNA molecule, RBP can potentially bind first to the RNA and prevent the binding of the miRNA that will degrade it. How can external stimuli alter post-transcriptional control of gene expression? A UV rays can alter methylation and acetylation of proteins. B RNA-binding proteins are modified through phosphorylation. C External stimuli can cause deacetylation and demethylation of the transcript. D UV rays can cause dimerization of the RNA-binding proteins. ******************************************************** ***********************************************359*** Solution 36 The solution is (B). External stimuli can modify RNA-binding proteins (i.e., through phosphorylation of proteins) to alter their activity. Protein modifications can alter gene expression in many ways. How can phosphorylation of proteins alter gene expression? A Phosphorylation of proteins can alter translation, RNA shuttling, RNA stability, or posttranscriptional modification. B Phosphorylation of proteins can alter DNA replication, cell division, pathogen recognition, and RNA stability. C Phosphorylated proteins affect only translation and can cause cancer by altering the p53 function. D Phosphorylated proteins affect only RNA shuttling, RNA stability, and posttranslational modifications. Solution 37 The solution is (A). Because proteins are involved in every stage of gene regulation, phosphorylation of a protein (depending on the protein that is modified) can alter accessibility to the chromosome; can alter translation (by altering the transcription factor binding or function); can change nuclear shuttling (by influencing modifications to the nuclear pore complex); can alter RNA stability (by binding or not binding to the RNA to regulate its stability); can modify translation (increase or decrease); or can change post-translational modifications (add or remove phosphates or other chemical modifications). Changes in epigenetic modifications alter the accessibility and transcription of DNA. How could environmental stimuli, such as ultraviolet light exposure, modify gene expression? A UV rays could cause methylation and deacetylation of the genes that could alter the accessibility and transcription of DNA. B UV rays could cause phosphorylation and acetylation of the DNA and histones, which could alter the transcriptional capabilities of the DNA. C UV rays could cause methylation and phosphorylation of the DNA bases, which could become dimerized, rendering no accessibility of DNA. D UV rays can cause methylation and acetylation of histones, making the DNA more tightly packed and leading to inaccessibility. Solution 38 The solution is (A). Environmental stimuli, such as ultraviolet light exposure, can alter the modifications to the histone proteins or DNA. Such stimuli may change an actively transcribed gene into a silenced gene by removing acetyl groups from histone proteins or by adding methyl groups to DNA. New drugs are being developed that decrease DNA methylation and prevent the removal of acetyl groups from histone proteins. How could these drugs affect gene expression to help kill tumor cells? ******************************************************** **************************360************************ A These drugs maintain the demethylated and the acetylated forms of the DNA to keep transcription of necessary genes on. B The demethylated and the acetylated forms of the DNA are reversed when the silenced gene is expressed. C The drug methylates and acetylates the silenced genes to turn them back on. D Drugs maintain DNA methylation and acetylation to silence unimportant genes in cancer cells. Solution 39 The solution is (A). These drugs will keep the histone proteins and the DNA methylation patterns in the open chromosomal configuration so that transcription is feasible. If a gene is silenced, these drugs could reverse the epigenetic configuration to re-express the gene. How can understanding the gene expression pattern in a cancer cell tell you something about that specific form of cancer? A Understanding gene expression patterns in cancer cells will identify the faulty genes, which is helpful in providing the relevant drug treatment. B Understanding gene expression will help diagnose tumor cells for antigen therapy. C Gene profiling would identify the target genes of the cancer-causing pathogens. D Breast cancer patients who do not express EGFR can respond to anti-EGFR therapy. Solution 40 The solution is (A). Understanding which genes are expressed in a cancer cell can help diagnose the specific form of cancer. It also can help identify treatment options for that patient. For example, if a breast cancer tumor expresses EGFR in high numbers, it might respond to specific anti-EGFR therapy. If that receptor is not expressed, it would not respond to that therapy. What is personalized medicine? How can it be used to treat cancer? A Personalized medicines would vary based on the type of mutations and the gene’s expression pattern. B The medicines are given based on the type of tumor found in an individual’s body. C The personalized medicines are provided based only on the symptoms of the patient. D The medicines tend to vary depending on the severity and the stage of the cancer. Solution The solution is (A). The design of drugs on the basis of gene expression patterns within individual tumors is the basis of personalized medicine. With an increased understanding of gene regulation and gene function, medicines can be designed specifically to target diseased cells without harming healthy cells. Cancer is a heterogeneous disease with many different mutations and gene-signaling pathways, leading to the development and progression of the disease. By identifying the gene expression patterns in individuals and within individual tumors, treatments can be ******************************************************** ***********************************************359*** developed and prescribed to target only those genes and pathways that are abnormal. TEST PREP FOR AP® COURSES 41 What is found in both prokaryotes and eukaryotes? A 3’ poly-A tails B 5’ caps C Promoters D Introns Solution 42 The solution is (C). In prokaryotic as well as in eukaryotic genes, regulation of transcription takes place with the help of promoters. The enzyme polyadenylate polymerase catalyzes the addition of adenosine monophosphate to the 3’ ends of mRNAs to form a poly-A tail. If the enzyme were blocked so that it could not function, the result would be — A increased mRNA stability in eukaryotes and decreased mRNA stability in prokaryotes B decreased mRNA stability in eukaryotes and no effect in prokaryotes C no effect in eukaryotes and increased mRNA stability in prokaryotes D no effect in eukaryotes and decreased mRNA stability in prokaryotes Solution 43 The solution is (B). In eukaryotes, poly-A tails provide mRNA with stability and protection against random endonucleases. In prokaryotes, this modification is absent. What are two ways in which gene regulation differs and two ways in which it is similar in prokaryotes and eukaryotes? A Prokaryotes show co-transcriptional translation, whereas eukaryotes perform transcription prior to translation; in both cell types, regulation occurs through the binding of transcription factors, activators, and repressors. B Prokaryotes perform transcription prior to translation, whereas eukaryotes show cotranscriptional translation—that is, the processes occur in the same organelle. ******************************************************** **************************360************************ C Prokaryotes show co-transcriptional translation that is regulated prior to translation, whereas eukaryotes perform transcription prior to translation that is regulated only at the level of transcription. In both domains, transcription factors, activators, and repressors provide regulation. D Prokaryotes show cotranscriptional translation that occurs in the nucleus whereas eukaryotes show transcription prior to translation. In both cell types, regulation occurs using transcription factors, activators, and repressors. Solution 44 The solution is (A). In prokaryotes, RNA transcription and protein formation occur almost simultaneously. In eukaryotes, RNA transcription occurs prior to protein formation and takes place in the nucleus. Translation of RNA to protein occurs in the cytoplasm. In prokaryotes, gene expression is regulated primarily at the transcriptional level, for example through operons. In eukaryotes, gene expression is regulated at many levels (epigenetic, transcriptional, post-transcriptional, translational, and post-translational). In both prokaryotes and eukaryotes, gene expression can be regulated through transcription-factor binding of promoters. In both prokaryotes and eukaryotes, repressors can suppress transcription and activators can increase transcription in response to external stimuli. Lactose digestion in E. coli begins with its hydrolysis by the enzyme galactosidase. The gene encoding - galactosidase, lacZ, is part of a coordinately regulated operon containing other genes required for lactose utilization. Which figure correctly depicts the interactions at the lac operon when lactose is NOT being utilized? A B Solution 45 C D The solution is (D). The correct configuration when lactose is not being utilized is RNA polymerase on promoter and repressor protein bound to lactose, not bound to DNA. What would be the result of a mutation in the repressor protein that prevented it from binding lactose? A The repressor will bind to lactose when it is removed from the operator. B The repressor will bind the operator in the presence of lactose. C The repressor will not bind the operator in the presence of lactose. ******************************************************** ***********************************************359*** D The repressor will not bind the operator in the absence of lactose. Solution 46 The solution is (B). The active repressor normally binds to lactose if it is present. If the repressor is mutated, then it will not be able to bind with lactose and will, in turn, bind to the operator site and suppress the operon and RNA synthesis. What type of modification might be observed in the GR gene in all newborn rats? A The DNA will have many methyl molecules. B The DNA will have many acetyl molecules. C The DNA will have few methyl groups. D The histones will have many acetyl groups. Solution 47 The solution is (A). DNA, when methylated at many locations, suppresses the expression of the gene. Therefore, the GR genes would not be expressed in the newborn rats. What type of modification will be observed in the GR gene in the highly nurtured rats? A The DNA will have many methyl molecules. B The DNA will have many acetyl molecules. C The DNA will have few methyl groups. D The histones will have few acetyl groups. Solution 48 The solution is (C). The highly nurtured pups will show a greater amount of GR gene expression, thereby showing very few methylated molecules in the DNA. The low methylation would be responsible for the higher expression of the GR gene. The level of transcription of a gene is tested by creating deletions in the gene as shown in the illustration. These modified genes are tested for their level of transcription: (++) normal transcription levels; (+) low transcription levels; (+++) high transcription levels. Which deletion is in an enhancer involved in regulating the gene? ******************************************************** **************************360************************ A Deletion 1 B Deletion 2 C Deletion 3 D Deletion 4 Solution 49 The solution is (C). Deletion 3 is an enhancer involving in regulating the gene. There are deletions in the gene sequence in Deletion 3 which reduces the transcription level. In the figure, which deletion is in a repressor involved in regulating the gene? A Deletion 1 B Deletion 2 C Deletion 3 D Deletion 4 Solution 50 The solution is (A). Deletion 1 is in a repressor, as there is a sudden increase in the level of transcripts when it is induced. If it were not a repressor, then the level of transcript would be lower, proving that is it a repressor. The diagram shows different regions (1–5) of a pre-mRNA molecule, a mature-mRNA molecule, and the protein corresponding to the mRNA. A mutation in which region is most likely to be damaging to the cell? ******************************************************** ***********************************************359*** A Region 1 B Region 2 C Region 3 D Region 5 Solution 51 The solution is (B). Region 2 seems to be encoding a gene. Any mutation in this region would likely produce a nonfunctional protein, damaging the cell. What do regions 1 and 5 correspond to? A Exons B Introns C Promoters D Untranslated regions Solution 52 The solution is (D). The untranslated regions are useful for post-transcriptional regulation. The 5’UTR (leader sequence) contains the ribosome-complex binding site and 3’UTR (trailer sequence) contains binding sites for regulatory proteins. What are regions 1 through 5 in the diagram? A Regions 1, 3, and 5 are exons; regions 2 and 4 are introns. B Regions 2 and 4 are exons; regions 1, 3, and 5 are introns. C Regions 1 and 5 are exons; regions 2, 3, and 4 are introns. D Regions 2, 3, and 4 are exons; regions 1 and 5 are introns. Solution The solution is (A). Regions 1, 3, and 5 are the exons. These are the DNA regions encoding a useful gene. In between every exon, larger introns are present, corresponding to regions 2 and 4 in the diagram. ******************************************************** **************************360************************ 53 A mutation results in the formation of the mutated mature-mRNA as indicated in the diagram. What type of mutation occurred, and what is the likely outcome of the mutation? A Mutation in the GU-AG sites of introns produced a nonfunctional protein. B A transversion mutation in the introns led to alternative splicing, producing a functional protein. C A transversion mutation in the GU-AG site mutated this mRNA, producing a nonfunctional protein. D Transition mutations in the introns could produce a functional protein. Solution 54 The solution is (A). The mutation caused a failure in recognition of the intron 2 end and instead appears to have recognized the end of the next intron (intron 4). This caused excision of introns 2, 3, and 4. The most likely outcome of this mutation is a truncated protein that will be nonfunctional. The diagram illustrates the role of p53 in response to UV exposure. What would be the result of a mutation in the p53 gene that inactivates it? ******************************************************** ***********************************************359*** A Skin will peel in response to UV exposure. B Apoptosis will occur in response to UV exposure. C No DNA damage will occur in response to UV exposure. D No peeling of skin will occur in response to UV exposure. Solution 55 The solution is (D). Mutation in p53 would restrict its function, causing no activation of p21 and also no apoptosis, leading to no peeling of the skin. What will NOT occur in response to UV exposure if a p53 mutation inactivates the p53 protein? A Damage to DNA, p53 activation, and p21 activation B p21 activation and apoptosis C p21 activation D p53 activation, p21 activation, and apoptosis Solution 56 The solution is (C). If p53 inactivates due to a mutation, then p21 will get inactivated too. As a result, apoptosis will not take place. What happens when tryptophan is present? A The repressor binds to the operator, and RNA synthesis is blocked. B RNA polymerase binds to the operator, and RNA synthesis is blocked. C Tryptophan binds to the repressor, and RNA synthesis proceeds. D Tryptophan binds to RNA polymerase, and RNA synthesis proceeds. Solution 57 The solution is (A). In the trp operon, the tryptophan binds to the inactive aporepressor and makes it active. This active repressor would bind to the operator site, blocking RNA synthesis. What happens in the absence of tryptophan? A RNA polymerase binds to the repressor. B The repressor binds to the promoter. C The repressor dissociates from the operator. D RNA polymerase dissociates from the promoter. Solution The solution is (C). The absence of tryptophan would inactivate the repressor, dissociating it from the operator. This apo-repressor now remains in the cell in an inactive form, allowing the operon to synthesize RNA. ******************************************************** **************************360************************ 58 Anabaena is a simple multicellular photosynthetic cyanobacterium. In the absence of fixed nitrogen, certain newly developing cells along a filament express genes that code for nitrogen-fixing enzymes and become non-photosynthetic heterocysts. The specialization is advantageous because some nitrogen-fixing enzymes function best in the absence of oxygen. Heterocysts do not carry out photosynthesis, but instead provide adjacent cells with fixed nitrogen and receive fixed carbon and reduced energy carriers in return. As shown in the diagram, when there is low fixed nitrogen in the environment, an increase in the concentration of free calcium ions and 2-oxyglutarate stimulates the expression of genes that produce two transcription factors (NtcA and HetR) that promote the expression of genes responsible for heterocyst development. HetR also causes production of a signal, PatS, that prevents adjacent cells from developing as heterocysts. Based on your understanding of the ways in which signal transmission mediates cell function, which prediction is most consistent with the information given? A In an environment with low fixed nitrogen, treating the Anabaena cells with a calciumbinding compound should prevent heterocyst differentiation. B A strain that overexpresses the PatS gene should develop many more heterocysts in a low nitrogen environment. C In an environment with abundant fixed nitrogen, free calcium levels should be high in all cells, preventing heterocysts from developing. D In environments with abundant fixed nitrogen, loss of the HetR gene should induce heterocyst development. Solution The solution is (A). As increased calcium stimulates the expression of heterocyst development genes, providing Anabaena cells with calcium-binding compound inhibits heterocyst development. ******************************************************** ***********************************************359*** 59 Which statement about Anabaena is false? A Decreasing the concentration of free calcium ions will prevent heterocyst development. B In the presence of fixed nitrogen, NtcA will not be expressed. C Low fixed nitrogen levels result in increased PatS levels. D A mutation in NtcA that makes it nonfunctional also will allow adjacent cells to develop as heterocysts. Solution The solution is (D). The NtcA gene is responsible for heterocyst development whereas the HetR gene promotes PatS gene, which controls the heterocyst development in adjacent cells. SCIENCE PRACTICE CHALLENGE QUESTIONS 16.4 Eukaryotic Transcriptional Gene Regulation 60 The operon model describes expression in prokaryotes. Describe this model and the essential difference in the way in which expression is regulated in eukaryotes. Solution Sample answer: An operon is a cluster of genes involved in the same biochemical pathway. The genes are transcribed together, and are all under the control of the same promoter. In eukaryotes, genes are not clustered into operons, and a different promoter regulates each gene. An operon includes structural genes involved in a single biochemical pathway which are under the same control and transcribed into a single mRNA. Either the operon is active and all structural genes are transcribed or the operon is off and none of the ******************************************************** **************************360************************ genes are transcribed. Upstream of the structural genes are located the regulatory sequences that include a promoter where RNA polymerase binds and operators where repressors bind. The genes encoding the repressor proteins are usually not part of the operon. Activator sequences can be located upstream of the promoter. Operons are found in bacteria with few exceptions in some organisms such as yeast and C. elegans. In eukaryotic cells, genes are generally transcribed individually. Each gene is preceded by its own promoter upstream (at the 5′ end) and a transcription terminator at the 3′ end. Genes encoding proteins involved in the same pathway are nevertheless expressed individually. ******************************************************** **********************************************************379*** 17 | BIOTECHNOLOGY AND GENOMICS REVIEW QUESTIONS 1 How are GMOs created? A Introducing recombinant DNA into an organism by any means B In vitro fertilization methods C Mutagenesis D Plant breeding techniques Solution 2 The solution is (A). Recombinant DNA is DNA that has been genetically modified in the laboratory. A genetically modified organism (GMO) is created by introducing recombinant DNA into an organism. Which technique used to manipulate genetic material results in a significant increase in DNA or RNA fragments? A Gel electrophoresis B Nucleic acid extraction C Nuclear hybridization D Polymerase chain reaction (PCR) Solution 3 The solution is (D). PCR is a method used to make many copies of DNA or RNA fragments from a small number of copies. What is the role of plasmids in molecular cloning? A They are used to create clones. B They are used as vectors to insert genes into bacteria. C They are a functional part of binary fission. D They contain the circular chromosome of prokaryotic organisms. Solution 4 The solution is (B). Plasmids are vectors that can be used to insert genes into bacteria. What is meant by a recombinant DNA molecule? A Chimeric molecules B Bacteria transformed into another species C Molecules that have been through the PCR process D The result of crossing over during cell reproduction Solution The solution is (A). Recombinant DNA is a chimeric molecule, as it has been genetically modified in the laboratory. ******************************************************** **************************380*********************************** 5 What is Bt toxin is considered to be? A A gene for modifying insect DNA B An organic insecticide produced by bacteria C A nerve toxin in humans D A strain of genetically modified tomatoes Solution 6 The solution is (B). Bt, or Botulinum toxin, is lethal to insects and is, therefore, an organic insecticide produced by a bacterium. What is one trait of the Flavr Savr Tomato? A Has a better shelf life B Is not a variety of vine-ripened tomato in the supermarket C Was not created to have better flavor D Undergoes soft rot Solution 7 The solution is (A). The Flavr Savr Tomato was genetically modified to resist rot and to ripen more slowly, which gives it a better shelf life. What is first step in isolating DNA? A Generating genomic DNA fragments with restriction endonucleases B Introducing recombinant DNA into an organism by any means C Overexpressing proteins in E. coli D Lysing the cells in the sample Solution 8 The solution is (D). The first step in isolating DNA is cell lysis, a process in which the cell membrane is broken. What is genomics? A Genomics is the study of entire genomes, including the complete set of genes, their nucleotide sequence and organization, and their interactions within a species and with other species. B Genomics is the process of finding the locations of genes on each chromosome. C Genomics is an illustration that lists genes and their location on a chromosome. D Genomics is a genetic marker is a gene or sequence on a chromosome that cosegregates (shows genetic linkage) with a specific trait. Solution The solution is (A). Genomics is the study of entire genomes, including the complete set of genes, their nucleotide sequence and organization, and their interactions within a species and with other species. ******************************************************** **********************************************************379*** 9 What is required in addition to a genetic linkage map to build a complete picture of the genome? A A genetic marker B A physical map C Linkage analysis of chromosomes D Plasmids Solution 10 The solution is (B). A physical map, which shows the arrangement of genes on the chromosome, is necessary to build a complete picture of the genome. Genetic recombination occurs by which process? A Crossing over B Chromosome segregation C Independent assortment D Sister chromatids Solution 11 The solution is (A). Genetic recombination occurs when homologous chromosomes exchange material in a process called crossing over. Individual genetic maps in a given species are — A genetically similar B genetically identical C genetically dissimilar D not useful in species analysis Solution 12 The solution is (A). Individual genetic maps in a given species are genetically similar but not identical. What procedure uses information obtained by microscopic analysis of stained chromosomes? A Cytogenetic mapping B Radiation hybrid mapping C RFLP mapping D sequence mapping Solution The solution is (A). A cytogenetic map is the visual appearance of a chromosome that has been stained. ******************************************************** **************************380*********************************** 13 What is true about linkage analysis? A It is used to create a physical map. B It is based on the natural recombination process. C It involves breaking and re-joining of DNA artificially. D It requires radiation hybrid mapping. Solution 14 The solution is (B). Genes far apart on a chromosome are separated by natural recombination more frequently than genes that are close together. Linkage analysis assesses the relative order of genes based on this natural recombination frequency. What does the chain termination method of DNA sequencing use to terminate polynucleotide elongation? A Labeled dideoxynucleotides B Unlabeled dideoxynucleotides C Labeled deoxynucleotides D Unlabeled deoxynucleotides Solution 15 The solution is (A). Chain termination occurs when a dideoxynucleotide is introduced into the DNA strand. Each dideoxynucleotide is labeled so that the strand can be visualized. What sequencing technique is used to identify regions of similarity between cell types or species? A Dideoxy chain termination B Proteins, DNA, or RNA sequence alignment C Shotgun sequencing D Whole-exome sequencing Solution 16 The solution is (B). Similarity among cell types or species can be assessed by aligning the sequence of proteins, DNA, or RNA. Whole-genome sequencing can be used for advances in what field? A Bioinformatics B Iron industry C Multimedia D The medical field Solution The solution is (D). Whole genome sequencing yields genetic information that can lead to medical advances. ******************************************************** **********************************************************379*** 17 Sequencing an individual person’s genome — A is currently impossible B helps identify genetic mutations associated with certain diseases C will not lead to legal issues regarding discrimination and privacy D will not help make informed choices about medical treatment Solution 18 The solution is (B). Sequencing an individual’s genome yields information about genetic mutations associated with disease. Genomics can be used in agriculture to do what? A Generate new hybrid strains B Improve disease resistance C Improve yield D Improve yield, resistance, and generate hybrids Solution 19 The solution is (D). Genomics yields a wide range of information that can lead to improved yield, resistance to disease, and the generation of new hybrid strains. What are the uses of metagenomics? A Identification of biofuels B Testing for multiple drug susceptibility in a population C Use in increasing agricultural yields D Identifying new species more rapidly and analyzing the effect of pollutants on the environment Solution 20 The solution is (D). Metagenomics is the study of genetic material recovered from the environment. This information can be used to identify new species and, over time, to assess the effect of pollutants on the environment. Genomics can be used on a personal level to do what? A Determine the risks of genetic diseases for an individual’s children B Increase transplant rejection C Predict the career success of a person D Produce antibodies for an antigen Solution The solution is (A). Genomics can be used to determine whether an individual has genes that might cause genetic disease in his or her children. ******************************************************** **************************380*********************************** 21 What is the percentage of single gene defects causing disease in developed countries? A 0.05 B 0.1 C 0.2 D 0.4 Solution 22 The solution is (A). Approximately 0.05, or 5 percent, of single gene defects cause disease in persons living in developed countries. The rapid identification of new species and the analysis of the effect of pollutants on the environment is a function of what? A Metagenomics B Linkage analysis C Genomics D Shotgun sequencing Solution 23 The solution is (A). Metagenomics involves the study and analysis of genetic material obtained from the environment. The risks of genetic diseases for an individual’s children can be determined through — A metagenomics B linkage analysis C genomics D shotgun sequencing Solution 24 The solution is (C). Genomics involves analyzing genes, which can determine the chances of genetic diseases appearing in an individual’s children. What is a biomarker? A The color coding of different genes B A protein uniquely produced in a diseased state C A molecule in the genome or proteome D A marker that is genetically inherited Solution 25 The solution is (B). A biomarker is a protein only produced when disease is present. What is a metabolome? A A provisional listing of the genome of a species B A unique metabolite used to identify an individual ******************************************************** **********************************************************379*** C A method used for protein analysis D The complete set of metabolites related to the genetic makeup of an organism Solution 26 The solution is (D). A metabolome is complete set of metabolites related to the genetic makeup of an organism. How would you describe a set of proteins with altered expression levels? A A group of biomarkers B A protein signature C The result of a defect in mRNA transcription D The results of crossing over during cell replication Solution 27 The solution is (B). A protein signature is a unique set of proteins present in a disease state. What is a protein signature? A A protein expressed on the cell surface B A unique set of proteins present in a diseased state C The path followed by a protein after it is synthesized in the nucleus D The path followed by a protein in the cytoplasm Solution 28 The solution is (B). A protein signature is a unique set of proteins present in a diseased state What describes a protein that is uniquely produced in a diseased state? A A genomic protein B A genetic defect C A chimeric molecule D A biomarker Solution 29 The solution is (D). A biomarker is a protein that indicates the occurrence of a particular disease. What are the metabolites that result from the anabolic and catabolic reactions of an organism called? A Genetic metabolic profile B Metabolic signature C Metabolome D Metagenomic Solution The solution is (C). Metabolome refers to the metabolites that are produced by catabolic and anabolic biochemical reactions. ******************************************************** **************************380*********************************** CRITICAL THINKING QUESTIONS 30 What is the process of Southern blotting? A Southern blotting is used to find particular DNA sequences. Fragments are separated on gel, incubated with probes to check for sequence of interest, and transferred to nylon membrane. B Southern blotting is used to find particular DNA sequences. Fragments are separated on gel, transferred to nylon membrane, and incubated with probes to check for sequence of interest. C Southern blotting is used to find particular RNA sequences. Fragments are separated on gel, transferred to nylon membrane, and incubated with probes to check for sequence of interest. D Southern blotting is used to find particular RNA sequences. Fragments are separated on gel, incubated with probes to check for sequence of interest, and transferred to nylon membrane. Solution 31 The solution is (B). Southern blotting is used to find particular DNA sequences. Fragments are separated on gel, transferred to nylon membrane, and incubated with probes to check for sequence of interest. A researcher wants to study cancer cells from a patient with breast cancer. Is cloning the cancer cells an option? A The cancer cells should be cloned along with a biomarker for better detection and study. B The cells should be screened first in order to assure their carcinogenic nature. C The cancer cells, being the clones of each other already, should directly be grown in a culture media and then studied. D The cancer cells should be extracted using the specific antibodies. Solution 32 The solution is (C). Cancer cells are by definition clones of each other. All the researcher needs to do is grow the patient’s cell through cell culture and study them. What are the uses of genome mapping? A Genome mapping is useful in identifying human disease-causing genes, developing microbes to clean up pollutants, and increasing crop yield. B Genome mapping is directly required to produce recombinants, in FISH detection, and detecting the methylated parts of genetic material. C Genome mapping is useful for knowing the pedigree of diseases in humans and tracing the movement of transposons in plants. D Genome mapping identifies human disease-causing genes only. ******************************************************** **********************************************************379*** Solution 33 The solution is (A). Human genome maps help researchers in their efforts to identify human disease-causing genes and can be used in a variety of other applications, such as using live microbes to clean up pollutants or even prevent pollution. Research involving plant genome mapping may lead to producing higher crop yields or developing plants that better adapt to climate change. If you had a chance to get your genome sequenced, what are some questions you might be able to have answered about yourself? A You can determine the drugs that can rectify the disease, symptoms of the disease, and its severity. B You can determine the ancestry and genetic origin of diseases and their susceptibility to drugs. C You can predict the symptoms of disease, the vectors to be used in gene therapy, and the causal organism of the disease. D You can determine the pedigree of a disease, produce recombinants, and detect the presence of extracellular genes using FISH. Solution 34 The solution is (B). It would be possible to determine ancestry, tendency to develop some diseases that are of genetic origin, or susceptibility to drugs. What is an example of a genomic mapping method? A The radiation mapping method is an example which uses radiations to break the DNA and is affected by the change in recombination frequency. B Cytogenetic mapping obtains information from microscopic analysis of stained chromosomes. It can estimate the approximate distance between markers. C In restriction mapping, the DNA fragments are cut by using the restriction enzymes and then stained fragments are viewed on gel. D Cytogenetic mapping obtains information from microscopic analysis of stained chromosomes. It can estimate the exact base pair distance between markers. Solution 35 The solution is (B). Cytogenetic mapping uses information obtained by microscopic analysis of stained sections of the chromosome. It is possible to determine the approximate distance between genetic markers using cytogenetic mapping, but not the exact distance (number of base pairs). What are three methods of gene sequencing? A Chain termination method – automated sequencers are used to generate sequences of short fragments; Shotgun sequencing method – incorporation of ddNTP during DNA replication; Next-generation sequencing – cutting DNA into random fragments, sequencing using chain termination, and assembling overlapping sequences B Chain termination method – incorporation of ddNTP during DNA replication; Shotgun sequencing method – cutting DNA into random fragments, sequencing using chain ******************************************************** **************************380*********************************** termination, and assembling overlapping sequences; Next-generation sequencing – automated sequencers are used to generate sequences of short fragments C Chain termination method – incorporation of ddNTP during DNA replication; Shotgun sequencing method – automated sequencers are used to generate sequences of short fragments; Next-generation sequencing – cutting DNA into random fragments, sequencing using chain termination, and assembling overlapping sequences D Chain termination method – automated sequencers are used to generate sequences of short fragments; Shotgun sequencing method – cutting DNA into random fragments, sequencing using chain termination, and assembling overlapping sequences; Next-generation sequencing – incorporation of ddNTP during DNA replication Solution 36 The solution is (B). The basic sequencing technique used in all modern-day sequencing projects is the chain termination method (also known as the dideoxy method), which was developed by Fred Sanger in the 1970s. The chain termination method involves DNA replication of a single-stranded template with the use of a primer and a regular dideoxynucleotide (ddNTP), which is a monomer of DNA. In the shotgun sequencing method, several copies of a DNA fragment are cut randomly into many smaller pieces (somewhat like what happens to a round shot cartridge when fired from a shotgun). All of the segments are then sequenced using the chainsequencing method. Next-generation sequencing is a group of automated techniques used for rapid DNA sequencing. These automated, low-cost sequencers can generate sequences of hundreds of thousands or millions of short fragments (25 to 500 base pairs) in the span of one day. What is the greatest challenge facing genome sequencing? A The lack of resources and use of chemicals for the sequencing of the DNA fragments B The ethical issues such as discrimination based on person’s genetics C The use of chemicals during the sequencing methods could incorporate mutations D The scientific issues such as conserving the human genome sequences Solution The solution is (B). The ethical issues surrounding genome sequencing are the most challenging. Humans have a responsibility to use this knowledge wisely. Otherwise, it could be easy to misuse the power of such knowledge, leading to discrimination based on a person's genetics, human genetic engineering, and other ethical concerns. This information also could lead to legal issues regarding health and privacy. ******************************************************** **********************************************************379*** 37 How is shotgun sequencing performed? A The DNA is cut into fragments, sequencing is done using chain termination method, fragments are analyzed to see the overlapping sequences, and the entire fragment is reformed. B The DNA is cut into fragments, overlapping sequences are analyzed using computer, sequencing is done using chain termination method, and the DNA fragment is reformed. C The DNA is cut into fragments, stained with fluorescent dye, and sequenced using the chain termination method; the fragments are analyzed to see the overlapping sequences; and the entire DNA fragment is reformed. D The DNA is cut into fragments, sequencing is done using the chain termination method, the DNA is stained with fluorescent dye, and a computer is used to analyze and reform the entire DNA fragment. Solution 38 The solution is (A). In the shotgun sequencing method, several copies of a DNA fragment are cut randomly into many smaller pieces (somewhat like what happens to a round shot cartridge when fired from a shotgun). All of the segments are sequenced using the chain-sequencing method. Then, with the help of a computer, the fragments are analyzed to see where their sequences overlap. By matching up overlapping sequences at the end of each fragment, the entire DNA sequence can be reformed. Coumadin is a drug frequently given to prevent excessive blood clotting in stroke or heart attack patients, which could lead to another stroke or heart attack. Administration of the drug also can result in an overdose in some patients, depending on the liver function of a patient. How could pharmacogenomics be used to assist these patients? A Pharmacogenomics could provide a counteracting drug to decrease the effect of Coumadin. B Pharmacogenomics could test every patient for their sensitivity to the drug. C Pharmacogenomics will not be able to provide any help to patients highly sensitive to the drug. D Pharmacogenomics could provide an overdose to each patient to test for the symptoms of the drug. Solution The solution is (B). Pharmacogenomics allows each patient to be tested for genotype associated sensitivity to drugs, thereby identifying patients who might experience an overdose of drugs prior to administration. ******************************************************** **************************380*********************************** 39 Why is so much effort being poured into genome mapping applications? A Genome mapping is necessary to know the base pair difference between the markers. B The mapping would help scientists understand the role of proteins in specific organelles. C The mapping technique identifies the role of transposons. D Genome mapping helps identify faulty alleles, which could cause diseases. Solution 40 The solution is (D). A genetic map of the human genome for multiple individuals could identify alleles of genes susceptible to cancer causing agents. The mapping could also identify allele variations resistant to changes resulting in cancer, thereby offering the opportunity for genetic therapy for the disorders. What is the reason for studying mitochondrial genomics that is most directly important for humans? A Mitochondria evolved from bacteria; therefore, their genome is important to study. B Mitochondria undergo rapid mutation; and it is essential that this pattern be studied. C Mitochondria contain DNA, and it is passed on from mother to the offspring, which renders it helpful in tracing genealogy. D Mitochondria are the only ATP-producing organelles of the cell, thus their genome is important. Solution 41 The solution is (C). Mitochondria are intracellular organelles that contain their own DNA. Mitochondrial DNA mutates at a rapid rate and often is used to study evolutionary relationships. Another feature that makes studying the mitochondrial genome interesting is mitochondrial DNA in most multicellular organisms only is passed on from the mother. For this reason, mitochondrial genomics often is used to trace genealogy. How can proteomics complement genomics? A The genes are responsible for producing proteins, which implies that proteomics complements genomics. B Genomics is responsible for deciding the structure of the proteins and, thereby, the result of proteomic studies. C The genome is constant, but proteome is dynamic as different tissues possess the same genes but express different genes, thereby complementing genomics. D The study of genes is incomplete without the study of their respective proteins, thus they complement each other. ******************************************************** **********************************************************379*** Solution 42 The solution is (C). Proteomics complements genomics and is useful when scientists want to test their hypotheses that were based on genes. Even though all cells of a multicellular organism have the same set of genes, the set of proteins produced in different tissues is different and dependent on gene expression. Thus, the genome is constant, but the proteome varies and is dynamic within an organism. How could a proteomic map of the human genome help find a cure for cancer? A A genetic map could help identify genes that could counteract the cause of cancer. B Metabolomics can be used to study the genes producing metabolites during cancer. C Proteomics detects biomarkers whose expression is affected by the disease process. D The mapping helps analyze the inheritance of cancer-causing genes. Solution 43 The solution is (C). Proteomic approaches are being used to improve screening and early detection of cancer. This is achieved by identifying proteins whose expression is affected by the disease process. An individual protein is called a biomarker, whereas a set of proteins with altered expression levels is called a protein signature. What contributions have been made through the use of microbial genomics? A Microbial genomics has provided various tools to study psychological behaviors of organisms. B Microbial genomics has been useful in producing antibiotics, enzymes, improved vaccines, disease treatments, and advanced cleanup techniques. C Microbial genomics has contributed resistance in other bacteria by horizontal and lateral gene transfer mechanisms. D Microbial genomics has contributed to fighting global warming. Solution The solution is (B). Microorganisms are used to create products, such as enzymes that are used in research, antibiotics, and other anti-microbial mechanisms. Microbial genomics is helping develop diagnostic tools, improved vaccines, new disease treatments, and advanced environmental cleanup techniques. TEST PREP FOR AP® COURSES 44 In separating DNA for genomic analysis, it is important to consider RNA contaminating the sample during the cell lysis step of a DNA extraction, which is likely to cause what to occur? A DNA separates into the supernatant. B The protease destroys the DNA. C RNA does not affect the DNA. D DNA combines with the RNA. Solution The solution is (C). DNA is unaffected by the RNA. ******************************************************** **************************380*********************************** 45 There are many techniques for investigating human genomic disorders. Western blotting looks for protein, eastern blotting looks for post-translational changes, northern blotting looks at mRNA, and Southern blotting looks at DNA. If you were to look at sickle cell anemia, a disorder affecting hemoglobin produced in red blood cells, which technique would be the most useful in detecting a polymorphism in a sample? A Northern blotting B Southern blotting C Western blotting D Eastern blotting Solution 46 The solution is (B). The polymorphism results in change (mutation) in the sequence of a gene, so the analyses of DNA will be most useful. A population of insects was formally distinguished by a mix of colors on their thorax and legs. This population now appears to be split into two subgroups, purple legged and orange legged. Researchers hypothesize that the purple-legged group may be increasing resistance to the Bt (Bacillus thuringiensis) toxin. Which idea supports this observation? A Transgenesis B Natural selection C Hybridization D Recombination Solution 47 The solution is (B). The resistance to certain environmental pressure is result of mutations that produce organisms able to survive and reproduce, transmitting the favorable trait, which in turn will increase the number of organisms carrying the trait. Which statement describes the process of molecular cloning? A The foreign DNA and plasmid are cut with the same restriction enzyme, and DNA is inserted within the lacZ gene, whose product metabolizes lactose. The foreign DNA and vector are allowed to anneal. The vector is transferred to a bacterial host that is ampicillin sensitive, and those with a disrupted lacZ gene show an inability to metabolize X-gal. B The foreign DNA and plasmid are denatured using high heat, and DNA is inserted within the lacZ gene, whose product metabolizes glucose. The foreign DNA and vector are allowed to anneal. The vector is transferred to a bacterial host that is ampicillin sensitive, and the disrupted lacZ gene will metabolize X-gal. ******************************************************** **********************************************************379*** C The foreign DNA and plasmid are cut with the same restriction enzyme, and DNA is inserted randomly in the plasmid. The foreign DNA and vector are allowed to anneal. The vector is transferred to a bacterial host that is ampicillin sensitive, and the disrupted lacZ gene shows an inability to synthesize X-gal. D The foreign DNA and plasmid are cut with the same restriction enzyme, and DNA is inserted within the lacZ gene, whose product metabolizes lactose. The foreign DNA and vector are allowed to anneal. The vector is transformed into a viral host that is ampicillin sensitive, and the disrupted lacZ gene shows an inability to synthesize X-gal. Solution 48 The solution is (A). The foreign DNA and plasmid are cut with the same restriction enzyme, which recognizes a particular sequence of DNA called a restriction site. The restriction site occurs only once in the plasmid, and is located within the lacZ gene, a gene necessary for metabolizing lactose. The restriction enzyme creates sticky ends, which allow the foreign DNA and cloning vector to anneal. Ligase, an enzyme, binds the annealed fragments together. The ligated cloning vector is transformed into a bacterial host strain that is ampicillin sensitive and missing the lacZ gene from its genome. Bacteria are grown in a medium containing ampicillin and X-gal, a chemical metabolized by the same pathway as lactose. The ampicillin kills bacteria without the plasmid. Plasmids lacking the foreign insert have an intact lacZ gene and are able to metabolize X-gal, releasing a dye that turns blue. Plasmids with an insert have a disrupted lacZ gene and produce white colonies. There are three methods of creating maps to evaluate genomes: cytogenetic (staining chromosomes), radiation hybrid maps (fragments with X-rays), and sequence maps (comparing DNA sequences). Which option accurately describes the three methods? A Cytogenetic mapping – stained sections of chromosomes are analyzed using microscope, the distance between genetic markers can be found; Radiation hybrid mapping – breaks DNA using radiation and is affected by recombination frequency; Sequence mapping – DNA sequencing technology used to create physical maps B Cytogenetic mapping – stained sections of chromosomes are analyzed using microscope, the approximate distance between genetic markers can be found; Radiation hybrid mapping – breaks DNA using radiation and is unaffected by recombination frequency; Sequence mapping – DNA sequencing technology used to create physical maps C Cytogenetic mapping – stained sections of chromosomes are analyzed using microscope, the distance in base pairs between genetic markers can be found; Radiation hybrid mapping – breaks DNA using radiation and is unaffected by recombination frequency; Sequence mapping – DNA sequencing technology used to create physical maps D Cytogenetic mapping – stained sections of chromosomes are analyzed using a telescope, the distance between genetic markers can be found; Radiation hybrid mapping – breaks DNA using radiation and is affected by recombination frequency; Sequence mapping – DNA sequencing technology used to create physical maps ******************************************************** **************************380*********************************** Solution 49 The solution is (B). Cytogenetic mapping uses information obtained by microscopic analysis of stained sections of the chromosome. It is possible to determine the approximate distance between genetic markers using cytogenetic mapping, but not the exact distance (number of base pairs). Radiation hybrid mapping uses radiation, such as X-rays, to break the DNA into fragments. The amount of radiation can be adjusted to create smaller or larger fragments. This technique overcomes the limitation of genetic mapping and is not affected by increased or decreased recombination frequency. Sequence mapping resulting from DNA sequencing technology allowed for the creation of detailed physical maps with distances measured in terms of the number of base pairs. How many cells with different genetic variations are possible after a single round of meiosis? A Two B Three C Four D Eight Solution The solution is (C). Complete meiosis produces four gametes which are genetically different due to events in meiosis I (recombination during prophase I, and the random alignment of the homologous chromosomes during metaphase I). SCIENCE PRACTICE CHALLENGE QUESTIONS 17.1 Biotechnology 50 Prokaryotes have an adaptive strategy to identify and respond to viral infections. This strategy uses segments of the cyclic DNA called CRISPRs and genes coding for CRISPRassociated (cas) proteins. When a virus enters the cell, a strand of the viral DNA is excised by a cas protein and inserted into the bacterial DNA in a CRISPR region. When the same viral DNA is encountered subsequently, this foreign DNA is targeted by cas proteins that carry RNA markers transcribed from the inserted segment. The cas proteins cleave the viral DNA. The bacteria “remember” the infectious agent, providing a form of immunity. A. Use the diagram to identify the components of a transcript-based response of bacteria to the presence of viral DNA by placing the corresponding number next to each feature of the diagram. blank ******************************************************** **********************************************************379*** ___ viral DNA blank blank blank blank ___ cellular DNA ___ excised viral DNA ___ cas protein–RNA–viral DNA complex blank blank blank blank ___ cas protein ___ cas protein–RNA complex ___ cell membrane ___ stored viral DNA template ___ degraded viral DNA The CRISPR system was discovered in cultures of yogurt in 2002. Subsequently, researchers developed a technology based on manipulation of this system. The code for the prokaryotic CRISPR/cas system is highly conserved and is found in the human genome. DNA sequences are known that encode proteins responsible for many heritable diseases. CRISPR/cas is a technology that allows DNA to be cleaved at the boundaries of a nucleotide sequence, making the protein dysfunctional. The break in the strand is then recognized and replaced with code for the functional protein. If the editing is done with zygote-forming cells, the change is inherited. Not only the patient, but all progeny of the patient is cured. This technology is the first to easily make genomic modifications of a germ line. In the words of a prominent molecular biologist, this technology, which was recognized as the Breakthrough of 2015 in the journal Science, “democratizes genetic engineering.” Just as PCR became a standard tool that is widely used, any molecular biology lab is now able to apply this technology. B. Pose three questions—whose pursuit would require an understanding of genetics— regarding the ethical and social issues that accompany the use of this medical technology. C. Explain the value of genetic variation within a population. Predict a possible effect that this technology could have, if unregulated, on human genetic variation. Solution Sample answer: A. 4 viral DNA 9 degraded viral DNA 3 cellular DNA 2 cas protein 5 excised viral DNA 6 cas Protein–RNA complex 8 cas protein–RNA–viral DNA complex 1 cell membrane 7 stored viral DNA template B. The nature of the problem elicits predictions of future events which cannot be tested now. However, the question should discriminate between those that are likely to eventually be answerable and those that will not be. For example, should the use of the technology be restricted or should the use depend on the ability to pay, are questions that will be decided without scientific reasoning. There are many possible questions that do require reasoning. Sample questions: ******************************************************** **************************380*********************************** How will disease be defined and differentiated from a phenotypic variation? What is the uncertainty associated with environmental factors that may or may not be required for expression? What is the likelihood of developing a disease? How do we weight the induction of expression due to environmental factors that may be avoided? If the gene in question is in a gene network, what is the relative importance of the targeted gene? At the time when a decision regarding the use of the technology is needed, how will expertise needed to answer these questions be accessed? C. Phenotypic variation within a population increases likelihood of survival of genes within that population. Genome editing will cause a reduction in variation. If unregulated, the genome could behave as if the population were small, since effective population size is a measure of diversity. Small populations have greater risk of extinction. 51 Gel electrophoresis of polymers and polymer fragments is an important element in many investigations. Samples of a solution are pipetted onto a gel. The gel is placed in a solution that maintains a constant pH, and an electric field is applied over the length of the gel. Separated components are placed on a substrate where they can be visualized and identified by comparison with samples of standards. Application of this method to DNA is called a Southern blot, named for the inventor of the technology. Application to RNA is called a northern blot, another demonstration that biologists have fun (there are also western, eastern, and far-eastern blots, but these techniques are not named for their inventor). A. Consider the three amino acids shown in the figure and explain how when placed on a gel in an electric field the amino acids would move, how the amino acids would be separated as they moved, and which would move further. B. A biologist wants to determine whether a new protocol is successful in constructing and amplifying a molecular clone of a segment of DNA introduced as a plasmid. After the ******************************************************** **********************************************************379*** procedure is complete, the bacterial cells containing plasmid with the inserted segment are lysed, and a gel is run on which spots of the lysate and the sequence to be cloned have been pipetted. Use the data displayed in the developed gel shown in the figure to evaluate the question of whether or not the protocol was successful. C. Design a plan to answer the question of whether the new DNA has been incorporated in the DNA of the host organism. Solution Sample answer: A. The electric field exerts a force on a charged molecule. Higher molecular mass molecules migrate less far, as do molecules with smaller charge and so the components will be separated and in this case it is uncertain whether histidine, with a larger charge but also a greater mass, will travel further than serine. Certainly alanine will travel further than serine. The rubric will award points for reasoning, not for correct answer. B. The sequence is indicted in the run of the lysate. However, this does not mean that the sequence is incorporated into the genome. C. To determine whether the sequence is heritable, the examination of lysate would be repeated after the passage of a few generations. So, the plan would be grow the cells on nutrient media, dilute, plate, grow, and repeat this sequence a few times. Then lyse the cells and run a gel. 17.3 Whole-Genome Sequencing 52 Genetic engineering can be applied to heritable information to produce what is referred to as a “knockdown organism.” Biotechnology also can be applied to produce nonheritable changes in a “knockdown gene.” Post-transcriptional strategies target the mRNA product of a gene. One such strategy uses the conserved genes that encode RNA interference (RNAi) proteins for the regulation of level of mRNA transcription. Some viral RNA is double stranded (dsRNA). A cell responds to the presence of double stranded RNA by attachment of the enzyme DICER which cuts dsRNA into short fragments. One strand of the fragment is transferred to the RNA induced silencing complex (RISC) which searches for mRNA with a sequence matching that of the fragment strand. When detected, this mRNA is degraded. ******************************************************** **************************380*********************************** A. Common in cancer cells is a mutation of the gene that codes for the protein p53, whose role is to detect and repair errors in DNA and, if repairs cannot be made, initiate apoptosis. Create a visual representation to explain how the DICER-RISC system within the cell can be used to suppress the translation of a mutated form of the gene coding for p53, potentially destroying a tumor. B. Whole genome sequences provide a library of potentially expressed proteins, but they do not provide information on the functions of each protein. In an approach called reverse genetics, investigations attempt to determine the function of the gene, often by silencing the gene using RNAi technology. Assume that you have the ability to synthesize dsRNA from a DNA segment taken from an organism whose whole genome has been determined. Design a plan for collecting data that could be used to assign a function to the protein encoded by this sequence. (Hint—Do not worry about the number of experiments that might need to be conducted to implement your plan. An automated technique called high throughput screening robotically supports thousands of simultaneous experiments.) Solution Sample answers: A. Because the sequence coding for the mutant p53 is known, a dsRNA complementing this template can be constructed and introduced with a viral vector. When the dsRNA is recognized, the DICER-RISC machinery will repress the mutant. This technology was applied to this problem in 2014. B. A dsRNA is constructed using the desired sequence as the template. Cells (probably using a surrogate such as yeast) are exposed. Any suspected function can be evaluated by presenting the appropriate substrate and looking for reaction products or growing the cells on nutrient media and looking for the absence of product. ******************************************************** ****************************************************************405**** 18 | EVOLUTION AND ORIGIN OF SPECIES REVIEW QUESTIONS 1 Which scientific concept did Charles Darwin and Alfred Wallace independently discover? A Mutation B Natural selection C Overbreeding D Sexual reproduction Solution 2 The solution is (B). Darwin and Wallace both independently described the process of natural selection, the mechanism by which evolutionary change is produced. Which statement about a natural principle that points to the inevitability of natural selection is false? A Most characteristics of organisms are inherited. B Offspring vary among each other in regard to their characteristics. C Some generations of offspring do not need to compete for resources. D Certain traits will be better represented in the next generation. Solution 3 The solution is (C). After reading the work of economist Thomas Malthus, Darwin recognized that all species over-reproduce. He reasoned that at each and every generation, then, there is competition for resources to survive. What is the best definition of adaptation? A A trait or behavior that aids an organism’s survival and reproduction B A heritable trait or behavior that aids an organism’s survival and reproduction C A trait or behavior that aids a population’s survival and reproduction D A heritable trait or behavior that aids a population’s survival and reproduction Solution 4 The solution is (B). An adaptation is a heritable characteristic that aids an organism’s survival and reproduction. What is an example of an adaptation? A The better nutrition of a human helps her grow taller. B The webbed feet of a duck help it swim. C The urban location of a raccoon helps it find food. D The large leaves of a desert plant require more water. Solution The solution is (B). An adaptation is a heritable trait or behavior, like the webbed feet of a duck, which helps it swim. ******************************************************** ***************************404***************************************** 5 Which process is divergent evolution? A Groups of organisms evolve in different directions from a common point. B A new species develops rapidly when an event cuts off a portion of a population. C Groups of organisms independently evolve to similar forms. D A species evolves when a few members move to a new geographical area. Solution 6 The solution is (A). When two species evolve in diverse directions from a common point, divergent evolution occurs. Which situation is most likely an example of convergent evolution? A Some fish that live in total darkness have eyes. B Hawks and other birds have feathers. C Worms and snakes both move without legs. D Flowers that look very different have the same reproductive organs. Solution 7 The solution is (C). Worms and snakes do not share a recent common ancestry, yet they move similarly. This is likely an example of convergent evolution. What are homologous structures? A Physical structures that have no apparent function B Parallel structures in diverse organisms C Physical structures that are used only occasionally D Similar structures in diverse organisms Solution 8 The solution is (B). Scientists call synonymous parts in diverse species homologous structures. What are two examples of vestigial structures? A Gills in fish and parts of the throat in humans B Butterfly wings and dragonfly wings C Hind leg bones in whales and leaves on some cacti D Shark fins and dolphin fins Solution The solution is (C). Vestigial structures are those with minimal to no utility for an organism. Whales do not use hind leg bones and leaves are of minimal use to cacti. They represent remnants of features useful to the organisms’ ancestors. ******************************************************** ****************************************************************405**** 9 Which statement best describes the relationship between the theory of evolution and the origin of life? A The theory includes an explanation of life’s origins. B The theory cannot explain the origin of life. C The theory does not try to explain the origin of life. D The theory does not contribute understanding to pre-life processes. Solution 10 The solution is (C). The theory of evolution describes how populations change over time and how life diversifies the origin of species; it does not try to explain life’s origins. Which statement best describes what happens when an antibiotic is applied to a population of bacteria? A The bacteria develop resistance to the antibiotic in direct response to its application. B The bacteria’s genetic material mutates in response to the antibiotic, resulting in resistance. C A gene for resistance, already present in the population, decreases in frequency. D A gene for resistance, already present in the population, increases in frequency. Solution 11 The solution is (D). A gene for resistance is in the population already. The antibiotic kills bacteria without the resistant gene, strongly selecting resistant individuals. The gene for resistance thus increases in frequency in the gene pool. Which option is the best definition of species? A A group of individual organisms with significant genetic similarities B A group of individual organisms with significant genetic similarities that share external and internal characteristics C A group of individual organisms that interbreed D A group of individual organisms that interbreed and produce viable, fertile offspring Solution 12 The solution is (D). The best definition is a group of individual organisms that interbreed and produce viable, fertile offspring. What do scientists focus on to distinguish between species? A Ecological niches B Morphological differences C Reproductive barriers D Genetic changes ******************************************************** ***************************404***************************************** Solution 13 The solution is (C). Biological species concept is widely used to distinguish between species. It states that species are groups of interbreeding natural populations that are reproductively isolated from other such groups. What are two primary sources of genetic variation? A Mutations and sexual reproduction B Isolation and sexual reproduction C Sexual reproduction and asexual reproduction D Migration and sexual reproduction Solution 14 The solution is (A). The accumulation of mutations and the genetic shuffling of sexual reproduction are two primary sources of genetic variation. Which statement best describes the relationship between genetic variation and speciation? A Without genetic variation, speciation would occur more slowly. B Without genetic variation, speciation would not be possible. C Genetic variation influences sympatric speciation, but not allopatric speciation. D There is no relationship between genetic variation and any form of speciation. Solution 15 The solution is (B). Two basic mechanisms of evolutionary change—natural selection and genetic drift—cannot operate without genetic variation. Other mechanisms of evolutionary change, migration and mutation, produce genetic variation. Genetic variation makes evolutionary change possible. Which statement about postzygotic barriers is false? A They occur after fertilization. B They include hybrids that are sterile. C They include hybrid organisms that do not survive the embryonic stage. D They include reproductive organ incompatibility. Solution 16 The solution is (D). Reproductive organ incompatibility prevents reproduction from taking place, so it is a prezygotic barrier rather than a postzygotic barrier. Which situation is an example of a prezygotic barrier? A Two species of fish produce sterile offspring. B Two species of flowers attract different pollinators. C Two species of insects mate, but the zygote does not survive. D Two species of lizards mate, but the offspring dies before reproducing. Solution The solution is (C). A postzygotic barrier takes place after zygote formation. An organism does not survive the embryonic stage or is born sterile. ******************************************************** ****************************************************************405**** 17 Which situation would most likely lead to allopatric speciation? A A flood causes the formation of a new lake. B A storm causes several large trees to fall down. C A mutation causes a new trait to develop. D An injury causes an organism to seek out a new food source. Solution 18 The solution is (A). Allopatric speciation occurs when a population becomes geographically isolated. The formation of a new lake is one way in which such isolation could occur. What is the main difference between an autopolyploid individual and an allopolyploid individual? A Number of extra chromosomes B Functionality of extra chromosomes C Source of extra chromosomes D Number of mutations in the extra chromosomes Solution 19 The solution is (C). In autopolyploids, the source of the extra chromosomes is the individual itself. A diploid parent produces a polyploid offspring. In allopolyploids, different species combine to produce polyploid offspring. What is unique about speciation due to adaptive radiation? A It leads to multiple species forming from one parent species. B It only occurs on or around island archipelagos. C It requires a population to disperse from its parent species. D It is a special kind of sympatric speciation. Solution 20 The solution is (A). Adaptive radiation occurs when many adaptations evolve from a single point of origin. This is a unique quality, as multiple speciations fuel greater evolutionary change. What is least likely to be a factor that increases the probability of speciation by adaptive radiation? A There are vacant ecological niches nearby. B Genetic drift in a population increases. C There are isolated regions with suitable habitats. D There are few competitor species. Solution The solution is (B). Genetic drift is not likely to be a factor in speciation by adaptive radiation. ******************************************************** ***************************404***************************************** 21 In a hybrid zone, in addition to interacting, what else do two closely related species do? A Compete B Reproduce C Transition D Fuse Solution 22 The solution is (B). Two closely related species that interact and reproduce to form hybrids do so in an area called a hybrid zone. Which situation means reinforcement is more likely to occur in the hybrid zone? A The hybrid offspring are more fit than the parent species. B Reproductive barriers weaken. C The hybrid offspring are about as fit as the parent species. D Reproductive barriers strengthen. Solution 23 The solution is (D). When reproductive barriers strengthen, hybrids are generally not as fit as the parent species, and reinforcement of speciation takes place. Which statement is false? A Gradual speciation and punctuated equilibrium both result in the divergence of species. B Punctuated equilibrium is most likely to occur in a large population in a stable environment. C In the punctuated equilibrium model, gradualism is not excluded. D In the gradual speciation model, traits change incrementally. Solution 24 The solution is (B). Punctuated equilibrium is most likely to occur in a smaller population and in an environment undergoing change. Which component of speciation would be least likely to be a part of punctuated equilibrium? A A division in populations B A change in environmental conditions C Ongoing gene flow D A number of mutations occurring at once Solution The solution is (C). Continued gene flow is likely to slow down the rate of speciation. This is less likely to be a part of the evolutionary change described by the model of punctuated equilibrium. ******************************************************** ****************************************************************405**** CRITICAL THINKING QUESTIONS 25 What conclusions can you draw about the relationship between the way in which the present-day theory of evolution developed and the credibility of the theory? Explain your thinking. A When the theory of evolution was first proposed, it was met with a lot of criticism and disbelief, but it is widely supported today. Theories that have withstood a larger amount of criticism are more credible than those that are accepted easily. B The theory of evolution has its foundation in both biological and geological observations, making it a more credible theory because it can explain more about the world. C The theory of evolution relies on the heritability of traits, but the mechanism of this inheritance was not understood when the theory was developed. This reduces the credibility of the theory because the people who created it did not understand how it worked. D It is meaningful that two naturalists working independently from each other offered the same explanation for the same set of phenomena. When two people independently look at the same evidence and come to the same conclusion, this reinforces the credibility of that conclusion. Solution 26 The solution is (D). I can conclude that the way in which the theory was developed says something about its credibility. It is meaningful that two naturalists who worked independently from each other offered the same explanation for the same set of phenomena. When two people independently look at the same (or similar) body of evidence and come to the same conclusion, this reinforces the credibility of that conclusion. This same idea is part of today’s scientific process, in which studies must be replicable to be considered valid. How does an adaptation, such as better running speed, relate to natural selection? A Natural selection produces beneficial adaptations, such as better running speed, in individuals that run more frequently. B Natural selection randomly mutates individuals’ genetic code until it produces beneficial adaptations, such as better running speed. C Natural selection produces adaptations, such as better running speed, to help individuals survive and reproduce. D Natural selection reproduces individuals with favorable genetic traits, such as the adaptation of better running speed, over time. Solution The solution is (C). An adaptation is a heritable trait that allows an individual organism to better survive and reproduce in its present environment. An animal that runs faster, for example, is better suited to catching prey and/or avoiding predators, and thus more likely to survive and pass on its ability to run quickly to its offspring. ******************************************************** ***************************404***************************************** Natural selection is the reproduction of individuals with favorable genetic traits— such as the adaptation of better running speed—over time. 27 What is an example of convergent evolution? How does it support the theory of evolution by natural selection? A An example of convergent evolution is the development of the same function, swimming, in organisms that live in different parts of the globe, such as Arctic beluga whales and Antarctic right whales. The fact that organisms that do not come in contact with each other have developed the same traits suggests that natural selection can produce similar adaptations in organisms who share a similar environment. B An example of convergent evolution is the set of adaptations, such as better running speed or more efficient hunting, developed by a species in response to competition with a new species that moves into the same region. The fact that a species adapts after it comes into contact with a competitor suggests that natural selection works more quickly with higher selective pressures. C An example of convergent evolution is the development of an ancestral structure, a limb, into two different modern structures, such as a hand and a flipper. The fact that natural selection can cause a structure to develop down two different pathways due to different environmental conditions supports the theory of evolution. D An example of convergent evolution is the development of the same function, flying, in organisms that do not share a recent common ancestry, such as insects and birds. The fact that wings that allow flight have developed from very different original structures suggests that the process of natural selection can produce similar adaptations in two very different types of organisms who share a similar environment. Solution 28 The solution is (D). An example of convergent evolution is the development of the same function, flying, in organisms that do not share a recent common ancestry, such as insects and birds. The fact that wings that allow flight have developed from very different original structures suggests that the process of natural selection can produce similar adaptations in two very different types of organisms who share a similar environment. Why do scientists consider vestigial structures evidence for evolution? A Vestigial structures are the result of convergent evolution, so they are good evidence that natural selection act similarly in similar environmental conditions. B Vestigial structures are the result of common ancestry, so they are good evidence that different populations of organisms evolved from a common point. C Vestigial structures are the result of convergent evolution, so they are good evidence for an end goal to evolution. D Vestigial structures are the result of common ancestry, so they are good evidence for a common origin of all life. ******************************************************** ****************************************************************405**** Solution 29 The solution is (A). Vestigial structures are, in fact, homologies—underlying patterns in form (if not exactly function) that are the result of common ancestry. Whether or not they are useful, the fact that anatomical similarities exist even in very different organisms is good evidence that different populations of organisms evolved from a common point. Furthermore, the actual uselessness of vestigial structures points to ongoing evolution—these structures are no longer part of the adaptive features of an organism, but likely once were. Reproduction in sexually-reproducing organisms occurs when two sex cells, or gametes, fuse. In fish, this occurs when sperm swim through the water to find the ovum. In flowers, pollen is dispersed through the air and carried to another flower. Explain what evolutionary adaptations for reproduction occur in humans, based on the fact that we are land-based animals. Solution Because humans are land-based animals, the female must provide an internal, fluid environment in which sperm will not be desiccated and can move. Thus, the male sperm is deposited directly into the internal reproductive tract of the female, which creates an insulated environment in which sperm can survive and travel to the ovum. 30 While examining the human genome, you find a gene that is not homologous to any other organisms known to man. You conclude that this gene must be unique to the human species and could not have evolved from another organism. Would this discovery suggest that humans do not share a common ancestor with all other organisms on Earth? Explain your answer. Solution No. Although all organisms descended from other organisms, it is possible for a new gene to form in the human genome, or any organism, through processes such as mutation or crossing-over during meiosis. Such a gene would not need to have been inherited from an ancestor and humans would still likely share many other genes with other Earth organisms. 31 Mutations in the glucose 6-phosphate dehydrogenase (G6PD) gene can cause a rare anemia when inherited. However, homozygotes with this mutation are less prone to malaria infection, a disease that historically was the most widespread deadly disease among humans. Predict how this mutation would affect the fitness of individuals living in countries where malaria is endemic. Solution 32 Despite the ability of G6PD to cause deadly anemia, the mutation is favored and will remain in the population because it causes resistance to malaria, a much more widespread disease among humans. How does the scientific meaning of theory differ from the common vernacular meaning? A A scientific theory is a hypothesis that needs to be tested, whereas people often use theory to mean a simple guess. B A scientific theory is a statement that has been proven correct, while people often use it to mean a statement that has not yet been verified. ******************************************************** ***************************404***************************************** C A scientific theory is a thoroughly tested set of explanations for a body of observations of nature, while people often use it to mean a guess or speculation. D A scientific theory is a random guess, while people often use it to mean a statement that is somewhat based in fact. Solution 33 The solution is (B). In science, a theory is a body of well tested and verified explanations for a set of observations about the natural world. Its use connotes rigorous examination of evidence and general consensus by experts in the field. In common vernacular, theory is used to mean a guess or suggested explanation. It does not connote rigorous examination of evidence. It is much more like the scientific term hypothesis, which is a reasoned explanation but has not yet been tested and verified. Why is having a way of defining species and distinguishing between them important for the study of evolution? A A distinction between species allows scientists to understand the common origin of all species. B A common definition of species allows scientists to agree on all aspects of the theory of evolution. C Divergence can only occur at the species level: It does not occur to larger taxa. Therefore, it is important to know which groups are distinct species. D In the study of evolution, the species is the unit over which change is measured. Solution 34 The solution is (A). Studying evolution means studying change among life forms over time. It is always important to have a unit, or benchmark, for measuring change. In the study of evolution, the species is that unit or benchmark. If a population stopped reproducing sexually, but still reproduced asexually, how would its genetic variation be affected over time? Could speciation occur in this situation? A Genetic variation would increase, and speciation would be possible. B Genetic variation would increase, and speciation would not be possible. C Genetic variation would decrease, and speciation would be possible. D Genetic variation would decrease, and speciation would not be possible. Solution 35 The solution is (C). Over time, its genetic variation would probably decrease over all, since sexual reproduction is a primary means of variation. There would still be the possibility of mutations introducing genetic variation, and depending on the species, migration could introduce different genes into the population. Some genetic variation is necessary for any evolutionary change to happen. Over time, speciation would still be possible, especially if a portion of the population was isolated from the rest and mutations occurred. What role do prezygotic and postzygotic barriers play in speciation? ******************************************************** ****************************************************************405**** A Prezygotic and postzygotic barriers allow for the formation of less-fit hybrids that reinforces speciation. B Prezygotic and postzygotic barriers prevent interbreeding of species such that there is no gene flow between them. ******************************************************** ***************************404***************************************** C Prezygotic and postzygotic barriers prevent migration of the two species, causing them to remain in contact with each other and begin to interbreed. D Prezygotic and postzygotic barriers are present only in newly formed species, allowing scientists to identify the time of divergence of the species. Solution 36 A population of flowers was separated into two subpopulations when a new river cut through the plain in which they were growing. The number of interbreeding events per year for the two subpopulations of flowers is shown in the graph below. Twenty-four years after they were separated, can you conclude that the two subpopulations of flowers have become new species? Why or why not? Solution 37 The solution is (B). Speciation is the formation of two separate species from one original species. Inherent in the definition of speciation is the development of reproductive isolation—the inability to interbreed. Both prezygotic barriers and postzygotic barriers prevent interbreeding of species such that there is no gene flow between them. No; just because interbreeding does not occur does not mean the flowers are incapable of interbreeding. Which type of speciation, allopatric or sympatric, is more common? Why? A Allopatric speciation is more common because it prevents gene flow between the species. B Allopatric speciation is more common because it involves stronger prezygotic barriers. ******************************************************** ****************************************************************405**** C Sympatric speciation is more common because it prevents gene flow between the species. D Sympatric speciation is more common because it involves stronger prezygotic barriers. Solution 38 The solution is (A). Allopatric speciation is more common than sympatric speciation. Geographic isolation (the defining characteristic of allopatric speciation) greatly reduces gene flow between populations, so speciation is more likely to occur. Continued gene flow is more likely in a sympatric situation, so this type of speciation is less common. Using adaptive radiation, how can the diversification of the finches Darwin observed in the Galapagos be explained? A The finches likely shared a common ancestor when they came to the island, but exhibited different traits. Each species of finch settled the island where its particular traits would be the most adaptive. B The finches likely originated as one parent species, but over time mutations caused them to develop reproductive barriers and separate into different species. To reduce competition, the species then radiated out to inhabit different islands. C The finches likely dispersed from one parent species, and natural selection based on different food sources in differing habitats led to adaptive changes, evidenced in the different beak shapes of the different species, each suited to a different food type. D It is likely that a series of cataclysmic events caused an original finch species to diverge into the many finch species that inhabited the islands when Darwin observed them. The different species then radiated out to the different islands and adapted to the different conditions on each. ******************************************************** ***************************404***************************************** Solution 39 The solution is (D). Adaptive radiation is speciation that occurs when one population radiates out and forms several new species, each in a different niche or isolated habitat. Island archipelagos, like the Galapagos, are ideal contexts for this type of speciation, because the water provides the geographical isolation for speciation to occur. Darwin observed finches with different beak shapes and related these shapes to different food sources. These finches likely dispersed from one parent species, and evolution in response to natural selection based on different food sources in differing habitats led to behavioral, and eventually genetic, changes. These changes are evidenced in the different beak shapes of the different species—each suited to a different food type. In which situation would hybrid reproduction cause two species to fuse into one? A Separate species cannot interbreed, so hybrid reproduction does not occur in nature. B If the hybrid offspring are more fit than the parents, reproduction would likely continue between both species and the hybrids, eventually bringing all organisms under the umbrella of one species. ******************************************************** ****************************************************************405**** C Two species that have recently diverged from each other can reproduce with each other, creating hybrid individuals that belong to the species of the parents’ common ancestor. D If two species occupy the same niche in the same area, they can either compete or they can collaborate and reproduce with each other, eventually fusing into a single species. Solution 40 The solution is (D). If two closely related species continue to produce hybrids that are as fit as the parent species, or even more fit, the species would be more likely to fuse. If there was an increased pressure for that fusion, such as an environmental change that pushed the two species to share habitat or food sources more closely, this might increase the possibility of fusion. What do both rate of speciation models have in common? A Both models ignore the influence of gene flow for simplicity’s sake. B Both models apply only to island chains. C Both models require the influence of cataclysmic events which precipitate rapid adaptation and speciation. D Both models conform to the rules of natural selection and the influences of gene flow, genetic drift, and mutation. Solution 41 The solution is (A). Both rate of speciation models result in the divergence of one species into two species. Both rely on the usual mechanism of change, natural selection, and both are influenced by factors such as environmental changes, rates of mutation, and population size. In which situation would hybrid reproduction would two species to continue divergence? A If two closely related species continue to produce hybrids, the hybrids will compete with both species, causing them to find new niches that will further their divergence. B If two closely related species continue to produce hybrids, they will develop reproductive barriers to prevent production of hybrids, to ensure that they remain separate species. C If two closely related species continue to produce hybrids that are less fit than the parent species, there would be reinforcement of divergence. D If two closely related species continue to produce hybrids they will always converge into a single species. Solution The solution is (B). If two closely related species continue to produce hybrids that are less fit than the parent species, there would be reinforcement of divergence. And, if there were an increased division between the two species, such as an environmental change that pushed the two species to move farther apart or increased reproductive barriers that decreased hybrid fitness even further, this would increase the possibility of divergence even more. ******************************************************** ***************************404***************************************** TEST PREP FOR AP® COURSES 42 Prior to 1800 in England, the typical moth of the species Biston betularia (peppered moth) had a light pattern. Dark-colored moths were rare. By the late nineteenth century, the light-colored moths were rare, and the moths with dark patterns were abundant. The cause of this change was hypothesized to be selective predation by birds (Tutt, 1896). During the Industrial Revolution, soot and other wastes from industrial processes killed tree lichens and darkened tree trunks. Thus, prior to the pollution of the Industrial Revolution, dark moths stood out on light-colored trees and were vulnerable to predators. With the rise of pollution, however, the coloring of moths vulnerable to predators changed to light. Which aspect of Darwin’s theory of evolution does the story of the peppered moth most clearly illustrate? A There is competition for resources in an overbred population. B There is great variability among members of a population. C There is differential reproduction of individuals with favorable traits. D The majority of characteristics of organisms are inherited. Solution 43 The solution is (C). Through natural selection, moths with favorable coloring survived to reproduce and pass on their coloring. Prior to 1800 in England, the typical moth of the species Biston betularia (peppered moth) had a light pattern. Dark-colored moths were rare. By the late nineteenth century, the light-colored moths were rare, and the moths with dark patterns were abundant. The cause of this change was hypothesized to be selective predation by birds (Tutt, 1896). During the Industrial Revolution, soot and other wastes from industrial processes killed tree lichens and darkened tree trunks. Thus, prior to the pollution of the Industrial Revolution, dark moths stood out on light-colored trees and were vulnerable to predators. With the rise of pollution, however, the coloring of moths vulnerable to predators changed to light. In the late 1900s, England cleaned up its air, and pollution decreased. The bark of trees went from dark to light. Which outcome to the populations of peppered moth would you expect, given this environmental change? A An increase in the number of dark moths and a decrease in the number of light moths B An increase in the number of moths overall C An approximately equal number of light moths and dark moths D An increase in the number of light moths and a decrease in the number of dark moths Solution The solution is (D). Natural selection would allow for differential reproduction of the more favorable trait. With decreased pollution and lighter trees, this would mean the light moth. ******************************************************** ****************************************************************405**** 44 Prior to 1800 in England, the typical moth of the species Biston betularia (peppered moth) had a light pattern. Dark-colored moths were rare. By the late nineteenth century, the light-colored moths were rare, and the moths with dark patterns were abundant. The cause of this change was hypothesized to be selective predation by birds (Tutt, 1896). During the Industrial Revolution, soot and other wastes from industrial processes killed tree lichens and darkened tree trunks. Thus, prior to the pollution of the Industrial Revolution, dark moths stood out on light-colored trees and were vulnerable to predators. With the rise of pollution, however, the coloring of moths vulnerable to predators changed to light. Commonly used in biology text books, the peppered moth is a classic example of evolutionary change in action. The example describes changes in a population’s allele frequencies: a small-scale change, evolutionarily speaking. The presence of both light and dark forms within the gene pool is demonstrated by the story, but the peppered moth stays a peppered moth. Which scenario, if it were to occur, would be a model for large-scale evolutionary change? A Conditions change such that the dark form of the moth is favored and the light form is diminished in the population due to predation. Conditions change again, the dark form is vulnerable, and the light form returns to prevalence. B Conditions change such that the dark form of the moth is favored and the light form is eradicated in the population due to predation. Conditions change again, the dark form is vulnerable, and the dark form is eradicated due to predation. C Conditions change such that dark form of the moth is favored and the light form is diminished in the population due to predation. Conditions change again, and both forms have equal prevalence. D Conditions change such that dark form of the moth is favored and the light form is eradicated in the population due to predation. Conditions change again, the dark form is vulnerable. It develops an adaptation that shields it from predation. Solution 45 The solution is (B). Conditions change such that the dark form of the moth is favored and the light form is eradicated in the population due to predation. Conditions change again, the dark form is vulnerable, and the dark form is eradicated due to predation. Given your understanding of evolutionary theory and the relationship between evolution and the genetic makeup of populations, which statement is false? A Homologous characteristics that have evolved more recently are shared only within smaller groups of organisms. B The genetic code is a homologous characteristic shared by all species because they share a common ancestor in the deep past. ******************************************************** ***************************404***************************************** C DNA sequence data would likely support any evolutionary tree drawn from anatomical data sets. D The degree of relatedness between groups of organisms is only sometimes reflected in the similarity of their DNA sequences. Solution 46 The solution is (D). The degree of relatedness between groups of organisms is only sometimes reflected in the similarity of their DNA sequences. Each of the following observations comes from a different scientific discipline. Which observation best supports Darwin’s concept of descent with modification? A Geologists provide evidence that earthquakes reshape life by causing mass extinctions. B Botanists provide evidence that South American temperate plants have more in common with South American tropical plants than temperate plants from Europe. C Zoologists provide evidence that fewer animal species live on islands than on nearby mainlands. D Ecologists provide evidence that species diversity increases closer to the equator. Solution 47 The solution is (B). Botanists provide evidence that South American temperate plants have more in common with South American tropical plants than temperate plants from Europe. Paleontologists have recovered a fossil for an organism named Archaeopteryx. It has many features in common with reptiles, but, like birds, shows evidence of feathers. For what aspect of evolutionary theory does this piece of evidence suggest support? A Modern species are distinct natural entities. B Modern species are not currently evolving. C Modern species share a common ancestor. D Modern species have both convergent and divergent traits. Solution 48 The solution is (C). A fossil that demonstrates common characteristics of now divergent groups suggests common ancestry among living things, both specifically and generally. Which piece of evidence illustrates evolution as an ongoing process? A Some genes from the bacterium E. coli have sequences that are similar to genes found in humans. B Marsupial mammals live in just a few places in the world today: Australia, South America, and part of North America. ******************************************************** ****************************************************************405**** C The fossil record shows that Rodhocetus, an aquatic mammal related to whales, had a type of ankle bone that is otherwise unique to a group of land animals. D In the 1940s, infections by the bacterium Staphylococcus aureus could be treated with penicillin; today populations exist that are completely resistant. Solution 49 The solution is (A). Ethnic Tibetans are unique among humans for having physiological mechanisms that help them live in places with low oxygen levels. The process of mutation, which generates genetic variation, is random. Thus, life has evolved, and continues to evolve, randomly. Which statement is an appropriately evidence-based refinement of the above? A The process of mutation, which generates genetic variation, is random. However, the process of natural selection, which results in adaptations like the fit between a flower and its pollinator, favors variants which are better able to survive and reproduce. Natural selection is not random, so the overall process of evolution is not random, either. B The process of mutation, which generates genetic variation, is random. However, the process of migration, which results in gene flow between populations, also generates genetic variation. Migration is not random, so the overall process of evolution is not random, either. C The process of mutation, which generates genetic variation, is random. However, the process of sexual reproduction, which also introduces genetic variance, is not random. Because sexual reproduction is not random, the overall process of evolution is not random, either. D The process of mutation, which generates genetic variation, is random. Whether mutations have a positive, negative, or neutral effect in terms of selective advantage is also random. Mutations and their effects are random, so the overall process of evolution is random. Solution 50 The solution is (A). The process of mutation, which generates genetic variation, is random. However, the process of natural selection, which results in adaptations like the fit between a flower and its pollinator, favors variants which are better able to survive and reproduce. Natural selection is not random, so the overall process of evolution is not random, either. The selective breeding of plants and animals that possess desired traits is a process called artificial selection. For example, broccoli, cabbage, and kale are all vegetables that have been selected from one species of wild mustard. How is artificial selection both similar to and different from Darwin’s conception of natural selection? Does artificial selection provide evidence for evolution by natural selection? Explain. ******************************************************** ***************************404***************************************** A Both artificial selection and natural selection are the differential reproduction of individual organisms with favored traits. In artificial selection, humans have actively modified plants and animals by selecting and breeding individuals with traits deemed desirable. In natural selection, the most successful individuals in a species are selected by the species to reproduce. B Both artificial selection and natural selection are processes that result in betteradapted individuals within a species. In artificial selection, humans have actively modified plants and animals by selecting beneficial genes from other organisms and inserting them into the target organisms. In natural selection, natural processes such as mutations and viruses introduce new genes to a population. C Both artificial selection and natural selection are processes that cause organisms to be better adapted over time. In artificial selection, humans have trained animals to be more successful in completing tasks that the humans want completed. In natural selection, organisms train the functions that they will need to survive and reproduce. D Both artificial selection and natural selection are the differential reproduction of individual organisms with favored traits. In artificial selection, humans have actively modified plants and animals by selecting and breeding individuals with traits deemed desirable. In natural selection, individuals are selected naturally as its traits deem it more fit for survival and reproduction. Solution 51 The solution is (A). Both artificial selection and natural selection are the differential reproduction of individual organisms with favored (selected) traits. Each process works by acting on heritable variations in a population. In artificial selection, humans have actively modified plants and animals by selecting and breeding individuals with traits deemed desirable. In natural selection, individuals are selected naturally as its traits deem it more fit for survival and reproduction. The “more fit” individuals pass on these “more fit” traits to their offspring; over time, the population’s traits change in accordance with its environment. Artificial selection provides a good model for understanding how natural selection works. Although it brings about dramatic change in a relatively short period of time, it provides evidence for how small changes multiplied over many generations can result in a species that bears little resemblance to its ancestors. Genes important in the embryonic development of animals have been relatively well conserved during evolution. This means they are more similar among different species than many other genes. What explains this genetic conservation across animal species? ******************************************************** ****************************************************************405**** A Changes in the genes that are important to embryonic development have been relatively minor because there are no selective pressures on an individual before it is born. B Changes in the genes that are important to embryonic development have been relatively minor because not much time has elapsed since the divergence of the various animal taxa. C Changes in the genes that are important to embryonic development have been relatively minor because early embryos are very fragile and even small mutations can result in death. D Changes in the genes that are important to embryonic development have been relatively minor because mutational tweaking in the embryo has magnified consequences in the adult. Solution 52 The solution is (D). Early embryonic development of animal species follows a similar pattern. For example, structures such as tails and slits such as gills are present in all vertebrate such as embryos, though they develop differently or disappear altogether in later forms. Since anatomical development is the result of genetic material and the machinery of DNA replication and expression, it makes sense that the genetic information that produces these basic embryonic structures would be the same or very similar in very different species. The differences between species are evidence of evolutionary changes in their genetic makeup. Changes in the genes that are important to embryonic development have been relatively minor, as mutational tweaking in the embryo has magnified consequences in the adult. The upper forelimbs of humans and cats have fairly similar structures. In contrast, the upper forelimbs of whales (their flippers) have bones with a different shape and proportion from both cats and humans. Interestingly, genetic data suggest that all three organisms have a common ancestor from about the same point in time. What is a likely explanation for these data? A Cats and humans are more closely related to each other than either is to whales. B The shape of the whale forelimb arose a result of disadvantageous mutations. C The whale flipper is an adaptive characteristic unique to its water environment. D The whale flipper is a vestigial structure. Solution The solution is (C). Given the fact that all three organisms diverged from a common ancestor at about the same point in time, and that evolutionary change is spurred by environmental change, it seems that the whale flipper is an adaptive characteristic unique to its water environment. Both humans and cats live on land, whereas whales live in water. It makes good sense that natural selection in an aquatic environment resulted in the differences we see in whale forelimb anatomy. ******************************************************** ***************************404***************************************** 53 Biogeography is the study of biological species as they relate to geographical space and geological time. The fossil record shows that dinosaurs originated about 200 to 250 million years ago. Would you expect the geographic distribution of early dinosaur fossils to be broad (on many continents) or narrow (on one or a few continents)? Explain. A Broad because dinosaurs originated before the breakup of Pangaea B Broad because some dinosaurs could fly between continents C Narrow because they went extinct too quickly to disperse very far D Narrow because they lived so long ago that the fossils have mostly broken down or disappeared Solution 54 The solution is (A). Given that the dinosaurs originated before or about the same time as the breakup of the supercontinent Pangaea (about 200 million years ago), I would expect that the geographic distribution of early dinosaur fossils to be quite broad. It is likely that extant species of dinosaurs lived on many parts of Pangaea, and it is also likely that when Pangaea broke apart, the fossils of dinosaurs would have moved with the rocks as the continent broke apart and moved. The term microevolution describes evolution on its smallest scale: the change in allele frequencies in a population over generations. DDT is a pesticide that was widely in use in the United States from the 1940s until 1972. The table summarizes a particular allele frequency in laboratory strains of the common fruit fly, Drosophila melanogaster. Strains Collected from Flies in the Wild in the 1930s Strains Collected from Flies in the Wild in the 1960s Frequency of Allele 0% 40% Conferring DDT Resistance Using this information, in which model did natural selection improve the match between D. mealanogaster and its environment through microevolution? A DDT killed off a large proportion of the population, and the alleles present in the surviving fruit flies differed from those in the original population. B Mutations from the application of DDT caused the allele conferring DDT resistance to appear in the population. C Female mosquitoes chose to mate with male mosquitoes that had the allele conferring DDT resistance because it would make their offspring more fit. D The wide use of DDT meant that fruit flies with DDT resistance were more evolutionarily fit than their counterparts without DDT resistance. Solution The solution is (B). The allele conferring resistance to the pesticide DDT either arose by mutation or was present in the population in the 1930s very rarely. The wide use of DDT meant that fruit flies with DDT resistance were more evolutionarily fit than ******************************************************** ****************************************************************405**** their counterparts without DDT resistance. The number of individuals with DDT resistance in the population of fruit flies grew due to differential reproduction associated with the increased evolutionary fitness of these individuals. DDT is a pesticide—a poison—and natural selection strongly selected for individuals who had the resistant allele. Thus, from the 1930s to the 1960s, the frequency of the allele conferring DDT resistance increased in the fruit fly population. 55 In 1795, a Scottish geologist named Charles Hutton suggested that Earth’s geologic features could be explained by gradual processes that were still operating. This was in direct contrast to other scientific thought at the time, which included well-accepted proposals that geologic layers were representative of catastrophic events caused by processes no longer operating in the present time. Hutton proposed geologic features as the result of slow and consistent change, such as valleys formed by rivers wearing through rock. Hutton’s ideas were incorporated in the work of Charles Lyell, a geologist working in Darwin’s time. Lyell advocated a principle called uniformitarianism, the consistency of mechanisms of change over time. In other words, Lyell argued that the same geologic processes operating in the present had operated in the past, and at the same rate. The ideas of Hutton and Lyell influenced the work of Charles Darwin. How do Hutton’s and Lyell’s ideas connect to and provide support for Darwin’s theory of evolutionary change? A The idea that the same processes that operate in the present also operated in the past, and at the same rate, supported Darwin’s hypothesis of natural selection because humans could select for desirable traits and produce change very rapidly, so natural selection would also be fast enough to produce the full range of diversity in living organisms. B The idea that the same processes that operate in the present also operated in the past, and at the same rate, connects to Darwin’s hypothesis of natural selection because he had observed it happening in the present. C The idea that geologic change is the result of slow, continuous processes rather than sudden, substantial change connects to Darwin’s support of gradualism rather than punctuated equilibrium as the process that guided evolution. D The idea that geologic change is the result of slow, continuous processes rather than sudden, substantial change connects directly to Darwin’s hypothesis that, given enough time, slow and subtle processes could produce substantial biological change. Solution The solution is (C). The idea that geologic change is the result of slow, continuous processes rather than sudden, substantial change connects directly to Darwin’s hypothesis that, given enough time, slow and subtle processes could produce substantial biological change. Further, the thesis that the Earth’s geologic features result from such slow and steady change suggests a very old Earth. A theory of a much, much older Earth supports Darwin’s theory of evolutionary change—the Earth would have to be much older than commonly thought at Darwin’s time to be old enough for life to have evolved from a single common ancestor. ******************************************************** ***************************404***************************************** 56 The human immunodeficiency virus (HIV) reproduces very quickly. A single virus can replicate itself a billion times in one 24-h period. In a hypothetical treatment situation, a patient’s HIV population consists entirely of drug-resistant viruses after just a few weeks of treatment. How can this treatment result best be explained? How does this explanation illustrate that evolution is an ongoing process? A The resistant viruses passed their genes to the nonresistant viruses so that 100 percent of the viruses became resistant. This illustrates evolution as an ongoing process because the genes of the population changed in real time. B The nonresistant viruses died, and the resistant ones survived and rapidly reproduced. This illustrates evolution as an ongoing process because the change in the HIV population is the result of natural selection. C The viruses developed resistance to the drug after repeated exposure to it. This illustrates evolution as an ongoing process because the viruses were able to adapt to changing conditions. D The drug-resistant viruses were more fit than their nonresistant counterparts to begin with, and over time they dominated the population. This illustrates evolution as an ongoing process because natural selection favored one phenotype over another. Solution 57 The solution is (B). The treatment result is best explained as the result of natural selection. A few drug-resistant viruses were in the original HIV population (at the start of treatment) and natural selection increased their frequency to 100 percent. The non-resistant viruses died, and the resistant ones survived and rapidly reproduced. This explanation illustrates evolution as an ongoing process because the change in the HIV population in the explanation is the result of evolutionary processes. Natural selection strongly selected for the drug-resistant virus in the presence of the drug. There was differential replication of the drug-resistant virus with the accumulation of favorable traits (drug resistance) over generations. A friend says: “Natural selection is about the survival of the very fittest in a population. The fittest are those that are strongest, largest, and fastest.” Would you agree with that statement? What evidence from scientific disciplines can you offer to support your agreement or your disagreement? A The statement is true. If an organism is not strong and fast, it will not survive long enough to reproduce and pass on its genes, and if it is not large and fitter than the other individuals around it, it will not be able to compete for a mate. Many seal species, for example, have only a single male who gets to mate. He must be the very fittest seal to win all the females. B The very fittest organisms are not necessarily the ones that survive. Sometimes it is the least fit organisms that survive and reproduce. For example, in one generation, the mice that are bad at foraging for seeds may reproduce prolifically and dominate the ******************************************************** ****************************************************************405**** mice that are good at foraging. In this case, natural selection will select for the less-fit phenotype and spread it in the population. C The definition of fitness is not correct. The strongest and fastest organisms are more fit than the weaker and slower ones, but large individuals are often at a disadvantage to smaller ones because they are easily spotted by predators. For example, a large rabbit will stick out on a field more than a small one and will get eaten by a hawk. D What is meant by fittest is not necessarily strong, large, and fast. Fitness, as defined in evolutionary terms, has to do with survival and the reproduction of genetic material. For example, a small but showy male bird may be selected by female birds to reproduce, while a large but less colorful one is not. Solution 58 The solution is (D). Although survival of the fittest is a commonly used catchphrase when discussion natural selection and evolution, the phrase is a bit misleading. It is not only the very fittest that survive—many organisms that seem less fit than others may survive and reproduce. Further, what is meant by fittest is not necessarily strong, large, and fast. Fitness, as defined in evolutionary terms, has to do with both survival and the reproduction of genetic material. A small but showy male bird may survive and reproduce while a large but less colorful one does not. A thin, shortliving plant may yield plentiful seeds while a larger, longer-living plant does not. Some of the largest organisms to walk Earth, the dinosaurs, were fit for a time. Not many organisms of that size exist today—and certainly not in those numbers. What is fit at one point in time is not what is fit in another—and some of the most successful and prolific organisms ever are found among the microorganisms and invertebrates. A student placed 20 tobacco seeds of the same species on moist paper towels in each of two petri dishes. Dish A was wrapped completely in an opaque cover to exclude all light. Dish B was not wrapped. The dishes were placed equidistant from a light source set to a cycle of 14 h of light and 10 h of dark. All other conditions were the same for both dishes. The dishes were examined after seven days, and the opaque cover was permanently removed from dish A. Both dishes were returned to the light and examined again at 14 days. The following data were obtained: ******************************************************** ***************************404***************************************** What best supports the hypothesis that the difference in leaf color is genetically controlled? A The number of yellow-leaved seedlings in dish A on day 7 B The number of germinated seeds in dish A on days 7 and 14 C The death of all the yellow-leaved seedlings D The existence of yellow-leaved seedlings as well as green-leaved ones on day 14 in dish B Solution 59 The solution is (D). The existence of yellow-leaved seedlings as well as green-leaved ones on day 14 in dish B. A student placed 20 tobacco seeds of the same species on moist paper towels in each of two petri dishes. Dish A was wrapped completely in an opaque cover to exclude all light. Dish B was not wrapped. The dishes were placed equidistant from a light source set to a cycle of 14 h of light and 10 h of dark. All other conditions were the same for both dishes. The dishes were examined after seven days, and the opaque cover was permanently ******************************************************** ****************************************************************405**** removed from dish A. Both dishes were returned to the light and examined again at 14 days. The following data were obtained: What best describes the usefulness of the yellow-leaved phenotype as a variation subject to natural selection? A The yellow-leaved phenotype can germinate in environments without light. B The germination of the yellow-leaved phenotype is unaffected by light intensity. C The germination of the yellow-leaved phenotype is accelerated as compared to the green-leaved phenotype. D The yellow-leaved phenotype cannot germinate in environments with light. Solution The solution is (A). The yellow-leaved phenotype can germinate in environments without light. ******************************************************** ***************************404***************************************** 60 A student placed 20 tobacco seeds of the same species on moist paper towels in each of two petri dishes. Dish A was wrapped completely in an opaque cover to exclude all light. Dish B was not wrapped. The dishes were placed equidistant from a light source set to a cycle of 14 h of light and 10 h of dark. All other conditions were the same for both dishes. The dishes were examined after seven days, and the opaque cover was permanently removed from dish A. Both dishes were returned to the light and examined again at 14 days. The following data were obtained: Yellow-leaved seedlings are unable to convert light energy to chemical energy. Which observation is most likely to be made on day 21? ******************************************************** ****************************************************************405**** A A few yellow-leaved seedlings alive in dish A, but none in dish B B A few yellow-leaved seedlings alive in dish B, but none in dish A C No yellow-leaved seedlings alive in dish A or dish B D A few yellow-leaved seedlings alive in dish A and dish B Solution 61 The solution is (C). If the seedlings are unable to convert light energy to chemical energy, none will survive. Populations of a nocturnal toad live along a long river. On the other side of a band of territory that is about 10 km wide, there are populations of a toad that appear similar. Which data would provide compelling evidence that the two populations represent different species? A The populations of toads on the other side of the banded territory are not completely nocturnal. B Fertile hybrid populations of toads are found between the two other populations. C There appear to be some hybrid toads between the two populations, but they are few and frail. D The two populations of toads enact very different mating behaviors. Solution 62 The solution is (C). There appear to be some hybrid toads between the two populations, but they are few and frail. A group of students summarized information on five great extinction events. The students are sampling a site in search of fossils from the Devonian period. Based on the chart, what would be the most reasonable plan for the students to follow? A Searching horizontal rock layers in any class of rock and trying to find those that contain the greatest number of fossils B Collecting fossils from rock layers deposited prior to the Permian period that contain some early vertebrate bones ******************************************************** ***************************404***************************************** C Looking in sedimentary layers next to bodies of water in order to find marine fossils of bivalves and trilobites D Using relative dating techniques to determine the geological ages of the fossils found so they can calculate the rate of speciation of early organisms Solution 63 The solution is (B). The most reasonable plan to follow would be collecting fossils from rock layers deposited prior to the Permian period that contain some early vertebrate bones. Populations of a plant species have been found growing in the mountains at altitudes above 2,500 m. Populations of a plant that appears similar, with slight differences, have been found in the same mountains at altitudes below 2,300 m. Which plan for collecting two kinds of data could provide a direct answer to the following question: Do the populations growing above 2,500 m and the populations growing below 2,300 m represent a single species? A Scientists could take the genetic code of a plant from each altitude and determine whether the two sets of DNA are identical. They could also insert genes from one plant into the cells from the other and see if the cells survive B Scientists could look in the fossil record to find the plants’ most recent common ancestor. They could also check the surrounding mountains to determine if the most recent common ancestor is still living. C Scientists could breed the two groups in the same environment and observe whether, over several generations, they begin to look more similar. They could also switch the groups, growing the high-altitude plants at low altitude and the low-altitude plants at high altitude, and observe whether the former begin to look like low-altitude plants and the latter begin to look like high-altitude plants. D Scientists could collect seeds and test whether they might be cross-pollinated to produce fertile offspring. They also could investigate the area between 2,500 m and 2,300 m to see if fertile hybrid populations might be found living between the two other populations of plants. Solution 64 The solution is (D). Scientists could collect seeds and test whether they might be cross-pollinated to produce fertile offspring. They could also investigate the area between 2,500 m and 2,300 m to see if fertile hybrid populations might be found living between the two other populations of plants. Populations of a plant species have been found growing in the mountains at altitudes above 2,500 m. Populations of a plant that appears similar, with slight differences, have been found in the same mountains at altitudes below 2,300 m. How do the two types of data from the previous exercise provide a direct answer to the question of whether speciation has taken place? ******************************************************** ****************************************************************405**** A If the plants become more similar when grown in the same environment, or if the high-altitude plants respond to low altitude in the same way that low-altitude plants have, and low-altitude plants respond to high altitude the same way that high-altitude plants have, then the two groups have the same underlying genetic structure and belong to one species. B If the seeds from the plants can be cross-fertilized and developed into fertile offspring, the two populations are not yet reproductively isolated and remain one species. If hybrid forms are found, the two populations are not reproductively isolated and hybrids are both viable and successful. C If the genetic codes of the two plants are identical, then they must belong to the same species. Also, if genes transplanted between the plants function successfully, then the plants must be similar enough to each other to belong to the same species. D If scientists are able to find the common ancestor of the two groups in the fossil record or in neighboring communities, then they can determine whether the plants have diverged into separate species or remain a single species. Solution 65 The solution is (B). Both of these data are consistent with definition of biological species. If the seeds from the plants can be cross-fertilized and developed into fertile offspring, the two populations are not yet reproductively isolated and remain one species. If hybrid forms are found this provides the same direct answer: the two populations are not reproductively isolated and hybrids are both viable and successful. Assuming a population that has genetic variation and is under the influence of natural selection, place the following events in the order in which they would occur: Genetic frequencies within the population change. A change occurs in the population’s environment. Phenotypic variations shift. Individuals who are well-adapted leave more offspring than individuals who are poorly adapted. Individuals who are poorly adapted do not survive at the same rate as individuals who are well adapted. A 1. A change occurs in the population’s environment. 2. Individuals who are poorly adapted do not survive at the same rate as individuals who are well adapted. 3. Individuals who are well adapted leave more offspring than individuals who are poorly adapted. 4. Genetic frequencies within the population change. 5. Phenotypic variations shift. ******************************************************** ***************************404***************************************** B 1. A change occurs in the population’s environment. 2. Genetic frequencies within the population change. 3. Phenotypic variations shift. 4. Individuals who are poorly adapted do not survive at the same rate as individuals who are well adapted. 5. Individuals who are well adapted leave more offspring than individuals who are poorly adapted. C 1. Phenotypic variations shift. 2. A change occurs in the population’s environment. 3. Genetic frequencies within the population change. 4. Individuals who are poorly adapted do not survive at the same rate as individuals who are well adapted. 5. Individuals who are well adapted leave more offspring than individuals who are poorly adapted. D 1. Individuals who are well adapted leave more offspring than individuals who are poorly adapted. 2. Individuals who are poorly adapted do not survive at the same rate as individuals who are well adapted. 3. Phenotypic variations shift. 4. Genetic frequencies within the population change. 5. A change occurs in the population’s environment. Solution The solution is (A). The events in order of occurrence are as follows: 1. A change occurs in the population’s environment. 2. Individuals who are poorly adapted do not survive at the same rate as individuals who are well adapted. 3. Individuals who are well adapted leave more offspring than individuals who are poorly adapted. 4. Genetic frequencies within the population change. 5. 5. Phenotypic variations shift. 66 A biologist studies a population of voles for 20 years. During almost the entire research period, the population stays between 50 and 75 individuals. Additionally, fewer than half of the voles born do not survive to reproduce, due to predation and competition for food. Then, in one generation, 80 percent of the voles born live to reproduce. The population increases to 110 individuals. What inferences about food and predation can you make for the singular generation in which 80 percent of offspring survived? What prediction can you make about the genetic and phenotypic variation of future populations for this group of voles? ******************************************************** ****************************************************************405**** A Either there was less food available or the degree of predation increased. The future generations of this group of voles should evidence less genetic variation. B Either there was less food available or the degree of predation increased. The future generations of this group of voles should evidence greater genetic variation. C Either there was more food available or the degree of predation decreased. The future generations of this group of voles should evidence less genetic variation. D Either there was more food available or the degree of predation decreased. The future generations of this group of voles should evidence greater genetic variation. Solution 67 The solution is (D). It is reasonable to infer that for the generation of voles in which 80 percent live and reproduce, there was more food available or the degree of predation decreased. The future generations of this group of voles should evidence greater genetic variation because there are more individuals surviving and passing on their genetic information. If that variation persists (the population stays larger or increases in size), it is likely that future populations will evidence a different variety of phenotypes, as well. There are years of drought in a small, relatively isolated community. During the drought, small seeds with thin shells become rare. Large seeds with hard cases become increasingly common. The large, tough seeds are successfully eaten by birds with large and broad beaks. Assuming that the drought continues and the population of birds in the community stays isolated, what predictions for the population can you make under the influence of natural selection? A The birds with small, thin beaks will grow larger, broader beaks to be able to eat the larger seeds. This will result in subsequent generations having a higher percentage of birds with large, broad beaks. B There will be more birds with small, thin beaks dying and more birds with large, broad beaks surviving. Differential reproduction of birds with large, broad beaks will result in subsequent generations having a higher percentage of birds with large, broad beaks. C The species will diverge into two species, one with small, thin beaks and one with large, broad beaks. The two species will then compete for resources. D There will be neither phenotypic nor genotypic changes in the population. Solution The solution is (B). There will be more birds with small, thin beaks dying and more birds with large, broad beaks surviving. Differential reproduction of birds with large, broad beaks will result in subsequent generations having a higher percentage of birds with large, broad beaks. If the drought and isolated conditions persist, it is possible that the genetic variation of the population will shift such that the configuration of the small, thin beaked bird is increasingly rare or even no longer present in the gene pool. ******************************************************** ***************************404***************************************** 68 At one time, avian researchers in the Sulawesi region of Indonesia described the Flowerpecker populations on the mainland and the Wakatobi archipelago as one species. A recent reassessment of the Wakatobi populations resulted in the suggested reclassification of these populations as a distinct species, the Wakatobi Flowerpecker. Which piece of evidence, if true, would be cause for this reclassification? A The populations have become dependent on the island food sources. B The populations have become morphologically distinct from the mainland species. C The populations have become adapted to the island habitat. D The populations have become reproductively isolated from the mainland species. Solution 69 The solution is (D). The populations have become reproductively isolated from the mainland species. What pattern in the fossil record would you expect to see to support the model of gradual speciation? How would you expect this pattern to differ from a pattern in the fossil record that supports the model of punctuated equilibrium? Explain. A In the case of gradual speciation, the fossil record would show only a few hybrid individuals, followed by individuals of the two distinct species. For the case of punctuated equilibrium, the fossil record would show many hybrid individuals persisting through several geological layers. B In the case of gradual speciation, the fossil record would show the parent species in a single location, such that the newly diverged species remained in contact with each other. For the case of punctuated equilibrium, the fossil record would show a geographic divide within the parent species that caused it to diverge into multiple new species. C In the case of gradual speciation, the fossil record would show many intermediate forms. For the case of punctuated equilibrium, the fossil record would show new forms that persist essentially unchanged through several geological layers, then disappear just as a new form appears. D Gradual speciation would be undetectable in the fossil record. For the case of punctuated equilibrium, the fossil record would show a steady progression of distinct forms. Solution The solution is (C). In the case of gradual speciation, the fossil record would show many intermediate forms. The changes between them would be notable, but they would be gradual and perhaps span many, many geological layers. In contrast, for the case of punctuated equilibrium, the fossil record would show the sudden appearance of a new form. This new form would persist essentially unchanged through several geological layers, then disappear just as another new form appears. ******************************************************** ****************************************************************405**** 70 Until recently, these three species of short-tailed pythons, Python curtus, Python brongersmai (middle), and Python breitensteini were considered one species. However, due to the different locations in which they are found, they have become three distinct species. What is this an example of? A Divergent evolution B Sympatric speciation C Allopatric speciation D Variation Solution 71 The solution is (C). This refers to speciation that occurs through geographic separation. Consider two species of birds that diverged while separated geographically but resumed their contact before reproductive isolation was complete. Which option describes the first step in what would happen over time if the two species mated extensively and their hybrid offspring survived and reproduced more poorly than offspring from intraspecies mating? A Natural selection would cause prezygotic barriers to reproduction between the parent species to strengthen over time. B The production of unfit hybrids would increase and the speciation process would complete. C The extensive mating between the species would continue to produce large numbers of hybrids. D The gene pools of the parent species would fuse over time, reversing the speciation process. Solution The solution is (A). Natural selection would cause prezygotic barriers to reproduction between the parent species to strengthen over time. SCIENCE PRACTICE CHALLENGE QUESTIONS 18.1 Understanding Evolution 72 In addition to biology, evidence drawn from many different disciplines, including chemistry, geology, and mathematics, supports models of the origin of life on Earth. In order to determine when the first forms of life likely formed, the rate of radioactive decay can be used to determine the age of the oldest rocks (see optional parts C and D) exposed on Earth’s surface. These are found to be approximately 3.5 billion years old. The age of rocks can be correlated to fossils of the earliest forms of life. ******************************************************** ***************************404***************************************** A. The graph compares times of divergence from the last common ancestor based on the fossil record with a “molecular time” constructed by comparing sequences of conserved proteins to determine a mutation rate (after Hedges and Kumar, Trends in Genetics, 2003). Explain how such a molecular clock could be refined to infer time for the evolution of prokaryotes. B. Using a molecular clock constructed from 32 conserved proteins, Hedges and colleagues (Battistuzzi et al., BMC Evol. Biol., 2004) estimated the times during which key biological processes evolved. A diagram based on their work is shown. Connect the time of the origin of life inferred from this diagram with the age of the oldest fossil stromatolites and the age of the oldest exposed rock to show how evidence from different scientific disciplines provides support for the concept of evolution. Evaluate the legitimacy of claims drawn from these different disciplines (biology, geology, and mathematics) regarding the origin of life on Earth. ******************************************************** ****************************************************************405**** The oldest known rocks are exposed at three locations: Greenland, Australia, and Swaziland. The following application of mathematical methods provides the essential evidence of the minimum age of Earth. The mathematics is appropriate for students who have completed a second year of algebra. However, it is not illustrative of the type of item that could appear on the AP Biology Exam. The exposed rocks contain a radioactive isotope of rubidium, 87Rb, which decays into a stable isotope of strontium, 87Sr. An 87Rb atom with 37 protons and 50 neutrons decays when a proton is converted into a neutron to produce an atom, 87Sr, with 36 protons and 51 neutrons. As time passed, the number of each isotope changed from its initial value. When a crystal containing 87Rb atoms formed from the molten surface of the hot, early Earth during the Hadean eon, the number of these atoms at that initial time can be represented as N87Rb,0. As time passed, the number of atoms of this isotope changed to N87Rb. C. Justify the relationship between the number of each isotope at any time and the number of each at the time that the molten rock solidified (denoted by the subscript 0): N87Sr N87Sr ,0 N87Rb ,0 N87Rb The decay of unstable radioisotopes is exponential with a half-life of T1/2, which for 87Rb is 4.88 1010 years: N87Rb N87Rb,0 e 0.693t /T1/2 This can be used to replace the initial number of 87Rb atoms, which cannot be measured, with the present-day value: N87Sr N86Sr N87Sr ,0 N86Sr e 0.693t /T1/2 1 N N87Rb 86Sr When the measurements of the numbers of 87Rb and 87Sr were made (Moorbath et al., Nature, 1972), measurements of a second stable isotope of strontium, 86Sr, also were made. The ratio of the initial number of 87Sr and 86Sr atoms is the same as today, since the isotopes are both stable. The value of this ratio is 0.71. This is a linear equation in the form y ax b, where a is the term in parenthesis containing the half-life of 87Rb. If Y N87 / N86 is graphed versus N87 / N86 , the slope Sr Sr Sr Sr can be used to determine the time, t, that has passed since the rock formed from melting: ae 0.683t /T1/2 1, so t ln a 1 T1/2 / 0.693. ******************************************************** ***************************404***************************************** D. Data on the rubidium and strontium isotopes at Isua in Greenland are provided in the table. Analyze these data to obtain the age of formation of these rocks. N87Rb/N86Sr N87Sr/N86Sr 0.212 0.711 0.214 0.711 0.223 0.712 0.259 0.714 0.268 0.714 0.267 0.715 0.290 0.716 0.394 0.720 0.434 0.723 The solidification of the molten surface of Earth at the end of the Hadean eon (4 to 4.6 billion years ago) and the condensation of liquid oceans provided a medium from which life emerged. The most ancient fossils are colonial, photosynthetic cyanobacteria called stromatolites. As climate change melted the perennial snow covering Greenland, new geologic evidence of the time of that origin was obtained (Nutman et al., Nature, 2016) with the discovery of the most ancient stromatolites. These fossils record communities of photosynthetic bacteria embedded in Isua sediments 3.7 billion years ago. Worldwide stromatolite fossils show a decline between 1 and 1.3 billion years ago. Solution Sample answer: A. By comparing sequence differences in conserved proteins among prokaryotes, using such a molecular clock, the time separating last common ancestors among prokaryote groups can be inferred. B. At the extrapolation of these estimates for the emergence of biological innovation lies the ultimate biological innovation, life. A reasonable interval is between 4.5 and 4.0 bya. Interestingly, this is significantly older than the oldest surface rocks, leading many to conjecture that life emerged during the Hadean. The oldest surface rocks only provide an upper bound on the formation of solid Earth since these are just those that are exposed. The proliferation of cyanobacteria is unlikely to be a very early phenomenon, so the 3.7 bya fossils. So, all three of these disciplines have produced results which are complementary and providing evidence of both the concept of evolution and time of the origin of life on Earth. C. The left-hand side of this equality is the number of 87Sr atoms in a sample of rock. These were either present initially, 87Sr0, or were formed from the decay 87Rb. The number of 87Rb that have decayed is the difference between the initial number, 87Rb , and the current number of 87Rb. 0 ******************************************************** ****************************************************************405**** D. A graph should be constructed such as the one below except that axes should be labeled. The independent variable is the ratio of 87Rb to 86Sr. The slope is the factor that depends on the time interval that has elapsed since the solidification of the melt to form these rocks. Using the half-life of 87Rb, the age of the rocks can be determined. (e 0.693t / T1/2 1) 0.0514 t ln(1.0514) 0.693 0,724 0,722 0,72 0,718 0,716 0,714 0,712 0,71 y = 0,0514x + 0,7005 0 73 4.88 1010 years 3.5 billion years 0,1 0,2 0,3 0,4 0,5 In 1952, the Miller-Urey experiment showed that an electrical discharge in a gas-phase mixture of ammonia, hydrogen, methane, and water produced five amino acids. When the experiment was conducted, evidence indicated that this mixture was representative of the Hadean (early Earth) atmosphere. The experiment was repeated in the presence of jets of hot steam, simulating Hadean volcanic eruptions and producing an even larger variety of amino acids. A. Consider the following criticisms of the “organic soup” model and justify the selection of data that other experiments might provide regarding the origin of life on Earth. Biopolymers on Earth have a left-hand symmetry at the carbon adjacent to the carboxylic acid carbon, and these experiments produced mixtures of both leftand right-hand symmetries. No peptide bonds between amino acids were observed. Early Earth’s atmospheric oxygen concentration is known to have been very low, implying the absence of an ozone layer to filter high-energy ultraviolet (UV) radiation. Ammonia decomposes when it absorbs high-energy UV radiation, but diatomic nitrogen does not. Models of the abiotic synthesis of biomolecules suffer from a “chicken and egg” dilemma. Proteins are needed to synthesize DNA and RNA, and DNA and RNA are needed to synthesize proteins. Which molecules came first? ******************************************************** ***************************404***************************************** B. In light of the following observations, evaluate the hypothesis that nucleotides arose from a prebiotic mixture. Nuclei acids are not found in experiments like those of Miller and Urey. Purines and pyrimidines decompose at high temperature, and Earth was bombarded by meteors and comets during the Hadean eon. Bonds in the purine and pyrimidine rings of nucleic acids are broken by highenergy UV radiation. Carl Sagan and colleagues synthesized ATP from a mixture of adenosine, ribose, and phosphate when exposed to UV radiation. Ribose has never been synthesized in experiments like those conducted by Miller and Urey. Ribose has a left/right symmetry, and the right-handed form occurs in Earth organisms. Continuing with the analogy, if neither the chicken nor the egg came first, then both must have arisen together. Some regard simultaneous innovations in both catalysis and information storage and retrieval as too improbable. In samples of meteorites, both amino acids and nucleic acids have been found. The amino acids are mixtures of left- and right-handed symmetries, although some have shown a significant bias toward the lefthanded form (J. Elisa et al., ACS Central Science, 2016). The arrival from space of the seeds of biomolecules is called panspermia. Carl Sagan (1966) and Francis Crick (1973), one of the first to describe the structure of DNA, regarded panspermia as the only plausible origin of life on Earth. In fact, their belief was in directed panspermia, the intentional seeding by intelligent aliens. C. Describe the questions that must be addressed for panspermia to be a scientific hypothesis about the origin of life on Earth and describe the reasons for the directed panspermia revision of this hypothesis. To avoid the conflicting chicken-and-egg claims that “protein catalyst was first” and “DNA information storage was first,” two alternatives have emerged regarding the origin of life on Earth. Consider two simple ideas: (1) water blocks UV radiation, and cracks in the ocean floor (hot vents) provide a temperature difference that generates a source of entropy; and (2) ribosomes are composed of RNA. D. Describe one of the following as a hypothesis concerning the origin of life on Earth: Reactions among molecules in the vicinity of hot vents became organized in space and time, eventually developing structures that foreshadow the proton gradient upon which metabolism is based. This alternative is the basis for what is referred to as the metabolism-first hypothesis. ******************************************************** ****************************************************************405**** Solution The catalytic properties of the ribosome reflect the self-catalytic polymerization of nucleotides with sequential structures conserved in modern DNA, the catalytic properties conserved in proteins, and the catalytic properties of the ribosome whose core structure is RNA. This alternative is the basis for what is referred to as the RNA-first hypothesis. Sample answer: A. N2 is even more likely to be the predominant nitrogen molecule in the Hadean atmosphere and less likely to react to form amino acids. The experiment should be conducted with N2 and perhaps even more energetic radiation, such as X-rays. In the gas phase, there is no mechanism to bias the result toward either left- or righthanded products. If a surface—such as clay, mica, or metals—were present then the left-handed form might be favored. Also, experimental designs need to probe the way that surfaces might lead to polymerization of the amino acids. B. Polymerization of nucleotides, once they formed, would a very low probability event and that means very long times would be required. Whenever an impact occurred, the clock would be reset. Somehow the purine and pyrimidine products would need to be hidden from the UV radiation—though a report (http://dx.doi.org/10.1016/j.jphotochemrev.2008.12.001) has suggested that the five nucleic acids found in DNA and RNA are products of selection under pressure from UV radiation. Ribose has a left/right symmetry so synthesis in solution is not likely to favor the right-handed form. Sagan (1973) suggested that prokaryotes protected the bases from radiation by sinking in the primordial oceans while preeukaryotes remained near the surface in protective cages of radiation-absorbing nucleic acids. This “geographic separation” led to the division. This probably was dinner conversation between Sagan and his wife, Lynn Margulis, who later published the accepted endosymbiotic explanation. C. To be a valid scientific hypothesis panspermia needs to be testable. Or one must restate the model in a way that can be tested. For example, one could make the claim that nucleic acids found in meteorites can be extracted under pre-biotic conditions and, given a pathway to the synthesis of ribose, can form nucleotides that polymerize in the presence of phosphates. This claim, given the challenges summarized in this problem, is not likely to be tested in the near future. So, those of us who want answers to questions beyond the scope of science in our day sometimes pose conjectures that satisfy this need, such as directed panspermia. D. The assessment rubric would require that they state the hypothesis and sketch the means by which the hypothesis would be tested. Metabolism-first: The free energy source of a hot vent creates a resource: the flow of energy through a system can lead to structure. This is similar to other ecosystems but in the pre-biotic world the components of the ecosystem are molecular. The proton gradient is the essential structure for all life on Earth—both in the synthesis and degradation of carbohydrates. It could be claimed that under laboratory ******************************************************** ***************************404***************************************** conditions with a mixture representative of a deep Hadean ocean reduction oxidation reaction pairs coupled by proton movement could be observed. RNA-first: RNA plays the dual role of information storage and catalysis. So rather than the simultaneous emergence of both functions, RNA specializes to fill each role with new structures. It could be claimed that given a mixture composed of an RNA sequence, the RNA core of a ribosome, the DNA bases as nucleotides, and amino acids but no protein catalysts the protein coded for by the RNA could be produced. 74 The radiant energy emitted by a star gradually increases after its birth. During the Hadean eon, while the molten Earth cooled and life emerged, the sun provided approximately 25 percent less radiant energy than it does now. Ignoring effects due to differences in the composition of Earth’s atmosphere between then and now, this means that the average surface temperature of the surface would be about 25 °C below the freezing temperature of water. Evidence of liquid water on Earth during the Hadean eon is provided by geologic structures known only to form in liquid water, such as lava pillows and the stromatolites that are the fossilized layers of photosynthetic cyanobacteria. Pose a scientific question that guides inquiry into early Earth conditions that supported the innovation of photosynthesis. Solution 75 Sample answer: Just as the Earth’s temperature is currently increasing as we increase the concentration of carbon dioxide in the atmosphere, could the composition of the early atmosphere be largely carbon dioxide and water, two greenhouse gases that trapped energy, elevating the temperature? Connect the techniques of radiometric measurement, anatomy, and molecular biology to the supporting evidence of the theory of evolution provided. Solution Sample answer: The AP Biology Curriculum Framework is very specific about the scope of essential content knowledge regarding three techniques that provide evidence of evolution: Fossils can be dated by a variety of methods that provide evidence for evolution. These include the age of the rocks where a fossil is found, the rate of decay of isotopes including carbon-14, the relationships within phylogenetic trees, and the mathematical calculations that take into account information from chemical properties and/or geographical data. The details of these methods are beyond the scope of this course and the AP Exam. Morphological homologies represent features shared by common ancestry. Vestigial structures are remnants of functional structures, which can be compared to fossils and provide evidence for evolution. Biochemical and genetic similarities, in particular DNA nucleotide and protein sequences, provide evidence for evolution and ancestry. ******************************************************** ****************************************************************405**** 76 Describe reasons for the revision of scientific hypotheses of the origin of life on Earth. Solution Sample answer: There are few places in the AP Biology Curriculum Framework where the scope of content targeted by a declarative claim (describe) is not spelled out for you. In this case, there is no reference to any particular theory and it would be out of scope to assume prior knowledge of, for example, the Aristotelian model. And there is no help provided in the description in the Framework of enduring understanding 1D. And there is no supporting discussion of the nature of science that would clarify the meaning or general features of “scientific hypothesis.” Some alternative hypotheses regarding RNA worlds, panspermia, and directed panspermia were described in problem 2 in this chapter. An alternative interpretation that might lead to an Exam item could be implied by this task: Describe reasons for the development of hypotheses of the origin of life on Earth that are scientific. This would assess the student’s understanding that evidence is required to justify a claim. That is explicit in the description of Science Practice 6 in the Framework. 77 Directed evolution is an inquiry strategy that is usually used to investigate gene expression or the function of proteins that are expressed. The investigator imposes a selection pressure and observes the evolution of a population. In one investigation, unicellular yeast was allowed to sediment in a column that contained nutrients at its bottom. Yeast that reached the nutrients at the bottom was removed, weighed, and examined under the microscope. After 60 generations, it was found that all of the removed yeast was multicellular. To test the claim that their selection pressure favored multicellularity, the investigators performed another experiment. In one column, they provided a strong selection pressure, in which they only allowed 5 min for yeast to settle before removing what had traveled the farthest toward the bottom. In a second column, they provided a weak selection pressure, where they allowed 25 min for the yeast to settle before removal. Strong selection resulted in more massive clusters of multicellular yeast among the removed cells. Weak selection resulted in less massive multicellular clusters among the removed cells. A. Evaluate the claim that the use of both a strong and weak selection demonstrates that evolution is an ongoing process that, under artificially imposed conditions, led to the emergence of multicellularity in a single-celled organism. B. In this directed evolution study, the selection pressure imposed by the investigators led to a new phenotype. Consider a situation in which there is a vertical variation in the density of nutritional resources. Analyze the advantages and disadvantages of cooperative behavior, including changes in the likelihood of replication of the individual and population genomes. ******************************************************** ***************************404***************************************** Solution Sample answer: A. Yes, because the results of the second experiment demonstrate that multicellularity in yeast responded differently to the selection pressure when it was strong versus weak. In both experiments, the highest mass multicellular clusters will settle fastest and, therefore, be favored by selection because they reach the nutrients fastest. In the second experiment, when yeast is allowed only 5 min to settle, the yeast removed will only be the largest multicellular clusters. When yeast is allowed 25 min to settle, smaller multicellular colonies and possibly single cells will have enough time to settle and be removed. Therefore, the yeast responded to strong versus weak selection pressure, suggesting that the selection pressure was favoring multicellularity. B. If the density of nutrients is highest at the bottom of the column, multicellularity will be favored because the most massive multicellular colonies will reach the nutrients first. This will increase the frequency of genes that cause cells to engage in multicellularity within this population of yeast. If the density of nutrients is highest on the top of the column, multicellularity will not be favored because cells that sink will not reach the nutrients. This will decrease the frequency of genes that cause cells to engage in multicellularity within this population of yeast. 18.2 Formation of New Species 78 Selection processes in changing and unchanging environments differ. Connect the effects of negative and positive selection pressures to changes in the environment. Solution 79 Sample answer: The concept of homeostasis is strongly emphasized in the AP Biology Curriculum as is the theory of evolution. Classroom activities will have supported the idea that many homeostatic mechanisms are conserved through evolution and environmental changes select for mutations that have an advantage as these changes occur. An item targeting this LO would provide an opportunity for the student to summarize an illustrative example from those activities. In biology, the word “race” is rarely used. It could be imagined to be synonymous with a subspecies. Species is well defined, at least when horizontal gene transfer is not taken into account, by reproductive isolation. Speciation may arise through geographic isolation. A. Aside from geographic isolation leading to reproductive isolation, predict two other mechanisms of speciation in a population and suggest how these mechanisms could lead to a scientific definition of a subspecies. B. If individuals of a species become separated by a natural disaster, they could become increasingly different. Assuming that they could still interbreed, what would you predict about the consequences if females, but no males, from one population (population A) were introduced to the other population (population B)? What would be the effects on autosomal, X, and Y chromosomes? ******************************************************** ****************************************************************405**** Solution Sample answer: A. In addition to geographic isolation, other factors that cause individuals to stop interbreeding could cause speciation. For example, disruptive selection favoring individuals that forage in the morning and those that forage at night could eventually result in two populations that do not come into contact to breed. Over time, they could become sufficiently different to be considered subspecies or eventually species. The same thing could happen through strong sexual selection if different groups preferred different traits. B. Females cannot introduce Y chromosomes, so all of the Y chromosomes in the newly mixed population B would be from the original population B. The autosomal and X-linked characteristics of population A would be mixed into population B. ******************************************************** **************************440************************************* 19 | THE EVOLUTION OF POPULATIONS REVIEW QUESTIONS 1 What is the ultimate source of all variation in and among populations? A Genetic mutations that result in viable offspring B Natural selection C Diverse habitats D Factors in the environment that may affect development Solution 2 The solution is (A). Genetic mutations lead to changes in alleles and the frequency of alleles, contributing greatly to variation in and among populations. When male lions reach sexual maturity, they are thrown out of their group, or pride, and must live on their own or with other males until they can take over their own pride. This can alter the allele frequencies of the population through which mechanism? A Natural selection B Gene flow C Random mating D Genetic drift Solution 3 The solution is (B). Gene flow is the movement of alleles into (or out of) a population when an individual moves into (or out of) a population. When a male lion moves into a pride, it brings in new alleles to that population. Which population has violated the conditions of Hardy-Weinberg equilibrium? A An infinitely large population B A population in which the allele frequencies do not change over time C A population in which the Hardy-Weinberg equation is equal to 1 D A population undergoing natural selection Solution 4 The solution is (D). During natural selection, allele frequencies of a population change. A population in equilibrium is in a stable, nonevolving state. What is the difference between micro- and macroevolution? A Microevolution describes the evolution of small organisms, such as insects, while macroevolution describes the evolution of large organisms, such as people and elephants. B Microevolution describes the evolution of microscopic entities, such as molecules and proteins, while macroevolution describes the evolution of whole organisms. ******************************************************** ************************************************************439*** C Microevolution describes the evolution of organisms in populations, while macroevolution describes the evolution of species over long periods of time. D Microevolution describes the evolution of organisms over their lifetimes, while macroevolution describes the evolution of organisms over multiple generations. Solution 5 The solution is (C). Microevolution describes how a population changes over time, whereas macroevolution describes how new species arise. What would be supported by Lamarck? A Natural selection leads to changes in organisms over time. B The strong arms of a gorilla are the result of its parents constantly climbing, lifting, and fighting. C Lack of resources led to the death of three of four fox cubs. D The founder effect is when a few individuals in a population are separated from the original population. Solution 6 The solution is (B). Lamarck believed that traits that were acquired by a parent (such as strong arms) would be passed along to their offspring. What is population variance influenced by? A Genetic structure B Environment C Diet composition D All of the above Solution 7 The solution is (D). Genetic structure and environment are both factors that influence population variance. Genetic structure determines how many genes are present in the population and at what frequencies, while the environment creates natural selection pressure that changes genetic variance. What is genetic variance? A The change in a population’s genetic structure B The effect of chance on a population’s gene pool C The diversity of alleles and genotypes within a population D The magnification of genetic drift as a result of natural events or catastrophes Solution The solution is (C). This is the definition of genetic variance, which often enters into discussions of breeding. In order to prevent inbreeding, breeders try to increase a population’s genetic variance. ******************************************************** **************************440************************************* 8 When closely related individuals mate with each other, or inbreed, the offspring are often not as fit as the offspring of two unrelated individuals. Why? A Inbreeding causes normally silent alleles to be expressed. B The DNA of close relatives reacts negatively in the offspring. C Inbreeding can bring together rare, deleterious mutations that lead to harmful phenotypes. D Close relatives are genetically incompatible. Solution 9 The solution is (C). Inbreeding depression occurs when two carriers of a deleterious recessive mutations mate. Related individuals are more likely to carry the same deleterious mutation. Carriers of the mutation do not phenotypically express the mutation, but when both parents have the mutation, it may be expressed in the offspring who will not be as fit as the parents. What could cause genetic drift to occur within a population? A Accidental deaths B Predators C Disease D Lack of gene flow Solution 10 The solution is (A). Genetic drift is the change in allele frequencies due to chance. Accidental death is a random, chance event, whereas deaths due to predators or disease are not random. What is the evolutionary mechanism that alters allele frequencies by chance called? A Genetic drift B Natural selection C Inbreeding D Migration Solution 11 The solution is (A). Genetic drift is the effect of chance on a population’s gene pool. What is assortative mating? A When individuals mate with those who are similar to themselves B When individuals mate with those who are dissimilar to themselves C When individuals mate with those who are most fit in the population D When individuals mate with those who are least fit in the population Solution The solution is (A). Assortative mating refers to an individual’s preference to mate with partners who are phenotypically similar to himself or herself. ******************************************************** ************************************************************439*** 12 What is an example of a cline? A A random fluctuation in a species’ gene frequencies B A mutation that spreads across the ecological range of a species C The females of a species preferring males that are orange in coloration instead of white D Species having greater cold tolerance in the colder parts of its range than in the warmer parts of its range Solution 13 The solution is (D). A cline is a type of geographical variation where a species varies gradually across an ecological gradient. Temperature is an ecological gradient and species subject to colder temperatures are more cold tolerant. Which type of selection results in greater genetic variance in a population? A Stabilizing selection B Directional selection C Diversifying selection D Positive frequency-dependent selection Solution 14 The solution is (C). Diversifying selection is a type of selection that favors more than one distinct phenotype, resulting in increased genetic variance as the population becomes more and more diverse. What types of phenotypes does negative frequency-dependent selection favor? A Advantageous B Rare C Common D Disadvantageous Solution 15 The solution is (B). Negative frequency-dependent selection favors rare phenotypes. It is a type of selection that increases the genetic variance within a population through the selection of rare phenotypes. The good genes hypothesis is a theory that explains what? A Why more fit individuals are more likely to have more offspring B Why alleles that confer beneficial traits or behaviors are selected for by natural selection C Why some deleterious mutations are maintained in the population D Why individuals of one sex develop impressive ornament traits ******************************************************** **************************440************************************* Solution 16 The solution is (D). The good genes hypothesis states that males show off their healthy genes through the development of ornaments. For example, male lions with darker, fuller manes are selected by females over males with lighter, thinner manes. The males with dark, full manes have displayed that they are in good health and are able to father healthy offspring. Which option describes when males and females of a population look or act differently? A Sexual selection B Diversifying selection C Sexual dimorphism D A cline Solution The solution is (C). Sexual dimorphism concerns differences between the sexes, such as the difference in color of male and female birds. CRITICAL THINKING QUESTIONS 17 What is natural selection? Give an example of natural selection at work in a population. A The process in which genes flow from one population to another; the beak size of Darwin’s finches changing as the availability of different-sized seeds changes. B The process in which genes flow from one population to another; the founder effect occurring among humans immigrating to a new country. C The process in which better-adapted organisms are able to survive and reproduce; the beak size of Darwin’s finches changing as the availability of different-sized seeds changes. D The process in which better-adapted organisms are able to survive and reproduce; the founder effect occurring among humans immigrating to a new country. Solution 18 The solution is (C). Natural selection occurs when individuals with favorable genetic traits are better able to survive and reproduce. Natural selection leads to evolutionary change. Natural selection of Darwin’s finches occurred because individuals with beaks better adapted to different-sized seeds were selected for and so different species arose. Imagine you are trying to test whether a population of flowers is undergoing evolution. You suspect there is selection pressure on the color of the flower: bees seem to cluster around the red flowers more often than the blue flowers. In a separate experiment, you discover that blue flower color is dominant to red flower color. In a field, you count 600 blue flowers and 200 red flowers. What would you expect the genetic structure of the flowers to be? ******************************************************** ************************************************************439*** A You would expect 300 homozygous dominant blue flowers, 300 heterozygous blue flowers, and 200 homozygous recessive red flowers. B You would expect 200 homozygous dominant blue flowers, 400 heterozygous blue flowers, and 200 homozygous recessive red flowers. C You would expect 100 homozygous dominant red flowers, 100 heterozygous red flowers, and 600 homozygous recessive blue flowers. D You would expect 14 homozygous dominant red flowers, 186 heterozygous blue flowers, and 600 homozygous recessive blue flowers. Solution 19 The solution is (B). Red is recessive, so all red flowers would be homozygous recessive (bb). The blue flowers would be either BB or Bb. What must occur in order for a new trait to appear in a population and then reach a steady, high frequency within that population? A New traits appear through gene mutations or through genetic drift. In order to reach a steady, high frequency in the population, there must be many mutagens, such as UV radiation, in the environment to produce many mutations. B New traits appear through gene mutations or through genetic drift. In order to reach a steady, high frequency in the population, there must be a consistent source of immigrant individuals with the allele. C New traits appear through gene mutations or through evolution. In order to reach a steady, high frequency in the population, the allele must code for a favorable adaptation. D New traits appear through gene mutations or through gene flow. In order to reach a steady, high frequency in the population, the trait associated with the gene must be favored by either natural or sexual selection. Solution 20 The solution is (D). New traits appear through gene mutation and also through gene flow, which is the introduction of new alleles into a population through immigration of individuals or gametes. These traits are then incorporated into the population only if they are selected for by natural or sexual selection. What is population variation? Identify an example. A Population variation is a description of the diversity of different forms of life. An example of population variation would be the different forms and functions of prokaryotes versus eukaryotes. B Population variation is the geographic distribution of different phenotypes in a population. An example of population variation would be the fact that warm-blooded mammals that live near the poles tend to be larger than their southern counterparts to conserve heat. ******************************************************** **************************440************************************* C Population variation is the distribution of phenotypes in a population. An example of population variation would be the many different fur colors and patterns found in domestic dogs. D Population variation is the distribution of genotypes in a population. An example of population variation would be Mendel’s pea plants that were homozygous dominant, heterozygous, and homozygous recessive for various traits. Solution 21 The solution is (C). Population variation is the distribution of phenotypes (how individuals look) within a population. Domestic dogs, with their different colors and patterns of fur, have different phenotypes and are therefore examples of population variation. People who breed domesticated animals try to avoid inbreeding even though most domesticated animals are indiscriminate. Why this is a good practice? A A breeder would not allow close relatives to mate because inbreeding increases the likelihood of fatal mutations in offspring. B A breeder would not allow close relatives to mate because inbreeding prevents gene flow which can bring new, successful genes into the population. C A breeder would not allow close relatives to mate because inbreeding causes diversifying selection, which dilutes the breeder’s desired genes in the population. D A breeder would not allow close relatives to mate because inbreeding can bring together deleterious recessive mutations that can cause abnormalities and susceptibility to disease. Solution 22 The solution is (D). A breeder would want to avoid inbreeding depression. Inbreeding depression occurs when two closely related individuals who carry the same recessive, deleterious mutation mate. This can result in offspring that have abnormalities or are more susceptible to disease. What is the founder effect? Identify an example. A The founder effect is an event that isolates part of a population, generating an allele frequency which is not typical of the original population. An example of the founder effect is the Amish population. The Amish population was established from about 200 German immigrants. Individuals of this founding population carried gene mutations that cause inherited disorders such as Ellis-van Creveld syndrome. This form of dwarfism is found in a large concentration in the Amish population today because the immigrants that established the population had a high concentration of the disorder in a very small population. B The founder effect is an event that kills off a significant proportion of a population, generating an allele frequency which is not typical of the original population. An example of the founder effect is the Northern elephant seal. At one point, hunting of these seals had reduced their numbers to less than 50 individuals. The population has ******************************************************** ************************************************************439*** since rebounded, but still contains less genetic variation than the related Southern elephant seal due to the loss of some alleles. C The founder effect is when only a few males within a population are selected by females to reproduce, generating an allele frequency which is different from the original population. An example of the founder effect is the reproductive pattern of mountain gorillas. Mountain gorillas tend to have a single dominant male, the silverback, who gets the vast majority of the mating in the population. This leads to the next generation expressing mainly genes from the silverback and very few genes from the other males, changing the genetic structure of the population. D The founder effect occurs when the selective pressure on a trait varies depending on the alleles expressed within the population, generating varying allele frequencies based on the genetic makeup of the original population. An example of the founder effect is the cyclical dominance of three throat-color patterns in side-blotched lizards. Solution 23 The solution is (A). The founder effect is an event that causes a change in the genetic structure of the population. The Amish are a good example of this because the small founding population of 200 people had a gene mutation that is now in higher frequency than in other populations. What is a cline? Give an example. A A cline is a type of geographic variation that is seen in populations of a given species that vary gradually across an ecological gradient. For example, endothermic animals tend to have larger bodies in the cooler climates closer to Earth’s poles, allowing them to better conserve heat. B A cline is a change in ecological conditions over a geographic distance. For example, a latitudinal cline is the decrease in temperature towards Earth’s poles, and an altitudinal cline is the decrease in temperature with increase in altitude. C A cline is the specific set of traits in a population of a given species that have been influenced by the local environment. For example, a population of warm-blooded animals that lived in a cooler climate closer to the North Pole would have larger bodies, allowing them to better conserve heat. D A cline is the specific set of ecological conditions in a geographic region. For example, towards the North Pole it is cold and there is little precipitation. This will influence the traits of the organisms that live there. Solution The solution is (A). A cline is a type of geographical variation in which the populations of a species change gradually over an ecological gradient. For example, the bodies of endothermic animals tend to be larger in colder climates than in warmer climates. ******************************************************** **************************440************************************* 24 The table below shows data for a small population of mice. The mice are either brown or white. Based on the data, is the population experiencing genetic drift? Explain. Solution 25 Generation Brown Mice Black Mice 1 14 32 2 20 26 3 24 22 4 21 28 5 19 30 6 24 29 No. The frequency of both phenotypes oscillate around the same numbers through six generations. The large alpha male elephant seal is constantly fending off the advances of medium-sized males. Small males are then able to sneak copulation with females and successfully pass on their genes. What is this an example of? Explain. A This is an example of sexual selection. The females are selecting the small males over the large male. B This is an example of genetic drift. Because there are so many medium-sized males to compete with the large alpha male, the small males are able to mate and cause the gene pool to shift toward smaller individuals. C This is an example of positive frequency-dependent selection, which is selection that favors phenotypes that are either common or rare. The sneaky males (rare) are favored in this case. D This is an example of directional selection. Because only the smallest males are mating, the next generation will have a higher proportion of alleles for small size, making the seals smaller over time. Solution The solution is (C). This is an example of positive frequency-dependent selection. This is selection that favors phenotypes that are either common or rare. The sneaky males (rare) will allow genetic variance to increase so the gene pool is not dominated by medium sized and large males. ******************************************************** ************************************************************439*** 26 Why is there no perfect organism despite natural selection? A Because natural selection works on a geographic level B Because natural selection works in a random manner like mutations C Because of limitations due to a population’s existing variation in genes D Because natural selection is limited to sexual dimorphism Solution 27 The solution is (C). Natural selection acts on the existing variation within a population. It does not create new genetic variation. Therefore, natural selection selects the fittest organisms, but it is limited by what natural genetic variation there is in the population. A new predator invades the habitat of a population of field mice. Individuals with larger body size are easier for the predator to capture then individuals with smaller body size. Draw a histogram of body sizes with two plot lines, one showing the former population and another showing the new population that indicates how this population will likely evolve. On your histogram, also indicate what type of natural selection is occurring here. Solution 28 Quinine is an antimalarial drug that is used to treat malaria in the Western Hemisphere. Scientists have noticed that this drug has become less effective over time. Based on the data below, what type of selection is being exerted on the malaria population? ******************************************************** **************************440************************************* Solution Directional selection. TEST PREP FOR AP® COURSES 29 A scientist is studying the genetics of a population of plants that she suspects is undergoing natural selection. After examining samples of the population’s DNA over several years, she reports the following data: Do these data provide evidence of natural selection in this population? Why or why not? A No, because the genotype frequencies, not allele frequencies, have to change for evolution to occur B No, because the allele frequencies are changing randomly, suggesting that genetic drift is occurring, not natural selection C Yes, because it shows that the previously favorable or neutral allele A is now being selected against in favor of allele B D Yes, because it is showing that the frequency of both alleles is changing over time Solution 30 The solution is (C). Natural selection influences the allele frequency in a population. In this example, allele B is being selected for and allele A is being selected against. A scientist is studying two large populations of deer that are centralized in nearby forests. She takes blood samples from all of the deer in each population and records in how many individuals she finds allele A. She then computes the frequency of the alleles in each population. The frequencies observed over r years are shown in the tables. ******************************************************** ************************************************************439*** Which forms of evolution are most likely occurring in populations A and B? Explain your answer. A In population A, genetic drift is likely occurring, causing allele A to become more prevalent than allele B. In population B, mutation apparently occurred, introducing allele A to population B. Allele A also appears to be increasing due to genetic drift in population B. B In population A, natural selection is likely occurring, with allele A being favored over allele B. In population B, gene flow apparently occurred, allowing allele A to become established in population B. Allele A also appears to be favored by selection in population B. C In population A, gene flow apparently occurred, allowing allele B to become established in population A. Allele A also appears to be favored by selection in population A. In population B genetic drift is likely occurring, causing allele A to become more prevalent than allele B. D In population A, mutation apparently occurred, introducing allele B to population A. Allele A also appears to be increasing due to genetic drift in population A. In population B natural selection is likely occurring, with allele A being favored over allele B. ******************************************************** **************************440************************************* Solution 31 The solution is (A). In population B, allele A was not present in year 1. There may have been a mutation in year 3 that introduced the allele to the population and genetic drift increased the frequency of the allele over time. In population A, genetic drift is likely causing the frequencies of allele to change. A land manager mows a section of annual grass at the end of July. Over the years, he recorded the date of flowering from the mown field as well as a similar grass field that was not mown. What is the most likely explanation for this trend? Year Mowed Field Flowering Date Unmowed Field Flowering Date 2010 7/29 7/28 2011 7/30 7/26 2012 7/31 8/1 2013 7/8 7/29 2014 7/1 8/2 2015 6/29 7/26 A Mowing stimulates flowering, so the grass adapts by flowering after mowing occurs. B Mowing stabilizes the flowering time, which follows a steady trend in the mowed field but not in the unmowed field. C The mowing is preventing the grass from reproducing, causing the mowed grass to adapt by flowering earlier. D The grass typically flowers earlier and earlier every year as it becomes older with each passing year. Solution 32 The solution is (C). Mowing is selecting for earlier-flowering individuals. Over time, earlier-flowering phenotypes will dominate the population. This is an example of directional selection. A scientist observed two populations of insects for 10 years. They took data on the length, in millimeters, of the insects’ mouthparts. Their data are shown in the graphs. ******************************************************** ************************************************************439*** How is this population evolving and what agent of evolution is most likely at work? A Inbreeding, because the gene distributions are becoming less similar among the population B Genetic drift, as the distribution of traits has become more random C Gene flow, as the population has likely gained new mouthpart traits through immigration D Natural selection, as insects that have mid-sized mouthparts are being favored Solution 33 The solution is (D). Natural selection is occurring as insects with mid-sized mouthparts are being selected for and are therefore becoming more prevalent in the population. Researchers believe that in a fish species, individuals with the recessive genotype aa are predisposed to disease. Homozygous dominant (AA) individuals and heterozygous (Aa) individuals are not believed to be susceptible to this disease. A pond was stocked with 100 fish of the AA genotype and 100 fish of the aa phenotype, and the fish were allowed to breed. In the next generation, 35 percent of the fish had the dominant (AA) phenotype. What does this result indicate? A The homozygous dominant phenotype is higher than expected, indicating that evolution has occurred. B The homozygous dominant phenotype is lower than expected, indicating that evolution has occurred. C The homozygous dominant phenotype is higher than expected, indicating that evolution has not occurred. D The homozygous dominant phenotype is lower than expected, indicating that evolution not occurred. ******************************************************** **************************440************************************* Solution 34 The solution is (A). In the parent population, 50/100 fish have the AA genotype, and 50/100 fish have the aa phenotype. This means that the frequency of each allele in the population is 0.5, so p = 0.5 and q = 0.5. If the population is in Hardy-Weinberg equilibrium, the frequency of homozygous dominant individuals in the offspring is expected to equal p2, or 0.25. In other words, 25 percent of the offspring are expected to have the AA genotype. The actual frequency is higher (35 percent), indicating that an evolutionary change has occurred. Heterozygote advantage is a condition in which heterozygotes are favored by natural selection. How would the value of 2pq likely change if the population was undergoing heterozygote advantage? A It would remain in equilibrium because the value of p and q would remain the same. B It would remain in equilibrium because the value of 2pq would remain the same. C It would not remain in equilibrium because the value of 2pq would likely increase. D It would not remain in equilibrium because the value of 2pq would likely decrease. Solution 35 The solution is (C). The value of 2pq, which is the predicted frequency of heterozygous offspring, would increase because it is at a selective advantage. The graph shows the change in gene frequency of the two alleles of a gene: A and B. The population being studied has no emigration or immigration. Which type of evolution is likely occurring here and is the allele selected for, neutral, or selected against by natural selection? A Nonrandom mating; Both alleles are favored. B Gene flow; Allele A is favored. C Genetic drift; Both alleles are neutral. D Natural selection; Allele A is not favored. Solution The solution is (C). The allele frequencies fluctuate up and down. Genetic drift is the random change in a population. In this population, the frequencies of alleles A and B are changing randomly over time, not due to an advantage of either allele. ******************************************************** ************************************************************439*** 36 The figure shows the change in gene frequency of the two alleles: A and B. These alleles are located on separate genes that do not influence each other in any way. The population being studied has no emigration or immigration. Which type of evolution is likely occurring here, if at all? Explain how you know. Solution 37 Sample answer: This graph shows that allele A is not evolving because its gene frequency is steady and unchanging over time. Allele B is undergoing genetic drift due to the random ups and down of the allele frequency. The graph shows the current frequencies of two genotypes of the same gene: AA and aa. What would most likely happen to the frequencies of A and a if heterozygous individuals were favored by natural selection? A Both AA and aa will drop in frequency by the same amount. B Both AA and aa will drop, but aa will drop more. C AA will increase in frequency and aa will drop in frequency. D Genotype aa will increase in frequency and AA will drop in frequency. Solution The solution is (A). The frequency of both AA and aa would decrease by the same amount. ******************************************************** **************************440************************************* 38 The diagram shows the frequency of alleles on two species of wind-pollinated plants, as well as the prevailing wind direction. These frequencies have been fairly stable for around 10 years. However, climate change has created a new prevailing wind direction, as shown in the diagram. How will the populations likely evolve in the future? A Natural selection will cause the frequency of B to increase in population 1. B Gene flow will cause the frequencies of A and B to drop in population 3. C Genetic drift will cause the frequencies of A and C to increase in populations 1 and 2. D Inbreeding will reduce the frequency of allele B in populations 2 and 3. Solution 39 The solution is (B). The new wind direction will introduce genes from population 1 that have lower frequencies of alleles A and B. When individuals from population 1 mate with individuals from population 3, the frequencies of alleles A and B will therefore decrease. The diagram shows two populations of a species that have been long separated by a river, which prevents interbreeding. The two populations differ in coloration, as shown in the diagram. Recent human activity has caused the river to dry, however, resulting in the two populations shown in the lower diagram. What is the most likely explanation for this change? ******************************************************** ************************************************************439*** A An increase in gene flow between the two populations B A decrease in gene flow between the two populations C An increase in nonrandom mating between the two populations D A decrease in nonrandom mating between the two populations Solution 40 The solution is (A). Gene flow is increased because the river is no longer a solid barrier between the populations. As a result, genes for blue individuals have been introduced to population 2. Antibiotics are medicines that are designed to kill disease-causing organisms, or pathogens. However, some pathogens evolve antibiotic resistance, where they gain traits that allow them to survive in the presence of antibiotics. The ability of bacteria to adapt to antibiotics so quickly has created a huge concern over whether antibiotics are being overused. What form of evolution is antibiotic resistance an example of, and why? ******************************************************** **************************440************************************* A Gene flow because the bacteria are passing on the resistant trait within their populations B Natural selection because the bacteria is adapting to a new environmental condition: the presence of the antibiotic C Genetic drift because medical workers cannot follow the randomly fluctuating gene frequencies of bacterial populations D Mutation because each bacterium must mutate to an antibody-resistant form in order to survive Solution The solution is (B). Natural selection is selecting for individuals that are resistant to antibiotics. As a result, the presence of antibiotic resistance increases in the bacteria population. SCIENCE PRACTICE CHALLENGE QUESTIONS 19.1 Population Evolution 41 Consider a polymorphic gene with three alleles: A, B, and C. A. If the frequencies of the alleles A and B are 0.2 and 0.3, the frequency of allele C is closest to — A 0.25 B 0.5 C 0.2 D 0.3 Consider a gene with only two alleles: dominant A and recessive a. In a population of 1,000 organisms, the fraction expressing the homozygous recessive phenotype is 0.37. B. The calculated allele frequencies p and q have values that are closest to — A 0.69 and 0.31 B 0.31 and 0.69 C 0.37 and 0.63 D 0.63 and 0.37 C. The calculated number of individuals in this population that are heterozygotes is closest to — A 240 B 230 C 430 D 476 ******************************************************** ************************************************************439*** Mountain pine beetles (Dendroctonus ponderosae) were collected from a 1-acre tract of lodge pole pine trees (Pinus contorta) in a region of British Columbia where the forests are under temperature stress. The beetles were crushed, and a cellulase enzyme was extracted. Three polymorphs of the enzyme were observed when separated by gel electrophoresis. The three proteins observed correspond to alleles labeled C1, C2, and C3. The numbers of beetles with each allele are shown in the table. Genotype C1 C1 C2 C2 C3 C3 C1 C2 C1 C3 C2 C3 Total Observed 120 230 112 175 198 165 1,000 D. The calculated allelic frequencies pC1, pC2, and pC3 are closest to — A pC1 = 0.57, pC2 = 0.57, pC3 = 0.59 B pC1 = 0.29, pC2 = 0.29, pC3 = 0.42 C pC1 = 0.61, pC2 = 0.80, pC3 = 0.59 D pC1 = 0.31, pC2 = 0.40, pC3 = 0.29 E. To investigate the presence of selection at the cellulase locus due to changing temperature, what should a biologist do? A The biologist should calculate the values of the sums pC1 + pC2 + pC3 and (pC1 + pC2 + pC3)2. If these numbers are not equal to 1, the gene is not in HardyWeinberg equilibrium, and the gene is evolving. B The biologist should return next year and repeat this examination of the enzyme, calculating frequencies of each allele each year. Then calculate the values of the sums pC1 + pC2 + pC3 and (pC1 + pC2 + pC3)2. If these numbers are not the same each year, the gene is not in Hardy-Weinberg equilibrium, and the gene is evolving. C The biologist should return each year for several years and repeat this examination of the enzyme, calculating frequencies of each allele each year. If the allele frequencies are changing, the gene is not in Hardy-Weinberg equilibrium, and temperature is exerting a selection pressure. D The biologist should return each year for several years and repeat this examination of the enzyme, calculating frequencies of each allele each year. If the allele frequencies are changing, the gene is not in Hardy-Weinberg equilibrium. Analysis of the dependence of allele frequencies on temperature could indicate selection. Solution A. The solution is (B). pA +pB +pC 1 pC 0.5 ******************************************************** **************************440************************************* B. The solution is (B). The fraction of the population with the homozygous recessive genotype, 0.37, is equal to q2. Taking the square root, q = 0.61 Since p + q = 1, p = 1 – 0.61 or 0.39. q2 0.37 q 0.61 p 0.39 C. The solution is (D). The number of heterozygous individuals is expected to equal to 2pq. The value of p is 0.39 and the value of q is 0.61, so 2pq 0.476. D. The solution is (D). Two copies of the C1 allele are present in the 120 C1C1 individuals, for 240 alleles total. One copy of the C1 allele is present in the 175 C1C2 individuals, and 165 copies are present in C1C3 individuals. Thus, the total number of C1 alleles is equal to 240 + 175 + 198, or 613. A total of 2,000 alleles are present in the 1,000 beetles tested, so the frequency of the C1 allele equals 613 divided by 2,000, as shown in equation 1. The frequency of the C2 allele is calculated in equation 2, and the frequency of C3 is calculated in equation 3. 2 120 175 198 613 0.31 2,000 2,000 2 112 165 198 587 equation 2: 0.29 2,000 2,000 2 230 175 165 800 equation 3: 0.4 2,000 2,000 equation 1: E. The solution is (D). A change in allele frequency indicates that selection is occurring, but selection could occur for many reasons. Scientists must assess whether the change in allele frequency depends on temperature to determine if a temperature change is driving the change in allele frequency. 19.2 Population Genetics 42 Calamus finmarchicus is the dominant copepod in the Gulf of Maine. The polymorphic aminopeptidase locus, Lap-1, has been shown to be useful for the genetic differentiation of populations of this organism. By examining the population dynamics of copepods, the dynamics of the fin fish on which they feed can be predicted. The map shows a landmass separating two coastal estuarine habits, the mud flats of Egypt Bay and the Mount Desert Narrows. For the past 40 years, transport between the two habits has been hindered by a dam over the Carrying Place Inlet. However, small volumes of water occasionally crest the dam. ******************************************************** ************************************************************439*** To evaluate the geographic isolation of invertebrate populations in these two habitats, copepods are sampled at the points labeled 1 and 2 on the photograph. These points lie at either ends of the Carrying Place Inlet. Enzymes encoded by three alleles, labeled A, B, and C, were determined by gel electrophoresis of equal numbers of the organisms collected at the two sites. Numbers of each genotype are given in the table. Site AA AB AC BB BC CC Total 1 82 114 102 75 98 30 500 2 96 108 92 54 110 40 A. Calculate the frequencies, f, of each allele and complete the table. Site 1 2 f(A) f(B) f(C) blank blank Blank blank blank Blank 500 B. Using a 2 test, evaluate these data to determine whether the aminopeptidase gene in these two populations is evolving. State your conclusion as claims supported by evidence at both the 95 percent and 99 percent confidence levels. The formula for the 2 test is provided on the AP Biology Exam. 2 ( e)2 e ******************************************************** **************************440************************************* This table of critical p values is also provided on the AP Biology Exam. P 1 2 3 4 5 6 7 8 0.05 3.84 5.99 7.82 9.49 11.07 12.59 14.07 15.51 0.01 6.64 9.32 11.34 16.81 18.48 20.09 13.28 15.09 Degrees of Freedom C. Based on these data, predict, with justification, changes over time in the aminopeptidase enzyme for these populations. D. The B form of this aminopeptidase is slightly more efficient at extracting nutritional leucine from a protein than the A and C forms but slightly less efficient at extracting valine and serine. Describe an investigation of the two habitats that could suggest a causal relationship between changes in allele frequency and characteristics of the environment. E. Single-nucleotide mutations are neutral when they encode changes in proteins that result in no significant differential selection. If differences in environmental factors between sites 1 and 2 are not observed, predict what other factors could result in departures from Hardy-Weinberg equilibrium for aminopeptidase. Solution Sample answer: f (A) A. 2¥ NAA NAB NAC 0.76 500 and so on 0.76 0.78 2 0.72 0.652 0.52 0.56 2 B. 500¥ 0.263 3.40 1.43 5.1 0.76 0.72 0.56 2 C. This result says that we must reject the statement that the population is changing if our criterion is 99 percent confidence. However, at the 95 percent level we can accept the claim. D. Differences between the amino acid content of the copepod diet in the two habits should be investigated. Copepods eat phytoplankton but that is not in-scope. Only that variation is diet may be suspected as the mechanism in the possible selective pressure is needed. E. The frequencies of the alleles of a gene that is in Hardy-Weinberg equilibrium are constant. The conditions of HW equilibrium are no migration, no mutation, random mating, no selection and no genetic drift. If we suppose that the mutations are negligible so that there is no selection, then we are left with genetic drift as the possible explanation of a changing allele frequency. 43 Bioluminescence is an example of convergent evolution; 30 distinct lineages have acquired this characteristic, and all involve some form of a class of molecules called luciferins. Sexual selection pressures are strong for light-emitting organisms. Ellis and Oakley (Curr. Biol., 2016) examined the number of species that lack luminosity in groups of closest evolutionary relation (sister linear) with those species that are luminous. ******************************************************** ************************************************************439*** Similarly, scientists made the same comparison between groups that use luminosity for concealment (counter-illumination) and their sister lineages. The graphs summarize their results, comparing the natural logarithm of the number of species in each lineage. Based on the data shown in the graphs, describe a model that can account for the increased speciation of bioluminescent lineages, including the mechanism of speciation. Solution Sample answer: The data show speciation increases when luminosity is a component of mating selection, but when luminosity is not a component of mating selection, there is no effect on speciation. The evidence for the former claim is that there is a positive slope for each lineage. The evidence for the latter claim is that the slopes are both positive (speciation increases) and negative (speciation decreases). A likely mechanism is that strong selection during mating leads to reproductive isolation and this leads to speciation. ******************************************************** **************************440************************************* 44 A biologist is using a simulation to model populations of African hornbills (Bycanistes spp. and Ceratogymna spp.), a keystone species of the savanna. Populations of the birds are declining due to habitat loss. The hornbill’s diet consists primarily of termites and fruit. A critical component of termite digestion is chitin deacetylase, an enzyme whose mutation rate is a model parameter. The other model parameter is population size, N. In the results of the simulation study shown in the graphs, there is no selection, and the mutation rate is fixed. Although both population size and mutation rate are fixed, randomness results in the five different outcomes shown in each graph. A. Select the graph displaying the results that are closer to Hardy-Weinberg equilibrium. Justify the selection of the graph. B. Based on these simulations, predict the future heterozygosity, 2pq, of the smaller populations, as shown in graph (a). C. Justify the use of a simulation study with no selection under environmental conditions in which the availability of both termites and fruit is high. D. If a change in the environment occurs suddenly, such as an increase in average temperature, where fruit production declines, analyze the effect of the change on allele frequency in the large and small populations. Solution Sample answer: A. The graph at the right closely approximates a constant allele frequency and so is closer to HW equilibrium. B. The simulation with smaller population size shows a drift toward fixation of one or the other allele. In either case, the heterozygosity tends to zero. C. When both resources are in abundance, the mutation is neutral. This is equivalent to the absence of selection. D. With suddenly increasing temperature, a small population of hornbills may or may not survive, depending on whether the functional chitin-deacetylase is fixed. ******************************************************** ***************************************************************************459*** 20 | EVOLUTIONARY RELATIONSHIPS OF LIFE ON EARTH REVIEW QUESTIONS 1 Who devised a commonly used classification system? A Carl Linnaeus B Darwin C Plato D Aristotle Solution 2 The solution is (A). Carl Linnaeus, a Swedish botanist, zoologist, and physician created the taxonomic classification system. What uses a hierarchical model to classify organisms? A Analogy B Taxonomic classification system C Order D Systematics Solution 3 The solution is (B). The taxonomic classification system uses a hierarchical model to classify organisms. Correctly list the hierarchy of taxonomy. A Kingdom, Domain, Phylum, Order, Class, Family, Genus, species B Domain, Kingdom, Class, Phylum, Order, Family, Genus, species C Domain, Kingdom, Phylum, Class, Order, Family, Genus, species D Domain, Kingdom, Class, Phylum, Order, Family, Genus, species Solution 4 The solution is (C). The correct taxonomy order is Domain, Kingdom, Phylum, Class, Order, Family, Genus, species. Which category, below the level of Kingdom, would have the next largest number of organisms? A Order B Phylum C Family D Class ******************************************************** **************************458**************************************************** Solution 5 The solution is (B). Each sublevel in the taxonomic classification system moves toward specificity with other organisms, causing the diversity of organisms to narrow. The Kingdom category has the widest diversity of organisms, and because the Phylum category is a subcategory of the Kingdom, it would have the next widest diversity of organisms. How is systematics related to phylogeny? A Systematics provides guidelines that scientists use to describe the relationships of organisms. B Scientists use systematics programs to put together the phylogeny of an organism. C In systematics, scientists use combined data based on evolutionary relationships from many sources to put together the phylogeny of an organism. D Systematics is a process used to put together the phylogeny of an organism. Solution 6 The solution is (C). Systematics is a field of organizing and classifying organisms based on evolutionary relationships. Combined data from many sources is used to put together the phylogeny of an organism. Scientists use combined data based on evolutionary relationships from many sources to put together the phylogeny of an organism. What is the best explanation of what systematists do? A Scientists in the field of systematics organize organisms by characteristics. B Scientists in the field of systematics provide information on how organisms are similar or different. C Scientists in the field of systematics contribute to building, updating, and maintaining the tree of life. D Scientists in the field of systematics collect data from fossils. Solution 7 The solution is (A). Systematics is a field of organizing and classifying organisms based on evolutionary relationships, which scientists use to put together the phylogeny of an organism. What is the purpose of a phylogenetic tree? A To organize and name organisms into specific categories B To show the evolutionary pathways and connections among organisms C To show geographic or behavioral factors Solution The solution is (B). The purpose of a phylogenetic tree is to show the evolutionary pathways and connections among organisms. Scientists use a tool called a phylogenetic tree to show the evolutionary pathways and relationships among organisms. ******************************************************* ***************************************************************************459*** 8 What does the term rooted mean on a phylogenetic tree diagram? A Relationships among species are not shown. B All organisms represented in the diagram relate to a single ancestral lineage. C A single lineage evolved into a distinct new one. D A lineage evolved early from the root and remains unbranched. Solution 9 The solution is (B). On a phylogenetic tree diagram, the term rooted means all organisms represented in the diagram relate to a single ancestral lineage. Scientists call trees rooted when there is a single ancestral lineage (typically drawn from the bottom or left) to which all organisms represented in the diagram relate. Phylogeny is important to everyday life in human society. How did the research team in China use phylogeny as a guide to discover new plants that can be used to benefit people? A The research team used DNA to uncover phylogenetic relationships in the legume family, and they found a compound in the plant that is effective in treating cancer. B The research team used DNA to uncover phylogenetic relationships in the legume family, and then they identified a newly discovered plant as Dalbergia sissoo. C The research team used DNA to uncover phylogenetic relationships in the legume family, and they found a DNA marker that can be used to screen for plants with potential medicinal properties. D The research team searched all the relatives of the newly discovered plant Dalbergia sissoo to find antifungal properties. Solution 10 The solution is (C). The research team used DNA to uncover phylogenetic relationships in the legume family, and they found a DNA marker that can be used to screen for plants with potential medicinal properties. The research supports the idea that DNA markers can be used to screen for plants with potential medicinal properties. Which animals in the figure belong to a clade that includes animals with hair? Which evolved first, hair or the amniotic egg? ******************************************************** **************************458**************************************************** A Rabbits have hair, which evolved before the amniotic egg. B Rabbits and lizards have hair, which evolved after the amniotic egg. C Rabbits have hair, which evolved after the amniotic egg. D Rabbits and lizards have hair, which evolved before the amniotic egg. Solution 11 The solution is (C). Rabbits have hair, as indicated on the phylogeny, and the amniotic egg evolved before hair, because it evolved in the common ancestor of lizards and rabbits. Hair is a derived trait that only evolved in rabbits in this phylogeny. What is the largest clade in the diagram of those listed? A Animals, Fungi, and Plants B Fungi C Diplomonads D Flagellates Solution 12 The solution is (A). All the organisms within a clade stem from a single point on the tree. The clade of Animals, Fungi, and Plants contains multiple groups. Why is it important for scientists to distinguish between homologous and analogous characteristics before building phylogenetic trees? A Phylogenetic trees are based on evolutionary connections, so scientists can use the analogous characteristics to build the phylogenetic trees. B Phylogenetic trees are based on evolutionary connections, so scientists can use the homologous characteristics to build the phylogenetic trees. C Phylogenetic trees are based on similar functions, so scientists can use the homologous characteristics to build the phylogenetic trees. D Phylogenetic trees are based on similar functions, so scientists can use the homologous characteristics to build the phylogenetic trees. ******************************************************* ***************************************************************************459*** Solution 13 The solution is (B). Scientists consider phylogenetic trees to be a hypothesis of the evolutionary past since one cannot go back to confirm the proposed relationships. As such, phylogenetic trees are often constructed based on evidence drawn from multiple disciplines to illustrate evolutionary pathways and connections among organisms. Which option describes an analogous structure? A A penguin and a seagull both have wings. The penguin uses wings to swim while the seagull uses wings to fly. B Lizards and whales have similar skeleton structures, but they have a different habitat and lifestyle. C Birds and butterflies have wings with similar characteristics for flight, even though their wings do not share an evolutionary relationship. D The bone structure in leg of a cat is very similar to the bone structure in the arm of a human, but the functions of the limbs are very different. Solution 14 The solution is (A). A penguin and a dolphin each use fins to navigate through water. Even though the penguin (a bird) is not related to a dolphin (a mammal), they both use fins for the same function. This is an example of an analogous structure because they both use the fins for the same function. What is the ring of life? A A phylogenetic model where all three domains of life evolved from a pool of primitive prokaryotes B An evolutionary history and relationship of an organism or group of organisms C A phylogenetic model that attempts to incorporate the effects of horizontal gene transfer on evolution D A field of organizing and classifying organisms based on evolutionary relationships Solution 15 The solution is (A). This is the definition of the ring of life model. It suggests that all three domains-Archaea, Bacteria and Eukarya-evolved from a single pool of geneswapping prokaryotes. Some individual prokaryotes were responsible for transferring the bacteria that caused mitochondrial development to the new eukaryotes, whereas other species transferred the bacteria that gave rise to chloroplasts. This statement best describes which model? A Ring of life B Tree of life C Branches of life D Web of life ******************************************************** **************************458**************************************************** Solution 16 The solution is (D). This statement best describes the web of life. The web of life model tries to incorporate the effects of horizontal gent transfer on evolution as described in the statement. Explain why the classic tree model needs to be revised. A The model is unable to incorporate DNA evidence. B The model is erroneously based on many analogous traits, which have no basis in evolutionary relationships. C The model cannot be experimentally verified. D The model does not consider the possibility that genes could be transferred between unrelated species. Solution 17 The solution is (D). Classical thinking on prokaryotic evolution suggests that species evolve through random mutations causing descent into the variety of species as it branches from extinct species to modern-day species. Sexual reproduction in eukaryotes makes this view complicated, but genetic variation is still ultimately thought to be the result of mutation. What this way of thinking does not consider is the possibility that genes could be transferred from unrelated species, causing genetic variation in that manner. What are three different ways that eukaryotic cells may have evolved? A Some hypotheses propose that mitochondria were acquired first. Others propose that the nucleus evolved first. Still others hypothesize that prokaryotes descended from eukaryotes by the loss of genes and complexity. B Some hypotheses propose that eukaryotic cells are a combination of bacterial and archaeal cells. Others propose that eukaryotic cells are a combination of bacterial and fungal cells. Still others hypothesize that eukaryotic and prokaryotic cells developed independently. C Some hypotheses propose that mitochondria developed from bacterial cells. Others propose that they developed from archaeal cells. Still others hypothesize that bacteria developed from mitochondria that had been released from eukaryotic cells. D Some hypotheses propose that eukaryotic cells developed from gram-negative bacteria. Others propose that they developed from gram-positive bacteria. Still others hypothesize that both gram-positive and gram-negative bacteria contributed to the eukaryotic genome through horizontal gene transfer. Solution 18 The solution is (C). Some hypotheses propose that mitochondria were acquired first, followed by the development of the nucleus. Others propose that the nucleus evolved first and that this new eukaryotic cell later acquired the mitochondria. Still others hypothesize that prokaryotes descended from eukaryotes by the loss of genes and complexity. Explain the ring of life model. ******************************************************* ***************************************************************************459*** A The ring of life model is a phylogenetic model where the three domains of life started as distinct groups that could swap genes horizontally with each other in all directions. B The ring of life model is a phylogenetic model where all three domains of life are said to have developed from a pool of primitive prokaryotes. C The ring of life model is a phylogenetic model where bacterial and archaeal cells fused to form eukaryotic cells. D The ring of life model is a phylogenetic model where there is only a single domain of life due to modern DNA analysis. Solution 19 The solution is (D). The ring of life model is a phylogenetic model where all three domains of life said to have developed from a pool of primitive prokaryotes. This model proposes that Archaea, Bacteria, and Eukarya evolved from this pool of geneswapping prokaryotes. This model takes into account horizontal gene transfer and genomic fusion to explain genetic variation and the evolution of the three domains of life. In a transformation experiment, a sample of E. coli bacteria was mixed with a plasmid containing the gene for resistance to the antibiotic ampicillin (ampr). Plasmid was not added to the second sample. Samples were plated on nutrient agar plates, some of which were supplemented with the antibiotic ampicillin. The results of E. coli growth are summarized in the figure. The shaded area represents extensive growth of bacteria; dots represent individual colonies of bacteria. Which plates have only ampicillin-resistant bacteria? A Plate I only B Plate III only C Plate IV only D Plates I and II Solution The solution is (C). Colonies formed on plate IV containing ampicillin. They are due to resistant bacteria that could grow in the presence of ampicillin. ******************************************************** **************************458**************************************************** CRITICAL THINKING QUESTIONS 20 How are organisms classified in the taxonomic classification system? A The taxonomic classification system uses a hierarchical model to organize living organisms. At each sublevel, the organisms are more similar. B The taxonomic classification system uses a hierarchical model to organize living organisms. At each sublevel, the number of organisms increases. C The categories in the taxonomic classification system are organized from smaller, more specific categories to larger categories. D In the hierarchal model for the taxonomic classification system, from the point of origin, the groups become less specific. Solution 21 The solution is (A). Scientists historically organized Earth’s millions of species into a hierarchical taxonomic classification system from the most inclusive category to the most specific: Domain, Kingdom, Phylum, Class, Order, Family, Genus, and species. What is the correct way to format a two-word scientific name? A Italicize both words. Both words are lowercase. B Italicize both words. The first word should be capitalized. The second word should be lowercase. C Italicize both words. Capitalize both words. D Underline both words. Capitalize both words. Solution 22 The solution is (B). Scientists generally refer to an organism only by its genus and species, which is its two-word scientific name, also referred to the as binomial nomenclature. The first name of this binomial, the genus name should always be written first, with only the first letter capitalized. The second name, the species, is written last and all letters are lowercase. Both names should be typed in italics. Some organisms that appear very closely related may NOT actually be closely related. Why is this? A There are cases where organisms used to be closely related but diverged from each other and no longer look closely related. B There are cases where organisms can interbreed making them look like a single species, when in fact they are not closely related at all. C There are cases where organisms evolved through convergence and appear closely related but are not. D There are cases when extremely distant taxa can recombine into a single group. Solution The solution is (C). In most cases, organisms that appear closely related actually are; however, there are cases where organisms evolved through convergence and appear closely related but are not. ******************************************************* ***************************************************************************459*** 23 How does a phylogenetic tree relate to the passing of time? A A phylogenetic tree relates to the passing of time because species branch off from each other at regular time intervals. B A phylogenetic tree is not related to the passing of time because speciation is based on geographic changes. C The phylogenetic tree only shows the order in which things took place. D A phylogenetic tree relates to the passing of time when the diagram also shows how long ago the divergence from the common ancestor occurred. Solution 24 Judeo-Christian religious texts explain that the Earth and all the organisms on Earth were created in seven days. Why is this not a scientific hypothesis? Solution 25 The solution is (D). A phylogenetic tree relates to the passing of time when the diagram also shows how long ago the divergence from the common ancestor occurred. Phylogenetic trees approximate the passing of time by the lengths of their branches. Longer branches mean that more time has passed since the organisms shared a common ancestor. Sample answer: This is not a good hypothesis because it cannot be tested. There is no way to gather scientific evidence that deities exist or that they created humans in the distant past. Scientists use the cladistics system to organize what? A Homologous traits B Homoplasies C Analogous traits D Monophyletic groups Solution 26 The solution is (A). After the homologous and analogous traits are sorted, scientists often organize the homologous traits using a system called cladistics. This system sorts organisms into clades: groups of organisms that descended from a single ancestor. How does a clade relate to a monophyletic group? A Clades vary in size depending on the number of branches. B All the organisms within a clade stem from a single point on the phylogenetic tree. C A clade shows branches that do not share a single point. D A clade shows groups that diverge at a different branch point. Solution 27 The solution is (B). Clades must include all of the descendants from a branch point. As such, all of the organisms in the clade or monophyletic group stem from a single point on the phylogenetic tree. Scientists apply the concept of maximum parsimony to do what? ******************************************************** **************************458**************************************************** A Describe phylogenies accurately. B Eliminate analogous traits. C Identify mutations to DNA codes. D Locate homoplasies. Solution 28 You discover a new species of animal in the rainforest. What characteristics could you use to distinguish this organism from the other organisms in the same clade based on Figure 11? Solution 29 Organisms on a cladogram are distinguished by morphological or molecular features. One could analyze the sequence of DNA in the organism’s genome to understand how it relates to other organisms in its clade. Based on the phylogenetic tree below, is the Jungle cat likely closer related to a tiger or a cougar? Why? Solution 30 The solution is (A). Scientists apply the concept of maximum parsimony to describe phylogenies accurately. The maximum parsimony concept is used to aid in the tremendous task of describing phylogenies accurately. The Jungle cat is separated from the Cougar by two common ancestors on the tree and from the Tiger by five common ancestors on the tree. Therefore, the Jungle cat is likely more closely related to the Cougar then the Tiger. Two cultures of bacteria are separated by a filter that blocks the movement of cells but allows free exchange of anything smaller than a bacterial cell. On one side of the filter, a sample of penicillin-resistant cells in culture broth is added, and on the second side of the tube, a culture of penicillin-sensitive cells in culture is added. After 24 h, resistant cells appear on the side with the cells sensitive to penicillin. Which three genetic mechanisms can account for appearance of the penicillin-resistant cells? A Transformation, transduction, and conjugation ******************************************************* ***************************************************************************459*** B Transformation, transduction, and mutation C Transformation, conjugation, and mutation D Transduction, conjugation, and mutation Solution The solution is (A). Transformation and transduction are mechanism that explain the appearance of resistant cells because the filter is permeable to viruses and free DNA. Conjugation is not possible because cells cannot cross the filter. The third mechanism is spontaneous mutation. TEST PREP FOR AP® COURSES 31 What evolutionary question is better addressed by the fig-shaped evolutionary tree (a) as opposed to the more typical, single-trunk phylogenetic tree (b)? (a) (b) A What was the single organism from which all other forms of life on Earth arose? ******************************************************** **************************458**************************************************** B Did animals evolve from fungi? C In which species of eukaryote did chloroplasts first appear? D Were chloroplasts and mitochondria transferred to eukaryotic cells through horizontal gene transfer? Solution 32 The solution is (D). The addition of multiple trunks and horizontal interconnections in the fig-shaped tree of life best answers the question “Were chloroplasts and mitochondria transferred to eukaryotic cells through horizontal gene transfer?” This is because the fig-shaped tree shows interconnections and multiple starting points that are missing from a single-trunk phylogenetic tree. Which question, relating to the endosymbiotic hypothesis and the evolution of eukaryotes, is NOT answered by the eukaryote-first hypothesis, based on the figures? A Which evolved first, the nucleus or prokaryotes? B Which evolved first, mitochondria or prokaryotes? C How and when did the nucleus evolve in eukaryotes? D How and when did prokaryotes evolve? Solution 33 The solution is (C). The eukaryote-first hypothesis does not answer the question, “How and when did the nucleus evolve in eukaryotes?” The phylogeny shows the evolution of traits in vertebrates. ******************************************************* ***************************************************************************459*** Based on this phylogeny, a student asks, “Does this mean that lizards, frogs and rabbits all possessed hair and an egg with amnion at some point in their evolution?” Based on the phylogeny, how should you respond to the student? A Hair and an amniotic egg were both possessed by all three species at some point in their evolution. B Hair is only a characteristic found in the rabbit evolutionary history. The amniotic egg was possessed by both the rabbit and lizard, but not frogs, at some point in their evolutionary history. C Hair is a characteristic only found in the rabbit evolutionary history. The amniotic egg was possessed by all three species at some point in their evolutionary history. D Hair was possessed by all three species at some point in their evolutionary history. The amniotic egg was possessed by both the lizard and frog, but not the rabbit at some point in their evolutionary history. Solution The solution is (B). The best response would be to show the student that “Hair” is a derived trait only possessed by the rabbit on this tree. The amniotic egg is possessed by both the rabbit and lizard, but not frogs. ******************************************************** **************************458**************************************************** 34 The tree shows the phylogenetic relationships between four species. A scientist wishes to perform a genetic analysis on all four species in which she determines the number of genetic similarities among all four species. What would she likely find regarding the genetic similarities among species A, B, D, and E? A Species D and E would share more genetic similarities with each other than with species A and B, and vice versa. B Species A and E would share more genetic similarities with each other than with species B and D, and vice versa. C Species D and A would share more genetic similarities with each other than with species A and B, and vice versa. D Species D and B would share more genetic similarities with each other than with species A and E. Solution 35 The solution is (D). Species D and B have fewer nodes between them than A and E and thus are more related. What is the aim of scientists applying the maximum parsimony concept when creating phylogenetic trees? A The scientists spend more time creating the phylogenetic table. B Scientists find the shortest tree with the smallest number of changes. C A complex, detailed phylogenetic tree diagram is created. D The scientists spend more time researching the data for evolutionary connections. Solution The solution is (B). Scientists look for the most obvious and simple order of evolutionary events, and doing so, they find the shortest tree with the smallest number of changes. ******************************************************* ***************************************************************************459*** 36 Dolphins and fish have similar body shapes. Is this feature more likely a homologous or analogous trait? Explain your answer. A Analogous: Dolphins are mammals and fish are not, thus their evolutionary paths are quite separate. They have similar body shapes because of their similar environment. B Analogous: Dolphins and fish are both vertebrates, thus they share an evolutionary history, causing them to have similar body shapes. C Homologous: Dolphins and fish are both vertebrates, thus they share an evolutionary history, causing them to have similar body shapes. D Homologous: Dolphins are mammals and fish are not, thus their evolutionary paths are quite separate. They have similar body shapes because of their similar environment. Solution 37 The solution is (A). Dolphins are mammals and fish are not, which means that their evolutionary paths (phylogenies) are quite separate. Dolphins probably adapted to have a similar body plan after returning to an aquatic lifestyle, and, therefore, this trait is probably analogous. What effect has the advancement of DNA technology had on determining phylogeny? A Morphologic and molecular information often disagree. B Scientists are struggling with molecular systematics. C Information is not reliable because organisms appear to be closely related when they are not. D Computer programs help determine relatedness using DNA sequencing, and morphologic and molecular information is more effective in determining phylogeny. Solution 38 The solution is (D). Computer programs help determine relatedness using DNA sequencing, and molecular data. Molecular information is more effective in determining phylogeny. New computer programs confirm earlier classified organisms and also uncover previously made errors. What is maximum parsimony used for in evolutionary biology? A Maximum parsimony hypothesizes that organisms that share the most traits are the most likely to share a common ancestor. B Maximum parsimony hypothesizes that organisms that share a common ancestor are more likely to have many traits in common. C Maximum parsimony hypothesizes that events occurred in the simplest, most obvious way, and the pathway of evolution probably includes the fewest major events that coincide with the evidence at hand. D Maximum parsimony hypothesizes that organisms that display homologous structures are closely related, while organisms that display analogous structures must have diverged much farther in the past. ******************************************************** **************************458**************************************************** Solution 39 The solution is (C). Maximum parsimony hypothesizes that events occurred in the simplest, most obvious way. To aid in defining phylogenies accurately, scientists decipher the pathway of evolution using the fewest major events that coincide with the evidence at hand. The emu in Australia and ostrich in Africa are flightless birds that look similar. One proposed hypothesis was the birds descend from an early common ancestor that spread when the continents were connected. DNA analysis shows that emus and ostriches share more genetic homology with flying birds, which live in the same region than with each other. What is the best explanation for these findings? A This is an example of an early shared ancestor B This is an example of convergent evolution. C This is an example of random DNA homology D This is an example of divergent evolution. Solution 40 The solution is (B). This is an example of convergent evolution because the emu and ostrich, despite their physical similarities, likely do not share a recent common ancestry because of their isolation on different continents. A scientist decides to investigate the evolutionary connection between closely related bacteria. Which gene would be a good choice to use for establishing relatedness, a very well conserved gene or a poorly conserved sequence? Explain your reasoning. a. A very well conserved gene would be a good choice, because well conserved genes undergo sufficient changes during relatively short times, which allows for the study of recent evolutionary events. Well-conserved genes do not undergo changes during short durations. b. A poorly conserved gene would be a good choice, because poorly conserved genes show sequence similarity, which is used as evidence of evolutionary relationships between sequences. c. A poorly conserved gene would be a good choice, because poorly conserved genes undergo sufficient changes during relatively short times, which allows for the study of recent evolutionary events. d. A very well conserved gene would be a good choice, because well conserved genes show sequence similarity, which is used as evidence of evolutionary relationships between sequences. Solution Sample answer: When recent evolutionary events are being studied, regions or genes that evolve much more rapidly are needed because they have undergone sufficient changes during this relatively short time. ******************************************************* ***************************************************************************459*** 41 In a hypothetical population of beetles, there is a wide variety of color, matching the range of coloration of the tree trunks on which the beetles hide from predators. The graphs illustrate four possible changes in the beetle population as a result of a change in the environment due to pollution that darkened the tree trunks. What would be the most likely change in the coloration of the beetle population after pollution and why? A The coloration range shifted toward more light-colored beetles, as in diagram I. The pollution helped the predators find the darkened tree trunks. B The coloration in the population split into two extremes, as in diagram II. Both the light-colored and the dark- colored beetles were able to hide on the darker tree trunks. C The coloration range became narrower, as in diagram III. The predators selected beetles at the color extremes. D The coloration in the population shifted toward more dark-colored beetles, as in diagram IV. The light-colored beetles were found more easily by the predators than were the dark-colored beetles. Solution 42 The solution is (D). Selection pressure will favor the dark-colored beetles which are well camouflaged and less likely to be seen and eaten by predators. A population of rodents settles on the shore of an island close to the Arctic Circle. The landscape consists mainly of rocks. If the individuals are too large, they cannot hide in crevices to escape hawks. On the other hand, small bodies do not maintain internal temperature in cold weather. Show diagrammatically the change in the population and explain what selective pressures took place. Solution ******************************************************** **************************458**************************************************** ******************************************************* ***************************************************************************459*** 43 Five new species of bacteria were discovered in Antarctic ice core samples. The nucleotides (base sequences of rRNA subunits) were determined for the new species. The table shows the number of nucleotide differences between the species. Species 1 2 3 4 5 1 – 3 19 18 27 – 19 18 26 – 1 27 2 Blank 3 Blank blank 4 – 27 Which phylogenetic tree is most consistent with the data? Blank blank blank C A B Solution 44 D The solution is (C). Species 5 has the most nucleotide changes compared to the other four species. Species 1 and 2 have three nucleotide differences between them, whereas there are 18 and 19 between 1 or 2 and the second group, species 3 and 4. There is only 1 nucleotide difference between 3 and 4. Draw the phylogenetic tree for the species in the table. Identify where on the tree each feature evolved. Species Amniotic Egg Endotherm Feathers Lungs Vertebrae Notochord Snake Y N N Y Y Y Ostrich Y Y Y Y Y Y Shark N N N N Y Y Frog N N N Y Y Y Lancelet N N N N N Y ******************************************************** **************************458**************************************************** A The ostrich branched off first, followed by the snake, then the frog, then the shark, and then the lancelet. B The shark branched off first, followed by the lancelet, then the frog, then the ostrich, and then the snake. C The lancelet branched off first, followed by the shark, then the frog, then the snake, and then the ostrich. D The lancelet branched off first, followed by the shark, then the ostrich, then the snake, and then the frog. Solution 45 The solution is (C). The organism with the fewest most common features branched off early and is the least related to the other organisms. The ostrich possesses a notochord (embryonic stage), vertebrae, lungs, amniotic eggs, feathers, and constant body temperature (warm-blooded is not used any longer to describe endothermic.) Barbara McClintock discovered transposons while working on maize genetics. What are the transposons composed of when they are able to shift from one location to another? A Segments of RNA B Plasmids C Segments of DNA D Proteins Solution 46 The solution is (C). Transposons are segments of DNA that can shift their location and can enter a cell with the assistance of a plasmid. What is horizontal gene transfer (HGT)? A The proposal that eukaryotes developed a nucleus first, and then their mitochondrion B The transmission of genetic material from one species to another through mechanisms other than from parent to offspring C The fusion of two prokaryotic genomes D The division of kingdom in the taxonomic classification ******************************************************* ***************************************************************************459*** Solution 47 The solution is (B). HGT is the introduction of genetic material from one species to another species by mechanisms other than the vertical transmission from parent(s) to offspring. These transfers allow even distantly related species to share genes, influencing their phenotypes. What is referred to as the transfer of genes by a mechanism that does not involve asexual reproduction? A Web of life B Meiosis C Gene fusion D Horizontal gene transfer Solution 48 The solution is (D). Horizontal gene transfer is the introduction of genetic material by mechanisms other than parent to offspring transfer as in sexual and asexual reproduction. Which options describes small, virus-like particles that act as a mechanism of gene transfer between prokaryotes? A Gene transfer agents B Horizontal gene transfer C Vertical gene transfer D Basal taxon Solution The solution is (A). Gene transfer agents are bacteriophage-like particle that transfers random genomic segments from one species of prokaryote to another. SCIENCE PRACTICE CHALLENGE QUESTIONS 20.1 Organizing Life on Earth 49 The figure shows a plot of the fraction, as a percentage, of families of marine organisms living at a particular point in time that became extinct (vanished from the fossil record) in the next geological moment. These mass extinctions mark the ends of geologic periods. For example, the Triassic period ended around 213 million years ago (Mya). ******************************************************** **************************458**************************************************** A. The fact that evolution is an ongoing process is illustrated by these data. Whether the process displays a pattern involving regular intervals is a question that has been raised. Of those who believe periodicity is present, a period of 26 million years is favored. A wave with this periodicity is drawn on the figure. Evaluate the evidence provided in terms of agreement and disagreement with the marine extinction record. The Cretaceous and Jurassic were periods of warm landmasses covered by a shallow sea. The ends of these periods are known to be due to asteroids that left a sedimentary trace. At the end of the Triassic, there is no evidence of an asteroid impact. Instead, there was massive volcanism associated with the opening of the Atlantic Ocean, a four-fold increase in carbon dioxide, and a 3–6 °C temperature rise (A. Marzoli et al., Science, 1999). Macrofossil, spore, and pollen data show that marine animal species declined much more than marine plant species (L. Mander et al., Proc. Natl. Acad. Sci., 2010). The cause of the end of the Permian period is less uncertain, but an 8°C temperature rise has been established (McElwain and Punyasena, Trends in Ecology and Evolution, 2007). Both terrestrial and marine taxa were affected. B. The graph estimates the number of distinct families, including both marine and terrestrial, as a function of time before the present. Note that the time scale for this graph is much longer than that of the other graph. ******************************************************* ***************************************************************************459*** Analyze this graph by Solution Identifying times of mass extinctions Posing a question regarding any difference between the graph of extinctions of marine life and the graph of family level diversity Explaining the slope of the graph of family diversity following a mass extinction event Sample answer: A. End-Permian, end-Triassic, end-Jurassic, and end-Cretaceous all fit the period. However, during the Jurassic and Tertiary the fit is poor. If there is periodicity, no other wavelength is better than the one suggested. Aside: A possible explanation involving the effect of dark matter on the frequency of asteroid collisions has been presented (Rampino, MNRAS, 2014). B. The Permian, Triassic, and Cretaceous boundaries show drops for family diversity, and these lines up with marine extinctions. After these extinction events, there is a rapid increase in speciation as shown by the steep slopes following these events. At the boundary marking the end of the Jurassic and beginning of the Cretaceous, there is no drop in family diversity, though there is a mass extinction of marine life. Is the increase in speciation in the warm, shallow seas of these periods compensating for the extinction of marine life? Did the increase in temperature and/or carbon dioxide have a more profound effect on marine species because of drops of oxygen ******************************************************** **************************458**************************************************** concentration or acidification of oceans than they had on terrestrial species, particularly plants? 20.2 Determining Evolutionary Relationships 50 Lactate dehydrogenase, an enzyme involved in glycolysis, from several species is compared using a Southern blot technique in the figure: (A) yeast; (B) snail; (C) mouse; (D) rat; and (E) human (after K. Webster, Journal of Experimental Biology 2003). A. Justify the claim that these data provide evidence that supports glycolysis as a conserved core property. There are three forms of lactate dehydrogenase whose roles vary within an organism: LDH-A, -B, and -C. The question arises as to the origins of each. The cladogram (after S. Tsuji et al., Proc. Natl. Acad. Sci., 1994) displays a proposed relatedness of variations in LDHs over many species. ******************************************************* ***************************************************************************459*** B. Based on these data, describe the differences between the likely evolutionary sequences of LDH-A, -B, and -C in the African frog and in humans. C. Evaluate the claim that in both species (African frog and human), LDH-C is the most recently evolved form of the enzyme. Solution Sample answer: A. Each of the species shows the same two DNA sequences. Since this enzyme is a component of glycolysis, we see that this is a core conserved process. B. The African frog LDH-A and LDH-B represent a divergence and the modification of the nucleotide sequence in LDH-B led to LDH-C. In the human LDH-B and LDH-A have been derived from LDH-C. C. LDH-C probably came last in the frog and first in the human. The evidence is the larger number of nucleotide substitutions. ******************************************************** **************************458**************************************************** 51 Nucleotide-repeat sequences often occur within the intron, and sequence variation is neutral; that is, there is no selection. For example, the nucleotide-repeat sequence labeled A might be ACGGGC, and the repeat sequence labeled B might be ACTGGT. Repeat sequences evolve by single-step duplication, deletion, and inversion, rather than by single nucleotide substitution. Because these repeat sequences can be used to infer phylogeny, a phylogenetic tree can be hypothesized based on the principle of parsimony—the simplest explanation is the best explanation. Consider the repeat sequences A, B, C, D, and E shown in the list in which only inversions have occurred among five different species. Species 1: ABCDE Species 2: ADCBE Species 3: BACED Species 4: DACBE Species 5: ABCED Species 6: DEBCA A. Pose three questions that can be used to infer the evolution of these five species. B. Draw lines between nearest relatives to construct a cladogram that displays the relationships inferred by answers to your questions. Carson (Drosophila Genetics 1983) used inversions in intron-repeat sequences of the fruit fly to infer evolution among the Hawaiian Drosophila. He further assigned the chronological sequence of islands on which the flies appeared by assuming (parsimoniously) that the geologically oldest of the volcanic islands was home to the oldest fly ancestor. When a fly or flies arrived on a newer island, speciation occurred, which, after a time, stabilized until another island hop occurred. C. Evaluate Carson’s reasoning for speciation and ongoing evolution. D. Pose two questions whose pursuit could provide additional evidence of Carson’s hypothesized evolutionary sequence. ******************************************************* ***************************************************************************459*** Solution Sample answer: A. Which species are most closely related? Which species are connected by a single inversion? Which sequences cannot be connected by a single inversion? How can I get from species 1 to 2, 1 to 3, 1 to 4, 1 to 5, 2 to 3, 2 to 4, 2 to 5, etc.? B. C. The concept is that introduction to a new island introduces new selection acting on mutations. Over time, the speciation that results undergoes a convergence to a stable population. This species then migrates to a new island and the process is repeated. D. Do other proteins or gene sequences display the same phylogeny? Where did the population on the oldest island originate? How can we be sure that only a single inversion occurred between island hops? The answer to the first question is yes, and Carson’s simple reasoning is regarded as very eloquent in this time of more bruteforce, next-generation sequencing to infer phylogeny. The answer to the second question was South America. The answer to the third is that we cannot; the frequent use of the word “hypothesis” is more common amongst evolutionary biologists than among other sub-disciplines, say, biochemists. Other evidence is always desirable until, as Stephen Jay Gould said, “It is pernicious to withhold consent.” ******************************************************** **************************498**************** 21 | VIRUSES REVIEW QUESTIONS 1 Viruses were first discovered after the development of the porcelain filter, called the Chamberland-Pasteur filter. How did the porcelain filter enable scientists to discover viruses? A The porcelain filter removed diseases from a liquid sample. B The porcelain filter removed virions from a liquid sample. C The porcelain filter removed bacteria from a liquid sample. D The porcelain filter removed a disease from tobacco plants. Solution 2 The solution is (C). The porcelain filter could remove all bacteria visible in the microscope from any liquid sample, while allowing the viruses to pass through the filter. In the late 1930s, scientists got their first good view of viruses. How did this happen? A The development of the light microscope helped scientists discover many viruses of all types of living organisms. B The development of the viral receptor helped scientists discover many viruses of all types of living organisms. C The development of the porcelain filter helped scientists discover many viruses of all types of living organisms. D The development of the electron microscope helped scientists discover many viruses of all types of living organisms. Solution 3 The solution is (D). When the electron microscope was developed in the late 1930s, it provided greater magnification and helped scientists see the structure of many viruses of all types of living organisms. Determining the origins of viruses is challenging. Which hypothesis proposes to explain the origin of viruses by suggesting that viruses evolved from free-living cells? A Escapist or the progressive B System of self-replication C Devolution or the regressive D Virus molecular systematics Solution The solution is (C). When the electron microscope was developed in the late 1930s, the development of the electron microscope helped scientists discover many viruses of all types of living organisms based on viral adaptations to surviving inside of host cells. ******************************************************** ***************************************499*** 4 Which statement best describes what biologists know about the evolution of viruses? A Scientists can look at fossil records and similar historic evidence. B Much about virus origins and evolution remains unknown. C Biologists have accumulated a significant amount of knowledge about how viruses originated. D Biologists know exactly when viruses emerged and from where they came. Solution 5 The solution is (B). Much about virus origins and evolution remains unknown. Although biologists have accumulated a significant amount of knowledge about how present-day viruses evolve, much less is known about how viruses originated in the first place. What is an individual virus particle outside a host cell that consists of a nucleic acid core, an outer protein coating, and sometimes an outer envelope? A A capsid B A virion C A capsomere D A viral receptor Solution 6 The solution is (B). A virion is a viral particle in its extracellular (outside of the cell) state. It consists of the nucleocapsid (combination of viral nucleic acid and the protenacious-capsid coat). For many viruses to penetrate the cell membrane and complete their replication inside the cell, the virus must attach to their host cells. How does a virus attach to a host cell? A A virus uses its cellular structure to attach to a host cell. B A virus uses a plasma membrane to connect to a host cell. C A virus uses matrix proteins to attach to a host cell. D Viruses use viral receptors to attach to a host cell. Solution 7 blank The solution is (B). Viruses exhibit high specificity for select host cells. Nonenveloped or enveloped viruses may enter cells in different ways. Proteins present in the viral capsid for nonenveloped viruses or on the envelope for enveloped viruses, binds to the receptor for which it is specific, in much the same manner as a “key” fits in a specific “lock.” _____ means that the genomic RNA can function as mRNA. A Double-stranded B Negative polarity C Positive polarity D Replica intermediates ******************************************************** **************************498**************** Solution 8 The solution is (C). Positive polarity is an ssRNA virus with a genome that contains the same base sequences and codons found in their mRNA. Viruses often are classified based on the type of genetic material and its structure. In the Baltimore classification scheme, which virus has a single-stranded RNA (–) genome? A Human immunodeficiency virus (HIV) B Rabies (rhabdovirus) C Canine parvovirus (parvovirus) D Common cold (pircornavirus) Solution 9 The solution is (B). The rabies (rhabdovirus) virus is in Group V, single-stranded RNA (–). What must scientist use to get a visual look at the internal structure of virions? A A scanning electron microscope B A transmission electron microscope C A porcelain filter D A light microscope Solution 10 The solution is (B). The internal structures of a virus can be observed in images from a transmission electron microscope. A beam of electrons is transmitted and an image is formed from the interaction of the electrons transmitted through the specimen. Which statement about the viral replication cycle is accurate? A The viral replication cycle does not affect the structure of the host cell. B The viral replication cycle cannot affect genetic material of the host cell. C The viral replication cycle has seven basic steps. D The viral replication cycle can change cell functions or even destroy the host cell. Solution 11 The solution is (D). The viral replication cycle can produce dramatic biochemical and structural changes in the host cell, which may cause cell damage. What happens in the lysogenic cycle of viral replication? A During the budding process, virions leave the host cell individually. B During the budding process, the host cell bursts. C During the budding process, the virus connects with a permissive host cell. D During the budding process, the host cell dies immediately. Solution The solution is (A). Budding is a method of exit from the cell where virions leave the host cell individually. ******************************************************** ***************************************499*** 12 In which cycle does the virus replicate and burst out of the host cell? A Lytic B Lysogenic C Cytopathic D Latency Solution 13 The solution is (A). The lytic cycle is a type of virus replication in which virions are released through lysis, or bursting, of the cell. How is the lytic cycle different from the lysogenic cycle? A The phage infects a cell in the lytic cycle. B The lytic cycle contains the formation of a prophage. C In the lytic cycle, new phages are produced; immediately in the lysogenic cycle phage DNA is merged into the host genome. D The phages move on to infect other cells in the lysogenic phase. Solution 14 The solution is (C). In the lytic cycle, new phages are produced in recurring cycles and can lead to host death; in the lysogenic cycle phage DNA is merged into the host genome and remains dormant. Which statement is false? A Enveloped viruses and naked viruses both may enter cells using the fusion method. B Many enveloped viruses enter the cell by receptor-mediated endocytosis. C Naked viruses enter the cell by receptor-mediated endocytosis. D Undergoing shape changes and creating channels in the host cell membrane is an alternative method of cell penetration used by naked viruses. Solution 15 The solution is (A). Naked viruses cannot enter cells using the fusion method. Fusion only occurs with enveloped virions. An apple tree has yellow splotches on the leaves. What is this is a symptom of? A Cell necrosis B Discoloration C Hyperplasia D Hypoplasia Solution The solution is (D). Hypoplasia, or decreased cell growth, causes thin, yellow areas to appear in the leaves of plants. ******************************************************** **************************498**************** 16 What happens during the release step in the lytic or lysogenic cycle of replication? A During the release step, genetic information is transferred through the lytic and lysogenic cycles. B During the release step, DNA is transcribed to messenger RNA. C During the release step, the nucleic acid is released from the viral capsid or envelope. D During the release step, the new virions are able to infect adjacent cells and repeat the replication cycle. Solution 17 The solution is (D). In the release step, the new virions leave the host cell and are able to infect adjacent cells and repeat the replication cycle. Why does the HIV virus use reverse transcriptase in the replication process? A The HIV virus uses reverse transcriptase to replicate cells and build proteins. B The HIV virus uses reverse transcriptase to erase mutated virions. C The HIV virus uses reverse transcriptase because it is a retrovirus. D The HIV virus uses reverse transcriptase because it has a DNA genome. Solution 18 The solution is (C). The HIV virus uses reverse transcriptase because it is a retrovirus. Reverse transcriptase converts the viral RNA into double-stranded DNA. What are the symptoms of the herpes simplex virus? A The herpes simplex virus causes eye infections. B The herpes simplex virus causes pneumonia. C The herpes simplex virus causes pancreatitis. D The herpes simplex virus can cause septicemia. Solution 19 The solution is (A). The herpes simplex virus, adenovirus, and cytomegalovirus all cause eye infections. Which statement accurately describes the measles virus? A The measles virus causes nasal and lung infections. B The measles virus causes pancreas and liver infections. C The measles virus causes mouth and gum infections. D The measles virus causes brain and skin infections. Solution The solution is (D). The measles virus infects the respiratory system and can cause brain and skin disorders. ******************************************************** ***************************************499*** 20 Which statement best describes vaccines? A Vaccines kill viruses. B Vaccines stimulate an immune response against future infections. C Vaccines inhibit the virus by blocking the action of key viral proteins. D Vaccines control and reduce symptoms. Solution 21 The solution is (B). Vaccines stimulate an immune response against future infections. Viral vaccines are designed to give protective immunity against future infections. Which kind of therapy attacks a stage of the virus replicative cycle? A Phage therapy B Antiretroviral therapy C Gene therapy D Cancer therapy Solution 22 The solution is (B). Antiretroviral therapy attacks a virus at different stages of its replicative cycle. Which virus causes parotitis? A Measles virus B Norovirus C HIV D Mumps virus Solution 23 The solution is (D). The mumps virus causes parotitis, which is inflammation of a parotid gland. Which statement about prions is true? A Prions are larger than viruses. B Prions contain DNA and RNA. C The PrPC is the normal form of the protein. D The PrPSC is folded abnormally. Solution The solution is (D). The PrPSC, a proteinaceous infectious particle, is the abnormally folded from of PrPC. ******************************************************** **************************498**************** 24 Kuru is a prion disease that affects both humans and animals. How is kuru spread? A Kuru disease is spread between cattle. B Kuru is passed from person to person. C Kuru is passed from cows with BSE to humans. D Kuru is a viroid that infects plants. Solution 25 The solution is (B). Kuru is spread by the consumption of meat, nervous tissue, or internal organs between members of the same species. Kuru, native to humans in Papua New Guinea, was spread from human to human via ritualistic cannibalism. Which statement about viroids is true? A Viroids are single-stranded RNA particles. B Viroids reproduce only outside of the cell. C Viroids produce proteins. D Viroids affect both plants and animals. Solution 26 The solution is (A). Viroids are small, single-stranded RNA particles. They are known to infect plants such as potatoes, cucumbers, tomatoes, chrysanthemums, avocados, and coconut palms. This results in crop failures and the loss of millions of dollars in agricultural revenue each year. On which industry can viroids have a severe impact? A Dairy B Poultry C Avocado D Livestock Solution 27 The solution is (C). Viroids can have a severe impact on the avocado industry because viroids are known to infect plants, including avocados. Which statement best explains how infected prions cause disease? A Infected prions cause disease by transmitting nucleic acids to normal prion proteins. B Infected prions cause disease by converting DNA to RNA in normal prion proteins. C Infected prions cause disease by converting the shapes of normal proteins. D Infected prions cause disease by replicating the normal form of the protein. Solution The solution is (C). Infected prions cause disease by converting the shapes of normal proteins. Prions convert the normal protein shape to an abnormal form. ******************************************************** ***************************************499*** CRITICAL THINKING QUESTIONS 28 How did the development of a porcelain filter, called the Chamberland-Pasteur filter, help scientists discover viruses? A After filtering a liquid plant extract, the scientists could see the virions using the light microscope. B After filtering a liquid plant extract, the disease was still transferred to a healthy plant. C After filtering a liquid plant extract, the virus cells multiplied. D After filtering a liquid plant extract, scientists were able to trace historical footprints. Solution 29 The solution is (B). After filtering a liquid plant extract, the disease was still transferred to a healthy plant. Years later, it was proven that the infectious agents were a new type of very small, disease-causing particle. Scientists have a few hypotheses about virus origins. Why might they develop and refine further hypotheses to explain the origin of viruses? A Advances in technology provide historic evidence. B Biochemical and genetic information provide historic evidence. C Advances in technology provide new information for scientists. D Advances in technology have proven that viruses have a single common ancestor. Solution 30 The solution is (C). New technology provides new avenues for investigation and stimulates new ideas. Why don’t dogs catch the measles? A Measles is a DNA virus, and DNA viruses cause human diseases. B Dogs do not have glycoproteins. C The virus can’t attach to dog cells. D Dogs do not get RNA viruses. Solution 31 The solution is (C). The virus cannot attach to dog cells, because dog cells do not have the receptors for the virus and/or there is no cell within the dog that is permissive for viral replication. The Baltimore classification system groups viruses according to how the mRNA is produced. When classified this way, the viruses in each group — A behave in a similar manner B look very similar C connect with living things D are based on the type of disease they cause ******************************************************** **************************498**************** Solution 32 The solution is (A). Each of the seven groups has a different replication strategy. Researchers have been able to develop a variety of anti-HIV drugs, such as the drug AZT. How does the drug AZT work? A AZT blocks the enzyme called HIV protease, which the virus uses to reproduce itself. B AZT blocks the HIV integrase enzyme, which the virus uses to insert its viral DNA into the DNA of the host cell. C AZT prevents reverse transcriptase and HIV protease enzyme from functioning inside the body. D AZT prevents reverse transcriptase from making DNA from the viral RNA genome. Solution 33 The solution is (D). AZT is a nucleic acid synthesis inhibitor that functions as an analog of thymidine. It works by preventing the viral enzyme reverse transcriptase from making DNA from the viral RNA genome. Without this step, HIV cannot replicate. Compare the lytic and lysogenic cycles. Which cycle has the potential to produce the most virions? A The lytic cycle can theoretically produce more virions as the viral genome takes over the host cell, resulting in the large-scale release of virions. B The lysogenic cycle can theoretically produce more virions as the reproductive cycle of viruses undergoing lysogeny is much faster than the reproductive cycle of viruses following lytic cycle. C The lysogenic cycle can theoretically produce more virions as the viral genome is incorporated into the host cell’s genome replicating along with the host cell. D The lytic cycle can theoretically produce more virions as the prophage following lysogenic cycle ultimately gets excised from the host cell’s genome and enter the lytic cycle. Solution 34 The solution is (C). The lysogenic cycle can theoretically produce more virions than the lytic cycle, because in the lysogenic cycle the viral genome is incorporated into the genome of the host cell and the virus continues to live. The virus’s nucleic acid is replicated as the host cell multiplies. The virus can live and replicate inside a host for a long time. Would a person who has never been in contact with the varicella-zoster virus be at risk of developing chickenpox or shingles if they come in close contact with a person with shingles? Explain your reasoning. A The person is at risk of developing chickenpox. Chickenpox is the first infection with the virus before it enters latency in the host. B The person is at risk of developing shingles. Shingles is the first infection with the virus before it enters latency in the host. ******************************************************** ***************************************499*** C The person is at risk of developing chickenpox. Chickenpox is the first infection with the virus that is already latent in the body. D The person is at risk of developing shingles. The virus enters the person and gets activated when a person with shingles comes in close contact. Solution 35 The solution is (A). Because this is the first time the person comes in contact with the varicella-zoster virus, they risk developing chickenpox. Chickenpox is the first infection with the virus before it enters latency in the host. Shingles is a secondary infection from virus that is already latent in the body. Which step in the replication cycle of viruses do you think is most critical for the virus to infect cells? Explain why. A The attachment step is the most critical, as infection cannot begin if the virus does not attach to the host cell. B The replication step is the most critical, as this step directs protein synthesis. C The assembly step is the most critical, because new virions are assembled to infect cells. D The entry step is the most critical, as nucleic acid of the virus needs to enter the host cell naked, leaving the capsid outside. Solution 36 The solution is (A). The attachment step is the most critical step in the replication cycle, because if the virus does not attach to the host cell, the infection process cannot begin. For most people, the measles virus does not cause a serious illness. Symptoms include fever and a rash, but the symptoms are usually gone in about a week. However, for some, the measles virus can be much more serious. How can the measles virus cause a potentially fatal illness? A Measles can cause meningococcal disease, which causes severe headaches, seizures, and in severe cases, can be life-threatening. B Measles can cause variant Creutzfeldt-Jakob disease, which causes severe headaches, seizures, and in severe cases, can be life-threatening. C Measles can cause encephalitis/meningitis, which causes severe headaches, seizures, and in severe cases, can be life-threatening. D Measles can cause Legionnaires’ disease, which causes severe headaches, seizures, and in severe cases, can be life-threatening. Solution The solution is (C). Measles is a contagious respiratory system virus that can cause encephalitis/meningitis, which is inflammation of the brain. Severe cases of encephalitis can be life-threatening. ******************************************************** **************************498**************** 37 Why is immunization after being bitten by a rabid animal so effective and why aren’t people vaccinated for rabies like dogs and cats are? A It takes a month for the virus to travel from the site of the bite to the central nervous system. People are not vaccinated beforehand as routine vaccination of domestic animals makes it unlikely that humans will contract rabies from an animal bite. B It takes a week for the virus to travel from the site of the bite to the peripheral nervous system. People are not vaccinated beforehand as routine vaccination of domestic animals makes it unlikely that humans will contract rabies from an animal bite. C It takes a week for the virus to travel from the site of the bite to the central nervous system. People are not vaccinated beforehand as routine vaccination of domestic animals makes it unlikely that humans will contract rabies from an animal bite. D It takes a week for the virus to travel from the site of the bite to the central nervous system. People are not vaccinated beforehand, as routine vaccination of domestic animals makes it fully sure that humans will contract rabies from an animal bite. Solution 38 The solution is (C). The rabies vaccine works after a bite because it takes week for the virus to travel from the site of the bite to the central nervous system, where the most severe symptoms of the disease occur. Adults are not routinely vaccinated for rabies for two reasons: first, because the routine vaccination of domestic animals makes it unlikely that humans will contract rabies from an animal bite; second, if one is bitten by a wild animal or a domestic animal that one cannot confirm has been immunized, there is still time to give the vaccine and avoid the often fatal consequences of the disease. Why don’t dogs and cats catch human colds from humans? A As cats and dogs have different proteins than humans, the virus that causes colds in humans cannot find receptors in dogs and cats. B As cats and dogs have different receptors than humans, the virus that causes colds in humans cannot find receptors in dogs and cats. C Cats’ and dogs’ immune systems attack the virus unlike humans’ immune systems, so the virus that causes colds in humans cannot find receptors in dogs and cats. D As natural killer cells of cats and dogs attack the virus, the virus that causes colds in humans cannot find receptors in dogs and cats. Solution The solution is (B). To survive and reproduce, the viruses need a host of living cells. Cats and dogs have different DNA than humans, so the virus that causes colds in humans cannot find receptors in dogs and cats. ******************************************************** ***************************************499*** 39 Prions are responsible for variant Creutzfeldt-Jakob Disease (vCJD). How has this disease been documented to spread from human to human? A Surgery with instruments previously used in a patient with vCJD that were not adequately sterilized and contaminated pineal growth hormones taken from human pineal glands from infected cadavers. B Through human consumption of infected meat and contaminated pituitary growth hormones taken from human pituitary glands from infected cadavers. C Surgery with instruments previously used in a patient with vCJD that were not adequately sterilized and contaminated pituitary growth hormones taken from human pituitary glands from unwell individuals. D Surgery with instruments previously used in a patient with vCJD that were not adequately sterilized and contaminated pituitary growth hormones taken from human pituitary glands from infected cadavers. Solution 40 The solution is (B). This prion-based disease is most often transmitted through human consumption of infected meat. However, humans can contract the disease following surgery with instruments previously used in a patient with vCJD that were not adequately sterilized. Also, transmission has been linked to contaminated pituitary growth hormones taken from human pituitary glands from infected cadavers. What characteristics do viroids and viruses have in common? A They both replicate within a host cell and contain nucleic acids. B They both replicate within a host cell and do not contain nucleic acids. C They both replicate within a host cell and contain proteins. D They both replicate within a host cell and contain only RNA. Solution 41 The solution is (A). These noncellular infectious agents both replicate within a host cell, and they both contain nucleic acid. Why is the transmission of a prion NOT reliant upon genes made of DNA or RNA? A DNA or RNA, though present, is not transmitted when a prion causes infection. B The prion does not contain DNA or RNA. C Only parts of DNA or RNA are transmitted in a prion. D More of protein and less of DNA or RNA is transmitted. Solution The solution is (B). Prions are infectious acellular particles that consist of abnormally folded proteins that replicate without DNA or RNA. ******************************************************** **************************498**************** TEST PREP FOR AP® COURSES 42 The table shows the Baltimore Classification used to classify viruses based on their genetic material. What is the difference between how Group I and Group III viruses reproduce? A In Group I, RNA is transcribed from an RNA genome while in Group III, RNA is transcribed from a DNA genome. B In Group I, RNA is transcribed from a DNA genome while in Group III, RNA is transcribed from an RNA genome. C In Group I, DNA is transcribed from a DNA genome while in Group III, RNA is transcribed from an RNA genome. D In Group I, DNA is transcribed from an RNA genome while in Group III, RNA is transcribed from a DNA genome. Solution The solution is (B). The Group I contains double-stranded DNA which requires the RNA being transcribed from a DNA genome while in Group III, which has doublestranded RNA, the RNA is transcribed from an RNA genome. ******************************************************** ***************************************499*** 43 The table shows the Baltimore Classification used to classify viruses based on their genetic material. What is similar or different between the genome of Group I and Group VI, as well as how the two virus types reproduce? A Group I and VI viruses use RNA as their genome. Group I viruses reproduce by transcribing RNA from their DNA genome, while Group VI viruses first synthesize their RNA genome using reverse transcriptase before they can reproduce. B Group I and VI viruses use DNA as their genome. Group I viruses reproduce by transcribing RNA from their DNA genome while group VI viruses first synthesize their DNA genome using reverse transcriptase before they can reproduce. C Group I and VI viruses use DNA as their genome. Group I viruses reproduce by transcribing RNA from their DNA genome, while group VI viruses first synthesize RNA genome using reverse transcriptase before they can reproduce. D Group I viruses use DNA as their genome while group VI use RNA. Group I viruses reproduce by transcribing RNA from their DNA genome while group VI viruses synthesize DNA from RNA using reverse transcriptase before they can reproduce. Solution The solution is (B). Group I and Group VI viruses both use DNA as their genome, however, Group I viruses reproduce by transcribing RNA from their DNA genome, while Group VI viruses have to first synthesize their DNA genome using reverse transcriptase before they can reproduce. ******************************************************** **************************498**************** 44 The diagram shows the stages during which a virus infects a host cell. During which of the numbered steps does the amount of viral genetic material begin to change within the host cell? A Step 1. Virus enters the cell. B Step 2. Virus RNA enters the nucleus. C Step 3. New viruses assemble within the cell. D Step 4. Viruses leave the cell. Solution The solution is (B). As viral RNA enters the host’s nucleus in step 2, the virus hijacks the host cell, causing the host to make more viral RNA. ******************************************************** ***************************************499*** 45 The diagram shows the stages during which a virus infects a host cell. How could the influenza virus change the function of a host cell? A Because it replicates its DNA within the cell and reproduces, which could interfere with cell processes B Because it replicates RNA within the cell and reproduces, which could interfere with cell processes C Because it attacks the immune system of the host cell, which in turn would interfere with cell processes D Because it replicates its protein within the cell and reproduces, which could interfere with cell processes. Solution The solution is (B). The influenza virus could change the function of the host cell because it replicates its RNA within the cell and reproduces. This could interfere with cell processes as more cell resources are used to make viruses instead of used to keep the cell alive. ******************************************************** **************************498**************** 46 The diagram models the lytic and lysogenic reproductive cycles of viruses. Which cycle would maintain the DNA of the virus over several generations, and why? A Lysogenic, because the viral DNA can be excised from the host cell’s DNA when under stress B Lytic, because the viral DNA can be excised from the host cell’s DNA when under stress C Lytic, because the viral DNA can be passed on when the host cell replicates D Lysogenic, because the viral DNA can be passed on when the host cell replicates Solution The solution is (D). Lysogenic, because the viral DNA can be passed on when the host cell replicates. ******************************************************** ***************************************499*** 47 The diagram models the lytic and lysogenic reproductive cycles of viruses. Based on the diagram, is the following statement is true or false? Explain. The lysogenic cycle allows viruses to preserve their genome during unfavorable conditions. A True, because when the host cell experiences unfavorable conditions, it stops dividing and stays in the same state B True, because the host cell in both the replication stage and during unfavorable conditions stays in the lysogenic cycle as it is more preferable over the lytic cycle. C False, because when the host cell experiences unfavorable conditions, the prophage exits the genome and enters the lytic cycle D False, because when the host cell experiences unfavorable conditions, the virus enters latency period Solution The solution is (C). The lysogenic cycle does allow viruses to preserve their genome during host cell replication. However, when the host cell experiences unfavorable conditions, the prophage exits the genome and enters the lytic cycle. ******************************************************** **************************498**************** SCIENCE PRACTICE CHALLENGE QUESTIONS 21.1 Viral Evolution, Morphology, and Classification 48 Influenza A virus is the most pathogenic of the human influenza viruses. Its envelope encloses a protein complex (vRNP) and eight, single-stranded, negative RNA (the complement of a positive RNA strand that can be transcribed by a ribosome) segments (vRNA). Each segment encodes one or two proteins that support viral replication. On the outer surface of the envelope are proteins that recognize and bind to host receptors. A. Annotate the representation to briefly describe each process associated with a numbered label. B. Describe influenza A viral replication as a process regulated by either positive or negative feedback, and justify your selection. C. The human acquired immunodeficiency syndrome (AIDS) and many cancers are cause by double-stranded RNA retroviruses. Contrast the processes of viral replication of HIV and influenza A virus. D. Explain the difference in the effects of infection by HIV and influenza A virus on host genetic variability. E. Measured mutation rates for influenza A virus and HIV are nearly identical (Sanjuan et al., Jour. Virology, 2010). Explain this observation even though host error-checking operates in one of these replication modes. Solution Sample answer: A. 1. Attachment/endocytosis 2. Envelope dissolution 3. Transport over nuclear membrane 4. Synthesis of RNA complement 5. Transport to cytoplasm 6. Budding to release progeny ******************************************************** ***************************************499*** B. A virion infects the cell and the cell amplifies, creating multiple copies. The mechanism of this one-to-many positive feedback is that the code for the ribosomeprotein complex vRNP increases the transcriptive capacity of the cell. The human acquired immunodeficiency syndrome and many cancers are caused by double-stranded RNA retroviruses. C. The details of HIV replication are defined as “in-scope” by the Framework, and so the student should be able to describe the process of retrovirus replication. In the HIV there is a double-stranded RNA that is used to produce a strand of DNA through action of reverse transcriptase that accompanies the viral contents. The DNA strand is then incorporated into the host DNA, and the host cell DNA polymerase is used to synthesize new copies of the virus RNA. As described in A, there is no comparable hijacking of cell replication apparatus in the case of influenza A. Consequently, the flu is an acute disease and HIV is a chronic disease. D. As described above the host DNA is altered by the retrovirus, while the host DNA is unaltered by the single strand RNA virus. E. The host error-checking apparatus (spliceosome) is used in replication of retrovirus, and there is no error checking in the single strand RNA virus. However, there is no error checking on the reverse transcriptase step, so the net mutation rates are comparable. 49 A. Three-dimensional (3-D) structures, or folding, of proteins have been shown to contain more information about evolutionary relationships than the sequences of DNA nucleotides that encode the proteins. Amino acid sequences of rabbit skeletal muscle actin (375 amino acids) and bovine ATPase (386 amino acids) have only 39 locations in common. However, the 3-D structure of these proteins is nearly identical (Flaherty et al., Proc. Natl. Acad. Sci. USA, 1991). As information about the 3-D folding of proteins and the number of sequenced whole genomes has increased, folding has been shown to be an evolutionarily conserved property. Analyze these data to refine the following model: The evolutionary history of life on Earth can be inferred from variations over time of the nucleotide sequence of a gene. ******************************************************** **************************498**************** B. By applying a classification scheme based on protein folding, Nasir and Caetano-Anollés (Sci. Adv., 2015) have determined the number of folding families that viruses share with the three domains. Approximately 60 percent of the folding patterns found in viruses were common to all three domains, as shown in the figure. Fewer than 10 percent were unique to viruses. Viruses are acellular, and, consequently, they lie outside of the three domains of cellular life. However, their exclusion is increasingly challenged. Since 2012, several very large viruses have been discovered, each a double-stranded DNA virus with more than one million bases, with some encoding nucleotides and amino acids. However, none encode ribosomes, so these viruses are still dependent on a marine bacteriovore (amoeba or flagellate) host for replication. Hypotheses regarding the origin of life on Earth need to account for the relationship between proteins and genetic information. Proteins are required to read and write genetic information, but genetic information is required to synthesize proteins. Which of these systems evolved first, and if neither came first, how could they evolve simultaneously? The RNA-first model is based on the idea that ribosomal RNA both encodes and synthesizes proteins. Describe a hypothesis for the origin of life on Earth that combines the dual functionality of RNA and the function of retroviral reverse transcriptase to propose a mechanism leading to an ancient, acellular lineage of very large, double-stranded DNA viruses and a first DNAbased cellular life form. C. Like viruses, the nucleus of a eukaryote uses the machinery of the cell to transcribe DNA and synthesize proteins. Evaluate the possibility of the origin of Eukarya by specialization of a very large double-stranded DNA virus. ******************************************************** ***************************************499*** Solution All are areas of rapid development and conjecture. They can be traced from the citations provided in the problem. Sample answer: A. The evolutionary history of life on Earth can be inferred from the variation over time of the molecular phenotype of a gene that includes the 3-D shape of the protein expressed. B. An RNA world is consistent with a precursor to the retrovirus that acquired an alternative (and more stable) information storage strategy through the construction of DNA using reverse transcriptase. The formation of complementary strands of DNA led to double stranded viruses. The accumulation of large genomes may indicate an ancient origin. The nature of the host (single celled marine eukaryotes) suggests a selective advantage when the information is enclosed within a lipid membrane. A test of the hypothesis would be the self-assembly of double-stranded DNA from a system composed of retrovirus and the formation of lipid vesicles containing these strands. C. Engulfment of a large double stranded virus by a prokaryote could lead to a specialized information storage and retrieval function for this virus within the cell— which continues to use the prokaryote systems for protein synthesis. 21.2 Virus Infections and Hosts 50 Viruses evolve but leave no fossil evidence that can be used to construct phylogenies. However, viral DNA, especially that of retroviruses, is commonly found in the host genome. By comparing sequences from the same virus integrated at different points in time, the evolutionary history of the virus can be constructed. The viral genomes are typically found incomplete, in segments, and interrupted by stop codons. In jawed vertebrates, retroviral sequences or sequences that have been derived from them are a significant fraction of the whole genome. A. Explain why retroviral DNA, rather than the genomes of single-stranded or doublestranded DNA or single-stranded RNA viruses, is found in host DNA. B. Exaptation occurs when gene expression provides a function that is independent of the selection pressures that have acted on the gene. For example, a pigment that provided selective advantage by reducing damage from solar radiation becomes an element of mating behavior. Feathers that evolved under selection to prevent heat loss become a means of flight. In a study of viral evolution within host genomes of primates, Katzuorakis and Gifford (PLOS Genetics, 2010) found that viral genomes within the host were surprisingly stable; with computer simulation, they estimated the probability of such constancy at 1 in 100,000. Explain in terms of selection how viral genetic information that no longer replicates the virus is maintained by the host. ******************************************************** **************************498**************** C. Distemper is an incurable disease of cats, dogs, and their sister lineages caused by a parvovirus. The virus exploits the host’s transferrin, a membrane-bound protein used for iron transport, to attach to the cell. The phylogeny of the Parvoviridae family has been constructed (J. Kaebler, PLOS Pathogens, 2012). That study revealed the evolution of both the virus and the host protein through selection to resist infection. About 54 million years ago when the lineage of cats (Feliformia) diverged from that of dogs (Caniformia), the parvovirus envelope diverged as well, conforming to changes in the host’s transferrin. In 1978, a worldwide disease in dogs due to a parvovirus suddenly appeared. Explain how this pandemic could have originated in the cat population. Solution Sample answer: A. Single- and double-strand DNA viruses and single-strand RNA viruses do not integrate their DNA into the host DNA, while retroviruses do and then use the host’s DNA transcription system for replication. This leaves the viral DNA in the host genome. If the cell becomes a zygote, the viral DNA is inherited. B. Since the viral segments are retained, there must be a negative selection pressure—their loss would disadvantage the organism relative to some selection. Estimates of the amount of retroviral DNA in the human genome are between 1 and 8 percent. This is an area of active research where not much is known yet. C. Mutations of the code for the envelope have been selected by the changing protein configuration of the receptor site—an arms race between host and virus. Small differences in transferring between the two divergent lineages arose. The highly pathogenic form of the virus that created the pandemic could have been a mutation of the code for the envelope in the Feliformia lineage. That is the conclusion drawn in the Kaebler paper. 51 A. A simple calculation of the rate of spread of a pox virus (virion) led researchers at Imperial College London to a new insight. Virions communicate with other virions. The researchers observed that the radius of an approximately circular plaque of infected cells grew to 1.45 mm in just 3 days. They measured the distance between adjacent cells to be 0.037 mm to obtain the apparent time for the lytic cycle (from infection to lysis). They compared this time to the actual rate at which new virions are formed: 5 to 6 h. Predict the radius of infection if the infection process involved a sequence of entry, replication, lysis, and infection of an adjacent cell. B. To account for this discrepancy between observed and predicted growth rates, the researchers examined the viral entry process and discovered that the actin protein on the host cell’s surface that provided the viral receptor was modified by attachment. They then found a mutant virus that did not modify the cell surface protein. The dependence of the growth of plaque radius on time for the wild type and mutant are shown in the graph. ******************************************************** ***************************************499*** Analyze these data and compare the infection rates calculated with those predicted in (A). C. Use the results of this experiment to support the claim that responses to information and communication of information affect natural selection. Solution Sample answer: A. A virus jumps a distance of 0.037 mm every 5 to 6 h. The distance traveled in 72 h then is between 0.4 and 0.5 mm in 3 days. The researchers observed a distance between three and four times greater than this. B. The rate for the wild type is approximately 0.35 mm/12 h or 0.029 mm/h. The rate for the mutant is 0.0063 mm/h. The rate per hour calculated in (A) is between 0.006 and 0.007 mm/h, which is comparable to the rate observed for the mutant. C. These data show that the virus is not infecting adjacent cells if that cell has already been infected. The signaling is accomplished by leaving a modified actin, to which a virion subsequently responds. This is communication that has evolved by the positive selection of increased replication rates. 52 Describe how viral replication introduces genetic variation in the viral population. Solution Sample answer: This declarative learning objective can be clarified by looking at the specific aspects of viral replication that create replication errors that are identified as assessable on the AP Biology Exam: Virus replication allows for mutations to occur through usual host pathways. RNA viruses lack replication error-checking mechanisms, and thus have higher rates of mutation. Related viruses can combine/recombine information if they infect the same host cell. ******************************************************** **************************536***************************************** 22 | PROKARYOTES: BACTERIA AND ARCHAEA REVIEW QUESTIONS 1 Which is the best evidence that prokaryotes evolved about 3 billion years ago? A Scientists believe photosynthesis evolved about 3.0 billion years ago. B There is fossil evidence of mammalian forms going back about 4.0 billion years. C Earth and its moon are thought to be about 4.5 billion years old. D There is fossil evidence of microbial mats—large multilayered sheets of prokaryotes— starting about 3.5 billion years ago. Solution 2 The solution is (D). There is fossil evidence of microbial mats starting about 3.5 billion years ago. Microbial mats are thought to represent the earliest forms of life on Earth. Which statement describing the environment of early Earth is false? A The atmosphere contained much less molecular oxygen. B Strong volcanic activity was common. C It was subject to mutagenic radiation from the sun. D There was little to no geologic activity. Solution 3 The solution is (D). There was little to no geologic activity. Due to fossil evidence indicating the presence of microbial mats approximately 3.5 million years ago, scientists believe that hot springs and hydrothermal vents may have been the environments in which life began, thereby suggesting that life on early Earth did have geologic activity. Which type of extremophile grows optimally at temperatures of −15 to 10 °C or lower? A Alkaliphiles B Thermophiles C Hyperthermophiles D Psychrophiles Solution The solution is (D). Psychrophiles are microbes that survive in extremely cold environments. Therefore, the optimal conditions needed for growth include temperatures of −15–10 °C (5–50 °F) or lower. ******************************************************** *****************************************************************535*** 4 What is an example of a relatively moderate environmental condition to which some prokaryotes are adapted and can survive as spores? A Extremely low temperature B Hypersalinity C High doses of radiation D Normal drought Solution 5 The solution is (D). Some soil bacteria are able to form endospores that are heatresistant and drought-resistant, which thereby allows for them to survive until favorable conditions improve or reoccur. More than how much bacteria and archaea cannot be successfully cultured in a laboratory setting? A 9% B 19% C 91% D 99% Solution 6 The solution is (D). Over 99 percent of bacteria and archaea are unculturable in a laboratory setting due to a lack of knowledge about the special nutritional requirements for growth that are needed by these organisms. Additionally, some bacteria cannot be cultured because they are obligate intracellular parasites and cannot be grown outside a host. The most substantial difficulty in culturing prokaryotes in laboratory settings is related to — A the lack of knowledge about their needs for growth B growth requirements that are too difficult to meet C inefficient methods for resuscitation of viable-but-nonculturable (VBNC) organisms D the expense of techniques such as polymerase chain reaction (PCR) Solution 7 The solution is (A). The most substantial difficulty in culturing prokaryotes in laboratory settings is related to the lack of knowledge about their needs for growth. What represents the earliest forms of life on Earth? A Hydrothermal vent B Microbial mat C Meteorite D Stromatolite ******************************************************** **************************536***************************************** Solution 8 The solution is (B). A microbial mat, or large multi-layered sheet of prokaryotes, may represent the earliest forms of life on Earth. Fossil evidence of their presence on Earth dates back approximately 3.5 billion years ago. Which statement best summarizes the conditions of early Earth at the time that life first evolved? A The atmosphere of early Earth was very different from today’s atmosphere, but most other conditions (such as geologic upheaval and volcanic activity) were very much the same. B The atmosphere of early Earth was very much like today’s atmosphere, but many other conditions (such as geologic upheaval and volcanic activity) were very different. C Early Earth had a very different atmosphere, was subject to extreme radiation, and had a lot of geologic upheaval and volcanic activity. D Early Earth had a very different atmosphere and was subject to extreme radiation, but there was very little geologic upheaval or volcanic activity. Solution 9 The solution is (C). Early Earth had a very different atmosphere, consisting of less molecular oxygen than is present in current times. The planet was also subject to strong radiation and volcanic activity due to geologic upheaval. Halophiles prefer conditions in which there is a — A high sugar concentration B salt concentration of at least 0.2 M C pH of 3 or below D high level of radiation Solution 10 The solution is (B). Halophiles have an absolute dependence on salt. As such, a salt concentration of at least 0.2 M is needed for their survival. These salt-tolerant microbes can form salt-tolerant bacterial mats such as those found in the Dead Sea. The presence of a membrane-enclosed nucleus is a characteristic of — A prokaryotic cells B eukaryotic cells C all cells D viruses Solution The solution is (B). Eukaryotic cells possess nuclei, the large membrane-bound organelle present within the cell, which serves as the location of their DNA. This organelle is absent within prokaryotes, which lack membranous intracellular structures such as organelles. ******************************************************** *****************************************************************535*** 11 All prokaryotic and eukaryotic cells have four structures in common: the plasma membrane, the cytoplasm, nucleic acid, and — A the cell wall B ribosomes C the nucleus D organelles Solution 12 The solution is (B). Ribosomes are nonmembranous organelles found in both cell types. The function of these structures is to serve as the site of protein synthesis for the cell. Which statement comparing the prokaryotes Bacteria and Archaea is false? A The cytoplasm of both bacterial and archaean prokaryotic cells has a high concentration of dissolved solutes. B Osmotic pressure in both types of prokaryotic cells is relatively high. C The domains Bacteria and Archaea differ in the use of fatty acids versus phytanal groups in their cell membranes. D The domains Bacteria and Archaea have very similar cell wall structure. Solution 13 The solution is (D). Unlike the cell walls of eubacteria in Domain Bacteria, the composition of cell walls for Domain Archaea bacteria do not include peptidoglycan. There are four different types of Archaean cell walls, one type of which is composed of pseudopeptidoglycan, which is similar to peptidoglycan in morphology but contains different sugars in the polysaccharide chain. Pseudopeptidoglycan is a characteristic of the walls of some — A eukaryotic cell B bacterial prokaryotic cell C archaean prokaryotic cells D bacterial and archaean prokaryotic cells Solution 14 The solution is (C). Pseudopeptidoglycan is a glycoprotein found in Archaean cell walls. It is similar to peptidoglycan in morphology but contains different sugars in the polysaccharide chain. The cell wall, a feature of most prokaryotes, is — A interior to the cell membrane B exterior to the cell membrane C a part of the cell membrane D interior or exterior, depending on the particular cell ******************************************************** **************************536***************************************** Solution 15 The solution is (B). The cell wall is located outside the cell membrane and prevents osmotic lysis. The chemical composition of cell walls varies between the two major bacterial Domains. Which statement summarizes what is known about macronutrient needs of prokaryotes? A Boron is required in small amounts by some prokaryotic organisms. B Manganese is required in small amounts by some prokaryotic organisms. C Iron is required in small amounts by some prokaryotic organisms. D Sulfur is needed in large amounts by prokaryotic organisms. It is part of the structure of some amino acids and is also present in some vitamins and coenzymes. Solution 16 The solution is (D). Macronutrients are nutrients that are required in large amounts. Only a few elements are considered to be macronutrients. Examples include carbon, hydrogen, oxygen, nitrogen, phosphorus, and sulfur. Which statement about the importance of particular nutrients is false? A Carbon is a macronutrient and major element in all macromolecules. B Nitrogen is a macronutrient and necessary component of proteins and nucleic acids. C Hydrogen is a macronutrient and key component of many organic compounds, including water. D Iron is a macronutrient necessary for the function of cytochromes. Solution 17 The solution is (D). This statement is false. Iron is not a macronutrient. While it is needed for the proper function of cytochromes, it is a micronutrient, which means that prokaryotes require it in small amounts. What are prokaryotes that obtain their energy from chemical compounds called? A Phototrophs B Autotrophs C Chemotrophs D Heterotrophs Solution 18 The solution is (C). Chemotrophs, also referred to as chemosynthetic organisms, obtain their energy from chemical compounds. Chemotrophs that can use organic compounds as energy sources are called chemoorganotrophs, while those that can also use inorganic compounds as energy sources are called chemolithotrophs. What uses organic compounds as both an energy source and as a carbon source? A Chemolithotrophs B Photoautotrophs ******************************************************** *****************************************************************535*** C Photoheterotrophs D Chemoorganotrophs Solution 19 The solution is (D). Chemotrophs that can use organic compounds as both an energy source and a carbon source are called chemoorganotrophs. A primary role of many prokaryotes in the carbon cycle is that of — A producers B decomposers C fixers D synthesizers Solution 20 The solution is (B). The most important contributor of carbon dioxide to the atmosphere is microbial decomposition of dead material, such as dead animals, plants, and humus, that undergo respiration. As decomposers, bacteria, along with fungi, are responsible for carrying out the decomposition process of the aforementioned materials. Ammonification is the process by which — A ammonia is released during the decomposition of nitrogen-containing organic compounds B ammonium is converted into nitrite and nitrate in soils C nitrate from soil is transformed to gaseous nitrogen compounds D gaseous nitrogen is fixed to yield ammonia Solution 21 The solution is (A). Ammonification is the process by which ammonia is released during the decomposition of nitrogen-containing organic compounds. Ammonia released to the atmosphere, however, represents only 15 percent of the total nitrogen released; the rest is as N2 and N2O. Which option is a macronutrient needed by prokaryotes? A Phosphorus B Iron C Chromium D Boron Solution The solution is (A). Phosphorus is needed for the synthesis of nucleotides and consequently DNA and RNA, as well as for post-translational modification of proteins. ******************************************************** **************************536***************************************** 22 A disease that is constantly present in a population is called — A pandemic B endemic C emerging D reemerging Solution 23 The solution is (B). An endemic disease is a disease that is constantly present, usually at low incidence, in a population. Which set of terms names diseases caused by bacteria? A Diphtheria, bubonic plague, yellow fever B Yellow fever, dengue fever, bubonic plague C Bubonic plague, diphtheria, cholera D Cholera, diphtheria, dengue fever Solution 24 The solution is (C). All three terms refer to diseases that are caused by bacteria. Specifically, diphtheria is caused by the bacterium Corynebacterium diptheriae, bubonic plague is caused by the bacterium Yersina pestis, and cholera is caused by the bacterium Vibrio cholerae. Which health issue is caused by biofilm colonization? A Dental plaque B Dry scalp C Skin rash D Prosthetic discomfort Solution 25 The solution is (A). A biofilm is a microbial community held together in a gummytextured matrix that consists primarily of polysaccharides secreted by the organisms, together with some proteins and nucleic acids. Biofilms grow attached to surfaces, such as teeth, where they can produce dental plague and eventually lead to tooth decay, if left untreated. Which statement about the loci of biofilm-related disease is false? A Biofilms are related to foodborne illnesses because they colonize food surfaces and food-processing equipment. B In healthcare environments, biofilms grow on ventilators, shunts, and other medical equipment. C Biofilms tend to colonize medical devices such as prostheses, contact lenses, and catheters. D Biofilms do not form in open wounds, burned tissue, or internal medical devices such as pacemakers. ******************************************************** *****************************************************************535*** Solution 26 The solution is (D). Biofilms do not form in open wounds, burned tissue, or internal medical devices such as pacemakers. Which statement best describes the crisis related to antibiotics? A It is becoming too expensive to manufacture effective antibiotics. B It takes too much time to develop effective antibiotics; infections spread before treatment is available. C Bacteria are increasingly resistant to antibiotics used to treat and eradicate infections. D People are increasingly allergic to antibiotics commonly used in treatment. Solution 27 The solution is (C). As result of mutations in bacteria some become resistant to certain antibiotic, and as such will be able to reproduce regardless of the presence of that antibiotic. Which statement about the cause of resistant bacteria is false? A The excessive use of antibiotics has resulted in the natural selection of resistant forms of bacteria. B Antibiotics are used by patients with colds or the flu, the treatment for which antibiotics are useless. C There is excessive use of antibiotics in livestock and in animal feed. D Antibiotics are used by patients of different ages and the fact that their ages differ increases resistance. Solution 28 The solution is (D). One of the main causes of resistant bacteria is the abuse of antibiotics. The imprudent and excessive use of antibiotics has resulted in the natural selection of resistant forms of bacteria. The antibiotic kills most of the infecting bacteria, and therefore only the resistant forms remain. These resistant forms reproduce, resulting in an increase in the proportion of resistant forms over non-resistant ones. Another major misuse of antibiotics is in patients with colds or the flu, for which antibiotics are useless. Another problem is the excessive use of antibiotics in livestock. The routine use of antibiotics in animal feed promotes bacterial resistance as well. Which statement about diseases is false? A An epidemic is a disease that occurs in a high number of individuals in a population at a time. B A pandemic is a widespread, usually worldwide, epidemic. C An endemic disease is a disease that is constantly present, usually at high incidence, in a population. D An emerging disease is a disease that has appeared in a population for the first time. ******************************************************** **************************536***************************************** Solution 29 The solution is (C). This statement is false. An endemic disease is a disease that is constantly present, usually at low incidence, not high incidence, in a population. Which statement best explains which organisms need nitrogen fixation and why? A Prokaryotes cannot use gaseous nitrogen to synthesize macromolecules, so it must be converted into ammonia. B Prokaryotes cannot use ammonia to synthesize macromolecules, so it must be converted into gaseous nitrogen. C Eukaryotes cannot use ammonia to synthesize macromolecules, so it must be converted into gaseous nitrogen. D Eukaryotes cannot use gaseous nitrogen to synthesize macromolecules, so it must be converted into ammonia. Solution 30 The solution is (D). Eukaryotes cannot use gaseous nitrogen to synthesize macromolecules, so it must be converted into ammonia. The largest pool of nitrogen available in the terrestrial ecosystem is gaseous nitrogen from the air, but this nitrogen is not usable by eukaryotes, such as plants. Gaseous nitrogen is transformed, or fixed into more readily available forms such as ammonia through the process of nitrogen fixation. Ammonia can be used by plants or converted into other forms. Which statement about nitrogen fixation is false? A It can be accomplished abiotically, as a result of lightning. B It can be accomplished abiotically, as a result of industrial processes. C It can be accomplished biologically, by algae. D It can be accomplished biologically, by cyanobacteria. Solution 31 The solution is (C). It can be accomplished biologically, by algae. Biological nitrogen fixation (BNF) is carried out exclusively by prokaryotes such as soil bacteria, cyanobacteria, and Frankia spp., filamentous bacteria interacting with actinorhizal plants such as alder, bayberry, and sweet fern. As eukaryotes, algae are unable to carry out nitrogen fixation. Which three foods use prokaryotes in their processing? A Cheese, yogurt, and milk B Cheese, yogurt, and bread C Wine, bread, and butter D Milk, wine, and beer Solution The solution is (B). Food products such as cheese, bread, and yogurt are produced via the use of bacteria and other microbes, such as yeast, a fungus. The use of such microbes for the use of food production is referred to as biotechnology. ******************************************************** *****************************************************************535*** 32 What was the initial benefit for humans in processing foods with prokaryotes? A The foods taste better. B Nutrients are preserved. C The food is less stable. D Nutrients were safer. Solution 33 The solution is (B). Scientists believe that the production of certain food products such as cheese began approximately 4,000–7,000 years ago. Fermentation, the anaerobic process whereby cheese is produced by microbes, results in the preservation of nutrients: Milk will spoil relatively quickly, but when processed as cheese, it is more stable. Which option best defines bioremediation? A The use of microbial metabolism to clean up oil spills B The use of microbial metabolism to ferment food C The use of microbial metabolism to remove pollutants D The use of microbial metabolism to fix nitrogen Solution 34 The solution is (C). Microbial bioremediation is the use of prokaryotes, via their metabolic processes, to remove pollutants from the environment. Bioremediation has been used to remove contaminants such as those arising from agricultural chemicals (i.e., pesticides, fertilizers). Which statement about bioremediation is false? A It includes removing agricultural chemicals. B It includes removing industrial by-products. C It includes cleaning up oil spills. D It includes cleaning up ammonia in soil. Solution 35 The solution is (D). Bioremediation has been used to remove agricultural chemicals—pesticides, fertilizers—that leach from soil into groundwater and the subsurface. Certain toxic metal