Cracking the AP Biology Exam


Answers and Explanations to Practice Test 1

1. E The resting potential depends on active transport (the Na+/K+ pump) and the selective permeability of the axon membrane to K+ than to Na+, which leads to a differential distribution of ions across the axonal membrane.

2. B The Krebs cycle occurs in the mitochondrial matrix. Don’t forget to review the site of each stage of aerobic respiration. (A), Glycolysis, the first step in aerobic respiration, occurs in the cytoplasm. (C), The electron transport chain occurs along the inner mitochondrial membrane. (E), Oxidative phosphorylation occurs as protons (H+ ions) move from the intermembrane space to the mitochondrial matrix.

3. A The earth’s major land biomes are classified according to the climate zones and organisms in a particular region. An individual traveling up a mountain will probably encounter four biomes. Tropical regions (tropical rain forests) are known for their abundant rain supply and high temperatures. Temperate deciduous forests are similar to regions in North America with cold winters, dry summers, and trees that shed leaves. This is followed by the taiga biome. Taiga includes the evergreen conifer trees and animals such as moose and black bears. The top of the mountain is a cold, treeless region with little vegetation—the tundra.

4. B In meiosis, the sister chromatids separate during the second metaphase of meiosis (Meiosis II) whereas the sister chromatids separate during metaphase of mitosis. (A), In meiosis, there are two rounds of cell division, whereas in mitosis, there is only one round of cell division. (C), Chromosomes are replicated during interphase in both meiosis and mitosis. (D), Spindle fibers form during prophase in both mitosis and meiosis. (E), In meiosis, there is a reduction in the number of chromosomes, whereas in mitosis, the number of chromosomes remains the same.

5. C (A), (B), and (D), Amino acids are organic molecules that contain carbon, hydrogen, oxygen, and nitrogen. Don’t forget to associate amino acids with nitrogen because of the amino group (NH2). (E), Amino acids do not contain phosphorus.

6. B Unlike eukaryotes, prokaryotes (which include bacteria) do not contain membrane-bound organelles. (A), (D), and (E), Bacteria contain circular double-stranded DNA, ribosomes, and a cell wall. (C), Bacterial cell membranes are made up of a bilipid layer with proteins interspersed.

7. D Populations can be described as the evolutionary unit because changes in the genetic makeup of populations can be measured over time. (A), Genetic changes occur only at the individual level. (B), Only under Hardy-Weinberg equilibrium does the gene pool remain fixed over time in a population. However, this statement does not explain why the population is the evolving unit. (C), This statement is true but does not address the question. (E), This statement is also true, but it does not address the question.

8. D The medulla controls basic involuntary actions such as breathing. Don’t forget to review the other parts of the brain. (A), The cerebellum controls and coordinates movement and balance. (B), The cerebrum controls higher-level thinking and reasoning. (C), The hypothalamus maintains homeostasis. (E), The thalamus integrates information from other parts of the brain. We did not discuss the thalamus. This was a great opportunity to use Process of Elimination.

9. C The vascular cambium includes actively dividing cells that lie between the xylem and phloem to produce secondary xylem and phloem. (A), The apical meristem refers to actively dividing cells located at the tip of roots and stems. (B), The epidermis is the outermost top layer in leaves. (D), Cork cambium is the outer layer of the meristem in woody stems. (E), Lenticels are the pores used for gas exchange in woody stems.

10. B In order to determine the genotype of the parents, use the ratio of the offspring given in the question and work backward. The ratio of black-haired to white-haired guinea pigs is 3,1. The offspring were therefore BB, Bb, Bb, and bb. Now draw a Punnett square to figure out the genotype of the parents. The parents are Bb and Bb.

11. D The mean weight of the offspring in the next generation will be heavier than the mean weight of the original population because all the lighter horses in the original population died off. The normal distribution for weight will therefore shift to the heavier end (to the right of the graph). You can therefore eliminate answer choices (C) and (E) because the mean weight should increase. (A) and (B), The mean weight of the offspring could be heavier or lighter than their parents.

12. D A fertilized egg undergoes a series of changes in the following order, cleavage (first divisions of the zygote), morula (solid ball), blastula (hollow ball), gastrula (formation of three germ layers), and neurula (formation of the nervous system).

13. A The primitive atmosphere lacked oxygen (O2). It contained methane (CH4), ammonia (NH3), hydrogen (H2), and water (H2O).

14. C Make a Punnett square to determine the probability that the couple has a child with blood type AB. The probability is whether it’s the first child or the third child.

15. B You’re looking for the best explanation for the low trophic level efficiency among elks. The best explanation is that large herbivores, such as elks, eat plants but only incorporate a small portion of the nutrients in plants because they cannot digest cellulose. (A), This answer choice sounds good, but we don’t know if elks consume less food than do other herbivores. (C), This response also sounds good, but we don’t know if elks use a lot more energy than do other herbivores as they search for food. (D), This response is clearly false. Eating producers is a very efficient way of getting energy. (E), We don’t know if elks lose more body heat than do other herbivores. Even if they did, this is not directly related to trophic level efficiency.

16. A Porifera sponges and coelenterates are more primitive than platyhelminthes (such as flatworms), so cross out (D) and (E). Echinoderms (such as starfish) and chordates are more complex than arthropods, so cross out (B) and (C).

17. B Pepsin works in the stomach (not the small intestine) to break down proteins to peptides. Complete digestion occurs in the small intestine. (A), Pancreatic lipase breaks down fats into three fatty acids and glycerol. (C), Pancreatic amylase breaks down carbohydrates into simple sugars. (D), Bile emulsifies lipids and makes them more accessible to lipase. (E), Food is absorbed by the villi in the small intestine.

18. E Ribosomes are the site of protein synthesis. Therefore, the correct answer should start with ribosome. So eliminate answer choices (A), (B), and (D). The polypeptide then moves through the rough ER to the Golgi apparatus, where it is modified and packaged into a vesicle. The vesicle then floats to the plasma membrane and is secreted.

19. E This statement is false because an individual with two identical alleles is said to be homozygous not heterozygous with respect to that gene. (A) and (B), Alleles are different forms of the same gene found on corresponding positions of homologous chromosomes. (C), More than two alleles can exist for a gene, but a person can have only two alleles for each trait. (Remember our discussion of blood groups in Chapter 11.) (D), One allele can mask another allele.

20. C To make multiple copies of a plasmid (a small circular DNA), it should be inserted into a bacterium. (A), A plasmid would not replicate if it were inserted into a virus. (B), If a plasmid were treated with a restriction enzyme, it would be cut into smaller fragments. This would not give us cloned versions of the plasmid. (D), If the plasmid were run on a gel (using gel electrophoresis), this would only tell us the size of the plasmid. (E), Plasmids are small circular DNA and cannot be infected. Only living organisms can be infected.

21. D The least likely explanation for why mutations are low is that mutations produce variability in a gene pool. Any gene is bound to mutate. This produces a constant input of new genetic information in a gene pool. This answer choice doesn’t give us any additional information about the rate of mutations. (A), Some mutations are subtle and cause only a slight decrease in reproductive output. (B), Some mutations are harmful and decrease the productive success of the individual. (C), Some mutations are deleterious and lead to total reproductive failure. The zygote fails to develop. (E), Some mutations back mutate. By the way, this is a tough question and should have been skipped on both a first and second pass!

22. C Desert plants (for example, cacti) have all the features of flowering plants (vascular tissues, multiple leaves, megaspores, and stomates). An adaptive feature in this type of plant is thick cuticles to prevent excessive water loss.

23. B Legume plants are able to live in nitrogen-poor soil because they obtain nitrogen from nitrogen-fixing bacteria. (A), (C), and (D), These plants cannot make their own proteins without nitrogen from nitrogen-fixing bacteria. (E), Legumes do not possess nitrogen-absorbing root hairs.

24. E Don’t forget to review the organs that arise from the three primary germ layers. (A), (B), (C), and (D), The brain, eyes, skin, and nervous system are derived from the ectoderm. (E), Ovaries, which are part of the reproductive system, are derived from the mesoderm.

25. D Deoxygenated blood from the vena cava enters the right atrium (2), then the right ventricle (1), and then enters the pulmonary arteries (4). The left atrium (5), left ventricle (6), and aorta (3) all carry oxygenated blood.

26. E The egg yolk provides food for the embryo. (A), The amnion protects the embryo. (B), The chorion is the outermost layer that surrounds the embryo. (C), The placenta provides nourishment for the embryo in mammals, not in birds. (D), The ovary is the reproductive organ in which eggs mature.

27. B Normal cells can become cancerous when a virus invades the cell and takes over the replicative machinery. (A), A pilus forms between two bacteria. (C), The host’s genome is not converted to the viral genome. (D), Spores are released by fungi, not viruses. (E), Incorporation of free-floating DNA would not necessarily produce cancerous cells.

28. C The primary difference between bone and cartilage is that cartilage is flexible and lacks blood vessels. (A), They are both part of the skeletal system. (B) and (E), Bone and cartilage are connective tissues made up of collagen and calcium salts. (D), Both cartilage and bones secrete a matrix.

29. D If potential mates have similar breeding seasons they will most likely mate. Answer choices (A), (B), (C), and (E) are all examples of prezygotic barriers that can prevent interbreeding. Use common sense to eliminate the other answer choices. (A), If the organisms don’t meet, they won’t reproduce. (B) and (C), If the potential mates do not share the same behaviors (such as courtship rituals), they may not mate. (E) This is an example of physiological isolation. If the gametes are biochemically different, the organisms won’t reproduce.

30. D There is no union of gametes in mitosis. (A) and (C), Asexual reproduction involves the production of two new cells with the same number of chromosomes as the parent cell. If the parent cell is diploid, then the daughter cells will be diploid. (B), The daughter cells are identical to the parent cell. (E), Chromosomes replicate during interphase.

31. D Arteries are thick-walled vessels that carry blood away from the heart. Blood moves by contracting muscles. (A), (B), (C), and (E) are all characteristics of veins. Veins are thin-walled vessels (with valves) that return blood to the heart.

32. B Auxins are plant hormones with many properties. They stimulate cell elongation, bud and fruit development, and play a role in phototropism. (A), Cytokinins stimulate cell division in plants. (C), Gibberellins promote stem elongation (especially in dwarf plants). (D), Abscisic acid causes growth inhibition. (E), Ethylene causes fruit ripening.

33. A After double fertilization, the ovule becomes the seed. (B) and (E), The ovary becomes the fruit and the fertilized egg becomes the embryonic plant. (D), The cotyledons are the leaves of the embryonic plant. (C), The pollen grains form the male gametes.

34. E The frequency of the homozygous dominant genotype (AA) is 0.49. To find the dominant allele frequency, we can use the formula provided by the Hardy-Weinberg theory, p2 + 2pq + p2 = 1, where p represents the dominant allele and q represents the recessive allele. Because we know that p2 represents the frequency of the homozygous dominant genotype, we can find the frequency of the dominant allele (p) by taking the square root of the frequency of the genotype. The square root of 0.49 is 0.7. Note that by using the formula p + q = 1, we can also determine the frequency of the recessive allele (q). It would be 0.3 (0.7 + q = 1).

35. A Crossing-over (exchange of genetic material) occurs in prophase I of meiosis. (B), During metaphase I, the tetrads line up at the metaphase plate. (C), During anaphase, the tetrads separate. (D), During prophase II, the chromosomes split at the centromere. (E), During metaphase II, the chromosomes line up at the metaphase plate.

36. D The microspores (male gametes) are produced in the pollen grains of the anther, not the stamens. All of the other statements are true. (A), Double fertilization occurs in the ovules. (B), A megaspore divides and becomes the female gametes. (C), The pollen tube grows down the style to the ovary. (E), The endosperm (3n) becomes the food for the embryo.

37. E Bryophytes are primitive plants that include mosses and liverworts. (A), They do not have a protective layer around their gametes. The gametes are found in the parent gametophyte. (B), They lack conductive tissues. (C) and (D), They lack true leaves. They therefore do not have stomates or waxy cuticles.

38. D The net primary productivity is the energy producers have left for storage after their own energy needs have been met. Energy stored = (Total energy produced) – (energy used in cellular respiration). (A), It is not the energy available to producers. (B), It is not the total chemical energy in producers. That’s the gross primary productivity. (C), It is not the biomass (total living material) among producers. (E), It is not the energy expended by producers.

39. C (A), (B), (E), These organisms exhibit the same behavior because they were subjected to the same environmental conditions and similar habitats. This is an example of convergent evolution. (C), However, they are not genetically similar. (One is an insect and the other a bird.) (D), They are analogous. They exhibit the same function but are structurally different.

40. C These characteristics are all found in insects, which include arthropods with wings. (A), A snail is a mollusk. (B), A sponge is a porifera. (D), An earthworm is an annelid. (E), A flatworm is a platyhelminthes.

41. D Beta cells secrete insulin. Destruction of beta cells in the pancreas will halt the production of insulin. Therefore, eliminate answer choices (A), (B) and (C). This will lead to an increase in blood glucose levels.

42. D The sympathetic division is active during emergency situations. This leads to a decrease in peristalsis in your gastrointestinal tract. (Your stomach shuts down.) Stimulation of the sympathetic nervous system leads to (A), pupils dilating, (B), peripheral blood vessels constricting, (C), sweating, and (E), heart rate increasing.

43. C This is an example of mutualism. Both organisms benefit. (A), Parasitism is an example of a symbiotic relationship in which one organism benefits and the other is harmed. (B), Commensalism is when one organism benefits and the other is unaffected. (D), Endosymbiosis is the idea that some organelles originated as symbiotic prokaryotes that live inside larger cells. (E), This is a behavior that is directly beneficial to other members of the species but at some risk to the organism.

44. A They both contain genetic material (DNA), a plasma membrane, and a cell wall. Use Process of Elimination. Unlike fungi, bacteria lack a definite nucleus. Therefore, eliminate (B) and (E). Bacteria are unicellular, whereas fungi are both unicellular and multicellular. Therefore, eliminate (C) and (D).

45. B Let’s use Process of Elimination. The two hormones that are responsible for the maintenance of Ca2+ in the blood are parathyroid hormone and calcitonin. Therefore, eliminate answer choices (A), (C), and (E). Now review the effects of these hormones. Parathyroid hormone increases Ca2+ in the blood while calcitonin decreases Ca2+ in the blood. A sustained decrease in circulating Ca2+ levels might be caused by decreased levels of parathyroid hormone.

46. B This question tests your understanding of what stage is responsible for the synthesis of new proteins for lactose utilization. The region of bacterial DNA that controls gene expression is the lac operon. Structural genes will be transcribed to produce enzymes, which produce an mRNA involved in digesting lactose.

47. C (A) and (B), Enzymes, which are proteins, are organic catalysts that speed up reactions without altering them. They are not consumed in the process. (D), The rate of reaction can be affected by the concentration of the substrate up to a point. (E), They denature (uncoil) when they’re exposed to high temperatures.

48. E The binomial classification system is based on a genus and species name. (Don’t forget the classification mnemonic). Therefore, the paramecia share the same genus (same first name) but different species (different second name).

49. C DNA polymerase, not RNA polymerase, is the enzyme that causes the DNA strands to elongate. (A), DNA helicase unwinds the double helix. (B), DNA ligase seals the discontinuous Okazaki fragments. (D) and (E), DNA strands, in the presence of DNA polymerase, always grow in the 5′ to 3′ direction as complementary bases attach.

50. B A voltage change from +50 to –70 is called repolarization. (A), A voltage change from –70 to +50 is called depolarization. (C), A voltage change from –70 to –90 is called hyperpolarization. (D), An action potential is a traveling depolarized wave. It refers to the whole thing, from depolarization, repolarization, hyperpolarization, and back to a resting potential. (E), Saltatory conduction refers to the skipping of an impulse from one node to another node in myelinated neurons.

51. C Electrons passed down along the electron transport chain from one carrier to another lose energy and provide energy for making ATP. (A), Glucose is decomposed during glycolysis, but this process is not associated with energy given up by electrons. (B), Glucose is made during photosynthesis. (D) and (E), NADH and FADH2 are energy-rich molecules, which accepted electrons during the Krebs cycle.

52. A The oxygen released during the light reaction comes from the splitting of water. (Review the reaction for photosynthesis.) Therefore water must have originally contained the radioactive oxygen. (B), Carbon dioxide is involved in the dark reaction and produces glucose. (C), Glucose is the final product and would not be radioactive unless carbon dioxide was the radioactive material. (D), Nitrogen is not part of photosynthesis. (E), Atmospheric oxygen has nothing to do with photosynthesis.

53. E Hydrophobic means “water fearing;” so water cannot be a hydrophobic solvent (A) and (C), Water is a polar solvent; it contains both negatively and positively charged ends that can form hydrogen bonds with other polar substances. (B), It can resist temperature changes because it has a high specific heat. (You need lots of heat to raise the temperature of water.) (D), Water can break up into H+ and OH ions.

54. C Pheromones act as sex attractants, alarm signals, or territorial markers. Use Process of Elimination for this question. (A), Auxins are plant hormones that promote growth. (B), Hormones are chemical messengers that produce a specific effect on target cells within the same organism. (D), Enzymes are catalysts that speed up reactions. (E), Coenzymes are organic substances that assist enzymes in a chemical reaction.

55. D Homologous structures are organisms with the same structure but different functions. The forelegs of an insect and the forelimbs of a dog are not structurally similar. (One is an invertebrate and the other a vertebrate.) They do not share a common ancestor. However, both structures are used for movement. All of the other examples are vertebrates that are structurally similar.

56. E (I), Muscle contractions require calcium ions. (II), In order to have a muscle contraction you need energy—ATP. (III), Actin is one of the proteins involved in muscle contractions.

57. B Speciation occurred in the Galapagos finches as a result of the different environments on the islands. This is an example of divergent evolution. The finches were geographically isolated. (A), Convergent evolution is the evolution of similar structures in distantly related organisms. (C), Disruptive selection is selection that favors both extremes at the expense of the intermediates in a population. (D), Stabilizing selection is selection that favors the intermediates at the expense of the extreme phenotypes in a population. (E), Directional selection is selection that favors one of the extremes in a population.

58. E Mutations produce genetic variability. All of the other answer choices are forms of asexual reproduction.

59. B These structures are homologous. They are structurally similar but have different functions. The swim (air) bladder is a gas-filled air sac that provides buoyancy for fish. It is homologous with the lungs of vertebrates.

60. A These structures are analogous. They have the same function, but they’re different structures. Spines and quills are both used for protection.

61. C These structures are vestigial. They are parts of an animal that are no longer functional.

62. D This is an example of convergent evolution. Convergent evolution is the independent evolution of similar structures in distantly related organisms that occur in similar ecological niches.

63. B Lenticels are pores in stems that facilitate gas exchange.

64. A Stomata are pores in the epidermis of leaves that allow the diffusion of gases.

65. C Don’t be thrown off by the term mesophyll. You know that the palisade layer contains photosynthetic cells.

66. D Binary fission is the equal division of a bacterial cell into two. It is a form of asexual reproduction.

67. C Regeneration is the growing back of a lost body part. Use Process of Elimination.

68. B Budding is a form of asexual reproduction in yeasts whereby smaller cells grow from a parent cell.

69. E Vegetative propagation is a form of asexual reproduction in which flowering plants can reproduce with the use of stems, leaves, or roots.

70. D A triglyceride is an organic molecule that consists of three fatty acids and glycerol.

71. E Glycogen is an animal starch. It is stored in the liver and muscles of animals.

72. B A protein is a macromolecule consisting of polypeptides (amino acids).

73. C Cellulose is a carbohydrate that can not be broken down by cows. That’s why they need bacteria in their stomachs.

74. E A bird has a four-chambered heart. Review the classification list in this book.

75. A A hydra is a simple animal with two layers, an ectoderm and an endoderm. You can therefore eliminate (B). Review the classification list in this book.

76. C A fish has a two-chambered heart. Review the classification list in this book.

77. D An amphibian has a three-chambered heart. Review the classification list in this book.

78. B Reptiles are cold-blooded and usually covered with dry scales. Review the classification list in this book.

79. D Homeotherms means warm-blooded. But even if you didn’t know that, the only animals with air-filled bones are birds. Review the classification list in this book.

80. B These are the characteristics of reptiles. Don’t forget you can use an answer choice more than once. The egg is surrounded by a tough leathery shell, which protects the embryo against desiccation. Review the classification list in this book.

81. E Amphibians have thin, moist skin. They were the first animals to conquer land, so they have both gills and lungs. Review the classification list in this book.

82. C Secondary consumers feed on primary consumers. If you set up a pyramid of numbers, you’ll see that the herrings belong to the third trophic level.

83. A The biomass is the total bulk of a particular living organism. The plankton population has both the largest biomass and the most energy.

84. C If the herring population decreases, this will lead to an increase in the number of crustaceans and a decrease in the plankton population. Reorder the organisms according to their trophic levels and determine which populations will increase and decrease accordingly.

85. D This question tests your ability to trace the neural pathway of a motor (effector) neuron. The nerve conduction will travel from the spinal cord (where interneurons are located) to the muscle.

86. D Because the brain is destroyed, it is not associated with the movement of the leg. (A), Reflex actions are automatic. (B) and (C), Both of these statements are true but are not supported by the experiment. (E), We do not have enough information from the passage to determine if this statement is true.

87. C Neurotransmitters are released from the axonal bulb of one neuron and diffuse across a synapse to activate a second neuron. The second neuron is called a postsynaptic neuron. A neurotransmitter can either excite or inhibit the postsynaptic neuron. (A), The myelin sheath speeds up the conduction in a neuron. (B), (D), and (E), Both sodium and potassium channels open during an action potential. Neurotransmitters are not involved in actions related to the axon membrane. They do not force potassium ions to move against a concentration gradient.

88. A If you look at the absorption spectrum, you’ll see that chlorophyll a has two peaks, one at 425 nm and 680 nm. Chlorophyll a maximally absorbs light at approximately 425 nm.

89. B Chlorophyll pigments absorb light in the violet-blue and orange-red range and reflect green light. That’s why plants appear green. Therefore, you can eliminate answer choices that contain the color green. That’s all of them except (B).

90. C (A), Based on the results of the absorption spectrum, the light absorbed by chlorophylls and carotenoids is involved in photosynthesis. (B) and (D), We know from our review of photosynthesis that only certain wavelengths of visible light can be used for photosynthesis. (E), The light that is reflected is primarily green.

91. C Draw a Punnett square for the couple and determine the probability of color-blindness for the boys. Individuals 6 and 7, (XcX and XY) will produce two males, one XcY and one XY. The probability of color-blindness is therefore 1/2.

92. D Because the genotype of the mother is unknown, we need to look at her children in order to determine her status. Because one of the sons is color-blind, she must have the color-blind gene. However, she is not color-blind, because she produced a normal male.

93. C The DNA template strand is complementary to the mRNA strand. Using the mRNA strand, work backward to establish the sequence of the DNA strand. Don’t forget that DNA strands do not contain uracil, so eliminate answer choices (B) and (D).

94. B Use the amino acid chart to determine the sequence after uracil is deleted. The deletion of uracil creates a frameshift.

95. D The mRNA is modified before it leaves the nucleus. It becomes smaller when introns (intervening sequences) are removed. (A) and (B), A poly (A) tail and a cap are added to the mRNA and would therefore increase the length of the mRNA. (C), Exons are the coding sequences that are kept by the mRNA. (E), Translation is when we produce a protein from a molecule of mRNA.

96. E Make sure you read the question carefully. You are asked to calculate the number of liters per week, not per day. The chart tells us that a coconut palm loses 75 liters a day, which would mean 525 liters a week (7 × 75 = 525).

97. A Water moves from the soil to the leaves via tracheids. (B) and (D), Sieve-tube cells and companion cells carry nutrients, not water. (C), Lenticels are pores that regulate gas exchange in woody stems. (E), The spongy layer consists of loosely organized cells in the lower layer of the leaf.

98. E Answer choices (A) through (D) all provide the force that raises water and solutes from the root to the stem. (E), Nutrients are carried through phloem and are not involved in transpiration.

99. B The Weddell seal stores twice as much oxygen as humans. Calculate the liters per kilograms weight for both the seal and man using the information at the bottom of the chart. The Weddell seal stores 0.058 liters/kilograms (25.9 liters/450 kilograms) compared to 0.028 liters/kilograms (1.95 liters/70 kilograms) in humans.

100. C The most plausible answer is that blood is redirected toward the central nervous system, which permits the seal to navigate for long durations. (A), The seal does not need to increase the number of red blood cells in the nervous system. (B), The seal does not need to increase the amount of oxygen to the skeletal system. (D), The diversion of blood does not increase the concentration of oxygen in the lungs. (E), The seal does not decrease pressure on itself if it diverts blood to the eyes, brain, and spinal cord.


In the following pages you’ll find two types of aids for grading your essays, checklists, and written paragraphs. You’ll recall from our discussion of the essays in Chapter 14 that ETS uses checklists just like these to grade your essays on the actual test. They’re actually quite simple to use.

For each item you mentioned, give yourself the appropriate number of points (1 point, 2 points, etc.). Remember that you can only get a maximum of 10 points for each essay. As you evaluate your work, don’t be kind to yourself just because you like your own essay, If the checklist mentions an explanation of structure and function and you failed to give both, then do not give yourself a point. ETS is very particular about this. You need to mention precisely the things they’ve listed, in the way they’ve listed them, in order to gain points. What if something you’ve mentioned doesn’t appear on the list?

Provided you know it’s valid, give yourself a point. If it was something you pulled out of your hat at the last minute, odds are it’s not directly applicable to the question. However, because you might come up with details even more specific than those contained in the checklist, it’s not unlikely that the example or structure you’ve cited is perfectly valid. If so, go ahead and give yourself the point. Remember that ETS hires college professors and high school teachers to read your essays, so they’ll undoubtedly recognize any legitimate information you slip into the essay.

The second part of the answer key involves short paragraphs. These are not templates, ETS does not expect you to write this way. They are simply additional tools to help illustrate the things you need to squeeze into your essay in order to rack up the points. They explain in some detail how the various parts of the checklist relate to one another, and may give you an idea about how best to integrate them into your own essays come test-time.

If you find that you didn’t do too well on the essay portion, go back and practice some essays from Chapter 14. You can use the chapters themselves as checklists. If you find it too difficult to grade your own essays, see if your teacher or a classmate will help you out. Good luck!


a. Cellular structures—4 points maximum

Neuronal plasma membranes and other cell membranes—4 points maximum
Bilipid layer (enriched with phospholipids and proteins)
Selective permeability (regulates the passage of certain ions across its membrane)

b. Ions associated with plasma membrane—4 points maximum

(1 point each)



Selectively permeable to K+ ions

(Concentrated on the inside of the cell)

Impermeable to Na+ ions

(Concentrated on the outside of the cell)

Build up of positively charged ions outside of the membrane

(Inside is negatively charged)

Sodium-potassium pump

(Maintains the ion gradient)

c. Role of the membrane—4 points maximum

(.5 point each)

Nerve impulses are electrochemical (associated with ion and electrical changes in plasma membrane)

Action potential (change in the membrane potential)

Resting stage (voltage charge is –70 millivolts)

Depolarization (Na+ moves into the cell down its cell gradient)

Repolarization (K+ ions move out of the cell down their cell gradient)

Sodium channels (membrane channel that is voltage-gated)

Potassium channels (membrane channel that is voltage-gated)

Sodium-potassium pump (membrane protein that maintains ion gradient)


A nerve cell is fundamentally similar in structure and function to other somatic cells. Like other cells, a nerve cell consists of a bilipid layer made up of phospholipids and proteins. This membrane is semipermeable; it regulates the passage of certain ions across its membrane. The cell body of a nerve cell also contains cellular structures such as a nucleus, which regulates the activities of the cell, and ribosomes, which make proteins.

The plasma membrane of a resting neuron is selectively permeable to K+ ions and impermeable to Na+ ions. Because of this selective permeability, the K+ ions are concentrated on the inside of the cell and are able to slowly diffuse outward, while the Na+ ions are concentrated on the outside of the cell. This leads to a build up of positively-charged ions outside of the neuronal membrane. The inside of the plasma membrane is therefore negatively charged. These chemical gradients are maintained by a sodium-potassium pump. This pump actively moves three Na+ ions out of the cell for every two K+ ions brought into the cell.

The neuronal membrane plays an important role in the conduction of a nerve impulse. Nerve impulses are electrochemical. This means, the forces that cause ions to move across a membrane are both a concentration gradient and an electrical gradient. The membrane is associated with two chemicals (Na+ ions and K+ ions) as well as a change in the voltage charge. When a nerve cell is undisturbed, the membrane is said to be in a resting stage. The membrane is polarized and the voltage charge is –70 millivolts. During a nerve impulse, there is a change in the membrane permeability, Na+ ions rush into the cell, and the inside becomes more positively charged. The membrane is now said to be depolarized. Na+ move into the cell down its concentration gradient. Next, the Na+ channels close and K+ channels open and K+ ions move out of the cell. The cell is now said to be repolarized. The original ion concentration is reestablished by the sodium-potassium pump.


I. Adaptational Problems—4 points maximum

(1 point each)

Plants required water on land.

Plants required specialized plant organs (roots, stems, and leaves).

Plants needed a method to carry out sexual reproduction on land.

Plants needed to develop physiological adaptations that would meet certain specific challenges (need for moisture via transpiration/need to withstand high temperatures via evaporation).

II. Structural adaptations to acquire and retain water in land plants—4 points maximum

(1 point each)

(1 point each)


Function described


Retard water loss


Pores in leaf that regulate water and gas exchange


Structures that absorb water and minerals

Root hairs

Extensions of the root that increase the surface area for water absorption


Transport water and minerals throughout plant

III. Other structural adaptations—4 points maximum


Transport nutrients


Resist desiccation; dormancy


Contain photosynthetic cells


Contain conductive system and support plant


Protection of embryo, food


Facilitate sexual reproduction


There were several adaptational problems that had to be solved before plants could survive on land without water. First, plants needed to develop different methods to retain water while on land. Second, plants required specialized organs such as roots and leaves. Third, plants needed to develop a method of conducting sexual reproduction in a terrestrial environment. Fourth, plants developed physiological adaptations to deal with their new environment. These include finding ways of keeping enough moisture and withstanding extreme temperatures.

In order to acquire and retain water, plants developed waxy cuticles on their outer surfaces, which prevent desiccation. They also developed stomates on their leaf surfaces, which controlled water loss (transpiration). Plants also contain root hairs, which absorb water and minerals. Plants also developed other structures to survive on land. Some cells specialized and became leaves with cells that carry out photosynthesis. Other cells specialized and became stems with vascular tissues, which include xylem (which conduct water) and phloem (which conduct food).


a. Type of Mutation—4 points maximum

(1 point each)

Base substitution (involves one nucleotide being replaced by another)

Change in DNA/codon of mRNA draws new tRNA molecule

New tRNA carries wrong amino acid, altering polypeptide

Change in polypeptide in turn alters hemoglobin protein

Distorted red blood cell cannot efficiently carry oxygen

b. Gel Electrophoresis—5 points maximum

How it works—3 points maximum

(1 point each)


DNA is put into wells of an agarose gel with a buffer


Electrical potential (electrical charge moves fragments)


Negatively charged fragments move toward the positive pole

Size/Molecular weight

Smaller fragments move faster across the gel

How it can be applied to identify the mutant gene—2 points maximum

(1 point each)

Compare the normal hemoglobin to the abnormal hemoglobin

Use restriction enzymes to cut the two hemoglobin proteins into several fragments

Normal hemoglobin can be used as a marker

Fragments of the two proteins should be identical except for the fragment containing the abnormal gene

c. Punnett Square—2 points maximum

A normal female who does not carry the X-linked allele mates with a male homozygous for the disease,

HH → hh → All the offspring are Hh (heterozygotes)

They are all carriers (sickle-cell trait)


Sickle-cell anemia is a disease in which the red blood cells have an abnormal shape because of a base substitution. Base substitution involves an error in DNA replication in which one nucleotide is replaced by another nucleotide. This causes the codon of an mRNA to contain an incorrect base. The codon therefore matches up with the anticodon of a different tRNA. This tRNA carries a different amino acid. The change in the amino acid alters the polypeptide. The polypeptide, in turn, alters the hemoglobin protein. In this case, the distorted red blood cell cannot efficiently carry oxygen.

Biologists must have determined the nature of the hemoglobin mutation by comparing the normal hemoglobin to the abnormal hemoglobin, using the technique gel electrophoresis. Gel electrophoresis identifies the difference between two molecules by examining the different rates each molecule moves across the gel. Substances move across a gel according to their molecular weight. For example, smaller fragments move faster than larger fragments. The biologists must have placed fragments of the two proteins on an agarose gel and used the normal hemoglobin as the marker—the source of comparison to the abnormal hemoglobin. (A restriction enzyme was used to cut the two proteins into several fragments prior to loading the gel). The fragments of the two proteins should have been identical except for the fragment that contained the abnormal gene.

If a normal noncarrier female mates with a male who is homozygous for the disease, these are the results using a Punnett square,

All of the offspring would be heterozygotes. For sickle-cell anemia, these offspring would be carriers.


a. Krebs cycle, the electron transport chain, oxidative phosphorylation as aerobic processes—2 points maximum

They require oxygen

They cannot occur under anaerobic conditions

b. Site of each step—3 points maximum



Krebs cycle

Mitochondrial matrix

Electron transport chain

Along the inner mitochondrial membrane

Oxidative phosphorylation

As hydrogens move from the intermembrane space to the mitochondrial matrix

c. Role of NADH and FADH2 in aerobic respiration—5 points maximum

NADH and FADH2 are electron/hydrogen ion carriers

Coenzymes NAD+ and FAD+ accept electrons (and hydrogen) to form NADH and FADH2

NADH and FADH2 shuttle electrons to the electron transport chain

The hydrogens dissociate into electrons ions and electrons

The electrons are passed down the electron transport chain to form water

The hydrogen ions are pumped out into the intermembrane space and cross back to produce ATP


a. The Krebs cycle, electron transport chain, and oxidative phosphorylation are all part of aerobic respiration. During aerobic respiration, glucose is completely converted to CO2, ATP, and water. These steps are considered aerobic processes because they cannot occur under anaerobic conditions; they require oxygen. Glycolysis, on the other hand, can occur under both aerobic and anaerobic conditions.

b. These three stages of aerobic respiration occur in different parts of the mitochondria. The Krebs cycle occurs in the mitochondrial matrix. Two acetyl CoA enter the Krebs cycle and produce NADH, FADH2, ATP, and CO2. The products of the Krebs cycle (NADH and FADH2) are sent to the electron transport chain. The electron transport chain occurs along the inner mitochondrial membrane. The final stage of aerobic respiration—oxidative phosphorylation—occurs as hydrogens move across from the intermembrane space to the mitochondrial matrix.

  1. NADH and FADH2 are hydrogen ion and electron carriers. When the coenzymes NAD+ and FAD+ accept hydrogens, they form NADH and FADH2. NADH and FADH2 shuttle the hydrogens to the electron transport chain, which have the potential to make a lot of ATP. Once the carriers drop off the hydrogens at the electron transport chain, the hydrogens dissociate into hydrogen ions and electrons. The electrons are sent down the electron transport chain and eventually make water. Meanwhile, the hydrogen ions are pumped out into the intermembrane space. When they cross back into the mitochondrial matrix, they produce ATP.