SAT Biology E/M Subject Test
Part IV The Princeton Review Practice SAT Biology E/M Subject Tests and Explanations
Chapter 17 Practice SAT Biology E/M Subject Test 1: Answers and Explanations
Answers and Explanations
1 B Prokaryotes (Domains Bacteria and Archaea) have no membrane-bound organelles, so all reactions and processes occur in the cytoplasm. You should have been able to eliminate choices C, D, and E because they did not describe locations.
2 D When muscle cells run out of oxygen and switch to anaerobic metabolism (glycolysis only) to make ATP, the end product is lactic acid. Yeast can also switch to anaerobic metabolism; their end product is ethanol. You should have been able to eliminate choices A and B because they are not products.
3 B Eukaryotes possess organelles and as such divide the location of their cellular processes among them. Glycolysis occurs in the cytoplasm, whereas the Krebs cycle and electron transport occur in the mitochondria. As in Question 1, you should have eliminated choices C, D, and E.
4 E Recombination occurs when the homologous chromosomes are paired and crossing over can take place. This occurs during prophase I of meiosis. Remember that recombination occurs during prophase, and this would help you eliminate choices A, B, and D.
5 B After prophase I, the homologous chromosomes remain paired and align at the center of the cell, on the “metaphase plate.” (The prefix meta means “middle.” Use this fact to help you elimate choices A, C, and E.) During metaphase II, the individual, unpaired chromosomes align at the cell center.
6 A During meiosis, the chromosomes remain replicated (i.e., remain as two joined sister chromatids) for the entire first set of divisions. The whole point to the second set of meiotic divisions is to separate the sister chromatids. This takes place during anaphase II.
7 D Habituation involves becoming accustomed to certain stimuli that are not harmful or important. For example, if you walk down the hallway and a friend jumps out at you and you get scared, that is a normal reaction to a startling stimulus. However, if this happens every time you walk down the hallway, you get accustomed to it and no longer are startled. You have become habituated to the stimulus. Note that for Questions 7–9, you just had to know the basic definitions of these types of learning. Most of the classification-type questions are like that.
8 B Some animals do not have an instinctive sense for who their mother is and will bond with any object they are exposed to during a certain time period after their birth. The object “imprints” on their minds, and thereafter, even if exposed to their real mother, they will still treat the object as Mom.
9 C Conditioning involves the association of and response to one stimulus with a second, different stimulus. The best example is Ivan Pavlov”s dogs. He rang a bell when he fed them, and the dogs salivated in response to the food. Soon, all he had to do was ring the bell, and the dogs would salivate, even in the absence of food.
10 A Most digestion and absorption occur in the small intestine. A very small amount of digestion (starch only) takes place in the mouth, and a very small amount of digestion takes place in the stomach (acid hydrolysis of food and some protein digestion). As with Questions 7–9, Questions 10–12 require the same type of knowledge—memorization of basic facts.
11 E Saliva contains the enzyme amylase, which breaks down starch.
12 C Cells in the stomach secrete hydrochloric acid, which keeps the pH of the stomach around 1–2. The other regions of the digestive tract maintain a fairly neutral pH.
13 A Divergent evolution occurs when the same ancestral organism is placed into different environments and must then adapt to function in these different environments. Thus the same original structures evolve separately and “diverge” from one another. Examples of homologous structures are the arm of a man, the wing of a bat, and the flipper of a whale. All have the same basic bone structure but vastly different functions. The opposite of divergent evolution is convergent evolution, in which vastly different organisms are placed into the same environment and must adapt to perform similar functions with different structures. Convergent evolution produces analogous structures, examples of which are the wings of bats, the wings of birds, and the wings of butterflies. Speciation is often the result of divergent evolution, not the cause of it.
14 C The job of the blood is to carry oxygen from the lungs, where it is plentiful, to the tissues, where it is not. Thus hemoglobin should have a high affinity for oxygen in the lungs so it can bind oxygen (choices A, B, and D could be eliminated) and a low oxygen affinity in the tissues (so it can release the oxygen where it is needed). The reverse is true for carbon dioxide. Hemoglobin has a high carbon dioxide affinity in the tissues and a low carbon dioxide affinity in the lungs.
15 D In RNA, the base thymine (T) is replaced with uracil (U), so choice A can be immediately eliminated. Further, A will always pair with U, and G will always pair with C. The only choice that has the bases paired correctly is choice D.
16 B Choices A, C, D, and E all describe characteristics of the phylum arthropoda. A water vascular system is a characteristic of the phylum echinodermata, the “spiny skinned” animals such as sea stars and sea urchins. Their water vascular system ends in tube feet that play a role in locomotion and feeding. Don”t forget your LEAST/EXCEPT/NOT technique of circling the word “EXCEPT” and drawing a vertical line through the answer choices to help you remember to choose the incorrect statement.
17 E The kidney”s primary role is to filter blood to remove wastes (statement I is true, and choice D is eliminated), but it is also involved to a fair extent in blood pressure regulation (through renin and aldosterone, so II is true and choices A and C are eliminated) and in pH regulation (through excretion of hydrogen ions, so III is true and choice B is eliminated).
18 D If a pure-breeding long-tailed chicken (TT) mates with a pure-breeding short-tailed chicken (tt), all of their offspring (the F1 generation) will have the genotype Tt (and have long tails). Thus all of them, if mated with the correct genotype (Tt or tt), could produce offspring with short tails. Draw some quick Punnett squares to prove it to yourself.
19 B Density-dependent factors are those that get more significant as the size of the population increases. Limited nutrients and water, toxic waste build-up, and predation are all issues that are of greater concern to a large population than to a small one. Only choice B, climate temperature, is not more worrisome to a large group than to a small one. It will affect all populations equally, regardless of their size. Remember your LEAST/ EXCEPT/NOT technique.
20 C Two populations are considered separate species when they are so different from one another that they can no longer mate and produce viable offspring. Thus, organisms that can mate with each other must be of the same species.
21 E Blood that is poor in oxygen returns from the body to the right side of the heart (I is true, so choices B and C are eliminated), then travels through the pulmonary artery (III is true, so choices A and D are eliminated) to get to the lungs, where it picks up oxygen again. This oxygen-rich blood returns to the left side of the heart through the pulmonary vein (II is false) and is pumped back out to the body through the aorta.
22 B The characteristics described are those of the phylum annelida, the best example of which is the earthworm. Mollusks have external shells (snails), echinoderms and arthropods have exoskeletons (sea stars, crustaceans, insects), and chordates have endoskeletons and, in any case, are not worms.
23 A The light reactions of photosynthesis convert solar energy to usable energy in the form of ATP (choices C and E are eliminated) and NADPH (a reduced electron carrier). The ATP and NADPH (i.e., energy) produced during these reactions are used later during the Calvin cycle to fix carbon dioxide into carbohydrates, like glucose. Because glucose is a product of the light-independent reactions, choice B could be eliminated. Remember that NADPH belongs with photosynthesis to eliminate choice D.
24 C When the concentration of substrate far exceeds the concentration of enzyme (remember, the question states that enzyme concentration is assumed to be constant), all the enzyme active sites are saturated with substrate, and the product is being formed at its maximum rate. The only way to increase product formation at this point is to increase the concentration of the enzyme. Note that enzymes should not be used up in the course of the reaction (A is wrong). Furthermore, product formation is still occurring, just at a stable rate (B and D are wrong). There is no reason to assume an inhibitor has been added; the rate of product formation remains constant.
25 D Berries are plant products (i.e., primary producers), thus any organism that eats berries is a primary consumer, or an herbivore (I is true, choices B, C, and E can be eliminated). Notice at this point that the only remaining choices (A and D) do not contain option III, therefore, option III is false. Secondary consumers, carnivores and omnivores (e.g., birds) eat primary consumers (e.g., bugs), thus II is also true, and choice A can be eliminated. Tertiary consumers are carnivores (e.g., cats) that eat other carnivores (e.g., birds, secondary consumers). III is false, as we saw earlier.
26 B Monosaccharides have the general molecular formula CnH2nOn, as in glucose, which is C6H12O6. The only formula that fits this rule is choice B.
27 C A change in a population that occurs over a long period of time is evolution. This alone is a good tip-off that choice C, the only choice that mentions evolution, is correct. Speciation has not occurred, only a change in the characteristic of the birds, thus choices A and D can be eliminated. Any change in the fitness of the insects would change the characteristics of the insect population, not the bird population (B is wrong), and random mutation would not produce a specific, directed effect (E is wrong). Birds with large beaks had greater fitness because they could more easily obtain food, thus they had an advantage over birds with smaller beaks, which died out as time passed.
28 E The other choices do not describe substrate binding sites on an enzyme.
29 D Break this question down one piece at a time. First, the question states that there is a high sodium concentration outside the cell. This means sodium wants to move into the cell (where concentration is lower). Choices A and C can be eliminated because they state that sodium would move out of the cell. Active transport requires ATP, and because there is no ATP involved, choice E can be eliminated. Last, because sodium moves across the membrane with the help of a channel, it is moving by facilitated diffusion (choice B is eliminated).
30 E Decomposers take organic material and break it down into its individual compounds, thus returning these compounds back to the earth. The other processes listed are carried out by other organisms: nitrogen fixing bacteria (choice A), heterotrophs and autotrophs (choice B), and other soil bacteria (choices C and D).
31 B The endocrine system is a body control system, but it is NOT rapid. The fastest hormone in the body is adrenaline, and even that takes a few seconds, compared to the nervous system”s milliseconds. Most hormones operate in the minutes to hours range. The other choices regarding the endocrine system are all true. Remember your LEAST/EXCEPT/NOT technique!
32 B This symbiotic relationship describes commensalism. In mutualism both partners benefit, in parasitism and predator-prey relationships one partner benefits while the other is harmed, and symbiosis is just a general term used to describe close living arrangements.
33 D Relationships that are directly proportional have linear graphs with a positive slope.
34 B Male haploid cells (pollen grains or microspores) are produced on the anther (#2 in the diagram), which is at the tip of the filament (#3 in the diagram).
35 A Pollen grains stick to the stigma (#1 in the diagram), which is supported by the style (#4 in diagram).
36 E The pollen fertilizes the ovule (female haploid cells, or megaspores, #6 in the diagram); once fertilized, the ovary (#8 in the diagram) develops into a fruit.
37 C The prefix photo refers to light; because the plant is growing toward light we can eliminate choices A, D, and E. “Positive” means growing toward, so choice B can be eliminated.
38 A The prefix geo refers to the earth, or soil; thus we can eliminate choices C, D, and E. Because the plant is growing away from the earth, this is a negative tropism and we can eliminate choice B.
39 D A surge in LH is what causes ovulation and is measured in the ovulation prediction kits. Estrogen and progesterone affect the uterus, not the ovary, and FSH causes development of a follicle (A, B, and C are wrong). Testosterone is a male hormone (E is wrong).
40 B Because we know hormone X is peaking at ovulation time, a quick look at the graph shows hormone X peaking at about Day 14 of the cycle.
41 A The rise in hormone Y occurs after ovulation (choice E is wrong) and coincides with formation of the corpus luteum (choices B and D are wrong). The primary hormone secreted by the corpus luteum is progesterone.
42 E Rapid cell division after fertilization is known as cleavage. Blastulation is the formation of a hollow ball of cells (A is wrong), gastrulation is formation of the three primary germ layers (B is wrong), neurulation is development of the nervous system (C is wrong), and implantation is when the morula (solid ball of cells) burrows into the uterine lining (D is wrong).
43 C You should remember the prefixes ecto, meso, and endo for the primary germ layers and therefore eliminate choices D and E right away. The ectoderm forms the skin, hair, nails, mouth lining, and nervous system. The mesoderm forms muscle, bone, blood vessels, and organs (B is wrong). The endoderm forms inner linings and glands (A is wrong).
44 B The development of a thriving ecosystem from a barren area is known as succession. Note that evolution usually has a much longer time frame than succession.
45 B The climax community is the final, stable community in succession. The key word here is “stable.” That should tip you off that this is the end of the process, or the “ climax.”
46 D The first organisms to colonize a barren area are known as the pioneer organisms.
47 A Diuretics help eliminate water (i.e., increase urine production) from the body. From the data tables, the only substance that increases urine production significantly is caffeine. Don”t forget the I, II, III technique here; even just knowing that option I is true allows you to eliminate two (choices B and D) of the five choices.
48 A Again, from the data tables, there is a directly proportional (i.e., linear) relationship between the amount of caffeine ingested and the volume of urine produced. As caffeine consumption increases, so does urine volume. The only graph that shows this relationship is choice A.
49 C Because the caffeine and the sodium chloride were dissolved in water, plain water was consumed as a control, to make sure the effects seen were due to the added substances and not the water. Questions about experimental controls come up fairly frequently on the SAT Biology E/M Subject Test; make sure you know the definition for a control and how to spot it in the experiment.
50 C From Table 2, an increase in sodium chloride of 0.9 g results in a decrease in urine volume of approximately 40 ml. When 3.6 g sodium chloride are ingested, 82 ml urine is produced; thus if 4.5 g sodium chloride were to be ingested, the expected urine volume would be 40 ml less, approximately 40 ml.
51 B As soon as you see “Hemophilia,” you should be thinking “X-linked disorder.” Then use the technique for avoiding the temptation trap, which is particularly dangerous here, because the passage and the questions are confusing, and it”s very tempting to just guess blindly. Resist! Take the paragraph apart piece by piece, sentence by sentence. Write out genotypes as you read through, and construct Punnett squares to help you see probability. Out of this family, the only members that express this condition are males. This is a tip-off for X-linked disorders, which are more common in males because they have only a single X chromosome. (In any case, you should remember the two most common X-linked disorders: hemophilia and color blindness.) John”s genotype is XhY. He passed his Y chromosome to Mark and Mike; they also received a normal X from Jane, thus they do not have hemophilia, nor can they pass it on to their kids. Molly and Mary received Xh from John but also received a normal X from Jane, thus they are carriers of hemophilia but do not display its symptoms.
52 A Because Mike does not carry the gene for hemophilia (see above), he cannot pass it on to his children, and they in turn cannot pass it on to their children.
53 D Mark and Mike do not carry the gene for hemophilia (see solution to 51 above), thus we can eliminate choices A, B, and E. Jane is normal, so choice C is eliminated as well.
54 C From Table 1, substance B caused an increase in serum calcium levels, which is the effect that parathyroid hormone has on the body.
55 E Again, from Table 1, substance A caused a decrease in serum glucose levels, which is the effect that insulin has on the body.
56 B Substance D causes an increase in serum sodium, which is the effect aldosterone has on the body. Aldosterone is released when blood pressure is low, because excess sodium will have the effect of causing water retention, which will increase blood volume, which will increase blood pressure. (Note: Even if you did not know this, you should have been able to eliminate the other choices.)
57 E The change in serum sodium after injection of Substance B is insignificant. All other choices cause significant change from the baseline values of the variables being measured. Don”t forget the LEAST/EXCEPT/NOT technique.
58 C Cell Type A has no nucleus. The only organisms that do not have nuclei are bacteria (Domains Bacteria and Archaea).
59 E Cell Type C has a nucleus, a cell wall, and chloroplasts and therefore most likely comes from a plant. Equations II and III are the equations for photosynthesis and would occur in plants (choices A, B, and C are eliminated), and equation I is the equation for cellular respiration, which also occurs in plants (choice D is eliminated).
60 D This is a great question to do some answer predicting on. At Time 1 the oxygen was removed from the cultures and cell Type C died. Clearly it is an obligate aerobe. Thus we can eliminate choices A, B, and C. Because cell Type B was growing well in the presence of oxygen, it cannot be an obligate anaerobe, thus choice E is eliminated. Cell Type B must be a facultative anaerobe, using oxygen when it is available and fermenting when oxygen is not available. The decrease in growth of cell Type B after Time 1 is most likely because energy is produced during fermentation than during aerobic metabolism.
61 E These are the characteristics of desert.
62 B These are the characteristics of taiga.
63 C These are the characteristics of tropical rain forest.
64 A These are the characteristics of tundra.
65 D Flowering plants are angiosperms. Gymnosperms are conifers (naked seed plants; C is wrong), bryophytes are mosses (A is wrong), and tracheophytes are non-seed producing plants (ferns; B is wrong). “Endosperm” is not a classification for plants.
66 A If an organism”s environment remains absolutely constant, and that organism still exhibits regular rhythms of activity, there must be some internal “clock” that keeps it on schedule (C and E are wrong). Roosters vary the time of their crow as the sun varies the time it rises (B is wrong). The magnetic field has nothing to do with internal clocks (D is wrong).
67 D Remember: “Dumb King Philip Came Over From Germany—So?” or make up your own!
68 A Rattlesnakes are clearly heterotrophs (only photosynthetic organisms are autotrophs), so we can eliminate choices B and C. Rattlesnakes are carnivores, not producers (D is wrong) or primary consumers (herbivores, so E is wrong).
69 E Remember your LEAST/EXCEPT/NOT technique. There are four features present in all chordates—dorsal nerve cords (choice A), notochords (choice B), gill slits (choice C), and postanal tails (choice D). Note that some of these features may be found only in embryonic or larval stages. The “wrong” answer is choice E—not all chordates (for example, sea squirts and lancelets) have a bony endoskeleton.
70 D The question asks for the average beak length in cm, but the graph gives it in mm. Average beak length is 30 mm. 10 mm = 1 cm; therefore, 30 mm = 3 cm. Read the questions carefully!
71 B Clearly the birds with 30 mm beaks were not surviving too well. There is no reason to assume they flew to another island; remember, they are on a remote, isolated island. There may not be another island near enough to fly to (A is wrong). If predators consumed birds with 20 mm or 40 mm beaks, they would not be the prevalent populations in Figure 2 (C and D are wrong), and if birds with 30 mm beaks were selected for, the population would not have been divided (E is wrong).
72 B The defining characteristic for speciation is an inability to interbreed.
73 E Just because we have some information about how the population changed in the last 200 years, it doesn”t tell us how it may change in the next 200 years. It would depend on how the environment changed during that time period.
74 D As the acidity increases (pH goes down), the average mass of the fish decreases. However, it does not decrease linearly; rather, it stays constant for a while, then gradually drops off, then rapidly drops off as acidity becomes severe. The best representation of this is choice D.
75 D The plant benefits by easier availability of water and phosphorus, and the fungi benefit by receiving amino acids and sugars. Another term for this type of relationship is mutualism.
76 E pH 7.0 is neutral, thus neither the plant nor the fungi would be harmed.
77 B The best way to prevent damage from acid rain would be to prevent its formation by reducing the burning of the fossil fuels that cause it. There is no guarantee that bigger fish will resist the acidity any better than smaller fish (A is wrong), supplying plants with excess phosphorus will not help them take it up any easier (C is wrong), supplying fungi with sugars and amino acids will not help them overcome the effects of acid soil (D is wrong), and, even if alkalines will neutralize acid, they might be just as harmful to the environment (E is wrong)!
78 C Fungi are eukaryotes (A and E are eliminated), and they are not photosynthetic (B and D are eliminated).
79 B Clearly P. aurelia can compete better and get more food that P. caudata; thus it will grow while P. caudata is competed to extinction. Choice A is highly unlikely, because the food source the paramecia prefer is bacteria, not each other. This is not a symbiotic relationship but a competitive one (C and D are eliminated), and the data contradict choice E.
80 D Paramecia are single-celled eukaryotes, members of kingdom Protista.
81 B RNA bases do not include thymine; they are adenine, guanine, cytosine, and uracil. All other statements about RNA are correct. Remember the LEAST/EXCEPT/NOT technique!
82 A Ribosomes synthesize protein (B is wrong), mitochondria function in respiration (C is wrong), vacuoles help store waste (D is wrong), and lysosomes function in digestion (E is wrong).
83 D Choices A and E are prokaryotic and can be eliminated. Chordates have no cell walls (B is wrong), and plants have chloroplasts (C is wrong).
84 A The defining characteristic for speciation is an inability to interbreed. Choices B, C, D, and E could all ultimately produce two different populations that lack the ability to mate. Choice A would not lead to an inability to mate. Remember this is a LEAST/EXCEPT/NOT question.
85 B The ratio of purines to pyrimidines should be constant because purines always bind with pyrimidines, no matter which ones they may be.
86 B The members that have the most in common with one another are the members near the bottom of the hierarchy. Of the choices given, genus is the closest to the bottom of the hierarchy.
87 D Anyone who has produced offspring has demonstrated their fitness. Regardless of how healthy a child is, he has not yet produced offspring to prove his fitness. Use the LEAST/EXCEPT/NOT technique.
88 C The only plate that Colony 3 cannot grow on is Plate C, which lacks proline. Thus Colony 3 requires proline to grow and is a proline auxotroph (pro-). This eliminates choices A, B, and D. Choice E is incorrect because Colony 3 does not require arginine or leucine to grow; it can grow just fine in the absence of these amino acids, as is indicated on plates A and B.
89 D Colony 1 can grow on any of the plates, thus it does not require any additional amino acids. Auxotrophs require additional supplements to their growth media.
90 B Because these are bacterial colonies, they would not have any membrane-bound organelles, and thus no nuclei (choices A, C, and E can be eliminated) or mitochondria (choice D can be eliminated). Bacteria do have ribosomes to synthesize proteins.
91 A It really helps to predict an answer BEFORE you look at the answer choices. Sometimes looking at the choices first can confuse your thinking and lead you to a trap, but if you have an idea of the correct answer before you look at the choices, you will be less tempted. Colony 4 cannot grow in the absence of arginine as is evidenced by Plate B. Thus, because the liquid medium does not contain arginine, no bacterial growth would be observed.
92 B Again, try to predict an answer first. Clear spots on lawns of bacteria are due to infection by viruses that cause lysis of the bacteria. Even if this was not obvious to you, you should have been able to eliminate the other choices. There is no reason to assume Colony 2 would prey on Colony 1 (A is wrong); strong acid would lyse and destroy the bacteria immediately, not after 24 hours (C is wrong); bacteria are not delicate, they can grow just about anywhere, under any conditions (D is wrong); and there is no data to support the fact that threonine may be toxic to the bacteria or that the “unknown organism” was producing it.
93 D Structures with similar functions but different underlying structures are the result of vastly different organisms being placed into similar environments and having to adapt to the same stresses with different starting materials. These are termed “analogous structures” and are the result of convergent evolution. (See also Question #13.)
94 B Because the percentage of black moths is increasing, they must be selected for. This eliminates C and E. Because the percentage of white moths is decreasing, they must be selected against, eliminating choice A. Choice D is wrong because white moths would not blend better against dark tree bark.
95 D The original parent had white bark. The change in bark color is the result of an accumulation of soot on the tree. This is an acquired characteristic and would not be passed on to offspring. Seedlings that grew far from the plant would not be exposed to soot in the air and would not experience discoloration of their bark.
96 C The reason the percentage of black moths increased was because they were no longer visible against the now darkened bark of the trees. Because the white moths were more easily seen by the birds, their numbers declined. However, because bats rely on sonar to locate prey instead of vision, darker coloring would not give the moths any advantage, and the population percentages would stay at the point they were at when the birds were replaced by bats, a fifty-fifty split.
97 C Sucrose cannot cross the dialysis membrane, so it cannot cause any effect on the mass of the tubes. This eliminates choices A, B, D, and E.
98 D Because movement across the membrane relies strictly on concentration gradients, the fact that there is no gradient in Tube 1 would prevent the movement of water into or out of the tube. Thus there would be no change in mass.
99 B The gradient in Tube 3 is much larger than the gradient in Tube 2, thus water would be expected to enter more rapidly. This is confirmed by the data in Table 1. A linear graph should show two lines with positive slopes, and the slope of Tube 3”s line should be greater than the slope of Tube 2”s line.
100 C The fact that the cell does not swell or shrivel in 0.9% NaCl implies that there is no concentration gradient. A cell in a 20% NaCl solution would experience similar stresses to a dialysis tube filled with water sitting in a beaker filled with a more concentrated solution, such as Tube 4. The data indicate that Tube 4 lost mass, thus water exited the tube, and the same would happen to the red blood cell.