Test Bank for Becker’s World of the Cell 9th Edition by Jeff Hardin

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Test Bank for Becker’s World of the Cell 9th
Edition by Jeff Hardin
Becker’s The World of the Cell, 9e (Hardin/Bertoni/Kleinsmith)
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Chapter 1 A Preview of Cell Biology
1.1 Multiple-Choice Questions
1) Robert Hooke coined the term cell when studying thin slices of cork. These ________
were the first cells observed because ________.
1. A) dead plant cells; the thick cells walls did not require high resolution or magnification to
view
2. B) dead animal cells; they were immobile and did not need to be fixed before viewing
3. C) compartments; they were actually the result of multiple cells that had merged and
died to form large compartments that were easy to view
4. D) immune cells; they produce antibodies that embed in the cell membrane to make it
visible
5. E) “little rooms”; they were 100 nm in diameter, much larger than most plant cells
Answer: A
Chapter Section: 1.1
Bloom’s Taxonomy: Comprehension
Learning Outcome: 1.1
Global LO: G1
2) The Latin phrase omnis cellula e cellula refers to a cellular principle. Which of the
following statements is the best interpretation of this phrase?
1.
2.
3.
4.
5.
A) Tissues are composed of similar cells.
B) Cells generally are found in clusters.
C) All cells arise only from preexisting cells.
D) Organs are composed of tissues and cells.
E) The cell is the basic unit of structure.
Answer: C
Chapter Section: 1.1
Bloom’s Taxonomy: Comprehension
Learning Outcome: 1.1
Global LO: G7
3) ________ improved the original light microscope in the late 1600s, allowing the
visualization of ________.
1. A) Theodor Schwann; the internal structures of cells, such as ribosomes, nuclei, and
golgi bodies
2. B) Robert Hooke; bacteria and viruses
3. C) Antonie van Leeuwenhoek; sperm cells, bacteria, algae, and other protists
4. D) Robert Brown; cell structures using fluorescent antibodies
5. E) Rudolf Virchow; collagen and muscle cells
Answer: C
Chapter Section: 1.1
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.2
Global LO: G1
4) Which organelle stores most of the DNA in plant and animal cells?
1.
2.
3.
4.
5.
A) Golgi complex
B) mitochondrion
C) chloroplast
D) nucleus
E) lysosome
Answer: D
Chapter Section: 1.2
Bloom’s Taxonomy: Comprehension
Learning Outcome: 1.2
Global LO: G1
5) Which of the following statements is false?
1.
2.
3.
4.
5.
A) All organisms consist of one or more cells.
B) All cells arise from preexisting cells.
C) The cell is the basic unit of structure for all organisms.
D) All cells have a membrane-bound nucleus.
E) Cells come in a wide variety of sizes and shapes.
Answer: D
Chapter Section: 1.2
Bloom’s Taxonomy: Analysis
Learning Outcome: 1.1
Global LO: G7
6) Which of the following is true of a nanometer?
1. A) A nanometer is about the size of a common bacterial cell.
2. B) A nanometer is one millionth of a meter.
3. C) A nanometer is equivalent to 10 Angstroms (Å).
4. D) The nanometer is the most common measurement used in measuring whole cells.
5. E) None of the above.
Answer: C
Chapter Section: 1.2
Bloom’s Taxonomy: Comprehension
Learning Outcome: 1.2
Global LO: G4
7) Which of the following is closest to a micrometer in size?
1.
2.
3.
4.
5.
A) the width of a strand of DNA
B) the length of a plant cell
C) the length of a chicken egg
D) a typical prokaryotic cell
E) the size of a ribosome
Answer: D
Chapter Section: 1.2
Bloom’s Taxonomy: Application
Learning Outcome: 1.2
Global LO: G4
8) Cell biology emerged from which of the following fields of biology?
1.
2.
3.
4.
5.
A) biochemistry
B) cytology
C) genetics
D) biochemistry, cytology, and genetics
E) cytology and biochemistry
Answer: D
Chapter Section: 1.2
Bloom’s Taxonomy: Comprehension
Learning Outcome: 1.1
Global LO: G1
9) Which of the following is smallest?
1.
2.
3.
4.
5.
A) ribosome
B) virus
C) protein
D) mitochondrion
E) prokaryote
Answer: C
Chapter Section: 1.2
Bloom’s Taxonomy: Application
Learning Outcome: 1.2
Global LO: G4
10) Early microscopes did not allow clear visualization of cells because they were
limited by
1.
2.
3.
4.
5.
A) magnification.
B) number of kernels.
C) resolution.
D) refraction.
E) both magnification and resolution.
Answer: E
Chapter Section: 1.1
Bloom’s Taxonomy: Comprehension
Learning Outcome: 1.2
Global LO: G4
11) You are working on a project that involves the direct observation of DNA molecules.
The microscope that would give you the best information at this time would be the
1.
2.
3.
4.
5.
A) light microscope.
B) phase-contrast microscope.
C) transmission electron microscope.
D) digital video microscope.
E) fluorescent microscope.
Answer: C
Chapter Section: 1.2
Bloom’s Taxonomy: Analysis
Learning Outcome: 1.2
Global LO: G4
12) The limit of resolution can best be defined as
1.
2.
3.
4.
5.
A) the distance that an object must be moved to be distinguished from its background.
B) the inverse of the wavelength of light; it is greatest for black light.
C) the distance that two objects must be apart to be distinguished as separate objects.
D) the solvent that must be available to remix a solution.
E) the magnification power of a microscope.
Answer: C
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.2
Global LO: G4
13) How does brightfield microscopy allow images to be visualized?
1. A) Specimens are illuminated with white light.
2. B) Electrons strike the specimen being examined.
3. C) Specimens are fixed and have bright fluorescent molecules attached to them.
4. D) Specimens are illuminated with blue light to visualize internal features of cells smaller
than 100 nm.
5. E) Specimens are viewed under phased light to improve magnification.
Answer: A
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.2
Global LO: G7
14) Which of the following is an application of immunofluorescence microscopy?
1.
2.
3.
4.
5.
A) Visualization of the natural fluorescence of a specimen under UV light.
B) Identification of specific components of the immune system.
C) Identifying which organelle or cellular compartment contains a particular protein.
D) Visualization of the surface structures of a specimen.
E) Construction of three-dimensional images of structures smaller than 10 nm.
Answer: C
Chapter Section: 1.2
Bloom’s Taxonomy: Application
Learning Outcome: 1.2
Global LO: G1
15) Which type of microscopy enhances and amplifies slight changes in the phase of
transmitted light?
1.
2.
3.
4.
5.
A) differential interference contrast microscopy
B) digital video microscopy
C) fluorescence microscopy
D) phase-contrast microscopy
E) both differential interference contrast microscopy and phase-contrast microscopy
Answer: E
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.2
Global LO: G1
16) Which type of microscopy has the greatest resolving power?
1.
2.
3.
4.
5.
A) electron microscopy
B) phase-contrast microscopy
C) fluorescence microscopy
D) digital video microscopy
E) confocal scanning microscopy
Answer: A
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.2
Global LO: G4
17) Which of the following can only be viewed by electron microscopy?
1.
2.
3.
4.
5.
A) frog eggs
B) DNA
C) nuclei
D) mitochondria
E) prokaryotes
Answer: B
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.2
Global LO: G4
18) Which of the following types of light microscopy improves the resolution of thick
specimens by illuminating one plane of the specimen at a time?
1.
2.
3.
4.
5.
A) fluorescence microscopy
B) phase-contrast microscopy
C) confocal microscopy
D) differential interference contrast microscopy
E) brightfield microscopy
Answer: C
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.2
Global LO: G4
19) A scientist is examining motile protist. He wishes to determine their direction of
movement. Which of the following microscopic techniques is least likely to be used to
view these cells?
1.
2.
3.
4.
5.
A) light microscopy
B) electron microscopy
C) differential interference contrast microscopy
D) fluorescence microscopy
E) phase-contrast microscopy
Answer: B
Chapter Section: 1.2
Bloom’s Taxonomy: Application
Learning Outcome: 1.2
Global LO: G2
20) Scanning electron microscopy (SEM) is especially suited to
1.
2.
3.
4.
5.
A) observing living specimens.
B) examining internal cellular structure.
C) creating a sense of depth.
D) both observing living specimens and creating a sense of depth.
E) simultaneously observing living specimens, examining internal cellular structure, and
creating a sense of depth.
Answer: C
Chapter Section: 1.2
Bloom’s Taxonomy: Application
Learning Outcome: 1.2
Global LO: G2
21) Melvin Calvin and his colleagues used which of the following to deduce the steps in
the Calvin cycle?
1.
2.
3.
4.
5.
A) negative staining
B) Drosophila melanogaster
C) electron microscopy
D) ultracentrifugation
E) radioisotopes
Answer: E
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.2
Global LO: G1
22) A microtome is used to
1. A) view microscopic organisms.
2. B) slice thin sections of specimens.
3. C) focus short wavelengths of light.
4. D) manipulate tiny objects.
5. E) dissect cellular organelles.
Answer: B
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.2
Global LO: G4
23) The classic work of Friedrich Wöhler (1828) that united the fields of biology and
chemistry was based on the
1.
2.
3.
4.
5.
A) discovery of yeast ferments.
B) production of urea in the laboratory.
C) discovery of ATP.
D) identification of nucleotide bases.
E) analysis of gene segregation.
Answer: B
Chapter Section: 1.2
Bloom’s Taxonomy: Comprehension
Learning Outcome: 1.2
Global LO: G1
24) You wish to obtain a purified sample of mitochondria from lysed cells. The best way
to obtain this sample would be
1.
2.
3.
4.
5.
A) centrifugation.
B) chromatography.
C) polyacrylamide gel electrophoresis.
D) agarose gel electrophoresis.
E) both centrifugation and polyacrylamide gel electrophoresis.
Answer: A
Chapter Section: 1.2
Bloom’s Taxonomy: Evaluation
Learning Outcome: 1.2
Global LO: G2, G7
25) 1 mm = ________ nm
1.
2.
3.
4.
5.
A) 1,000,000
B) 1000
C) 10
D) 1/1000
E) 1/1,000,000
Answer: A
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.2
Global LO: G4
26) The outcome of the joining of cytology and biochemistry yielded a better
understanding of the cell by
1.
2.
3.
4.
5.
A) identification of cellular structures.
B) identification of cellular biochemical pathways.
C) creating bioinformatics.
D) identification of cellular structures and biochemical pathways.
E) identification of biochemical pathways and creating bioinformatics.
Answer: D
Chapter Section: 1.2
Bloom’s Taxonomy: Synthesis
Learning Outcome: 1.1
Global LO: G1
27) Wöhler revolutionized biology through his demonstration that biological molecules
are governed by the ordinary laws of physics and chemistry. He demonstrated this
principle by
A) synthesizing urea in the laboratory from ammonium cyanate.
B) developing techniques for isolating, purifying, and analyzing subcomponents of cells.
C) defining the laws of heredity.
D) discovering active agents in cell extracts that were specific biological catalysts that
have since come to be called enzymes.
5. E) inventing mass spectrometry which is commonly used to determine the size and
composition of individual proteins.
1.
2.
3.
4.
Answer: A
Chapter Section: 1.2
Bloom’s Taxonomy: Comprehension
Learning Outcome: 1.1
Global LO: G1
28) Gregor Mendel was most influential in which field of biology?
1.
2.
3.
4.
5.
A) genetics
B) chromatography
C) biochemistry
D) prokaryotic transformation
E) cytology
Answer: A
Chapter Section: 1.2
Bloom’s Taxonomy: Comprehension
Learning Outcome: 1.1
Global LO: G1
29) The scientific work that established DNA, rather than protein, as the molecule of
heredity was performed prior to
1.
2.
3.
4.
5.
A) Mendel’s work on heredity.
B) the elucidation of the double helix structure of DNA.
C) Antonie van Leeuwenhoek’s observation of internal cell structures.
D) the description of the enzymatic steps of glycolysis.
E) the formation of the chromosome theory of heredity.
Answer: B
Chapter Section: 1.2
Bloom’s Taxonomy: Synthesis
Learning Outcome: 1.1
Global LO: G1
30) Jacques Monod and François Jacob deduced the mechanism responsible for the
regulation of prokaryotic gene expression. They are, therefore, responsible for
launching the era of
1.
2.
3.
4.
5.
A) the scientific method.
B) molecular genetics.
C) biochemistry.
D) light microscopy.
E) radioisotopes.
Answer: B
Chapter Section: 1.2
Bloom’s Taxonomy: Synthesis
Learning Outcome: 1.1
Global LO: G1
31) Which of the following biochemical techniques uses an electrical field to separate
macromolecules based on their mobility through a semisolid gel?
1.
2.
3.
4.
5.
A) light microscopy.
B) ultracentrifugation.
C) chromatography.
D) electrophoresis.
E) mass spectrometry.
Answer: D
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.2
Global LO: G1
32) To which of the following do Mendel’s observations relate?
1.
2.
3.
4.
5.
A) thermodynamics
B) gravity
C) ideal gas laws
D) heredity
E) diffusion
Answer: D
Chapter Section: 1.2
Bloom’s Taxonomy: Synthesis
Learning Outcome: 1.1
Global LO: G1
33) The steps of the scientific method, in the correct order, are
1. A) design experiments, draw conclusions, collect data, interpret results, make
observations, and test the hypothesis.
2. B) make observations, formulate the hypothesis, design experiments, collect data,
interpret results, and draw conclusions.
3. C) collect data, interpret results, test the hypothesis, design experiments, make
observations, and draw conclusions.
4. D) collect data, interpret results, test the hypothesis, make observations, and design
experiments.
5. E) none of the above.
Answer: B
Chapter Section: 1.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.3
Global LO: G1
34) Scientists use various terms to describe conclusions reached through the scientific
method. Which of the following terms conveys the least degree of certainty?
1.
2.
3.
4.
5.
A) theory
B) hypothesis
C) law
D) both hypothesis and theory
E) both theory and law
Answer: B
Chapter Section: 1.3
Bloom’s Taxonomy: Comprehension
Learning Outcome: 1.3
Global LO: G1
35) Once a scientific theory becomes a law, it
1.
2.
3.
4.
5.
A) cannot be changed.
B) cannot be challenged.
C) becomes static.
D) is subject to modification.
E) is irrefutable.
Answer: D
Chapter Section: 1.3
Bloom’s Taxonomy: Comprehension
Learning Outcome: 1.3
Global LO: G1
36) You are studying the response of macrophages infected with the intracellular
bacterium Brucella, specifically by examining which gene products are being expressed.
You would be studying the macrophage ________ to obtain this information.
1.
2.
3.
4.
5.
A) proteome
B) genome
C) transciptome
D) amplicon
E) metabolome
Answer: C
Chapter Section: 1.2
Bloom’s Taxonomy: Application
Learning Outcome: 1.2
Global LO: G1
37) Which of the following is an important characteristic for a model organism?
1.
2.
3.
4.
5.
A) marginally characterized
B) difficult to manipulate in the laboratory
C) prone to random changes that alter primary characteristics
D) widely studied
E) all of the above
Answer: D
Chapter Section: 1.3
Bloom’s Taxonomy: Comprehension
Learning Outcome: 1.3
Global LO: G1
38) All of the following are model organisms, except
1.
2.
3.
4.
5.
A) Saccharomyces cerevisiae.
B) Drosophila melanogaster.
C) Caenorhabditis elegans.
D) Arabidopsis thialana.
E) Homo sapiens.
Answer: E
Chapter Section: 1.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.3
Global LO: G1
39) In studying osteoporosis in humans, you wish to test a newly designed treatment for
efficacy. Your best choice for a model organism would be
1.
2.
3.
4.
5.
A) Escherichia coli.
B) Mus musculus.
C) Caenorhabditis elegans.
D) Arabidopsis thaliana.
E) Pisum sativum.
Answer: B
Chapter Section: 1.3
Bloom’s Taxonomy: Application
Learning Outcome: 1.3
Global LO: G1
40) Which of the following is mismatched?
1. A) Escherichia coli – genetics
2. B) Drosophila melanogaster – embryogenesis
3. C) Mus musculus – immunology
4. D) Caenorhabditis elegans – photosynthesis
5. E) Arabidopsis thaliana – plant gene function
Answer: D
Chapter Section: 1.3
Bloom’s Taxonomy: Analysis
Learning Outcome: 1.3
Global LO: G1
1.2 Matching Questions
Match each scientist or group of scientists on the left with the appropriate phrase to the
right.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
A) fruit fly
B) DNA double helix
C) cell theory
D) transfer RNA
E) Calvin cycle
F) “one gene—one enzyme”
G) transformation
H) translation
I) chromosome theory of heredity
J) embryonic bacteria
K) “ferments” of yeast
L) oral prokaryotes
M) urea
N) hereditary factors
O) pollen grain
P) dog saliva
Q) transcription
1) Gregor Mendel
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.1
Global LO: G1
2) Walter Sutton
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.1
Global LO: G1
3) Matthias Schleiden
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.1
Global LO: G1
4) Oswald Avery, Colin MacLeod, and Maclyn McCarty
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.1
Global LO: G1
5) George Beadle and Edward Tatum
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.1
Global LO: G1
6) James Watson and Francis Crick
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.1
Global LO: G1
7) Thomas Hunt Morgan
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.1
Global LO: G1
8) Friedrich Wöhler
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.1
Global LO: G1
9) Louis Pasteur
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.1
Global LO: G1
Answers: 1) N 2) I 3) C 4) G 5) F 6) B 7) A 8) M 9) K
Match the type of microscopy with the appropriate characteristic.
1.
2.
3.
4.
5.
6.
A) amplifies variations in density
B) light passes directly through specimen
C) detects electrons deflected from the surface of the specimen
D) shows specific molecules
E) uses a laser to view a single plane of a specimen
F) detects electrons passing through a specimen
10) brightfield
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.2
Global LO: G4
11) fluorescence
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.2
Global LO: G4
12) phase-contrast
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.2
Global LO: G4
13) confocal
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.2
Global LO: G4
14) transmission electron microscopy
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.2
Global LO: G4
15) scanning electron microscopy
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.2
Global LO: G4
Answers: 10) B 11) D 12) A 13) E 14) F 15) C
1.3 Short Answer Questions
1) To be useful to scientists, a hypothesis must be ________; in other words, the
hypothesis must be able to be confirmed or discredited.
Answer: testable
Chapter Section: 1.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.3
Global LO: G1
2) A scientific ________ must be so thoroughly confirmed that virtually no doubt
remains about its accuracy.
Answer: law
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.3
Global LO: G1
3) Glycolysis is also called the ________ pathway after the scientists who did most of
the work to define it.
Answer: Embden-Meyerhof
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.1
Global LO: G1
4) ________ synthesized urea in the laboratory from inorganic starting materials. Much
of what is now called ________ dates from this discovery.
Answer: Friedrich Wöhler; biochemistry
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.1
Global LO: G1
5) Melvin Calvin used ________, a specific ________, to deduce the Calvin cycle of
photosynthesis.
Answer: 14C; radioisotope
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.1
Global LO: G1
6) A(n) ________ is an instrument used to separate subcellular structures and
macromolecules on the basis of size, shape, and density. ________ developed this
instrument in Sweden during the period 1925–1930.
Answer: ultracentrifuge; Theodor Svedberg
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.2
Global LO: G1
7) Around 1914, ________ determined that DNA was an important component in
________ by using a staining technique that is still in use today.
Answer: Robert Feulgen; chromosomes
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.1
Global LO: G1
8) Because of the low penetration power of electrons, samples for transmission electron
microscopy must be extremely thin. A(n) ________ is able to cut sections as thin as 20
nm.
Answer: ultramicrotome
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.2
Global LO: G4
9) In 1880, Walther Flemming identified ________, threadlike bodies seen in dividing
cells.
Answer: chromosomes
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.1
Global LO: G1
10) The ________ was developed in the late 1920s by Theodore Svedberg. He
originally used it to determine the sedimentation rate of proteins.
Answer: ultracentrifuge
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.2
Global LO: G1
11) ________ is a biochemical technique that allows one to separate biological
molecules based on size, shape, and/or affinity for specific molecules or functional
groups.
Answer: Chromatography
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.2
Global LO: G1
12) The total protein content of the cell is called the ________.
Answer: proteome
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.2
Global LO: G1
13) ________ is the ability to distinguish two objects that are close together as
separate. In any microscope, this ability is determined by ________.
Answer: Resolution; wavelength
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.2
Global LO: G4
1.4 Inquiry
Scientific discoveries have had great impact in human history. The people who make
these discoveries and the circumstances that surround these discoveries are very
important to our understanding of science. Can you identify the individuals as they might
have described themselves?
1.
2.
3.
4.
5.
6.
7.
A) Antonie van Leeuwenhoek
B) Melvin Calvin
C) Alfred Hershey and Martha Chase
D) Theodor Svedberg
E) Friedrich Wöhler
F) Robert Hooke
G) James Watson and Francis Crick
1) I am a seventeenth-century shopkeeper from Holland. My hobby involves handpolishing glass to make lenses, some of which can magnify almost 300-fold. I was the
first to observe living cells and am known as the “Father of Microbiology.”
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.1
Global LO: G1, G7
2) I was the Curator of Instruments for the Royal Society of London in 1665. I developed
a microscope that could magnify around 30-fold. I examined plant material and
observed many small chambers that I called cells.
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.1
Global LO: G1, G7
3) At the University of California, Berkeley, I worked with radioisotopes. In the late
1940s and early 1950s, I used 14C to identify the most common pathway for
photosynthetic carbon metabolism.
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.1
Global LO: G1, G7
4) We worked out the double helix model of DNA structure in 1953. We later received
the Nobel Prize for this work.
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.1
Global LO: G1, G7
5) I am a nineteenth-century German chemist. By synthesizing an organic molecule
from inorganic components, I dispelled the idea that biological processes were exempt
from the laws of chemistry.
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.1
Global LO: G1, G7
6) My colleague and I worked with bacterial viruses. We were able to demonstrate that
DNA–not protein–was the genetic material of the cell.
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.1
Global LO: G1, G7
7) I am a Swedish scientist. I developed the ultracentrifuge to determine sedimentation
rates of proteins. The ultracentrifuge was later used to isolate subcellular fractions.
Chapter Section: 1.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 1.1
Global LO: G1, G7
Answers: 1) A 2) F 3) B 4) G 5) E 6) C 7) D
8) The following paragraph describes the activities of hypothetical scientists. After
reading this paragraph, list the steps of the scientific method, and list the activities that
correspond to the steps of the scientific method.
A rancher noticed that several grazing animals had become sick after grazing in a new
area. The rancher asked a team of scientists to analyze this problem. They visited the
area and found that the food available to the animals was similar to the food they had
been eating. The water supply in the area was adequate but limited to a single spring.
Some of the scientists felt that the water might be contaminated with a pathogen.
Therefore, they collected water samples from the spring in the new area and compared
them with water samples taken from previous grazing sites. The scientists noticed that
water from the new area was cloudier than water obtained from other areas. Culturing
this water revealed that a pathogenic strain of bacteria was present. This bacterial strain
was found to be identical to a strain obtained from sick animals. This strain was not
present in healthy animals. They concluded that a contaminated water supply in the
new area was responsible for the problem and instructed the rancher to avoid the water
supply. The disease was not found in the rancher’s livestock again.
Answer: (Answers may vary.)
Observation. The rancher and the scientists made initial observations regarding the
food and water that the livestock consumed.
Hypothesis. The water supply was contaminated with a pathogen.
Experimentation. Water was collected, examined, and cultured.
Collect data. The turbidity of the water was examined. The cultures were positive for a
pathogenic strain of bacterium.
Interpret results. The data was compared to other water samples. The cultures were
compared to those obtained from livestock.
Draw conclusion. The water was contaminated and responsible for the outbreak.
Chapter Section: 1.3
Bloom’s Taxonomy: Evaluation
Learning Outcome: 1.3
Global LO: G1
9) A number of different types of microscopy exist. Each type of microscopy has
advantages and disadvantages. Can you identify the microscope that would be most
advantageous for the situations below?
1. A cell biologist wishes to visualize the ribosomes of a cell.
2. A microbiologist wishes to examine the motility of a bacterium.
3. An immunologist wishes to determine if a lymphocyte possesses a certain surface
protein.
4. A virologist is trying to determine the three-dimensional shape of a virus.
5. A pathologist is trying to examine the cytoplasm of a cell for changes that result from
viral infection.
Answer: (Answers may vary.)
1.
2.
3.
4.
5.
Electron microscopy, preferably transmission electron microscopy, should be used.
Phase contrast or differential-interference-contrast would be most helpful.
Fluorescence microscopy is often used.
Scanning electron microscopy should be used.
Transmission electron microscopy will enable the pathologist to visualize the interior.
Chapter Section: 1.2
Bloom’s Taxonomy: Evaluation
Learning Outcome: 1.2
Global LO: G4
10) You have identified a new molecule associated with the immune system that
drastically reduces cell division by tumor cells in vivo. Develop a hypothesis and design
an experiment to test your hypothesis using a model organism. Include an explanation
as to why it is the best model for your experiment.
Answer: Answers will vary; however, the hypothesis would indicate the utility of the
cytokine for lymphoma treatment. The obvious model organism would be the mouse
model. It shares a great many similarities to humans at the cellular, anatomical, and
physiological levels. It is well characterized, and the genome has been sequenced.
Further, there is a mouse model of lymphoma currently available. Mice are easy to care
for and require a relatively small amount of space to maintain.
Chapter Section: 1.3
Bloom’s Taxonomy: Synthesis
Learning Outcome: 1.3
Global LO: G1
11) You have been given a sample of Mimivirus, which has the largest capsid diameter
of all currently known viruses (600 nm) and has the form of a 20-sided polyhedron (an
icosahedron). Based on your knowledge of microscopes, what would you be able to
see/determine about mimiviral structure using each of the following microscopes?
1. simple compound (light) microscope
2. fluorescent microscope using fluorescently labeled antibodies to a novel capsid protein
3. scanning electron microscope
Answer:
1. Light microscope: will be able to see basic viral shape, especially if particles are stained
2. Fluorescent microscope: should illuminate the outside of the viral particles
3. Scanning electron microscope: would allow imaging of the surface structure of the virus
Chapter Section: 1.2
Bloom’s Taxonomy: Analysis
Learning Outcome: 1.2
Global LO: G4
Becker’s The World of the Cell, 9e (Hardin/Bertoni/Kleinsmith)
Chapter 3 The Macromolecules of the Cell
3.1 Multiple-Choice Questions
1) What are the three general types of amino acids?
1.
2.
3.
4.
5.
A) α helices, β sheets, and looped segments
B) covalent, noncovalent, and van der Waals forces
C) positive, negative, and noncharged
D) hydrophobic, polar (noncharged), polar (charged)
E) acidic, basic, and neutral
Answer: D
Chapter Section: 3.1
Bloom’s Taxonomy: Knowledge
Learning Outcome: 3.1
Global LO: G7
2) You are researching a cytoplasmic protein associated with a nerve disorder. The
native form of the enzyme appears to be globular protein; however, when a sample of
the purified protein is treated with a chemical that reduces disulfide bonds, the
enzymatic activity decreases dramatically and multiple globular proteins can be
detected in the sample. What does this tell you about the protein?
1. A) The primary structure of the protein contains multiple cysteine residues that are
hydrolyzed by the chemical reductant.
2. B) The protein is most likely composed of multiple polypeptide chains that are held
together by disulfide bonds.
3. C) The protein is most likely composed of α helices that are held together by disulfide
bonds.
4. D) The protein is most likely composed of β sheets that are held together by disulfide
bonds.
5. E) The primary and secondary structure of the protein depends on disulfide bonds.
Answer: B
Chapter Section: 3.1
Bloom’s Taxonomy: Evaluation
Learning Outcome: 3.1
Global LO: G2, G5, G7
3) A peptide bond
1. A) is a covalent bond between the carboxyl carbon of one amino acid and the amino
nitrogen of a second amino acid.
2. B) is a covalent bond between the functional R groups of adjacent amino acids.
3. C) is a covalent bond between the NH group of one polypeptide and the CO group of an
adjacent polypeptide that holds together multimeric proteins.
4. D) is a noncovalent bond that dictates the tertiary structure of a protein.
5. E) is a covalent bond between adjacent glucose molecules in a peptide.
Answer: A
Chapter Section: 3.1
Bloom’s Taxonomy: Knowledge
Learning Outcome: 3.1
Global LO: G7
4) Disulfide bonds are often found to stabilize which of the following levels of protein
structure?
1.
2.
3.
4.
5.
A) primary
B) secondary
C) tertiary
D) primary, secondary, and tertiary
E) None of these structures involve disulfide bonds.
Answer: C
Chapter Section: 3.1
Bloom’s Taxonomy: Comprehension
Learning Outcome: 3.1
Global LO: G7
5) The primary structure of a protein
1.
2.
3.
4.
5.
A) is important for determining the secondary and tertiary structure of a protein.
B) is simply the order of amino acids from one end of the protein to another.
C) is important both genetically and structurally.
D) is the linear sequence of amino acids that are linked together by peptide bonds.
E) All of these statements are true.
Answer: E
Chapter Section: 3.1
Bloom’s Taxonomy: Comprehension
Learning Outcome: 3.1
Global LO: G2, G7
6) Proline is referred to as the “helix breaker” because
1.
2.
3.
4.
5.
A) its only found in the L form, which is incompatible with helical protein structure.
B) it lacks the hydrogen atom needed for hydrogen bonding.
C) it lacks a charged functional groups for ionic bonding.
D) it is hydrophobic.
E) it has a polar functional group.
Answer: B
Chapter Section: 3.1
Bloom’s Taxonomy: Comprehension
Learning Outcome: 3.1
Global LO: G2, G7
7) Which of the following accurately describes the structure of fibrous proteins?
1. A) Fibrous proteins usually contain a number of different domains with different structural
motifs.
2. B) Fibrous proteins are composed of an equal mixture of α helices and β sheets with
interconnecting looped segments.
3. C) Fibrous proteins have an extensive tertiary and quaternary structure that affects the
strength and elasticity of each fiber.
4. D) Fibrous proteins have a simple primary structure and very little secondary structure,
resulting in long, thin fibers.
5. E) Fibrous proteins are usually composed of either α helices or β sheets throughout the
molecule, giving them a highly ordered, repetitive structure.
Answer: E
Chapter Section: 3.1
Bloom’s Taxonomy: Knowledge
Learning Outcome: 3.1
Global LO: G7
8) Hydrogen bonding is most important in stabilizing the ________ structure of many
proteins.
1.
2.
3.
4.
5.
A) primary
B) secondary
C) tertiary
D) quaternary
E) primary, secondary, tertiary, and quaternary
Answer: B
Chapter Section: 3.1
Bloom’s Taxonomy: Comprehension
Learning Outcome: 3.1
Global LO: G7
9) Which of the following statements is false?
1.
2.
3.
4.
5.
A) There are more than 60 different kinds of amino acids present in cells.
B) The R group of amino acids differs from one amino acid to another.
C) Only around 20 amino acids are used in protein synthesis.
D) Equal amounts of D- and L-amino acids are found in cells.
E) An amino acid has an N-terminus, a C-terminus, and an R group.
Answer: D
Chapter Section: 3.1
Bloom’s Taxonomy: Knowledge
Learning Outcome: 3.1
Global LO: G7
10) Which of the following is not a major functional class of proteins?
1.
2.
3.
4.
5.
A) hereditary proteins
B) enzymes
C) motility proteins
D) regulatory proteins
E) structural proteins
Answer: A
Chapter Section: 3.1
Bloom’s Taxonomy: Comprehension
Learning Outcome: 3.1
Global LO: G7
11) Two proteins associated with a rare neurodegenerative disorder have been
sequenced. Protein A contains many polar amino acids with small regions containing
nonpolar, hydrophobic amino acids. Protein B is rich in nonpolar, hydrophobic amino
acids with only two small regions containing polar amino acids. What might this suggest
about the two proteins?
1. A) The two proteins may have different secondary structures.
2. B) Protein A is fibrous and Protein B is globular.
3. C) Protein A may be a cytoplasmic protein and Protein B may be a membrane associate
protein.
4. D) Protein A and Protein B are complementary parts of a supramolecular structure.
5. E) Protein A is most likely and enzyme and Protein B is most likely a storage protein.
Answer: C
Chapter Section: 3.1
Bloom’s Taxonomy: Evaluation
Learning Outcome: 3.1
Global LO: G2, G5, G7
12) Which of the following is a possible function of a terpene?
1.
2.
3.
4.
5.
A) cell surface receptor
B) vitamin
C) enzyme
D) motility
E) structure
Answer: B
Chapter Section: 3.4
Bloom’s Taxonomy: Knowledge
Learning Outcome: 3.2
Global LO: G7
13) Which of the following pairs correctly matches the monomer with its polymer?
1.
2.
3.
4.
5.
A) peptides; proteins
B) amino acids; polysaccharides
C) glucose; proteins
D) terpenes; nucleic acids
E) nucleotides; nucleic acids
Answer: E
Chapter Section: 3.1, 3.2, 3.3, 3.4
Bloom’s Taxonomy: Knowledge
Learning Outcome: 3.1
Global LO: G7
14) Cellulose belongs to which of the following groups of macromolecules?
1.
2.
3.
4.
5.
A) lipids
B) carbohydrates
C) proteins
D) nucleic acids
E) none of these
Answer: B
Chapter Section: 3.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 3.1
Global LO: G7
15) Which of the following has the greatest number of glycosidic bonds?
1.
2.
3.
4.
5.
A) glucose
B) triacylglycerol
C) amylose
D) DNA
E) vitamin A
Answer: C
Chapter Section: 3.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 3.1
Global LO: G7
16) To which of the following classes of sugars does glucose belong?
1.
2.
3.
4.
5.
A) pentose
B) hexose
C) tetrose
D) heptose
E) triose
Answer: B
Chapter Section: 3.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 3.1
Global LO: G7
17) You are investigating the structure of the seeds of a newly discovered tropical plant.
There is storage material inside the seed. You treat the seed with peptidase (an enzyme
that breaks peptide bonds), glycoside hydrolases (an enzyme that breaks β glycosidic
bonds), and amylase (an enzyme that breaks α glycosidic bonds). Only the amylase
appears to dissolve the storage material in the seed. What does this tell you about the
identity of the storage material?
1.
2.
3.
4.
5.
A) The seed contains fibrous proteins to store carbon and energy.
B) The seed contains lipids to store carbon and energy.
C) The seed contains starch to store carbon and energy.
D) The seed contains cellulose to store carbon and energy.
E) The seed contains globular proteins to store carbon and energy.
Answer: C
Chapter Section: 3.3
Bloom’s Taxonomy: Application
Learning Outcome: 3.1
Global LO: G1, G2, G7
18) Fatty acids are ________; they function in the cell as ________.
A) short chains of double-bonded carbon molecules; storage lipids
B) short chains of double-bonded carbon molecules; vitamins and cofactors
C) four-ringed hydrocarbon molecules; key components of membranes
D) long, unbranched hydrocarbon chains with a carboxyl group at one end; building
blocks for other lipids
5. E) short chains of double-bonded carbon molecules; vitamins, cofactors, and storage
lipids
1.
2.
3.
4.
Answer: D
Chapter Section: 3.4
Bloom’s Taxonomy: Knowledge
Learning Outcome: 3.2
Global LO: G7
19) The two strands of DNA are held together by ________; thus ________.
1. A) covalent bonds; double-stranded DNA is very stable at a range of temperatures
2. B) ionic bonds; double-stranded DNA separates into two separate strands in water
3. C) hydrogen bonds; double-stranded DNA separates into two separate strands at high
temperatures
4. D) antiparallel bonds; double-stranded DNA is amphipathic
5. E) hydrophobic interactions; double-stranded DNA separates into two separate strands
when dissolved in a hydrocarbon (hydrophobic) solvent
Answer: C
Chapter Section: 3.2
Bloom’s Taxonomy: Comprehension
Learning Outcome: 3.1
Global LO: G7
20) RNA and DNA differ
1.
2.
3.
4.
5.
A) in that RNA contains ribose and DNA contains deoxyribose.
B) in that RNA contains nucleosides and DNA contains nucleotides.
C) in that RNA contains uracil and DNA contains thymine.
D) both A and C.
E) All of these are correct.
Answer: D
Chapter Section: 3.2
Bloom’s Taxonomy: Analysis
Learning Outcome: 3.1
Global LO: G7
21) Complementary relationships between purines and pyrimidines
1. A) allow adenine to form two hydrogen bonds with thymine (or uracil) and guanine to
form three hydrogen bonds with cytosine to form double-stranded nucleic acids.
2. B) allow the interaction of the oppositely charged amino acids to form the tertiary
structure of proteins.
3. C) allow adjacent bases in a nucleotide chain to stack tightly, stabilizing the DNA double
helix.
4. D) provide highly ordered, repetitive bonding to form α helices and β sheets within
proteins.
5. E) Both A and C are correct.
Answer: A
Chapter Section: 3.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 3.1
Global LO: G7
22) The components of a nucleotide are
1.
2.
3.
4.
5.
A) a carboxyl group, an amine group, and a variable R group.
B) a five-carbon sugar, a phosphate group, and a nitrogen-containing aromatic base.
C) a three-carbon alcohol with a hydroxyl group on each carbon and three fatty acids.
D) two six-carbon sugars attached with an α(1→4) glycosidic bond.
E) a six-carbon sugar, an ester linkage, and a four-ringed hydrocarbon.
Answer: B
Chapter Section: 3.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 3.1
Global LO: G7
23) The chemical nature of each amino acid is determined by which of the following
groups?
1.
2.
3.
4.
5.
A) amino
B) carboxyl
C) hydroxyl
D) R
E) hydrogen
Answer: D
Chapter Section: 3.1
Bloom’s Taxonomy: Comprehension
Learning Outcome: 3.1
Global LO: G7
24) The nucleoside triphosphate molecules in DNA are linked together in the 5’→3′ by
a(n) ________ bridge.
1.
2.
3.
4.
5.
A) phosphate
B) covalent
C) phosphodiester
D) peptide
E) phosphatidyl
Answer: C
Chapter Section: 3.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 3.1
Global LO: G7
25) The function of triglycerides is
1. A) to store energy.
2.
3.
4.
5.
B) to form semipermeable membranes.
C) to transport substances in and out of cells.
D) store information.
E) Both B and C are correct.
Answer: A
Chapter Section: 3.4
Bloom’s Taxonomy: Knowledge
Learning Outcome: 3.2
Global LO: G7
26) A general trend in the structure of many biomolecules is
1. A) that the order and bonding of monomers form the basis for the secondary and tertiary
structure of the polymer.
2. B) that they are all soluble in water independent of the size of the polymer.
3. C) that each class of biomolecule forms one type of secondary structure independent of
the order of the monomers in the polymer.
4. D) that each class of biomolecule can form either fibrous or globular conformations
depending on the chemical conditions inside the cell.
5. E) that four different monomers form the basis for the functional and structural properties
of each polymer.
Answer: A
Chapter Section: 3.1, 3.2, 3.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 3.1
Global LO: G7
27) Cholesterol is a ________, which ________.
1.
2.
3.
4.
5.
A) terpene; is the basis for many animal and plant vitamins
B) fatty acid; functions in energy storage
C) steroid; is a component of eukaryotic membranes
D) steroid; is the basis for many animal and plant hormones
E) steroid; is a component of eukaryotic membranes and is the basis for many animal
and plant hormones
Answer: E
Chapter Section: 3.4
Bloom’s Taxonomy: Knowledge
Learning Outcome: 3.2
Global LO: G7
28) Which of the following is found exclusively in RNA?
1.
2.
3.
4.
5.
A) thymine
B) guanine
C) uracil
D) adenine
E) cytosine
Answer: C
Chapter Section: 3.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 3.1
Global LO: G7
29) Which of the following is true of purines?
1.
2.
3.
4.
5.
A) Cytosine is a purine.
B) Adenine’s bonding to thymine is stronger than is guanine’s to cytosine.
C) Purines have a double-ringed structure.
D) Both adenine and thymine are purines.
E) Purines bind readily to deoxyribose but not to ribose.
Answer: C
Chapter Section: 3.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 3.1
Global LO: G7
30) Which of the following contributes to the stability of the DNA double helix?
1. A) hydrophobic interactions between aromatic bases at the center of the double helix
2. B) hydrogen bonding between the phosphate and sugar groups in the backbone of the
double helix
3. C) covalent bonding between complementary purine and pyrimidine bases.
4. D) ionic bonds between the negatively charged phosphate groups and the positively
charged pyrimidine bases.
5. E) hydrophobic interactions between aromatic bases at the center of the double helix
and ionic bonds between the negatively charged phosphate groups and the positively
charged pyrimidine bases
Answer: A
Chapter Section: 3.2
Bloom’s Taxonomy: Comprehension
Learning Outcome: 3.1
Global LO: G2, G7
31) Which of the following is false?
1.
2.
3.
4.
5.
A) Phospholipids are important in membrane structure.
B) Serine is a molecule that may be part of a phosphoglyceride.
C) Phosphatidic acid contains two fatty acids and a phosphate group.
D) Sphingolipids are the predominant phospholipid in membranes.
E) Phospholipids are amphipathic.
Answer: D
Chapter Section: 3.4
Bloom’s Taxonomy: Knowledge
Learning Outcome: 3.2
Global LO: G7
32) The term amphipathic describes the characteristic of some molecules that have
1.
2.
3.
4.
5.
A) two polar regions.
B) only a single polar region.
C) both a polar and a nonpolar region.
D) no polar regions.
E) two nonpolar regions.
Answer: C
Chapter Section: 3.1, 3.4
Bloom’s Taxonomy: Knowledge
Learning Outcome: 3.1, 3.2
Global LO: G7
33) Which of the following is not one of the six classes of lipids?
1.
2.
3.
4.
5.
A) steroids
B) terpenes
C) fatty acids
D) triacylglycerols
E) pectins
Answer: E
Chapter Section: 3.4
Bloom’s Taxonomy: Knowledge
Learning Outcome: 3.2
Global LO: G7
34) Which of the following statements is true?
1. A) Fatty acids are synthesized by the stepwise addition of three carbon units.
2. B) Unsaturated fatty acids are usually branched.
3. C) Fatty acids with 24 carbons are most common.
4. D) Saturated fatty acids have no double bonds between carbons.
5. E) Hormones are unsaturated fatty acids.
Answer: D
Chapter Section: 3.4
Bloom’s Taxonomy: Knowledge
Learning Outcome: 3.2
Global LO: G7
35) Which of the following is true of glycolipids?
1.
2.
3.
4.
5.
A) Glycolipids are usually found on the exterior surface of the plasma membrane.
B) Glycolipids are found in plastids and are used to store energy.
C) Fructose and sucrose are often part of glycolipids.
D) Glycolipids contain steroids.
E) Usually more than 10 sugar units are attached to the glycolipid.
Answer: A
Chapter Section: 3.4
Bloom’s Taxonomy: Knowledge
Learning Outcome: 3.2
Global LO: G7
36) Which of the following is not a steroid?
1.
2.
3.
4.
5.
A) testosterone
B) estradiol
C) cortisol
D) aldosterone
E) phenylalanine
Answer: E
Chapter Section: 3.4
Bloom’s Taxonomy: Knowledge
Learning Outcome: 3.2
Global LO: G7
37) Which of the following is a terpene?
1.
2.
3.
4.
5.
A) testosterone
B) vitamin A
C) glycerol
D) estrogen
E) chitin
Answer: B
Chapter Section: 3.4
Bloom’s Taxonomy: Knowledge
Learning Outcome: 3.2
Global LO: G7
38) Which of the following is not a polymer of numerous monomer units?
1.
2.
3.
4.
5.
A) a phospholipid
B) a polypeptide
C) an RNA molecule
D) cellulose
E) starch
Answer: A
Chapter Section: 3.4
Bloom’s Taxonomy: Comprehension
Learning Outcome: 3.2
Global LO: G7
39) Trans fats
1. A) are unsaturated fatty acids.
2.
3.
4.
5.
B) resemble saturated fatty acids in shape.
C) are associated with an increased risk of heart disease.
D) are present in small amounts in meat and dairy products.
E) All of these statements are true.
Answer: E
Chapter Section: 3.4
Bloom’s Taxonomy: Knowledge
Learning Outcome: 3.2
Global LO: G7
40) Lipid rafts are
1. A) regions of the membrane that are high in sphingolipids, which facilitate
communication with the external environment of the cell.
2. B) important regions of membrane structure comprised of phospholipids.
3. C) regions not typically associated with signal transduction.
4. D) regions where greater concentrations of sphingolipids are on the inner side of the
membrane.
5. E) rafts of lipids inside of the cell that serve to store energy.
Answer: A
Chapter Section: 3.4
Bloom’s Taxonomy: Knowledge
Learning Outcome: 3.2
Global LO: G7
3.2 Matching Questions
Match the choice on the left with the choice on the right.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
A) spatial structure
B) linear structure
C) AMP
D) monosaccharide
E) adenine
F) quaternary structure
G) starch
H) cytosine
I) collagen
J) amino acid
K) enzyme
L) steroids
M) guanine
1) monomer of protein
Chapter Section: 3.1, 3.2, 3.3
Bloom’s Taxonomy: Analysis
Learning Outcome: 3.1
Global LO: G2, G7
2) polymer of glucose
Chapter Section: 3.1, 3.2, 3.3
Bloom’s Taxonomy: Analysis
Learning Outcome: 3.1
Global LO: G2, G7
3) pyrimidine
Chapter Section: 3.1, 3.2, 3.3
Bloom’s Taxonomy: Analysis
Learning Outcome: 3.1
Global LO: G2, G7
4) nucleotide
Chapter Section: 3.1, 3.2, 3.3
Bloom’s Taxonomy: Analysis
Learning Outcome: 3.1
Global LO: G2, G7
5) multimeric complex
Chapter Section: 3.1, 3.2, 3.3
Bloom’s Taxonomy: Analysis
Learning Outcome: 3.1
Global LO: G2, G7
6) Haworth projection
Chapter Section: 3.1, 3.2, 3.3
Bloom’s Taxonomy: Analysis
Learning Outcome: 3.1
Global LO: G2, G7
7) a fibrous protein
Chapter Section: 3.1, 3.2, 3.3
Bloom’s Taxonomy: Analysis
Learning Outcome: 3.1
Global LO: G2, G7
Answers: 1) L 2) G 3) H 4) C 5) F 6) A 7) I
List all the functions that match with each biomolecule. Note that the functions may
match with more than one biomolecule and each biomolecule may have multiple
functions.
1.
2.
3.
4.
5.
6.
7.
8.
9.
A) carbon and energy storage
B) enzyme cofactor
C) hormone
D) enzyme
E) vitamin
F) structural component of cells
G) cell receptor
H) informational molecule
I) motility
8) protein
Chapter Section: 3.1, 3.2, 3.3, 3.4
Bloom’s Taxonomy: Comprehension
Learning Outcome: 3.1, 3.2
Global LO: G2, G7
9) DNA
Chapter Section: 3.1, 3.2, 3.3, 3.4
Bloom’s Taxonomy: Comprehension
Learning Outcome: 3.1, 3.2
Global LO: G2, G7
10) lipids and steroids
Chapter Section: 3.1, 3.2, 3.3, 3.4
Bloom’s Taxonomy: Comprehension
Learning Outcome: 3.1, 3.2
Global LO: G2, G7
11) terpenes
Chapter Section: 3.1, 3.2, 3.3, 3.4
Bloom’s Taxonomy: Comprehension
Learning Outcome: 3.1, 3.2
Global LO: G2, G7
12) polysaccharides
Chapter Section: 3.1, 3.2, 3.3, 3.4
Bloom’s Taxonomy: Comprehension
Learning Outcome: 3.1, 3.2
Global LO: G2, G7
13) RNA
Chapter Section: 3.1, 3.2, 3.3, 3.4
Bloom’s Taxonomy: Comprehension
Learning Outcome: 3.1, 3.2
Global LO: G2, G7
Answers: 8) A, C, D, F, G, I 9) H 10) C, F 11) B, E 12) A, F 13) H
3.3 Short Answer Questions
1) ________ is the major storage polysaccharide for animals, while ________ is the
major storage polysaccharide for plants.
Answer: Glycogen; starch
Chapter Section: 3.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 3.1
Global LO: G7
2) A ________, a ________, and a ________ are the major parts of a nucleotide,
whereas a ________ and a ________ are the major parts of a nucleoside.
Answer: phosphate group, pentose sugar (ribose or deoxyribose), aromatic base (in
any order); pentose sugar, aromatic base (either order)
Chapter Section: 3.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 3.1
Global LO: G7
3) Nucleotides are to nucleic acids as amino acids are to ________.
Answer: proteins
Chapter Section: 3.1, 3.2
Bloom’s Taxonomy: Comprehension
Learning Outcome: 3.1
Global LO: G7
4) Triglycerides are composed of three fatty acids attached to a molecule of ________.
Answer: glycerol
Chapter Section: 3.4
Bloom’s Taxonomy: Knowledge
Learning Outcome: 3.2
Global LO: G7
5) ________ RNA is a component of the ribosomes that serve as the site of protein
synthesis.
Answer:
Ribosomal (or r)
Chapter Section: 3.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 3.1
Global LO: G7
6) Cellulose is a polymer of ________, a common monosaccharide.
Answer: glucose
Chapter Section: 3.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 3.1
Global LO: G7
7) ________ and ________ are the two kinds of sugars found in bacterial cell walls.
Answer:
N-acetylglucosamine (GlcNAc); N-acetylmuramic acid (MurNAc)
Chapter Section: 3.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 3.1
Global LO: G7
8) The ________ projection illustrates saccharides as linear, whereas the ________
projection suggests the spatial relationships of saccharides.
Answer: Fischer; Haworth
Chapter Section: 3.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 3.1
Global LO: G7
9) Structurally, starch exists as both unbranched polysaccharide chains called
________ and branched polysaccharide chains called ________.
Answer: amylose; amylopectin
Chapter Section: 3.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 3.1
Global LO: G7
10) In 1953 the structure of DNA was determined to be a ________.
Answer: double helix
Chapter Section: 3.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 3.1
Global LO: G7
11) For the amino acid glutamine, the three-letter abbreviation is ________ and the
one-letter abbreviation is ________.
Answer: Gln; Q
Chapter Section: 3.1
Bloom’s Taxonomy: Knowledge
Learning Outcome: 3.1
Global LO: G7
12) ________ fatty acids are liquid at room temperature, while ________ fatty acids are
solid.
Answer: Unsaturated; saturated
Chapter Section: 3.4
Bloom’s Taxonomy: Knowledge
Learning Outcome: 3.2
Global LO: G7
13) Because of their structure, trans fats resemble ________ fatty acids in shape and
are produced artificially during the manufacture of shortening and margarine.
Answer: saturated
Chapter Section: 3.4
Bloom’s Taxonomy: Knowledge
Learning Outcome: 3.2
Global LO: G7
3.4 Inquiry
1) As a protein is being synthesized, the correct folding of the protein is aided by the
movement of nonpolar amino acids toward the inner areas of the protein. How can this
phenomenon be explained? How will this influence other levels of protein structure?
Answer: The exclusion of hydrophobic groups from the aqueous surface is called the
hydrophobic effect. This then influences the other levels in orientation. (Many correct
answers are possible.)
Chapter Section: 3.1
Bloom’s Taxonomy: Synthesis
Learning Outcome: 3.1
Global LO: G7, G8
2) A transmembrane protein has 1000 amino acids. The fifth amino acid is found on the
external surface of the cell membrane. It interacts with the aqueous environment
outside the cell. Amino acids 25 and 205 are covalently bonded to each other and are
required to give the protein its three-dimensional shape. Amino acid 554 is found on the
outer surface of the protein, but it is deep in the interior of the membrane. Amino acid
979 is found on the external surface of the protein, where it forms a weak ionic bond
with a chloride ion. Can you identify each of the numbered amino acids by structural
group and name? Please note that there may be more than one specific amino acid
possible in some cases.
Amino acids:
5
25
205
554
979
Answer: 5: polar, charged or uncharged (hydrophilic) amino acid; any amino acid in this
group
25: polar, uncharged (hydrophilic) amino acid; cysteine
205: polar, uncharged (hydrophilic) amino acid; cysteine
554: nonpolar (hydrophobic) amino acid; any amino acid in this group
979: polar, charged (hydrophilic) amino acid; any positively charged amino acid in this
group (Lys, Arg, or His)
Chapter Section: 3.1
Bloom’s Taxonomy: Evaluation
Learning Outcome: 3.1
Global LO: G2, G7, G8
3) 5′-ATAGGGCTT-3′ is a short sequence of a strand of DNA. What will be the
complementary sequence of the other strand? What will be the complementary mRNA
sequence derived from the second strand of DNA? Be sure to indicate the 5′ or 3′ ends
of each sequence.
Answer: Complementary DNA: 3′ TATCCCGAA5′ or 5’AAGCCCTAT3′
Complementary mRNA: 5’AUAGGGCUU3′
Chapter Section: 3.2
Bloom’s Taxonomy: Comprehension
Learning Outcome: 3.1
Global LO: G7
4) Amylose, cellulose, and glycogen are all polysaccharides composed of glucose.
What makes each of these polysaccharides unique in spite of the fact that all are
composed solely of glucose?
Answer: Amylose: is produced by plants and is a nutritional form; linear structure with
glucose molecules connected by α(1→4) linkages.
Cellulose: is produced by plants and is a structural form; linear structure with glucose
molecules connected by β(1→4) linkages; mammals do not have the enzyme needed to
break this type of linkage.
Glycogen: is produced by animals and is a nutritional form; branched form with α(1→4)
linkages with α(1→6) linkages at the branch points.
Chapter Section: 3.3
Bloom’s Taxonomy: Comprehension
Learning Outcome: 3.1
Global LO: G7, G8
5) Give an example of both fibrous and globular proteins and explain how their structure
relates to their unique functions.
Answer: Fibrous proteins such as (silk, collagen, elastin, or keratin) have extensive
secondary structure (either helix or b sheet) throughout the molecule that gives them a
highly ordered, repetitive structure. Fibrous proteins usually have an extended,
filamentous structure. Often multiple proto-filaments will interact to create thicker,
stronger filaments. Small hydrophobic amino acids that pack tightly are common in
fibrous proteins. In contrast, globular proteins, such as (enzymes, transport proteins, or
transmembrane proteins) have extensive tertiary structure wherein β sheets and/or α
helices are connected by loops that fold into a compact structure. Globular proteins
often have multiple regions, or domains, that have different functions. The diverse
shapes of globular proteins create binding pockets for substrates, regions of
hydrophobicity that interact with membranes, and other domains that allow globular
proteins to act as enzymes, receptors, and a diversity of other functions in the cell.
Chapter Section: 3.1
Bloom’s Taxonomy: Synthesis
Learning Outcome: 3.1
Global LO: G7, G8
Becker’s The World of the Cell, 9e (Hardin/Bertoni/Kleinsmith)
Chapter 11 Phototrophic Energy Metabolism: Photosynthesis
11.1 Multiple-Choice Questions
1) Photoheterotrophs are best described as organisms that obtain energy to make ATP
1.
2.
3.
4.
5.
A) from organic compounds but use sunlight to produce carbon sources.
B) and organic compounds from sunlight.
C) from sunlight but cannot make organic compounds from CO2.
D) from NADH.
E) from organic compounds.
Answer: C
Chapter Section: 11.1
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
2) Carbon dioxide generally enters a plant through pores called
1.
2.
3.
4.
5.
A) mesophylls.
B) stomata.
C) cuticles.
D) connexons.
E) bacteriochlorophylls.
Answer: B
Chapter Section: 11.5
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
3) The process by which photosynthetic organisms use light energy for ATP production
is known as
1.
2.
3.
4.
5.
A) photorespiration.
B) photosynthesis.
C) phototaxis.
D) photophosphorylation.
E) photoreduction.
Answer: D
Chapter Section: 11.1
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
4) The endosymbiont theory states that chloroplasts originated
1.
2.
3.
4.
5.
A) by spontaneous generation in the cytoplasm of an ancestral cell.
B) by fusion of several prokaryotes.
C) from cyanobacteria entering and remaining in the cytoplasm of another cell.
D) by fusion of mitochondria and peroxisomes.
E) through evolutionary change in a mitochondrion.
Answer: C
Chapter Section: 11.1
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.3
Global LO: G2
5) Each of the following is true regarding bacterial anoxygenic photosynthesis except
A) O2is released.
B) Hydrogen sulfide or thiosulfate can act as electron donors.
C) Bacteriochlorophyll b absorbs light energy in the far red range of the light spectrum.
D) Electron flow is coupled to unidirectional proton pumping across the bacterial plasma
membrane.
5. E) Following electron flow, the electron is transferred from cytochrome c to
bacteriochlorophyll b.
1.
2.
3.
4.
Answer: A
Chapter Section: 11.4
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
6) In a chloroplast, the stroma is
1.
2.
3.
4.
5.
A) a pore in the outer membrane.
B) a stack of thylakoids.
C) the region between in the inner and outer membranes.
D) a connection between the grana.
E) a gel-like matrix between the inner chloroplast membrane and the thylakoid
membrane.
Answer: E
Chapter Section: 11.1
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
7) Of the following plastids, which is not paired with its proper function?
1.
2.
3.
4.
5.
A) amyloplasts; sugar synthesis
B) chromoplasts; contain non-chlorophyll pigment that give flowers distinctive colors
C) chloroplasts; photosynthesis
D) proteinoplasts; protein storage
E) elaioplasts; lipid storage
Answer: A
Chapter Section: 11.1
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
8) The plastids that give flowers and fruits their distinctive colors are
1.
2.
3.
4.
5.
A) chloroplasts.
B) elaioplasts.
C) amyloplasts.
D) chromoplasts.
E) proteinoplasts.
Answer: D
Chapter Section: 11.1
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
9) Which of the following is not an accessory pigment used to absorb photons that
cannot be captured by chlorophyll?
1.
2.
3.
4.
5.
A) β-carotene
B) phycoerythrin
C) lutein
D) phycochlorophyll
E) phycocyanin
Answer: D
Chapter Section: 11.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
10) Having various types of chlorophylls and accessory pigments for the photosynthetic
process
1.
2.
3.
4.
5.
A) allows for a variation in color among plants.
B) allows many wavelengths of light to be captured to provide energy for photosynthesis.
C) allows light to be harvested specifically in a stepwise manner.
D) shifts the equilibrium of photosynthesis toward the formation of products.
E) increases the efficiency of the Calvin cycle.
Answer: B
Chapter Section: 11.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
11) Which of the following initially traps solar energy in the process of photosynthesis?
1.
2.
3.
4.
5.
A) water
B) chlorophyll
C) triose phosphate
D) NADP+
E) ATP
Answer: B
Chapter Section: 11.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
12) Which of the following pigments enables red algae to utilize the dim light that
penetrates ocean surface water as a source of energy?
1.
2.
3.
4.
5.
A) chlorophyll a
B) chlorophyll c
C) lutein
D) phycoerythrin
E) β-carotene
Answer: D
Chapter Section: 11.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
13) Each of the following is part of the energy transduction from solar energy to
chemical energy except
1. A) light absorption by chlorophyll.
2. B) electron flow through an electron transport system.
3. C) unidirectional proton pumping across a membrane.
4. D) an electrochemical proton gradient.
5. E) All are involved in the energy transduction.
Answer: E
Chapter Section: 11.1
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
14) Most reactions involving NADP+ are ________ reactions, whereas most reactions
involving NAD+ are ________ reactions.
1.
2.
3.
4.
5.
A) anabolic; catabolic
B) catabolic; anabolic
C) metabolic; anabolic
D) anabolic; both catabolic and anabolic
E) both catabolic and anabolic; catabolic
Answer: A
Chapter Section: 11.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
15) When a photon is absorbed by a pigment (light-absorbing molecule), such as
chlorophyll, the energy of the photon is transferred to
1.
2.
3.
4.
5.
A) D1 protein to activate PSII.
B) ATP synthase to make ATP.
C) an electron to move it from a low to a high energy orbital.
D) Plastocyanin to be passed on to PSI.
E) NADP+to make NADPH.
Answer: C
Chapter Section: 11.2
Bloom’s Taxonomy: Comprehension
Learning Outcome: 11.1
Global LO: G2
16) The Emerson enhancement effect
1.
2.
3.
4.
5.
A) explains the lack of O2release during anoxygenic photosynthesis.
B) results from the activity of three photosystems.
C) explains the synergistic effect of two wavelengths of light on photosynthesis.
D) makes it possible for plants to avoid the glyoxylate pathway.
E) accounts for the major difference between C3and C4
Answer: C
Chapter Section: 11.2
Bloom’s Taxonomy: Comprehension
Learning Outcome: 11.1
Global LO: G2
17) The relationship between photosystem I (PSI) and photosystem II (PSII) is correctly
described by which of the following statements?
1. A) PSI precedes PSII in photosynthesis.
2. B) Electrons are excited by PSI or PSII, but not both.
3. C) Chlorophyll molecules in the reaction centers of PSI and PSII are designated P700
and P680, respectively, to reflect the wavelengths of their light absorption maxima.
4. D) PSII was discovered before PSI.
5. E) Only PSI contains a special pair of chlorophyll a
Answer: C
Chapter Section: 11.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
18) Plastocyanin (PC)
1.
2.
3.
4.
5.
A) is a mobile electron carrier.
B) is functionally similar to plastoquinol.
C) carries only one electron at a time.
D) carries electrons to PSI.
E) All of the above are true of PC.
Answer: E
Chapter Section: 11.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
19) The plastoquinol (Q8H2) transfers electrons from
1.
2.
3.
4.
5.
A) cytochrome b6/f to PSII.
B) PSII to the cytochrome b6/f
C) water to PSII.
D) light harvesting complex II to PSII.
E) cytochrome b6/f to plastoquinone.
Answer: B
Chapter Section: 11.3
Bloom’s Taxonomy: Comprehension
Learning Outcome: 11.1
Global LO: G2
20) The ATP synthase complex associated with the thylakoid membrane
1.
2.
3.
4.
5.
A) contains a CF1proton channel and a CFoATPase complex.
B) is identical to the FoF1complex of mitochondria.
C) performs photophosphorylation in oxygenic phototrophs.
D) is a soluble enzyme complex peripherally associated with PSI.
E) accepts electrons from PSI.
Answer: C
Chapter Section: 11.1
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
21) Each of the following is an essential feature of electron transfer from water to
NADP+ except
1.
2.
3.
4.
5.
A) PSII.
B) PSI.
C) coenzyme Q.
D) plastocyanin (PC).
E) cytochrome b6/f
Answer: C
Chapter Section: 11.1
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
22) The transfer of energy directly from a photoexcited electron in a pigment molecule
to another electron in an adjacent pigment molecule occurs by
1.
2.
3.
4.
5.
A) photoexcitation.
B) photochemical reduction
C) photophosphorylation.
D) resonance energy transfer.
E) photorespiration.
Answer: D
Chapter Section: 11.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
23) Your house plant is growing slowly and lacks its normal bright green color, so you
call a botanist for advice. She suggests that the plant may be deficient in magnesium
(Mg2+). Why are these symptoms associated with Mg2+ deficiency?
1.
2.
3.
4.
5.
A) Mg2+is important in the structure of many amino acids.
B) Lack of Mg2+in the soil allows pathogenic bacteria to grow.
C) Mg2+is a cofactor for NADPH function.
D) Mg2+ is an important component in chlorophyll.
E) Cytochromes require Mg2+ to function properly.
Answer: D
Chapter Section: 11.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
24) When energy captured by antenna pigments of LHCII is funneled to the reaction
center it changes the reduction potential of a P680 chlorophyll a molecule to ________,
making it a better electron ________.
1.
2.
3.
4.
5.
A) −0.80 V; donor
B) −0.80 V; acceptor
C) +0.80 V; donor
D) +0.80 V; acceptor
E) None of the above are correct.
Answer: A
Chapter Section: 11.3
Bloom’s Taxonomy: Comprehension
Learning Outcome: 11.1
Global LO: G2
25) Each of the following statements accurately describes the Calvin cycle except
1. A) It is found in all oxygenic and most anoxygenic phototrophs.
2. B) Carbon dioxide is fixed by reduction to form organic sugars.
3. C) Its products enter a variety of metabolic pathways including sugar and starch
biosynthesis.
4. D) It involves ribulose-1,5-bisphosphate (RuBP).
5. E) It is confined to the cytoplasm in plants and algae.
Answer: E
Chapter Section: 11.5
Bloom’s Taxonomy: Comprehension
Learning Outcome: 11.2
Global LO: G2
26) The Calvin cycle occurs in the chloroplast
1.
2.
3.
4.
5.
A) stroma.
B) grana.
C) thylakoid lumen.
D) thylakoid membrane.
E) outer membrane.
Answer: A
Chapter Section: 11.5
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.2
Global LO: G2
27) In the Calvin cycle, the enzyme that catalyzes the capture of carbon dioxide and the
formation of 3-phosphoglycerate is
1.
2.
3.
4.
5.
A) phosphoglycerokinase.
B) ribulose-1,5-bisphosphate carboxylase/oxygenase (rubisco).
C) triose phosphate isomerase.
D) ferredoxin-NADP+
E) glyceraldehyde-3-phosphate dehydrogenase.
Answer: B
Chapter Section: 11.5
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.2
Global LO: G2
28) Which of the following statements correctly describes cyclic electron flow pathways?
1.
2.
3.
4.
5.
A) Water molecules are split.
B) Both PSI and PSII are required.
C) Proton gradients are never formed.
D) No exogenous electron source is required.
E) O2is released.
Answer: D
Chapter Section: 11.4
Bloom’s Taxonomy: Comprehension
Learning Outcome: 11.1
Global LO: G2
29) Continuous carbon assimilation in the Calvin cycle is made possible by the
regeneration of
1. A) glyceraldehyde-3-phosphate.
2.
3.
4.
5.
B) dihydroxyacetone phosphate.
C) ribulose-1,5-bisphosphate.
D) glycerate-1,3-bisphosphate.
E) 3-phosphoglycerate.
Answer: C
Chapter Section: 11.5
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.2
Global LO: G2
30) Which molecule in the Calvin cycle is used to synthesize starch and sucrose?
1.
2.
3.
4.
5.
A) 3-phosphoglycerate
B) glyceraldehyde-3-phosphate
C) ribulose-1,5-bisphosphate
D) 1,3-bisphosphoglycerate
E) ribulose-5-phosphate
Answer: B
Chapter Section: 11.5
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.2
Global LO: G2
31) Which three organelles are involved in the glycolate pathway?
1.
2.
3.
4.
5.
A) lysosome, chloroplast, rough endoplasmic reticulum
B) mitochondrion, nucleus, lysosome
C) peroxisome, smooth endoplasmic reticulum, chloroplast
D) peroxisome, chloroplast, mitochondrion
E) Golgi complex, peroxisome, lysosome
Answer: D
Chapter Section: 11.8
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.2
Global LO: G2
32) Biosynthesis of sucrose occurs in the
1.
2.
3.
4.
5.
A) cytosol.
B) stroma.
C) thylakoid lumen.
D) chloroplast intermembrane space.
E) vacuole.
Answer: A
Chapter Section: 11.1
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
33) Which of the following is the site of photosynthesis in cyanobacteria?
1.
2.
3.
4.
5.
A) peroxisomes
B) mitochondria
C) photosynthetic membranes
D) outer walls
E) thylakoid lumens
Answer: C
Chapter Section: 11.1
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
34) Bacteriochlorophyll differs from chlorophyll a and b in structure and function. Which
of the following statements correctly contrasts these types of chlorophyll?
1. A) Bacteriochlorophyll has a heme ring, but chlorophyll a and b each have a porphyrin
ring.
2. B) Chlorophyll a and b each have a phytol side chain, which is missing from
bacteriochlorophyll.
3. C) Chlorophyll a has a methyl group attached to the porphyrin ring, but
bacteriochlorophyll has a formyl group.
4. D) Bacteriochlorophyll has a saturated carbon–carbon bond at one location in the
porphyrin ring that is not saturated in chlorophyll a and b.
5. E) Chlorophyll a and b porphyrin rings bind Mg2+, but the bacteriochlorophyll ring binds
Fe3+.
Answer: D
Chapter Section: 11.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G7
35) Each of the following compounds is a source of electrons for anoxygenic
photosynthesis except
1.
2.
3.
4.
5.
A) S2- (sulfide).
B) S2O32-(thiosulfate).
C) H2O (water).
D) succinate.
E) All of the above are sources of electrons for anoxygenic photosynthesis.
Answer: C
Chapter Section: 11.1
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
36) Based on the comparisons of anoxygenic and oxygenic photosynthesis, the entry of
oxygen into the atmosphere is most directly attributed to
1.
2.
3.
4.
5.
A) PSI activity.
B) Calvin cycle activity.
C) photophosphorylation.
D) Hatch–Slack pathway activity.
E) Q cycle activity.
Answer: C
Chapter Section: 11.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
37) Which of the following types of organisms makes use of the Hatch–Slack cycle?
1.
2.
3.
4.
5.
A) algae
B) C3plants
C) C4 plants
D) cyanobacteria
E) photosynthetic bacteria
Answer: C
Chapter Section: 11.8
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.2
Global LO: G7
38) C3 and C4 plants get their classification from the number of carbons in the first
detectable product of carbon dioxide fixation. The C3 product is ________, whereas the
C4 product is ________.
1.
2.
3.
4.
5.
A) glycerol; malate
B) triose phosphate; oxaloacetate
C) 3-phosphoglycerate; malate
D) ribose; phosphoenolpyruvate
E) 3-phosphoglycerate; oxaloacetate
Answer: E
Chapter Section: 11.8
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
39) Each of the following can be used in the Calvin cycle except
1.
2.
3.
4.
5.
A) phosphoglycolate.
B) NADPH.
C) 3-phosphoglycerate.
D) CO2.
E) ribulose-1,5-bisphosphate.
Answer: A
Chapter Section: 11.8
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.2
Global LO: G2
40) In crassulacean acid metabolism (CAM) plants
A) stomata open during the day.
B) malate is stored in large vacuoles.
C) CO2 is transported directly into mesophyll cells.
D) the amount of carbon assimilated for each water molecule transpired is only 25
percent of that in C3
5. E) carboxylation and decarboxylation reactions occur in the same location at the same
time.
1.
2.
3.
4.
Answer: B
Chapter Section: 11.8
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
41) Each of the following directly regulates the Calvin cycle activity to ensure maximum
efficiency except
1.
2.
3.
4.
5.
A) stromal pH.
B) Mg2+
C) chlorophyll photon absorption.
D) NADPH concentration.
E) ATP concentration.
Answer: C
Chapter Section: 11.6
Bloom’s Taxonomy: Comprehension
Learning Outcome: 11.2
Global LO: G2
11.2 Matching Questions
Match the plant process or characteristic with the location in which it occurs in the cell.
1.
2.
3.
4.
A) matrix
B) cytoplasm
C) nucleolus
D) thylakoid membrane
5. E) ribosomes
6. F) thylakoid lumen
7. G) stroma
1) sucrose synthesis
Chapter Section: 11.1
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G7
2) starch synthesis
Chapter Section: 11.1
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G7
3) electron transport
Chapter Section: 11.1
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G7
4) concentration of protons
Chapter Section: 11.1
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G7
Answers: 1) B 2) G 3) D 4) F
Match the organelle with its function.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
A) protein storage
B) lipid storage
C) photosynthesis
D) starch storage
E) nucleic acid production
F) flower color
G) chloride storage
H) protein production
I) chromosome movement
J) plastid precursor
5) proplastid
Chapter Section: 11.1
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
6) amyloplast
Chapter Section: 11.1
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
7) chloroplast
Chapter Section: 11.1
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
8) chromoplast
Chapter Section: 11.1
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
9) elaioplast
Chapter Section: 11.1
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
Answers: 5) J 6) D 7) C 8) F 9) B
Match the pathway with the characteristic or example.
1.
2.
3.
4.
5.
6.
7.
8.
A) C3plants
B) nucleic acid replication
C) severe water restriction
D) lysosome passage
E) fungi
F) salvage pathway
G) C4plants
H) carbon assimilation
10) Calvin cycle
Chapter Section: 11.5, 11.8
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1, 11.2
Global LO: G2
11) glycolate pathway
Chapter Section: 11.5, 11.8
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1, 11.2
Global LO: G2
12) Hatch–Slack cycle
Chapter Section: 11.5, 11.8
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1, 11.2
Global LO: G2
13) crassulacean acid metabolism (CAM)
Chapter Section: 11.5, 11.8
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1, 11.2
Global LO: G2
Answers: 10) H 11) F 12) G 13) C
Match the term on the left with its definition or description on the right.
1.
2.
3.
4.
5.
6.
7.
A) light-dependent generation of NADPH
B) electron donors such as sulfide and thiosulfate
C) electron donor is water
D) energy from sunlight, carbon from organic sources
E) energy from carbon, carbon from sunlight
F) electron donor is carbon dioxide
G) energy from sunlight, carbon from CO2
14) photoreduction
Chapter Section: 11.1, 11.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
15) photoheterotrophs
Chapter Section: 11.1, 11.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
16) photoautotrophs
Chapter Section: 11.1, 11.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
17) oxygenic phototrophs
Chapter Section: 11.1, 11.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
18) anoxygenic phototrophs
Chapter Section: 11.1, 11.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
Answers: 14) A 15) D 16) G 17) C 18) B
11.3 Short Answer Questions
1) The light-dependent generation of NADPH is referred to as ________.
Answer: photoreduction
Chapter Section: 11.1
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
2) 3-phosphoglycerate is to C3 plants as ________ is to C4 plants.
Answer: oxaloacetate
Chapter Section: 11.8
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.2
Global LO: G2
3) ________ organisms such as green and purple bacteria derive electrons from sulfide,
thiosulfate, and succinate to generate NADPH.
Answer: Anoxygenic phototrophs
Chapter Section: 11.1
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
4) Light behaves as a stream of discrete particles called ________, each carrying a
________ (indivisible packet) of energy.
Answer: photons; quantum
Chapter Section: 11.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
5) ________ proteins stabilize the arrangement of the chlorophyll within a photosystem
and modify the absorption spectra of specific chlorophyll molecules.
Answer: Chlorophyll-binding
Chapter Section: 11.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
6) The Calvin cycle in C3 plants occurs in ________ cells, whereas the Calvin cycle in
C4 plants occurs in the ________.
Answer: mesophyll; bundle sheath
Chapter Section: 11.8
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.2
Global LO: G2
7) Crassulacean acid metabolism (CAM) plants open their stomata only during the
________ to minimize water loss.
Answer: night
Chapter Section: 11.8
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.2
Global LO: G2
8) Plastocyanin (PC) is a copper-containing protein that, like plastoquinol, is a mobile
________ carrier in photosynthesis.
Answer: electron
Chapter Section: 11.3
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
9) Chlorophyll is comprised of a central ________ containing a(n) ________ ion and
a(n) ________ side chain.
Answer: porphyrin ring; Mg; phytol
Chapter Section: 11.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
10) Diatoms have both chlorophyll a and chlorophyll ________.
Answer: c
Chapter Section: 11.2
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
11) When atmospheric CO2 concentrations decrease, the ________ cycle gives
C4 plants an advantage over C3 in carbon fixation.
Answer: Hatch–Slack
Chapter Section: 11.8
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.2
Global LO: G2
12) The glycolate pathway performs the process of ________ by consuming O2 and
releasing CO2 in a light-dependent manner.
Answer: photorespiration
Chapter Section: 11.8
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.2
Global LO: G2
13) The energy transduction reactions of photosynthesis convert solar energy into
chemical energy in the form of ________ and ________.
Answer:
NADPH; ATP
Chapter Section: 11.1
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
11.4 Inquiry
1) Rubisco is an important enzyme in the Calvin cycle because it acts as a carboxylase
to catalyze addition of carbon dioxide and water to ribulose-1,5-bisphosphate, which
then forms two molecules of 3-phosphoglycerate. Rubisco also acts as an oxygenase to
catalyze addition of molecular oxygen rather than carbon dioxide to ribulose-1,5bisphosphate.
1. a) What is (are) the product(s) of the rubisco oxygenase reaction activity and why is this
a problem for plants?
2. b) Explain why rubisco’s much lower affinity for oxygen than for carbon dioxide does not
mitigate the rubisco oxygenase problem for plants.
3. c) Identify the pathway plants use to metabolize the problematic product of rubisco
oxygenase and the final fate of its carbons.
4. d) Explain why the oxygenase activity of rubisco is particularly problematic for plants in
arid environments.
5. e) What strategies do plants in arid environments use to deal with this problem?
Answer:
1. a) Rather than the carboxylase activity yielding two molecules of 3-phosphoglycerate to
feed into the Calvin cycle, rubisco oxygenase activity yields one three-carbon product (3phosphoglycerate) and one two-carbon product (phosphoglycolate) that cannot be used
directly in the next step of the Calvin cycle. Producing the two-carbon product appears to
be a wasteful diversion of material from carbon assimilation.
2. b) The Earth’s atmosphere contains a much higher concentration of O2than CO2, so
oxygenase activity could have a significant negative impact on carbon fixation.
3. c) The glycolate pathway transforms ~75 percent of the phosphoglycolate carbon into 3phosphoglycerate and the remaining 25 percent is released as CO2.
4. d) Higher temperatures decrease carbon dioxide solubility relative to oxygen solubility. In
addition, these plants close their stomata to reduce water loss, which further reduces the
CO2concentration in the leaf.
5. e) C4plants confine rubisco to cells containing high carbon dioxide levels. CAM plants
only open their stomata at night.
Chapter Section: 11.8
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.2
Global LO: G7, G8
2) The highest intensity of the Sun’s radiation at the Earth surface is in the visible light
wavelength ranges of between ~400 to 700 nm.
Questions:
1. a) Explain how plants maximize the uptake of sunlight energy. In your answer, consider
chlorophyll and antenna pigment structures and functions.
2. b) Why does the photosynthetic part of plants appear green?
3. c) The leaves of some plants lose their green color in the fall, often appearing red or
yellow. Use your knowledge of plant pigments to propose a possible reason for these
color changes.
Answer:
1. a) Plants expand the energy wavelengths they can absorb by expressing several
chlorophylls that have different absorption spectra. Chlorophyll a, which has a methyl (CH3) group on the porphyrin ring, has a broad absorption spectrum, with maxima at
about 420 and 660 nm, whereas chlorophyll b, which has a formyl (-CHO) group instead
of the methyl group, absorbs more toward the center of the visible spectrum. Antenna
pigments are arranged in the photosystem to collect light energy much like a radio
antenna collects radio waves. The antenna pigments absorb photons and pass the
energy to a neighboring chlorophyll molecule or accessory pigment by resonance energy
transfer until it eventually reaches the reaction center chlorophyll.
2. b) Chlorophylls absorb mainly blue and red light, so plants appear green.
3. c) Some plants stop synthesizing chlorophylls as they become more dormant in the fall.
Losing chlorophyll can expose the colors of other plant pigment molecules that are
present longer.
Chapter Section: 11.1
Bloom’s Taxonomy: Knowledge
Learning Outcome: 11.1
Global LO: G2
3) There has been and remains great interest in developing safe and effective
herbicides. In some situations, it would be desirable to have a herbicide that killed a
wide variety of different plants, but in other cases it would be best to have a herbicide
that was much more selective.
1. a) What plant activity would you target to kill the widest variety of plants?
2. b) What plant activity would you target to kill C4plants?
3. c) What plant activity would you target to kill CAM plants?
Be as specific as possible.
Answer:
1. a) The best general herbicide would target an activity of virtually all plants, such as
photosynthesis. Atrazine blocks D1 protein electron transfer in the PSII complex.
Another good target might be the mobile soluble plastocyanin protein that transfers
electrons to PSI.
2. b) C4plants would be more specifically killed by an inhibitor of the Hatch–Slack pathway,
such as an inhibitor of the specific cytosolic PEP carboxylase form that is abundant in
C4mesophyll cells.
3. c) CAM plants would be more specifically killed by an inhibitor that targets conversion of
oxaloacetate to malate or by an activator PEP carboxylase activity, which would cause a
futile cycle.
Chapter Section: 11.1, 11.3, 11.8
Bloom’s Taxonomy: Synthesis
Learning Outcome: 11.1, 11.2
Global LO: G1, G7, G8
4) Some herbicides, such as paraquat, cause oxidation of NADPH to NADP+. How does
paraquat interfere with plant energy metabolism?
Answer: By oxidizing NADPH to NADP directly, the amount of NADPH available for the
Calvin cycle is minimized, and the plant literally starves to death.
Chapter Section: 11.5
Bloom’s Taxonomy: Application
Learning Outcome: 11.2
Global LO: G7, G8
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