SAT Biology E/M Subject Test
Part IV The Princeton Review Practice SAT Biology E/M Subject Tests and Explanations
Chapter 19 Practice SAT Biology E/M Subject Test 2: Answers and Explanations
ANSWERS AND EXPLANATIONS
1 D The guard cells are found near the openings in the leaf called stomates. Stomates allow gas exchange to occur (notably oxygen and carbon dioxide), and while they are open water leaves through transpiration. You should have immediately eliminated choices A and E, because they are not leaf structures.
2 B Start by eliminating structures that are not layers, such as choices A, D, and E. Because humans do not have a cuticle as an outer layer (that”s a plant layer), choice C can also be eliminated.
3 C Using the same strategy as above, eliminate choices A, D, and E, which are not layers. Plants have an epidermis, but the layer that restricts evaporation is the cuticle.
4 E As always, eliminate first. Choices C and D are not found in humans and can be eliminated. A thermoregulatory structure is one that helps control temperature. The epidermis (choice B) prevents loss of water, but does not really aid in thermoregulation (that”s the job of the dermis), so eliminate choice B as well. You may be tempted by choice A, because hair plays a minor thermoregulatory role in humans (its role is much more important in other animals), but a sweat gland is a major thermoregulatory structure.
5 C For this question, you really have to know the definition of translate and know which type of nucleic acid gets translated. Eliminate choice E first, because it is not a nucleic acid (it”s an enzyme). You should also be able to eliminate choice A fairly easily, if you know that RNA, and not DNA, is the nucleic acid that is translated. Finally, you should know that mRNA (messenger RNA) is the nucleic acid that is read (translated) to synthesize protein.
6 B You should again be able to eliminate choices E and A, as above. The type of RNA that carries amino acids is tRNA. Remember t for “transport.”
7 A The only type of nucleic acid that is passed on to offspring (progeny—don”t fall into the camouflage trap) is DNA.
8 B Remember that the choices given for the classification questions can be used more than once. tRNA includes the anticodon that base-pairs with a codon in mRNA.
9 D For these diagram questions, start at the right-hand side of the diagram and work toward the left until you get to a letter (answer choice) that separates the animals described in the question. Beginning on the right, find the animals with nails (apes and humans) and the animals with claws (lions and bears). Then work toward the left until you find the common branch from which these species originated. The split into orders Carnivora and Primates separated these animals.
10 B On the right, find segmented worms (earthworm) and nonsegmented worms (roundworm), and again work toward the left. The split occurs at phylum Nematoda (nonsegmented worms) and phylum Annelida (segmented worms).
11 E On the right, the animals with feathers are birds and animals with hair include lions, bears, apes, and humans. Working toward the left, the split that separates birds from the rest is the one that separates class Mammalia from class Aves.
12 A This is the most difficult of the four questions. Animals with exoskeletons are ants and spiders; all the rest have endoskeletons (lions, bears, apes, humans, and birds) or no skeletons (roundworms and earthworms). If you work toward the left, you find the choices D and E only divide up the animals with endoskeletons but do not separate them from the animals with exoskeletons. Choice C only subdivides phylum Arthropoda; it does not make the separation asked for in the question. This leaves only choices A and B, and choice B separates segmentation from no segmentation; neither of these animals has a skeleton at all. The correct answer is A, which divides phylum Arthropoda (animals with exoskeletons) from phylums Nematoda and Annelida (animals with no exoskeletons).
13 E If the albino plant is draining food from the green plant, the green plant is being harmed (being deprived of nutrition). Therefore, you can eliminate choices that do NOT cause harm: choice B (commensalism) and choice D (mutualism). Choice C can be eliminated because it discusses competition between two different species (interspecific) and the plants are of the same species. The choice falls between A and E. E is a better choice because the green plant is not being killed by the albino plant and predators generally kill their prey.
14 B Because the turtle is not being harmed, eliminate choices that cause harm to one of the parties involved, such as choices A and E. Choice C may not directly cause harm to one of the species involved, but neither are the algae and turtle competing for anything; choice C can be eliminated. Eliminate choice D, because in mutualism both parties must benefit from the association, and while the algae may benefit, the turtle does not.
15 E Again, the human is being harmed by the tapeworm, so eliminate choices in which no harm is done (choices B and D). The human is not preyed on by the tapeworm (choice A is eliminated), and neither are the human and tapeworm competing (choice C is eliminated).
16 C For a population to prevent another population from using a resource, the first population must be a better competitor. Choice C is the only one that deals with competition.
17 E Maples are deciduous trees (they lose their leaves); this is the best tip-off for choice E. You should be able to eliminate choice A (tundra has only small shrubby plants) and choice D (taiga is coniferous forest) easily. Tropical rain forest (choice B) and temperate grasslands (choice C) do not have black bears.
18 B This is a typical genetics probability question with a twist. Make sure you read the questions carefully! You should know that color blindness is caused by an X-linked recessive allele. The only way the color blind man can produce a daughter is if he passes on his X chromosome; therefore, the daughter, even though normal, must be a carrier of color blindness. The daughter marries a normal man, meaning that he does not carry the allele for color blindness on his X chromosome. For them to produce a son, he must pass on his Y chromosome (so it doesn”t really matter what his X is like, anyway) and the probability of him passing on his Y is 50%. The final probability lies with the daughter and whether she passes on her normal X or her color blind X to her son; this is also a 50% probability. It”s very tempting to choose choice C; however, here”s the twist: The question asks for the probability that BOTH events (male child AND color blind) occur TOGETHER. The easiest way to solve this is to a the Punnett square.
The four possibilities in offspring are XXc (carrier daughter), XX (normal daughter), Xc Y (color blind son), and XY (normal son). The probability of producing a color blind son is 1 out of 4, or 25%.
19 B First of all, remember that this is a LEAST/EXCEPT/NOT question. Circle the word “EXCEPT” and draw a vertical line through the answer choices to remind yourself that you are looking for the “wrong” answer, in this case, the word that doesn”t fit with the description in the question. Basically, the process of photosynthesis involves using a green pigment called chlorophyll (choice C) to capture light energy (choice A). Water (choice D) is used to convert the light energy into chemical energy (ATP and NADPH). This chemical energy is used to convert carbon dioxide (choice E) into carbohydrates. Glucose (choice B) is an end product of photosynthesis. It isn”t needed in the course of the reactions.
20 D Again, make sure you read the question carefully and know exactly what you”re being asked. If black coat color in horses is the dominant phenotype, and both parental horses are black, then they must have at least one allele for black coat color. Because white coat color is the recessive phenotype, it can be caused only by a homozygous recessive genotype. The fact that the parental horses produced a white foal indicates that they must BOTH be heterozygous for coat color, each having an allele for white coat color as well as black. In the Punnett square below, B represents the dominant allele (causing black coat color) and b represents the recessive allele (causing white coat color). The probability of these horses producing a foal with a black coat is 75%, or 3 out of 4.
Note that the fact that we”re asked about the second foal produced doesn”t really matter. Each time the parental horses mate they have the same probability of producing black foals (75%) or white foals (25%). The coat color of previous offspring does not change the probability.
21 B You should immediately recognize ribose as being the sugar used in RNA nucleotides, thereby eliminating choice A. Because the question asks about a single ribose molecule, you can eliminate choice C as well. Nucleic acids are made up of many nucleotides, containing many ribose molecules. Choices D and E are very tempting because they have adenine in them, and the question asks about ribose bonded to adenine. However, ATP stands for adenosine triphosphate—which means it has three phosphates—and ADP stands for adenosine diphosphate—which means it has two phosphates. The question specifies that only a single phosphate is included. The best answer is choice B—a nucleotide.
22 D First, the disorder skips generations (the parents of affected individuals do not display the disorder) so it is recessive, and choices A and C can be eliminated. Furthermore, all affected individuals are male, indicating that the disorder is sex-linked, and choice B can be eliminated. Finally, because the disease does not pass from father to son, it must be X-linked, and choice E can be eliminated.
23 D Remember your I, II, III technique—eliminate answer choices as you determine options I, II, or III to be true or false. Let”s look at option I. Normal blood pressure in a resting human is about 120/80 mm Hg. The higher number (120 mm Hg) is the systolic blood pressure. Option I is false. 180 mm Hg is much too high a systolic blood pressure for a resting human. Knowing that I is false, we can eliminate choices A, C, and E. Of the remaining choices (B and D), both contain option II, so option II must be true. When you”re taking the real SAT II, don”t waste time thinking about option II; move right ahead to option III. For the purposes of this discussion, however, let”s take a look. Option II is a heart rate of 60–80 beats per minute. This is true for a resting human. Continuing with option III: Humans do maintain a body temperature of around 37oC at rest (it increases during exercise). Because option III is true, choice B can be eliminated, leaving D as the correct answer choice.
24 C Whenever you”re trying to determine the genotype of an organism that has a dominant phenotype, the best thing to do is something called a test cross. A test cross is a mating between the organism with the unknown genotype and an organism with the recessive phenotype. (Organisms that display the recessive phenotype MUST have the homozygous recessive genotype; the genotype of this organism is known.) Then, you just look at the offspring produced. If the organism with the unknown genotype were homozygous dominant, all the offspring would have the dominant phenotype. If, however, the organism with the unknown genotype were heterozygous, half the offspring would have the dominant phenotype and half would have the recessive phenotype. Look at the two possible Punnett squares:
This is the Punnett square that would result if the genotype of the unknown organism were homozygous dominant. Note that all offspring from this test cross display the dominant phenotype.
This is the Punnett square that would result if the genotype of the unknown organism were heterozygous. Note that half the offspring from this test cross display the dominant phenotype and half display the recessive phenotype.
Knowing this, let”s take a look at the answer choices. You should be able to eliminate choice D quickly—all chromosomes look physically alike, regardless of the alleles they carry. Choice B should also be eliminated relatively quickly—just because there are a few white hairs on the guinea pig doesn”t mean that it carries the recessive allele. There could be other causes for the occasional white hair. Choice E would give us some information, but not enough to determine if our unknown guinea pig were heterozygous or homozygous. In other words, if the guinea pig did have white siblings, it would tell us that the guinea pig”s parents were heterozygous (that”s the only way they could produce white offspring), and it tells us that the guinea pig has a chance of carrying the recessive allele for white hair, but it doesn”t tell us for sure that it is carrying the recessive allele. It comes down to a choice between answer choice A and answer choice C. Choice C describes a test cross and is the correct answer. The problem with choice A is that you”d be mating your unknown guinea pig with another unknown guinea pig—not a great way to figure something out. The idea is to mate your unknown with a known—then you know what to expect and can use those results to determine the unknown genotype. Note that you should be familiar with a test cross and what it can tell you—and now you are.
25 D This question is intimidating because it seems so complex. Remember to take the question apart piece by piece—don”t guess at an answer until you are sure what the question is asking! The best way to tackle this question is to draw a diagram of the interactions of the populations with one another. It might look something like this:
The question goes on to ask which population is most likely to evolve after a change—in other words, which population can best adapt to its new environment. You should know that evolution requires genetic variability, which is achieved through random mutation and through exchange of genetic material with other organisms (sexual reproduction). Clearly, the population that will be most likely to evolve is the one in which there is the greatest genetic variability. In this situation, that population is Population 5, because it interacts with the most other populations.
26 A Infertility can be caused by anything that prevents the meeting of sperm and egg, including things that prevent sperm or egg survival. Remember also that this is a I, II, III style question—eliminate answer choices as you work through the options. Option I is true. The egg must travel through the uterine tube toward the uterus, and the sperm must travel through the uterine tube toward the egg. If the tube is blocked, travel cannot occur and the sperm and egg would not meet. (You can now eliminate choices B and D.) Options II and III are false. Large amounts of FSH and LH are normally released prior to ovulation; the LH surge is, in fact, what triggers ovulation. These could not be reasons for infertility. (Eliminate choices C and E.)
27 B Note that the plasmid has EcoRI sites, a HindIII site, and a PstI site. Because only EcoRI and HindIII are used, only those sites will be cut; the PstI site will effectively be ignored. To “digest to completion” means that the plasmid will be cut completely at those restriction sites. Three fragments will be produced, each corresponding to the distance (in kb) between the EcoRI and HindIII sites.
28 C Remember your LEAST/EXCEPT/NOT technique. You should be familiar with the rules for blood typing in humans, but if you aren”t, this question gives you the basics. Because the alleles IA and IB are codominant, a man with blood type AB can only have the genotype IAIB. The woman with blood type A could have either IAIA or IAi as her genotype. Here are the Punnett squares:
This is the Punnett square that would result if the genotype of the type A woman was IA IA. Offspring would have either type A or type AB blood.
This is the Punnett square that would result if the genotype of the type A woman was IA i. Offspring would have type A, type B, or type AB blood.
Because type O blood is the recessive phenotype, it requires the homozygous recessive genotype (in this case, ii) in order to be displayed. This is the only blood type not possible in their children. A quick way to solve this question is to recognize that type O requires the homozygous recessive genotype, and for that to happen, each parent must donate a recessive i allele. Because the type AB man does not carry the recessive allele, this is not possible.
29 B Remember your LEAST/EXCEPT/NOT technique. Lizards, turtles, alligators, and snakes are all members of the class Reptilia. Frogs, however, are members of class Amphibia.
30 C This is a I, II, III–style question, so eliminate choices as you eliminate options. Option I is true. Eukaryotes have nuclei and prokaryotes do not. You can eliminate choices B and D. Option II is also true. Eukaryotes have mitochondria, and prokaryotes do not. You can eliminate choice A. Option III is false. Prokaryotes do have ribosomes; they need to synthesize proteins just as eukaryotes do. Eliminate choice E. The thing to remember about prokaryotes is that they have no membrane-bound organelles. Ribosomes are not membrane-bound and ARE found in prokaryotes.
31 E You need to know the pathway that blood takes through the heart and some basic heart anatomy (names of chambers and vessels). Beginning at the right atrium, the next chamber blood enters is the right ventricle, which on this diagram is labeled “6.” Because choices A and C do not start with “6” they can be eliminated. All of the remaining choices then contain structure 2 (the pulmonary artery), which is the next structure in sequence. From the pulmonary artery, blood flows to the lungs, then returns to the heart through the pulmonary veins, labeled “4” on the diagram. This eliminates choices B and D. Note that you don”t have to worry about the rest of the sequence, because you have already eliminated all other answer choices (although “5,” the left atrium, and “7,” the left ventricle, are the correct structures in sequence).
32 D Oxygen-poor blood returns to the right side of the heart through the superior and inferior vena cavae. The superior vena cava is labeled “3,” so choices A and B can be eliminated. The right side of the heart (which includes the right ventricle, “6”) sends oxygen-poor blood to the lungs through the pulmonary artery, structure “2” (choices C and E are eliminated). Structures 1 (the aorta), 5 (the left atrium), and 7 (the left ventricle), all carry blood rich in oxygen.
33 C As mentioned in Question 32 above, blood is returned to the heart through the super-ior and inferior vena cavae. The superior vena cava in this diagram is labeled “3.”
34 A You should immediately recognize this molecule as a protein.
35 D When a new amino acid is added to a protein, it is added at the carbon atom on the far right of the molecule. The OH group already attached to this carbon is lost as part of a water molecule.
36 B As described in Question 35 above, to attach two amino acids together, the OH group attached to the carbon of the first amino acid and the H attached to the nitrogen of the second amino acid leave as water.
37 D The area of least productivity in the ocean is the area that does not receive any sunlight, so photosynthesis cannot occur. The area without any sunlight is called the aphotic zone.
38 B The abyssal zone is characterized by extremely high water pressure, extreme cold, darkness, and very low nutrient density. Organisms that live here must be able to tolerate all of those conditions. Well-developed eyes would not be useful, because there is no light in this deepest part of the ocean. Remember, this is a LEAST/EXCEPT/NOT question. Circle the word “EXCEPT” and draw a vertical line to remind yourself to choose the “wrong” answer.
39 E The intertidal zone receives plenty of sunshine, periodically dries up (when the tide goes out), and has large fluctuations in temperature because of the changes in water depth. Organisms that live here may have characteristics that help them survive those conditions. Options I, II, and III are all true. Don”t forget to eliminate answer choices as you decide whether an option is true or false.
40 B Pineapple appears to have some effectiveness from pH 2 up to pH 12. None of the other fruit extracts operate over such a wide range.
41 A Pineapple extract”s maximum effectiveness is approximately 10. 50% of that is 5. The pH range where pineapple extract operates at at least 5 or better is pH 4 to pH 10.
42 D Pineapple is maximally effective at pH 7, which is a neutral pH. pHs from 0 to 7 are acidic, and pHs from 7 to 14 are basic, or alkaline.
43 D The graph shows relative effectiveness of the extracts when compared with distilled water. If an extract had a relative effectiveness of 1, that means it is equally as effective as the distilled water.
44 C Upon reading this experiment, you should note that the plant with orange flowers produced plants with red flowers, plants with orange flowers, and plants with yellow flowers. The blended phenotype (orange flowers) is a tip-off for incomplete dominance in flower color, and the parent plant, with orange flowers, must have both the red allele and the yellow allele (is heterozygous). The plant produced 25% red flowers, 50% orange flowers, and 25% yellow flowers. If the allele for yellow flower color were recessive, the heterozygous parent plant would have produced 75% orange flowers and 25% yellow flowers (choice A is eliminated). If it were dominant, the resulting offspring would be 75% yellow flowers and 25% orange flowers (choice B is eliminated). There is no such thing as “incompletely recessive” (choice D is eliminated), and codominance between red and yellow would not produce a blended phenotype, it would produce plants that had both red and yellow flowers, but no orange.
45 A The only way plants can have yellow flowers is if they are homozygous for yellow allele. If a plant is homozygous for a particular trait and is self-pollinated, it can only produce plants with that trait.
46 D We know that the plant must be heterozygous for flower color, because that”s the only way you can get orange flowers (the blended phenotype). Choices B and C can be eliminated. Furthermore, the plant must be homozygous for leaf arrangement because all progeny plants look like the parent plant. If the self-pollinated parent plant had been heterozygous for leaf arrangement, we would have expected to see different phenotypes in the offspring (choices A and E can be eliminated).
47 E This question really needs to be broken down a step at a time, and the I, II, III technique applied. First, let”s consider the genotype of the parent plant. If it has red flowers, its genotype for flower color must be homozygous (RR). Furthermore, a cross between a plant homozygous for alternating leaves (the original plant) and this plant (with nonalternating leaves) produces plants that all have non-alternating leaves. This tells us two things: First, the allele for non-alternating leaves must be dominant (otherwise all progeny plants would have alternating leaves), and second, the plant with non-alternating leaves must be homozygous (otherwise some progeny plants would have non-alternating leaves, and some would have alternating leaves). Let”s look at the Punnett squares. You can draw two separate Punnett squares, one for flower color and one for leaf arrangement:
This is the Punnett square for flower color.
This is the Punnett square for leaf arrangement. N represents non-alternating leaves, and n represents alternating leaves.
Now let”s look at option I. Looking at the Punnett square for flower color, we see that some of the offspring will have orange flowers, so option I is false. You can eliminate answer choices A and D. Option II is true (we discussed that above); you can eliminate choice C. Option III is also true. Looking again at the Punnett square for flower color, about half the offspring should have red flowers. Choice B is eliminated.
48 C The experiment started with groups of 20 termites each, and 50% of that is 10 termites. The dose that allows 10 termites (or 50% of the termite population) to survive is 25 PPM 2-butoxyethanol.
49 E This question is an excellent example of the temptation trap. The experiment description really doesn”t tell us anything about how this substance kills the termites, just that it does kill the termites. Choices A, B, and C may be true, but we don”t know for sure. The tempting choice is choice D, because that is a true statement. Expecting the graph to continue the pattern it shows, a dose of 40 PPM would leave only 4 termites alive, or 20% of the original population. However, even though it”s a true statement, it doesn”t answer the question. The best choice is choice E.
50 B A dose of 20 PPM kills 40% of the population. 40% × 2.5 = 100%, so 20 PPM × 2.5 = 50 PPM.
51 C Plant 1, the intact plant, grows at a rate of about 3 mm per day. Removal of the tip, as in Plant 2, reduces that growth to only about 1 mm per day. However, returning the auxin to the cut plant (as in Plant 4) restores the growth to about 3 mm per day.
52 D First, this is a I, II, III–style question, so you will be eliminating answer choices as you work through the options. Second, you should always be able to identify the control or controls in an experiment; this comes up very frequently on the SAT Biology E/M Subject Test. Option I is true. Plant 1 was left in its natural state. You can eliminate choices B and C. Option II is also true. Neither Plant 2 nor Plant 3 is treated with the test substance (auxin), but they are subjected to the other conditions the test plants are subjected to. You can eliminate choice A. Option III is false. Plant 5 is testing the effects of auxin when it is applied at a location other than the tip of the plant. You can eliminate choice D. NOTE: Don”t get careless and mix up your option numbers with your plant numbers! It”s an easy mistake to make.
53 B First go through the answer choices and eliminate obviously wrong ones, such as choices C and D. The data table given shows that Plant 5 grew more slowly than Plant 1, and that the growth of Plants 2 and 3 are about the same. To prove that auxin is necessary for plant growth, we have to show that when it is present plant growth occurs, and when it is absent, growth does not occur, or is much slower. Choice A is true, but it doesn”t show the necessity for auxin, you can eliminate it. Choice E is true, but again, doesn”t show the necessity for auxin, because auxin is present on both plants.
54 D Because light is not mentioned in the question at all, we can eliminate choices that have light as the reason for growth, such as choices A and E. We can also eliminate choice B, because movement toward the earth is not described. For a plant to bend to the right, cells on the left side of the plant must grow faster than cells on the right. For a plant to bend to the left, the opposite must occur; cells must grow faster on the right than on the left. In the question, auxin applied to half the plant causes the plant to bend in the opposite direction; for example, auxin on the left stimulates bending to the right. Bending to the right means growth on the left; therefore, auxin must stimulate growth on the same side of the plant to which it is attached.
55 B Radiolabeled amino acids would be incorporated into proteins.
56 E Remember that a control is subjected to everything the experimental group is, except for the item being tested. The experiment in this case is testing the effect of fertilization on protein synthesis. A control group should be exposed to everything the experimental group is except for the actual fertilization.
57 C Of the choices available, you should be able to eliminate choice E (endocytosis) first. There is no evidence to support this. As for the remaining choices, be careful not to come to conclusions too quickly based on your prior knowledge. In other words, you know that after fertilization occurs the zygote undergoes rapid cell division. This means that DNA will be replicated, RNA will be transcribed, proteins will be synthesized, and ultimately, organogenesis will occur. So in effect, choices A, B, C, and D are all occurring. The question, however, asks specifically for the process that is most likely occurring based on the results of the experiment. Because the experiment measures the amount of a radiolabeled amino acid present in the cells after fertilization, it is most directly measuring the rate of protein synthesis, making C a better choice than the others.
58 E The experiment description explains that the oocytes were placed into Frog Ringer”s solution to prevent changes in volume caused by to osmosis. The only way that osmosis can be prevented is to make sure that the solution the cells are suspended in is equally concentrated to the cells themselves.
59 A Remember that any type of behavior in newborn animals that depends on exposure to an object during a critical time period (such as right after hatching) is called imprinting.
60 A For imprinting to occur, the only critical thing is the time period of exposure. The object the chicks are exposed to doesn”t matter; whatever it is, they will attempt to imitate it. Because the chicks in this question were exposed to a pig during the critical period, they would exhibit piglike behavior.
61 D A sparrow is the only animal out of that group that is endothermic.
62 C Remember the I, II, III technique! Statement I is true; parasites (including viruses) harm their hosts (eliminate choices B and D). Not all parasites kill their hosts—you didn”t die from the last viral cold you had, did you? Statement II is false (eliminate choice E). Last, statement III is true; a parasite requires a host for something, in the case of a virus it is to reproduce (eliminate choice A).
63 B The easiest choice to eliminate is E; primary producers do not feed at any trophic level. They support all other trophic levels by using solar energy to produce carbohydrates from carbon dioxide. Choices C and D mean the same thing. Primary consumers eat the primary producers (essentially plants) and therefore are called herbivores. Because there cannot be two correct answer choices, both C and D can be eliminated. Of the two remaining choices, both could be correct. Carnivores as well as omnivores can feed at several trophic levels. However, we know for a fact that all omnivores feed at at least two trophic levels (they eat primary producers as well as primary or secondary consumers), making choice B better than choice A.
64 E The definition of fitness describes an organism that is successful at passing its genes on to the next generation. Because yeast reproduce by budding, the yeast that produce the most buds are the most successful at passing their genes on and are therefore the most fit.
65 C Mushrooms belong to kingdom Fungi, as do yeast; therefore, mushrooms and yeast will have the most in common. Mosses (choice A), ferns (choice B), and pines (choice D) are all members of kingdom Plantae. Seaweeds (choice E) are members of kingdom Protista.
66 B If 100,000 kcal are available at the first trophic level (the primary producers), then 10,000 kcal are available to the second trophic level, 1,000 kcal to the third trophic level, and 100 kcal to the fourth trophic level.
67 A The biosphere is the highest level of organization, so choices D and E can be eliminated. Populations represent a lower level of organization than ecosystems, so choice B can also be eliminated. Finally, a community is the next step up from populations, choice A is correct and C is wrong.
68 D Because the human hemoglobin and the gibbon hemoglobin differ by only two amino acids, these two organisms (of the ones listed) are the most closely related.
69 C The defining factor in considering two organisms as separate species is their inability to interbreed. If humans and gorillas cannot interbreed, then they are separate species, regardless of the similarity in their hemoglobin amino acid sequence.
70 D Looking at the diagram, we see that plants take up nitrogen from the soil or obtain it through nitrogen-fixing bacteria in their roots, and animals obtain nitrogen by eating plants. Choice A is true; animals and plants do take in nitrogen during respiration. However, choice A should be eliminated, because this nitrogen is in a form that is not usable by animals and plants. Choice B is true but doesn”t really answer the question and should be eliminated. Choice C is false; plants cannot take nitrogen directly from the atmosphere. It must first be fixed by symbiotic root bacteria or converted into ammonia. Choice E is true for plants but not for animals, and can therefore be eliminated.
71 E Bacteria anywhere are essentially decomposers. The only exceptions are the photosynthetic cyanobacteria.
72 A Structure #1 is a human arm and hand, which can be used for grasping.
73 D Divergent evolution produces structures that are similar in basic structure (look at the skeletal pattern in each of the limbs) but that may differ in function. Note that mutation (choice A) may have produced the changes that ultimately led to evolution, but choice D is a better, more accurate answer. Succession (choice B) describes the change from barren land to a stable climax community. Convergent evolution (choice C) produces organisms with similar functions but with different basic structures. Regression (choice E) makes no sense at all. It is not a term that describes anything to do with evolution; in fact, it means the opposite of evolution.
74 D The A. afarensis skull and the P. troglodytes skulls have the same forward-jutting jaw, the same enlarged brow ridge, and the same general skull shape.
75 B Having forward-facing eyes (choice A) was certainly an advantage as it increased visual depth perception. However, because all the skulls have forward-facing eyes, this does not represent a difference between ancient primates and modern humans. Loss of the canine teeth, loss of the brow ridge, and a reduction in jaw size (choices C, D, and E) are also apparent differences, but none is as helpful as the increased brain capacity.
76 E Remember this is a LEAST/EXCEPT/NOT question—you”re looking for the “wrong” answer. Choices A through D all describe possible reasons for the failure of embryo development. However, choice E is false, because four pairs (of the six pairs of bald eagles released) successfully nested and raised young.
77 A Because the higher trophic levels feed on the lower trophic levels, toxins tend to accumulate in the higher trophic levels by a process called biomagnification. Eagles are top carnivores and would exhibit a lot of toxin accumulation, more so than plants or insects. Choice B may be true but doesn”t answer the question—don”t fall into the temptation trap. Choice C is not mentioned in the passage, and the plants do take up PCB (choice D is eliminated). The PCB in the eggs came from the eagles themselves, not from absorption.
78 D Because the population is obviously growing bigger, choices C and E, which depict declining populations, can be eliminated. Take a look at the number of rabbits in each year—it”s growing exponentially (choice A can be eliminated) and hasn”t yet leveled off (choice B can be eliminated).
79 B If the resources are unlimited, the population can continue growing exponentially. In the first year, there were 4 rabbits. In the second year, there were approximately 4 × 4 (42), or 16 rabbits. In the third year, there were 4 × 4 × 4 (43) rabbits, about 64. In the fourth year, there were approximately 4 × 4 × 4 × 4 (44) rabbits, or about 256. Following this pattern, we can assume in the fifth year there would be approximately 4 × 4 × 4 × 4 × 4 (45), or 1024 rabbits. This number is closest to choice B.
80 D This is a I, II, III–style question, so remember to eliminate answer choices as you work through the options. If the nutrients and other resources are “limiting,” they will limit the population size and it will reach the “carrying capacity” of the environment. Option I is therefore true, and choice B can be eliminated. Option II directly contradicts option I; if option I is true, then option II must be false; choices C and E can be eliminated. Option III is also true. If nutrients and other resources are limited, the rabbits will compete with one another (intraspecific competition) for these limited nutrients; choice A can be eliminated. The stronger rabbits will survive, the weaker will die off, and the population will stabilize at its carrying capacity.
81 A You should know that animal cells are approximately as concentrated as (isotonic to) a 0.9% solute solution. If the cells are placed in such a solution, there will be no osmosis, so the cell will neither swell nor shrink.
82 E The centromere is the structure that holds the sister chromatids (the replicated DNA) together. As soon as the centromere splits (in anaphase), the sister chromatids (new chromosomes) are pulled to the opposite sides of the cell in preparation for cytokinesis.
83 D Convergent evolution is the evolution of two totally separate species along similar lines, so that they both produce features that have similar functions. It can result in everything described, but it cannot make the two different species into a single species. Remember, the species were originally very different, and their underlying structures are still different. They just share some common behaviors and features. (Remember also that this is a LEAST/EXCEPT/NOT question—you”re looking for the “wrong” answer.)
84 B All cell types have a plasma membrane. Chloroplasts (choice A) are found only in photosynthetic organisms, cell walls (choice C) are found in plants, bacteria, and fungi only, mitochondria (choice D) are found in all eukaryotic cells but not in prokaryotic cells, and flagella (choice E) is only found on certain cells to aid in their movement, such as sperm cells or paramecia.
85 B Remember, the key to fitness is not how many children you produce or how long you live, but how much you contribute to the next generation”s gene pool. Because the buck with smaller antlers sired three surviving offspring, and the buck with larger antlers sired only two surviving offspring, the buck with smaller antlers is more “fit” in the evolutionary sense.
86 B Evolution will always occur faster in an organism that has short generation times, simply because it is replicating its DNA more frequently and, thus, has a greater chance of mutations occuring.
87 E Know your terms for genetic variability in bacteria! Transduction is the transfer of DNA (and therefore of new traits, such as resistance to an antibiotic) through viral infection. Choice A, evolution, is true—the bacterial population has evolved. However, the question asks for the more specific answer. The bacteria are not a new species; they have simply acquired a new trait (choice B is eliminated). Conjugation (choice C) is the transfer of DNA between two different strains of bacteria and requires that the two different strains be mixed together at some point. This did not occur in the experiment. Transformation (choice D) is a situation in which bacteria take up naked DNA from the environment; again, this did not occur here.
88 D Organisms that use oxygen when it is available and survive by fermenting when oxygen is not available are classified as facultative anaerobes. You should have at least been able to eliminate choices A and B; the word obligate implies that they MUST be a certain way (e.g., always aerobic or always anaerobic), and these bacteria are able to switch.
89 B Remember the I, II, III technique! Again, you need to know the definitions of these terms. Conjugation is the transfer of DNA between two strains of bacteria and leads to genetic diversity. Therefore, option I is true, and choices C and D can be eliminated. Transformation means the bacteria acquire DNA from the environment; this also leads to genetic diversity. Because option II is also true, choice A can be eliminated. Crossing over is a phenomenon associated with meiosis, which bacteria do not undergo. Option III is false, eliminating choice E.
90 A Even without looking at Figure 2, the experiment description states that AAT is an enzyme, and enzymes are proteins. Figure 2 confirms this, because when an inhibitor of protein synthesis was added to the mix, no AAT was made.
91 B By looking at the two figures, you should be able to eliminate choices C, D, and E, because these processes are not monitored in the experiment. Because mRNA transcription is the first process to occur after addition of hormone, it seems likely that this is the process the hormone is stimulating.
92 E An inhibitor of protein synthesis will affect the production of new proteins only, not previously existing proteins. Because the enzyme needed to make the mRNA is already present, it is available to transcribe the AAT mRNA. Choices A and C are contradicted by the question, which states that mRNA production is run by an enzyme and that the enzyme is a protein. Choice B is true; however, the enzyme needed to make mRNA is a protein, and this is what the question is asking about. Don”t fall into the temptation trap—be sure you know what the question is asking before you choose an answer! Choice D is false—if this enzyme still needed to be translated, no mRNA production could occur and the figure shows that mRNA production is occurring.
93 B Radiolabeled uracil would be incorporated into mRNA and so would be detected as soon as mRNA were present—at approximately one minute after the addition of hormone.
94 B Choice A is false; clearly the bacteria are able to produce b-gal as shown in the figure. Choice C is also false. The bacteria cannot continue to use glucose because they were switched to a lactose-based medium. There is no reason to assume that the switch to lactose-based medium would induce DNA replication (eliminated choice D), and there has been no infection by a virus (eliminate choice E). It takes a little bit of time for the enzyme b-gal (which is a protein) to be produced, because the mRNA for this enzyme is not normally present. The mRNA is transcribed when the enzyme is needed (such as when the bacteria are in lactose), and the time it takes the mRNA to be transcribed is the delay seen in production of b-gal.
95 A Because humans consume lactose-containing foods irregularly, it makes sense for the bacteria to expend the energy only to make this enzyme when it is needed. None of the choices B, C, D, or E make any sense. Producing the enzyme b-gal would not prevent buildup of lactic acid. It”s not as though the lactose spontaneously converts to lactic acid and the enzyme would prevent this. b-gal cannot be harmful to bacteria (choice C). If it were, they might die. Over time, the ability to make b-gal would be lost, because the bacteria that had this ability would die out. The bacteria are still able to use glucose as a nutrient source (choice D). They were grown in glucose originally and were returned to glucose at the end of the experiment. Finally, lactose is a nutrient. It is not harmful to bacteria (choice E).
96 E An error in any of these processes could result in a nonfunctional protein. Option I, an error in DNA replication, could cause a mutation in the b-gal gene, leading to defective mRNA and a defective protein. Option II, an error in RNA transcription, could produce a defective mRNA, resulting in a defective protein. Option III, an error in protein synthesis, could definitely produce a defective protein. Remember the I, II, III technique.
97 C This is the definition of a species!
98 A Of the choices given, the only one the two groups have in common is that they are now reproductively isolated—in other words, they are no longer able to interbreed. Genetic drift (choice B) occurs when random events eliminate certain genes from a population; this has not occurred in either situation. Increased fitness (choice C) may have occurred with the moths in the first situation, but not with the frogs in the second. Geographic separation (choice D) occurred with the frogs, but not with the moths, and competition (choice E) is not mentioned in either situation.
99 C Divergent evolution often leads to speciation (C is correct, and B, the opposite, can be eliminated). Survival of the fittest (choice A) may have occurred but is not stated so specifically in the passage. Stabilizing selection (choice D) and directional selection (choice E) happen to a single population and usually do not separate it into two new species.
100 B One of the main issues that separates frogs from reptiles is that reptile eggs have a shell that prevents them from dehydrating and can therefore live away from a water source. Be careful of tempting, correct-sounding answer choices! The first part of choice A is true; frogs are amphibians, but that means they can live both in water and land. Choice C is true, but doesn”t address the issue of reproduction. Choice D is false; frogs lack the scaly skin of reptiles, which is why they need to be around water, and choice E is unlikely. Maybe over a long period of time the frogs could adapt, but switching from a partially aquatic to a completely terrestrial lifestyle would require many significant changes and would take a very long time.