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. CO2
enrichment had to be combined with increased temperature to have its
maximum effect on growth. d. Yes, especially at 30C.
Again, boosting all factors that contribute to photosynthetic rate
produced the maximum effect. e. Yes, an increase in CO2
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 CO2
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
CO2
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 CO2
and temperature. This may influence total CO2
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 C6
molecule results.; c. Oxidation results in NADH and CO2.;
d.
ATP is produced by substrate-level and ATP synthesis.; e. Oxidation
produces more NADH and FADH2;
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. G1;
b. S; c. G2;
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. XbY;
b. XBXb;
c.
XBY;
d. XBY;
e. XbY;
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.
XaXa,
father XaY,
mother XAXa
or XaXa.
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 under 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 with 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 believed 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 cells. 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