Appendix A

Answer Key

Chapter 1

Thinking Scientifically

1. Plant a section of your garden in the same variety of tomato and plant seedlings that are the same age and size. One possibility would be to plant the seedlings in
square groups of four, with space between each group. It would be best to plant the groups in a block. If you have room, you could plant 16 groups in a 4 by 4 square.
Then, randomly assign 8 groups to receive the name-brand fertilizer and 8 groups to receive the generic brand. Treat all the groups the same otherwise, by watering and
weeding them uniformly. Fertilize them at the same time and with the same amount of fertilizer. At the end of the growing season, weigh the fruit that you harvest from
each plot and compare the total weight of the name-brand treatment with that of the generic treatment. 2. Biscoe showed treatment effects that increased consistently
from 21 to 7 days before harvest. The growth regulator did not work equally effectively in both cultivars. It worked better for Biscoe because the difference between the
untreated and treatment was greater. The most effective treatment was 7 days before harvest.

Testing Yourself

1. c; 2. a. small molecules; b. large molecules;
c. cells; d. tissues; e. organs; f. organ systems;
g. complex organisms; see Fig. 1.1, p. 2, in text.;
3. b; 4. c; 5. d; 6. b; 7. b; 8. d; 9. c; 10. a; 11. c;
12. a; 13. g

Understanding the Terms

a. hypothesis; b. model; c. adaptation;
d. species; e. taxonomy; f. biosphere; g. data

Chapter 2

Thinking Scientifically

When the bird enters the water, surface tension produced by hydrogen bonding keeps the surface smooth and continuous. When the bird flies out of the water, the
cohesiveness of water molecules causes drops of water to be pulled out with the bird.

Testing Yourself

1. b; 2. c; 3. c; 4. d; 5. b; 6. a; 7. b

Understanding the Terms

a. electrons; b. matter; c. polar; d. hydrophilic;
e. isotopes; f. tracer; g. ion

Chapter 3

Thinking Scientifically

1. a. plasma membrane; b. increases, because the plants are improving their ability to function at cold (but not freezing) temperatures by increasing the proportion of
linolenic acid in their plasma membranes. Chilling injury results mainly from a loss of membrane function. The plasma membrane is composed of proteins floating in a
―sea‖ of phospholipids. If those proteins can not flow freely, then the membrane can not function properly.
c. Canola oil remains liquid; butter is solid. d. no;
e. fatty acids; f. The fatty acids in the plasma membrane of cold tolerant plants, such as temperate plants, are mainly unsaturated. They remain fluid at cold temperatures,
allowing the membrane to function properly. The fatty acids in the plasma membrane of plants that are not cold tolerant, such as many tropical plants, have high levels of
saturated fatty acids. At cold temperatures, the membrane lipids solidify, inactivating membrane-bound proteins. This results in a loss of membrane function and the cells
are damaged.
2. A protein’s ability to function correctly is a factor of its shape. A substituted amino acid with a polar R group may be more likely to bind to other amino acids in the
protein, and hence change its shape, than substituted amino acids with nonpolar R groups.

Testing Yourself

1. c; 2. c; 3. d; 4. d; 5. b; 6. b; 7. d; 8. a; 9. c; 10. b; 11. d; 12. b; 13. d; 14. c; 15. a; 16. a; 17. c; 18. b; 19. d; 20. e; 21. b; 22. a; 23. d; 24. c; 25. d; 26. c; 27. d; 28. d; 29. c;
30. d; 31. a; 32. c; 33. a; 34. d; 35. a; 36. d; 37. b

Understanding the Terms

a. isomer; b. polymer; c. steroid; d. enzyme;
e. phospholipid; f. denatured

Chapter 4

Thinking Scientifically

1. a. A dietician may be able to suggest a diet that would not include the offending long-chain fatty acids. b. Lysosomes are similar to peroxisomes in that they are single
membrane organelles containing enzymes that break down cellular compounds. 2. The plastid must be necessary to the life of the parasite because the parasite dies if
the plastid no longer functions.

Testing Yourself

1. e; 2. a; 3. c; 4. d; 5. c; 6. f; 7. b; 8. d; 9. c; 10. b; 11. c; 12. b; 13. a; 14. b; 15. c; 16. d; 17. c;
18. a. vesicle; b. centrioles; c. mitochondrion;
d. rough endoplasmic reticulum; e. smooth endoplasmic reticulum; f. lysosome; g. Golgi apparatus; h. nucleus; see Fig. 4.7, p. 54, in text.

Understanding the Terms

a. cell theory; b. endomembrane system;
c. apoptosis; d. granum; e. Golgi apparatus;
f. cytoskeleton; g. capsule

Chapter 5

Thinking Scientifically

1. On a calm, dark night on the ocean, collect a water sample and count the number of dinoflagellate-eaters and predators of the dinoflagellate-eaters in the sample.
Shine a light into the water for a period of time and then take another sample. Determine whether the number of dinoflagellate-eaters has decreased and whether the
number of predators of the dinoflagellate-eaters has increased. 2. A normal channel protein allows chloride to leave the cells. When chloride leaves, water will follow,
keeping the outer cell surfaces hydrated. If chloride does not leave cells, then water will not exit and the cell surface will be covered with a thick mucus. The cholera
bacterium produces a toxin that opens chloride channels in cells of the small intestine. As salt leaves the intestinal cells, water follows, resulting in diarrhea. The defective
cystic fibrosis protein closes chloride channels, so the toxin can not open the channels.

Testing Yourself

1. d; 2. e; 3. b; 4. d; 5. c; 6. a,d; 7. b,c; 8. b,c;
9. a,d; 10. a. without enzyme; b. with enzyme; see Fig. 5.8, p. 75, in text; 11. a; 12. a,b,c; 13. b,c,d; 14. d; 15. d; 16. a; 17. b; 18. c; 19. b; 20. a; 21. c; 22. d; 23. a; 24. c;
25. b; 26. a; 27. c; 28. b; 29. b; 30. c; 31. e

Understanding the Terms

a. entropy; b. plasmolysis; c. metabolic pathway;
d. enzyme; e. substrate

Chapter 6

Thinking Scientifically

1. a. No. b. At first it did, but above a certain level

—approximately 1/3 of the final light intensity—additional light did not increase the photosynthetic rate; c. Yes. CO

2

enrichment had to be combined with increased temperature to have its maximum effect on growth. d. Yes, especially at 30•C. Again, boosting all factors that contribute to
photosynthetic rate produced the maximum effect. e. Yes, an increase in CO

2

levels resulted in a dramatic increase in photosynthetic rate. 2. a.

–d. below are suggested

answers to this question. a. There may be other limiting factors for growth, such as mineral nutrients and water. In a greenhouse study, plants are typically supplied with
adequate levels of water and nutrients.
b. High levels of CO

2

may allow soil microorganisms to grow faster and outcompete plants for limiting nutrients, such as phosphorus or potassium. In greenhouse studies,

plants are usually grown in pasteurized soil, so microorganisms are not a variable. c. Mature trees may not take up CO

2

as effectively as young, herbaceous greenhouse

plants. Trees store large amounts carbon that can be tapped for growth, so not as much needs to be taken in via photosynthesis. Greenhouse studies focus on
non-

woody plants that don’t have much capacity for carbon storage. d. Changes in plant community composition may result from competition among species that differ in

their response to elevated CO

2

and temperature. This may influence total CO

2

utilization in the ecosystem. Greenhouse studies typically focus on individual species. 3.

The alga was producing oxygen as a byproduct wherever its chlorophyll was able to absorb the light energy needed for photosynthesis. The bacteria were using the
oxygen for cellular respiration.

Testing Yourself

1. c; 2. a; 3. d; 4. a; 5. a; 6. a; 7. b; 8. d; 9. a. stroma; b. electron transport chain; c. ATP;
d. NADPH; e. ATP synthetase complex; f. thylakoid membrane; g. thylakoid space; see Fig. 6.7, p. 89, in text; 10. d; 11. c; 12. a, b, c; 13. b; 14. b; 15. c

Understanding the Terms

a. stroma; b. stomata; c. ATP synthase; d. thylakoid; e. coenzymes; f. carbon dioxide fixation;
g. photosystem; h. carotenoid; i. CAM

Chapter 7

Thinking Scientifically

1. Physiologically active kernels and microbes carry out respiration, producing heat and water as by-products. The warmer temperatures and additional moisture increase
physiological activity (including respiration) even more, producing ever-increasing levels of heat and moisture. Flammable gases produced by physiological activity will
eventually ignite as the temperature rises above their ignition point. 2. The mitochondria of older people exhibit lower levels of metabolic activity than those of younger
people. Since the mitochondria are less active, they are less able to break down fatty acids in muscle cells. This leads to the accumulation of fatty acids in cells, resulting
in insulin resistance. To reduce the risk of diabetes, the mitochondrial activity in muscle cells must be increased. This could be accomplished by increasing either the
number or activity of mitochondria. Recent research indicates that exercise increases the number of mitochondria in cells. These cells, therefore, are better equipped to
metabolize fatty acids.

Testing Yourself

1. a; 2. c; 3. a; 4. d; 5. b; 6. a; 7. b;

8. a. Pyruvate is broken down to an acetyl group.; b. Acetyl group is taken up and a C

6

molecule results.; c. Oxidation results in NADH and CO

2

.;

d. ATP is produced by substrate-level and ATP synthesis.; e. Oxidation produces more NADH and FADH

2

; see Fig. 7.6, p. 103, in text.; 9. c; 10. a, c, d; 11. a; 12. a, b, c;

13. d; 14. c; 15. a; 16. b, c, d; 17. c; 18. d; 19. c; 20. b; 21. b; 22. b; 23. b;
24. c; 25. c; 26. a; 27. c; 28. a. citric acid cycle;
b. anaerobic; c. pyruvate; d. fermentation

Understanding the Terms

a. glycolysis; b. preparatory reaction; c. fermentation; d. intermembrane space; e. citric acid cycle

Chapter 8

Thinking Scientifically

1. Cancer results from a series of muations (―hits‖) that accumulate in somatic cells over a person’s lifetime. People exposed to high levels of radiation are likely to have a
higher number of somatic mutations than the general population. They would then require fewer additional random somatic mutations to occur before they develop
cancer. A reasonable explanation for the data above is that some cancers require a larger number of hits in order to be expressed. That is, more genes need to be
mutated in a pathway before cancer develops. A type of cancer that requires more hits (random somatic mutations) would generally develop later in life. 2. Sister
chromatids of chromosomes (inside the nucleus) must link to microtubules produced by centrosomes (outside the nucleus) in order to be pulled apart during mitosis.
Therefore, microtubules cannot attach to and separate chromatids if the nuclear membrane is present. The current hypothesis is that cytoplasmic dynein, a molecular
motor protein, tears the nuclear envelope and transports pieces of the membrane away from the area along microtubules. Once pieces are torn away, the nuclear
envelope loses shape and cytoplasmic proteins flow into the nuclear region. The rush of cytoplasmic proteins is believed to stimulate chromosome condensation and
spindle formation. So, dismantling of the nuclear membrane is coordinated with chromosome behavior at mitosis.

Testing Yourself

1. b; 2. c; 3. a; 4. c; 5. a; 6. b; 7. a. G

1

; b. S; c. G

2

;

d. prophase; e.metaphase; f. anaphase; g. telophase; h. cytokinesis; see Fig. 8.3, p. 114, in text; 8. a;
9. d; 10. c; 11. c; 12. b; 13. a, b; 14. b; 15. b; 16. c; 17. a; 18. a; 19. c; 20. c; 21. d; 22. e; 23. e; 24. d

Understanding the Terms

a. histones; b. centromere; c. spindle;
d. centrosome; e. signal

Chapter 9

Thinking Scientifically

1. a. Normal • mutant • 10 • 20 • 30 (That is, a triploid plant, with 3 copies of every chromosome.) b. Mutant • mutant • 20 • 20 • 40 (That is, a tetraploid plant, with 4
copies of every chromosome.) 2. a. To disprove the hypothesis, look at the incidence of Down syndrome in families of various sizes. Down syndrome is correlated with
maternal age, but not family size. b. Hypotheses for the relationship between maternal age and incidence of Down syndrome include: (1) The older a woman is, the longer
her oocytes have been arrested in meiosis. Exposure to mutagens during this time may result in nondisjunction; (2) A woman may have a pool of oocytes resulting from
nondisjunction, but is better able to prevent them maturing at a young age; (3) Estrogen levels (which control the rate of meiosis in developing oocytes) drop with maternal
age and may slow down the rate of meiosis. This might allow nondisjunction to happen more frequently in older women.

Testing Yourself

1. a; 2. d; 3. d; 4. a; 5. b; 6. a; 7. c; 8. d; 9. a; 10. f; 11. h; 12. c; 13. g; 14. b; 15. e; 16. a. homologues condense and undergo synapsis, chromosomes condense and do
not pair; b. tetrads at spindle equator, dyads at spindle equator; c. homologues separate, daughter chromosomes separate; d. two daughter cells from meiosis I; (bottom)
four daughter cells following meiosis II, two daughter cells following mitosis; see Fig. 9.7, p.135, in text.; 17. a; 18. c

Understanding the Terms

a. oogenesis; b. dyad; c. synapsis; d. crossing-over; e. nondisjunction; f. Turner syndrome; g. Barr body

Chapter 10

Thinking Scientifically

1. a. CcPp • CcPp would produce 9:16 purple (C__P__) and 7:16 white offspring; b. All white offspring. 2. That chromosomal segment contained the SRY gene. XX men
have the SRY gene on one of the X chromosomes, so they have the male phenotype. XY women contain a Y chromosome in which SRY has been lost, so they have the
female phenotype. The SRY gene probably moved from the Y chromosome to the X chromosome following a rare recombination event.

Testing Yourself

1. b; 2. c; 3. d; 4. d; 5. b; 6. b; 7. e; 8. a. TG; b. tg;
c. TtGg; d. TTGG; e. TtGg; f. TTgg; g. Ttgg; h. TtGG; i. ttGG; j. TtGg; k. ttGg; l. ttgg; see Fig. 10.6, p. 146, in text.; 9. b; 10. c; 11. c; 12. b; 13. a; 14. c, 15. d; 16. e; 17. d;
18. a; 19. a, b; 20. a, b, c, d; 21. c, d; 22. a, b, d; 23. c

Understanding the Terms

a. alleles; b. homozygous; c. phenotype;
d. wild-type; e. rule of multiplication; f. incomplete dominance; g. linkage group

Chapter 11

Thinking Scientifically

1. DNA replication occurs as a part of the cell cycle and replication errors sometimes lead to mutations. Skin cells undergo the cell cycle more frequently than nerve cells
in the brain; therefore, skin cells may be mutating faster than nerve cells. Some of these mutations may be cancer-causing. This hypothesis would be supported if there are
more differences in DNA base sequences in skin cells compared to brain cells. However, you need a standard to compare your base sequence data to. You should choose
another tissue, like muscle tissue, whose cells also don’t divide often. 2. A vertebrate has tens of thousands of genes. Many of the genes are for basic cellular functions,
and would have no role in the specific developmental program of the animal. So even with 30 complete genes it is highly unlikely that the genes would have much to do
with development. Supposing that you did have developmental genes, they could only function properly if they were active at specific points in devleopment. This would
require other specific genes that would turn the developmental genes on and off at appropriate times. Lastly, even if one did have the luck to possess developmental
genes and regulatory genes, these developmental genes would require specific target molecules which would probably not exist in a modern reptile. Therefore, even in

the highly improbable circumstance that one had 100 intact genes it is extremely unlikely that these genes could be used to recreate a dinosaur.

Testing Yourself

1. a. sugar-phosphate backbone; b. complementary base pairing or rungs; c. hydrogen bonds; d. sugar; see Fig. 11.5, p. 163, in text.; 2. a; 3. d; 4. d;
5. a, b, c; 6. b; 7. a; 8. c; 9. c; 10. b; 11. a. DNA template strand; b. RNA polymerase; c. mRNA; see Fig. 11.11, p. 167, in text.; 12. d; 13. b; 14. c; 15. a ;
16. b

Understanding the Terms

a. adenine, guanine; b. complementary base pairing; c. template; d. DNA polymerase; e. promoter

Chapter 12

Thinking Scientifically

1. Not all genes on the second X chromosome are inactivated. Expression of genes on the second X chromosome is apparently necessary for normal growth and sexual
development. 2. A tortoiseshell male must have two X chromosomes (one carrying the orange allele and one carrying the black allele) and a Y chromosome (so it is male).
This happens only if nondisjunction of the sex chromosomes happened in one of its parents. The cat will be sterile because the three sex chromosomes do not separate
normally at meiosis, producing nonviable gametes.

Testing Yourself

1. b; 2. c; 3. a; 4. a. regulatory gene; b. promoter; c. operator; d. lactose metabolizing genes; e. mRNA; f. repressor; g. RNA polymerase; see Fig. 12.5,
p. 182, in text.; 5. b; 6. c; 7. d; 8. c; 9. a; 10. b;
11. c; 12. a; 13. d; 14. d; 15. a; 16. e; 17. c; 18. c; 19. e; 20. e; 21. b; 22. b; 23. a

Understanding the Terms

a. operon; b. repressor; c. heterochromatin;
d. transcription factor; e. enhancer; f. growth factor; g. telomere

Chapter 13

Figure Legends

Fig. 13.7. The individual is nomal, but she has a child that is affected. Fig. 13.8. The individual is affected, but he has a child that is normal.

Thinking Scientifically

1. The gene for CFTR is large, and many different mutations can produce an altered protein. Some mutations result in a protein that is somewhat functional. People carrying
these mutations will have mild forms of cystic fibrosis. Other mutations prevent the CFTR gene from functioning at all, resulting in severe expression of cystic fibrosis. A genetic
test will identify a specific mutation. People with that mutation will test positive, however, other people will have mutations in different parts of the CFTR gene. These mutations
will not be detected by the same genetic test. 2. The adenovirus probably inserted itself into a gene that regulates the cell cycle. The gene may have been a tumor-suppressor
gene or a proto-oncogene. The insertion of the virus into the cell cycle gene resulted in a mutation, preventing that gene from producing a normal product. In fact, the virus was
found to have inserted itself in the proximity of a proto-oncogene called LMO-2.

Testing Yourself

1. a; 2. c; 3. d; 4. a. aa; b. A?; c. Aa; d. Aa; e. Aa;
f. Aa; g. aa; h. aa; i. A?; see Fig. 13.7, p. 200, in text.; 5. a. Aa; b. Aa; c. aa; d. Aa; e. aa; f. Aa; g. Aa; h. aa; see Fig. 13.8, p. 200, in text.; 6. a. X

b

Y; b. X

B

X

b

;

c. X

B

Y; d. X

B

Y; e. X

b

Y; see Fig. 13.9, p. 201, in text.; 7. b; 8. a; 9. b, d; 10. a, c; 11. b; 12. e, f; 13. g;

14. f; 15. i; 16. b; 17. h, i, j; 18. a; 19. c; 20. d; 21. b; 22. c; 23. b; 24. 50%, 0%; 25. parents Aa, child aa

; 26. Yes, the son may inherit the father’s recessive allele.; 27. X

a

X

a

,

father X

a

Y, mother X

A

X

a

or X

a

X

a

.

Understanding the Terms

a. karyotype; b. chorionic villi sampling;
c. translocation; d. pedigree; e. proteomics; f. gene therapy

Chapter 14

Thinking Scientifically

1. The genes involved in male reproduction are directly and immediately affected by natural selection. These genes are under continuous pressure to outcompete other
males and father offspring. Natural selection acts on other traits, such as the ability to compete for food and other resources. However, these traits are likely to be influenced
by the environment and not consistently selected in every generation. 2. Some genes simply can not tolerate much change. These genes, called housekeeping genes,
produce products necessary for basic metabolic processes. Mutations would be so catastrophic to the organism that they would be lethal. Therefore, they would not be
perpetuated. Other genes may code for products in which some variation is tolerated. These genes would accumulate mutations faster than housekeeping genes.

Testing Yourself

1. b; 2. b; 3. d; 4. e; 5. c; 6. b; 7. b; 8. a. Originally, giraffes had short necks.; b. Giraffes stretched their necks in order to reach food.; c. Today most giraffes have long
necks.; d. Originally, giraffe neck length varied.; e. Struggle to exist causes long-necked giraffes to have the most offspring.; f. Today most giraffes have long necks.; see
Figure 14.8, p. 222, in text.; 9. e; 10. d; 11. c; 12. c; 13. b; 14. a; 15. c; 16. b; 17. c; 18. c; 19. e; 20. e; 21. e; 22. e

Understanding the Terms

a. homologous structure; b. uniformitarianism;
c. natural selection; d. vestigial structure;
e. adaptation

Chapter 15

Thinking Scientifically

There are several reasonable answers. First of all, although the disease is lethal, individuals with CF may reproduce (and pass on the defective allele) before they die.
Second, the recessive (defective) allele may be hidden and perpetuated by carriers. Finally, the defective allele may confer a selective advantage in certain environments.
Heterozygotes actually survive cholera epidemics better than homozygous dominant individuals because they resist the opening of chloride channels by the cholera
bacterial toxin.

Testing Yourself

1. a; 2. d; 3. c; 4. a; 5. a; 6. c; 7. d; 8. b; 9. a;
10. a; 11. b; 12. b; 13. a; 14. a; 15. a; 16. c;
17. c; 18. b; 19. d; 20. a; 21. a. Wave will be in the center with shaded area in center of rise; b. Wave will shift to the right with small shaded area to
the left of rise.; see Fig. 15.10, p. 239, in text.; 22. a. Wave has narrowed at base.; b. Wave has narrowed still with light shading; see Fig. 15.11,
p. 240, in text.; 23. a. Wave has dipped in the middle separating the shaded area from the nonshaded area.; b. Waves are two distinct waves, one shaded and not; see
Fig. 15.12, p. 240, in text.

Understanding the Terms

a. gene pool; b. assortative mating; c. sexual selection; d. bottleneck effect

Chapter 16

Thinking Scientifically

1. a. Ferns; b. seed plants; c. naked seeds; d. needle-like leaves, Conifers; e. fan-shaped leaves, Gingkos;
f. enclosed seeds; g. one embryonic leaf, Monocots; h. two embryonic leaves, Eudicots. 2. The Keys are too close to the mainland for populations to become isolated and
undergo speciation. Immigrants from the mainland are constantly mating with individuals on the islands, allowing continuous gene exchange. In addition, the Keys do not
exhibit the environmental diversity seen on the Hawaiian Islands. Dramatic habitat differences do not exist on each island, so specialization to unique environments does
not exist.

Testing Yourself

1. c; 2. c; 3. b; 4. f; 5. a; 6. g; 7. d; 8. e; 9. h;
10. a. species 1; b. geographic barrier; c. genetic changes; d. species 2; e. genetic changes; f. species 3;
see Fig. 16.7, p. 250, in text.; 11. b; 12. a; 13. c;
14. a, b, c; 15. d; 16. b; 17. e; 18. e; 19. a. three; designated by color; b. all animals: vertebra; only snakes and lizards: amniotic egg and internal fertilization; c. the
animals in a clade; they share the same derived characters; 20. b; 21. a

Understanding the Terms

a. adaptive radiation; b. paleontology; c. mass extinction; d. binomial name; e. evolutionary tree; f. domain; g. homologous structure

Chapter 17

Thinking Scientifically

1. The genetic engineers remove the genes that cause disease and replace them with the beneficial genes that they would like to introduce into the plant. Then, they
infect the plant with the modified bacterium and allow the bacterium to introduce the foreign genes into the plant genome. 2. It appears that the plant-produced viral capsid
blocks an early stage of the viral infection process. This capsid may prevent the virus from uncoating when it first enters the plant cell. If it cannot uncoat, then it cannot
express the viral genome and cause disease. Since infection with the RNA alone can cause disease, the expression of the capsid does not appear to inhibit the expression
of the viral genome.

Testing Yourself

1. c; 2. a. capsid; b. protein unit; c. DNA; d. spike; see Fig. 17.1, p. 266, in text.; 3. d; 4. b; 5. a; 6. d; 7. d; 8. c; 9. a; 10. c; 11. a; 12. b; 13. a; 14. d;
15. b; 16. d; 17. a. contractile vacuole (partially full); b. pellicle; c. cilia; d. food vacuoles; e. oral groove; f. macronucleus; g. micronucleus; h. gullet; i. anal pore; j.
contractile vacuole (full); see Fig. 17.21, p. 278, in text.

Understanding the Terms

a. lysogenic cycle; b. retrovirus; c. prion;
d. peptidoglycan; e. binary fission

Chapter 18

Thinking Scientifically

1. An open field is not likely to contain the mycorrhizal fungi that colonize pine tree roots. However, in areas where pine trees are already growing, those fungi are likely to
be prevalent in the soil. Therefore, if you transplant a native seedling along with some of its soil, you introduce valuable mycorrhizal fungi to the open field. The seedling
now has the opportunity to establish a mycorrhizal association that will help it to obtain nutrients and water. 2. The males are fooled into believing that the orchid flowers
are female wasps, so they fly to the flowers and try to copulate with them. This is a good strategy because it attracts only one type of pollinator. Each male wasp will fly
from flower to flower within a single plant species, carrying only pollen from that species.

Testing Yourself

1. b; 2. a; 3. a. sporophyte; b. meiosis; c. spore;
d. gametophyte; e. gametes; f. zygote; g. 2n; h. n; see Fig. 18.3, p. 286, in text.; 4. b, c, d; 5. a, b, c, d; 6. c, d; 7. a; 8. c; 9. a, b; 10. d; 11. a. anther;
b. filament; c. stigma; d. style; e. ovary; f. ovule;
g. sepals (calyx); h. petals (corolla); see Fig. 18.15, p. 292, in text.; 12. b; 13. b; 14. a; 15. b; 16. e;
17. e; 18. c; 19. c; 20. e; 21. b; 22. b; 23. e

Understanding the Terms

a. seed; b. anther; c. xylem; d. fronds; e. pollen grains; f. mycelium; g. saprotrophs; h. sporophyte; i. fruit

Chapter 19

Thinking Scientifically

1. The recent findings support the idea that a constant lineage exists in Africa from primitive to modern humans. However, it does not prove that this lineage exists only in
Africa. The discovery of similar fossils in Asia and Europe would provide support for the multiregional continuity hypothesis.
2. Animals that are sessile tend to be radially symmetrical because their food comes to them from all directions. There is no need to have anterior and posterior body
regions. Animals that move through their environment are bilaterally symmetrical, with the anterior portion containing sensory organs. This allows the animal to sense
and respond to the environment as it travels through it.

Testing Yourself

1. b; 2. a. ectoderm; b. mesoderm; c. endoderm;
d. ectoderm; e. mesoderm; f. endoderm;
g. pseudocoelom; h. ectoderm; i. mesentery;
j. endoderm; k. coelom; l. mesoderm; see Fig. 19.5, p. 309, in text.; 3. d; 4. g; 5. a, b, c, d, e, g; 6. a, b, c; 7. a; 8. e; 9. c; 10. f; 11. b; 12. e; 13. c; 14. b; 15. a; 16. a; 17. c;
18. a; 19. d; 20. d; 21. a; 22. b; 23. a; 24. e; 25. a; 26. e; 27. a; 28. b; 29. d; 30. d; 31. b; 32. d

Understanding the Terms

a. cephalization; b. coelom; c. hermaphrodite;
d. radial symmetry; e. endothermic

Chapter 20

Thinking Scientifically

1. Only the nitrogen-fixing bacteria produce polysaccharide derivatives that the plant recognizes and latches onto. Then, root hairs adjacent to the bacteria fold around
them and the plant engulfs the bacteria. 2. The plants get most of their water from fog that rolls off the nearby ocean at night. Therefore at night, the plants open all their
stomata and take in both moisture and carbon dioxide for photosynthesis. The stomata are closed during the day. The large number of stomata, in this unusual case,
actually helps the plant to survive in a very dry environment.

Testing Yourself

1. c; 2. e; 3. a; 4. b; 5. a. terminal bud; b. blade (leaf); c. axillary bud; d. node; e. shoot system;
f. vascular tissues; g. root system; see Fig. 20.1,
p. 338, in text.; 6. d; 7. a; 8. a, b, c, d; 9. b;
10. a, d; 11. b; 12. c; 13. a; 14. a. cuticle; b. upper epidermis; c. palisade mesophyll; d. leaf vein;
e. spongy mesophyll; f. lower epidermis; g. guard
cell; see Fig. 20.9, p. 343, in text.; 15. a. epidermis; b. cortex; c. vascular bundle; d. xylem; e. phloem; f. pith; see Fig. 20.11a, p. 344, in text.;
16. a. endodermis; b. pericycle; c. xylem; d. phloem; e. cortex; f. epidermis; g. root hair; h. apical meristem protected by root cap; i. root cap; j. zone of cell division; k. zone
of elongation; l. zone of maturation; see Fig. 20.13, p. 346, in text.

Understanding the Terms

a. perennial; b. phloem; c. stomata; d. wood;
e. meristem

Chapter 21

Thinking Scientifically

1. When the plants are shipped long distances, they produce ethylene as a result of the stress. The ethylene concentration builds up in the truck and leaf abscission
is stimulated. The plastic sleeves help to retain high concentrations of ethylene near the plant, increasing the abscission response. 2. You could determine whether
flowers or stems are responsible by cutting off the flowers and looking to see whether the stems alone exhibit heliotropism. If they do, then the stems rather than the
flowers sense the sun. You could determine which portion of the stem is responsible for heliotropism by shading different portions of the stem and determining which
shading treatment blocks the heliotropic response. You could determine whether auxin is responsible by looking at a stem unde r a microscope to see whether the
differential cell elongation has occurred.

Testing Yourself

1. d; 2. a; 3. e; 4. b; 5. a; 6. e; 7. b; 8. c; 9. b; 10. c; 11. d; 12. d; 13. a. sporophyte; b. meiosis;
c. microspore; d. megaspore; e. male gametophyte; f. female gametophyte; g. egg; h. sperm; i. zygote; see Fig. 21.10, p. 362, in text.; 14. a; 15. c; 16. d; 17. b; 18. c; 19.
d; 20. d; 21. b; 22. d; 23. a; 24. c; 25. e; 26. c; 27. e; 28. b; 29. a. stigma; b. style;
c. carpel; d. ovary; e. ovule; f. receptacle; g. peduncle; h. sepal; i. petal; j. stamen; k. filament; l. anther; see Fig. 21.11, p. 362, in text.; 30. b; 31. b; 32. e; 33. d;
34. a. seed coat; b. embryo; c. cotyledon; d. seed coat; e. first true leaves; f. cotyledons (two); g. seed coat; see Fig. 21.19, p. 368, in text.; 35. d; 36. a; 37. b

Understanding the Terms

a. auxin; b. abscisic acid; c. photoperiod;
d. phytochrome; e. microspore

Chapter 22

Thinking Scientifically

1. Both are diseases of muscle tissue. Researchers have found that all cardiomyopathy gene mutations also lead to skeletal myopathy. They are involved in
fundamental pathways that lead to muscle development. 2. One possibility is that socioeconomic factors such as compliance wit h post-operative procedures and

availability of health insurance may differentially influence survival. To test this hypothesis,

you could look at patients’ records to determine whether there is a

relationship between, for example, availability of health insurance and survival rate. Another explanation is that there is a genetically-based physiological difference
between African Americans and white Americans. To test this hypothesis, you would need to perform inheritance studies to determine whether there is a genetic
basis for liver transplant success.

Testing Yourself

1. a. cell; b. tissue; c. organ; d. organ system;
e. organism; see Fig. 22.1, p. 378, in text.; 2. a; 3. b;
4. a, b; 5. a; 6. a, b, c, d; 7. a, b; 8. d; 9. c;
10. a. plasma; b. blood cells; c. white blood cells;
d. platelets; e. red blood cells; see Fig. 22.6, p. 383,
in text.; 11. d; 12. b; 13. a. skeletal, b. cardiac,
c. smooth; 14. a; 15. a. dendrite; b. cell body;
c. nucleus; d. nucleus of Schwann cells; e. axon; see Fig. 22.8, p. 385, in text.; 16. e; 17. d; 18. c; 19. c; 20. d; 21. d; 22. b; 23. a; 24. d; 25. a; 26. d; 27. b; 28. e; 29. e; 30. c;
31. a; 32. e; 33. c; 34. e; 35. b; 36. e

Understanding the Terms

a. ligaments; b. bone; c. platelets; d. skeletal muscle; e. striated; f. neuroglia; g. integumentary system; h. endocrine system; i. homeostasis;
j. negative feedback

Chapter 23

Thinking Scientifically

1. Endothermic animals use ten times more energy than ectothermic animals of similar size, so their cardiovascular systems need to deliver large amounts of oxygen and
fuel and remove large amounts of carbon dioxide and waste products. The separation of the systemic and pulmonary systems, along with a large, powerful heart provides
the capacity to carry the large volume of blood necessary to maintain an endothermic lifestyle. 2. At high altitudes, there is less oxygen in the air, so the body must find a
way to deliver more oxygen to tissues. In response to low oxygen levels, the kidneys secrete erythropoietin. This stimulates red blood cell production in the red bone
marrow, resulting in an increase in the number of oxygen-carrying red blood cells. When the athlete returns to a lower altitude, he or she has a higher than normal red blood
cell count, providing a high oxygen-carrying capacity at normal atmospheric oxygen levels.

Testing Yourself

1. c; 2. a. pulmonary vein; b. aorta; c. renal artery; d. lymphatic vessel; e. pulmonary artery; f. superior vena cava; g. inferior vena cava; h. hepatic vein;
i. hepatic portal vein; j. renal vein; see Fig. 23.9,
p. 402, in text.; 3. a; 4. a. superior vena cava;
b. aortic semilunar valve; c. pulmonary semilunar valve; d. right atrium; e. tricuspid valve; f. right ventricle; g. inferior vena cava; h. aorta; i. pulmonary artery; j. pulmonary
arteries; k. pulmonary veins; l. left atrium; m. bicuspid mitral valve; n. septum; o. left ventricle; see Fig. 23.4, p. 399, in text.; 5. d; 6. b; 7. c; 8. e; 9. d; 10. b; 11. a; 12. a, b; 13. a;
14. b; 15. c, d; 16. b; 17. d; 18. a; 19. b; 20. b; 21. c; 22. b; 23. c; 24. e; 25. a; 26. c; 27. a; 28. d; 29. c; 30. e; 31. c; 32. d; 33. a. blood pressure; b. osmotic pressure; c.
blood pressure; d. osmotic pressure; see Fig. 23.17, p. 408, in text.

Understanding the Terms

a. systemic circuit; b. atria; c. heart murmur; d. SA (sinoatrial) node; e. capillary; f. aorta, venae cavae;
g. lymph; h. plasma; i. macrophages; j. platelets

Chapter 24

Thinking Scientifically

1. A camel maintains its plasma volume at the expense of tissue fluid. Therefore, circulation is not impeded by a lack of water. The red blood cells of camels are oval-shaped,
allowing them to circulate when the blood becomes more viscous. Camels can drink up to 20 gallons of water at a time and slowly absorb it from their stomach and intestines.
In addition, red blood cells can ―store‖ water by swelling up to more than double their normal size without bursting. Camels’ kidneys can concentrate urine to reduce water loss.
Finally, water can be extracted from their fecal pellets. 2. According to Dr. Atkins, overweight people eat too many carbohydrates. If we severely restrict carbohydrates in our
diet, our bodies will begin to burn stored body fat more efficiently, leading to weight loss. However, a recent study published in the New England Journal of Medicine indicated
that, while initial weight loss was rapid in Atkins dieters, there was no difference in weight loss between Atkins dieters and conventional dieters after one year. One explanation
for this observation is that the initial weight loss is due to water loss. The body responds to a shortage of glucose by storing it as glycogen. The synthesis of glycogen results in
the production of water. Negative side effects of the diet include increased risk of colorectal cancer (due to the high intake of red meat), increased incidence of heart disease
(due to high levels of saturated fat), reduced kidney function (due to the processing of large amounts of protein), and osteoporosis (due to loss of urinary calcium).

Testing Yourself

1. a. salivary glands; b. mouth; c. liver; d. gallbladder;
e. duodenum; f. cecum; g. appendix; h. anus;
i. pharynx; j. esophagus; k. diaphragm; l. stomach; m. pancreas; n. small intestine; o. large intestine; p. anal canal; see Fig. 24.2b, p. 415, in text.; 2. b; 3. a; 4. d; 5. e; 6. b;
7. a. nostril; b. nasal cavity; c. pharynx; d. epiglottis; e. glottis; f. larynx; g. trachea; h. bronchus; i. bronchiole; j. capillary network; k. lung; l. diaphragm; see Fig. 24.14, p.
423, in text.; 8. d; 9. d; 10. a. kidney; b. ureter; c. urinary bladder; d. urethra; see Fig. 24.23a, p. 427, in text.; 11. c; 12. d; 13. b; 14. a; 15. a; 16. d; 17. b; 18. a; 19. c; 20.
b; 21. d; 22. c; 23. c; 24. d

Understanding the Terms

a. pharynx; b. pepsin; c. chyme; d. gallbladder;
e. larynx; f. trachea; g. alveoli; h. heme; i. nephrons

Chapter 25

Thinking Scientifically

Soda Then: Total calories 84; Soda Now: Fat 0; Carbohydrate 93; Protein 0; Total calories: 372; Calorie increase 288; Fries Then: Total calories 223; Fries Now: Fat 25;

Carbohydrate 64; Protein 7; Total calories 509; Calorie increase 286; Bagels Then: Total calories 188; Bagels Now: Fat 1; Carbohydrate 100; Protein 15; Total calories
469; Calorie increase 281

Testing Yourself

1. b; 2. a; 3. f; 4. a; 5. b; 6. e; 7. c; 8. b; 9. d; 10. e; 11. c; 12. a; 13. a. grains; b. 6 oz.; c. vegetables;
d. 2.5 cups; e. fruit; f. 2 cups; g. oils; h. 1

–2 Tbs.; i. milk; j. 3 cups (2 for kids aged 2–8); k. meat and beans; l. 5.5 oz.; see Fig. 25.14, p. 449, in text.;

14. d; 15. c; 16. a; 17. d; 18. a

Understanding the Terms

a. nutrient; b. fiber; c. triglycerides; d. cholesterol; e. vitamins; f. antioxidants; g. body mass index; h. anorexia nervosa; i. bulima nervosa; j. muscle dysmorphia

Chapter 26

Thinking Scientifically

1. Each antibody gene is composed of several discrete segments. The antibody sites that recognize antigens are pieced together from various combinations of these
segments. The ability to combine multiple segments into many different combinations allows the genome to produce antibodies that interact with a multitude of antigens.
Mutations in antibody genes produce additional variability. 2. Drugs such as cyclosporine inhibit IL-2, therefore suppressing the production of natural killer cells and
cytotoxic T cells. However, it does not affect other components of the immune or healing systems, including the production of other types of white blood cells.

Testing Yourself

1. c; 2. a. red bone marrow; b. thymus gland;
c. lymph nodes; d. spleen; e. lymphatic vessels; see Fig. 26.1, p. 456, in text.; 3. e; 4. a; 5. c; 6. a; 7. b; 8. d; 9. b; 10. c; 11. d; 12. c; 13. d; 14. e; 15. a;
16. b; 17. e; 18. c; 19. d; 20. c; 21. b; 22. c;
23. a. hair shaft; b. oil (sebaceous) gland; c. sweat glands; d. dermis; e. epidermis; see Fig. 26.2, p. 458, in text.; 24. e; 25. a; 26. a; 27. a; 28. e; 29. b; 30. c; 31. b; 32. d;
33. a; 34. b; 35. a; 36. c; 37. a

Understanding the Terms

a. appendix; b. red bone marrow; c. macrophage; d. histamine; e. B cells; f. agglutination; g. cytotoxic T cells; h. vaccine; i. allergens; j. Autoimmune disease

Chapter 27

Thinking Scientifically

1. Parkinson’s patients do not exhibit a constriction of blood vessels, so their blood pressure slowly decreases. The loss of function of the sympathetic division;
prevents the body from regulating blood pressure, so rapid changes in blood pressure can occur, leading to orthostatic hypotension. 2. Diabetes type 1 is believ ed to
result when an environmental agent such as a virus causes T cells to destroy the pancreatic islets. Vaccines that use a live virus may introduce that agent. For
example, the mumps and rubella viruses may remain in the body for years after vaccination and can infect pancreatic islet cel ls. Infection lowers the levels of insulin
secreted by these cells. In addition, exposure to the viruses may lead to an autoimmune disorder that causes the destruction of pancreatic islet cells.

Testing Yourself

1. d; 2. c; 3. b; 4. c; 5. a; 6. b; 7. a. cerebrum;
b. skull; c. corpus callosum; d. diencephalon;
e. pineal gland; f. pituitary gland; g. cerebellum;
h. spinal cord; i. medulla oblongata; j. pons;
k. midbrain; l. brain stem; see Fig. 27.9b, in text.;
8. b; 9. a,c; 10. a; 11. d; 12. a; 13. d; 14. d; 15. b;
16. a. pineal gland; b. pituitary gland (hypophysis); c. hypothalamus; d. parathyroid gland; e. thyroid gland; f. thymus gland; g. adrenal gland; h. pancreas; see Fig. 27.15, p.
483, in text.; 17. d; 18. a; 19. d; 20. d; 21. a; 22. a; 23. c; 24. d; 25. e; 26. c; 27. d; 28. d; 29. b

Understanding the Terms

a. axon; b. synaptic cleft; c. acetylcholinesterase;
d. spinal cord; e. prefrontal area; f. nerve;
g. gonadotropic hormones; h. adrenal medulla;
i. Addison disease; j. pancreatic islets

Chapter 28

Thinking Scientifically

1. The blood vessels interfere with the transmission of light to the back of the eye, consequently impairing the signal sent to the brain. Both cancer and eye diseases
involve the growth of new blood vessels. Anti-angiogenesis drugs that have been developed for cancer treatment are currently being studied for the treatment of eye
diseases. 2. Our olfactory system allows us to recognize and discriminate between many different odors. Therefore, we can detect the difference between a dangerous
odor, such as smoke, and a harmless one, such as flower fragrance. On the other hand, every natural toxin tastes bitter, so it is more important to be able to sense
bitterness than to discriminate between bitter tastes. Therefore, humans can identify bitter compounds by taste, but cannot distinguish between different compounds.

Testing Yourself

1. c; 2. a. cranial nerve; b. cochlea; c. tympanic membrane; d. auditory tube; e. auditory canal;
f. ossicles; g. semicircular canals; h. outer ear;
i. middle ear; j. inner ear; see Fig. 28.3a, p. 496;
3. c; 4. b; 5. b; 6. c; 7. a. sclera; b. retina; c. optic nerve; d. vein; e. artery; f. fovea; g. ciliary muscle; h. iris; i. pupil; j. cornea; k. lens; see Fig. 28.8, p. 499, in text.; 8. a; 9. a;
10. c; 11. a, b; 12. c; 13. b;
14. b, c; 15. a

Understanding the Terms

a. spiral organ; b. statocysts; c. iris; d. fovea;
e. vertebral column; f. osteoblast; g. myofibrils;
h. myosin; i. synovial joint; j. meniscus

Chapter 29

Thinking Scientifically

1. The two most obvious potential differences are that the sperm can tolerate higher temperatures or that the body temperature of elephants is lower than that of humans.
To test the high temperature hypothesis, compare the temperatures at which sperm are viable in elephants with that in humans. To test the low body temperature
hypothesis, compare the body temperatures of elephants with that of humans. (The latter is true.) 2. The most logical (although controversial) explanation is that
menopause in women developed early in human evolution. It allowed women to channel their efforts into caring for their existing children, increasing the survival rate of
these children. This would give the woman the best chance to pass her genes through generations, so it increased her fitness. Since men were not the main caregivers
for their children, their fitness would not be increased if they evolved the ability to undergo menopause.

Testing Yourself

1. c; 2. b; 3. a; 4. b; 5. d; 6. a; 7. a. ureter;
b. bladder; c. vas deferens; d. seminal vesicle;
e. ejaculatory duct; f. prostate gland; g. bulbourethral gland; h. urethra; i. epididymis;
j. penis; k. foreskin; l. testis; m. scrotum; see Fig. 29.4b, p. 514, in text.; 8. a. oviduct; b. ovary; c. rectum; d. cervix; e. bladder; f. vagina; g. urethra; see Fig. 29.6b, p. 516, in
text.; 9. a; 10. d; 11. c;
12. a; 13. b; 14. d; 15. c

Understanding the Terms

a. copulation; b. bulbourethral glands;
c. testosterone; d. uterus; e. endometrium;
f. ovulation; g. fertilization; h. morula; i. induction; j. umbilical cord

Chapter 30

Thinking Scientifically

1. Current growth rate • 1.3%; doubling rate • 53 years; growth rate at 10 births per 1,000 • 0.4%; new doubling rate • 172 years. 2. At the 19th week, the field will be
half full. It will double during the 19th week to fill the field. At the 18th week, it will be one quarter full and will double to fill half of the field by the 19th week. Therefore, a
significant population of dandelions will not be observed until the last week.

Testing Yourself

1. a; 2. d; 3. b; 4. d; 5. b; 6. b; 7. d; 8. c; 9. d;
10. b; 11. a. increasing, b. decreasing, c. stable; see
Fig. 30.7, p. 539, in text. 12. a; 13. c; 14. a; 15. a; 16. d; 17. b; 18. c; 19. c; 20. b; 21. c; 22. e; 23. c

Understanding the Terms

a. community; b. resources; c. range; d. population density; e. biotic potential; f. carrying capacity;
g. competition; h. equilibrium population;
i. extinction; j. more-developed countries

Chapter 31

Thinking Scientifically

1. First, insects probably found the sap on the wind-pollinated flowers and used it as a food source. Because it was nutritious and easy to find, the insects gradually
became dependent on it. The insects accidentally picked up and carried pollen as they moved from flower to flower in search of food. Plants that were pollinated by
these insects were more successful (produced more seeds) than those that were wind-pollinated. Therefore, plants that evolved mechanisms to encourage visits by
insects were more fit. Nectaries evolved to bring insects to flowers, and then showy petals and fragrance evolved to advertise the presence of flowers to insects.
Insects continued to evolve to utilize flowers of certain species, and those plant species evolved flower features to attract and reward those insects. 2. Their burrows
act as homes to many other animals, while their burrowing loosens soil, increasing the ability of plants to grow. Their foraging and feeding habits encourage a diversity of
plants to grow. The plants that grow as a result of prairie dog activity are food for other animals. Finally, prairie dogs themselves are major food sources for several
predatory species.

Testing Yourself

1. a; 2. a; 3. e; 4. a; 5. c; 6. b; 7. d; 8. c; 9. c; 10. e; 11. a. energy; b. nutrients; c. heat; d. heat;
e. producers; f. consumers; g. inorganic nutrient pool; h. decomposers; i. heat; see Fig. 31.16,
p. 561, in text.; 12. c; 13. a; 14. a; 15. a; 16. a;
17. d; 18. c; 19. a. top carnivores; b. carnivores;
c. herbivores; d. producers; see Fig. 31.20, p. 563, in text.; 20. c; 21. c; 22. c; 23. c; 24. b; 25. a;
26. d; 27. a; 28. a; 29. b; 30. e; 31. a; 32. g; 33. d; 34. f; 35. b; 36. a; 37. c; 38. d

Understanding the Terms

a. community; b. parasitism; c. ecological niche;
d. character displacement; e. keystone species;
f. detritus; g. trophic level; h. transfer rate; i. biosphere

Chapter 32

Thinking Scientifically

1. Canal option. Positive environmental impacts are: refilling the Aral Sea; combating desertification that has been occurring in recent years; providing ample water for
irrigation, and therefore maintaining the livelihood of farmers. Negative environmental impacts are: Water flow downstream of the diversion canal on the second two rivers
will be lessened. Dam removal option. Positive environmental impacts are: refilling the Aral Sea; combating desertification that has been occurring in recent years.
Negative environmental impacts are: Farmers who rely on irrigation will no longer be able to grow their crops. 2. These would most likely be high priorities: regions with
the highest levels of biodiversity; regions containing species used for traditional medicine; regions containing species used for medicine by indigenous people; regions
containing species that are related to species with known medicinal value; regions containing species that produce chemical compounds related to those of known
medicinal value.

Testing Yourself

1. e; 2. b; 3. d; 4. c; 5. b; 6. a; 7. b; 8. b; 9. c, d, e;
10. a; 11. a, d, e; 12. e; 13. a; 14. Use of only renewable energy sources, reuse of heat and waste materials, and maximal recycling of products. See Fig. 32.20b, p. 592,
in text.; 15. c; 16. e; 17. d;
18. b; 19. d; 20. b; 21. e; 22. e

Understanding the Terms

a. pollution; b. deforestation; c. aquifer;
d. subsidence; e. saltwater intrusion; f. salinization; g. greenhouse gases; h. photovoltaic cell; i. mineral; j. chlorofluorocarbon