Solution Manual for Campbell Biology in Focus, 4th edition Lisa A. Urry Michael L. Cain Peter V. Min

Answers for Instructors Campbell BIOLOGY IN FOCUS

by Lisa A. Urry, Michael L. Cain, Steven A. Wasserman, Peter V. Minorsky, and Rebecca B. Orr Appendix A of the book includes answers for students for Figure Questions, Concept Check Questions, Summary of Key Concepts Questions, end-of-chapter multiple-choice questions, and Draw It Questions. This document for instructors includes suggested answers and teaching tips for the Scientific Skills Exercises and suggested answers for the Problem-Solving Exercises, Interpret the Data Questions, and the short-answer essay questions at the end of each chapter.

The Scientific Skills Exercises, Problem-Solving Exercises, Interpret the Data Questions, and additional questions for the Visualizing Figures can be assigned in Mastering Biology, 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 instructors, 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 Theme 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 themes introduced in Chapter 1 and featured throughout the book. The Focus on a Theme Questions 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 Theme Questions are provided for instructors and TAs 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 instructors from assigning them. Using a grading rubric, however, can streamline the process. Some instructors have found they can train TAs or even students to grade short essays accurately. (Students can grade their own essays or those of their classmates.) A suggested grading rubric for the Focus on a Theme essays is shown at the end of these tips and in the Study Area of Mastering Biology.

This rubric can also be modified to use with the other end-of-chapter 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).

To train TAs to grade essays in a large class, the instructor should 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) can be passed out to TAs, asking them to grade the essays based on the rubric and a 0-4 grading scheme. Subsequent discussion with the TAs about their essay rankings should clarify the standards, after which they can be given a few ―test‖ essays to grade to ensure consistency in grading practices. This training exercise should take no more than 30-45 minutes.

Using a similar rubric and training scheme, the Montgomery County Public School System in Maryland has been able to train a team of instructors to grade thousands of short essays consistently in a relatively short time.

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 instructor should point out the rubric to students (in the Study Area of Mastering Biology) 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. Instructors should also encourage students to read the Writing Tips provided under ―Additional Resources / Writing Tips and Rubric‖ in the Study Area of Mastering Biology, which also includes the suggested grading rubric.

Suggested Grading Rubric for “Focus on a Theme” Short-Answer Essays Understanding of Use of Supporting Appropriate Use Quality of Writing Theme and Examples or Details of Terminology Relationship to Topic Examples well Accurate scientific Excellent 4 Evidence of full and complete understanding chosen, details terminology organization, accurate and applied enhances the essay sentence structure, to theme and grammar Examples or details Terminology is Good sentence flow, 3 Evidence of good understanding are generally well correctly used sentence structure, applied to theme and grammar Supporting examples Terminology used Some organizational 2 Evidence of a basic understanding and details are is not totally and grammatical adequate accurate or problems appropriate Examples and details Appropriate Poorly organized; 1 Evidence of limited understanding are minimal terminology is not grammatical and present spelling errors detract from essay Examples lacking or Terminology Essay is very poorly 0 Essay shows no 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 Mastering Biology.

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, while the total caught on nights without moonlight was about 95. This suggests that owls hunting on moonlit nights are slightly less successful than are owls hunting on nights without moonlight.

Interpret the Data Figure 1.21 In the beach habitat, approximately 14 light models and 36 dark models were attacked. In the inland habitat, approximately 38 light models and 12 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 2016 forward. The range of abstracts from which students might choose will grow as the Hoekstra lab generates additional publications.

9. Focus on a Theme: 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 a Theme: 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 Mastering Biology.

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 CO3-2 (carbonate) ions in units of micromoles (mol) of CO3-2 per kilogram of seawater. (b) The y-axis shows the calcification rate in units of millimoles of CaCO3 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 CO3-2 in the seawater. As the concentration of CO3-2 increases, the rate of calcification increases.

3. If the seawater CO3-2 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].

4. (a) The final step of the process shown in Figure 2.25, 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 2.25, more CO2 entering the ocean will push the reactions toward formation of more HCO3- (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. (As an example, if you do the same calculation as in question 3 for a CO3-2 concentration of 250 mol/kg, you will see it takes 2.5 days instead of 1.6 days to accumulate the same amount of calcium carbonate at this lower CO3-2 concentration.)

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 cellsurface 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 a Theme: 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 a Theme: 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 Mastering Biology.

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 -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.

Problem-Solving Exercise Teaching objective: In this exercise, students will use short DNA sequences from three species of salmon to try to determine the species identification of a sample.

Teaching tip: A version of this Problem-Solving Exercise can be assigned in Mastering Biology.

1.

2.

(a) 3

(b) 5

(c) 0

3.

(a) O. kisutch 17/20 = 85% identical to sample (b) O. keta: 15/20 = 75% identical to sample

(c) S. salar: 20/20 = 100% identical to sample 4. Hypothesis: My fish sample is Salmo salar, Atlantic salmon, not O. kisutch as it was labeled.

Reasoning: The DNA sequence of the sample matches the standard sequence of S. salar but not the standard sequences of O. kisutch or O. keta. The DNA from individuals of the same species will have more similar DNA than the DNA from individuals of different species. Therefore, it appears that the sample is S. salar, based on these data 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.

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 a Theme: 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 a Theme: 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 Mastering Biology.

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 × 1 µm = 4 µm. The diameter of the new cell is 3 scale bars long, and 3 × 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 3 4 cell with radius 1.5 µm, the volume is π(1.5 µm) = 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.

7. 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.

8. Focus on a Theme: 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.

9. Focus on a Theme: 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.

10. 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 Mastering Biology.

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 20 30 45 60

Concentration of Radioactive Glucose (mM) in Red Blood Cells from a 15-Day-Old Guinea Pig 50 72 85 97 104

Concentration of Radioactive Glucose (mM) in Red Blood Cells from a 1-Month-Old Guinea Pig 22 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 (increase in H+ concentration) 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, as [H+] decreases, 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 a Theme: Evolution One might expect protists 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 a Theme: 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 Mastering Biology.

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.

Interpret the Data Figure 6.16(b) The optimal pH for pepsin is 2.0. Natural selection likely resulted in pepsin’s optimal temperature because 2.0 is the pH inside the stomach, as seen in Figure 2.24. The optimal pH for trypsin is 8.0.

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/Draw It

A. The substrate molecules are entering the pancreatic 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 a Theme: 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 a Theme: 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 Mastering Biology.

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 Test Your Understanding #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. It makes sense that evolutionary processes would have resulted in this specific allosteric regulation 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 AMP stimulates activity of PFK. This makes sense because when AMP and ADP are high, more ATP would be needed, thus phosphofructokinase would be more active, moving

sugar along in the breakdown pathway. Therefore, PFK activity is higher when the concentrations of AMP and ADP are high, and that 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 a Theme: 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 a Theme: 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: A version of this Scientific Skills Exercise can be assigned in Mastering Biology.

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 × 100]. For velvetleaf, the dry mass would increase by about 32% [(50 g – 38 g) / 38 g × 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 Light wavelengths at about 400–450 nm and 670–680 nm, in the violet-blue and red portions of the spectrum, respectively, 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. Scientific Inquiry/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 a Theme: Evolution Since the endosymbiont theory proposes that the ―ancestor‖ of the chloroplast was a photosynthetic prokaryotic cell, 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 a Theme: 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 a Theme: 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 outside the thylakoids.  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 outside the thylakoids. 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 Mastering Biology.

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 as indicated by the values on the x-axis, and thus has 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) See above for the labeling of the cell cycle phases in the treated sample graph. 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 a Theme: 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. If the proposed alternative occurred, these chromosomes would be divided equally in number, five chromosomes distributed into each of the two daughter cells. Chromosome duplication would then restore the number of chromosomes in each cell to ten. Each daughter cell would end up with ten chromosomes. but each would have duplicate copies of the genetic information on the five chromosomes it happened to receive from the parent cell. We have noted in this chapter that the chromosomes are made up of two sets, one from each parent. (You’ll learn more about this in Chapter 10.) 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 (duplicate) 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 a Theme: 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 Mastering Biology.

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) × (9.78 x 105 base pairs per fg) = 1.2 × 107 or 12 × 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 × 107 base pairs (haploid value) × 2 (for diploid value) divided by the number of minutes of the S phase. For the example shown, (1.2 × 107 base pairs)(2)/120 minutes = 200,000 (or 2.0 × 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 a Theme: 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. Thus, the cells of these species that had evolved the ability to switch to sexual reproduction under adverse conditions would have survived better and reproduced more, passing on the alleles of whatever genes allow this switch to happen.

9. Focus on a Theme: 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 (e.g., egg and sperm), 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 (growth from cuttings) 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 version of this Scientific Skills Exercise can be assigned in Mastering Biology.

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.

In our simplified model, 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.

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. 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 make predictions 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 using a diagram to make 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. 16. Focus on a Theme: 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.

17. Focus on a Theme: 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: that the two genes are not linked and are assorting independently. (They will not reject the hypothesis, concluding that the genes are assorting independently.)

Teaching tips: A version of this Scientific Skills Exercise can be assigned in Mastering Biology.

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. Using the same results to test this alternative hypothesis, the χ2 value will fall in the significant zone, so that the hypothesis (of

being linked) would be rejected, leading to the same conclusion as above, 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 when two genes are on different chromosomes (meaning different homologous pairs, e.g. ―chromosome 2‖ and ―chromosome 6‖) so that the alleles of the two genes move independently of each other. The other way is crossing over between two linked genes (that are therefore on the same pair of homologs). 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. A common misconception is that ―parental phenotype‖ refers to the phenotypes of the parents in the cross, which are the F1 dihybrids.

For the situation where the two genes are on the same chromosome (linked), the P parent’s phenotype reflects the genotype of either of its chromosomes. (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 were 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, and conclude that they are most likely linked.

Answers: 1.

Offspring from testcross of AaBb (F1) × aabb Expected ratio if the genes are unlinked Expected number of offspring (of 900) Observed number of offspring (of 900)

Purple stem/ short petals (A-B-) 1

Green stem/ short petals (aaB-) 1

Purple stem/ long petals (A-bb) 1

Green stem/ long petals (aabb) 1

(1/4)×900=225

(1/4)×900=225

(1/4)×900=225

(1/4)×900=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 (that the genes are assorting independently) 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/Draw It (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 a Theme: 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 a Theme: 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 X0. The XXX cell would have given rise to the male half of the butterfly, and the X0 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 Mastering Biology.

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 13.30 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/Draw It

12. Focus on a Theme: 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 a Theme: 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.

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: A version of this Scientific Skills Exercise can be assigned in Mastering Biology.

The consensus sequence determination (based on the predominant base at each position) 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: https://alum.mit.edu/www/toms/papers/hawaii/hawaii.pdf To download a program that shows how a sequence logo "evolves" with more and more sequence information, go to: https://alum.mit.edu/www/toms/papers/ev 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 in the sequence logo 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 in the sequence alignment 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.

Problem-Solving Exercise Teaching objective: Students will identify differences between wild-type and mutant insulin gene coding regions by comparing patient sequences to the wild-type sequence and reading a codon table to determine which amino acids are changed. Students will then predict what effect, if any, the change in sequence will have on the function of the insulin protein.

Teaching tips: A version of this Problem-Solving Exercise can be assigned in Mastering Biology.

Genome sequencing holds much promise for medical advances and personalized medicine. This exercise introduces students to the basics of how the process works. There are several important points to make.

First, there needs to be a reference sequence—the wild-type cDNA sequence encoding a normally functioning protein. (The cDNA sequence is the same as the mRNA sequence except that thymine [T] in the cDNA takes the place of uracil [U] in the mRNA.) Given the natural genetic variation among healthy individuals, determining the reference sequence is not as straightforward as it appears, so the sequence is ultimately derived by analyzing large numbers of individuals and coming up with a consensus sequence. These large data sets tell us that some loci (nucleotide positions) in a gene can have large natural variation with no effect on health, while other loci are strictly conserved in healthy individuals, indicating that they are essential for normal protein function. It is these conserved loci that are most valuable for medical comparative genome sequence analysis.

Second, there has to be a way to analyze a particular region of a gene from among the approximately 3 billion base pairs in the human genome. The entire human genome has been mapped and sequenced, so the region of interest can be found in the computer database using its unique coordinates. Scientists who build the bioinformatics databases make use of statistical and

other mathematical algorithms to locate and align multiple sequences; the small section of the insulin gene shown in this exercise was aligned using these methods. In this exercise, students will identify the mismatches by eye, but in practice, it would be done by computer.

Third, the fact that all mutations are not equally ―bad‖ is a concept hard for some students to internalize. For example, silent mutations are not expected to affect health, hence the name ―silent.‖ In this exercise, none of the mutations is silent, but the mutant codon in each case results in a different degree of change to the chemical properties of the resultant amino acid.

Because protein folding is determined in large part by the chemical properties of its amino acids, these changes can have a ripple effect: different folding leads to a different shape, which can lead to a different function, which can lead to disease.

Answers: 1.

2.

Patient 1: The codon with the mutation (TGC) codes for the amino acid cysteine (Cys). The corresponding codon in the wild-type sequence (TTC) codes for the amino acid phenylalanine (Phe). This patient has a missense mutation because the wrong amino acid will be incorporated into this position in the insulin protein.

Patient 2: The codon with the mutation (TCC) codes for the amino acid serine (Ser). The corresponding codon in the wild-type sequence (TTC) codes for the amino acid phenylalanine (Phe). This patient has a missense mutation because the wrong amino acid will be incorporated into this position in the insulin protein.

Patient 3: The codon with the mutation (TTG) codes for the amino acid leucine (Leu). The corresponding codon in the wild-type sequence (TTC) codes for the amino acid phenylalanine (Phe). This patient has a missense mutation because the wrong amino acid will be incorporated into this position in the insulin protein.

None of these mutations were silent because different amino acids were specified by the mutant codon compared to the wild type. None were nonsense, because the mutant codons did not cause an early stop to occur.

3. For all three patients, the wild-type codon that corresponds to the mutation codes for the amino acid phenylalanine (Phe). As shown in Figure 3.18, phenylalanine has a nonpolar side chain and is hydrophobic.

Patient 1: The mutant codon codes for cysteine (Cys), which has a weakly polar side chain, so it could change the shape of the insulin protein.

Patient 2: The mutant codon codes for serine (Ser), which is polar and hydrophilic, so it is likely to change the shape of the insulin protein.

Patient 3: The mutant codon codes for leucine (Leu), which is nonpolar and hydrophobic, so it might not have much effect on the shape of the insulin protein.

In general, the more the shape of the insulin protein is changed, the more likely its function is also changed, contributing to the severity of the patient’s neonatal diabetes.

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 a Theme: 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 a Theme: 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. (This is similar to the albino donkeys shown in Figure 14.1.) 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 Mastering Biology.

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 that is 8-9 kb long 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 y-axis 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 the same 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. Using qRTPCR will reveal the relative amounts of mRNAs in each stage.

9. Focus on a Theme: 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 a Theme: 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 Mastering Biology.

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 represents 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, a more extreme reduction than when only one or two segments are deleted.

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, but only in the digit-forming 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 estrogenreceptor 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 a Theme: 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 a Theme: 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 Mastering Biology.

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 a Theme: 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 rapidly. 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 ―shape-shifters.‖ 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 a Theme: Organization 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: 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: A version of this Scientific Skills Exercise can be assigned in Mastering Biology.

It will be useful to remind the students that the values in the table were calculated based on pairwise 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 (alpha 2)

 (zeta)

β (beta)

 (delta)

 (epsilon)

A

 (gamma A)

 (gamma 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

A



42

42

41

73

71

80

------

99



42

42

41

73

72

80

99

------

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 Identity

Pairs

100

α2 - α1

99

G

93

-β

80

A

76

-β

73

-

72

G

71

A

60

 - α1

 - α2

45

β - α1

β - α2

44

 - α1

 - α2

42

A

G

41

-

A

40

-

39

 - α1

38

β-

 - A

-

-

G

-β

A

-β

G

- -

 - 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 Concept 12.3).

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 a Theme: 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 a Theme: 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 mis-expression 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 mis-expression 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 mis-expression 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 could have 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 Mastering Biology.

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 also 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 (predators of 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. (a) 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. (b) 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.

6. Scientific Inquiry/Draw It (a)

(b) The rapid rise in the percentage of mosquitoes resistant to DDT was most likely caused by natural selection in which mosquitoes resistant to DDT could survive and reproduce while other mosquitoes could not. (c) In India—where DDT resistance first appeared—natural selection would have caused the frequency of resistant mosquitoes to increase over time. If resistant mosquitoes then migrated from India (for example, transported by wind or in planes, trains, or ships) to other parts of the world, the frequency of DDT resistance would increase there as well. In addition, if resistance to DDT were to arise independently in mosquito populations outside of India, those populations would also experience an increase in the frequency of DDT resistance.

7. Focus on a Theme: 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 a Theme: 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 Mastering Biology.

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. (a) In the partial polypeptides shown, the sequence from Ustilago has the most amino acids (38) in common with the aphid sequence. (b) 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. (a) 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. (b) 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.

8. Scientific Inquiry/Draw It (a)

(b)

(c) The tree in (a) requires seven evolutionary changes, while the tree in (b) requires nine evolutionary changes. Thus, the tree in (a) is more parsimonious because it requires fewer evolutionary changes.

9. Focus on a Theme: Evolution Darwin’s suggestion 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 a Theme: 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 evolutionary lineage leading to manatees 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 Mastering Biology.

Answers: 1. 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:

p

(2  49) (1111) (0  56)  0.484 .

432

Using p = 0.484, we calculate that q = 1-p = 0.516.

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 × 0.484 × 0.516) = 0.499, and 0.5162 = 0.266.

3. (a) 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. (b) 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. (a) 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. (b) 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, it did not appear that homozygous CYCY had been selected against, most likely because they had not used up the supply of food stored in the seed. However, by day 21, it did appear that CYCY individuals had been 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.

6. 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 natural 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.

7. Focus on a Theme: 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).

8. Focus on a Theme: 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.

9. 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 Mastering Biology.

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.

Problem-Solving Exercise

Teaching objective: In this exercise, students will analyze DNA results from two species of mosquito to determine whether alleles encoding insecticide resistance has been transferred.

Teaching tip: A version of this Problem-Solving Exercise can be assigned in Mastering Biology.

1. (a) In A. gambiae, the pre-2006 frequencies of kdr genotypes were 30% +/+, 50% +/r, and 20% r/r. In 2006, these frequencies were 34.8% +/+, 34.8% +/r, and 30.4% r/r , and post-2006 they were 4.8% +/+, 4.8% +/r, and 90.4% r/r. (b) Overall, from pre-2006 to post-2006, the frequency of the +/+ and +/r genotypes decreased, while the frequency of the r/r genotype increased dramatically. A hypothesis accounting for these observations is that natural selection caused the frequency of individuals with genotype r/r to increase over time because individuals with that genotype are resistant to insecticides.

2. In A. coluzzii, the pre-2006 frequencies of kdr genotypes were 100% +/+, 0% +/r, and 0% r/r.

In 2006, these frequencies were 90.9% +/+, 9.1% +/r, and 0% r/r , and post-2006 they were 26.3% +/+, 42.3% +/r, and 31.4% r/r. Overall, from pre-2006 to post-2006, the frequency of the +/+ genotype decreased, while the frequency of the +/r and r/r genotypes increased. A hypothesis accounting for these observations is that in 2006, the formation of A. gambiae x A. coluzzii hybrids (and subsequent matings between these hybrids and A. coluzzii individuals) led to the transfer of the r allele from A. gambiae to A. coluzzii. Once the r allele had been transferred to A. coluzzii populations, the frequency of that allele increased because it confers resistance to insecticides.

3. Yes. When humans apply insecticides, the r allele is adaptive. Moreover, as the data summarized in question 2 indicates, the r allele was not present in A. coluzzii before 2006, but was present in A. coluzzii populations once hybrid individuals carrying the r allele were observed in 2006. This suggests that A. gambiae × A. coluzzii hybrids mated with A. coluzzii indivduals, leading to the transfer of an adaptive allele into A. coluzzii populations.

4. Prior to 2006, A. coluzzii mosquitoes were not resistant to insecticides. Hence, their populations could be reduced in size by applying insecticides, making it less likely that A. coluzzii mosquitoes could contribute to the spread of malaria. In the years following 2006, resistant r/r genotypes had appeared in A. coluzzii populations, making it more likely that malaria could be spread by A. coluzzii mosquitoes—thereby potentially leading to an increase in the number of malaria cases.

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. 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.

8. Scientific Inquiry/Draw It Here is one possibility:

9. Focus on a Theme: 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 a Theme: 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 Mastering Biology.

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, in millions of years, of more ancient (top) to more recent (bottom) fossil snail species: Volute species with nonplanktonic larvae 2.2 2.2 5.4 1.4 3.2 3.2 0.9 2.0 2.0 5.6 3.6 3.6 2.3 2.3 2.3 2.3 1.3 1.3 1.3 1.3

Volute species with planktonic larvae 6.7 6.7 6.7 1.6 1.8 7.0 1.4 10.4 3.1 5.6 5.6 8.1 4.0

1.3 1.3 1.3 1.0 1.0 1.0 1.0 2.9

2. The mean persistence time for species with planktonic larvae is 5.3 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 23.12 The blue curve is for marine animal families. Families often contain many species.

As a result, a family can survive even if many of the species in the family become extinct.

Hence, 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 in fact 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 a Theme: 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 a Theme: 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. Although there often is a good match between form and function, this match does not indicate that the structure was 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 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 calculating and interpreting means and standard errors using data from a scientific experiment. Specifically, students are provided the formula for the mean ( x ) and asked to use that formula to calculate the mean for each treatment in the experiment. They are also provided the formula for the standard deviation (s) and asked to calculate the standard deviation and standard error (SE) for each treatment. Students then use the mean and standard error for each treatment as a way to evaluate the extent to which the treatment means differ from one another.

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Teaching tips: A version of this Scientific Skills Exercise can be assigned in Mastering Biology.

Antibiotic-resistant bacteria pose a growing health risk to humans. Many species of bacteria are resistant to multiple antibiotics, prompting scientists to search for new sources of antibiotics.

Naturally occurring soil bacteria are a potential source for new antibiotics since these species synthesize their own antibiotics, which they use against other species that attack or compete with them. However, the vast majority of soil bacteria cannot be cultured using standard laboratory techniques and hence, researchers have had limited success in their efforts to obtain new antibiotics from soil bacteria.

To determine whether soil bacteria harbor antibiotics that would be effective against resistant bacteria, researchers first developed a technique for growing hard-to-culture bacteria in the laboratory. Their technique involved drilling small holes in a plastic chip, filling the holes with a dilute solution containing soil bacteria and agar, and covering the chip with soil in which the bacteria could grow. The researchers used this approach to isolate a promising new antibiotic, teixobactin, from a soil bacterium. The researchers then tested whether teixobactin was effective against MRSA (methicillin-resistant Staphylococcus aureus), a potentially lethal bacterium that is resistant to multiple antibiotics; they also compared the effectiveness of teixobactin to that of vancomycin, an existing antibiotic.

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This exercise enables students to calculate and interpret means, standard deviations, and standard errors. Many students will be familiar with using the mean as a way to estimate the ―typical‖ value found in a set of data—as when a student calculates the mean of the exam scores they have received over the course of a semester. Students may be less familiar, however, with using the standard deviation as a measure of the variation found in a set of data. Clarify for students the steps used in calculating the standard deviation, emphasizing that xi represents the value of the ―ith‖ observation for the variable x and giving them practice in using the summation sign, .

Also describe how the standard error (SE) provides a measure of how greatly the mean would likely vary if an experiment were to be repeated. In particular, if an experiment were to be repeated, the new mean would typically lie within two standard errors of the value for the mean ( x ) that was obtained in the original experiment; expressed mathematically, the new mean would typically lie within the range x  2SE. This range provides a rough but useful indication of whether two means differ from one another. For example, if the interval surrounding the mean for one treatment overlaps the interval surrounding the mean for another treatment, this would 

suggest that the two means were approximately equal and hence that the treatments from which the means were calculated were roughly similar in their overall effect.

Answers: 1. On a log scale, the mean number of colonies for each treatment are as follows. control: 1 mg/kg vancomycin: 1 mg/kg teixobactin: 5 mg/kg vancomycin: 5 mg/kg teixobactin:

9.1 8.4 7.2 5.3 4.7

2. Overall, the results for question 1 indicate that teixobactin reduced the number of MRSA colonies at least as effectively as did vancomycin. For example, at a dose of 1 mg/kg, teixobactin reduced the number of MRSA colonies by more than 10-fold compared to the control; vancomycin also reduced the number of MRSA colonies, but it did so by less than a 10-fold reduction. At a dose of 5 mg/kg, both vancomycin and teixobactin greatly reduced the number of MRSA colonies (roughly 10,000-fold) compared to the control.

3. The standard deviations are as follows. control: 1 mg/kg vancomycin: 1 mg/kg teixobactin: 5 mg/kg vancomycin: 5 mg/kg teixobactin:

0.27 0.15 1.42 0.6 0.62

4. The standard errors are as follows. control: 1 mg/kg vancomycin: 1 mg/kg teixobactin: 5 mg/kg vancomycin: 5 mg/kg teixobactin:

0.14 0.09 0.71 0.35 0.31

Using these values for the standard error, along with the mean values calculated in question 1, the x 2SEvalues for each treatment are as follows. control: 1 mg/kg vancomycin:  1 mg/kg teixobactin: 5 mg/kg vancomycin: 5 mg/kg teixobactin:

9.1 ± 0.28 8.4 ± 0.18 7.2 ± 1.42 5.3 ± 0.7 4.7 ± 0.62

The interval that surrounds the mean for the control does not overlap with any of the intervals surrounding the means for either antibiotic; this suggests that both antibiotics reduce the number of MRSA colonies compared to the control. In addition, for a dose of 1.0 mg/kg, the interval that surrounds the mean for vancomycin overlaps with the interval that surrounds the mean for teixobactin; this suggests that the two antibiotics have roughly similar effects at this dose. The two antibiotics also have roughly similar effects at a dose of 5.0 mg/kg.

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 a Theme: 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 a Theme: 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 Mastering Biology.

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 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. (a) The value of the cell that represents the comparison of C. testosteroni and E. coli is 61. (b) 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. (c) It is unnecessary to make the comparisons

represented by a dash, as they would pair up a species with itself, which is pointless. (d) 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. (a) Agrobacterium tumefaciens, an alpha proteobacterium, is identical to the wheat mitochondrion at 48% of the 617 nucleotide positions compared in the matrix. (b) This suggests that alpha proteobacteria are more closely related to ancestor of mitochondria than are 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 a Theme: Evolution/Draw It

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 a Theme: 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 (or a cell that was closely related to archaea)—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, tubulinid amoebas are not more closely related to all other protists than they are to plants, fungi, or animals. 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 Mastering Biology.

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 a Theme: 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 a Theme: 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 Mastering Biology.

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: (a) Crabs are more successful in preying on northern periwinkles. (b) 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.

Problem-Solving Exercise Teaching objective: In this exercise, students will examine whether an amphibian population can acquire immunological resistance to a pathogen from prior exposure.

Teaching tip: A version of this Problem-Solving Exercise can be assigned in Mastering Biology.

1. This figure shows that the abundance of Bd on a frog’s skin is higher for naïve frogs (those with 0 prior exposures to Bd) than it is for frogs with prior exposure to Bd. The figure also shows that when naïve frogs are exposed to Bd, they survive at lower rates than do frogs with prior exposure to Bd. Both of these results indicate that frogs with prior exposure to Bd have developed some degree of resistance to the pathogen.

2. (a)

(b) Based on these data, a hypothesis that explains the results discussed in question 1 is that prior exposure to Bd stimulates the frog’s immune response, enabling the frogs to acquire resistance to the pathogen.

3. A strategy to repopulate Cuban tree frog populations could consist of breeding these frogs in captivity; exposing the captive frogs to dead Bd (thereby enabling them to acquire resistance to the pathogen); and then releasing these resistant frogs to regions where their populations were hard-hit by Bd. 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 a Theme: 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 a Theme: 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 VASCULAR 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 Mastering Biology.

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 a Theme: 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 a Theme: 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 Mastering Biology.

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.

Interpret the Data Figure 29.12 (a) The endophytic bacteria within roots are least like the bacteria in the other two

locations (in the rhizosphere and outside the rhizosphere). As shown in the diagram, the endophytic bacteria are a separate branch from the other two communities. (b) The first branch in the diagram divides the samples by location inside versus outside of roots, indicating that these locations are the main factor in distinguishing groups of bacteria. The next set of branches divides the samples by soil type, indicating that soil type is the second-most important factor in determining the composition of root-associated bacterial communities. Finally, younger and older roots have the highest percentage similarity, indicating that this factor is the least significant.

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) × (0.05) mol/liter × (0.00831 liter MPa/mol K) × (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 a Theme: 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 a Theme: 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 Mastering Biology.

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 a Theme: 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 a Theme: 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 Mastering Biology.

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.

Problem-Solving Exercise Teaching objective: Students will learn about the effects of global climate change on crop productivity and analyze the resulting human impacts in the 20 most populous nations.

Teaching tips: A version of this Problem-Solving Exercise can be assigned in Mastering Biology.

The data presented can be incorporated into a discussion on the possible geopolitical consequences of global climate change as well as ethical considerations relating to the ecological footprints of different nations. The data can be compared with Figure 53.25 on the annual per capita energy use around the world.

Answers: 1. Increases in suitable days for plant growth will occur mainly in the northern parts of North America, Europe, and Asia. Almost all of South America, Africa, and southern Asia will have decreases, particularly in tropical regions. So overall, climate change will have a harmful effect on plant growth.

2. All of the nations with projected decreases have economies that rely mainly on agriculture, so their populations will be more vulnerable to reduced crop productivity.

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 a Theme: 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 a Theme: 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 Mastering Biology.

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.

Problem-Solving Exercise Teaching objective: In this exercise, students will analyze levels of thyroid hormones in a patient suffering from episodes of paralysis.

Teaching tip: A version of this Problem-Solving Exercise can be assigned in Mastering Biology.

1.

Test

Lab comments

Serum total T3

high

Free thyroxine T4

high

TSH levels

high

TSH receptor autoantibody

normal

2. The fact that the levels of T3, T4, and TSH are all greater than normal indicates that the patient is hyperthyroid.

3. It is not likely that the patient has Graves’ disease. A patient with Graves’ disease would be expected to have a level of TSH receptor-activating autoantibodies above the normal range.

4. A pituitary tumor, but not a thyroid tumor, could explain the observed results. A pituitary tumor could cause increased levels of TSH, which in turn would cause increased production of T3 and T4. If there were a thyroid tumor TSH levels should instead drop in response to negative feedback from the elevated levels of T3 and T4 (the same would be true for Graves’ disease).

Interpret the Data Test Your Understanding #9 (a)

(All values are average daily water gain or loss in mL/day.) (b) 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).

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. Focus on a Theme: 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 a Theme: 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 Mastering Biology.

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.

Interpret the Data Figure 33.14 Actinobacteria are abundant in the stomach, but rare in the intestinal tract.

Although the digestive tract leads directly from the stomach to the intestines, the environments are very different. In the stomach the pH is low and little chemical digestion has occurred. These conditions favor Actinobacteria (which are known to breakdown complex nutrients in soil). In the intestine, the neutral pH and abundant nutrients allow other bacteria to be more abundant, decreasing the fraction of the microbiome represented by Actinobacteria.

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 a Theme: 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.

9. Focus on a Theme: 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.

10. Synthesize Your Knowledge In regurgitating indigestible material, the owl resembles the hydra (see Figure 33.5). Rabbits produce pellets, but they are eliminated through the anus. Cows regurgitate, but the cud is then swallowed again for further digestion. With regard to compartments, pellets are produced in the gizzard, which carries out mechanical breakdown of ingested food (in this case prey animals).

Passing feathers or bones along to the intestine would not be beneficial because they have no extractable nutrients and because they could damage the epithelium that carries out absorption.

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 Mastering Biology.

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. Yes. A larger fraction of the individuals with an inactivating mutation in the PCSK9 gene 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.)

3. If the researchers had compared the ranges of LDL levels for the two groups, they would not have observed a significant difference. Therefore, they would likely have concluded that the PCSK9 gene does not have an effect on this variable.

4. The PCSK9 enzyme is not the only factor affecting plasma LDL levels. Other factors tend to contribute to variability in the data, causing the two histograms to overlap.

5. (a) The two individuals have the same risk of developing cardiovascular disease. (b) This prediction is based on the fact that the two individuals’ LDL levels are equal. The histograms revealed an effect of PCSK9 mutations on the average LDL level across a group of individuals, but it is the LDL level in a particular individual that predicts risk, regardless of the group (carrier or control) to which the individual belongs.

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 a Theme: 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 a Theme: Interactions 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: A version of this Scientific Skills Exercise can be assigned in Mastering Biology.

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. The first column (day of infection) represents the independent variable, and the second column (millions of parasites/mL of blood) represents the dependent variable.

2. The amount of parasites in the blood cycles up and down, with a peak approximately every 7 days.

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.

8. 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.

9. Focus on a Theme: 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.

10. Focus on a Theme: 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.

11. 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 Mastering Biology.

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. (a) The normal process that was blocked was the signaling from the gonads to the structures that form the genitalia. (b) 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.

9. Focus on a Theme: Evolution Hermaphroditism is more conducive to sessile species because, unless they form colonies, they have limited opportunities to encounter members of the opposite sex.

10. Focus on a Theme: 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.

11. 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 Mastering Biology.

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. (a) Morphine, methadone, and levorphanol blocked naloxone binding. (b) 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 a Theme: 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 a Theme: 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 Mastering Biology.

Answers: 1. (a) The experimental variable was the genotype of the transplanted tissue. (b) 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. (c) 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. (a) 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. (b) These data indicate that the SCN determines the period of the circadian rhythm.

4. (a) 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. (b) 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 a Theme: 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 a Theme: Organization 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 You might predict that an increased sensitivity to odorants would be associated with a greater number of olfactory receptor cells and a larger surface area for olfactory detection. You might also predict relative increases in the number of axons conveying olfactory information to the brain and the size of the brain region dedicated to processing olfactory information. Observations confirm each of these predictions.

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 Mastering Biology.

Students may need to review the topic of logarithms. To help them appreciate the scale change with each tick 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. (a) The energy cost is about 10 times greater for the smaller animal. (b) 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. (a) From highest energy cost to lowest: running, flying, swimming. (b) 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 39.23 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 (a) 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.) (b) 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 a Theme: 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 a Theme: Information 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 briefly exceeds K. Teaching tips: A version of this Scientific Skills Exercise can be assigned in Mastering Biology.

Encourage the students to use the example of Table 40.2 as the organizational 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 40.21b). 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 40.21b, the population declines between approximately 60100 days, drops below the carrying capacity between about 100160 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 Temperature. Because tundra is much cooler than deserts, less water evaporates during the growing season and the tundra retains more moisture.

Test Your Understanding #6

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.

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.

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 a Theme: 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: Due to the ways in which a species responds to differing abiotic and biotic environments, some habitats within its continent of origin will be suitable for the species, while others will not.

9. Focus on a Theme: 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 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.

10. Synthesize Your Knowledge At sea level, habitats near the equator and in the tropics in general tend to be warm (see Figures 40.3 and 40.7). However, Mt. Kilimanjaro is tall (5,895 meters or 19,341 feet), and as described in this chapter, every 1,000-m increase in elevation produces an average temperature drop of 6°C—a change in temperature equivalent to that produced by an 880-km increase in latitude. As a result, average temperatures decrease substantially from the base of the mountain to its top, which holds permanent glaciers. The wide range of temperatures found at different elevations on the mountain enables diverse habitats to found at this one location near the equator.

CHAPTER 41 ECOLOGICAL COMMUNITIES 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 Mastering Biology.

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. (a) Dependent variable: percentage reduction in snake swimming speed. Independent variable: number of years cane toads were present in the area.

(b) 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 the swimming speed of snakes living in areas with cane toads for 50–60 years is less, on average, than the reduction in the 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 41.5 The smallest beak depth observed for G. fortis on Floreana and San Cristóbal Islands is about 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 a Theme: 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 such as a beak or a metabolic feature such as 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 their choice of where or when to forage, or their preferred food-morsel size.

8. Focus on a Theme: 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 bee may have been attracted to the flower to collect pollen or nectar, and while collecting these sources of food, it would have pollinated the flower—thus, both the bee and the plant would have benefited, making the interaction a mutualism. The highest trophic level is represented by the spider, a predator, which shows evidence of cryptic coloration; the spider appears to have ambushed the bee when it visited the flower to collect nectar or pollen.

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 Mastering Biology.

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%‖ energy pyramid in Figure 42.11. 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. (a) 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%). (b) The proportion incorporated into net primary production = 6,585 kcal/m2 • yr ÷ 600,000 kcal/m2 • yr = 0.011 (or 1.1%).

2. (a) 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. (b) 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%).

Problem-Solving Exercise Teaching objective: In this exercise, students examine the effect of an outbreak of the mountain pine beetle on NEP in a forest ecosystem.

Teaching tip: A version of this Problem-Solving Exercise can be assigned in Mastering Biology.

1. Before the insect outbreak, NEP = NPP – Rh = 440 – 408 = 32 g per m2 per year. Because NEP > 0, the ecosystem was a carbon sink, absorbing more CO2 than it released. After the insect outbreak, NEP < 0 (NEP = 400 – 424 = -24). Hence, after the outbreak, the ecosystem was a carbon source, releasing more CO2 than it absorbed.

2. NEP is often defined as NEP = GPP – RT, where RT = Ra + Rh. As also stated in Question 2: NPP = GPP – Ra . Rearranging the terms of this equation gives us: GPP = NPP + Ra .

This equation for GPP can be substituted into the relation for NEP given above: NEP = GPP – RT = GPP – (Ra + Rh) (because RT = Ra + Rh) = (NPP + Ra) – (Ra + Rh) (GPP=NPP + Ra , so we can replace GPP with NPP + Ra) = NPP – Rh .

3. The mountain pine beetle outbreak converted the forest ecosystem from a carbon sink to a carbon source. A forest that is a carbon source releases more CO2 to the atmosphere than it absorbs. Thus, by contributing to a rise in atmospheric CO2 concentrations, the mountain pine beetle outbreak could have feedback effects on the global climate.

Interpret the Data Table 42.1 Molybdenum limits primary production in this system because the addition of N + P + Mo stimulates growth 6-fold, whereas the addition of N + P alone has a minor effect (stimulates growth 1.1-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 42.10 Approximately 17% of the energy in the caterpillar’s food is used for secondary production (33 J/200 J = 0.165 = 0.165 × 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.

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 a Theme: 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 genome composed exclusively of DNA sequences whose entire line of descent can be traced back from one living organism to another. 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 many other 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 far-fetched after all.

12. Focus on a Theme: 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 CONSERVATION BIOLOGY AND GLOBAL CHANGE 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 Mastering Biology.

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 5–8 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 (for example, 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. 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 a Theme: 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 a Theme: 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.