Answers for Teachers
Campbell Biology in Focus, AP* Edition
by Lisa A. Urry, Michael L. Cain, Steven A. Wasserman,
Peter V. Minorsky, and Jane B. Reece
*Advanced Placement, Advanced Placement Program, AP, and Pre-AP are registered trademarks of the College Board,
which was not involved in the production of, and does not endorse, this product.
Appendix A of the book includes answers for students for figure legend questions, Concept
Check questions, Summary of Key Concepts questions, multiple-choice questions, and Draw It
questions. This document for teachers includes suggested answers and teaching tips for the
Scientific Skills Exercises and suggested answers for the Interpret the Data questions and the
short-answer essay questions at the end of each chapter. The Scientific Skills Exercises can be
assigned in MasteringBiology, where they are graded automatically.
Tips for Grading Short-Answer Essays
The ability to communicate clearly in writing is essential for almost any profession your students
choose to pursue. As teachers, it is often frustrating to be faced with a large class full of students
who have had inadequate preparation in writing skills, knowing that you don’t have the resources
to help your students develop these skills.
The Focus on a Big Idea questions at the end of each chapter are an attempt on the part of
the authors to partner with you in this endeavor. At the end of each chapter, we ask the student to
write a short essay of 100-150 words that relates the material they learned in the chapter to one
of the Big Ideas of biology that are introduced in Chapter 1 and featured throughout the book.
The Focus on a Big Idea exercises can be used as in-class or outside-of-class assignments.
For ease of grading, sample key points and sample top-scoring answers for the Focus on a
Big Idea questions are provided for teachers in this document. The list of key points provides a
guide to the ideas that students should include in their essays. In addition, suggested answers to
all of the end-of-chapter essay questions can be found in this document.
The time necessary to grade writing exercises has prohibited many teachers from
assigning them. Using a grading rubric, however, can streamline the process. A suggested
grading rubric for the Focus on a Big Idea essays is shown at the end of these tips and in the
Study Area of MasteringBiology. This rubric can also be modified to use with the other end-ofchapter essay questions.
The simplest way to use the rubric is to read through each essay and determine how well
the writer has accomplished the four aims listed at the top of the columns. The essay can then be
graded as a 4, 3, 2, 1, or 0 based on the overall quality of the essay. Alternatively, you could
assign 0 to 4 points for each of the aims, and then total the points out of 16 possible points.
You can also weight one of the aims more highly. For example, if you want to focus
primarily on writing skills (aim #4: Quality of Writing) with the other aims weighted equally, the
score for each aim can be multiplied by a “weighting factor.” Aim #4 could be assigned 40% of
the total points, with aims # 1, 2, and 3 each worth 20%. The score (out of 4) obtained for aim #4
is multiplied by 40, and each of the others multiplied by 20, giving a total of 400 points (160 +
80 + 80 + 80 = 400).
Using a similar rubric, the Montgomery County Public School System in Maryland has
been able to train a team of teachers to grade thousands of short essays consistently in a
relatively short time. To train teachers, the lead teacher first read through some of the essays,
looking for a representative example of each of the five scores (4, 3, 2, 1, and 0 for the simplest
grading scheme described above). Copies of the five representative essays (with scores hidden)
were passed out to the teachers, asking them to grade the essays based on the rubric and a 0-4
grading scheme. Subsequent discussion with the teachers about their essay rankings clarified the
standards, after which they were given a few “test” essays to grade to ensure consistency in
grading practices.
There is also a web-based program called Calibrated Peer Review (CPR) (developed at
UCLA with funding from the National Science Foundation and the Howard Hughes Medical
Institute) that trains students to evaluate their own work or that of their classmates (“peers”). The
program is described at http://cpr.molsci.ucla.edu/.
When assigning essays, the teacher should point out the rubric to students (in the Study
Area of MasteringBiology) or provide a customized rubric to students. Students can then refer to
the rubric before writing to see what is expected of them. They can also check their essay before
submitting it to make sure they have met all the criteria in the rubric. Teachers should also
encourage students to read the Writing Tips provided under “Writing Tips and Rubric” in the
Study Area of MasteringBiology, which also includes the suggested grading rubric.
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3
2
1
0
Suggested Grading Rubric for “Focus on a Big Idea” Short-Answer Essays
Understanding of
Use of Supporting
Appropriate Use
Quality of Writing
Theme and
Examples or Details
of Terminology
Relationship to Topic
Evidence of full and
Examples well
Accurate scientific Excellent
complete understanding chosen, details
terminology
organization,
accurate and applied
enhances the essay sentence structure,
to theme
and grammar
Evidence of good
Examples or details
Terminology is
Good sentence flow,
understanding
are generally well
correctly used
sentence structure,
applied to theme
and grammar
Evidence of a basic
Supporting examples Terminology used
Some organizational
understanding
and details are
is not totally
and grammatical
adequate
accurate or
problems
appropriate
Evidence of limited
Examples and details Appropriate
Poorly organized;
understanding
are minimal
terminology is not
grammatical and
present
spelling errors
detract from essay
Essay shows no
Examples lacking or
Terminology
Essay is very poorly
understanding of theme incorrect
lacking or incorrect written
Suggested Answers and Teaching Tips
CHAPTER 1 INTRODUCTION: EVOLUTION AND THE FOUNDATIONS OF
BIOLOGY
Scientific Skills Exercise
Teaching objective: Students build scientific skills by interpreting data in a pair of bar graphs
and relating the data to the biological system it came from.
Teaching tips: A version of this Scientific Skills Exercise can be assigned in MasteringBiology.
If this is the first exercise the students are doing related to interpreting graphs, then you will need
to spend time reviewing independent and dependent variables. If the students are confused by
having two independent variables on one graph, have them cover one set of data while they look
at the other (for example, cover the "full moon" portion of graph A while analyzing the "no
moon" portion of it).
In these graphs, there are no statistical significance values given for comparisons between
treatments. In the original paper, there was a statistical difference between predation levels of
light brown versus dark brown mice in light-colored soil enclosures with no moon and in darkcolored soil enclosures under a full moon. The other two combinations, light-colored soil under a
full moon and dark-colored soil with no moon, had no statistically significant difference between
light and dark mice.
Answers:
1. (a) The independent variables for each graph are the coat color of the mice (light or dark
brown) and the presence or absence of moonlight (full moon or no moon). These are on the xaxis. Taking both graphs together, a third independent variable is the color of soil in the
enclosure. (b) The dependent variable is the amount of predation, measured as the number of
mice caught. The dependent variable is on the y-axis of the two graphs.
2. (a) About 19. (b) About 12. (c) Based on the data, the mouse would be more likely to escape
on dark soil. This might be because in the moonlight, a dark mouse on light soil would be more
noticeable than one on dark soil.
3. (a) Under a full moon (12 were caught vs. 20 under no moon). (b) Under no moon (11 were
caught vs. 18 under a full moon).
4. (a) Dark soil field with a full moon. (b) Light soil with no moon.
5. (a) No moon plus dark brown coat had the highest predation level in the light soil enclosure
(38 mice were caught). (b) Full moon plus light brown coat had the highest predation level in the
dark soil enclosure.
6. Being on the contrasting soil is most deadly for both colors of mice.
7. The total number of mice caught on moonlit nights was about 77 and on nights with no moon
was about 95, so the dark nights seem to be slightly better overall for hunting for owls.
Interpret the Data
Figure 1.21 In the beach habitat, approximately 27 light models and 73 dark models were
attacked. In the inland habitat, approximately 76 light models and 24 dark models were attacked.
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and a suggested rubric at the
beginning of this document.
7. Scientific Inquiry
Many legitimate hypotheses could be proposed to extend the investigation. Here is one example.
If the camouflage color has arisen through the processes of natural selection due to visual
predators, then you might wonder what would happen if a population of beach mice lived in an
area where predators were absent. It might be possible to do a long-term study in an area where
you excluded predators. Mice have fairly short generation times, so if predation is “naturally
selecting” lighter colored mice, then in the absence of predation you might predict the coat color
would not remain predominantly light in such an experimental population.
8. Scientific Inquiry
Students are asked to use a PubMed search to identify an abstract of an article authored or coauthored by Hopi Hoekstra from 2014 forward. It is therefore expected that the range of abstracts
from which students might choose will grow as the Hoekstra lab generates additional
publications.
9. Focus on Evolution
Sample key points:
 Darwin used reasoning based on observations to develop his theory of natural selection as a
mechanism for evolution.
 His observations included:
o Heritable variations exist in each population.
o A population has more individuals than can be supported by the environment.
o Each species seems suited for its particular environment.
 He proposed that the best-adapted individuals in a population would outcompete others for
resources and disproportionately survive and produce more offspring, leading to an increase
in the adaptations seen in the population.
Sample top-scoring answer:
Based on many observations of different species, Darwin proposed his theory that evolution by
means of natural selection accounts for both the unity and diversity of life on Earth. He noticed
that variations existed among the individuals in a population and that these variations seemed to
be heritable. He also saw that populations could grow larger than could be supported by the
resources around them. Finally, he observed that species (like the different species of finches)
seemed to suit their environment. He proposed that the best-suited individuals in a population
would survive and reproduce more successfully that those less adapted to their environment, and
he called this “natural selection.” In Darwin’s view, this mechanism could account for both the
unity and diversity of features among species. The descent of organisms from a common
ancestor explains similar features, while the force of natural selection in different environments
accounts for differences between organisms.
10. Focus on Information
Common ancestry explains this observation. The thousand-some-odd genes shared by humans
and prokaryotes originated in early prokaryotes. They have been retained, with some
modification, over the billions of years of eukaryotic evolution. These genes no doubt code for
proteins and RNAs whose functions are essential for survival—for example, the genes that code
for ribosomal RNA, which is important for protein synthesis in both prokaryotes and eukaryotes.
11. Synthesize Your Knowledge
It’s difficult to pick out this gecko against the background of the tree trunk, because the gecko
itself looks like mossy bark. This coloration likely makes it harder for the gecko to be seen by
predators, thus enhancing its survival. This cryptic coloration pattern probably evolved over
generations. The members of a gecko population that more closely resembled their background
would have been less visible to predators, thus more likely to survive, reproduce, and leave
offspring. The offspring would inherit the genes that generated the mossy bark coloration, and
the offspring that blended in better would survive better and reproduce more successfully. Over
generations, the coloration would become a closer and closer match to the tree bark. (The mossy
leaf-tailed gecko is endemic to Madagascar, meaning it is found only there and nowhere else in
the world. Many endemic species live in Madagascar. This is because it is an island with land
features and climatic factors that have allowed evolution of many species in isolation.)
CHAPTER 2 THE CHEMICAL CONTEXT OF LIFE
Scientific Skills Exercise
Teaching objective: This exercise is designed to give students practice in figuring out what is
shown on a graph, how to describe the major trend(s) in the data, and extracting values from the
graph to calculate related information. The student is then led back to the biological context of
the data to draw a conclusion.
Teaching tips: A version of this Scientific Skills Exercise can be assigned in MasteringBiology.
Most students can look at a graph and describe the slope of the data line. However, many
struggle with writing out what the trend means in terms of the relationship between what was
reported on one axis relative to the other axis. Thus, while a student may respond that the data
line has a positive slope, they may also respond that a higher calcification rate results in a higher
carbonate ion concentration. Helping them sort out dependent and independent variables should
clear up the problem. Visual learners will benefit from drawing a mock-up of 1 square meter of
the reef, with dots in the water to represent carbonate ions and arrows to indicate calcification.
In this example, students will need to make the additional mental step of reading the trend line
right to left, instead of left to right (the natural tendency), to reach a conclusion about the effect
of decreased carbonate ion concentration on calcification rate and reef growth.
Answers:
1. (a) The x-axis shows the concentration of carbonate ions in units of micromoles of carbonate
ions per kilogram of seawater. (b) The y-axis shows the calcification rate in units of millimoles
of calcium carbonate accumulated per square meter of reef per day. (c) Carbonate ion
concentration is the independent variable. (d) Calcification rate is the dependent variable.
2. The data show that the rate of calcification is positively related to the concentration of
carbonate ions in the seawater. As the concentration of carbonate ions increases, the rate of
calcification increases.
3. (a) If the seawater carbonate ion concentration was 270 µmol/kg, the calcification rate would
be approximately 19 mmol CaCO3/m2·day. It would take 1 square meter of reef approximately
1.6 days to accumulate 30 mmol of CaCO3 [(30 mmol of CaCO3/m2) / (19 mmol CaCO3/m2·day)
= 1.6 days]. (b) If the seawater carbonate ion concentration was 250 µmol/kg, the calcification
rate would be approximately 12 mmol CaCO3/m2·day. It would take 1 square meter of reef 2.5
days to accumulate 30 mmol of CaCO3 [(30 mmol of CaCO3/m2) / (12 mmol CaCO3/m2·day) =
2.5 days]. (c) If carbonate ion concentration decreases, the rate of calcification decreases, and it
takes coral longer to grow.
4. (a) The final step of the process shown in Figure 2.24, the rate of conversion of CO32- and Ca2+
into CaCO3, is measured in this experiment. (b) The results do support the hypothesis that
increased concentration of atmospheric CO2 could lead to slower growth of coral reefs. It
supports it because, according to the chemistry shown in Figure 2.24, more CO2 entering the
ocean will push the reactions toward formation of more bicarbonate ions, decreasing the amount
of CO32- available for formation of CaCO3. The results in the graph show that, under the
experimental conditions, the lower the concentration of CO32-, the lower the rate of calcification,
and thus the slower the growth of coral reefs (for example, 2.5 days versus 1.6 days to
accumulate the same amount of calcium carbonate at a lower carbonate ion concentration,
calculated in question 3).
Interpret the Data
Table 2.1 As you probably know, the human body is made up in large part of water, H2O. The
atoms of oxygen in water, one per water molecule, likely account for the high percentage of
oxygen (65.0%) found in the human body.
Figure 2.19 The inland temperatures (100F, 96F, 106F) are much higher than those along the
coast (73F, 75F, 72F) because oceans are large bodies of water that can absorb or release heat,
moderating the climate nearer the coast.
Concept Check 2.5 #5 A liter of blood would contain 7.8 × 1013 molecules of ghrelin (1.3 ×
10–10 moles per liter × 6.02 × 1023 molecules per mole).
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and the suggested rubric at the
beginning of this document.
11. Scientific Inquiry
The complex shapes of biological molecules determine the great specificity with which they
interact with one another and form weak or strong bonds.
(a) Hypothesis: Receptor cells on the filaments of the male luna moth’s antennae contain
cell-surface molecules that are complementary in shape to sex attractant molecules
(pheromones) produced by the female luna moth.
(b) This hypothesis leads to several testable predictions. (1) Luna moth pheromones will bind
to specific sites on the cells of the filaments of the male’s antennae. (2) If it is possible to
synthesize molecules that are very similar in shape to luna moth pheromones, these
molecules will also attract male luna moths. (3) Chemical or temperature treatments that
modify the molecular shape of luna moth pheromones will reduce the attractiveness of
these molecules to male luna moths.
(c) An experiment could be designed to test the third prediction. A number of male luna
moths could be exposed to two separate treatments. In the first treatment, unaltered
pheromones would be released near male luna moths, and the response of the moths
would be noted. The second treatment would be identical in every way except that the
pheromone would be heated to permanently modify its molecular shape before it was
released.
12. Focus on Evolution
It would be surprising if the percentages of naturally occurring elements in most organisms were
not roughly the same, because all organisms evolved on Earth (with its unique elemental
composition) and all are genetically related to one another. (Species living under unusual
conditions might differ more than others, though.) Further, we might predict that the more
similar the percentages of naturally occurring elements are in two species, the more closely
related those two species are.
13. Focus on Organization
Sample key points:
 Water’s versatility as a solvent arises from the polar covalent bonds of water molecules.
 Water molecules form hydrogen bonds with atoms that are part of polar covalent bonds in
other molecules.
 The partially charged regions of water molecules are attracted to oppositely charged ions.
Sample top-scoring answer:
Water is the solvent of life, a function emerging from the polar covalent bonds of water
molecules. A water molecule consists of an oxygen atom bonded to two hydrogen atoms. Due to
oxygen’s high electronegativity, the shared electrons are attracted closer to the oxygen at the
apex of this V-shaped molecule. The resulting partial negative charge associated with oxygen
and partial positive charge associated with each hydrogen result in hydrogen bonding between
adjacent water molecules.
Water molecules also form hydrogen bonds with atoms in polar covalent bonds in other
molecules, dissolving those molecules. The partial positive and negative regions of water
molecules are also attracted to negatively and positively charged ions, respectively, forming
hydration shells around ions that separate them from each other and dissolve them. Most of the
chemical reactions of life involve solutes that are dissolved in water, so the properties of water
that allow it to form hydrogen bonds are crucial to life on Earth.
14. Synthesize Your Knowledge
The water adheres to the molecules on the cat’s tongue, drawing it upward. The column of water
forms due to both cohesion of water molecules within the column and the surface tension along
the sides of the column. Adhesion, cohesion, and surface tension are possible because of
extensive hydrogen bonding that takes place between water molecules, which in turn is because
of the structure of the water molecule. The oxygen region of the water molecule has a partial
negative charge while the hydrogens each carry a partial positive charge. This leads to an
attraction between the hydrogen of one water molecule and the oxygen of an adjacent molecule.
Hydrogen bonds constantly break and re-form between water molecules. Although they are
individually weak, the large number of them means that water sticks together very well—as well
as to other hydrophilic molecules, such as those on a cat’s tongue—allowing cats to drink in this
way.
CHAPTER 3 CARBON AND THE MOLECULAR DIVERSITY OF LIFE
Scientific Skills Exercise
Teaching objective: In this exercise, students will examine aligned amino acid sequences from
three different species to find any and all differences between them. They will calculate percent
identity and create supportable hypotheses about relatedness among the species.
Teaching tips: A version of this Scientific Skills Exercise can be assigned in MasteringBiology.
The sequences are too long to fit on one line, so some students may get confused and think there
are three different sequences shown in the data, although you can point out that this is explained
in the paragraph introducing the data. Also, a review of how to calculate percentage may be
needed.
A key aspect of this exercise is using the data to support a hypothesis about evolutionary
relationships. The “molecular genealogy” approach introduced in Chapter 3, and used in this
exercise, may not be intuitive for students who have not yet studied evolution. The group game
“Telephone” is an analogy that might help them to understand. In this game, one person whispers
a sentence to the next person, and so on down the line, until the last person says it out loud. The
farther away a person is from the source, the more errors will have been introduced into the
sentence. Similarly, the more generations between species separations, the more changes there
will be to the DNA and protein sequences.
If you would like to show your students the gorilla sequence of â globin (mentioned in the text),
it is at http://www.ncbi.nlm.nih.gov/protein/P02024.2.
Answers:
1. (a) There are eight amino acid differences between the human and monkey sequences (S/N,
A/T, V/L, T/S, A/N, T/Q, R/K, P/Q). (b) There are two amino acid differences between the
gibbon and the human (T/Q, P/Q).
2. For the rhesus monkey, 94.5% (138/146) of its amino acids are identical to the human
sequence of  globin. For the gibbon, 98.6% (144/146) of its amino acids are identical to the
human sequence.
3. Based solely on the -globin amino acid sequences, one can hypothesize that gibbons are
more closely related to humans than rhesus monkeys are. The reasoning is that the more similar
the amino acid sequences, the more closely related two species are. This is based on the premise
that there has been less evolutionary change since gibbons and humans shared a common
ancestor—in other words, fewer mutations have occurred and thus there has been less divergence
of the DNA and polypeptide sequences of these species. Conversely, a more distantly related pair
of species (here, human and monkey) would have a more divergent sequence because there has
been more time for evolutionary change since they shared a common ancestor.
4. Other evidence could include amino acid sequences of other proteins, DNA sequences, and
morphological and biochemical similarities and differences. The same principles of
similarity/divergence and relatedness would apply.
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and the suggested rubric at the
beginning of this document.
12. Scientific Inquiry
DNA has many phosphate groups along the backbone of each strand, so the molecule has many
negative charges. Therefore, you might expect a DNA-binding protein to have many amino acids
with positively charged side chains. These are the basic amino acids in Figure 3.18: lysine,
arginine, and histidine. (In fact, DNA-binding proteins do have regions that are rich in these
three amino acids.)
13. Focus on Evolution
Some cellular functions are more essential than others to the survival of the organism. Proteins
perform most cellular functions, so some proteins could be considered more essential than
others. You should therefore expect the amino acid sequences of more essential proteins to be
more highly conserved (retained with little or no change) than those of less essential proteins.
Moreover, different species often live in different habitats and experience different selection
pressures. Consequently, you should expect different degrees of divergence among the proteins
shared by species that live in different habitats.
14. Focus on Organization
Sample key points:
 Amino acids share common chemical groups but have unique side chains that allow for
variation.

Common components allow the formation of polypeptides, whose repeating units interact to
establish secondary structure.
 Interactions of the varying side chains determine tertiary structure.
 The combination allows for an almost infinite number of possible different structures, each
with a different function.
Sample top-scoring answer:
The structure of an amino acid, including both common and varying groups, is key to its
function. The common groups (amino and carboxyl groups attached to an  carbon) allow amino
acids to link together into a polypeptide of any length via peptide bonds. The repeating subunits
of the polymer can interact with each other, forming  helices and  pleated sheets that establish
secondary structure. The varying groups are the amino acid side chains. Each polypeptide
therefore has a unique sequence of side chains attached to the common backbone. Interactions
between the side chains determine the tertiary structure of the polypeptide. A combination of
secondary and tertiary structure establishes the overall 3-D structure of a protein (or one of its
subunits), which determines its function. Amino acid structure thus allows for an almost infinite
number of possible protein structures, accounting for the diversity of proteins functioning in a
cell.
15. Synthesize Your Knowledge
A developing chick is growing rapidly, increasing its number of cells. To build new cells it needs
large stores of cell membrane components, including cholesterol and lipids, and other cell
components such as proteins. It also requires energy to fuel all this construction, and that is
available in the form of fats. Fat molecules are a rich source of energy when metabolized.
CHAPTER 4 A TOUR OF THE CELL
Scientific Skills Exercise
Teaching objective: This exercise is designed to give students practice using a scale bar to
measure the size of a cell in a micrograph and to use the measurement to calculate volume and
surface area of the cell.
Teaching tips: A version of this Scientific Skills Exercise can be assigned in MasteringBiology.
Most students have difficulty grasping the relationship between the scale bar and the size of the
object shown in the image. It may help to sketch (or photocopy) the picture at one size, then
again in a larger size, to emphasize that the scale bar is enlarged along with the object. For both
sizes, you can have the student make a small “ruler” the same size as the scale bar and use it to
manually measure the cell. This will illustrate that even though the image size and ruler length
have changed, the cell remains the same size. There will be some variation in measurements
using the scale bar, according to where students delineate the edges of the cells in the image. As
long as they are consistent between the two cells, the relationships will hold.
Answers:
1. If the 1-µm scale bar is 5 mm long, and the diameter of the image of the parent cell is 20 mm;
dividing 20 by 5 yields a cell diameter of 4 µm. Since the diameter of the image of the new cell
is 15 mm, the actual diameter of that cell is 15 ÷ 5 = 3 µm. Another way to put it is that the
diameter of the parent cell is 4 scale bars long, and the scale bar is 1 µm long, so 4 x 1 µm = 4
µm. The diameter of the new cell is 3 scale bars long, and 3 x 1 µm = 3 µm.
2. (a) For the parent cell with radius 2 µm, the volume is 4 π(2 µm)3 = 33.5 µm3. For the new
3
4
cell with radius 1.5 µm, the volume is π(1.5 µm)3 = 14.1 µm3. (b) The new cell will need to
3
synthesize 19.4 µm3 of cytoplasm to reach its mature volume (33.5 µm3 – 14.1 µm3 = 19.4 µm3).
3. (a) For the parent cell with radius 2 µm, the surface area is 4π(2 µm)2 = 50.2 µm2. For the new
cell with radius 1.5 µm, the surface area is 4π(1.5 µm)2 = 28.3 µm2. (b) The new cell will need to
synthesize 21.9 µm2 of plasma membrane to reach its mature size (50.2 µm2 – 28.3 µm2 = 21.9
µm2).
4. When it is mature, the new cell will be approximately 2.4 times its current volume (33.5 µm3 ÷
14.1 µm3 ≈ 2.4) and 1.8 times its current surface area (50.2 µm2 ÷ 28.3 µm2 ≈ 1.8).
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and the suggested rubric at the
beginning of this document.
9. Scientific Inquiry
You could start by examining the organism and especially its mystery organelle by TEM at high
magnification, looking closely for structural similarities with known organelles. Then, starting
with a suspension of the organism in an aqueous solution, you could carry out cell fractionation:
successive centrifugations at higher and higher speeds, saving the pellet at each stage. Using
electron microscopy on material from the pellets, you could identify the pellet containing the
organelles of interest. You could then re-suspend these organelles in solution, and test them for
the activities of various enzymes, which might reveal the organelle's function in the cell. You
could also attempt to isolate DNA from the organelles and compare any you find with DNA from
normal mitochondria and chloroplasts. If, for example, the organelles contain enzymes known to
participate in cellular respiration and also contain DNA similar to known mitochondrial DNA,
you could conclude that the mystery organelle is most likely a mitochondrion. An organelle
unrelated or more distantly related to known organelles might take much additional study to
figure out.
10. Focus on Evolution
(a) Unity is best revealed by cell structures that are shared. Almost all biological membranes are
phospholipid bilayers. This feature arose early in the evolution of life on Earth and has been
consistently retained by all subsequent life-forms. Ribosomes also originated early and are found
in all cells. Mitochondria arose early in the origin of eukaryotic cells, all of which seem to
contain mitochondria or related organelles.
(b) The less commonly shared a structure is, the more likely it is a specialized adaptation. Cilia
and flagella of the “9 + 2” variety are seen only in motile eukaryotes and in eukaryotic cells that
move fluid substances past them.
11. Focus on Organization
Sample key points:
 Life emerges from the integration of the functions performed by individual cell components.
 Organelles and cellular structures are not alive on their own; their activities must be
integrated with those of other components.
 Some of the characteristics of life include the ability to transform energy, to respond to the
environment, and to reproduce and pass along characteristics to offspring.
 Although the nucleus contains instructions for all cell components, other cell components are
required to carry them out.
 “Life” as a phenomenon is an emergent property of the parts of a cell.
Sample top-scoring answer:
The phenomenon we call “life” emerges from the integration of all the functions performed by
the individual structures and organelles of a cell. Not one of the cellular components could exist
indefinitely on its own. Integrating the activity of all cellular components allows a cell to display
the fundamental characteristics of life: the ability to transform energy, to respond to the
environment and interact with other organisms, and ultimately to reproduce and pass along
characteristics to offspring. The nucleus houses the genetic instructions for the proteins needed
for a cell’s functions, but it could not carry out those instructions without the cooperation of
ribosomes on which to build proteins, the endoplasmic reticulum and Golgi apparatus to modify
them, and mitochondria to convert fuel energy to ATP to drive the process. Furthermore, the
arrangement of all these organelles in relation to each other is crucial. Life is an emergent
property that results from the intricate organization and orchestrated functioning of these and
other cellular components.
12. Synthesize Your Knowledge
Epithelial cells in the small intestine function both in the absorption of nutrients from partially
digested food and as a barrier between the substances in the small intestine and the blood supply.
The microvilli—the many hair-like projections of the plasma membrane at the top of each cell in
the SEM—function to increase the surface area of the cell to allow more area for absorption (see
Figure 4.24). The microvilli are supported internally by microfilaments, filaments made of actin
monomers. The cells in the epithelial sheet must be tightly joined to each other so that the sheet
can act as a barrier. This is accomplished by tight junctions, specialized regions on each cell
where it is bound by proteins to cells surrounding it (see Figure 4.27; the red dashed arrow at the
top of the cell shows that substances can’t get across). Tight junctions among the entire sheet of
cells hold together the whole sheet. The cells are also anchored together by desmosomes, further
strengthening the sheet.
CHAPTER 5 MEMBRANE TRANSPORT AND CELL SIGNALING
Scientific Skills Exercise
Teaching objective: The main objective of this exercise is to give students practice in
interpreting a graph, especially in regard to recognizing and comparing the major trends in two
sets of data. Secondary objectives include encouraging students to deal with a real-life activity of
one of the membrane transport proteins they read about in this chapter and, as they're thinking
about hypothesis formulation and testing, to realize that what they're learning in this chapter
connects to topics in other chapters.
Teaching tips: A version of this Scientific Skills Exercise can be assigned in MasteringBiology.
The instructor may want to point out that the rate of glucose "uptake" was not being measured
directly; rather, the experimenters were measuring the concentration of radioactive glucose in the
cells at successive time points, with the results reflecting the differing rates of glucose uptake. In
answering question 3, some students may observe that, for both data sets, the uptake of glucose
increases more slowly as incubation time increases—that is, the slopes of the curves are
decreasing with time—and that the uptake by cells from the older animal is leveling off sooner
than uptake by the cells from the younger animal. This could be explained by invoking the
hypothesis in the question 4 answer, that the older cells have fewer glucose transporter
molecules: Presumably, the lower number of such protein molecules on the older cells are more
quickly saturated with glucose. Regarding question 5, at this point in the book students don't yet
have the background to design an experiment in detail, but the allusions to studying specific
(purified) proteins in Concept 3.5 and the mentions of labeling in Chapter 2 (radioactive tracers
in Concept 2.2) and Chapter 4 (fluorescent labeling of molecules for microscopy; see Figure 4.3)
should enable them to come up with a general answer.
Answers:
1. (a) Incubation time (in minutes) is the independent variable. (b) Concentration of radioactive
glucose (in millimoles) is the dependent variable. (c) The red dots indicate the data points for red
blood cells from a 15-day-old guinea pig. (d) The blue dots indicate data points for red blood
cells from a 1-month-old guinea pig.
2. Note: Numbers for concentration will vary slightly in student answers since these numbers are
estimates from the graph.
Incubation Time
(min)
10
Concentration of Radioactive
Glucose (mM) in Red Blood Cells
from a 15-Day-Old Guinea Pig
50
Concentration of Radioactive
Glucose (mM) in Red Blood Cells
from a 1-Month-Old Guinea Pig
22
20
30
45
60
72
85
97
104
33
40
48
53
3. The increasing concentrations of radioactive glucose at successive time points indicate that
glucose is being taken up by the cells throughout the experiment. Careful study of how much the
numbers are increasing reveals that the rate of glucose uptake is slowing down as time
progresses. The graph shows that the cells from the older guinea pig took up less glucose and
took it up less rapidly than the cells from the younger animal, at all incubation times.
4. Answers will vary but should involve the idea that glucose enters cells through glucose
transporter proteins, which must be different in some way between the cells of 15-day-old guinea
pigs and 1-month-old guinea pigs. One possible hypothesis for the lower glucose uptake by the
cells from the older guinea pig is that the red blood cells of older guinea pigs have fewer
molecules of glucose transporter protein. (This trait may have evolved due to natural selection
because the cells of older animals need less glucose, and there is a cost to making more glucose
transporter proteins than needed.)
5. Answers will vary. One way to test the first hypothesis mentioned in answer 4 would be to
isolate glucose transporter proteins from the two kinds of cells and compare the amounts. More
practically, one could tag the transporter proteins with radioactive or fluorescent labels and
measure levels of the labels to compare the amounts of protein in cells from animals of the two
ages.
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and a suggested rubric at the
beginning of this document.
7. Scientific Inquiry
(a) Evaluating the data reveals that the proton pump builds up an H+ gradient across the
membrane by pumping protons out of the cell, represented by the initial decrease in pH seen in
the solution. Once the gradient is sufficient, H+ can move down its electrochemical gradient into
the cell, and sucrose appears to move with it. The proton pump, using ATP, continues to pump
protons out of the cell, maintaining a steady gradient with a higher concentration of H+ outside
(lower pH than the starting pH). This leads to the following hypothesis: Sucrose and H+ move
together into the plant cell through a cotransport protein, with sucrose moving up a concentration
gradient and H+ moving down an electrochemical gradient. The H+ gradient is maintained by a
proton pump, which requires ATP. This hypothesis is consistent with the data.
(b) If an inhibitor of ATP regeneration were added, the proton pump would no longer be able to
work, and the surplus of H+ outside the cell would continue to drive cotransport until the gradient
ran down and the outside H+ concentration was the same as that inside the cell. Thus, you would
see the pH in the solution rise until the gradient was no longer sufficient to drive cotransport, at
which point sucrose transport would cease and the pH would remain steady.
8. Science, Technology, and Society
If a plant cell is placed in a hypertonic solution such as that found in wet saline soils, water will
be lost by osmosis from the cytoplasm and central vacuole of the cell. The cell may plasmolyze
as the plasma membrane pulls away from the cell wall, causing the cell to shrivel and die. Even
if soil salinity is not at lethal levels, water becomes less available to plants as soil salinity
increases because of the tendency of water to leave the plants’ cells.
9. Focus on Evolution
One might expect unicellular eukaryotes that live in hypertonic environments to have the
opposite sort of adaptations from those seen in Paramecium. An example would be membranes
that impede water loss to the environment. An organism that regularly experiences fluctuations
in environmental salt concentrations (such as organisms in shallow tide pools) might be expected
to have both types of adaptations, plus the ability to activate and inhibit them as its needs change.
10. Focus on Interactions
Sample key points:
 The plasma membrane is selectively permeable and regulates the passage of materials.
 Nonpolar molecules diffuse through the lipid bilayer.
 Polar molecules require transport proteins.
 Cholesterol enters by receptor-mediated endocytosis.
 Hormones bind to receptors and trigger the secretion of enzymes by exocytosis.
 Ions move through specific ion channels, which may be gated.
 The sodium-potassium pump actively transports Na+ and K+.
 All of these mechanisms enable a cell to exchange materials with its environment.
Sample top-scoring answer:
The selectively permeable plasma membrane mediates the passage of substances into and out of
the cell via several mechanisms. Small nonpolar molecules, such as O2 and CO2, can easily
diffuse through the hydrophobic interior of the lipid bilayer, which interferes with entry of polar
substances, even very small ones. Larger polar molecules, such as glucose and amino acids,
require specific carrier proteins that bind and move them across the membrane. Cholesterol
(carried by LDLs) is brought into the cell by receptor-mediated endocytosis. Hormones bind to
cell surface receptors that trigger the secretion of digestive enzymes: the enzymes are packaged
in vesicles and released by exocytosis. Ions move through specific ion channels, which may be
gated. The energy-requiring sodium-potassium pump actively transports Na+ and K+. Together,
the hydrophobic interior of the lipid bilayer, the activities of transport proteins and receptor
proteins on the cell surface, and the expenditure of ATP to drive active transport allow the
plasma membrane to regulate the cell’s exchange of materials with its environment.
11. Synthesize Your Knowledge
The water is hypotonic to the plant cells, so the plant cells take up water. Thus, the cells of the
vegetable remain turgid rather than plasmolyzing, and the vegetable (for example, lettuce or
spinach) remains crisp.
CHAPTER 6 AN INTRODUCTION TO METABOLISM
Scientific Skills Exercise
Note: In this document, inorganic phosphate is symbolized as “(P)i” since we can’t type a circled
P as done in the text.
Teaching objective: Students will practice making a line graph in this exercise. The students are
then encouraged to think of a biological explanation for the data pattern.
Teaching tips: A version of this Scientific Skills Exercise can be assigned in MasteringBiology.
Some students may have a difficult time understanding why the concentration of (P)i goes up
over time. A common misconception is that the y-values represent the enzyme activity itself,
such that the steep rise near the start of the line shows that the enzyme is becoming activated and
then it stays at a constant high level after the plateau. An analogy of snowfall rate versus snow
accumulation may help them visualize what is going on in the buffer: The cells producing (P)i
are analogous to the clouds dropping snowflakes, and the accumulation of the (P)i in the buffer
over time is like having the snow get deeper over time. The snow started falling from the clouds
high in the sky long before they eventually reached ground, analogous to the lag time for initial
accumulation of (P)i in the buffer due to the time it takes for the substrate to enter the cells and
the product to be leave the cells. And even after the snow clouds stop dropping flakes, the snow
stays the same depth on the ground, like the (P)i concentration in the buffer stays the same after
the enzyme has stopped functioning.
Answers:
1. (a) By taking samples every 5 minutes, the researchers intentionally varied the amount of time
elapsed since the addition of substrate. Thus, time is the independent variable and should be on
the x-axis. (b) The units for time are “min” (the abbreviation for minutes). (c) The variable that is
being measured, which depends on the time since substrate addition, is the concentration of (P)i
in the buffer due to its release by the cells. This is the dependent variable and should go on the yaxis. (d) The units for the concentration of (P)i are μmol/mL, which means micromoles of (P)i
per milliliter of sampled buffer.
2. (a) The largest value to go on the x-axis is 40 min. It is reasonable to label tick marks every 5
min, and the highest should be at 40 min. (It is also acceptable to label ticks every 10 min—this
is a matter of choice.) (b) The largest value to go on the y-axis is 355 μmol /mL. It is reasonable
to label tick marks every 50 μmol /mL, so the highest tick should be at 400 μmol /mL. (You
could also choose to label every 100 μmol /mL, but it would be more difficult to figure out where
to place the points.)
3.
4. (a) The concentration of (P)i does increase over time, but not in a linear pattern. The
concentration is low for the first 5 minutes, then increases steadily for the next 15 minutes, then
the rate of increase slows down, and finally the concentration stays the same for the rest of the
experiment. (b) The highest rate of enzyme activity occurred where the line has the steepest
slope, from 5 to 20 minutes. The slope is calculated as Δy/Δx; for the steepest region the slope is
(270-10) µmol/mL divided by (20-5) minutes = 17.3 µmol/mL ∙ min. (c) When the substrate is
first added, there is a lag time before product appears in the buffer because the substrate has to be
transported into the cells and processed by the enzyme, and then the product has to be released
from the cells into the buffer. After the rate of transport becomes steady, the concentration of (P)i
reflects the enzyme activity level. As the rate slows down (where the line plateaus), the
concentration is unchanging—there is no additional (P)i being released into the buffer because
the cells have used up the substrate.
5. If your blood sugar level is low, this reaction will occur in your liver cells:
Glucose 6-phosphatase
Glucose 6-phosphate ----------------------------> Glucose + (P)i
Your blood sugar level will increase as glucose is released from your liver cells into your
bloodstream.
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and a suggested rubric at the
beginning of this document.
8. Scientific Inquiry
A. The substrate molecules are entering the cells, so no product is made yet.
B. There is sufficient substrate, so the reaction is proceeding at a maximum rate.
C. As the substrate is used up, the rate decreases (the slope is less steep).
D. The line is flat because no new substrate remains and thus no new product appears.
9. Science, Technology, and Society
This answer depends on your personal viewpoint. Some might argue that improving crop yields
is of paramount importance in a world where many are starving, considering that rapid
breakdown of the compounds means the risk of harm to humans is low. Others might argue that
although short-lived, organophosphates are highly toxic to humans who work in agriculture and
not worth the risk. Also, other organisms, including bees and other pollinating insects, birds, and
small mammals, could be affected. It is often impossible to predict the negative effects on
ecosystems and therefore not worth the risk.
10. Focus on Evolution
For a given metabolic pathway, its current function is accomplished by a complex series of
reactions catalyzed by a specific set of enzymes. However, earlier versions of the pathway may
have existed that accomplished the same or different functions. Organisms with cellular enzymes
for carrying out the earlier pathway may have had mutations that allowed new enzymes to arise.
These new enzymes could have participated in the earlier pathway, modifying the pathway in a
way advantageous to the organism. If natural selection caused the new genes to be passed on
because the organism survived better, this would represent one step in a gradual modification of
a biochemical pathway. With additional steps taking place over long periods of time, this is a
plausible model for how evolutionary processes could have resulted in any biochemical pathway
seen in organisms today. This evolutionary process could have happened differently in distinct
lineages of organisms. The existence of diverse pathways to produce the same or similar
products argues in support of this model.
11. Focus on Energy and Matter
Sample key points:
 Energy enters an animal cell in the form of the chemical potential energy of fuel molecules
such as glucose.
 Energy enters a plant cell in the form of light, which is converted to the chemical energy of
sugars.

The transfer and transformation of energy increase entropy, and a cell returns heat and the
low-energy molecules of CO2 and H2O to its surroundings.
 In energy coupling, the free energy released by exergonic chemical reactions is used to drive
endergonic reactions.
 ATP represents a store of energy that links catabolic pathways to anabolic pathways.
 The hydrolysis of ATP also drives the mechanical and transport work of a cell.
 Enzymes are required for all the chemical reactions of a cell. By lowering the activation
energy, enzymes enable specific reactions to occur more quickly.
Sample top-scoring answer:
The work of a cell requires energy. Energy enters an animal cell as the chemical energy of fuel
molecules such as glucose. In a photosynthesizing cell, light energy drives the synthesis of
sugars, which in turn provide the energy for metabolism. The free energy released by the
exergonic reactions of catabolic pathways is stored in ATP. The energy released by the
hydrolysis of ATP to ADP and (P)i drives the endergonic reactions of anabolic pathways, which
synthesize the complex molecules of a cell. Also, during the hydrolysis of ATP in mechanical
and transport work, potential energy is transformed into kinetic energy. In these transfers and
transformations, some energy is lost as heat, and the cell returns the less ordered molecules of
CO2 and H2O to its surroundings. All the reactions of a cell require enzymes, which ensure
efficient cell function by binding substrates and lowering activation energy, thus speeding up
metabolic reactions.
11. Synthesize Your Knowledge
On top of the glacier the penguins have potential energy, which is transformed into kinetic
energy as they slide down the slope and dive into the water. Climbing back up the glacier and
gaining potential energy requires the activity of muscles. To power the muscle function that
drives their motion, the penguins burn fuel from food they consume, such as the proteins and
lipids in fish that they feed on while they are in the water. They digest the food by means of
enzymes, which break down the large energy-rich biomolecules into smaller molecules with less
free energy. Ultimately, the ATP molecules that are generated from catabolic food breakdown
will be used to cause muscle contraction. In addition to producing ATPs, catabolic processes
result in the generation of heat, which keeps the penguins’ body temperatures elevated above the
freezing polar temperatures.
CHAPTER 7 CELLULAR RESPIRATION AND FERMENTATION
Scientific Skills Exercise
Teaching objective: Students will practice making a bar graph with categorical data in this
exercise. The students are then encouraged to think of a biological explanation for the data
pattern.
Teaching tips: A version of this Scientific Skills Exercise can be assigned in MasteringBiology.
When faced with data that fall into categories instead of along a continuum, many students will
“connect the dots” of data points instead of keeping them separate in bar graph format. Having
students make the graph several times, changing the order that the categories align on the x-axis,
will demonstrate that the pattern will change if points are connected but not if left separate. Some
astute students may argue that in this example, the cells with low, normal, and elevated thyroid
hormone levels represent a continuum of thyroid hormone levels and can be treated as
continuous data. In this case, challenge them to assign numerical values to the hormone levels
and they will see that in the absence of actual hormone data, the cell types must be treated as
non-continuous categories. However, it is logical to put the categories in order from low to high
hormone levels.
Answers:
1. (a) The researchers intentionally varied the thyroid hormone level, which is therefore the
independent variable and should be put on the x-axis. The data should be shown in the categories
Low, Normal, and Elevated. Putting the categories in the order Low, Normal, and Elevated is
logical, but the categories could go in a different order. (b) The researchers measured the oxygen
consumption rate, which is therefore the dependent variable and should be put on the y-axis. (c)
The units on the y-axis are nanomoles of oxygen gas (O2) per minute per milligram of cells
(abbreviated as “nmol O2/min ∙ mg cells”). The largest value on the y-axis is 8.7, so numbers
from 0.0 to 9.0 should be used as tick mark labels on the y-axis.
2.
A bar graph is more appropriate than a scatter plot or line graph because the thyroid hormone
levels are categories, not continuous values. (As mentioned in the “Teaching Tips” section,
above, numerical values for “low,” “normal,” and “elevated” are not given.)
3. (a) The cells with elevated thyroid hormone levels had the highest rate of oxygen
consumption, and the cells with low thyroid hormone levels had the lowest rate of oxygen
consumption. (b) Thyroid hormone may stimulate the rate of mitochondrial electron transport in
these cells. This hypothesis is supported because the state of thyroid hormone of each cell type
(low, normal, or elevated) is correlated with the rate of mitochondrial oxygen consumption. Note
that the experiment does not give us any information about the mechanism of action of thyroid
hormone. You might also want to discuss the fact that correlative evidence is not as strong as
some other types of evidence, such as those referred to as “loss-of-function” or “gain-offunction” evidence. For example, if an agent was used that blocked the action of thyroid
hormone in mice and O2 consumption decreased, or a different agent was used that increased the
activity of thyroid hormone and O2 consumption increased, then both of these would provide
stronger evidence. (c) Because the rate of oxygen consumption is a measure of the rate of
mitochondrial electron transport, and electron transport rate is related to how much heat is
produced by the mitochondria, one can predict that the rats with low thyroid levels will have the
lowest body temperature and the rats with elevated thyroid levels will have the highest body
temperature. (This does turn out to be the case.)
Interpret the Data
9. Under both conditions, the phosphofructokinase enzyme increases in activity as the
concentration of the substrate increases, up until a point at which the rate remains the same.
Under conditions of low ATP concentration, however, phosphofructokinase enzyme activity
increases faster and reaches maximum rate sooner. At higher concentrations of ATP the enzyme
is less active, which makes sense because if ATP is abundant, there is less need for ATP to be
made by the processes of cellular respiration, beginning with glycolysis. At low ATP
concentrations in the cell, the concentrations of AMP and ADP are higher, and it makes sense
that more ATP would be needed, thus phosphofructokinase would be more active, moving sugar
along in the breakdown pathway. Therefore, phosphofructokinase activity is higher when the
concentration of ATP is low.
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and a suggested rubric at the
beginning of this document.
10. Scientific Inquiry
In a person treated with uncoupling agents like DNP, the proton gradient established during
respiratory electron transport is no longer tied to ATP synthesis. As a result, oxidation of glucose
during the citric acid cycle yields very little ATP, since ATP is normally produced as protons
flow back through the ATP synthase sites in the inner mitochondrial membrane, while still
generating heat. Without large amounts of ATP available, biosynthesis cannot take place and
new organic molecules cannot be synthesized. Low ATP levels would signal the body to
continue breaking down its own molecules and feeding them into the cellular respiration
pathway, leading to excessive weight loss and severe overheating and dehydration. One or a
combination of these factors can cause death.
11. Focus on Evolution
(a) The presence of ATP synthases in all of these places suggests the likelihood of common
ancestry. As described in Chapter 4, the endosymbiont theory says that mitochondria evolved
from a free-living, aerobically respiring, nonphotosynthetic prokaryote that was engulfed by an
ancestral eukaryotic cell. It further holds that chloroplasts evolved from a free-living,
photosynthetic prokaryote engulfed in the same way. (b) One would expect to observe a high
degree of similarity when comparing the amino acid sequences for the ATP synthases of
mitochondria, chloroplasts, and prokaryotes living today. Observation of this similarity would
support the hypothesis that these eukaryotic organelles and prokaryotes share common ancestry.
Observation of a relatively low degree of similarity between the amino acid sequences of these
ATP synthases would fail to support the stated hypothesis.
12. Focus on Organization
Sample key points:
 An electron transport chain consists of complexes of membrane proteins that transfer
electrons from carrier to carrier in a series of redox reactions.
 Redox reactions release energy as electrons move to lower energy levels, while the protein
complexes shuttle protons across the membrane.
 Thousands of electron transport chains embedded in the inner mitochondrial membrane
establish a concentration gradient of protons.
 Through the energy-coupling process of chemiosmosis, the resulting proton-motive force
drives diffusion of H+ through ATP synthase.
 The organization of proteins in the electron transport chain and of ATP synthase in the inner
mitochondrial membrane are key to this process.
 The arrangement of proteins in the ATP synthase complex allow them to function.
Sample top-scoring answer:
Novel properties emerge from the arrangement and interaction of parts at each level of the
biological hierarchy. An electron transport chain consists of a sequence of electron carrier
molecules embedded in the inner mitochondrial membrane that transfer electrons through a
series of redox reactions, while shuttling protons across the membrane. The thousands of
electron transfer chains create a concentration gradient of H+. This proton-motive force drives
ATP formation by ATP synthase via chemiosmosis. It is the organization of the electron carrier
proteins in the electron transport chain and the location of ATP synthase that allows ATP
synthesis to occur. The function of the ATP synthase complex is also due to the specific, ordered
arrangement of the proteins that make up the complex. Thus, oxidative phosphorylation is an
emergent property resulting from the structural and functional organization of component parts,
none of which would be able to synthesize ATP on its own.
13. Synthesize Your Knowledge
Coenzyme Q (“Q” in Figure 7.14) is a small hydrophobic molecule that is part of the electron
transport chain in mitochondria, involved in shuttling electrons during oxidative
phosphorylation. This process occurs at the end of cellular respiration and helps produce ATPs
for use by the cell. As you might imagine, cells that are particularly active—such as heart muscle
cells—require additional coenzyme Q. The product shown in the photo therefore functions as a
nutritional supplement by providing extra molecules for the electron transport chain. (Although
the body can synthesize coenzyme Q, various disorders, including heart disease, are correlated
with a deficiency of this molecule.) However, coenzyme Q is not a fuel because it is not
consumed by cells; instead, it is used over and over again as an electron shuttle.
CHAPTER 8 PHOTOSYNTHESIS
Scientific Skills Exercise
Teaching objective: The primary objective of this exercise is to give students practice in
working with scatter plots and regression lines. Students are asked to describe trends shown by
the regression lines, to use the regression lines to estimate values, and to calculate the percentage
change of values. The student is further asked to consider the implications of the trends
expressed by the regression lines.
Teaching tips: In plotting data in the scatter plot, students may wish to connect the data points.
However, because the objective is to determine trends rather than to see the rate of change
between each individual point, regression lines should be used. Students can try “eyeballing” a
line, and then may want to check it by using Excel. They can input data into cells using the tools
within Excel to plot the data and insert regression lines into their scatter plot. Tutorials for
graphing in Excel can be found online. This exercise has been simplified in order to teach about
regression lines. In actuality, a scientist would draw a regression line for a set of raw data points,
rather than points representing averaged data.
Answers:
1. (a) The researchers intentionally varied the levels of CO2 concentration. Therefore, CO2
concentration is the independent variable and should go on the x-axis. The researchers measured the
dry mass of corn and velvetleaf plants, so mass is the dependent variable and should go on the yaxis. (b) See the scatter plot in answer #2.
2.
3. (a) As shown by the regression lines in the scatter plot, the dry mass of corn decreased with
increasing concentrations of CO2. In contrast, increasing concentrations of CO2 produced a marked
increase in the dry mass of velvetleaf as compared to the decrease in the mass of corn. (b) Because
velvetleaf is invasive to cornfields, one would expect that velvetleaf would compete with corn for
water and nutrients, and therefore would negatively impact the productivity of corn plants, reducing
crop yield.
4. (a) At 390 ppm CO2, the average dry mass of a corn plant is approximately 91 g, and velvetleaf is
about 38 g. At 800 ppm CO2, the average dry mass of a corn plant would be approximately 84 g, and
velvetleaf would be about 50 g. (b) At a CO2 concentration level of 800 ppm, the dry mass of corn
would decrease by about 8% [(84 g - 91 g)/91 g x 100]. For velvetleaf, the dry mass would increase
by about 32% [(50 g - 38 g)/38 g x 100]. (c) These results support the conclusion from other
experiments that C3 plants grow better under increased CO2 concentration than C4 plants because
velvetleaf, a C3 plant, grew better at higher CO2 levels than corn, a C4 plant.
Interpret the Data
Figure 8.9 About 400–450 nm and 670–680 nm drive the highest rates of photosynthesis. (The first
peak of the graph in (b) is quite broad.)
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and a suggested rubric at the
beginning of this document.
9. Science, Technology, and Society
Trees in tropical rain forests carry out photosynthesis, fixing CO2 as sugar and storing carbon as
they synthesize new biomass. However, trees also respire, using sugar as an energy source and
releasing CO2. While photosynthesis takes place in green leaves, respiration takes place
throughout the tree, in its leaves, stems, branches, and roots. Thus, living trees are not only sinks
for storing CO2, but also sources for the release of CO2. In addition, when a tree dies or its leaves
are eaten by an herbivore, microbes or animals digest its biomass and the sugars are broken
down by respiration to release CO2. Furthermore, mass cutting and burning of forests adds
significantly greater amounts of CO2 to the atmosphere, not offset in any way by the CO2
absorption that a living tree would be carrying out. (So the net result of cutting and burning
forests is to increase the atmospheric CO2.)
10. Draw It
The ATP would end up outside the thylakoid. The thylakoids were able to make ATP in the dark
because the researchers set up an artificial proton concentration gradient across the thylakoid
membrane; thus, the light reactions were not necessary to establish the H+ gradient required for
ATP synthesis by ATP synthase.
11. Focus on Evolution
Because the “ancestor” of the chloroplast was a photosynthetic prokaryotic cell according to the
endosymbiont theory, you would predict that the sequences of chloroplast DNA for ribosomal
RNAs in a chloroplast would be more similar to those of a photosynthetic prokaryote than to
those in the nucleus of a plant cell. If this prediction is correct, it adds to the considerable body
of evidence supporting the endosymbiont theory and tells us that photosynthesis first evolved in
prokaryotes, and at least one eukaryotic cell may have taken up a photosynthetic prokaryote,
becoming the ancestor of eukaryotic cells that contain chloroplasts. (In fact, when genes
encoding ribosomal RNAs are sequenced, the chloroplast's ribosomal RNA genes are much more
similar to those from photosynthetic prokaryotic species than to those from the nuclei of plant
cells.)
12. Focus on Evolution
Assuming that modern soybeans are the result of breeding programs meant to maximize the size
(mass) of the harvested soybeans, and assuming that soybean mass is due mostly to stored
carbohydrate (or matter ultimately derived from carbohydrate), it would seem logical that crop
scientists would have selected strains with reduced levels of photorespiration. This means that
the wild relatives of modern soybeans would be expected to have even higher levels of
photorespiration than 50%.
13. Focus on Energy and Matter
Sample key points:
 The light reactions take place in the thylakoids, where photosystems and electron transport
chains are embedded in membranes that enclose the thylakoid space.
 The energy of light is used to pump electrons from water through photosystem II, an electron
transport chain, and photosystem I to NADPH.
 In redox reactions in the electron transport chain, protons are pumped into the thylakoid
space. ATP synthase uses this proton-motive force to make ATP.

The Calvin cycle, which occurs in the stroma, uses the energy of ATP and the reducing
power of NADPH produced by the light reactions to reduce CO2 to sugar.
 The three stages of the Calvin cycle are carbon fixation (using rubisco to attach CO2 to
RuBP), reduction (using ATP and electrons from NADPH), and regeneration of the CO2
accepter (using energy from ATP). The three-carbon sugar, G3P, exits the cycle and is
converted to glucose and other organic molecules.
Sample top-scoring answer:
The energy transformations of photosynthesis take place in chloroplasts. The light reactions
occur in the thylakoids, where photosystems containing light-harvesting complexes of pigment
molecules, a reaction-center chlorophyll a, and a primary electron acceptor are embedded. When
a pigment absorbs a photon, an electron is boosted to a higher-energy state. The energy of light is
used to pump electrons from water through photosystem II, an electron transport chain, and
photosystem I to NADPH. During energy-releasing redox reactions in the electron transport
chain, protons are pumped into the thylakoid space. ATP synthase uses this proton-motive force
to make ATP. The Calvin cycle occurs in the stroma. CO2 is fixed into organic compounds,
which are reduced using the energy of ATP and electrons from NADPH. The three-carbon sugar
G3P is produced and can be converted to sugars and other organic molecules, ultimately
providing the chemical energy and carbon source for all of life.
14. Synthesize Your Knowledge
This green alga is photosynthetic, so it must have chlorophyll, which appears green. However, in
addition to chlorophyll, it must have accessory pigments that absorb a different wavelength,
including one at high levels that appears reddish pink in visible light. Given that UV light levels
are so high in the locations where these algae are located, this might be a carotenoid pigment that
has a photoprotective function. (In fact, this is one of the carotenoid pigments that has been
proposed to provide photoprotection, consistent with its high levels in these algal cells.
Carotenoids and other accessory pigments are also responsible for the brilliant reds, oranges, and
yellows of fall foliage on deciduous trees. In the summer, these carotenoids and accessory
pigments are at lower levels than the chlorophyll, and their colors are revealed only in the fall
when leaves stop photosynthesizing and the chlorophyll is broken down.)
CHAPTER 9 THE CELL CYCLE
Scientific Skills Exercise
Teaching objective: Students will practice reading a histogram in this exercise. The students
then relate the data to phases of the cell cycle based on relative amounts of DNA.
Teaching tips: A version of this Scientific Skills Exercise can be assigned in MasteringBiology.
The method used to measure DNA content in this experiment is flow cytometry, also sometimes
called fluorescence-activated cell sorting (FACS). The fluorescent dye (propidium iodide in this
example) binds to the cell’s DNA, so the strength of the fluorescence signal corresponds to the
amount of DNA the cell contains. The flow cytometer instrument passes individual cells through
a small tube that counts a cell while recording the cell’s size and fluorescence level. The analysis
software calculates how many cells in each sample had a given range of fluorescence and graphs
the relationship as cell frequency (or cell count) versus fluorescence interval. In the experiment
producing the results shown here, the researchers ran 10,000 cells through the flow cytometer for
each treatment.
The concept of dependent and independent variables is not obvious in this exercise. The x-axis,
fluorescence level, is the independent variable: The researcher specifically set the instrument to
count cells within each small interval of fluorescence along the x-axis. The number of cells thus
becomes the dependent variable, as the value on the y-axis depends on the fluorescence interval
within which cells are being counted. The treatment versus control is a separate level of analysis
and is presented in two histograms for that reason.
Students may need some help with the idea of frequency data in the form of cell counts. Students
may be used to seeing DNA content of cells over time, and may tend to fall into thinking of the
x-axis as representing time and the height of the peak as representing DNA content. A clearer
way for students to think of it is that in the case of the cell cycle data seen here, there are
basically two states that most cells will be in: G1 (less DNA) and G2 (more DNA). The spread of
cells in between these two states represents cells in S phase, transitioning from the low to high
amount of DNA. After a cell enters mitosis, it will have the doubled DNA content until it goes
through cytokinesis, at which point it will be counted as two cells, and those cells would
immediately show up in the G1 peak to the left along the x-axis. An analogy to use: If you were
to count the number of students in two sequential classrooms along a hallway at any one time,
you get relatively high head counts for each classroom location. But there will be some students
walking from classroom G1 to classroom G2; they are in transition between locations and will
represent a low head count spread out along the hallway between the two rooms (S phase). The
M to G1 transition would be like teleporting back to the first classroom and never showing up in
the hallway at all!
Students may note that while there are distinct peaks for G1 and G2, there is still a range of DNA
content for each state. This most likely represents differences in dye uptake and binding
efficiency of individual cells in the cell population, or damaged cells.
Answers:
1. (a) The relative amount of DNA is shown on the x-axis. Because the fluorescent dye is binding
to DNA in each cell, the amount of fluorescence corresponds to the amount of DNA (the more
DNA, the more fluorescence). (b) The second peak (region C) represents the group of cells with
the higher amount of DNA per cell because the cells at the second peak have a higher level of
fluorescence. (Note that the peak in C is shorter, because it has fewer cells, but each cell has a
higher level of fluorescence, thus more DNA.)
2. (a) The first peak (region A) represents cells in G1, when the cells have only one copy of
DNA; the second peak (region C) represents cells in G2—as the cell prepares to divide, there are
two copies of the DNA; in S (region B), the cells are synthesizing DNA, and so have a quantity
of DNA between the G1 and G2 phase.
(b) The S-phase group of cells does not have a distinct peak because as the cells transition from
G1 to G2, they are all “caught” at different stages of DNA synthesis and thus have different
amounts of fluorescing dye/DNA present.
3. (a) The G1 phase of the cell cycle has the greatest number of cells because the highest peak in
the treated sample corresponds to the lower fluorescence level, comparable to the G1 cells in the
control sample. (b) Most of the treated cells are in G1 instead of in S or G2, which strongly
suggests that treatment causes the cell cycle to arrest at G1. (c) The G1 checkpoint might be
inhibited; molecules could be affected at any stage of the relevant signal transduction pathway
(signaling molecules, receptors, second messengers, or proteins involved in the cellular
response). For example, the inhibitor could block action of a G1 checkpoint protein or a cyclin.
The inhibitor could block DNA replication at an early step while allowing other later events to
occur, so the cells “look” like they’re still in G1 on the graph when they are really past the G1
checkpoint. Or the inhibitor could be acting as a signal that triggers the cell to enter G0.
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and the suggested rubric at the
beginning of this document.
8. Scientific Inquiry
(a) The motor proteins at the kinetochore ends of kinetochore microtubules walk the
chromosomes toward the poles, so these motor proteins must be minus end-directed. In cells in
which the “reeling-in” mechanism is observed, plus end-directed motor proteins must be present
at the poles. (b) The motor proteins between associated nonkinetochore microtubules must move
toward the middle of the spindle to push the microtubules away from each other, so they are plus
end–directed.
9. Focus on Evolution
Natural selection favors organisms with attributes that lead to most offspring, and cell division is
a crucial function for both unicellular and multicellular eukaryotes. Although the number of
chromosomes would be correct in each daughter cell, the two cells would not contain identical
genetic information. Imagine that a cell had ten chromosomes and that they were divided equally
in number, five each into two daughter cells, followed by chromosome duplication. Each
daughter cell would end up with ten chromosomes but would have double copies of whatever
genetic information was on the five chromosomes it happened to receive from the parent cell.
We have noted in this chapter, though, that the chromosomes are made up of two sets, one from
each parent. (You’ll learn more about this in Chapter 10.) Therefore, if the random division of
chromosomes resulted in one parental set going into each daughter cell, the daughter cells might
each end up with two complete sets of genetic information. However, the probability of this
happening is very low—much like putting five differently colored pairs of socks into a bag,
blindly reaching in to pull out five socks, and having them be five different colors. As a
consequence, the vast majority of the daughter cells would not have a complete set of genetic
information. They would therefore most likely not be viable, and would not be able to survive
and reproduce.
10. Focus on Information
Sample key points:
 The continuity of life depends on cells dividing and faithfully passing heritable information
to daughter cells.
 The chromosomes of a eukaryotic cell duplicate in the S phase.
 The stages of mitosis include prophase, prometaphase, metaphase, anaphase, and telophase.
 In this sequence of stages, the chromosomes condense, a spindle forms, chromosomes line up
 on the metaphase plate, sister chromatids separate and move to opposite ends, and nuclei reform.
 Cytokinesis divides the cytoplasm, forming two genetically identical daughter cells.
Sample top-scoring answer:
The continuity of life—reproduction of unicellular organisms and the development, growth, and
repair of multicellular organisms—depends on the transmission of identical copies of heritable
information as cells divide. The DNA of a eukaryotic cell is divided among several
chromosomes. In the S stage of the cell cycle, DNA is replicated, and each duplicated
chromosome now consists of identical sister chromatids. Following a G2 phase of growth, the
cell divides. In the highly orchestrated stages of mitosis (prophase, prometaphase, metaphase,
anaphase, and telophase), the chromatin of the duplicated chromosomes condenses, a mitotic
spindle forms, and spindle microtubules attach to the kinetochore of each sister chromatid,
moving the duplicated chromosomes to the metaphase plate. The sister chromatids separate and
move to opposite ends of the cell. Nuclear envelopes re-form, completing mitosis, and
cytokinesis divides the cytoplasm. A cell has divided to form two genetically identical cells.
11. Synthesize Your Knowledge
Cancer cells divide without being subject to the usual cell cycle controls. They can divide in the
absence of growth factors or other molecules that are necessary for normal cells to pass
particular checkpoints. Furthermore, cancer cells are not inhibited by density and do not require
anchorage to a surface in order to divide. They may not respond appropriately to signals that
normally trigger apoptosis, such as those resulting from mistakes in DNA replication. The
underlying basis for this altered behavior is a series of genetic and cellular changes, including
mutations in genes whose protein products normally regulate the cell cycle. These gene products
are often proteins that function in cell signaling pathways.
CHAPTER 10 MEIOSIS AND SEXUAL LIFE CYCLES
Scientific Skills Exercise
Teaching objective: Students will identify the independent and dependent variables in a table of
data and graph the data to identify phases of meiosis in a yeast culture. They will also get
practice doing calculations with exponents, such as calculating the haploid genome size for this
species of yeast (Saccharomyces cerevisiae, commonly known as budding yeast).
Teaching tips: A version of this Scientific Skills Exercise can be assigned in MasteringBiology.
This is a basic graphing exercise with time-related data. Students may have difficulty with the
extremely small units—femtograms (fg)—used for the DNA. A review of metric units will be
helpful to them. If you want your students to have more practice with calculations, you could
extend question 4 by asking questions such as how many base pairs are in a nanogram (ng) or
how many kilobases (kb) are in a femtogram, etc.
The points at which samples were taken are not all 1 hour apart. This means that the students will
have to decide what interval to use on the x-axis for the tick marks. It doesn’t matter whether
they use 0.5 or 1 hour intervals so long as they are consistent and put the dots in the correct
locations. You can explain that the researchers collected samples more frequently during the
stages when the DNA content was changing, to better resolve when the events occurred.
Students may question why the line on the graph is not flat for each stage of meiosis. This is
because the data come from a population of cells, and there is some variation among the yeast
cells in their timing of progression through the process. This also explains why the amount of
DNA does not drop instantly at the end of MI and/or MII phase: Not all of the cells divide at
exactly the same time and so the average includes some pre-division cells, some dividing cells,
and some post-division cells for the relevant time points. This subtlety is addressed in question
3(d). The more advanced students will be able to think this through and realize that because the
y-axis is the amount of DNA per cell, the cell will have the duplicated amount of DNA (48 fg) in
one cell until two cells form at the end of meiosis I. At this point, the DNA content drops in half.
If you were graphing the DNA content of a single cell, this line would drop vertically to about 24
fg. Because this is a population of cells and they are completing MI at slightly different times,
given that the line represents the average of the population, it gradually decreases as a diagonal
rather than a vertical drop.
Students will not be able to precisely delineate each stage on the graph due to the population
variability. Instruct them to approximate based on what they expect for DNA content at each
stage.
Answers:
1. (a) The independent variable is time, which goes on the x-axis, measured in hours (hr). The
dependent variable is average amount of DNA per cell, which goes on the y-axis and is measured
in femtograms (fg). (b) The x-axis has data points over 14 hours, so it makes sense to label a tick
for each hour. (It would also be fine to label a tick for each half-hour, as long as all were labeled,
but this could look cluttered.) The y-axis has data points as high as 48.0 fg, so a logical decision
would be to label tick marks every 5 fg up through 50 or 55 fg. See the graph in the answer to
question 3.
2. See the graph in the answer to question 3.
3. (a) There are 24 fg of DNA in G1 (at 0.0 hr). (b) There should be 48 fg in G2; 24 fg at the end
of MI; and 12 fg at the end of MII. (c)
(d) Because the y-axis is the amount of DNA per cell, the cell will have the duplicated amount of
DNA (48 fg) in one cell until two cells form at the end of meiosis I. Therefore, the “corner” in
the data line represents the stage at which the cell divides at the end of meiosis I (cytokinesis). At
this point, the DNA content in each cell drops in half. If you were graphing the DNA content of a
single cell, this line would drop vertically to about 24 fg. Because this is a population of cells—
and they are completing MI at slightly different times—given that the line represents the average
of the population, it gradually decreases as a diagonal rather than a vertical drop.
4. (a) Each haploid cell has 12 fg of DNA. (12 fg of DNA) x (9.78 x 105 base pairs per fg) = 1.2
x 107 or 12 x 106 base pairs, therefore there are 12 megabase pairs (Mb) of DNA in each haploid
cell. (b) The answer for this will depend on the length of the S phase in the student’s graph. The
answer is 1.2 x 107 base pairs (haploid value) x 2 (for diploid value) divided by the number of
minutes of the S phase. For the example shown, (1.2 x 107 base pairs)(2)/120 minutes = 200,000
(or 2.0 x 105) base pairs per minute.
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and a suggested rubric at the
beginning of this document.
7. Scientific Inquiry
The alleles for freckles or no freckles are on all four chromatids at the F locus, and those for
black or blond hair are at the H locus. Let’s assume that the alleles on the red chromatids are for
freckles and black hair and the alleles on the blue chromatids are for no freckles and blond hair.
By drawing the chromatids, you can follow the alleles through meiosis. The chromatids that did
not participate in crossovers will donate either freckles and black hair alleles or no freckles and
blond hair alleles (which you can determine by the color of the chromatid). The chromatids that
were involved in crossovers have the other possible combinations, so all four possibilities will
exist in the gametes from this meiotic division:
 freckles, black hair (If you drew the chromosomes in the gametes after meiosis II, the
chromosome in this gamete will be red at both F and H.)
 no freckles, blond hair (blue at both F and H)
 freckles, blond hair (red at F and blue at H)
 no freckles, black hair (blue at F and red at H)
No other possible combinations exist, so no additional combinations will be seen in gametes
from other meioses in this individual. (If we tried the other assumption, that the alleles on the
blue chromatids were for freckles and black hair and those on the red chromatids were for the
other traits, the results would be the same with the colors reversed.)
8. Focus on Evolution
When the environment becomes unfavorable, this means it is unfavorable for a particular
genotype or genome that had been successfully adapted to the previous environment. At such
times, it becomes advantageous to switch from asexual reproduction, which results in genetically
identical daughter cells (all of which would be at a disadvantage), to sexual reproduction, which
introduces genetic variation into the daughter cells. Doing so increases the likelihood that at least
some of the individuals in the next generation will have genetic combinations that allow them to
survive and reproduce under the new conditions.
9. Focus on Information
Sample key points:
 Genes are stretches of DNA in chromosomes that encode traits and are passed on from parent
to offspring.
 Chromosome behavior during sexual reproduction accounts for both similarity and variation
among offspring.
 Inheritance of genes accounts for offspring exhibiting traits similar to those of each parent.
 In sexual reproduction, meiosis alternates with fertilization.
 Crossing over and independent assortment of chromosomes during meiosis lead to genetic
variation among gametes.
 Random fertilization increases the genetic variation among diploid offspring.
Sample top-scoring answer:
Genes encoding traits exist as stretches of the DNA molecule in a chromosome. Chromosome
behavior during meiosis and fertilization—processes that alternate in sexual reproduction—
accounts for both the inheritance of traits and the generation of genetic variation. Meiosis in each
parent generates haploid gametes with a single set of chromosomes. During this process,
crossing over and independent assortment of chromosomes result in many gametes with novel
combinations of alleles. Any pair of gametes, each containing chromosomes from one parent,
can fuse during fertilization to make a diploid cell. This random fertilization further augments
the number of different possible combinations of traits that can be obtained. The zygote develops
through mitosis into a multicellular diploid offspring. Because the offspring inherits genes from
both parents, it exhibits traits of each, in new combinations. Thus, both meiosis and fertilization
ensure inheritance of parental traits as well as genetic variation among offspring.
10. Synthesize Your Knowledge
With three sets of chromosomes, there is no way for the sets to be separated equally to gametes
during meiosis, and so seeds (which are embryos formed after meiosis and fertilization) cannot
form. Thus, taking cuttings from the plants and rooting them is the only way new plants can be
grown. Sexual reproduction generates genetic diversity, whereas asexual reproduction generally
does not. Genetic diversity would allow more genetic variants to be present. Some of the genetic
variants might have a particular combination of genes that would allow them to resist infection
by the fungus. These genetic variants would not be possible in a population of plants that
reproduces (or is reproduced by growers) asexually. (In fact, this is one argument against
agricultural “monocultures” that are propagated by cloning—they are often less resistant to
bacterial, fungal, and other pests.)
CHAPTER 11 MENDEL AND THE GENE IDEA
Scientific Skills Exercise
Teaching objective: Students will create a histogram showing the distribution of phenotype
frequencies in a population. The data are normally distributed in a typical bell curve.
Teaching tips: A histogram is related to a bar graph. A bar graph visually summarizes
categorical data, whereas a histogram is used to summarize numeric data. From a practical
perspective, a histogram is really a bar graph of numeric data that has been "binned" into
categories (numeric ranges; in this case the data themselves are integer values so each bin
represents the range from 0 to 0.99, 1.0 – 1.99, etc.). One visible difference between the two is
that one usually creates gaps between bars in a bar graph to make it easy to see the distinct
categories, while in a histogram, bars often touch to indicate the natural ordering and continuum
of the numeric values on the x-axis.
Drawing the curve that fits the histogram in this exercise introduces the concept of normal
distribution around a mean value. With a bell curve, one half of the curve is a mirror image of
the other with the peak in the center. (Note also that the area below the curve should equal the
total area of all the bins, so the curve should intersect each bar in the middle, and the middle bar
will be intersected twice.) In general, the more variation there is in the data set, the wider and
shorter the “bell” will be in the graph; the less variation, the higher the central peak and the
narrower the bell will be. If all of the genes contribute equally to the phenotype, and no
combinations are favored, then the offspring should show a normal distribution. If these
conditions are not met, then the histogram and the corresponding curve will deviate from normal.
The students can “visualize” the skin colors on each side of the two halves of their graph by
drawing a diagonal line from the upper right corner to the lower left corner of the Punnett square.
Most of the offspring have skin color that is close to those on the line; the corners farthest from
the diagonal line show the colors with lowest frequency. Note that this method works for this
particular example because of the order that the parental alleles are placed along the top and side,
and entered into the Punnett square cells, situating the triply homozygous individuals at opposite
corners.
When the lethal bb genotypes are removed from the population in question 3 (a purely
hypothetical exercise), the resulting distribution deviates from normal. (Determining whether the
distribution is “far enough” from what is expected in order to conclude that the distribution is not
normal requires statistical analysis, which is not covered in this exercise.) The skew of the
distribution in this hypothetical exercise is visually evident because more genotypes were
removed from one side of the diagonal line across the Punnett square than from the other; see the
answer to question 3.
Answers:
1.
2. (a) The skin color that results from three dark-skin alleles is most common (has the highest
frequency), with 20 out of 64 offspring carrying this genotype.
(b) It has a normal distribution, or bell curve. The shape of the curve is consistent with what
would be expected of a mating between AaBbCc heterozygotes, where a majority of the
offspring would have intermediate phenotypes.
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and the suggested rubric at the
beginning of this document.
15. Focus on Evolution
It should reduce the frequency of the lethal dominant allele in the population because if waiting
until a later age to reproduce allows a person to become aware that he or she possesses a lethal
dominant allele, there is increased likelihood that such a person will choose not to reproduce,
16. Scientific Inquiry
(a) TTAA, TTAa, TtAA, TtAa
(b) Testcross the mystery plant with a homozygous recessive plant (ttaa, with dwarf stems and
terminal flowers), using the procedure shown in Figure 11.2. Plant the seeds, and when the
plants grow, analyze them for the two characters of stem length and flower position.
(c) You can do this by drawing Punnett squares. Performing a cross is the process of mating two
varieties of organism, as shown in Figure 11.2. Drawing a Punnett square is simply making a
prediction of the possible outcomes of an actual cross.
(d) Matching the actual phenotypic ratios of the offspring to one of the Punnett squares indicates
the genotype of the mystery plant. You can analyze the results separately for each gene. If all
offspring are tall, the mystery plant must be TT; if half are dwarf, it must be Tt. If all
offspring have axial flowers, the mystery plant must be AA; if half have terminal flowers, it
must be Aa.
17. Focus on Information
Sample key points:
 Genes are stretches of DNA encoding characters that are passed on from parent to offspring.

The behavior of genes during sexual reproduction accounts for both similarity and variation
among offspring.
 The two alleles of a gene segregate during meiosis, with each gene assorting independently
of other genes.
 Inheritance of genes accounts for offspring exhibiting traits similar to those of each parent.
 Segregation and independent assortment result in new combinations of alleles in gametes,
and random fertilization results in even more in offspring, accounting for variation.
Sample top-scoring answer:
The behavior of genes accounts for both similarity and variation in inheritance. Genes encoding
characters exist as stretches of DNA that behave as “particles” during sexual reproduction. The
genotype of each diploid parent includes two alleles of each gene, each allele coding for a variant
trait; the two may be the same (homozygous) or different (heterozygous). The two alleles
segregate into different gametes during meiosis, and the alleles of one gene assort independently
of those of many other genes, resulting in gametes with unique allele combinations. Random
fertilization then increases the number of possible genetic combinations in offspring. Because an
offspring inherits genes from both parents, it exhibits traits of each. If different alleles are present
in the two parents, offspring have novel combinations of alleles (and of traits) not seen in either
parent. Thus, the behavior of genes during sexual reproduction ensures inheritance of parental
traits as well as genetic variation among offspring.
18. Synthesize Your Knowledge
The gene for “shirt-striping” has two alleles: SH for horizontal and SV for vertical. Because both
traits show up in the offspring, the alleles are codominant. The male is SHSH, the female is SVSV,
and the child is SHSV.
CHAPTER 12 THE CHROMOSOMAL BASIS OF INHERITANCE
Scientific Skills Exercise
Teaching objective: Students will calculate the χ2 value for a given data set and determine
whether or not to reject a hypothesis about the data.
Teaching tips: A version of this Scientific Skills Exercise can be assigned in MasteringBiology.
Students often stumble on the step of calculating expected numbers of offspring for each
category. It will be useful to remind them that “1:1:1:1” actually means “1/4:1/4:1/4:1/4” and
“3:1” means “3/4:1/4”, etc., so they see that the numbers in the ratios are fractions of the whole
population of offspring.
Note that the same data set can be used to test the opposite hypothesis that the two genes are
tightly linked on the same chromosome. The expected ratios would be 1:1:0:0 as shown in Figure
12.9, and the rest of the χ2 test would be carried out the same way. The χ2 value will fall in the
significant zone, so that the hypothesis (of being linked) should be rejected, leading to the same
conclusion that the genes are assorting independently. This is a good exercise for your more
thoughtful students.
The term “recombination” can be confusing because it represents two different ways of obtaining
recombinant (non-parental) phenotypes: One way is for genes on two different chromosomes to
move independently of each other and the other is the result of crossing over between two linked
genes. Make sure your students are clear about use of the terms “parental type” and
“recombinant type”. A parental phenotype is the same as the phenotype of either one of the P
generation parents, while a recombinant phenotype is a phenotype that is different from both.
For the situation where the two genes are on the same chromosome (linked), the P parent’s
phenotype reflects the genotype of that chromosome. (Both of a P parent’s chromosomes are the
same, since the P generation parents are true-breeding; that is to say, homozygous.) If that
chromosome undergoes no crossing over in the F1 dihybrid that is testcrossed, the F2 offspring
will show that “parental” phenotype because the testcross parent contributes only recessive
alleles. If crossing over occurs in the F1 dihybrid parent, a new (recombinant) chromosome will
be present in the offspring, which will thus have a new (recombinant) phenotype, different from
that of both P generation parents.
Another way of explaining the χ2 test is to say that χ2 may be used to measure the probability that
this data set could have been obtained by random sampling from a population for which the
hypothesis is correct. If the probability is less than 5%, it is very unlikely that the observed data
could have resulted randomly if the hypothesis were true, so the hypothesis must be rejected.
This exercise avoids getting into the fine details of the χ2 test. You can elaborate on its strengths
and weaknesses, and underlying assumptions, according to class level and your preference.
Some important considerations, mentioned briefly in the exercise, are sample size and the
strength of the linkage (how close the two genes are, if they are linked). With small sample sizes,
only very strong linkages (genes that are very close) will be detectable, while with larger sample
sizes we are capable of detecting even weak linkages. For this reason, we must be cautious with
our conclusions when we obtain large χ2 probabilities. We rarely have sufficient evidence to
prove our hypothesis, but rather we simply fail to disprove it. That is, our failure to detect any
linkage may simply be due to our “not looking hard enough”. Numerically, for samples of size
900 (as in our example), if there was fairly strong linkage between the genes (i.e., an expected
2:1:1:2 ratio for the population, where the “2”s represent the parental classes and the “1”s the
recombinants), the χ2-test would almost certainly detect the difference. However, if the linkage
were weaker (e.g., an expected 10:9:9:10 population ratio), we would only have about a 25%
chance of detection. That is, with genes that are weakly linked, there would only be about a 25%
chance that the χ2-test would yield a probability less than 0.05 so that we could (rightfully) reject
the hypothesis that the genes are unlinked.
Answers:
1.
Offspring from
testcross of AaBb
Purple stem/
short petals
Green stem/
short petals
Purple stem/
long petals (A-
Green stem/
long petals
(F1) x aabb
Expected ratio if the
genes are unlinked
Expected number of
offspring (of 900)
Observed number of
offspring (of 900)
(A-B-)
1
(aaB-)
1
bb)
1
(aabb)
1
(1/4)x900=225
(1/4)x900=225
(1/4)x900=225
(1/4)x900=225
220
210
231
239
2.
Testcross offspring
(A-B-)
(aaB-)
(A-bb)
(aabb)
Expected
(e)
225
225
225
225
Observed
(o)
220
210
231
239
Deviation
(o - e)
-5
-15
6
14
(o - e)2
25
225
36
196
(o - e)2/e
0.11
1.00
0.16
0.87
χ2 = SUM 2.14
3. (a) χ2 = 2.14 lies between 1.42 and 2.37. (b) If the hypothesis were rejected, there would be a
50–70% probability of being incorrect, which is beyond the standard cut-off (i.e., p > 0.05).
Because the χ2 value is in the range of “nonsignificant” differences between observed and
expected values, the hypothesis should be accepted. (We must, however, keep in mind the caveat
that this exercise has been simplified by not evaluating sample size—in an actual study, the
sample size would have to be increased about 10-fold to be able to accept the hypothesis with
confidence.)
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and the suggested rubric at the
beginning of this document.
10. Scientific Inquiry
(a) For each pair of genes, you had to generate an F1 dihybrid fly; let’s use the A and B genes as
an example. You obtained homozygous parental flies, either the first with dominant alleles of the
two genes (AABB) and the second with recessive alleles (aabb), or the first with dominant alleles
of gene A and recessive alleles of gene B (AAbb) and the second with recessive alleles of Gene A
and dominant alleles of gene B (aaBB). Breeding either of these pairs of P generation flies gave
you an F1 dihybrid, which you then testcrossed with a doubly homozygous recessive fly (aabb).
You classed the offspring as parental or recombinant, based on the genotypes of the P generation
parents (either of the two pairs described above). You added up the number of recombinant types
and then divided by the total number of offspring. This gave you the recombination percentage
(in this case, 8%), which you can translate into map units (8 map units) to construct your map.
(b)
11. Focus on Evolution
It has been suggested that the regions of internal homology within the Y chromosome may pair
with each other during meiosis I, exchanging genetic information. This exchange could allow the
genes on the Y chromosome to change or “evolve” even though the Y chromosome doesn’t have
a true homologous partner during meiosis. This recent discovery has changed our view of the Y
chromosome and its future prospects.
12. Focus on Information
Sample key points:
 A chromosome consists of one DNA molecule, carrying the heritable information, along with
associated proteins.
 Genes are stretches of DNA along the chromosome.
 In asexual reproduction, the behavior of chromosomes during mitosis ensures faithful
passage of genetic information to both daughter cells.
 In sexual reproduction, the behavior of chromosomes accounts for both the generation of
variation (crossing over and independent assortment) and faithful passage of the exact
complement of genetic information in a specific number of chromosomes to offspring.
Sample top-scoring answer:
Chromosomes contain the DNA that carries the heritable information. One chromosome is made
up of one DNA molecule and associated proteins. Genes, in different forms called alleles, are
present as stretches of DNA at specific locations along the chromosomal DNA molecule. The
behavior of chromosomes during mitosis and meiosis is responsible for the distribution of
genetic information during inheritance. In asexually reproducing species, mitosis ensures faithful
passage of exact genetic information to each of the two daughter cells. In sexually reproducing
species, meiosis produces new genetic combinations in haploid gametes as a direct consequence
of chromosomal behavior: Crossing over between homologs and independent assortment of
homologous pairs both lead to variation in allele combinations. At the same time, the meiotic
process ensures that each organism inherits an exact number of chromosomes and thus the
complete complement of necessary genetic information. By carrying and passing along the DNA
in a precise manner, chromosomes and chromosomal behavior ensure the continuity of life.
13. Synthesize Your Knowledge
A male butterfly zygote would be XX. During the first cell division, if nondisjunction occurred
for one of the X chromosomes, the two daughter cells would have been XXX and XO. The XXX
cell would have given rise to the male half of the butterfly, and the XO cell would have given
rise to the female half. (This is in fact the case.)
CHAPTER 13 THE MOLECULAR BASIS OF INHERITANCE
Scientific Skills Exercise
Teaching objective: Students will do predictions of base composition in DNA from different
species based on Chargaff’s rules. The calculations include solving for one to three unknowns.
Teaching tips: A version of this Scientific Skills Exercise can be assigned in MasteringBiology.
Because we now know the structure of DNA, students can be puzzled by why the number of A
bases does not exactly match the number of T bases, or C with G bases. This seeming disparity is
because the analytical methods available to do the measurements were not 100% accurate when
the data for many species were being reported. Once full genome sequences became available,
the values showed the predicted 1:1 ratios.
The calculations in this exercise are simple algebraic equations based on a few basic rules: A + T
+ C + G = 100%; A = T and C = G.
Answers:
1. Each species’ DNA has a slightly different percentage of a given base. For example, the
percentage of A is 32.8% for sea urchin and 29.7% for salmon DNA. This illustrates Chargaff’s
rule that the DNA of different species varies in its base composition. Chargaff’s other rule states
that in any given species, the percentage of A is roughly equal to that of T, and the percentage of
C is roughly equal to that of G. For example, sea urchin has about 32% each of A and T and
about 17% of G and C.
2. Correct answers are in italics in the table.
Base Percentage
Source of DNA
Adenine Guanine
Cytosine
Thymine
Sea urchin
32.8
17.7
17.3
32.1
Salmon
29.7
20.8
20.4
29.1
Wheat
28.1
21.8
22.7
28.1 (or
27.4*)
E. coli
24.7
26.0
26.0
24.7
Human
30.4
19.8
19.8
30.1
Ox
29.0
21.0
21.0
29.0
* For wheat: Predicted %T is 28.1 if the student calculates T = A = 28.1. Predicted %T is 27.4 if
the student calculates T = 100% – (A + G + C) = 100% - 72.6% = 27.4%. (Either is correct.)
For E. coli: C = G = 26.0; T = A = 24.7
For human: C + G = 100% - (A + T) = 100% - 60.5% = 39.5%; divided evenly between C and G
is 19.75% each, rounded to 19.8%.
For ox: T = A = 29.0%; C + G = 100% - (A + T) = 100% - 58.0% = 42.0%; divided evenly
between C and G is 21.0% each.
3. Average for A = 29.1, average for G = 21.2, average for C = 21.2, and average for T = 28.9 (or
28.7*). Yes, the average percentage of each base for all species still shows Chargaff’s A-T and
C-G equivalence relationship.
Source of DNA
Base Percentage
Adenine Guanine
Cytosine
Thymine
Sea urchin
32.8
17.7
17.3
32.1
Salmon
29.7
20.8
20.4
29.1
Wheat
28.1
21.8
22.7
28.1 (or
27.4*)
E. coli
24.7
26.0
26.0
24.7
Human
30.4
19.8
19.8
30.1
Ox
29.0
21.0
21.0
29.0
Total
174.7
127.1
127.2
173.1 (or
172.4*)
Average
29.1
21.2
21.2
28.9 (or
28.7*)
*Predicted %T for wheat is 28.1 if the student calculates T = A = 28.1. Predicted %T for wheat
is 27.4 if the student calculates T = 100% – (A + G + C) = 100% - 72.6% = 27.4%. (Either is
correct.) Answers for the total and average for T can also vary as shown depending on how T
was calculated for wheat.
Interpret the Data
Figure 13.29 The height of the peak reflects the number added, consecutively, of a particular
base. The first 25 nucleotides in the sequence are TCAGCGTAAGGTGATGTATAGGGGC.
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and the suggested rubric at the
beginning of this document.
11. Scientific Inquiry
12. Focus on Evolution
This could occur in several ways. The bacteria could decrease their synthesis of protective
pigments (or break down those already produced). This would be analogous to a human
decreasing the production of melanin, a skin pigment that absorbs UV radiation. Alternatively,
the bacteria might decrease the number of DNA repair enzymes in their cytoplasm. A third
possibility is that bacteria might have a mechanism for inactivating the proofreading region of
DNA polymerase. In all three cases, the result would be an increase in the frequency of
mutations. (You can probably think of some other valid possibilities as well.) When the
environment becomes unfavorable for the phenotype resulting from a given genetic constitution
or genotype, it is beneficial to increase variation and generate new genotypes, a small number of
which might make certain individuals better adapted to the new environment.
13. Focus on Organization
Sample key points:
 The heritable information in the DNA double helix is in the form of a sequence of nucleotide
bases along one strand, matched by complementary bases along the other strand.
 This structure allows the molecule to be copied accurately and efficiently.
 There are many possible sequences along the length of a DNA molecule.
 At the same time, the structure of the sugar-phosphate backbone provides some uniformity,
so common sets of proteins can interact with all DNA molecules.
 The combination of uniformity and variation in the structure of the DNA molecule makes it
very well suited for its function of carrying heritable information.
Sample top-scoring answer:
The structure of the DNA molecule is well correlated with its function in carrying heritable
information. DNA is a double helix, carrying information in the form of a sequence of
nucleotides along one polynucleotide strand, matched by nucleotides with complementary bases
along the other strand. The structure of DNA provides an efficient mechanism for accurately
copying each molecule. During replication, the two strands separate, and each acts as a template
for the synthesis of a complementary strand, a process that copies the molecule exactly. While
the number of possible combinations of bases along the length of the molecule is immense, the
overall structure, with the uniform sugar-phosphate backbones on the outside, is consistent from
molecule to molecule. This allows common sets of cellular proteins to interact with the DNA.
The combination in a DNA molecule of a uniform molecular structure with variable elements is
well suited to the job of carrying heritable information for the diversity of species that have
evolved over time.
14. Synthesize Your Knowledge
The uniqueness of a stretch of DNA is based on its sequence of nucleotide bases, which are on
the interior of the double helix. The protein appears to have regions that are closely associated
with the space between the two strands of the DNA double helix, the region between the two
backbones, thus nestled against the base pairs. It is likely that the amino acids in these regions of
the protein are uniquely suited for binding (through weak interactions like hydrogen bonds and
van der Waals forces; see Chapter 3) with specific nucleotide sequences, thus “recognizing”
them specifically. (Scientists hope to exploit these proteins by altering the amino acids, so that
the researchers can direct what sequence will be bound by a particular TAL protein.)
CHAPTER 14 GENE EXPRESSION: FROM GENE TO PROTEIN
Scientific Skills Exercise
Teaching objective: Students will compare DNA sequences to identify conserved sequences and
then relate these sequences to their function. The data are presented in a graphical form called a
sequence logo, a way to visually compare sequences that is particularly information-rich.
Teaching tips: The consensus sequence determination should not be difficult for students. The
concept of “consensus” in this usage is the same as when voting on something: whichever base
has the most “votes” is the one chosen to represent that position in the consensus sequence. Once
the full consensus sequence is recorded, all of the information about how many “votes” each
base got at each position is lost (so you don’t know if it was a close “race” or a landslide).
Sequence logos are used to retain some of the lost information because scientists are often more
interested in the range of possible “votes” in different parts of the sequence than just the
consensus.
A highly readable, short introduction to sequence logos written by Shaner, Blair, and Schneider
is available at:
http://www.ccrnp.ncifcrf.gov/~toms/paper/hawaii/latex/paper.pdf .
Many students will recognize the similarity between sequence logos and Web 2.0 tag or data
clouds. Both use type size and/or color to represent frequency or “popularity” of words or terms
within a given set of information. In the case of sequence logos, the “words” are the bases, but
instead of being randomly arranged in the cloud display, their position along the horizontal and
vertical axes conveys important information regarding sequence and relative frequency. It will be
impossible to distinguish the individual bases in the shortest sequence logo stacks; at best they
can discern the color of the top base to figure out the consensus sequence.
In question 5, students may mistake the ATG as the ribosome binding site and choose it because
it has the tallest sequence logo stacks. The further question about positions 0-2 should help.
Review with them that prokaryotic mRNAs have ribosome binding sites upstream and distinct
from the start codon.
Answers:
1. (a) In the 10 sequences in the sequence alignment at position -9, there are 7 Gs, 2 As, and 1 C,
matching their relative size and placement in the sequence logo—G is larger and on top, A is
next, in the middle, and C is smallest and on the bottom. (b) At position 0, there are 8 As, 1 G,
and 1 T; in the sequence logo, A is largest on the top, and G and T are equally sized below the A
(G is in the middle). At position 1 in the aligned sequences, all 10 are Ts; in the sequence logo, T
is the only base at position 1 and it is quite tall.
2. (a) Positions 1 and 2 have the most predictable bases. Bases T and G, respectively, would
most likely be at those positions in a newly sequenced gene. (b) Positions -18, -17, -16, -12, -11,
-6, -5, -4, -2, 5, 6, and 8 have no bases shown so they have the lowest predictive power. The lack
of bases, or shortness of stack, reflects the much more similar frequencies of the bases at each of
these positions. For example, at position -18 there are 2 Gs, 3 As, 2 Ts, and 3 Cs; and at position
-17 there are 3 Gs, 4 As, 2 Ts, and 1 C.
3. (a) Positions 0, 1, and 2 have the most predictable bases; the bases are A, T, and G,
respectively. (b) Positions -18, -15, -5 and 8 have the least predictable bases; these positions
have the shortest stacks of bases, indicating that they are the least predictable.
4. (a) To write a consensus sequence, you would read off the base at the top of each stack at each
position, putting a dash where the base is too small to read. The consensus sequence for this
sequence logo is:
–AA–AAAGGGGAT–AATAATGACTAA–
(b) The sequence logo gives more information than the consensus sequence. In the consensus
sequence, the relative frequencies of the bases are not shown. Only the most frequent base is
shown, obscuring how strongly the base at each position can be predicted.
5. (a) Positions -12 to -8 have the tallest stacks of bases—with the clearest pattern for being able
to predict the base at that position in a new sequence—in the 5′ UTR region. Therefore, they may
represent the most likely candidate sequence that corresponds to the ribosome binding site. (b)
The bases in positions 0–2 represent the DNA codon corresponding to the mRNA start codon.
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and the suggested rubric at the
beginning of this document.
10. Scientific Inquiry
The β-globin gene contains introns, which could not be spliced out by the bacterial cell. The
translation machinery may have initiated translation at 5-AUG-3, but one of the introns
(probably the first) contained a stop codon, which caused premature termination of translation,
resulting in a shortened, nonfunctional protein.
11. Focus on Evolution
Given the essential nature of a triplet genetic code, we can assume that each of the 20 amino
acids was originally coded for by one unique triplet codon. Most amino acids are coded for by a
set of codons that share the same first two nucleotide bases. Evolutionary accidents (mutations)
that modified the third nucleotide base in the codon might have been selected for if they coded
for the same amino acid because of the selective advantage of having to make fewer tRNAs. This
could account for wobble. (Since one tRNA anticodon can bind to several mRNA codons, there
is an advantage for all those codons to code for the same amino acid. Otherwise, different tRNAs
would be required.)
12. Focus on Information
Sample key points:
 The fidelity with which DNA is copied and inherited guarantees the continuation of a
species.
 Variation is the raw material on which natural selection works.
 The ultimate source of changes in DNA sequence is mutation.
 Each species must have a mutation rate that is low enough to allow survival of individuals,
yet high enough to generate the diversity of traits on which natural selection acts.
Sample top-scoring answer:
DNA, containing the genetic information, is passed along from cell to cell and from generation
to generation. On one hand, the fidelity with which DNA is inherited guarantees the continuation
of each species, and many systems in the cell ensure that the DNA is copied accurately and that
identical copies of the genome are inherited by daughter cells. On the other hand, variation is the
raw material on which natural selection works. While genetic recombination during sexual
reproduction rearranges variant genes, the ultimate source of changes in DNA sequence is
mutation, which plays an important evolutionary role. Most often, mutations have negative
effects; only rarely do they lead to an improved gene product. Over long stretches of
evolutionary time, each species must have a mutation rate that is low enough to allow survival of
individuals, yet high enough to generate the diversity of traits on which natural selection acts—
an important balance.
13. Synthesize Your Knowledge
Siamese cats are “albinos,” with a temperature-sensitive mutation in a gene that codes for an
enzyme in the melanin pathway, such as tyrosinase. At the higher temperatures of the central part
of the body, the defective enzyme doesn’t function so those areas are cream-colored
(unpigmented) because melanin is not made there. In the extremities (the paws, ears, nose, and
tail), temperatures are cooler and the enzyme is able to function. Melanin is made and deposited
there, so these areas are pigmented. (The characteristic blue eye color in Siamese cats is a result
of albinism; albino animals always have either pink or blue eyes.)
CHAPTER 15 REGULATION OF GENE EXPRESSION
Scientific Skills Exercise
Teaching objective: Students will analyze a figure that combines gene construct diagrams and a
bar graph showing gene expression for the control DNA sequence and three experimental DNA
constructs.
Teaching tips: A version of this Scientific Skills Exercise can be assigned in MasteringBiology.
There are two conceptual challenges in this exercise. The first is to interpret the relationship
between the different kinds of graphical data that are presented together in the figure. The gene
construct portion of the diagram shows schematics of control DNA segments that have the
specified deletion. (To conserve space, the large region between the promoter and the enhancer
region is omitted and indicated with slashes, as mentioned in the exercise.) These constructs are
spatially arranged on the y-axis to correspond to the other kind of graphical data: the horizontal
bars in the graph showing the relative amount of reporter gene mRNA in the cells that contain
that particular gene construct. (The exercise introduces the term “construct,” commonly used for
a genetically engineered DNA sequence.) The placement of the independent variable on the yaxis may cause some confusion: It may help to have the students rotate the figure 90°
counterclockwise to see the more familiar vertical bar graph. This type of data is usually
represented as shown in the figure because we are used to thinking about DNA sequences in a
horizontal orientation.
The second challenge is to understand the idea of a reporter gene, which can be difficult for
students. The cultured cells in which the experiment is done are cells that normally express the
mPGES-1 gene. This choice of cells is made to ensure that the cells have the necessary specific
transcription factors for expression of that gene. On the other hand, the fact that the cells are
expressing the mPGES-1 gene themselves would complicate the results if the constructs had the
mPGES-1 gene in them, under control of the various deletions. In this case, researchers would be
measuring mPGES-1 mRNA levels, which would be added to levels of this mRNA already being
made by the cells. So, instead, a different gene is used as a “reporter” that is otherwise not
expressed in those cells. The reporter gene will only be expressed if the appropriate control
elements are present in the experimental construct. Comparison of reporter gene mRNA levels
gives information about the relative power of the control elements in different constructs to
upregulate expression of the reporter gene. (The reporter gene used in this experiment, although
not mentioned in the exercise, is the gene encoding human growth hormone, for which the
mRNA levels are measured.)
Students may wonder how this relates to the idea that each control element regulates a specific
gene and only that gene. In this experiment, researchers are altering the normal placement of
control elements. Normally, a group of control elements only regulates a given gene because
they are upstream of that gene at the appropriate distance for regulation of that gene to occur.
There is nothing control elements “recognize” about a specific gene. In a sense they are dumb
and blind: they regulate whatever gene is at the appropriate location downstream from them.
The illustrations of enhancers in the textbook (Figures 15.8, 15.10, and 15.11) focus on control
elements that act as activators. In this exercise, two of the control elements act as repressors of
gene expression, so students may need to be directed to re-read the section on “Enhancers and
Specific Transcription Factors” to review how control elements, in association with the proteins
that bind them, can work in either capacity.
Note that these data are the relative amount of gene expression, displayed as percentages relative
to the control, which is set at 100%. The actual level of reporter gene mRNA cannot be
determined from the graph.
Answers:
1. (a) The independent variable (the variable that was manipulated by the researchers) was the
identity of the DNA element that was deleted from the experimental DNA constructs. (b) The
dependent variable (the variable that varied as the independent variable changed) was the amount
of reporter gene mRNA produced relative to that in cells with the intact DNA sequence. (c) The
experimental control was the intact DNA construct (the one with all three possible control
elements). The top row of the left side of the diagram should be labeled “Control.”
2. The data suggests that all three possible control elements are actual control elements because
the deletion of each one had an effect on gene expression.
3. (a) Yes; deletion of control element #3 led to a reduction in gene expression. The bar graph
shows that cells with DNA that lacked control element #3 had only about 40% of the mRNA
compared to the control level. (b) If the loss of a control element causes a reduction in gene
expression, then the normal role of that control element must be to activate gene expression by
binding activators. If such a control element is deleted, then activators are not able to bind to that
control element, and the level of gene expression decreases.
4. (a) Yes; deletion of control element #1 or #2 led to an increase in gene expression relative to
the control DNA. The cells with DNA that lacked control element #1 produced about 180% of
the mRNA relative to the control, and the cells with DNA that lacked control element #2
produced about 140%. (b) If the loss of a control element causes an increase in gene expression,
then the normal role of that control element must be to repress gene expression, for example by
binding repressors. If such a control element is deleted, then repressors are not able to bind to
that control element, and the level of gene expression increases.
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and the suggested rubric at the
beginning of this document.
8. Scientific Inquiry
You could use the cloned gene to synthesize a labeled probe and carry out in situ hybridization
on a series of mouse embryos of different stages. Alternatively, you could isolate mRNAs from
each stage and make cDNA from them, then use the cDNAs to carry out RT-PCR. Running the
products on the gel will reveal the relative amounts of mRNAs in each stage.
9. Focus on Evolution
The highly conserved regions may code for sequences that help regulate gene expression (such
as proximal or distal control elements, including groups of elements in enhancers) using
mechanisms that are widespread among species. Alternatively, such regions may specify
important small noncoding RNAs or their precursors or larger noncoding RNA transcripts that
play important roles in gene regulation. These RNAs may operate at the level of chromatin
structure, transcription of a particular gene, or translation. There is evidence in support of at least
the first model, suggesting that regulatory systems are evolutionarily conserved across at least
some species.
10. Focus on Interactions
Sample key points:
 To survive, bacterial cells must efficiently use the available resources.
 They can synthesize most of the compounds they need, but when a needed compound is
available in their environment, they can stop synthesizing it.
 There are two mechanisms for this feedback regulation:
o Inhibition of the first enzyme by the final product
o Turning off expression of genes encoding pathway enzymes
 Together these mechanisms of feedback regulation ensure efficient use of resources and
energy by bacterial cells
Sample top-scoring answer:
To survive and reproduce, a bacterial cell must efficiently use the cellular resources that are
available. Bacterial cells can synthesize most of the compounds they need, such as amino acids,
from chemical building blocks in the cell. Some of the needed compounds are occasionally
found in the environment, however. In this case, bacteria can use feedback to regulate their own
synthesis of the compound so as not to squander energy and resources. Figure 15.2 shows two
mechanisms of this regulation. The first is feedback inhibition in which the final product
(tryptophan) inhibits the first enzyme in the synthetic pathway. The second is a longer term
response using gene regulation to turn off expression of the genes encoding pathway enzymes so
these enzymes are not made in the first place. These two mechanisms allow bacterial cells to use
only the energy and resources that are absolutely necessary in order to grow and reproduce.
11. Synthesize Your Knowledge
The lux genes are likely to be grouped in an operon, under the control of a single promoter and
operator, so that the genes are all turned on at the same time (or all turned off at the same time).
Since a certain density of bacterial cells must be reached before light is emitted, there must be
some signaling molecule that is present when the density is high enough. Perhaps this signaling
molecule would bind to a repressor allosterically, changing it into an inactive shape, so that it is
no longer able to bind an operator. With no repressor bound, RNA polymerase would be able to
bind and begin transcribing all the lux genes, resulting in light emission.
CHAPTER 16 DEVELOPMENT, STEM CELLS, AND CANCER
Scientific Skills Exercise
Teaching objective: Students will compare quantitative data with spatial data related to gene
expression during animal development.
Teaching tips: A version of this Scientific Skills Exercise can be assigned in MasteringBiology.
The two approaches to studying gene expression shown in this exercise complement each other.
The mRNA measurements (done by quantitative reverse-transcription PCR in this case) are
quantitative but have low spatial resolution because the entire digit-forming area is dissected out
and used for a single analysis. The in situ hybridizations are advantageous for showing spatial
patterns of expression, but are not accurate for measuring the amount of gene expression at a
resolution that allows samples to be compared. In this experiment, amount of Hoxd13 mRNA is
used to represent Hoxd13 gene expression, commonly the case in research on gene regulation.
While this is a good measure, students may bring up the question of whether this actually
measures gene expression since it doesn’t measure actual protein levels. This is a good point, and
a correct (although subtle) one.
A key part of this exercise is correlating the results from both approaches. You’ll want to
impress upon students the value of carefully observing subtle aspects of the patterns of mRNA
distribution in the paws. Note that the control embryo shows fairly even dark staining across the
whole digit-forming area of the paw and has the highest amount of mRNA, while deleting all
three possible regulatory segments (A, B, and C) eliminates most of the mRNA and staining,
again in a fairly even pattern. The other two deletions, however, show intermediate levels of
mRNA but an uneven pattern of expression. When C is eliminated there is one patch of lighter
staining near the thumb region that accounts for most of the reduction in overall mRNA level.
When both B and C are deleted, the expression is fainter overall, with darker narrow bands along
the forming digits. A simplistic way to help students make the connection between the
quantitative and qualitative data is to set up two piles of blocks on the table (or books, or any
other moveable object). Count the blocks in each pile to make sure they are the same number
(this represents equal amounts of mRNA). Now spread one pile over a larger area of the desk—
the stack gets shorter (lighter staining) but covers a larger area than the other pile. You can
arrange the blocks in ridges to represent the banded pattern of expression. If you remove a few
blocks from one set (less mRNA) you can still make some of the ridges the same height in both
piles—there will just be less area covered by the blocks in the lower-level mRNA set.
If you want to challenge your students, you can ask them what type of control element they
conclude might be present in region C. You can help them think through the scientific logic to
arrive at the idea that there must be one or more control elements that specifically up-regulate
expression in the pre-thumb area. (This is, in fact, what the researchers have concluded.)
Answers:
1. (a) The top sample, with all three DNA regulatory segments and a 100% level of mRNA, was
an experimental control. (b) Yes, their hypothesis is supported because when all three DNA
regulatory segments are deleted, the Hoxd13 mRNA drops to 6% of the control.
2. (a) There is a loss of almost 70% of the Hoxd13 mRNA in the embryonic paw (there is just
over 30% of the control amount). (b) The effect is visible because there is less blue stain in the
digit-forming area of the paw than in that location in the control. The pattern is relatively evenly
lightened, with dark stain around each digit zone, and lighter areas between the digits, as
compared to the control. (Note that although the difference is visible, it is not possible to
estimate a percentage decrease of expression by eye. This is the benefit of the quantitative
measurements of mRNA expression.)
3. (a) There is a loss of about 40% of the Hoxd13 mRNA in the embryonic paw (there is about
60% of the control amount). (b) The effect is visible because there is less blue stain in the digitforming area. The pattern is uneven, with dark stain across the top but no stain in the pre-thumb
region. The slight digit patterning visible in the control is also missing.
4. The differences in spatial patterns of Hoxd13 gene expression would have been missed
without the in situ hybridization results. Regulation of patterning controlled by the examined
regions may be an important part of digit development, especially the thumb. For the converse,
as noted above, although differences in mRNA level are discernible by eye in the in situ
hybridization experiment, this is a qualitative observation (“Staining is fainter”). Quantitative
data are contributed only by the approach that measures mRNA levels.
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and a suggested rubric at the
beginning of this document.
6. Scientific Inquiry
Hypothesis: Estrogen is activating genes normally controlled by testosterone and other
androgens in prostate cancer cells.
This hypothesis could be tested in several ways. A treatment could be used to eliminate
estrogen, and the effect on the proliferation of prostate cancer cells could be noted. If the genes
thought to be activated by estrogen are known, it might be possible to look for the presence of an
estrogen-receptor complex in the nuclei of prostate cancer cells or to look for evidence of
transcription or translation of the genes of interest in the absence of androgens.
7. Focus on Evolution
Because of a series of genetic changes such as mutations in the ras or p53 genes, cancer cells
evade the normal cell cycle controls and are able to divide in an unregulated fashion. During the
progression of cancer, further mutations accumulate. These mutations may cause genes whose
products promote cell division to be more active, or genes whose products hold cell division in
check to be nonfunctional. Cells with mutations that confer on their host cell a faster growth rate
will be selected for, and will divide and leave more daughter cells. Another type of mutation that
would be “beneficial” to cancer cells would be those that allow cancer cells to move through the
body and establish themselves in another tissue (metastasis). The process of natural selection will
ensure that advantageous mutations persist in the cancer cell population.
8. Focus on Organization
Sample key points:
 Apoptosis is essential to embryonic development and the destruction of damaged cells in the
adult.
 Either external or internal “death” signals initiate signal transduction pathways by binding
with receptor proteins.
 Interactions and sequential activation of relay molecules in such pathways lead to the cellular
response.
 Apoptosis involves the activation of enzymes that break down cellular molecules and
package them into vesicles, which are engulfed by scavenger cells.
Sample top-scoring answer:
Apoptosis plays a crucial role in embryonic development, including formation of the nervous
system and the separation of digits. It is also part of the removal of damaged cells in the adult.
Extensive damage to a cell’s DNA can also trigger a built-in cell suicide program. Timely and
orderly cell death results from the integration of well-coordinated signaling pathways. For
example, the binding of an extracellular death signal to a transmembrane receptor protein
activates that protein. The resulting shape change of the protein activates cytoplasmic molecules
that initiate signal transduction pathways. The death signal is passed through relay molecules that
eventually activate enzymes that break down cell components and package them in membranebounded “blebs” for disposal. Thus, the emergent property of controlled cell death arises from
the structural arrangement and interactions of the components of apoptotic signaling pathways.
9. Synthesize Your Knowledge
Plant stem cells would give rise to plant tissues (such as grape vines and grapes, in this case)
under the appropriate conditions. The advertisement is claiming somehow that rubbing grape
plant stem cells on the skin will replenish our own skin stem cells. This is very unlikely to occur.
First of all, grape plant stem cells would not replace skin stem cells, and no human would want
grape plants sprouting from their skin, even if the conditions did exist for such a differentiation
to occur. Second, if plant stem cells secrete factors that stimulate stem cell growth, it is very
doubtful that any receptors would exist on the surface of human skin stem cells that would
recognize, be able to bind and respond to such plant growth factors.
CHAPTER 17 VIRUSES
Scientific Skills Exercise
Teaching objective: Students will use information from a phylogenetic tree to trace the
evolution of the H1N1 virus in the population of Taiwan.
Teaching tips: A version of this Scientific Skills Exercise can be assigned in MasteringBiology.
If you haven’t already covered evolution and phylogenetic trees, you will need to spend some
time walking the students through the structure of a tree. (For this purpose, the exercise refers
them to Figure 20.5.) The value of using a phylogenetic tree in combination with the graph for
this exercise is two-fold. First, by relating genetic change to time in the last few years, it
demonstrates how fast evolution is taking place and that it is occurring currently—it is not
limited to the distant past. Second, it shows how infectious disease experts can relate patterns of
evolution in a particular gene sequence to the infectious potential of the virus. The resulting
changes in amino acid sequence are not mentioned in the exercise, but you can discuss this with
students if you wish, asking them to consider the type of proteins in which there might be
mutations that alter the infectious potential.
Note that the researchers analyzed 4,703 isolates, but there are only 36 variants shown in the
phylogenetic tree. This is because there are multiple isolates of each variant; i.e., the same
variant caused illness in many people, the “wave” of infection. The frequency of infection of
each group of variants is represented by the number of isolates on the graph. The data do not
show the abundance of each individual isolate on the tree, as they are grouped in the graph.
Answers:
1. A/Taiwan1018/2011 and A/Taiwan/552/2011 are more closely related to each other than
A/Taiwan1018/2011 and A/Taiwan/8542/2009 because the branch length of the former pair,
from the point at which they separated from each other, is shorter compared to the length of the
latter pair’s branches from the point of their divergence. This means the former pair of variants
has more identical nucleotides in the DNA sequence than does the latter pair of variants, so the
former two are grouped together while the latter two are not. In fact, A/Taiwan/8542/2009 is on a
different branch in a different group.
2. (a) No, the nodes lead to different numbers of branches or branch tips. For example, one node
may have two branches, one of which leads directly to a tip while the other leads to another
node. (b) No, the branches are of many different lengths. (c) These results indicate that the
different lineages, or variants, have accumulated different numbers of mutations and are of
variable relatedness to each other. Not all lineages had the same amount of mutation over the
time period studied.
3. (a) Group 7 variants caused the first wave of illness in over 100 patients. (b) No, each group of
variants caused only one wave of illness. (c) Judging by the low numbers of isolates, the group 1
vaccine might have been effective at preventing infection by members of that same group, but
was apparently ineffective against variants from other groups.
4. No, their hypothesis was still supported by the data, because these three groups are clearly
different from each other. All three groups had variants that were more recently evolved than
those in the earlier waves, and the three groups just happened to cause serious infection at the
same time. They all arose along different lineages but emerged to cause illness during the same
wave.
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and a suggested rubric at the
beginning of this document.
7. Scientific Inquiry
Each bacterium in the infected animal is capable of cell division by binary fission. Before the
animal mounts an immune response, the number of bacterial cells increases in an exponential
fashion, as shown in graph A.
Following infection by a virulent virus (one reproducing with only a lytic cycle), the virus
binds to cells of the host animal and injects its DNA. The viral DNA directs production of viral
proteins and genomes by host enzymes, and viruses self-assemble within the host cell. During
this time, there is no increase in viral number. Finally, the virus produces an enzyme that causes
the host cell to burst, releasing hundreds or thousands of assembled viruses. Each episode of
viral release causes a stepped increase in number of viruses, as shown in graph B.
8. Focus on Evolution
Evolution “in microcosm” is really no different from evolution in other habitats. Even in the
absence of antiviral drugs, animal viruses exist in an environment where an immune system is
actively engaged in their recognition and elimination. RNA viruses are especially adept at
escaping recognition because their polymerases do not have proofreading capability, meaning
that mutations accumulate. Thus, a large variety of genetically different viral progeny is created
that are no longer recognized by immune system cells. Such viruses are, in effect, tiny “shapeshifters.” This same feature enables them to rapidly develop strains that are no longer able to be
bound by (thus are resistant to) antiviral drugs. Those viral lineages that predominate are the
ones whose genetic constitutions permit them to avoid elimination by the immune system or
drugs, and thus to persist and reproduce in the current environment.
9. Focus on Organization
See the general information on grading short-answer essays and the suggested rubric at the
beginning of this document.
Sample key points:
 Viruses consist of a nucleic acid surrounded by a protein coat (and, for animal viruses,
sometimes a viral envelope).
 The nucleic acid carries genetic information, and the protein coat serves to protect the genetic
information from damage.
 The function of a virus is to make more viruses.
 Each structural part of the virus is well suited to its function.
Sample top-scoring answer:
A virus has a very simple structure, well correlated with its function, which is to produce more
viruses. A virus consists of a nucleic acid, either DNA or RNA, surrounded by a protein coat.
The nucleic acid carries genetic information that is capable of reprogramming the machinery in a
host cell to produce viral components, which then automatically assemble into viruses. The
protein coat surrounding the nucleic acid serves to protect the essential genetic information from
damage by outside agents, and also facilitates viral entry into host cells. Some animal viruses are
further surrounded by a viral envelope derived from a membrane of the host cell. In these
viruses, the envelope helps the virus enter the cell during infection and also evade the immune
system when not inside a cell. Each component of the virus’s structure plays a vital role in the
infectious process, which functions in propagating viruses.
10. Synthesize Your Knowledge
Neuraminidase is a molecule on the surface of the flu virus that acts as an enzyme to help release
new virus particles from an infected cell. Inhibiting this enzyme blocks the release of new
viruses from infected cells. If taken early enough by a person exposed to flu, oseltamivir
(Tamiflu®) can prevent viruses from building up to a level that would elicit an immune system
response and the symptoms of the flu. In a person already infected, stopping further production
of virus particles would mean a shorter duration of the illness, as the immune system would have
to respond to and eliminate only those viruses that were produced before the Tamiflu was taken.
CHAPTER 18 GENOMES AND THEIR EVOLUTION
Scientific Skills Exercise
Teaching objective: A version of this Scientific Skills Exercise can be assigned in
MasteringBiology.
Students will examine amino acid identity percentages in a table and see how these data are used
to propose evolutionary relationships. In this example, they will also see how the evolutionary
relationships among globins are consistent with the model presented in Figure 18.13.
Teaching tips: It will be useful to remind the students that the values in the table were calculated
based on pair-wise sequence alignments, not a consensus sequence, and that the globins in the
table are all contemporary (currently existing) proteins. The ancestral globins shown in Figure
18.13 are proposed based on the model and we do not have the actual sequences for those.
Globins proteins are 142 or 149 amino acids long. To get the percentages in the table, the
number used as the denominator is the number of amino acid positions, counted after the
alignment is done and including the spaces (shown as dashes) that were introduced to optimize
the alignment. That is why in the example presented there are 142 amino acids but 143 is used as
the denominator to calculate the percent identity.
You may want to introduce the notion that not every mutation in the duplicated globin genes will
result in altered amino acids, and not all amino acid changes will result in altered function. It is
only the mutations that do cause altered function (or expression patterns) that will be acted upon
by natural selection.
Answers:
1. (a) The table cells with dashed lines indicate self-pairs (for example, -). Because the globins
in these pairs are identical, the dashed lines signify 100% identity. (b) The cells were left blank
because they represent duplicates of the pairs in the upper right half of the table, so the data
would be redundant. (After filling them in, you’ll see the lower left is a mirror image—across the
diagonal—of the upper right of the table.)
Amino Acid Identity Table
β Family
α Family
α Family
β Family
α1
(alpha
1)
α2
(alph
a 2)

(zeta)
β
(beta)

(delt
a)

(epsilo
n)
A

(gamm
a A)

(gamm
a G)
α1
------
100
60
45
44
39
42
42
α2
100
------
60
45
44
39
42
42

60
60
------
38
40
41
41
41
β
45
45
38
------
93
76
73
73

44
44
40
93
------
73
71
72

39
39
41
76
73
------
80
80

42
42
41
73
71
80
------
99

42
42
41
73
72
80
99
------
A
G
G
2. (a) β and  are most divergent from each other, with a percent identity of 38%. (b) α1 and α2
are the two globin genes most recently duplicated, with a percent identity of 100%.
3. In general, the members of the α branch of the globin gene family are more similar to each
other than to those in the β branch, and the same holds true for within the β branch versus the α
branch.
4.
Percent
Pairs
Identity
100
α2 - α1
 - A
99
G
93
-β
80
A
76
-β
73
-
72
G
71
A
-
-
-
-
G
-β
A
-β
G
60
 - α1
 - α2
45
β - α1
β - α2
44
 - α1
 - α2
42
A
G
41
-
A
40
-
39
 - α1
38
β-
 - α1
 - α1
A
 - α2
-
G
 - α2
G
-
 - α2
(a) Yes. Globins that are most alike are drawn with close connections on the tree in Figure 18.13,
and the less alike they are, the further away their point of divergence is on the tree (in other
words, the more distant they are). (b) All of the within-group pairs are at or above 60% identical,
and all between-group pairs are at or below 45% identical.
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and the suggested rubric at the
beginning of this document.
5. Scientific Inquiry
SNPs act as genetic markers, and just like STRs, they exhibit genetic linkage (see Chapter 12).
Therefore, the closer they are to each other, the less likely they are to recombine in other
combinations. Furthermore, comparison of the human and chimpanzee genomes has shown that
particular sites in the genome are prone to recombination and are used as recombination sites
over and over again. Presumably, these sites are not in the haplotype blocks but instead are
between them.
6. Focus on Evolution
Genes whose functions are fundamental, and thus essential, tend to be highly conserved because
the existing gene sequences are well adapted to coding for functional gene products. Any
mutation that alters the early embryo is likely to have many harmful effects. If development is
able to proceed, harmful mutations of fundamental genes are likely to produce organisms that
cannot survive and reproduce; that is, such genes are not transmitted to the next generation. For
any mutation to be transmitted to the next generation, of course, it would have to occur in a germ
cell that would give rise to a gamete.
7. Focus on Information
Sample key points:
 Mutations can affect protein-coding genes or regulatory sequences.
 A mutation in a protein-coding gene usually has a harmful effect, if any, on protein function.

A mutation in a regulatory sequence may cause misexpression of the genes it regulates,
which usually has a harmful effect, if any, on the organism.
 In very rare cases, either type of mutation may confer extra fitness on the organism, leading
to a greater number of offspring that have the mutation.
 Although mutations in protein-coding genes and regulatory sequences act by different
mechanisms on the organism, they contribute to evolution in a similar way.
Sample top-scoring answer:
Mutations occur randomly along the chromosomal DNA and can affect any type of DNA
sequence. If the mutation occurs in a germ cell, it can be passed along to offspring. A mutation in
a protein-coding gene usually causes the protein to function less well or equally well, and a
mutation in a regulatory sequence that causes misexpression of the gene(s) it regulates usually
has harmful effects, or none, on the organism. In very rare cases, however, a mutation may
improve protein function or cause misexpression in a way that increases the fitness of the
organism, resulting in a greater number of offspring that will inherit the mutated gene. Although
mutations in protein-coding genes and regulatory sequences act via different mechanisms, the
way they contribute to evolution is generally the same: If either confers greater fitness on an
organism, more individuals in the next generation are likely to have that mutation.
8. Synthesize Your Knowledge
A change in gene regulation by the Hox gene product occurred, such that “wings” were no longer
inhibited and could form. The insect already had a functional pair of wings on the second
segment, and so random mutations over time could have led to the wings on the first segment
being fused together and then malformed in various ways. Ultimately, the structure could come
to resemble thorns. These modifications must have benefitted the organism’s survival and
reproduction, helping to disguise it from predators, and, due to natural selection, caused the
presence of the structure to persist over time.
CHAPTER 19 DESCENT WITH MODIFICATION
Scientific Skills Exercise
Teaching objective: Students build scientific skills by making and testing predictions on how
natural selection will affect color patterns in guppies.
Teaching tip: A version of this Scientific Skills Exercise can be assigned in MasteringBiology.
Answers:
1. (a) Can predation result in selection for color patterns in guppies? (b) Predation can result in
selection for less noticeable (more drab) coloring in prey populations of guppies. (This
hypothesis implies that reduced predation would result in a more brightly colored guppy
population, since females prefer brightly colored males.) (c) When guppies with drab colors are
transferred to a pool with only killifish (less active predators of guppies that prey mostly on
juvenile guppies), eventually the descendants of these drab guppies in the killifish would be
more brightly colored than the source population (because predation would not be counteracting
the selection pressure from the female guppies’ preference for brightly colored males) (d) the
source population in the pike-cichlid pools (e) the transplanted population in the killifish pools
2. Tracking the number of bright-colored spots and the total area of those spots provided a
quantitative way for the researchers to compare how bright or drab the study populations were.
3. Because both the number and total area of the bright-colored spots were greater in the
transplanted population than in the source population, the data suggest that brighter-colored
guppies are more successful in the transplanted population—in other words, that reduction of
predation on adults leads to a more brightly colored guppy population. Since the source
population is under heavier predation (especially of adult guppies displaying their adult
coloration), these results do support the hypothesis that predation is leading to selection for more
drab coloring in the source population.
4. If the hypothesis is correct, then the transplanted guppies that are returned should experience a
higher rate of predation than the guppies already in the source pool, because their brighter colors
would make them more visible to predators. Since they are less fit, they should leave fewer
descendants and over time, the average number and area of colored spots in the source pools
should again resemble those of the original source population. This prediction could be tested by
transplanting guppies into pike-cichlid pools from which the guppies were removed before
adding the transplanted guppies. Researchers would compare the predation rate of the
transplanted guppies compared to the predation rate in pike-cichlid pools containing the original
population of drab guppies. Researchers would also compare the color patterns of the new
transplant populations over time compared with those in the killifish pools.
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and a suggested rubric at the
beginning of this document.
7. Focus on Evolution
Organisms share many characteristics from their common descent, and more closely related
organisms share more characteristics than distantly related organisms. This reasoning applies to
all of an organism’s characteristics; hence, anatomical and molecular features often fit a similar
nested pattern because of common descent. Convergent evolution can break this nested pattern
for characteristics that evolved as distantly related organisms adapted to similar environments in
similar ways.
8. Focus on Interactions
Sample key points:
 All organisms interact with the living and nonliving components of their environment.
 An organism whose inherited traits cause it to be particularly successful in its environmental
interactions will tend to leave more offspring than other organisms.
 However, traits that are successful in one environment may be less successful or even
detrimental in another environment.

Hence, if a feature of the environment changes, that change may result in evolutionary
change in one or more of the species affected by the change in environmental conditions.
Sample top-scoring answer: All organisms interact with the living and nonliving components
of their environment. Furthermore, as a result of descent with modification by natural selection,
an organism whose inherited traits cause it to be particularly successful in its environment will
tend to leave more offspring than other organisms. However, natural selection often favors
different traits in different environments. In fact, a trait that is favorable in one environment can
be less successful or even harmful in another environment. Thus, if a feature of the environment
changes, selective pressures may change as well. For example, an increase in temperature may
select against cold-adapted genotypes, or the introduction of a new species of predator may cause
the evolution of new defenses in some prey species. Overall, it is likely that a change in
environmental conditions will result in evolutionary change in some of the species affected by
the environmental change.
9. Synthesize Your Knowledge
The expandable abdomen of the honeypot ants is an adaptation that allows them to store food
when it is available. As is apparent from the overall shape of their body (except for their unique
abdomen), honeypot ants share many of the characteristics found in other ants, thus illustrating
the unity of life that has resulted from descent with modification. Their unique abdomen
illustrates the diversity of life, which also has resulted from descent with modification.
CHAPTER 20 PHYLOGENY
Scientific Skills Exercise
Teaching objective: The main objectives of this exercise are to give students practice in
working with protein (polypeptide) sequence data and to help them appreciate how such data can
be used to suggest and test hypotheses about evolution. A secondary objective is to reinforce
what students have read about the idea of horizontal gene transfer.
Teaching tips: A version of this Scientific Skills Exercise can be assigned in MasteringBiology.
Note that the amino acid sequence for this comparison comes from the enzyme carotenoid
desaturase in A. thaliana.
The instructor may want to point out that the data shown in this exercise are very limited: only
part of one polypeptide is shown and only for six species. For this reason, any conclusions
students may draw from the data can only be tentative. However, this exercise does provide an
opportunity to discuss the different ways that horizontal gene transfer might occur and the
possible role of horizontal gene transfer in evolution.
Answers:
1.
2. In the partial polypeptides shown, the sequence from Ustilago has the most amino acids (38)
in common with the aphid sequence. In decreasing order of similarity to the aphid sequence,
Gibberella has 35 similar amino acids, Staphylococcus has 25, Pantoea has 20, and Arabidopsis
has 18.
3. These data—in particular, the similarity in amino acid sequence between the aphid
polypeptide and the polypeptides from the other, only distantly related organisms—do support
the hypothesis that aphids acquired the gene for the polypeptide by horizontal gene transfer.
From the fact that the fungal polypeptides are the most similar to the aphid polypeptide, we can
infer that a fungus is most likely to have been the source of the transferred gene.
4. Sequence data from the remainder of this polypeptide and others involved in carotenoid
biosynthesis could support the close similarity between aphid and fungal genes for this group of
proteins. Finding an extremely similar gene or polypeptide in two organisms known to be only
very distantly related—such as an animal and a fungus—suggests that the gene may have been
transferred “horizontally” long after the divergence of the two organisms in the evolutionary
past.
5. The similarities to the aphid sequence shown by the bacterial and plant sequences could be
explained in a number of ways. For example, the sequence for this carotenoid-biosynthesis gene
may have originally arisen in a bacterium, which passed it by horizontal gene transfer to a
fungus; a descendant of the fungus may have passed a mutated version to the aphid. Similarly, an
ancestor of the plant Arabidopsis may have acquired an earlier version of the gene from a
bacterium or other unicellular organism horizontal gene transfer.
Interpret the Data
Figure 20.13 The zebrafish lineage; of the five vertebrate lineages shown, its branch is the
longest, indicating that it has undergone the most change since diverging from the common
vertebrate ancestor.
Figure 20.19 The molecular clock indicates that the divergence time is roughly 45–50 million
years.
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and a suggested rubric at the
beginning of this document.
9. Focus on Evolution
Darwin’s suggestion directly anticipates phylogenetic bracketing, a method that uses shared
characteristics of organisms in a clade to make predictions about the common ancestor of the
clade and all of its descendants. His suggestion is also similar to how outgroups are used to help
differentiate between ancestral characters (present in the outgroup) and derived characters (new
features that originated with the ingroup).
10. Focus on Information
Sample key points:
 All organisms are composed of cells.
 DNA is the hereditary material.
 By descent with modification, newly formed species share many similarities with parent
species—including similarities in cell structure and DNA sequence.
 The parent species, in turn, share many similarities with their parent species, and so on.
 Thus, all organisms contain evidence of their evolutionary past, making it possible to
construct phylogenies.
Sample top-scoring answer:
All organisms are composed of cells and have their genetic information encoded in DNA. One
result of this fact is that by descent with modification (see Chapter 19), the cells and DNA of
newly formed daughter species will share many similarities with the cells and DNA of their
parent species—which in turn share many such similarities with their parent species, and so on.
As a result, all organisms contain both morphological evidence (for instance, cell structure) and
genetic evidence (for example, DNA sequence) of their evolutionary history. As we’ve seen,
untangling evolutionary history from morphological and genetic data requires that homologous
characters be sorted from analogous characters and that shared derived characters be
distinguished from shared ancestral characters. Although challenging, the existence of wellsupported phylogenies that cover hundreds of millions of years of evolution indicates that
organisms contain enough evidence of their evolutionary history to accomplish such tasks.
11. Synthesize Your Knowledge
Organisms are classified within groups such as the tetrapods based on their evolutionary history
rather than on whether they possess all of the traits typically found in other members of the
group. Amphibians, reptiles, and mammals all descended from the tetrapod common ancestor.
Since manatees are mammals, they also descended from the tetrapod common ancestor; hence
manatees are considered tetrapods. We can infer from Figure 20.12b that manatees (like
leopards) have hair, a shared derived character found in the mammals, as well as a vertebral
column, hinged jaws, and an amnion. It is likely that early members of the manatee lineage had
four legs, and that over time, their front legs became modified into flippers, while their hind legs
became reduced in size and ultimately lost.
CHAPTER 21 THE EVOLUTION OF POPULATIONS
Scientific Skills Exercise
Teaching objective: Students build scientific skills by interpreting data to determine whether a
soybean population is evolving.
Teaching tip: A version of this Scientific Skills Exercise can be assigned in MasteringBiology.
Answers:
Each individual has two alleles, so the total number of alleles at day 7 is 432. To calculate the
frequency of the CG allele, note that each of the 49 individuals of genotype CGCG has two CG
alleles, and each of the 111 individuals of genotype CGCY has one CG allele. The 56 individuals
of genotype CYCY have zero CG alleles. Thus, the frequency of the CG allele (p) is:
1.
(2  49)  (1  111)  (0  56)
 0.484 .
432
Using p = 0.484, we calculate that q = 1-p = 0.516.
p
2. Plugging the values of p and q from question 1 into the Hardy-Weinberg equation, we find
that the expected frequencies of genotypes CGCG, CGCY, and CYCY are, respectively, 0.4842 =
0.234; (2 x 0.484 x 0.516) = 0.499, and 0.5162 = 0.266.
3. At day 7, the observed genotype frequency of CGCG = 49/216 = 0.227, while the frequency of
CGCY = 111/216 = 0.514 and the frequency of CYCY= 56/216 = 0.259. These frequencies are
close to those expected for a population in Hardy-Weinberg equilibrium as calculated in question
2, suggesting that evolution is not taking place.
4. At day 21, the observed genotype frequency of CGCG = 47/173 = 0.272, while the frequency of
CGCY = 106/173 = 0.613 and the frequency of CYCY= 20/173 = 0.116. These results differ from
those expected for a population in Hardy-Weinberg equilibrium as calculated in question 2,
suggesting that evolution is occurring. Comparing these results to the observed frequencies at
day 7 suggests that genotype CYCY is being selected against: Of the 56 individuals of this
genotype alive on day 7, only 20 (36%) survived to day 21. In contrast, 96% of genotype CGCG
and 95% of genotype CGCY survived from day 7 to day 21.
5. When a population is in Hardy-Weinberg equilibrium, its allele and genotype frequencies do
not change over time; thus, if the soybean population is in Hardy-Weinberg equilibrium, its allele
and genotype frequencies should not change between days 7 and 21. At day 7 in this experiment,
homozygous CYCY individuals were not appearing to be selected against, most likely because
they had not used up the supply of food stored in the seed. However, by day 21, CYCY individuals
were appearing to be selected against, likely because by that time they had run out of stored food
and hence many were not surviving. Based on this hypothesis, we would predict that after day 21
the observed frequency of the CY allele would decrease whereas the observed frequency of the
CG allele would increase.
Interpret the Data
Figure 21.16 At each of three time periods when the phenotypes of breeding adults were
assessed, a majority of the adults that reproduced (represented by green dots) had the opposite
phenotype of that which was most common in the population (represented by red dots). This
suggests that left-mouthed individuals were favored by selection when right-mouthed individuals
were more common, and vice versa.
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and a suggested rubric at the
beginning of this document.
7. Scientific Inquiry/Interpret the Data
(a) and (b)
The frequency of the lap94 allele increases as one moves from southwest to northeast across
Long Island Sound. The frequency of the lap94 allele is higher at sites 9-11 (open ocean) than it
is at sites 1-7 (within the Sound); the northeast edge of the Sound (site 8) has about the same
frequency of the lap94 allele as do the open ocean sites. (c) A hypothesis that explains the shape
of the graph and accounts for the observations stated in the question is that the frequency of the
lap94 allele at different sites results from an interaction between selection and gene flow. Under
this hypothesis, in the southwest portion of the Sound, salinity is relatively low, and selection
against the lap94 allele is strong. Moving toward the northeast and into the open ocean, where
salinity is relatively high, selection favors a high frequency of the lap94 allele. However, because
mussel larvae disperse long distances, gene flow prevents the lap94 allele from becoming fixed in
the open ocean or from declining to zero in the southwestern portion of Long Island Sound.
8. Focus on Evolution
Although natural selection can improve the match between organisms and their environments,
the evolutionary process can also lead to imperfections in organisms. A central reason for this is
that evolution does not design organisms from scratch to match their environments and ways of
life but works instead by a process of descent with modification: Organisms inherit a basic form
from their ancestors, and that form is modified by natural selection over time. As a result, a
flying mammal such as a bat has wings that are not perfectly designed, but rather represent
modifications of forelimbs that bat ancestors used for walking. Imperfections in organisms result
from a variety of other constraints, such as a lack of genetic variation for the trait in question,
and the fact that adaptations often represent compromises (since organisms must do many
different things, and a “perfect” design for one activity might impair the performance of another
activity).
9. Focus on Organization
Sample key points:
 In sickle-cell heterozygotes, red blood cells that contain relatively large amounts of sicklecell hemoglobin are prone to sickling.
 Sickled cells are rapidly destroyed by the body.
 When sickling occurs, malaria parasites inside sickled red blood cells tend to be killed
rapidly. This reduces the density of the parasites and hence the severity of the disease—an
emergent property at the level of the individual.
 Natural selection can lead to an emergent property at the population level—a rise in the
frequency of the sickle-cell allele due to heterozygote advantage in regions where malaria is
common.
Sample top-scoring answer:
In a sickle-cell heterozygote, red blood cells that contain relatively large amounts of sickle-cell
hemoglobin are prone to sickling, and sickled cells are rapidly destroyed by the body. Malaria
parasites spend part of their life cycle inside red blood cells. If a heterozygote has malaria,
parasites that live inside red blood cells that become sickled are killed when the sickled cell is
destroyed. These events lead to two emergent properties: At the individual level, heterozygotes
receive protection from the most severe effects of malaria (since parasite density is lowered
when sickled cells are destroyed); at the population level, partial protection against the effects of
malaria can lead to an increase in the frequency of the sickle-cell allele in regions where malaria
is common.
10. Synthesize Your Knowledge
Members of a species that colonize a new environment often face environmental conditions that
differ from those found in their home (original) population. For example, in the kettle lake
shown here, the depth, temperature, and quality of the water might differ from those in other
nearby lakes, and at first there might be few or no competitor species for the colonists. As a
result, it is likely that natural selection would cause evolutionary change in animal populations
that colonized the lake. Over time, as more species colonized the lake and as other features of the
lake changed, the alleles favored by selection might change and hence natural selection could
continue to affect animal populations in the lake. Allele frequencies would probably also change
as a result of genetic drift, especially shortly after a given population colonized the lake; this is
the case because typically, few colonists reach new environments and so the size of the
population would be very small initially. In addition, populations founded by a small number of
colonists might initially have low levels of genetic variation. Over time, mutation and ongoing
gene flow would introduce new alleles into the populations; some of these alleles would be
favored by selection, others would likely change in frequency due to genetic drift.
CHAPTER 22 THE ORIGIN OF SPECIES
Scientific Skills Exercise
Teaching objective: Students build scientific skills by analyzing experimental design and
interpreting a scatter plot.
Teaching tip: A version of this Scientific Skills Exercise can be assigned in MasteringBiology.
Answers:
1. The researchers’ hypothesis was that reproductive isolation increases with geographic
distance between dusky salamander populations. The independent variable in this study
was the geographic distance between dusky salamander populations, and the dependent
variable was the reproductive isolation values for pairs of dusky salamander populations.
The researchers set up four mating combinations for each pair of populations in order to
compare the proportion of successful matings within populations to the proportion of
successful matings between populations, testing both males and females from each
population.
2. (a) The reproductive isolation value for two populations is the sum of the proportion of
successful matings of each type within populations (AA + BB) minus the sum of the
proportion of successful matings of each type between populations (AB + BA). In this
case, (1 + 1) – (0 + 0) = 2. (b) In this case, (AA + BB) = (AB + BA), so (AA + BB) –
(AB + BA) = 0.
3.
4. (a) The scatter plot helps reveal that there is a general trend upward from left to right in
the plotted data—that is, that reproductive isolation values tend to increase with
geographic distance. (b) Reproductive isolation is more likely to occur as populations
become separated by greater distances. One hypothesis for this observation is that
geographically separated populations of dusky salamanders gradually diverge, and the
divergence increases with geographic distance because ongoing gene flow is less likely to
occur between populations that are far from one another.
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and a suggested rubric at the
beginning of this document.
7. Scientific Inquiry
Here is one possibility:
8. Science, Technology, and Society
It is possible to argue both sides of this question based on the underlying biology. For example,
if the biological species concept is followed strictly, one could argue that red wolves and coyotes
are the same species since they can interbreed. Because coyotes are not rare, this line of
argument would suggest that red wolves should not be protected under the Endangered Species
Act. On the other hand, because red wolves and coyotes differ in many ways, they can be viewed
as distinct species according to alternative species concepts (or to a less strict application of the
biological species concept). This approach would suggest that red wolves should be protected
because they are rare.
9. Focus on Evolution
All fertile humans alive today can interbreed to produce viable, fertile offspring. Thus, they
conform to the biological species concept and are assigned to a single species. For a new human
species to originate, a small breeding population would have to be geographically isolated from
all other populations for a long enough time that genetic drift, mutation, or natural selection (or
all of them together) would change the gene pool of this isolated population. If so much genetic
change occurred that the population became reproductively isolated from all other populations,
even when reunited with them in the same range, it would constitute a new human species. This
is unlikely given the increasing size and mobility of the human population.
10. Focus on Information
Sample key points:
 Overall, 50% of the DNA in F1 hybrids comes from each parent species.
 However, on a per-chromosome basis, 100% of the DNA in one member of each
homologous pair is from one parent species, while 100% of the DNA in the other homolog is
from the other parent species.
 In later-generation hybrid offspring, genetic recombination will produce chromosomes that
contain a mixture of DNA from each parent species.
 At one genetic locus, natural selection may favor DNA from one parent species, while at
another locus, selection may favor DNA from the other parent species.
Sample top-scoring answer:
When parent species with homologous chromosomes produce hybrid offspring, overall 50% of
the DNA in the F1 hybrids comes from each parent species. On a per-chromosome basis,
however, 100% of the DNA in one homolog is from one parent species, while 100% of the DNA
in the other homolog is from the other parent species. In later-generation hybrid offspring,
genetic recombination will produce chromosomes that contain a mixture of DNA from each
parent species. This mixing is similar to what happens when members of a single species have
offspring, where recombination produces chromosomes that contain a mixture of maternal and
paternal DNA. In addition, at one genetic locus natural selection may favor DNA from one
parent species, while at another locus selection may favor DNA from the other parent species.
Selection can therefore cause chromosomes to consist primarily of DNA from one parent species
or the other.
11. Synthesize Your Knowledge
In allopatric populations of D. pumilio, genetic drift may have led to the fixation of one color
preference over another; sexual selection (perhaps combined with natural selection for intense
warning colors) then increased the intensity of coloration. In so doing, sexual selection would
have had the indirect result of increasing the reproductive isolation between the populations. In
sympatric populations, sexual selection/mate choice could have acted as a reproductive barrier
(like the example in the text of cichlids in a lake).
CHAPTER 23 BROAD PATTERNS OF EVOLUTION
Scientific Skills Exercise
Teaching objective: In this scientific skills exercise, students consider how the fossil record may
be used to explore the link between ecological factors and species longevity. Specifically,
students are asked to estimate quantitative data from a graph to determine whether differing
modes of larval dispersal explain species longevity within one taxon of marine snails. Students
then propose a hypothesis to explain differences in longevity of species with planktonic and
nonplanktonic larvae.
Teaching tips: A version of this Scientific Skills Exercise can be assigned in MasteringBiology.
Hansen (1978) analyzed the fossil record of the Volutidae (volutes), a taxon of widely
distributed, burrowing predatory or scavenging marine snails with distinctive spiral shells. Some
extinct volute species had swimming larvae that lived as plankton for long periods, allowing
them to disperse over great distances on ocean currents. Hansen points out that in living snails,
such larvae commonly disperse across the Atlantic Ocean. Other extinct volutes underwent direct
development. The dispersal of such snails is limited to the distance they can crawl as adults.
Because their larvae are not carried by ocean currents, they are incapable of crossing ocean
basins.
Hansen studied the stratigraphic distribution of volutes in outcrops of sedimentary rocks
located along North America’s Gulf Coast. These rocks, which formed during the early
Paleogene period from 66–37 million years ago, are an excellent source of well-preserved snail
fossils. Hansen was able to classify each fossil species of volute snail as having planktonic or
nonplanktonic larvae based on features of the earliest formed whorls of the snail’s shell.
Encourage your students to consider how the life histories of snails with planktonic and
nonplanktonic larvae would be expected to differ. When snails have long-lived and planktonic
larvae, the larvae and young adults will likely encounter conditions that are very different from
those to which their parents are adapted. Snails with planktonic larvae would have had much
higher fecundity than snails with nonplanktonic larvae, because the mortality rate of planktonic
larvae and the young adults that metamorphose from them were likely to be very high. Snails
with nonplanktonic larvae disperse only as far as the adult snail can crawl. Both larval and adult
snails will survive well, as they will find themselves in conditions similar to the ones that their
parents lived in.
Although snail species with nonplanktonic larvae had high larval and young adult
survival, they formed isolated populations with limited distribution. As a result, these snails were
vulnerable to changes in local conditions. If environmental conditions changed in ways that did
not support snail survival, snails with nonplanktonic larvae were prone to extinction. In contrast,
snail species with planktonic larvae likely had high larval and young adult mortality. However,
because these species had much broader distribution, they were not vulnerable to local extinction
and thus had higher species persistence.
Answers:
1. Here are the persistence times for the snails:
Volute species with
nonplanktonic larvae
1.7
1.7
1.7
4.1
2.8
2.4
2.4
0.9
1.5
Volute species with
planktonic larvae
5.9
5.9
5.9
1.8
1.8
6.3
1.7
9.0
2.8
1.5
4.1
2.8
2.8
1.8
1.8
1.8
1.8
2.8
2.8
2.8
2.8
2.8
2.8
2.8
1.1
1.1
1.1
1.1
2.3
2.7
2.7
7.1
3.6
2. The mean persistence time for species with planktonic larvae is 4.4 million years, while
the mean persistence time for species with nonplanktonic larvae is 2.2 million years.
3. 25 volute snail species with nonplanktonic larvae appear between 60 and 37 million years
ago, while only 10 volute snail species with planktonic larvae appear during the same
period.
4. One hypothesis is that the difference in longevity is related to snails with nonplanktonic
larvae having more limited distributions than snails with planktonic larvae. Because of
their limited distributions, snail species with nonplanktonic larvae are likely to be isolated
by local barriers and may be vulnerable to changes in local conditions. If local conditions
change in ways that are unsuitable for the snails, species with nonplanktonic larvae may
become extinct. In contrast, snails with planktonic larvae have a broader distribution and
are less prone to extinction.
Interpret the Data
Figure 23.6 There are two speciation events and five extinctions in lineage A, while there are
five speciation events and one extinction in lineage B during the last 2 million years.
Figure 23.11 The blue curve is for marine animal families. Families often contain many species,
so we would expect the percentage of families that became extinct to be lower than the
percentage of species that became extinct.
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and a suggested rubric at the
beginning of this document.
6. Scientific Inquiry
The information presented in this problem suggests that the evolution of herbivory in moths and
butterflies may have contributed to their adaptive radiation. To test this hypothesis, other
comparisons could be made between a group of insects in which herbivory evolved and their
sister group in which it did not. If groups of herbivorous insects consistently have more species
than their nonherbivorous sister groups, we could conclude that the evolution of herbivory
promotes adaptive radiation in insects. [This hypothesis has been tested; see C. Mitter et al., The
phylogenetic study of adaptive zones: has phytophagy promoted insect diversification? The
American Naturalist 132:107–128 (1988).]
7. Focus on Evolution
A factor (such as the separation of continents by continental drift) that reduces gene flow in
many groups of organisms promotes speciation on a grand scale, thereby influencing the
diversity of life seen in the fossil record. As discussed in Chapters 21 and 22, genetic drift can
lead to the fixation of alleles, causing reproductive isolation and again promoting speciation and
influencing patterns of diversity seen in the history of life. Adaptive evolution by natural
selection at a wide range of loci—including those that influence development—is a major cause
of the evolution of new morphological forms as well as the high rates of speciation that occur
during adaptive radiations.
8. Focus on Organization
Sample key points:
 There is generally a good match between form and function.
 However, structures are not designed to perform their function in the best possible way. If
they were, bird wings might have arisen from scratch, leaving the forelimbs free to perform
other tasks.
 Instead, new structures evolve by “tinkering,” a process in which new forms arise by the
modification of existing forms.
Sample top-scoring answer:
The structure of a feature often gives clues as to its function—there is, in general, a good match
between form and function. However, this match does not indicate that the structure has been
designed to perform its function in the best possible way. For example, although it might have
been advantageous to do so, bird ancestors did not sprout wings de novo from the middle of their
backs, leaving the forelimbs free to perform other tasks. Evolution does not work in that way.
Instead, new structures often arise as modifications (exaptations) of existing structures by a
“tinkering” process. Assuming that the necessary genetic variation exists, natural selection can
favor some structural variants over others, leading to improvements in function. Thus, wings in
birds originated as the forelimbs of bird ancestors were modified over time. In principle, wings
might have originated in some other way, but they could not have arisen completely from
scratch. Some existing structure would have to have been modified over time.
9. Synthesize Your Knowledge
The episode of volcanic eruptions at the end of the Permian period are thought to have produced
enough carbon dioxide to warm the global climate by 6°C, harming many temperature-sensitive
species. The eruptions are also thought to have contributed to ocean acidification (harming many
shell-building and reef-building species) and to have caused oxygen concentrations to drop
throughout the oceans (harming oxygen-breathers). At roughly the same time, continental drift
led to the formation of the supercontinent Pangaea. The congregation of Earth’s landmasses into
a one giant continent had a huge impact on the physical environment and climate, which drove
some species to extinction and provided new opportunities for groups of organisms that survived
the crisis. Overall, the mass extinction that occurred at the time of the end-of-the-Permian
volcanic eruptions and the formation of Pangaea greatly altered the history of life on Earth –
96% of marine animal species became extinct, and it took millions of years for the diversity of
life to recover.
CHAPTER 24 Early Life and the Diversification of Prokaryotes
Scientific Skills Exercise
Teaching objective: Students build scientific skills by interpreting data in a bar graph and by
identifying elements of experimental design. Specifically, students are asked to state the
hypothesis tested in an experiment; to identify the independent variable, dependent variable, and
controlled variables; and to explain how the experimenters provided replication. They draw a bar
graph from a table of data and are asked to explain whether the experimental results support or
disprove the experimental hypothesis.
Teaching tips: A version of this Scientific Skills Exercise can be assigned in MasteringBiology.
The rhizosphere, the soil layer surrounding plant roots, is not a passive layer of dirt but rather a
complex community in which archaea, bacteria, fungi, and plants interact with one another. Plant
roots may secrete over 20% of the carbon-containing compounds synthesized by the plant at the
root-soil interface. As a result, rhizosphere bacteria and archaea are far more metabolically active
than those in surrounding soils.
When crop plants are attacked by fungal or bacterial pathogens, such disease outbreaks
may lead to the development of disease-suppressive soil. Plants grown in these soils are
protected from pathogen attack. Mendes et al (2011) investigated whether the protective effects
of disease-suppressive soils were due to microbial action. The soil was obtained from an
agricultural field in the Netherlands in which sugar beet crops had previously been attacked by
Rhizoctonia solani, a fungal pathogen that also afflicts potatoes and rice. The researchers
collected disease-suppressive soil at a depth of 0-30 cm from 25 random sites in the Dutch sugar
beet field. Other soil was collected from the same depth at the grassy margin of the field. This
soil did not offer protection against pathogens.
Eight beet seedlings were grown in four pots in each of five different soil treatments: (1)
disease-suppressive soil, (2) soil from the margin of the field, (3) soil from the field margin
supplemented with 10% suppressive soil, (4) disease-suppressive soil heated to 500C for one
hour, and (5) disease-suppressive soil heated to 800C for one hour. All sugar beet seedlings were
grown in greenhouses at 240C, 70% relative humidity, and a 16-hour light: 8-hour dark regime.
The soil was inoculated with mycelia from the fungal pathogen Rhizoctonia solani. 20 days after
inoculation, the number of infected sugar beet seedlings was counted and the percentage disease
incidence was determined for each soil treatment.
Many general biology students need practice in preparing and interpreting graphs. Students who
have difficulty in identifying dependent and independent variables will be unable to set up graph
axes correctly or to explain graphical information in words. Clarify the difference between
dependent and independent variables for your students and check that they have plotted the
independent variable on the x-axis and the dependent variable on the y-axis. Explain that they are
graphically representing how incidence of fungal disease in sugar beet seedlings varies with five
different soil treatments.
Answers:
1. The researchers were testing the hypothesis that the disease-suppressive properties of soils are
due to the activities of soil microorganisms. The independent variable is the soil treatment, and
the dependent variable is the percentage of the sugar beet seedlings with fungal disease for each
soil treatment.
2. Total number of pots = 5 treatments x 4 pots each = 20 pots.
4 pots per treatment x 8 plants per pot = 32 plants.
Using multiple pots per treatment and multiple plants per pot gave the researchers a larger
sample size and reduced the chances that conditions in one particular pot or the condition or
characteristics of one particular plant would drive the results of the experiment in a
nonrepresentative way.
3.
Disease-suppressive soil was very successful in protecting sugar beet seedlings from fungal
disease: Only 3% of the seedlings grown in this soil had fungal disease. In contrast, 62% of sugar
beet seedlings grown in soil from the margin of the field have fungal disease. When 10%
disease-suppressive soil was added to soil from the margin of the field, sugar beet seedlings were
provided with some protection against fungal disease: 39% of the seedlings had fungal disease.
When disease-suppressive soil was heated to 50°C for one hour, it lost some of its diseasesuppressive properties: 31% of sugar beet seedlings grown in this soil had fungal disease.
Heating disease-suppressive soil to 80°C for one hour eliminated all of its disease-suppressive
properties: 70% of sugar beet seedlings grown in this soil had fungal disease.
4. Yes, the statement that “…disease suppressiveness [of soil] toward Rhizoctonia solani was
microbiological in nature” is supported by the results shown in the graph. Heating the diseasesuppressive soil killed any microorganisms living in that soil. Because the infection rates for
seedlings grown in heated soil were higher than the rates for seedlings grown in unheated
disease-suppressive soil, the researchers concluded that microorganisms must account for the
soil’s disease-suppressive properties. When disease-suppressive soil was heated at 50°C for an
hour, the property of disease suppression was partially lost. This is consistent with partial
mortality of disease-suppressive microbes and partial denaturation of microbial enzymes at 50°C.
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and a suggested rubric at the
beginning of this document.
8. Scientific Inquiry/Interpret the Data
(a)
(b) Some Rhizobium strains are much more effective at promoting plant growth than other
Rhizobium strains; the most ineffective strains have little positive effect (plant growth with these
strains differs little from plant growth in the absence of Rhizobium). The ineffective strains may
transfer relatively little nitrogen to their plant host, limiting plant growth.
9. Focus on Evolution
Discontinuing treatment while bacteria are still present frees the bacteria from the inhibiting
effects of the drugs, and their population is likely to rebound, returning the disease symptoms
along with them. When antibiotic therapy is started, the nonresistant strains are the first to be
eliminated, leaving the resistant strains still alive. Even resistant strains may succumb if high
levels of the drug are present for a long enough time. However, if the antibiotic therapy is not
completed, then it will be these resistant strains that repopulate the host and, potentially, get
spread to other individuals.
10. Focus on Energy and Matter
Sample key points:
 The activities of life require energy.
 There is no sunlight in hydrothermal vent communities, and hence there are no
photoautotrophs.
 Instead, the producers in the ecosystem are chemoautotrophs that obtain energy by oxidizing
chemicals such as H2S.
 Energy is transferred when consumers utilize sugars produced by chemoautotrophic bacteria
or eat other consumers.
 Energy is also transformed into other forms, such as kinetic energy.
Sample top-scoring answer:
The activities of life require energy. In hydrothermal vent ecosystems, no sunlight is available
and hence there are no photoautotrophs. Instead, the producers on which other organisms rely for
energy are chemoautotrophic bacteria. These bacteria obtain energy by oxidizing chemical
compounds such as hydrogen sulfide (H2S), and they use that energy to synthesize sugar
molecules. Additional transfers of energy occur when consumers utilize sugars produced by the
chemoautotrophic bacteria or when consumers eat other consumers. As in other ecosystems,
organisms in hydrothermal vent communities also transform chemical energy into other forms,
such as kinetic energy (the energy of motion).
11. Synthesize Your Knowledge
As a result of their small size, bacterial populations can consist of trillions of individuals that live
in small regions, such as a few handfuls of soil or the digestive tract of a person. Many species of
bacteria have rapid rates of reproduction, which can enable their populations to reach large
numbers in short periods of time (hours or days). Their often rapid rates of reproduction also can
cause mutations to accumulate in their populations after short periods of time. This contributes to
the high levels of genetic variation found in their populations. Additional genetic diversity arises
from genetic recombination—bacteria reproduce asexually by binary fission, but genetic
recombination occurs in their populations as a result of transformation, transduction, and
conjugation.
CHAPTER 25 THE ORIGIN AND DIVERSIFICATION OF EUKARYOTES
Scientific Skills Exercise
Teaching objective: In this exercise, students interpret a comparison matrix of rRNA gene
sequences for five bacterial species and a plant mitochondrion. Students assess the similarity of
the six rRNA gene sequences and interpret the significance of their findings.
Teaching tips: A version of this Scientific Skills Exercise can be assigned in MasteringBiology.
Explain to your students that DNA comparisons provide information that can be used to
reconstruct the evolutionary history of related groups of organisms. As a general rule, the greater
the similarities between DNA sequences of living organisms, the more recently the lineages
leading to these taxa separated. When students analyze a comparison matrix for cloned rRNA
gene sequences of five bacterial species and the mitochondrial rRNA gene of wheat, they should
recognize that bacterial sequence that is most similar to the mitochondrial sequence identifies the
bacterium that is most closely related to the ancestral bacterium that evolved into a
mitochondrion.
Many alpha proteobacterial species are intimately associated with eukaryotic hosts, establishing
intracellular associations with their hosts. For example, Rhizobium species live in nodules in the
roots of legumes and fix atmospheric N2 into ammonia, which can be used by the host plant to
make proteins. Agrobacterium species live within plant tissues, where they grow as tumors. This
feature of alpha proteobacteria is likely relevant to the origin of mitochondria, which arose as
mutualistic endosymbionts and evolved into essential eukaryotic organelles.
You can expand on this Scientific Skills Exercise, if you wish. Once students have completed the
exercise, ask them whether the rRNA gene of A. tumefaciens appears to be more closely related
to the rRNA gene of the wheat mitochondrion or to the other bacterial species. They should note
that the similarities between A. tumefaciens and the other bacterial species are all over 50%,
whereas there is only 48% similarity between the rRNA genes of A. tumefaciens and the wheat
mitochondrion. Next, ask students to propose a hypothesis to explain the greater similarity
among bacterial rRNA gene sequences than between the rRNA gene sequences of a wheat
mitochondrion and the most similar bacterial species. There are two possible hypotheses that
may be proposed to explain this finding. The first hypothesis is that the bacterial ancestor to
mitochondrion arose before the other bacterial lineages diverged from one another. If this were
true, the greater similarity between the rRNA genes of alpha proteobacterium and the wheat
mitochondrion is coincidental. The second hypothesis is that the mitochondrial rRNA gene
sequence has evolved more rapidly than those of the bacterial species, due to a higher rate of
accumulation of mutations.
Now tell students that researchers also compared the rRNA genes of the bacterial species and
the wheat mitochondrion with the rRNA gene of the archaean Methanococcus vannielii. The
researchers found that all of the bacterial sequences are equally similar to the archaeal sequence,
while the mitochondrial sequence is significantly less similar to the sequence of the archaean.
Ask students whether this finding supports their hypothesis. When Yang et al (1985) compared
the rRNA genes of the five bacteria and the wheat mitochondrion to the rRNA gene of the
archaean Methanococcus vannielii, they found that all of the bacterial rRNA gene sequences are
equally similar to the archaeal sequence, while the mitochondrial sequence was significantly less
similar. This does not support the first hypothesis outlined above but does support the second
hypothesis. Archaea and bacteria diverged long before the origin of the mitochondrion.
All of the species included in the comparison matrix – the five bacteria and the wheat
mitochondrion – have had an equally long evolutionary history since the bacterial and archaeal
lineages diverged. All bacterial rRNA gene sequences have accumulated an equivalent number
of mutations since this divergence. It is important to clarify for your students that rates of
evolutionary change may differ among lineages. The rRNA gene sequence of the wheat
mitochondrion is significantly less similar to that of the archaeal cell, suggesting a more rapid
rate of accumulation of mutations.
Answers:
1. The value of the cell that represents the comparison of C. testosteroni and E. coli is 61. This
means that C. testosteroni and E. coli are identical at 61% of the 617 nucleotide positions in
the rRNA gene that are compared in the matrix. It is unnecessary to make the comparisons
represented by a dash, as they would pair up a species with itself, which is pointless. The
gray cells in the bottom left half of the matrix would duplicate the comparisons – and thus
the numbers – from the top right half.
2.
The researchers wished to compare the gene that codes for the small-subunit rRNA
molecule in a plant mitochondrion to the same gene in a range of bacterial species in order to
determine which bacterium is most closely related to ancestor of mitochondria.
3.
Agrobacterium tumefaciens, an alpha proteobacterium, is identical to the wheat
mitochondrion at 48% of the 617 nucleotide positions compared in the matrix. This suggests
that alpha proteobacteria are more closely related to ancestor of mitochondria than the other
bacterial taxa represented in the comparison matrix.
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and a suggested rubric at the
beginning of this document.
8. Focus on Evolution
Pathogens that share a relatively recent common ancestor with humans will likely also share
metabolic and structural characteristics with humans. Because drugs target the pathogen’s
metabolism or structure, developing drugs that harm the pathogen but not the patient should be
most difficult for pathogens with whom we share the most recent evolutionary history. Working
backward in time, we can use the phylogenetic tree to determine the order in which humans
shared a common ancestor with pathogens in different taxa. This process leads to the prediction
that it should be hardest to develop drugs to combat animal pathogens, followed by
choanoflagellate pathogens, fungal and nucleariid pathogens, amoebozoans, other protists, and
finally prokaryotes.
9. Focus on Interactions
Sample key points:
 Organisms interact with other organisms and the physical environment.
 Diatoms decrease in abundance when nutrient concentrations drop.
 When there are fewer diatoms, the abundance of many other organisms also changes.
 Drops in diatom populations also reduce carbon export to the ocean floor, potentially
affecting CO2 concentrations.
 CO2 concentrations can affect global temperatures and that, in turn, can affect other features
of the physical environment and other organisms.
Sample top-scoring answer:
Organisms interact with each other and the physical environment. Diatom populations decrease
in size when nutrients become scarce. A drop in diatom abundance affects other organisms, such
as the consumers that depend on diatoms for food, either directly (by eating them) or indirectly
(by eating organisms that eat diatoms). Thus, a change in a physical feature of the environment
(a drop in the concentration of available nutrients) can alter the abundance of diatoms and many
other organisms. Furthermore, a drop in diatom abundance can reduce the amount of carbon
exported to the ocean floor, thereby affecting a physical feature of the environment, the
concentration of CO2. Carbon dioxide levels affect global temperatures. As a result, other aspects
of the physical environment and other organisms (including those that do not depend on diatoms
for food) ultimately might be affected by a drop in the concentration of nutrients used by
diatoms.
10. Synthesize Your Knowledge
The amoeba changes shape as it engulfs its ciliate prey. Eukaryotic cells can change shape (or
maintain an asymmetric form) because they have a well-developed cytoskeleton that provides
structural support. Endosymbiosis has played a large role in the evolutionary history of the
eukaryotes. For example, current evidence indicates that mitochondria are descended from a
bacterium that was engulfed by a cell from an archaeal lineage—a primary endosymbiosis event
that led to the origin of the eukaryotes. Current evidence also suggests that plastids evolved from
a cyanobacterium that was engulfed by a heterotrophic eukaryote—a second primary
endosymbiosis event that ultimately led to the origin of red and green algae. Later, some red and
green algae were engulfed by heterotrophic eukaryotes (secondary endosymbiosis), leading to
the origin of still other major groups of eukaryotes (including the alveolates, to which ciliates
belong). Tubulinid amoebas share a more recent common ancestor with other amoebozoan
protists than they do with plants, fungi, or animals. As a result, tubulinids are more closely
related to other amoebozoans than they are to plants, fungi, or animals. However, this is not true
for all protists. For example, tubulinids are as closely related to fungi and animals as they are to
nucleariids and choanoflagellates. Likewise, tubulinids are as closely related to plants as they are
to protistan members of the Archaeplastida.
CHAPTER 26 THE COLONIZATION OF LAND
Scientific Skills Exercise
Teaching objective: Students build scientific skills by reading and interpreting two tables that
provide information on the genome and gene expression patterns of mycorrhizal and
nonmycorrhizal fungi.
Teaching tip: A version of this Scientific Skills Exercise can be assigned in MasteringBiology.
Answers:
1. (a) L. bicolor has 505 such genes, which is more than the other species studied. (b)
Mycorrhizal fungi exchange nutrients with the roots of their plant hosts and would be expected
to use membrane transporters in this exchange. The fungi supply the plants with minerals such as
phosphate ions, and the plants supply the fungi with organic nutrients such as sugars.
2. (a) The number of genes for small secreted proteins (SSPs) is much higher for Laccaria than
for any of the other fungi tested. (b) Genes for secretory proteins would have a signal-peptide
coding sequence at the leading end. The remainder of these proteins would presumably lack
transmembrane sequences. (c) A reasonable hypothesis is that SSPs play important roles in the
mycorrhizal interaction. Not enough information is given to suggest a more specific hypothesis.
3. (a) The number 22,877 indicates that the amount of RNA transcript of this gene made in
mycorrhizal parts of the fungus was 22,877 times as high as the amount made in soil mycelium
away from plant roots. (b) The finding that two of the four most upregulated L. bicolor genes
encode SSPs is consistent with the hypothesis that SSPs are important for mycorrhizae. (c) The
four genes listed are all highly upregulated in the mycorrhizae of both types of tree, but the
relative amounts of upregulation differ. For example, the most upregulated gene in fir is the first
one listed, but in poplar the second. One hypothesis would be that a tree’s genome and its
expression influence gene expression in a symbiotic fungus (and vice versa). Another
hypothesis might be that differences in the trees’ environments affect which genes are expressed
most.
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and a suggested rubric at the
beginning of this document.
7. Scientific Inquiry/Interpret the Data
As indicated by the raw data and bar graph, grass plants with endophytes (E+) produced more
new shoots and had greater biomass than did grass plants that lacked endophytes (E–). These
differences were especially pronounced at the highest soil temperature, where E– grass plants
produced no new shoots and had a biomass of zero (indicating they were dead).
8. Focus on Evolution
Plants, being immobile, can store large amounts of food in their tissues, such as the storage of
starch in their roots; seed plants also have a supply of food in their seeds. Such supplies may help
plants survive during times when photosynthetic rate cannot match the rate of cellular
respiration. Some plants also have stems with meristem underground, which may provide the
ability to resprout after an aboveground conflagration. Moreover, sporopollenin, seed coats, and
tough cell walls may provide a higher degree of protection from harsh physical conditions.
9. Focus on Interactions
Sample key points:
 Tree lycophytes had small leaves called microphylls.
 These leaves would not have intercepted much sunlight, causing the forest floor to get a lot
of light.
 Hence, there might be competition for light among the small plants living beneath tree
lycophytes.
 Ferns and seed plants have larger leaves (megaphylls), so the forest floor beneath trees of
these species would receive less light.
 In such a forest, ecological interactions might favor small species that thrived in
comparatively shady or cool conditions.
Sample top-scoring answer:
Because the leaves (microphylls) of tree lycophytes were small, a considerable amount of light
would have reached the ground in this type of forest. In such an environment, competition for
light might occur among species living beneath the trees. Among the small species, those that
were taller or that had broader leaves, for example, might outperform other species. In a forest
where the trees were ferns or seed plants, the much larger leaves (megaphylls) of these species
would have intercepted much of the sunlight, causing less light to reach the forest floor. Among
the small plants living beneath these trees, ecological interactions might favor species that
thrived in relatively shady or cool conditions (rather than favoring species that could shade
others; the trees towering above the forest floor were doing that already).
10. Synthesize Your Knowledge
Stomata are specialized pores that support photosynthesis by allowing the exchange of CO2 and
O2 between the plant body and the outside air. Because plants are always exposed to the air, they
run a much greater risk of drying out than did their algal ancestors. One way that plants cope
with this risk is by closing the stomata in hot, dry conditions, thereby reducing water loss. Many
plants also have a waterproof covering, the cuticle, which helps to reduce water loss. Other
adaptations in plants that facilitate life on land include the formation of mycorrhizae (aiding in
the uptake of nutrients from the soil), sporopollenin-enriched spore walls (protecting against
harsh conditions), a multicellular dependent embryo (with the mother plant providing protection
and nourishment for the embryo), and the origin (in vascular plants) of an extensive vascular
system (allowing for the transport of water and nutrients throughout the plant body). The waterconducting cells in vascular plants are strengthened by lignin, which provided support against
gravity and enabled plants to transport water and nutrients high above the ground. Tall plants
could also outcompete short plants for access to the sunlight needed for photosynthesis. Vascular
plants also had branched sporophytes, which enabled their bodies to become more complex. As
plant bodies became both taller and increasingly complex, competition for space and sunlight
probably increased. That competition may have stimulated still more evolution in vascular
plants, eventually leading to the formation of the first forests.
CHAPTER 27 THE RISE OF ANIMAL DIVERSITY
Scientific Skills Exercise
Teaching objective: This exercise provides students with the opportunity to analyze
experimental design and interpret graphed results. Students are asked to identify the hypothesis
that is tested in an experiment and to identify dependent and independent variables. They are
asked to explain how the experimental treatments test the stated hypothesis, to summarize the
graphed results of the experiment in words, and to identify whether or not these results support
the initial hypothesis. Finally, students are asked to interpret the significance of the experimental
results in an evolutionary context.
Teaching tips: A version of this Scientific Skills Exercise can be assigned in MasteringBiology.
Point out to students that predators are important agents of natural selection that select for
successful behavioral or structural defenses in prey species. In turn, prey species act as agents of
natural selection in selecting for increased weaponry in predator species. The result is escalating
predator-prey arms races over evolutionary time. The invertebrate paleontologist Geerat Vermeij
studies the fossil record of evolutionary arms races, focusing on molluscs and their predators. He
notes that present-day molluscs tend to have shells with thick walls, a narrow opening, and other
protective characteristics, whereas many of these features do not appear in fossils from the first
few hundred million years of mollusc history. Ask students whether and how evolutionary arms
races can be investigated by looking at modern populations of molluscs that are under attack by
new, invasive predators. Rochette et al (2007) studied the question by looking at a predatory crab
and its snail prey in the Gulf of Maine. Carcinus maenas, the European green crab, is included
on the global invasive species database as one of the world’s 100 worst invasive alien species.
The crab is native to the coast of Europe and North Africa. It first appeared in Massachusetts in
1817 and has now spread along the Atlantic coast of North America from Virginia to
Newfoundland. Littorina obtusata, the flat periwinkle, is a widely distributed marine snail that is
abundant in the Gulf of Maine.
Answers:
1. The researchers’ hypothesis was that southern periwinkle populations with a history of
predation by crabs have become less vulnerable to crab predation through evolution.
Independent variables were the source of the periwinkles and the crab populations (northern
or southern sites in the Gulf of Maine). The dependent variables were the number of northern
and southern periwinkles killed by the crabs.
2. The four treatments enabled the researchers to compare each combination of crab and
periwinkle origins in order to test whether the origin (northern site or southern site) of the
crabs and periwinkles affected their interactions.
3. The researchers wanted to rule out the possibility that northern crabs didn’t like the taste of
southern periwinkles or that southern crabs didn’t like the taste of northern periwinkles. This
experiment allowed researchers to conclude that differences in numbers of southern and
northern periwinkles killed were due to differing vulnerability to crab predation, not due to
crab preferences.
4. Sample answer: Crabs are more successful in preying on northern periwinkles. Yes, this
supports the hypothesis in Question 1, which predicts that southern periwinkles have become
less vulnerable to predation by the crabs.
5. Sample answer: Southern populations of flat periwinkles have accumulated traits that make
them less vulnerable than northern populations to predation by European green crabs.
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and a suggested rubric at the
beginning of this document.
7. Scientific Inquiry/Draw It
(a) Because brain size tends to increase consistently in such lineages, we can conclude that
natural selection favored the evolution of larger brains and hence that the benefits outweighed
the costs. (b) As long as the benefits of brains that are large relative to body size are greater than
the costs, large brains can evolve. Natural selection might favor the evolution of brains that are
large relative to body size because such brains confer an advantage in obtaining mates and/or an
advantage in survival.
(c)
Adult mortality tends to be lower in birds with larger brains.
8. Focus on Evolution
The circled clade should include the birds, the ornithischian dinosaurs, and the saurischian
dinosaurs other than birds, plus the common ancestor of the dinosaurs. The phylogeny shows that
dinosaurs other than birds are nested between crocodilians and birds. Since crocodilians and
birds differ with respect to whether they are endothermic, we cannot use phylogenetic bracketing
to predict whether dinosaurs other than birds were endothermic (or not). However, we can
conclude that the dinosaur that gave rise to birds was endothermic, as are all living birds.
9. Focus on Organization
Sample key points:
 Early tetrapods diversified into a wide range of species.
 Early tetrapods could not run for long distances because their sprawling gait prevented
normal breathing while running.
 Tetrapod lineages that solved this problem (such as dinosaurs) would have had a competitive
advantage.
 A predator that could run without compressing its lungs could pursue a sprawling-gait prey
until its victim ran out of breath.
 This could lead to increases in the abundances of species that could breathe while running
and declines in sprawling-gait lineages.
 A change in limb structure that enabled individuals to run and breathe at the same time could
have led to emergent changes in the biological community—the rise of one major group of
organisms and the decline of another.
Sample top-scoring answer:
Although early tetrapods diversified into a wide range of species, their sprawling gait prevented
them from breathing while running. Tetrapod lineages that solved this problem, such as the
dinosaurs, had a competitive advantage. For example, a predator that could run without
compressing its lungs could breathe while running and hence pursue a sprawling-gait prey until
its victim ran out of breath; similarly, prey that could breathe while running might be more
difficult for predators with a sprawling gait to catch. The outcomes of such interactions could
have led to increases in the abundances of species that could breathe while running and decreases
in species with a sprawling gait. Thus, over time, a change in limb structure that enabled
individuals to breathe normally while running could have led to emergent changes in the
biological community—the rise of one group of organisms (for example, the dinosaurs) and the
decline of others (for example, early tetrapod lineages).
10. Synthesize Your Knowledge
No. Invertebrate species in phylum Chordata are more closely related to vertebrates than they are
to other invertebrates. Thus, a group comprised of all of Earth’s invertebrate species would not
be monophyletic since it would exclude some descendants (the vertebrates) of the group’s most
recent common ancestor. In terms of evolutionary history, the first animals were invertebrates.
Early in animal evolution, some invertebrate groups such as sponges and (later) cnidarians
diverged from all other animal groups. Most invertebrates are bilaterians, an enormous clade that
originated after cnidarians diverged from other animals and that has radiated into the three major
clades shown in Figure 27.10: Lophotrochozoa, Ecdysozoa, and Deuterostomia. The bilaterian
radiation of marine invertebrates that began about 535 million years ago radically altered life in
the oceans. Prior to that time, virtually all large animals had soft bodies. During the Cambrian
explosion, large predators emerged that had claws and other features for capturing prey;
simultaneously, new defensive adaptations, such as sharp spines and heavy body armor,
appeared in their prey. Earlier, soft-bodied animals declined in abundance, most likely because
they would have been easy prey for the newly evolved large invertebrate predators.
CHAPTER 28 PLANT STRUCTURE AND GROWTH
Scientific Skills Exercise
Teaching objective: This exercise is designed to give students practice in constructing and
interpreting a bar graph, and to introduce them to the concept of phenotypic plasticity, the
capacity of an organism with a specific genotype to produce different phenotypes when exposed
to different environments.
Teaching tips: A version of this Scientific Skills Exercise can be assigned in MasteringBiology.
Many students labor under the misconception that morphological traits are hardwired by genes.
They think that if an organism has a certain genotype, it will look a certain way. More often,
however, an organism’s appearance is a product of genes and environment. This exercise can
also be used to introduce the concept of proxy data, data obtained from organisms that are
sensitive to climatic phenomena. In fact, the “toothiness” of leaf fossils of known age has been
used as proxy data by paleoclimatologists to estimate the temperature of climates in the past.
Answers:
1.
Moving from north to south, the leaves of Acer rubrum trees have larger but fewer teeth.
2. Genetics plays a large role in determining the phenotypes of the leaves. Regardless of where
the seeds were grown, there is generally an increase in teeth size and a decrease in teeth number
that corresponds to the latitudinal origins of the collected seeds, from north to south. There is
also some indication that leaf morphology is partially attributable to the latitude at which it is
grown, but this effect is small. For example, there is no dramatic difference in average tooth area
between seeds grown in the north and south for three of the four seed collections. Only the seeds
collected in Florida showed a dramatic difference when grown in Florida versus Rhode Island.
3. If a 10,000-year-old fossil of an A. rubrum leaf with many teeth in its margin were discovered
in South Carolina, we could infer that South Carolina was much colder 10,000 years ago than it
is today because 4.2 teeth per cm2 of leaf area is similar to the number of teeth in leaves of trees
from Ontario, Canada, where the climate is much colder than in South Carolina.
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and a suggested rubric at the
beginning of this document.
7. Scientific Inquiry
Grazing animals that crop plants close to the ground have a more detrimental effect on eudicots
than on monocots because the removal of the lowest axillary buds prevents eudicots from
recovering. In contrast, the underground stems of grasses and the intercalary meristems of their
leaves are less affected by grazing. Thus, the presence of grazing animals selects for the survival
of grasses.
8. Focus on Evolution
Plant organs exhibit many exaptations. For example, in addition to their role as the primary
photosynthetic organs, leaves have evolved to have other roles in some species, such as tendrils
(which help the plant climb), spines (which deter herbivores), and bracts (which attract
pollinators). Stems, which normally enable plants to hold their leaves closer to the sun, have
evolved to have other roles, too. Stems can play a role in asexual reproduction (stolons and
rhizomes) and food storage (tubers). In some species, branches have been modified to become
thorns. In some cases, such as bulbs, the entire shoot has become modified for storage. All four
parts of the flower (sepals, petals, stamens, and carpels) are modified leaves.
9. Focus on Organization
Sample key points:
 Lignin is a tough polymer found in the cell walls of sclerenchyma cells and vascular cells,
such as tracheids and vessel elements.
 Fibers are sclerenchyma cells that give rigidity to stems during primary growth, enabling
plants to grow taller.
 As photosynthetic organisms, plants need to reach sunlight to compete successfully against
other plants.
 Vessels and tracheids provide both strength to plant organs and a low-resistance pathway for
the long-distance transport of water—a problem that arose as plants became increasingly
taller.
Sample top-scoring answer:
Lignin is a tough, polymeric cell-wall substance found in sclerenchyma cells, such as fibers, and
in some vascular cells, such as tracheids and vessel elements. Lignified cells make plant tissues
stronger and more rigid, making it possible for plants to grow taller. Because plants are
photoautotrophs, their growth and survival depend on their shoots, particularly their leaves,
being exposed to optimal sunlight. Since all plants share this requirement for sunlight, over the
course of evolution there was competition to be tall to avoid shading by other plants. Moreover,
as plants evolved and became taller, the increasing separation of the leaf canopy (the site of most
water loss) from the root system (the source of most water uptake) made the long-distance
transport of water more problematic. Lignified vascular cells such as tracheids and vessel
elements became adapted over the course of evolution for the long-distance transport of water.
10. Synthesize Your Knowledge
(a) As shown in Figure 28.14, a root has a solid vascular cylinder, in contrast to the three
separate vascular bundles in this micrograph. Also, this organ has stomata, which are not found
in roots. This evidence shows that the organ is not a root. Regarding whether it is a stem or a
leaf, there are layers of cells that look like the mesophyll cells in Figure 28.17, with many being
elongated and some loosely arranged with air spaces between them. This arrangement differs
from the tightly packed ground tissue cells in the stem micrographs in Figure 28.18. This
evidence indicates that the organ is a leaf rather than a stem. (b) This unusual cylindrical leaf has
a lower surface area-to-volume ratio than a typical flat leaf. This adaptation is well-suited for dry
conditions because there is less surface area for the evaporative loss of water.
CHAPTER 29 RESOURCE ACQUISITION, NUTRITION, AND TRANSPORT IN
VASCULAR PLANTS
Scientific Skills Exercise
Teaching objective: Generally, so much emphasis is placed on enzymes and metabolism in
introductory biology that students may lose sight of the fact that not every biological process
depends on metabolism. In this exercise, students will gain experience in calculating Q10 values
and arriving at mechanistic insights into how temperature affects life processes.
Teaching tips: A version of this Scientific Skills Exercise can be assigned in MasteringBiology.
Given the ease and relative inexpensiveness of measuring seed swelling, students might be
encouraged to design and execute individual experiments aimed at elucidating some aspect of
seed imbibition. Another possible exercise would be to have the students search the primary
literature and find the Q10 values for other biological processes.
Answers:
1. The data show that the initial uptake of water by radish seeds does vary slightly with
temperature. As temperature increases, so does water uptake.
2. (a) For 35ºC and 25ºC, Q10 = 36.2%/31.0% = 1.17
For 25ºC and 15ºC, Q10 = 31.0%/26.0% = 1.19
For 15ºC and 5ºC, Q10 =26.0%/18.5% = 1.41
(b) The average Q10 is 1.26. (c) Because Q10 for the process is closer to 1 than to 3, it is more
likely to be a physical process. (d) Yes, because their Q10 values are similar, it is possible that the
slight temperature dependence of water uptake by seeds could be a reflection of the slight
temperature dependence of the viscosity of water.
3. Answers will vary, but should involve a treatment that inhibits metabolism. Sample answers:
The effects of a chemical inhibitor of metabolism such as cyanide could be assessed. Seeds
could be subjected to temperatures that would denature proteins (and inhibit metabolism)
before they were placed in water.
4. The Q10 of plant growth would be expected to be closer to 3. Growth requires energy (ATP)
for the anabolic synthesis of complex chemicals and thus depends on metabolism.
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and the suggested rubric at the
beginning of this document.
9. Scientific Inquiry
Acid precipitation will deposit more H+ in the soil solution. This H+ will displace cations that are
bound to soil particles by cation exchange. These released cations will now be in the soil
solution. The rainwater will mix with the soil solution, diluting it but also increasing its volume.
Some of this excess volume will leave the soil as runoff, carrying the released cations out of the
soil. One test of this hypothesis would be to lower the pH of artificial rain and determine whether
more cations leach out of the soil than normal.
10. Scientific Inquiry
The solute potential S is equal to -(2) x (0.05) mol/liter x (0.00831 liter MPa/ mol K) x (293 K)
= -0.243 MPa. Since  = S + P , the  of the soil will be more negative (lower) near the
walkway. The more negative water potential of the soil will promote water loss from the root
system and slow the growth of roots.
11. Focus on Evolution
At a depth of 25 m, there may not be sufficient light penetration for the photosynthetic rate to
exceed the rate of cellular respiration. Thus, to prevent holdfast cells from “starving” to death,
the photosynthetic organs (the blades) must be linked to the holdfasts by sugar-transport cells of
the stalk. Without a holdfast, the kelps might be dispersed to unsuitable environments, as
happens when a storm breaks the stalks and the kelps wash up on the shore.
12. Focus on Interactions
Sample key points:
 Plants form mutualisms with many forms of soil microorganisms.
 Plants provide leaf litter for earthworms to eat, while earthworms help aerate the soil.
 Plant roots support the rhizosphere by secreting carbohydrates into the soil.
 Rhizobacteria help plants grow by a variety of mechanisms.
 Plants also form mutualistic associations with mycorrhizal fungi that aid in mineral uptake.
 Legumes form highly specific mutualisms with N2-fixing rhizobial bacteria.
Sample top-scoring answer:
Plants form mutualistic relationships with many types of soil organisms. For example, plants
provide leaf litter that contributes to the diet of earthworms. Earthworms, in turn, help aerate the
soil, allowing more oxygen to reach the roots of plants. Many plants also form mutualistic
relationships with soil fungi, most notably mycorrhizal fungi. Plants provide mycorrhizal fungi
with carbohydrates, while the mycorrhizal fungi provide roots with minerals, most notably
phosphorous. The interactions between plant roots and bacteria are manifold. The carbohydrates
secreted by plants into the soil provide energy for the rhizosphere, which teems with many types
of helpful rhizobacteria. Some rhizobacteria help by producing chemicals that stimulate plant
growth or by protecting roots from disease. Still others make nutrients more available to roots or
absorb toxic metals. Legumes form a special mutualism with Rhizobium bacteria that involves
the production of root nodules. Legume plants provide Rhizobium with carbohydrates, while
Rhizobium can provide these plants with nitrogen by converting N2 to NH3.
13. Synthesize your Knowledge
As a water molecule in the soil solution, I am bound to other water molecules by hydrogen
bonds. We are being tugged toward a plant root by a negative pressure. We are taken up through
an aquaporin channel into the cytoplasm of a root hair or cortical cell. We move through
plasmodesmata into the stele past the Casparian strip and are quickly carried upward with the
xylem sap to the vicinity of mesophyll cells in a leaf. The stomata of the leaf are open, and some
water molecules on the surface of the mesophyll cells are breaking their hydrogen bonds and
being released into the vapor phase. As each one breaks off, I am drawn closer to the air-water
interface. At the interface, the negative pressure is intense. My hydrogen bonds break, and I
diffuse through a stoma to the air outside the leaf.
CHAPTER 30 REPRODUCTION AND DOMESTICATION OF FLOWERING PLANTS
Scientific Skills Exercise
Teaching objective: This exercise is designed to give students practice in applying the concept
of negative versus positive correlations in interpreting data.
Teaching tips: A version of this Scientific Skills Exercise can be assigned in MasteringBiology.
Although most students understand the concept of positive correlation, some think that a
negative correlation means “no correlation.” This exercise will serve as a diagnostic to identify
students who have this misconception. It also helps identify students who are adept at passively
understanding the results of studies discussed in class but who lack experience in grasping the
significance of data sets.
Answers:
1. (a) The number of seeds per flower and the number of visits per flower are positively
correlated with nectar volume. (b) The number of rooted branches per gram fresh weight is
negatively correlated with nectar volume. (c) No clear relationship exists between nectar
volume and nectar concentration.
2. (a) M. rupestris and M. eastwoodiae appear to be primarily asexual reproducers because they
put relatively more energy into making rooted branches and relatively less energy into
making nectar and seeds than do the other three species. (b) M. verbenaceus, M. cardinalis,
and (to a slightly lesser extent) M. nelson put relatively more energy into making nectar and
seeds than they put into making rooted branches; hence, they appear to be primarily sexual
reproducers.
3. (a) If a pathogen were to indiscriminately infect these five Mimulus species, the sexually
reproducing species would be more likely to avoid extinction because the genetic
recombination that occurs with each generation would increase the likelihood of a few
individuals having the special traits necessary to avoid or counteract the disease caused by
the pathogen. (b) Conversely, if hummingbird populations dwindled, then the asexually
reproducing Mimulus species would be at a relative advantage because pollination and seed
production would be low in the sexually reproducing species.
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and the suggested rubric at the
beginning of this document.
11. Scientific Inquiry
As the chief scientific advisor, I would reassure critics that GM foods, including canola, cotton,
maize, and soy, have been ingested for years by both humans and farm animals. In many
countries, the potential risks of GM foods were carefully assessed for genetic stability, allergic
response, toxic effects, and nutritional value before they were approved for use. Despite
widespread scrutiny, no scientific evidence has suggested that GM crops are more dangerous—
for people or animals—than conventional foods.
12. Science, Technology, and Society
Traditional breeding manipulates the genetic makeup of organisms indirectly, by selecting for
variants in existing populations, crossbreeding closely related species, and identifying and
breeding novel mutants. Genetic engineering involves the introduction of specific genes or
groups of genes into single cells, using recently developed biotechnologies. A genetically
modified (GM) plant is then regenerated from the GM cell. Some opponents are concerned with
potential risks to human health, including allergic reactions to foreign genes introduced into
plants or toxic effects of intermediary products. Other opponents are concerned about the
environmental effects of GM organisms; for example, crops that grow outdoors could potentially
cross-pollinate wild plants, introducing new genes into wild populations. Concerns about human
health from ingestion of GM organisms are currently unsupported by scientific evidence. The
environmental effects of GM organisms are not yet fully understood and require more study.
13. Focus on Evolution
As a result of their size, small populations have significant challenges in avoiding extinction. The
fact that self-incompatible species cannot produce seeds without a partner merely adds to the list
of challenges, particularly if members of the initial population are widely scattered. Long-term
survival, however, would be enhanced by the genetic mixing afforded by self-incompatibility.
Thus, the best adaptation for initial and long-term survival might be partial self-incompatibility.
14. Focus on Organization
Sample key points:
 Emergent properties in structures are those that are not obvious unless a structure is looked at
in its entirety.
 Petals attract animal pollinators.
 The sexual organs are the stamens, which produce pollen, and the carpels, which receive
pollen.
 Sepals protect the entire floral bud from herbivores.
Sample top-scoring answer:
The basic idea of emergent properties is that “the whole is greater than its parts.” This concept is
well exemplified by a flower. The petals, by means of their bright colors and scents, attract
animal pollinators. Without petals, insects would be much less likely to visit a flower and
transfer pollen from flower to flower. Stamens and carpels are clearly necessary for sexual
reproduction. Without stamens, a flower could not pollinate other flowers of the same species,
and without a carpel and its receptive stigma, a flower could not receive pollen from another
flower. Sepals, which are typically green and leaflike in appearance, enclose and camouflage the
immature floral buds, thereby protecting them from herbivores.
15. Synthesize Your Knowledge
A pollen grain is the haploid male gametophyte of a seed plant, along with a surrounding spore
wall. A pollen grain develops from haploid microspores by the process of mitosis and at maturity
consists of two cells, the generative cell and the tube cell. The function of a pollen grain is to
transport sperm to the female gametophyte. A pollen grain is transported from an anther of a
flower to the stigma of a flower (typically from one plant to another) by wind, water, or animals.
After a pollen grain is deposited on a stigma, the tube cell gives rise to a pollen tube that grows
through the long style toward the micropyle of the female gametophyte, or embryo sac. As the
pollen tube elongates, the generative cell divides by mitosis to form two sperm. These sperm are
deposited near the micropyle. In a process called double fertilization, one sperm fertilizes the
egg, forming a diploid zygote, while the other combines with the two polar nuclei of the female
gametophyte, giving rise to the triploid endosperm. In an evolutionary context, the advent of
pollen grains was important because pollen grains allow for the sperm delivery over great
distances by wind or animals, eliminating the need for water the transport of flagellate sperm.
CHAPTER 31 PLANT RESPONSES TO INTERNAL AND EXTERNAL SIGNALS
Scientific Skills Exercise
Teaching objective: This exercise is designed to give students practice in interpreting a bar
graph and in understanding experimental design and interpretation.
Teaching tips: A version of this Scientific Skills Exercise can be assigned in MasteringBiology.
Many students are surprised by the extent of chemical communication between plants. You may
wish to relate these findings to other examples in the book, such as the ability of lima beans to
warn their neighbors of attack by mites.
Answers:
1. The stomatal openings in plants 6-8 are smaller than those of the other plants; the stomatal
openings of plants 9 and 10 are intermediate. This indicates that plants 6-8 are experiencing
drought stress and have closed their stomata to reduce water loss. Plants 9 and 10 are starting
to close their stomata.
2. Pea plants apparently do signal their drought-stressed condition to their neighbors, as
evidenced by the decrease in stomatal openings seen in plants 7 and 8, which are neighbors
to plant 6. No effect was seen in plants 5 and 4, even though their shoots were the same
distance from plant 6 as those of plants 7 and 8. This suggests that communication of the
drought signal is through the root system, not the shoot system.
3. Preventing the chemicals from moving through the soil ensured that the mannitol did not flow
from the soil in pot 6 into the soil in pots 7 through 11 and directly cause drought stress in
these plants.
4. The fact there was a delay in stomatal closure in plants 9 through 11 relative to plants 6
through 8 suggests that there is a chain reaction. As the drought response is achieved in each
plant, it signals, after a delay, “news” of this fact to its immediate neighbor.
5. Water was added as a control to mimic the handling of the plants without a chemical that is
known to cause drought stress. The control experiment indicated no major effect of water
alone.
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and the suggested rubric at the
beginning of this document.
8. Scientific Inquiry
The biologist could employ a fan to see how the direction of the wind affects the unpalatability
effect: plants upwind should be rejected less frequently than plants downwind. Another
experiment would be to block movement of the “alarm” signal out of a damaged leaf by
enclosing the leaf in a plastic bag.
9. Focus on Evolution
A small seed has few food reserves to feed the germinating seedling. Thus, small seeds have to
become phototropically self-sufficient soon after germination. If they germinate in an
environment with too little light, they will perish. If the environment is too shaded, it is better for
small seeds to remain dormant until light conditions improve.
10. Focus on Interactions
Sample key points:
 Phytochromes are photoreversible pigments that switch between a red-absorbing form (Pr)
and far-red-absorbing form (Pfr), depending on light quality.
 Phytochromes are major determinants of shoot architecture. Shaded conditions lower the
Pfr:Pr ratio, which promotes shoot elongation and inhibits branching. Sunny conditions raise
the Pfr:Pr ratio, slowing shoot elongation and promoting branching.
 Phytochromes control the etiolation response that enables dark-grown seedlings to emerge
more quickly and become photosynthetic.
 Phytochromes regulate light-dependent seed germination.
Sample top-scoring answer:
Because plants are photoautotrophs, their growth and survival depend on their shoots,
particularly their leaves, being exposed to optimal sunlight. Phytochromes are photoreversible
pigments that play a major role in altering plant development in response to light. Phytochrome
pigments switch between a red-absorbing (Pr) and far-red-absorbing form (Pfr), depending on
light quality. The ratio of these two forms profoundly affects plant growth and development.
Many seeds remain dormant until light conditions are conducive for growth and survival. The
Pfr:Pr ratio determines whether such seeds can germinate or not. Phytochrome also controls the
etiolation response (rapid shoot elongation and suppressed leaf expansion) that enables darkgrown seedlings to emerge from the soil and become photoautotrophic more quickly. Finally,
shaded conditions lower the Pfr:Pr ratio of plant shoots, which promotes elongation and inhibits
branching. Sunny conditions raise the Pfr:Pr ratio, slowing shoot elongation and promoting
branching.
11. Synthesize Your Knowledge
The mechanical removal of the shoot tips by this grazing mule deer will decrease the plant’s
biomass and its ability to harvest resources from its environment. Wounding of the plant will
also make it more susceptible to pathogen invasion. The hormone balance of the plant will also
be affected. Since auxin is produced in shoot tips, removal of the shoot tips by the deer will
decrease export of auxin from the shoot that, in turn, will lower the synthesis of strigolactones.
The resulting decrease in strigolactones will induce the remaining lateral buds to begin
elongating until one of the resulting branches becomes dominant: the result will be a bushier
shrub.
CHAPTER 32 THE INTERNAL ENVIRONMENT OF ANIMALS: ORGANIZATION
AND REGULATION
Scientific Skills Exercise
Teaching objective: Students build scientific skills by analyzing quantitative data in a table.
Specifically, students will quantitatively compare osmolarity and urea concentration in blood and
urine from two experimental treatment groups of mice.
Teaching tips: A version of this Scientific Skills Exercise can be assigned in MasteringBiology.
The topic fits with teaching about blood osmoregulation and mammalian kidney function.
Students will interpret mean and standard deviation values provided. An option is to ask students
to draw bar graphs of the data sets. (These graphs are presented in the source publication.)
Another option is to have students calculate and compare the urine : blood ratios of osmolarity
and urea concentration for the two treatment groups.
Answers:
1. (a) large difference; (b) small difference; (c) large difference; (d) medium difference; (e) The
urine osmolarity and urea concentration are very different between the unlimited-water and nowater conditions, whereas the blood osmolarity and urea concentration are relatively stable. This
suggests that the mice are regulating their blood osmolarity by altering their urine composition.
2. (a) 1.4; (b) 14.7; (c) The mice have an adaptation that allows them to produce highly
hyperosmotic urine in low-water conditions.
3. Not at all. The extent to which filtrate is concentrated will determine the volume of urine
produced, but the calculation is not dependent on knowing the urine volume.
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and a suggested rubric at the
beginning of this document.
9. Interpret the Data
Compared with the human, the kangaroo rat gains a much larger share of its water from
metabolism and loses a much larger share of its water through evaporation (during gas
exchange).
10. Focus on Evolution
If kangaroo rats are adapted to moist conditions, they can better afford to lose water to perform
evaporative cooling than can kangaroo rats that are adapted to deserts. This hypothesis could be
tested by separately placing members of moist-adapted and dry-adapted populations in sealed
terrariums that contain humidity sensors. Subject both populations to the same high temperature
(one that would make evaporative cooling beneficial), and after some time has passed, determine
whether there are differences in the humidity of the air within the terrariums.
11. Focus on Organization
Sample key points:
 The descending limb of the loop of Henle is freely permeable to water.
 The ascending limb of the loop of Henle is impermeable to water.
 The permeability of the collecting duct is regulated.
 Water cannot be actively transported.
 Aquaporin channels allow water to diffuse across a membrane.
Sample top-scoring answer:
Kidney function in water recovery relies on structural differences in the membranes lining
particular portions of the nephron. Where aquaporin channels are present, as in the descending
limb of the loop of Henle, water can readily diffuse across the membrane in response to a
gradient of osmolarity, substantially reducing the volume of filtrate. Where aquaporins are
absent, the membrane is effectively impermeable to water. This condition is found in the
ascending limb of the loop of Henle. Here filtrate loses salt but not water, becoming more dilute
and hence enabling a further reduction in volume as filtrate passes through the collecting duct. In
response to increases in blood osmolarity, the hormone ADH is released, triggering insertion of
additional aquaporin channels into the collecting duct membrane. These additional channels
result in the recapture of water necessary to reduce blood osmolarity.
12. Synthesize Your Knowledge
The macaque has thick fur that provides insulation, reducing the rate of heat loss to the cold air.
As an endotherm, the macaque can use heat generated by metabolism for thermoregulation. In
addition, it displays a behavioral adaptation, taking advantage of the hot springs to warm its body
and reduce its need to generate heat metabolically.
CHAPTER 33 ANIMAL NUTRITION
Scientific Skills Exercise
Teaching objective: Students build scientific skills by interpreting data from an experiment
investigating the roles of appetite regulation genes using mice with various genetic mutations.
Teaching tip: A version of this Scientific Skills Exercise can be assigned in MasteringBiology.
Answers:
1. In pairings (b)–(d), a mouse lacking ob gene activity was paired with a mouse that was either
(b) also lacking ob gene activity, (c) wild type, or (d) lacking db gene activity. Pairings (a) and
(b), which each involved two mice with the same genotype, were designed to reveal whether the
experimental procedure by itself would have an effect on the phenotype of the subject mice.
Pairing (c) was designed to test whether wild-type mice could produce a circulating factor that
compensated for the lack of ob gene activity in the subject mice. Pairing (d) was designed to
clarify the separate roles of the ob and db genes in appetite regulation.
2. The ob/ob subject mice in (b) gained more than four times as much mass as the wild-type
subject mice in (a). If the average change in body mass had been the same in (a) and (b), it would
have indicated that surgically connecting pairs of mice affects their growth and masks the effect
of genotypic differences. Such a result would have necessitated a new experimental system or
design.
3. Whereas the ob/ob subject mice in (b) gained much more mass than the wild-type subject mice
in (a), the ob/ob subject mice in (c) had the same mass gain as wild-type mice. This result
indicates that the ob+ gene product appears to suppress appetite. Sharing circulation with a wildtype mouse counteracted the effect of lacking ob gene function in the subject mice. The product
of the wild-type (ob+) gene was transferred to the ob/ob mice through the blood circulation,
preventing the ob/ob mice from overeating.
4. As in (b), the pairing in (d) affected mass gain in ob/ob mutants, but in (d) the result was a
severe and destructive mass loss. This indicates that db/db mutant mice not only produce the
circulating factor but produce more of it than wild-type mice. A hypothesis that explains the
difference between the results in (d) and (b) is that db/db mutant mice produce the circulating
factor when their caloric intake is sufficient but cannot respond to the factor. To test this
hypothesis, you could pair a wild-type mouse with a db/db mutant mouse. You would expect the
wild-type mouse to exhibit the same extent of mass loss seen in pairing (d) and the db/db mutant
mouse to become obese.
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and a suggested rubric at the
beginning of this document.
8. Scientific Inquiry
Excess iron uptake results in blood iron levels that are higher than normal. You might
hypothesize that women suffer symptoms less frequently because menstruation reduces iron
levels in the body. In effect, a monthly menstrual cycle has the same effect as the common
treatment for the condition, the drawing of blood from patients on a periodic schedule.
9. Focus on Evolution
The high oxygen demands of active endotherms, such as mammals, require nearly continuous
breathing. Most other amniotes cannot breathe while chewing food because the connection
between their external nostrils and their esophagus is in the mouth. During mammalian
evolution, the connection between the nasal cavity and the esophagus was relocated far back in
the throat. As a result, breathing can continue through the nostrils while chewing takes place in
the mouth. This adaptation solved one problem (the need to breathe while chewing) but not
another (the need to keep air and food in separate paths). Because of this “imperfection,” the
paths of air and food cross each other, and choking sometimes occurs.
10. Focus on Organization
Sample key points:
 Cells synthesize proteins from amino acids, not from whole proteins in food or other sources.
 Cell structures are assembled by living cells and cannot be replaced by molecules applied to
the cell surface.
Sample top-scoring answer:
The emergent properties of proteins in an animal body result from the synthesis of particular
proteins at a particular time and location. All proteins are assembled from amino acids, which
come from materials broken down in the digestive system. Some amino acids are derived
directly from plant or animal protein; others are synthesized from a range of carbon and nitrogen
sources in the diet. To make hair, the cells in the scalp synthesize keratin by assembling
individual amino acids in a particular order into a polypeptide. Regardless of the type or source
of the protein in shampoo, it will merely coat the hair, rather than replacing any keratin in hair
that has been damaged.
11. Synthesize Your Knowledge
Hummingbirds, like other animals, are heterotrophs and need to obtain from their diet chemical
energy, carbon and nitrogen containing organic building blocks, and essential nutrients. A diet of
nectar provides all of the chemical energy the hummingbird requires, but little more. By
supplementing its diet with insects and spiders, the hummingbird obtains a complete diet. In
particular, animal products in the diet provides some amount of all essential nutrients, as well as
proteins and other macromolecules that can be broken down to provide raw materials for
biosynthesis.
CHAPTER 34 CIRCULATION AND GAS EXCHANGE
Scientific Skills Exercise
Teaching objective: Students practice reading and interpreting a histogram showing data from a
study on genetic factors affecting susceptibility to cardiovascular disease.
Teaching tips: A version of this Scientific Skills Exercise can be assigned in MasteringBiology.
In answering question 4, have students brainstorm what other factors might affect individuals’
plasma LDL levels. Discuss whether it would be possible to design an experiment that would
control for any of these factors.
Answers:
1.
58% of the study group but only about 18% of the control group had an LDL of 100 mg/dL or
less. (In each case, the value equals the sum of the heights of the three bars representing LDL
levels of 100 mg/dL or less.)
2. On average, individuals with an inactivating mutation in the PCSK9 gene have lower plasma
LDL levels than wild-type individuals.
3. You would predict that carriers have a reduced risk for cardiovascular disease because a much
larger fraction of them have a plasma LDL level of 100 mg/dL or less. Lower levels of LDL
cholesterol are associated with a lower risk of cardiovascular disease. (A reduced risk in carriers
has in fact been found experimentally.)
4. The PCSK9 enzyme is not the only factor affecting plasma LDL levels. Other factors tend to
contribute to variability in the data, making the two histograms less distinct.
5. Although the study group (carriers) had a lower average LDL level than the control group,
there is a range of values for both groups. For these two individuals, the LDL levels are equal
and therefore so should be their risk for cardiovascular disease.
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and the suggested rubric at the
beginning of this document.
9. Scientific Inquiry/Interpret the Data
The oxygen dissociation curves indicate that fetal hemoglobin has a higher affinity for oxygen
than maternal hemoglobin does at any partial pressure of oxygen. This higher affinity promotes
the movement of oxygen from the mother’s blood to the blood of the fetus, ensuring that the
baby developing in the uterus has an adequate supply of oxygen.
10. Focus on Evolution
One reason gas exchange would be a challenge for giant insects is the absence of an efficient
system for ventilation. As an insect body increases in size, the tracheae have to become
increasingly long to reach all the cells in the interior. Since diffusion time is proportional to the
square of the distance traveled, at some point the tracheae would be too inefficient to provide
oxygen throughout the body (especially for an insect as active as Mothra). In addition, as body
size increases, body mass and volume, which determine oxygen demand, increase more rapidly
than surface area, which governs gas exchange. At some point, surface area becomes inadequate
for exchange.
11. Focus on Interactions
See the general information on grading short-answer essays and the suggested rubric at the
beginning of this document.
Sample key points:

PO2 decreases with increasing altitude.

Obtaining adequate O2 under conditions of low PO2 is difficult.

Supplemental O2 enables activity at low PO2.

Low-PO2 conditions trigger a compensatory increase in erythrocyte number.
Sample top-scoring answer:
Because adequate distribution of O2 throughout the body is essential for survival, physiological
control systems have evolved that maintain a close match between the O2 carrying capacity of
the body and environmental conditions. As altitude increases, PO2 decreases. Under conditions of
moderately high altitude, the body compensates adequately. The production of erythropoietin in
response to low PO2 stimulates stem cells in the bone marrow to produce additional erythrocytes,
increasing the O2 carrying capacity of the blood. A tent with low PO2 mimics these conditions,
triggering erythropoietin production and an increase in the number of red blood cells. The athlete
gains a temporary increase in aerobic capacity similar to that obtained by spending time at high
altitude or by blood doping. Above a certain altitude, however, the body is unable to make
sufficient adjustments to maintain activity. Under these circumstances, a mountaineer may rely
on an artificial atmosphere rich in O2 to enable him or her to climb up and, hopefully, down.
12. Synthesize Your Knowledge
Air has a higher O2 content than water and is less dense and less viscous. As a result, being able
to use an air supply underwater is advantageous for the spider. As the spider depletes O2 from the
trapped air, O2 in the surrounding water diffuses into the bubble. At the same time, CO2
undergoes net diffusion out of the bubble, keeping the air supply fresh. (Due to net diffusion of
nitrogen from the bubble to the water, which causes the bubble to shrink, the spider must
periodically make a trip to the surface to refill the bubble with a small amount of air.)
CHAPTER 35 THE IMMUNE SYSTEM
Scientific Skills Exercise
Teaching objective: The goal of this exercise is to lead students to see that graphing data can
reveal mechanisms in a way that is not readily apparent from the raw numerical data, and to see
that sometimes it is the pattern of the data that is particularly informative, rather than any single
bits of information.
Teaching tips:
Many students who know how to read simple line graphs have a much harder time recognizing
patterns and trends in more complex graphs. This exercise leads the students through the data
once with a table of information only on total parasite load and a second time with a table of
concurrent data on antibody abundance. After working through the graphing and analysis of
these two types of data, students are led through the experiment a third time by being asked to
predict the values for the appearance and disappearance of particular parasite variants. Through
this iterative approach, students learn to appreciate how both analysis and understanding can be
layered phenomena in scientific investigation.
Answers:
1.
2. The total parasite load cycles in abundance with a new peak roughly every week.
3. Trypanosomes have great potential variation in their coat genes. Even though each coat
variant is attacked in a strong immune response and dies out, the parasites undergo antigenic
shift and a new strain of parasite emerges. The person’s immune system successfully fights one
antigen, but when there is a shift to a new antigen, there must be a brand new immune response.
4.
5. The antibody levels support the hypothesis, since antibodies to each variant appear
sequentially, one week apart, and peak in abundance as the number of parasites in circulation
dips. As parasites with Variant A of the surface glycoprotein appear in the blood, B cells with
receptors for that variant are activated and clonally selected. Plasma B cells secrete antibodies,
which appear in the blood. The circulating antibodies produced in the immune response bind to
the parasites and lead to their death and removal. As the parasite number drops, parasites
expressing Variant B of the surface glycoprotein gene appear and rapidly proliferate. Their
numbers increase steadily, leading to an immune response that builds over time as B cells
specific for this new variant are activated, proliferate, and secrete antibodies. The cycle
continues, with new variants of the parasites appearing each time an effective immune response
is mounted against the previous variant.
6. You would predict that parasites recognized by antibodies of type A would peak in abundance
on day 8 and disappear by about day 14. Similarly, parasites recognized by antibodies of type B
would not appear until about day 12, would peak on day 14, and be largely absent by day 22.
Interpret the Data
Figure 35.12 The concentration of antibodies to A at day 36 is about 400 times that at day 16 (8
× 103 / 2 × 101). If the number of B cells producing a specific antibody is proportional to the
concentration of that antibody, then the number of B cells producing antibodies to A at day 36 is
about 400 times the number at day 16, and the frequency of B cells specific for A at day 36 is
about 400 times the frequency at day 16: 400 in 105, or 1 in 250.
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and a suggested rubric at the
beginning of this document.
9. Make Connections
Lamarck’s discredited idea was that organisms changed their form to fit challenges and then
somehow passed those changes on to their descendants. In clonal selection, heritable differences
that give rise to variation arise prior to any challenge. An encounter with a particular antigen
results in proliferation of the variants best suited to recognize and respond to that challenge.
10. Scientific Inquiry
You could inject LPS into the blood of each strain of mice. You predict that mice mutant for the
TLR4 gene will not develop septic shock, but mice mutant for a different TLR gene will. If the
results are as you predict, a drug that blocked TLR4 signaling might be useful in treating or
preventing septic shock.
11. Focus on Evolution
Lysozyme is an enzyme of invertebrate digestive systems that digests the peptidoglycan of
bacterial cell walls. It has been retained in vertebrates. In humans, for example, lysozyme is
secreted in sweat, saliva, tears, mucus, and breast milk.
12. Focus on Information
Sample key points:
 DNA loss in B and T cells is limited to specific regions of the genome.
 DNA loss in B and T cells is essential for the adaptive immune response.
 Change in DNA structure in B and T cells is heritable in that daughter cells inherit change.

Maintenance of the entire genome from one organismal generation to another is essential for
the survival and reproduction of species.
Sample top-scoring answer:
The inheritance of information encoded in DNA is the basis for the maintenance and propagation
of both cells and organisms. Animal populations reproduce through the fertilization of eggs by
sperm, the joining together of maternal and paternal DNA to generate a unique genome that can
be transferred to all cells of the developing animal by mitosis. B and T lymphocytes, like all
nucleated body cells other than eggs and sperm, are born with a full complement of DNA. At one
point in their life history, however, they recombine a specific portion of their DNA, eliminating
portions of one copy of the antigen receptor gene. This recombination event transforms a cell
with the capacity to synthesize a vast number of antigen receptors to a cell with a unique genome
that produces a single antigen receptor. When such a lymphocyte becomes activated by exposure
to antigen and interaction with a helper T cell, it undergoes mitotic divisions that propagate the
recombined genome, passing the change in the antigen receptor gene on to all daughter cells.
Thus, DNA conveys heritable information in B and T cells, just as it does in eggs and sperm.
13. Synthesize Your Knowledge
In a cell-mediated response, cytotoxic T cells might eliminate a polio infection by killing
infected neurons, but substantial damage to the nervous system would occur as a result. In a
humoral response, however, neutralizing antibodies could block most viruses from infecting host
neurons, thus minimizing nervous system damage.
CHAPTER 36 REPRODUCTION AND DEVELOPMENT
Scientific Skills Exercise
Teaching objective: Students build scientific skills by making inferences based on data from an
experiment on sex development in rabbits.
Teaching tips: A version of this Scientific Skills Exercise can be assigned in MasteringBiology.
In other experiments, Jost transplanted an undifferentiated gonad from a male embryo close to an
undifferentiated gonad in a female embryo. The transplanted male gonad developed into a testis,
and the female gonad near it developed into an ovary. However, the tissues surrounding that
ovary, which would normally develop into an oviduct and other female genital structures, instead
developed into male structures. These results indicate that the presence or absence of a Y
chromosome in the embryo determines whether an embryonic gonad becomes a testis or an
ovary, but the relative amount of a hormone released by the embryonic gonad after
differentiation determines whether the embryonic genitalia develop as male or female. This
explanation accounts for an observation in cattle: A female calf born as a twin of a male calf is
occasionally sterile as a result of the masculinizing effects of hormones produced by the male
twin.
Answers:
1. The normal process that was blocked was the signaling from the gonads to the structures that
form the genitalia. The development of male genitalia requires some signal from the male
gonads. In the absence of this signal, all embryos develop as female, regardless whether they are
genetically male (XY) or female (XX).
2. Perform a sham surgery. For example, open the embryo but don’t remove the gonads, or
remove the gonads and immediately place them back in the embryo.
3. The result of the surgery would have been the same for both sexes—an absence of sexual
differentiation in the genitalia.
4. Hypothesis: The signal that controls male development is the hormone testosterone. Data
collection plan: Surgically remove the gonad precursors in rabbit embryos. Implant a source of
testosterone (experimental group) or the same material or tissue without a source of testosterone
(control group). When the baby rabbits are born, note the appearance of the genital structures in
the male (XY) rabbits. Prediction: The XY rabbits in the experimental group will have male
genitalia, whereas the XY rabbits in the control group will have female genitalia. In fact, Jost did
perform this experiment, replacing the gonads with a crystal of testosterone.
Interpret the Data
Figure 36.4 A graph of the gray lizard’s estradiol level would be high when the other lizard’s
estradiol level is low, and the gray lizard’s progesterone level would be high when the other
lizard’s progesterone level is low. Said another way, females in the population all go through the
same cycles, but the cycles start and end at different times, resulting in behavioral differences
that are complementary with regard to mating behavior.
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and a suggested rubric at the
beginning of this document.
10. Scientific Inquiry
You could collect individual eggs from worms raised apart from one another. If such an egg
develops into a worm, if the cells outside the gonad of the new worm contain five chromosome
pairs, and if the worm produces oocytes and sperm, then there would be good evidence for selffertilization.
11. Focus on Evolution
Hermaphroditism is more conducive to sessile species because, unless they form colonies, they
have limited opportunities to encounter members of the opposite sex.
12. Focus on Energy and Matter
Sample key points:
 Female animals invest energy in producing offspring regardless of whether fertilization or
development is external or internal.
 There tends to be an inverse relationship between the number of eggs produced and the
energy invested in each egg.
 Energy can be invested in reproduction in the nutritional stores of the egg or in the feeding,
care, and protection of the offspring.
Sample top-scoring answer:
Female animals expend energy toward reproductive success in egg production and, in some
cases, in nurturing offspring directly or indirectly. The amount of energy invested varies with the
size and number of eggs and with the extent of investment made after egg production. Frogs
typically produce large numbers of relatively small eggs, with little energy invested after
fertilization. Human females make a very small investment in the egg itself and produce very
few eggs, but expend substantial energy before and after fertilization in generating an internal
environment that protects and feeds the developing fetus.
13. Synthesize Your Knowledge
Since the diploid offspring are homozygous for all genes, a simple hypothesis would be that
meiosis produced haploid eggs, as normally occurs, but the genomes of the eggs then underwent
a duplication. This duplication generated a diploid genome, allowing development to occur. This
is one of several ways progeny can arise by parthenogenesis.
CHAPTER 37 NEURONS, SYNAPSES, AND SIGNALING
Scientific Skills Exercise
Teaching objective: Students build scientific skills by interpreting data expressed in scientific
notation from an experiment on opiate receptors in the mammalian brain.
Teaching tips: A version of this Scientific Skills Exercise can be assigned in MasteringBiology.
Methadone is a pain reliever used in drug addiction treatment to prevent withdrawal symptoms.
Have students investigate how treatment centers use methadone, gradually reducing its dosage
during recovery.
Answers:
1. Morphine 0.000000006 M, atropine 0.0001 M
2. The concentration for phenobarbital (10–4 M) is higher. It’s 5,000 (5 × 103) times the
concentration for methadone (2 × 10–8 M).
3. No. These three drugs had no effect at a concentration of 10–4 M, which is higher than 10–5 M.
If they had no effect at the higher concentration, then they would have had no effect at the lower
concentration, too.
4. Morphine, methadone, and levorphanol blocked naloxone binding. These results indicate that
the receptors are specific for opiate drugs.
5. It suggests that there are no opiate receptors in mammalian muscle tissue.
Interpret the Data
Summary of Key Concepts, Concept 37.3 The action potential lasts from the beginning of the
rising phase to the point of maximal undershoot, about 4.8-2.5=2.3 msec in the graph shown.
The fastest rate at which a neuron could produce action potentials (assuming continuous
depolarization sufficient to reach threshold) would be 1,000/2.3=430 events per second (Hz).
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and a suggested rubric at the
beginning of this document.
10. Scientific Inquiry
Despite their wide use, the exact mechanism of action of most anesthetics is not well understood.
Most general anesthetics are a combination of drugs, including opiates, antianxiety medications,
muscle relaxants, and volatile gases. Here are three hypotheses for how various anesthetics
might block pain:
 Local anesthetics inhibit sodium channels in nerve cells, preventing transmission of
sensations, including pain, from a specific region of the body to the brain.
 Some general anesthetics reduce synaptic transmission. Most are lipid-soluble and may
bind reversibly to the receptor portions of ligand-gated ion channels on the postsynaptic
membrane, preventing neurotransmitter uptake.
 Some general anesthetics enhance the effect of inhibitory neurotransmitters such as
GABA. Others reduce the effect of excitatory neurotransmitters.
11. Focus on Evolution
Generation of an action potential requires a stimulus of sufficient intensity or duration that the
“threshold potential” is reached. If action potentials were not “on/off” (all-or-none) events,
neurons and their downstream effectors would respond continuously to even the most minor
stimuli. Consequently, a great deal of ATP would be consumed for no real purpose.
12. Focus on Organization
Sample key points:
 All animal cells have a membrane potential.
 Neurons differ from most other animal cells in having voltage-gated ion channels.
 Neurons differ from other animal cells in having long extensions for receiving and
transmitting signals.
 Myelination of some vertebrate neurons enhances their electrical properties for signal
transmission.
Sample top-scoring answer:
Like other animal cells, neurons have a negative membrane potential established by the pumping
action of the sodium-potassium ATPase and the selective permeability of the plasma membrane.
Neurons, however, differ from other animal cells in several fundamental ways that are important
for nervous system function in long-distance communication. First, a neuron has short extensions
(dendrites) for receiving stimuli and a long extension (axon) for transmitting nerve impulses.
Second, the presence of voltage-gated ion channels in the plasma membrane allows a neuron to
generate action potentials, all-or-none changes in the membrane potential that can propagate
along an axon without loss of strength. Third, many vertebrate neurons are myelinated, a
specialization that increases conduction speed along the axon. Thus, a combination of shared and
unique properties is responsible for the remarkable behavior that characterizes neurons.
13. Synthesize Your Knowledge
Nerve impulses called action potentials are propagated along axons by voltage-gated channels.
The opening of sodium channels initiates the action potential, whereas the inactivation of sodium
channels and opening of potassium channels returns the membrane potential toward (and
temporarily past) the resting potential. Signaling across synapses of the vertebrate neuromuscular
junction, where axon terminals transmit signals that cause muscle contraction, is mediated by
ligand-gated channels. A neurotransmitter (acetylcholine) released by the presynaptic cell binds
ligand-gated channels on the post-synaptic cell, triggering a change in membrane potential that
leads to muscle contraction.
CHAPTER 38 NERVOUS AND SENSORY SYSTEMS
Scientific Skills Exercise
Teaching objective: In this exercise, students explore how genetic mutants can be used as
experimental animals and how we can learn about physiology through genotype comparisons in
combination with experimental manipulations. Students evaluate experimental design and
suggest additional controls and manipulations.
Teaching tip: A version of this Scientific Skills Exercise can be assigned in MasteringBiology.
Answers:
1. The experimental variable was the genotype of the transplanted tissue. The researchers used
more than one hamster for each procedure to assess whether the differences in cycle period
before and after transplant were greater than variation in cycle period from animal to animal. The
age and sex of the hamsters would likely have been held constant.
2. One control would have been to transplant the SCN from one wild-type hamster to another
wild-type hamster. Another control would have been to do a “sham” operation to determine if the
surgical intervention itself had any effect: Surgically expose the brain of a wild-type hamster
without removing or damaging the SCN or transplanting the SCN from another animal.
3. The circadian cycle period of the transplant recipients was very similar to that of the transplant
donors, regardless of whether the recipient was wild type or τ mutant. These data indicate that
the SCN determines the period of the circadian rhythm.
4. The lack of restoration of rhythmic activity may have been due to individual differences in
circadian response or differences in surgical procedures. For example, the transplanted SCN may
have failed to establish functional connections with the recipient’s brain in 20% of the hamsters.
The fact that the other 80% showed clear changes in cycle period after transplant and little
variation in cycle period within each treatment group suggests that you can safely draw
conclusions from those hamsters. Scientists could replicate this experiment to be even more
confident about the conclusions.
5. (a) Transplant the SCN of a wild-type hamster into the no-rhythmic-activity mutant.
Prediction: After the transplant, the recipient would show an activity pattern with a period of
about 24 hours. (b) Transplant the SCN of a τ hamster into the no-rhythmic-activity mutant.
Prediction: After the transplant, the recipient would show an activity pattern with a period of
about 20 hours.
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and a suggested rubric at the
beginning of this document.
8. Scientific Inquiry
Hypothesis 1: The injury eliminated the ability to generate “speech” in the brain. Hypothesis 2:
The injury eliminated motor control of the hands. To distinguish between these hypotheses, you
could examine use of the hands in motor activities other than sign language. If the individual had
been able to hear and speak before the injury, you could also examine whether spoken language
was affected.
9. Focus on Evolution
As humans, we define intelligence based on our thought processes, just as we define the visible
spectrum based on the wavelengths of light that our eyes detect. Yet there are insects that choose
a particular flower as a food source based on their ability to “see” ultraviolet light. It seems
reasonable to assume that natural selection acting on other groups of animals may also have
resulted in types of information processing that we lack and consequently don’t recognize.
10. Focus on Organization
See the general information on grading short-answer essays and the suggested rubric at the
beginning of this document.
Sample key points:

The lens is clear, allowing transmission of light.

The lens has a convex shape that focuses light.

The lens is flexible, allowing for focusing at different distances.
Sample top-scoring answer:
The lens of the human eye, a disk formed from protein, exhibits three adaptations that are crucial
for vision. First, unlike most other structures in the body, the lens is transparent to light across a
range of wavelengths. This property allows light striking the lens to reach the photoreceptors on
the surface of the retina. Second, the convex shape of the lens captures and focuses light, much
as the lens of a camera focuses an image onto film or a detector grid. Third, the lens is flexible.
Changing the shape of the flexible lens causes it to focus on objects at different distances from
the eye.
11. Synthesize Your Knowledge
Having more olfactory receptor cells is likely to provide an increased sensitivity to odorants. If
there is a greater variety in the cell types as well, there could be an increased ability to
discriminate among different odorants. If so, you would expect that brain region dedicated to
processing olfactory information would occupy a larger fraction of the sensory processing
regions in a bloodhound brain than in a human brain.
CHAPTER 39 MOTOR MECHANISMS AND BEHAVIOR
Scientific Skills Exercise
Teaching objective: Students build scientific skills by reading and interpreting a graph with log
scales on the energy costs of locomotion.
Teaching tips: A version of this Scientific Skills Exercise can be assigned in MasteringBiology.
Students may need to review the topic of logarithms. To help them appreciate the scale change
with each tic mark, have them convert the exponential notation to the long form of each number
on both axes.
Answers:
1. To make the small data values distinct, you would have to use scale with small intervals for
each axis; to fit the large data values on the graph with such small scales, you would have to
make both axes very long. A log scale allows you to present data with a wide range of values
on the same graph with axes that are a reasonable length because the scale is expanded at the
small end and compressed at the large end.
2. The energy cost is about 10 times greater for the smaller animal. A larger animal travels
more efficiently.
3. If a 1,000-fold difference in body mass correlates to a 10-fold difference in the energy cost of
flying (the answer to question 2), and if the slopes of the flying and swimming lines are very
similar, then the energy cost of a 2-kg swimming animal should be about one-tenth that of a
2-g swimming animal: 0.12 cal/kg•m.
4. From highest energy cost to lowest: running, flying, swimming. Students may have expected
the energy costs of flying or swimming to be higher than that of running because humans are
not adapted for flying or swimming. The animals tested for each form of locomotion were
adapted for that form. Running requires a constant expenditure of energy to overcome
gravity.
5. Salmon are better adapted for swimming, having fins and a streamlined body.
Interpret the Data
Figure 39.20 A difference between two individual population lines could arise by a process,
such as genetic drift, that occurs randomly rather than in response to selection. Since the study
found that all three populations in one set behaved consistently and in a distinct manner from the
other populations in the other set, it is extremely unlikely that the difference arose randomly.
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and a suggested rubric at the
beginning of this document.
8. Scientific Inquiry
Sample hypothesis: During sustained contraction, paramyosin prevents binding of new
molecules of ATP to the myosin head, preventing release of the myosin head from actin and
maintaining contraction.
This hypothesis could be tested by experiments on isolated clam muscle. Experimental
treatments could test clam actin and myosin alone as well as clam actin, myosin, and
paramyosin. Binding of ATP to myosin and muscle relaxation would be measured.
9. Scientific Inquiry
One likely hypothesis is that the helper is closely related to one or both of the birds in the mated
pair. Because closely related birds share many genes, the helper bird is indirectly enhancing its
own fitness by helping its relatives raise their young. (In other words, this behavior evolved by
kin selection.) The easiest way to test the hypothesis would be to determine the relatedness of the
birds by comparing tissue samples using molecular methods. If birds are closely related, their
DNA and proteins should be more similar than those of more distantly related or unrelated birds.
10. Focus on Evolution
The two explanations for falling in love (subjective feeling or reproductive fitness) are not
necessarily incompatible. Ultimately, the behaviors and feelings associated with falling in love
may have a strong tie to reproductive fitness. Among modern humans, falling in love is a much
more complex phenomenon than preparing to beget and raise offspring. Indeed, falling in love
may never result in offspring (consider, for example, elderly people who get married, infertile
couples, and gay couples). Falling in love may be as much about emotional intimacy, affection,
and pleasure as it is about offspring. One might consider falling in love an exaptation: It
originally served one function but consequently evolved several others. Not only falling in love,
but all of life has an evolutionary basis. As Darwin said at the end of The Origin of Species,
“There is grandeur in this view of life.”
11. Focus on Information
See the general information on grading short-answer essays and the suggested rubric at the
beginning of this document.
Sample key points:
 Imprinting is a type of learning strictly constrained by an inherited program of response.
 The information encoded by DNA establishes the basic organization of the nervous system.
 The connections made during associative learning are constrained by the organization of the
nervous system.
Sample top-scoring answer:
Information inherited in the form of DNA establishes the capacity for learning. For example,
imprinting is based on an innate (and thus inherited) tendency to make a connection with an
object having particular properties. A program of gene expression establishes the nervous
system, which is then modified through learning to establish memories of information and
connections between particular bits of knowledge. Laboratory experiments on associative
learning reveal that some animals have a limited capacity to learn in the sense that not all
possible connections can be made. Rather, the genome of those animals has evolved in a way
that enables learning that is most likely to be beneficial for survival and reproduction in the
animal’s natural setting.
12. Synthesize Your Knowledge
Regarding proximate causation, a researcher might ask the extent to which the cooperative
behavior is learned, whether learning is individual or social, and when and under what
circumstances a helper transitions to the breeding population. Regarding ultimate causation, a
researcher investigating the evolutionary basis of these behaviors might, for example, ask
whether kin selection is involved and whether the fraction of birds in the population acting as
helpers correlates with the sex ratio or with the availability of acorns.
CHAPTER 40 POPULATION ECOLOGY AND THE DISTRIBUTION OF
ORGANISMS
Scientific Skills Exercise
Teaching objective: The goal of this exercise is for students to use and interpret a mathematical
model. Many students will have considered, and possibly analyzed, population numbers between
0 and K using the logistic growth model. This exercise asks them to use the model in cases where
N > K, something they likely will not have considered. The exercise also encourages students to
think about model “stability”. The population does not shoot up to infinity if N somehow briefly
exceeds K.
Teaching tips: A version of this Scientific Skills Exercise can be assigned in MasteringBiology.
Encourage the students to use the example of Table 40.2 as the organization tool for their
calculations. A few students with weaker math backgrounds may have trouble with
understanding what to do when the term “(K - N)” is negative. Just show them that they can do
the calculation as before but retain the negative sign for the final number. They will end up
adding a negative number (the same as subtraction) from the population size in the previous
interval.
The exercise provides a key opportunity to discuss model “stability” with students. The
population numbers approach K both from below and from above, when N > K (under the rarer
conditions in which N temporarily is greater than K).
In Part 3, the students are asked to compare the behavior of the population that they modeled in
Parts 1 and 2 to actual data for a Daphnia population (Figure 40.20b). Between about 70 and 105
days, the Daphnia population is declining, just as the model predicts, which will likely be a
surprise to some students. Note also that the population drops below K slightly from
approximately 105 to 150 days. In real-world populations, there will be some small variation
around K. That variation is normal.
Answers:
1. Using a population size of 1,510 as an example, if r = 1.0 and K = 1,500, then
(K  N )
(1,500  1,510)
dN
 rN
 1(1,510)
dt
K
1,500
and the population “growth” rate is –10 individuals per year. The rate is –107 when N = 1,600; –
292 when N = 1,750; and –667 when N = 2,000. The growth rate is negative for all of these
population sizes, but the greatest growth rate (the least negative value) occurs when N = 1,510.
2. Doubling r will double the population “growth” rate (actually, the rate of population decline).
If r = 2.0, the rate is –20 individuals per year when N = 1,510; –213 when N = 1,600; –583 when
N = 1,750; and –1,333 when N = 2,000.
3. In Figure 40.20b, the population declines between approximately 60–100 days, drops below
the carrying capacity between about 100—160 days, and stabilizes thereafter at the carrying
capacity. One hypothesis for why the population drops below the carrying capacity briefly is that
overcrowding causes the birth rate to decrease, and a lower birth rate is maintained for a while
after the overcrowding is eliminated.
Interpret the Data
Figure 40.8 Because tundra is much cooler than deserts, less water evaporates during the
growing season and the tundra retains more moisture.
Figure 40.13 When only urchins were removed, limpets may have increased in abundance and
reduced seaweed cover somewhat (the difference between the purple and green lines on the
graph).
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and the suggested rubric at the
beginning of this document.
6. Interpret the Data
Based on what you learned from Figure 40.13 and on the positive relationship you observed in
the field between kelp abundance and otter density, you could hypothesize that otters lower sea
urchin density, thereby reducing the feeding of urchins on kelp.
7. Scientific Inquiry
You could plant both sets of seeds at low and high elevations. If the differences are strictly
genetic, then seeds from low-elevation plants would still produce tall plants at high elevations,
and seeds from high-elevation plants would still produce short plants at low elevations. If only
abiotic factors are important, then all of the plants that grew at high elevations would be short,
and all of the plants that grew at low elevations would be tall. In practice, you would likely see
both genetic and abiotic effects in such an experiment. Another experiment would be to grow
both sets of seeds in a growth chamber, where you could vary a specific abiotic factor, such as
temperature, while holding other factors constant.
8. Focus on Evolution
Where a species lives today can be affected by a combination of its evolutionary history and
ecological factors. For example, a terrestrial species may be restricted to the continent on which
it or its ancestors originated. This will be especially likely if the species has limited dispersal
capabilities and if the species originated at a time when its continent of origin was not connected
to other landmasses. Even for such species, current ecological conditions will also be important:
Some habitats within its continent of origin will be suitable for the species, while others will not.
9. Focus on Interactions
Sample key points for a possible answer:
 Many diseases are important environmental concerns.
 Global travel makes it more likely that diseases can spread quickly.
 Disease may become a weapon for bioterrorism or warfare.
Sample top-scoring answer:
Answers will vary. One possible answer is that disease may ultimately be the most important
density-dependent factor regulating the size of the human population. In recent decades, many
diseases, including AIDS and the disease caused by the West Nile virus, have emerged as
important environmental concerns. An influenza pandemic caused by a new strain of virus, such
as the H1N1 flu virus, could also cause hundreds of millions of deaths in the future, particularly
as the human population continues to grow and crowding becomes more common. The advent of
global travel makes it more likely that new or once-localized diseases can spread quickly across
Earth. An additional concern is the chance that disease could become a weapon for bioterrorism
or warfare.
10. Synthesize Your Knowledge
Competition and predation are the two factors most likely to affect the locust population.
Individual locusts are likely to compete for food and in fact can defoliate large areas. Predation
will also play an important role in reducing locust numbers. However, one potential evolutionary
advantage of swarming is that the population “saturates” any predators that are present; the
predators can eat only so many of the locusts before they become full and stop feeding.
CHAPTER 41 SPECIES INTERACTIONS
Scientific Skills Exercise
Teaching objective: This exercise reinforces the ability of students to make and interpret bar
graphs and scatter plots. As described below, it also provides an excellent opportunity to teach
students about evolution as well as invasive species ecology.
Teaching tips: A version of this Scientific Skills Exercise can be assigned in MasteringBiology.
Along with the technical skills it presents, the exercise provides a good opportunity to teach
about evolution, using invasive species ecology as an example. Through the answer to question
3, students should reason out that snakes that can produce an enzyme nullifying the toad toxin
would then be expected to feed on the cane toads (i.e., the “0” in Table 1 might rise to 50% or
more). This ability would likely increase the fitness of snakes with this ability compared to
snakes without it.
Teachers can also highlight evolution through the results of Part 2. Snakes found in areas where
the cane toads have been present longer appear to have evolved the ability to detoxify at least
some of the toad toxin. Compare, for instance, the 52% drop in swimming efficiency for the 5year group of snakes compared to the modest 5-22% drop for the 50-year group (see row 2 of the
table in Part 2).
Answers:
1.
2. Black snakes will not prey on cane toads in areas where cane toads have been present for 40–
60 years. In areas with and without cane toads, black snakes are more likely to prey on native
frogs than on cane toads.
3. The black snakes in areas where cane toads have been present might be expected to eat 100%
of the cane toads offered to them, just as the snakes did for the native frog.
4. Dependent variable: percentage reduction in snake swimming speed. Independent variable:
number of years cane toads were present in the area.
Exposure to cane toads appears to be selecting for black snakes that are less affected by the
toxin. This conclusion is supported by the fact that the reduction in swimming speed of snakes
living in areas with cane toads for 50–60 years is less, on average, than the reduction in
swimming speed of snakes living in areas with cane toads for 5–20 years.
5. In part 1, the independent variable consists of two nonnumerical categories: cane toads present
in the area for 40–60 years and no cane toads in the area. A bar graph is useful for comparing the
value of the dependent variable in each category. In part 2, both the dependent variable
(percentage reduction in snake swimming speed) and independent variable (number of years
cane toads were present in the area) represent continuous numerical data. A scatter plot shows
the general trend in the dependent variable as the independent variable changes.
Interpret the Data
Figure 41.4 The smallest beak depth observed for G. fortis on Santa María and San Cristóbal
Islands is 10 mm. Therefore, the predicted beak length is 10 mm × 1.12 = 11 mm.
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and the suggested rubric at the
beginning of this document.
6. Scientific Inquiry
Hypothesis: The kangaroo rat is a keystone species in the desert. (Apparently, herbivory by the
rats kept the one plant from outcompeting the others; removing the rats reduced plant diversity.)
Additional supporting evidence: Observations of the rats preferentially eating dominant plants;
finding that the dominant plant recovers from herbivore damage faster.
7. Focus on Evolution
Character displacement means that there has been disruptive selection in two closely related
species with the result that some feature shared by both (perhaps an anatomical feature like a
beak or a metabolic feature like an enzyme) becomes more different over generational time.
There are other ways for competing species to adapt besides character displacement. One of
these is emigration. It may be possible for one species to leave the area of overlap and thereby
avoid interspecific competition altogether. Another way is resource partitioning, whereby
organisms change their behavior, such as habitat choice or preferred food-morsel size.
8. Focus on Information
Sample key points:
 On the island without predators, the bright coloration would offer no benefit, so it might fade
or disappear over time.
 On the island with predators, there would be selection for flies that are less visible, so the
bright coloration should fade even more quickly.
 On the island with both predators and similarly colored, unpalatable species, natural selection
would likely favor the continued bright coloration.
Sample top-scoring answer:
Based on this case of aposematic coloration and Batesian mimicry, evolutionary theory would
predict different outcomes for this fly species on different islands. On the first island, a lack of
predators suggests that there would be no apparent benefit to being brightly colored. Over many
generations, the brightly colored phenotype might fade or disappear. On the second island, the
bright coloration should fade even more quickly; as predators learned that the species is
palatable, there would be selection for flies that are less visible to the predators. Finally, on the
third island, where both predators and the similarly colored, unpalatable species exist, natural
selection would likely favor the continuation of the brightly colored phenotype in the Batesian
mimic.
9. Synthesize Your Knowledge
Two ecological interactions, pollination (a mutualism) and predation are suggested by the scene
in this photograph. The highest trophic level is represented by the spider, a predator, which
shows evidence of cryptic coloration.
CHAPTER 42 ECOSYSTEMS AND ENERGY
Scientific Skills Exercise
Teaching objective: Students build scientific skills by reading and interpreting quantitative data
from a table. The exercise uses an analysis of energy content and flow in a salt marsh as the
specific example.
Teaching tips: A version of this Scientific Skills Exercise can be assigned in MasteringBiology.
The math in this exercise is straightforward. The primary confusion for students with a relatively
weak math background will be the difference between a proportion and a percentage.
To broaden the relevance of the exercise, teachers could compare the energy in the consumer
trophic level with the average “10%” production pyramid in Figure 42.10. Students may notice
that the insect production is less than 10% of primary production. This observation provides an
opportunity to show that the typical responses are just averages and, more importantly, to ask,
“Are there other primary consumers not accounted for in the table?” The answer is “yes.” The
exercise mentions crabs, for instance. Are there other organisms that might not even have been
measured by Teal in the 1962 study?
Another teaching opportunity is a discussion of how to measure energy content. Some students
will not be familiar with a “calorimeter” or how it works. The organic material is burned
completely to assess its energy content accurately. Advanced students could be shown chemical
reactions and the net energy content released when organic matter is oxidized. Different
chemical bonds have different energy contents, of course.
Answers:
1. The proportion of solar energy incorporated into gross primary production = 34,580 kcal/m2 •
yr ÷ 600,000 kcal/m2 • yr = 0.058 (or 5.8%). The proportion incorporated into net primary
production = 6,585 kcal/m2 • yr ÷ 600,000 kcal/m2 • yr = 0.011 (or 1.1%).
2. The energy used by primary producers for respiration is the difference between gross and net
primary production (see Concept 42.2): 34,580 kcal/m2 • yr – 6,585 kcal/m2 • yr = 27,995
kcal/m2 • yr. For the insects: 305 kcal/m2 • yr – 81 kcal/m2 • yr = 224 kcal/m2 • yr.
3. The proportion of NPP leaving the marsh as detritus = 3,671 kcal/m2 • yr ÷ 6,585 kcal/m2 • yr
= 0.56 (or 56%).
Interpret the Data
Table 42.1 Molybdenum limits primary production in this system because the addition of N + P
+ Mo stimulates growth six fold. There is also a multiplicative benefit in adding iron and
molybdenum together. They stimulate primary production 72-fold, compared with 6- and 12-fold
individually.
Figure 42.9 Approximately 17% of the energy in the caterpillar’s food is used for secondary
production (33J/200J = 0.165 = 0.165 x 100% = 16.5%)
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and the suggested rubric at the
beginning of this document.
9. Interpret the Data
(a)
(b) On average, the ratio is 1, with equal amounts of water moving from the ocean to land as
precipitation and moving from land to ocean in runoff.
(c) During an ice age, the amount of ocean evaporation falling on land as precipitation would be
greater than the amount returning to the oceans in runoff; thus, the ratio would be >1. The
difference would build up on land as ice.
10. Scientific Inquiry
Nutrient availability may limit primary production in aquatic ecosystems. Decomposing fallen
leaves are a source of nutrients in aquatic ecosystems, increasing primary production of rooted
vegetation and phytoplankton.
Hypothesis: Primary production is higher in forest freshwater ponds that receive fallen leaves
than in ponds without fallen leaves.
Two adjacent ponds in the forest are used as the study site. Leaves are allowed to fall onto one
pond, while they are removed or prevented from falling on the other pond. Other significant
variables, such as initial nutrient availability, light intensity, initial concentration of
photosynthetic organisms, and temperature, are controlled. There are a number of ways to
measure primary production, such as chlorophyll concentration per milliliter or phytoplankton
density in millions of cells per milliliter. Primary production should be measured once a week
over the course of the experiment and the changes in primary production of the two ponds
compared.
11. Focus on Evolution
Certainly ecosystems and the biosphere change over time, but is such change evolution? If we
define evolution as changes to genomes over generational time, it is a stretch to consider Earth
an evolving entity. If, on the other hand, we expand the concept to include the joint changes
occurring in all life-forms (and exclude changes in abiotic factors), perhaps we can think of a
superorganism called Earth that evolves. Human beings, after all, do not have a single genome.
Rather, our DNA includes contributions from various viruses (especially retroviruses and
herpesviruses). Moreover, our bodies are actually communities of microorganisms, many of
which are so well adapted that they can live nowhere else, and we cannot survive without them.
Our cells depend on their mitochondria, which have their own genes, mutating at their own rates
(higher than the rates at which our nuclear genes mutate). The point is that human beings (and
pretty much all eukaryotes) are superorganisms, and the human genome, with its many
contributors, evolves over generational time. Taken together, the gene pools of Earth change
over generational time, so the idea of Earth as a superorganism that evolves may not be so farfetched after all.
12. Focus on Energy and Matter
Sample key points:
 Tropical peatlands typically have high rates of primary production.
 Lack of oxygen in the soil slows microbial activity and rates of decomposition, leading to a
long-term buildup of soil organic matter.
 Draining the peatland exposes peat to conditions more favorable to decomposers, increasing
the rate of decomposition and leading to net carbon loss in the ecosystem (negative net
ecosystem production).
Sample top-scoring answer:
Like other moist tropical forests, tropical peatlands have relatively high rates of primary
production. Unlike their moist forest counterparts, however, peatlands have reduced oxygen
availability in the soil due to waterlogging. As a result, decomposition is relatively slow, causing
soil organic matter to build up over time. By definition, a system that is accumulating carbon has
a positive net ecosystem production, with gross primary production outpacing respiration of both
autotrophs and heterotrophs. If a landowner drained the forest, exposing the peat to conditions
more favorable to decomposers, then decomposition would accelerate, temporarily outpacing
primary production and creating a negative net ecosystem production.
13. Synthesize Your Knowledge
By collecting and burying the dung, the beetle speeds up the process of decomposition. The dung
decomposes more quickly in the relatively moist soil than it would have if it were left exposed to
the dry air. Decomposition by soil fungi and other detritivores releases nitrogen, phosphorus, and
other nutrients into the soil, some of which become available to primary producers such as
plants.
CHAPTER 43 GLOBAL ECOLOGY AND CONSERVATION BIOLOGY
Scientific Skills Exercise
Teaching objective: Through this exercise, students will plot and interpret graphical data. They
will also gain some understanding of the dynamics of carbon dioxide in Earth’s atmosphere,
including seasonal changes and longer-term trends showing CO2 concentrations increasing
because of the burning of fossil fuels for energy production.
Teaching tips: A version of this Scientific Skills Exercise can be assigned in MasteringBiology.
Students may be surprised to see how much the concentration of CO2 in the atmosphere changes
seasonally in the northern hemisphere where there is a lot of land. A good follow-up exercise
would be to get them to calculate the amplitude of change within each year for the three years
shown. The range is 5 to 8 ppm CO2 across the three years. Have the students note the lack of
overlap in any values across the three years shown, an indication of how quickly atmospheric
CO2 is increasing.
This exercise can also be used to help students understand how Earth’s atmosphere mixes. For
instance, which would provide more information, a second station at the same latitude
somewhere else (e.g., on an island in the Atlantic Ocean) or at the same longitude but in the
Southern Hemisphere? The Southern Hemisphere station is the correct answer because the
rotation of dominant winds on Earth mixes the air relatively quickly in east to west directions but
much more slowly in north to south directions.
Answers:
1. One way to graph the data is to plot connected points for each year on a line graph, using the
same horizontal axis for all three years. Since the changes within a year are small, students will
have to choose an appropriate scale for the CO2 concentration. The graph should look similar to
this:
2. Within each year, the CO2 concentration is highest in April–May and lowest in September–
October. This cyclical pattern occurs because new plant growth, which begins during the spring,
removes CO2 from the atmosphere through photosynthesis, lowering the CO2 concentration. This
decline continues until the fall, when the leaves die, photosynthesis is reduced, and
decomposition of the leaves returns CO2 to the atmosphere, raising the CO2 concentration.
3. In the Southern Hemisphere, the CO2 concentration would be lowest in April–May and highest
in September–October, reflecting the fact that the seasons are reversed in the Southern
Hemisphere compared to the Northern Hemisphere.
4. The CO2 concentration increased between 1990 and 2010. The average CO2 concentration was
354.5 ppm in 1990, 369.4 ppm in 2000, and 389.8 ppm in 2010. The average increased by (369.4
– 354.5)/354.5 = 0.042 or 4.2% from 1990 to 2000 and by (389.8 – 354.5)/354.5 = 0.10 or 10%
from 1990 to 2010.
Suggested Answers for End-of-Chapter Essay Questions
See the general information on grading short-answer essays and the suggested rubric at the
beginning of this document.
8. Scientific Inquiry/Draw It
To minimize the area of forest into which the cowbirds penetrate, you should locate the road
along the west edge of the reserve (since that edge abuts deforested pasture and an agricultural
field). Any other location would increase the area of affected habitat. Similarly, the maintenance
building should be in the southwest corner of the reserve to minimize the area susceptible to
cowbirds.
9. Focus on Evolution
Although extinctions occur on a regular basis, life on Earth has undergone five mass extinction
events through time. The largest of these was the Permian extinction 252 million years ago.
When a mass extinction occurs, 50% or more of all species are driven to extinction, and many
groups of organisms (such as terrestrial dinosaurs during the Cretaceous mass extinction)
disappear forever as well. The fossil record shows that after a mass extinction it typically takes
5-10 million years for the diversity of life to recover to previous levels, but it can take as long as
100 million years (which is how long it took the number of marine families to rebound following
the Permian mass extinction).
These data suggest that if a human-caused sixth mass extinction occurs, it will take millions of
years for the diversity of life to recover. In fact, more than 1,000 species have already become
extinct in the last few centuries, a rate 100 to 1,000 times the typical background rate apparent in
the fossil record. If humans cause a 6th mass extinction, not only will Earth lose millions of
species, but humanity will lose priceless opportunities for food, medicines, and other beneficial
products. In addition, with the extinction of large number of species throughout the globe, many
ecosystem services would function poorly. As a result, all aspects of life on Earth would be
negatively affected—including human societies, which ultimately depend on the natural world
for their existence.
10. Focus on Interactions
Sample key points for one possible answer:
 Many diseases are important environmental concerns.
 Global travel makes it more likely that diseases can spread quickly.
 Disease may become a weapon for bioterrorism or warfare.
Sample top-scoring answer:
Answers will vary. One possible answer is that disease will ultimately have the greatest effect in
regulating the size of the human population. In recent decades, many diseases, including AIDS
and the disease caused by the West Nile virus, have emerged as important environmental
concerns. An influenza pandemic caused by a new strain of virus, such as the H1N1 flu virus,
could also cause hundreds of millions of deaths in the future, particularly as the human
population continues to grow and crowding becomes more common. The advent of global travel
makes it more likely that new or once-localized diseases can spread quickly across Earth. An
additional concern is the chance that disease could become a weapon for bioterrorism or warfare.
11. Synthesize Your Knowledge
The most important conservation tool that would help big cats survive is habitat preservation. An
effective conservation strategy would therefore include identifying biodiversity hot spots for big
cats and establishing new protected areas based on that and other information. A more specific
approach would be needed for species such as the snow leopard. This species lives at high
elevations and its range is shrinking as a result of global warming; thus, a conservation strategy
for snow leopards should include steps to mitigate climate change. Analyzing the effective
population sizes and genetic diversity of the different populations and species would be another
tool to help set priorities for habitat preservation and protection from hunting. In establishing
reserves, you would also need to consider issues of habitat fragmentation and dispersal corridors
to make sure that there were blocks of sufficient size and that the cats could move among
habitats. Human considerations of safety and economic use would also be a key part of any
strategy, however. A combination of zoned reserves would be needed for multiple land uses, and
conflicting demands on habitat use would need to be weighed.