Biology E/M Subject Test 1 PDF

BIOLOGY E/M SUBJECT TEST 1 Your responses to the Biology E/M Subject Test 2 questions must be filled in on the Test 2 part of your answer sheet (at the back of the...

BIOLOGY E/M SUBJECT TEST 1 Your responses to the Biology E/M Subject Test 2 questions must be filled in on the Test 2 part of your answer sheet (at the back of the book). Marks on any other section will not be counted toward your Biology E/M Subject Test score. When your supervisor gives the signal, turn the page and begin the Biology E/M Subject Test. There are 100 numbered ovals on the answer sheet. There are 60 questions in the core Biology test, 20 questions in the Biology-E section, and 20 questions in the Biology-M section. Therefore, use ONLY ovals 1-80 (for Biology-E) OR ovals 1-60 plus 81-100 (for Biology-M) for recording your answers. Cracking_SAT_Bio_21_Ch 16_1314_111612_DAS.indd 396 11/16/2012 3:53:40 PM  BIOLOGY E/M SUBJECT TEST 1—Continued FOR BOTH BIOLOGY-E AND BIOLOGY-M, ANSWER QUESTIONS 1-60 Directions: Each set of lettered choices below refers to the numbered statements immediately following it. Select the one lettered choice that best answers each question or best fits each statement, and then fill in the corresponding oval on the answer sheet. A choice may be used once, more than once, or not at all in each set. Questions 1-3 Questions 7-9 (A) Mitochondria (A) Reasoning/insight (B) Cytoplasm (B) Imprinting (C) Pyruvate (C) Classical conditioning (D) Lactic acid (D) Habituation (E) Glucose (E) Instinct 1. Location of cellular respiration in prokaryotes 7. A simple kind of learning involving loss of sensitivity to unimportant stimuli 2. End product of anaerobic metabolism in muscle cells 8. Geese recognize a ticking clock as “mother” if exposed to it during a critical period shortly after hatching 3. Location of glycolysis in eukaryotes Questions 4-6 9. Fish are given food at the same time as a tap on their glass bowl and soon learn to approach when (A) Anaphase II a tap sounds even in the absence of food (B) Metaphase I (C) Prophase II (D) Metaphase II Questions 10-12 (E) Prophase I (A) Small intestine (B) Large intestine (C) Stomach 4. Stage of meiosis during which recombination of (D) Esophagus genetic material occurs (E) Mouth 5. Stage of meiosis during which pairs of homologous 10. Structure where most digestion and absorption of chromosomes align at the center of the cell nutrients occurs 6. Stage of meiosis during which sister chromatids 11. Structure where starch digestion first takes place are separated 12. Structure with the lowest pH Practice SAT Biology E/M Subject Test 1 | 397 Cracking_SAT_Bio_21_Ch 16_1314_111612_DAS.indd 397 11/16/2012 3:53:41 PM  BIOLOGY E/M SUBJECT TEST 1—Continued Directions: Each of the questions or incomplete statements below is followed by five suggested answers or completions. Some questions pertain to a set that refers to a laboratory or experimental situation. For each question, select the one choice that is the best answer to the question and then fill in the corresponding oval on the answer sheet. 13. Homologous structures, which have similar 16. Arthropods can be characterized by all of the fol- underlying structures but may have different lowing EXCEPT functions, are formed by (A) a hard exoskeleton (A) divergent evolution (B) a water vascular system (B) speciation (C) jointed appendages (C) segregation (D) molting (D) convergent evolution (E) segmented body (E) stabilizing selection 17. Which of the following are functions of the 14. Hemoglobin is a protein in red blood cells that kidney? binds and carries oxygen and some carbon dioxide. I. filtration of blood to remove wastes Its affinity for oxygen changes as blood travels from the lungs to the body tissues and back to the lungs II. blood pressure regulation again. One could expect hemoglobin to have III.     pH regulation (A) a high carbon dioxide affinity in the lungs and a low oxygen affinity in the tissues (A) I only (B) a low carbon dioxide affinity in the lungs and (B) I and II only a high oxygen affinity in the tissues (C) I and III only (C) a high oxygen affinity in the lungs and a low (D) II and III only oxygen affinity in the tissues (E) I, II, and III (D) a low oxygen affinity in the lungs and a high oxygen affinity in the tissues 18. In chickens, the allele for long tail feathers (T) is (E) a high oxygen affinity in the lungs and a high dominant over the allele for short tail feathers (t). carbon dioxide affinity in the lungs If a pure-breeding long-tailed chicken (TT) mates with a pure-breeding short-tailed chicken (tt), 15. Which of the following RNA sequences would be what percentage of their offspring (if mated with transcribed from the DNA sequence the correct genotype) could give rise to chickens ATGCCTAGGAC? with short tails? (A) TACGGATCCTG (A) 25% (B) UAGCGAUCCUG (B) 50% (C) AUGCCUAGGAC (C) 75% (D) UACGGAUCCUG (D) 100% (E) GCAUUCGAAGU (E) unable to determine from the information given 398 | 3Cracking 98 the SAT Biology E/M Subject Test Cracking_SAT_Bio_21_Ch 16_1314_111612_DAS.indd 398 11/16/2012 3:53:41 PM  BIOLOGY E/M SUBJECT TEST 1—Continued 19. All of the following could be considered density- 23. Which of the following substances are produced dependent factors affecting population growth by the light reactions of photosynthesis? EXCEPT (A) ATP and NADPH (A) limited nutrients (B) ATP and glucose (B) climate temperature (C) NADH and glucose (C) build-up of toxins (D) ATP and NADH (D) predation (E) NADPH and glucose (E) limited water 24. Consider the following graph of substrate 20. The best definition of a species is concentration vs. product formation. Assume (A) a group of organisms that occupy the same enzyme concentration to be constant. Why does niche the graph level off at high substrate (B) a population that works together to defend concentrations? itself from predators (C) a group of organisms that can mate with each other (D) a population that preys on other populations (E) a population in which all members benefit from the association in some way product formation 21. Which of the following contains blood poor in oxygen? I. Right ventricle II. Pulmonary vein III. Pulmonary artery substrate concentration (A) I only (B) II only (C) III only (A) All the enzyme is used up, and product (D) I and II only formation cannot occur without it. (E) I and III only (B) There is no more substrate to be converted into product. (C) Substrate concentration, exceeds enzyme 22. An organism appears to be a segmented worm. concentration, and all active sites are Upon observation it is determined that the organism saturated. has a closed circulation, a mouth and an anus, and (D) The reaction has run to completion. does NOT have an exoskeleton. The organism (E) An inhibitor has been added, and it most likely belongs to the phylum has slowed down the rate of product (A) mollusca formation. (B) annelida (C) echinodermata (D) arthropoda (E) chordata Practice SAT Biology E/M Subject Test 1 | 399 Cracking_SAT_Bio_21_Ch 16_1314_111612_DAS.indd 399 11/16/2012 3:53:41 PM  BIOLOGY E/M SUBJECT TEST 1—Continued 25. A bird that feeds on both insects and berries 29. Human cells maintain concentration gradients would be classified as a across their plasma membranes, such that there I. primary consumer is a high sodium concentration outside the cell and a high potassium concentration inside the II. secondary consumer cell. Suppose that within the cell membrane are III. tertiary consumer sodium “leak” channels. These channels would allow sodium to (A) I only (A) move out of the cell by simple diffusion (B) II only (B) move into the cell by simple diffusion (C) III only (C) move out of the cell by facilitated diffusion (D) I and II only (D) move into the cell by facilitated diffusion (E) II and III only (E) move into the cell by active transport 26. Which of the following chemical formulas could 30. The role of decomposers in the nitrogen cycle is to represent a monosaccharide? (A) fix atmospheric nitrogen into ammonia (A) C6H6O6 (B) incorporate nitrogen into amino acids and (B) C3H6O3 organic compounds (C) C6H12O3 (C) convert ammonia to nitrate, which can then (D) C5H10O10 be absorbed by plants (E) CH2O4 (D) denitrify ammonia, thus returning nitrogen to the atmosphere 27. A population of birds lives in an area with many (E) release ammonia from organic compounds, insects upon which it feeds. The insects live thus returning it to the soil inside trees, burrowing into the bark. Over many hundreds of years, the average beak size in the 31. All of the following are true about the endocrine bird population has increased. This is due to system EXCEPT (A) increased fitness of the birds, leading to (A) it relies on chemical messengers that travel speciation through the bloodstream (B) decreased fitness of the insects, allowing the (B) it is a control system that has extremely rapid birds to catch them more easily effects on the body (C) increased fitness of large-beaked birds, (C) the hormones affect only certain “target” leading to evolution organs (D) decreased fitness of small-beaked birds, (D) it is involved in maintaining body leading to speciation homeostasis (E) random mutation and genetic recombination (E) its organs secrete hormones directly into the bloodstream, rather than through ducts 28. The location on an enzyme where a substrate binds is called the (A) binding site (B) reaction center (C) allosteric site (D) lock-and-key model (E) active site 400 | 4Cracking 00 the SAT Biology E/M Subject Test Cracking_SAT_Bio_21_Ch 16_1314_111612_DAS.indd 400 11/16/2012 3:53:42 PM  BIOLOGY E/M SUBJECT TEST 1—Continued 32. Two organisms live in close association with each Questions 34-36 refer to the following diagram. other. One organism is helped by the association, whereas the other is neither helped nor harmed. 1 Which of the following terms best describes this 2 relationship? (A) Mutualism 5 (B) Commensalism 3 (C) Symbiosis 6 (D) Parasitism 4 (E) Predator-prey relationship 7 8 33. Cardiac output (the amount of blood pumped out 9 of the heart in one minute) and blood pressure are directly proportional. Which of the following graphs best depicts the relationship between cardiac output and blood pressure? 34. Location where male haploid cells are produced (A) 1 (B) 2 (A) (B) (C) 3 cardiac cardiac (D) 6 output output (E) 8 blood pressure blood pressure 35. Sticky structure where pollen grains can attach and germinate (C) (D) (A) 1 (B) 2 cardiac cardiac output output (C) 4 (D) 6 (E) 8 blood pressure blood pressure 36. Structure which, when fertilized, develops into (E) fruit cardiac (A) 1 output (B) 2 (C) 5 blood pressure (D) 6 (E) 8 Practice SAT Biology E/M Subject Test 1 | 401 Cracking_SAT_Bio_21_Ch 16_1314_111612_DAS.indd 401 11/16/2012 3:53:42 PM  BIOLOGY E/M SUBJECT TEST 1—Continued Questions 37-38 39. The hormone labeled X in the diagram is often used in over-the-counter diagnostic tests Tropisms refer to movements made by plants toward to determine when ovulation has occurred. This or away from certain stimuli. “Positive” tropisms refer hormone is specifically to movements toward a stimulus, whereas (A) estrogen “negative” tropisms refer to movements made away (B) progesterone from a stimulus. (C) FSH (D) LH 37. A plant growing on the shady side of a building (E) testosterone bends around the corner of the building toward the sunlight. This is an example of 40. Based on the peak levels of hormone X, on what (A) negative geotropism day of the cycle is ovulation most likely to occur? (B) negative phototropism (A) Day 21 (C) positive phototropism (B) Day 14 (D) positive hydrotropism (C) Day 12 (E) negative hydrotropism (D) Day 25 (E) Day 28 38. The stem and leaves of the plant grow up, away from the soil. This is an example of 41. The hormone labeled Y in the diagram is (A) negative geotropism (A) progesterone, secreted by the corpus luteum (B) positive geotropism after ovulation has occurred (C) negative phototropism (B) progesterone, secreted by the ovary after (D) positive hydrotropism ovulation has occurred (E) negative hydrotropism (C) estrogen, secreted by the corpus luteum after ovulation has occurred Questions 39-43 refer to the following diagram. (D) estrogen, secreted by the ovary after ovulation has occurred (E) estrogen, secreted by the follicle before ovulation occurs se ha e as ep n ph tio iv y ua t ra or str ife et 42. Immediately after fertilization, the zygote begins en cr ol Se M Pr to undergo rapid cell division. This process is Uterine lining X known as Y (A) blastulation Hormone levels (B) gastrulation (C) neurulation (D) implantation 7 14 21 28 (E) cleavage Day of cycle 402 | 4Cracking 02 the SAT Biology E/M Subject Test Cracking_SAT_Bio_21_Ch 16_1314_111612_DAS.indd 402 11/16/2012 3:53:43 PM  BIOLOGY E/M SUBJECT TEST 1—Continued 43. From which of the primary germ layers does the Questions 47-50 refer to the following experiment. nervous system develop? (A) Endoderm Diuretics are substances that help eliminate water (B) Mesoderm from the body. The effects of various substances were (C) Ectoderm tested on several volunteers. All volunteers had a mass (D) Enteroderm of 70 kg. They drank nothing for eight hours before (E) Epidermis the test and urinated just prior to ingesting the test substance. The three substances (water, caffeine, and Questions 44-46 salt) were tested on three separate days. The results are shown in the tables below. A barren, rocky community near a lake has virtually no vegetation or animal life. After a period of approximately 75 years, the community boasts a wide Table 1 variety of flora and fauna, including deciduous trees, am o u n t caf f e i n e i n g e s te d v o l u m e u ri n e deer, and raccoon. v o l u n te e r (i n 100 m l w ate r) co l l e cte d af te r 1 h o u r A 50 m g 302 ml 44. The process which has taken place can best be B 100 m g 492 ml described as C 150 m g 667 ml (A) progression D 200 m g 863 ml (B) succession (C) evolution (D) habitation (E) colonization Table 2 am o u n t s o d i u m ch l o ri d e i n g e s te d v o l u m e u ri n e v o l u n te e r 45. The stable community of deciduous trees and (i n 100 m l w ate r) co l l e cte d af te r 1 h o u r animals is known as the A.9 g 201 ml (A) final community B 1. 8 g 162 ml (B) climax community C 2.7 g 125 ml (C) apex community (D) summit community D 3.6 g 82 ml (E) composite community Table 3 46. Usually the first organisms to colonize rocky areas are lichen. These are known as the v o l u m e u ri n e v o l u n te e r v o l u m e w ate r i n g e s te d (A) primary community co l l e cte d af te r 1 h o u r (B) starter community A 100 m l 230 m l (C) colony organisms B 200 m l 240 m l (D) pioneer organisms (E) settler organisms C 300 m l 252 m l D 400 m l 263 m l Practice SAT Biology E/M Subject Test 1 | 403 Cracking_SAT_Bio_21_Ch 16_1314_111612_DAS.indd 403 11/16/2012 3:53:43 PM  BIOLOGY E/M SUBJECT TEST 1—Continued 47. Which of the following substances could be 50. Based on the results in Table 2, if a volunteer classified as a diuretic? were to ingest 4.5 g sodium chloride dissolved I. Caffeine in 100 ml water, what would be the approximate predicted urine volume collected after one hour? II.     Sodium (A) 20 ml III.   Water (B) 30 ml (A) I only (C) 40 ml (B) II only (D) 50 ml (C) I and II only (E) 60 ml (D) II and III only (E) I, II, and III Questions 51-53 refer to the following information on heredity. 48. Which graph best represents the change in urine Hemophilia is a disorder in which blood fails to clot. volume when ingesting caffeine? John, a male hemophiliac, marries Jane, a normal woman, and together they have four children, two boys (Mark and Mike) and two girls (Molly and (A) (B) Mary). None of the children display the symptoms of hemophilia. Mark, Mike, Molly, and Mary all marry mg caffeine mg caffeine normal individuals and have children. None of Mark’s or Mike’s children, male or female, display symptoms of hemophilia, but the sons of Molly and Mary all urine volume urine volume display symptoms of hemophilia while the daughters of Molly and Mary do not. (C) (D) mg caffeine mg caffeine 51. Which of the following best explains the reason that Mark, Mike, Molly, and Mary do not display symptoms of hemophilia, even though urine volume urine volume their father, John, is a hemophiliac? (A) Hemophilia is an X-linked disorder, and John (E) can pass on only his Y chromosome. mg caffeine (B) Hemophilia is an X-linked disorder and even though Molly and Mary received a hemophiliac X chromosome from John, urine volume Jane gave them a normal X chromosome. (C) Hemophilia is a Y-linked disorder and therefore cannot be displayed in females. 49. The purpose of ingesting the plain water (D) Hemophilia is a Y-linked disorder and (Table 3) was to Mark and Mike must have received an X (A) rehydrate the volunteers chromosome from John. (B) dissolve the substances (E) Hemophilia is an X-linked disorder, and (C) act as a control even though Mark and Mike received a (D) flush out the kidneys hemophiliac X chromosome from John, (E) act as a positive test substance Jane gave them a normal X chromosome. 404 | 4Cracking 04 the SAT Biology E/M Subject Test Cracking_SAT_Bio_21_Ch 16_1314_111612_DAS.indd 404 11/16/2012 3:53:44 PM  BIOLOGY E/M SUBJECT TEST 1—Continued 52. If one of Mike’s daughters marries a normal man, 55. Based on the information in Table 1, which of the what is the probability that one of their children following is most likely substance A ? will display symptoms of hemophilia? (A) Glucagon (A) 0% (B) Aldosterone (B) 25% (C) Calcitonin (C) 50% (D) Parathyroid hormone (D) 75% (E) Insulin (E) 100% 56. Under what conditions might substance D be 53. Which of the following individuals are heterozygous released normally? for hemophilia? (A) Soon after a meal (A) John, Mark, and Mike (B) When blood pressure is low (B) Mark, Mike, Molly, and Mary (C) Between meals (C) John and Jane (D) When there has been limited intake of (D) Molly and Mary dietary calcium (E) Mark and Mike (E) When dietary calcium is in excess Questions 54-57 57. All of the following changes in variable values A volunteer was injected intravenously with several are significant EXCEPT test substances to determine the effect of each (A) the change in serum glucose when substance substance on normal body variables. The results are A is injected shown in Table 1. Assume that enough time was (B) the change in serum Na+ when substance D allowed between injections so that the substances do is injected not interfere with one another. (C) the change in serum Ca++ when substance B is injected Table 1 (D) the change in serum glucose when substance C is injected v al u e s af te r v al u e s af te r v al u e s af te r v al u e s af te r (E) the change in serum Na+ when substance B is b as e l i n e v ari ab l e v al u e s i n je cti n g s u b s tan ce A i n je cti n g s u b s tan ce B i n je cti n g s u b s tan ce C i n je cti n g s u b s tan ce D injected s e ru m Ca++ 2.3 m m o l/ L 2.3 m m o l/ L 3.0 m m o l/ L 2.3 m m o l/ L 2.3 m m o l/ L s e ru m N a+ 135 m m o l/ L 135 m m o l/ L 136 m m o l/ L 135 m m o l/ L 147 m m o l/ L s e ru m 5.6 m m o l/ L 3.3 m m o l/ L 5.6 m m o l/ L 7.4 m m o l/ L 5.6 m m o l/ L g lu co s e 54. Based on the information in Table 1, which of the following is most likely substance B ? (A) Calcitonin (B) Insulin (C) Parathyroid hormone (D) Glucagon (E) Aldosterone Practice SAT Biology E/M Subject Test 1 | 405 Cracking_SAT_Bio_21_Ch 16_1314_111612_DAS.indd 405 11/16/2012 3:53:44 PM  BIOLOGY E/M SUBJECT TEST 1—Continued Questions 58-60 59. Which of the following equations is cell Type C Three different cell types were observed under the able to run? microscope. The observations are summarized in I. C6H12O6 + 6 O2 → 6 CO2 + 6 H2O + ATP Table 1. II.     H2O + light → O2 + ATP + NADPH Table 1 III.   6 CO2 + 6 H2O + ATP + NADPH → C6H12O6 Ce l l ty p e N u cl e u s ? Ce l l w al l? Ch l o ro p l asts? (A) I only A No Yes No (B) II only (C) I and III only B Ye s Ye s No (D) II and III only C Yes Ye s Yes (E) I, II, and III The three cell types were grown in separate cultures 60. Consider Figure 1. Which of the following with plenty of oxygen and nutrients available. Figure statements best describes the reason for the 1 shows their rates of growth. At Time 1, oxygen was difference between the curves for cell Type B and no longer available to the cells. cell Type C ? (A) Cell Type B is unable to survive in the Figure 1 presence of oxygen, while cell Type C can ferment. (B) The products of fermentation in cell Type C are toxic to the cells and they are dying. Cell Type A (C) Cell Type B is an obligate aerobe while cell Cell Type B Type C is able to ferment. Size of cell (D) Cell Type B is a facultative anaerobe, while population cell Type C is an obligate aerobe. (E) Cell Type C is an obligate aerobe, while cell Type B is an obligate anaerobe. Cell Type C Time 1 Time (hours) 58. Based on the information in Table 1, which of the following is the most likely classification of cell Type A ? (A) Fungi (B) Plant (C) Bacteria (D) Animal (E) Protist If you are taking the Biology-E test, continue with questions 61-80. If you are taking the Biology-M test, go to question 81 now. 406 | 4Cracking 06 the SAT Biology E/M Subject Test Cracking_SAT_Bio_21_Ch 16_1314_111612_DAS.indd 406 11/16/2012 3:53:45 PM  BIOLOGY-E TEST Directions: Each of the questions or incomplete statements below is followed by five suggested answers or completions. Some questions pertain to a set that refers to a laboratory or experimental situation. For each question, select the one choice that is the best answer to the question and then fill in the corresponding oval on the answer sheet. Questions 61-64 66. Which of the following indicates that animals (A) Tundra have internal biological clocks? (B) Taiga (A) A mouse kept in constant darkness shows a (C) Tropical rain forest daily rhythm of activity. (D) Deciduous forest (B) A rooster crows whenever the sun rises in (E) Desert both winter and summer. (C) An owl kept in constant light drifts away 61. The driest of all terrestrial biomes, characterized from a 24-hour cycle. by low and unpredictable precipitation (D) Some species of birds can sense fluctuations in the Earth’s magnetic field. (E) A squirrel whose night and day are 62. Coniferous forests, characterized by long, cold artificially reversed soon adapts to its new winters and short, wet summers schedule. 67. Which of the following correctly lists the 63. Biome characterized by great diversity of flora phylogenic hierarchy? and fauna and high levels of precipitation (A) Domain, kingdom, phylum, family, class, order, genus, species 64. Northern areas, characterized by permafrost, (B) Phylum, family, order, domain, class, extremely cold temperatures, and few trees kingdom, species, genus (C) Kingdom, domain, family, order, class, phylum, genus, species 65. Plants that have true roots, stems, and leaves, as (D) Domain, kingdom, phylum, class, order, well as flowers and seeds enclosed in fruit, are family, genus, species classified as (E) Family, kingdom, order, domain, phylum, (A) bryophytes genus, class, species (B) tracheophytes (C) gymnosperms 68. A rattlesnake would be classified as a (D) angiosperms (E) endosperms (A) tertiary consumer and a heterotroph (B) secondary consumer and an autotroph (C) producer and an autotroph (D) producer and a heterotroph (E) primary consumer and a heterotroph 69. At some point in their development, chordates possess all of the following EXCEPT (A) a dorsal hollow nerve cord (B) a notochord (C) gill slits (D) postanal tail (E) an exoskeleton Practice SAT Biology E/M Subject Test 1 | 407 Cracking_SAT_Bio_21_Ch 16_1314_111612_DAS.indd 407 11/16/2012 3:53:45 PM  BIOLOGY-E TEST 1—Continued Questions 70-73 71. What is the most likely reason for the difference in distribution of beak lengths between the data A population of birds (Population A) on a remote, plotted in Figure 1 and the data plotted in isolated island is studied to determine beak length. Figure 2 ? The resulting data are plotted in Figure 1. (A) All birds with beaks of 30 mm flew to a new island over the 200-year time span. Figure 1 (B) Birds with beaks of 30 mm were selected against. (C) Predators consumed birds with beaks of 40 mm. (D) Predators consumed birds with beaks of 20 mm. Number (E) Birds with beaks of 30 mm were selected of birds (Population A) for extinction. 72. Suppose that a researcher studying Population B found that birds with beaks of 20 mm were 10 20 30 40 50 unable to mate with birds that had 40 mm beaks. These two groups of birds would now be Beak length (mm) classified as (A) occupying different niches Suppose that 200 years later, the beaks of the birds on (B) separate species the island were again measured (Population B). The (C) competitors data, when plotted, yielded a graph as in Figure 2. (D) predators (E) separate populations Figure 2 73. How would beak length in the bird population change after another 200-year time span? (A) The average beak length would return to 30 mm. (B) The average beak length would shift to Number of birds 40 mm. (Population B) (C) The average beak length would shift to 20 mm. (D) The differences in beak length would be more pronounced. (E) It is not possible to determine how beak 10 20 30 40 50 length might change. Beak length (mm) 70. What is the average beak length (in cm) of the birds in Figure 1 ? (A) 30 cm (B) 15 cm (C) 5 cm (D) 3 cm (E) 1 cm 408 | 4Cracking 08 the SAT Biology E/M Subject Test Cracking_SAT_Bio_21_Ch 16_1314_111612_DAS.indd 408 11/16/2012 3:53:46 PM  BIOLOGY-E TEST 1—Continued Questions 74-78 75. The relationship between mycorrhizal fungi and plants can best be described as one in which Acid rain is formed after the burning of fossil fuels (A) one partner benefits from the association and releases compounds containing nitrogen and sulfur the other partner is harmed into the atmosphere. Sunlight and rain bring about (B) one partner benefits from the association and chemical reactions that convert these compounds into the other partner is neither harmed nor nitric acid and sulfur dioxide, which combine with helped water droplets to form acid rain. Acid rain typically (C) one partner preys upon the other partner has a pH of approximately 5.5. (D) both partners benefit from the association (E) neither partner benefits from the association The higher acidity of soil and water affects many living organisms adversely. As the pH of lake water falls, fish become ill and die. Table 1 shows the effects 76. If the pH of the soil were 7.0, what would be the of pH on the size of adult fish. effect on the mycorrhizal fungi and plants? (A) The fungi would survive but the plant would Table 1 be harmed. (B) The fungi would be harmed but the plant p H o f l ak e A v e rag e l e n g th o f f i s h (cm ) A v e rag e m as s o f f i s h (g ) would survive. 7.5 30 cm 454 g (C) The fungi would be slightly harmed and the plant would be slightly harmed. 7.0 28 cm 450 g (D) Neither the fungi nor the plant would 6.5 29 cm 453 g survive. 6.0 25 cm 401 g (E) Neither the fungi nor the plant would be harmed. 5.5 20 cm 288 g 5.0 17 cm 127 g 77. What might be the best strategy to prevent 4.5 a ll fis h d e a d a ll fis h d e a d ecological damage due to acid rain? (A) Stock the lakes with bigger fish so that they can resist the effects of the acid better Mycorrhizal fungi, which form a mutualistic association (B) Reduce the amount of fossil fuels that are with many plant roots, are particularly sensitive to the burned effects of acid rain. These fungi facilitate the absorp- (C) Supply plants with excess phosphorus and tion of water and nutrients by the plants; in turn, the water plants provide sugars and amino acids without which (D) Supply fungi with excess sugars and amino the fungi could not survive. acids (E) Add alkalines to soil and water to neutralize the acid 74. The effect of acid rain on fish size is best 78. Fungi are classified as represented by which of the following graphs? (A) prokaryotic decomposers (B) eukaryotic producers (A) (B) (C) (C) eukaryotic decomposers mass mass mass (D) eukaryotic autotrophs of fish of fish of fish (E) prokaryotic consumers increasing increasing increasing acidity acidity acidity (D) (E) mass mass of fish of fish increasing increasing acidity acidity Practice SAT Biology E/M Subject Test 1 | 409 Cracking_SAT_Bio_21_Ch 16_1314_111612_DAS.indd 409 11/16/2012 3:53:46 PM  BIOLOGY-E TEST 1—Continued Questions 79-80 80. Paramecia are members of the kingdom (A) fungi The following graphs show the growth of two closely (B) animalia related species of paramecia, both when grown alone (C) archaea (Figure 1) and when grown together (Figure 2). Both (D) protista species consume bacteria as their food source and (E) plantae reproduce by binary fission as often as several times a day. P. aurelia P. caudata P. aurelia Size of Size of population population P. caudata Days in culture Days in culture Figure 1 Figure 2 79. The data in Figure 2 indicate that (A) P. aurelia is preying on P. caudata (B) P. aurelia is a better competitor than P. caudata (C) P. aurelia and P. caudata are in a symbiotic relationship (D) P. aurelia is a parasite of P. caudata (E) P. aurelia grew better when combined with P. caudata than it did when grown alone STOP IF YOU FINISH BEFORE TIME IS CALLED, YOU MAY CHECK YOUR WORK ON THE ENTIRE BIOLOGY-E TEST ONLY. DO NOT TURN TO ANY OTHER TEST IN THIS BOOK. 410 | 4Cracking 10 the SAT Biology E/M Subject Test Cracking_SAT_Bio_21_Ch 16_1314_111612_DAS.indd 410 11/16/2012 3:53:47 PM  BIOLOGY-M TEST If you are taking the Biology-M test, continue with questions 81-100. Be sure to start this section of the test by filling in oval 81 on your answer sheet. Directions: Each of the questions or incomplete statements below is followed by five suggested answers or completions. Some questions pertain to a set that refers to a laboratory or experimental situation. For each question, select the one choice that is the best answer to the question and then fill in the corresponding oval on the answer sheet. 81. All of the following are true about RNA EXCEPT 85. The base composition of DNA varies from one (A) it is single-stranded species to another. Which of the following ratios (B) its bases are adenine, thymine, guanine, and would you expect to remain constant in the DNA? uracil (A) Cytosine : Adenine (C) it has a sugar-phosphate backbone (B) Pyrimidine : Purine (D) its sugar is ribose (C) Adenine : Guanine (E) it is found in the both the nucleus and the (D) Guanine : Deoxyribose cytoplasm of the cell (E) Thymine : Guanine 82. The function of the Golgi apparatus is to 86. Which of the following groups have the most in (A) package and store proteins for secretion common with one another? (B) synthesize proteins (A) Members of the same kingdom (C) function in cellular respiration (B) Members of the same genus (D) help the cell expel waste (C) Members of the same phylum (E) digest foreign substances (D) Members of the same class (E) Members of the same family 83. A eukaryotic cell that has a cell wall but lacks chloroplasts would be classified as a 87. Which of the following individuals is the LEAST (A) bacteria fit in evolutionary terms? (B) chordate (A) A 45-year-old male with a terminal disease (C) plant who has fathered three children (D) fungus (B) A 20-year-old man who has fathered one (E) bacteria child (C) A 35-year-old woman with four children 84. All of the following could give rise to new (D) A healthy 4-year-old child species EXCEPT (E) A 25-year-old woman with one child, who has had a tubal ligation to prevent future (A) variations in antler size between male and pregnancies female reindeer (B) an earthquake that physically separates a population of lizards into two separate groups (C) divergent evolution (D) evolution of a population of cats such that they can no longer mate with their ancestors (E) a massive flood that separates a population of frogs onto opposite sides of a large lake Practice SAT Biology E/M Subject Test 1 | 4 11 Cracking_SAT_Bio_21_Ch 16_1314_111612_DAS.indd 411 11/16/2012 3:53:47 PM  BIOLOGY-M TEST 1—Continued Questions 88-92 88. Referring to Figure 1, what is the genotype of Colony 3 ? Most bacteria can be grown in the laboratory on agar (A) arg+, leu+, pro+ plates containing glucose as their only carbon source. (B) arg+, leu-, pro+ Some bacteria require additional substances, such as (C) arg+, leu+, pro- amino acids, to be added to the growth medium. Such (D) arg-, leu-, pro+ bacteria are termed auxotrophs. These bacteria are (E) arg-, leu-, pro- denoted by the amino acid they require followed by a “-” in superscript (e.g., arg-). Bacteria that do not 89. Is Colony 1 an auxotroph? require that particular amino acid can be indicated by a “+” in superscript. (A) Yes, it is able to grow in the presence of the three amino acids being tested. (B) Yes, it can only grow if glucose is present. Different strains of bacteria were grown on several (C) No, it is able to grow in the absence of plates containing a variety of nutrients. Figure 1 glucose. shows the colonies (numbered) that grew on each (D) No, it is able to grow in the absence of any plate. The supplements in each plate are indicated. additional amino acids. Figure 1 (E) The data available are insufficient to determine the answer. 1 1 1 90. Which structures could be observed in a sample of Colony 2? 2 2 3 3 I. Nuclei 4 4 II. Ribosomes III. Mitochondria Plate A: Glucose Plate B: Glucose Plate C: Glucose Arginine Leucine Arginine Proline Proline Leucine (A) I only (B) II only (C) I, and III (D) II and III only In a second experiment, Colony 1 was mixed with soft (E) I, II, and III agar and spread over a plate so that an even lawn of bacteria grew. Bacterial lawns appear cloudy on agar plates. A single drop of an unknown organism was placed in the center of the bacterial lawn, and after 24 hours, a clear area known as a “plaque” appeared at that spot. The clear area continued to expand at a slow rate. Although new colonies could be grown from samples taken from the lawn, attempts to grow new colonies from samples taken from the plaque area were unsuccessful. 412 | 4Cracking 12 the SAT Biology E/M Subject Test Cracking_SAT_Bio_21_Ch 16_1314_111612_DAS.indd 412 11/16/2012 3:53:48 PM  BIOLOGY-M TEST 1—Continued 91. If a liquid culture medium containing glucose, 92. What is the most likely reason for the clearing leucine, and proline was inoculated with Colony (the plaque) in the lawn of bacteria in the second 4, would bacterial growth be observed? experiment? (A) No, Colony 4 is an arginine auxotroph (arg-). (A) The unknown organism is bacterial Colony (B) No, Colony 4 cannot grow in the presence of 2, and these bacteria are eating the leucine. bacteria from Colony 1 forming the lawn. (C) Yes, Colony 4’s genotype is leu-, pro-. (B) The unknown organism is a virus that is (D) Yes, Colony 4 requires only glucose to grow. infecting the bacteria and causing them to (E) The data available are insufficient to make a lyse (killing them). prediction. (C) The drop placed in the center of the lawn contained a strong acid that destroyed the bacteria at that spot. (D) Bacteria are very delicate and the disturbance caused them to die. (E) The unknown organism began producing threonine, which is toxic to Colony 1. Practice SAT Biology E/M Subject Test 1 | 413 Cracking_SAT_Bio_21_Ch 16_1314_111612_DAS.indd 413 11/16/2012 3:53:48 PM  BIOLOGY-M TEST 1—Continued 94. What is the most likely explanation for the shift Questions 93-96 in the percentage of black moths in the population? In 1910, a small town on the East Coast of the United States relied primarily on agriculture to support its (A) The white moths no longer blended with economy. In the mid-1930s, a steel mill was built, the color of the tree bark and, thus, were and the economy shifted from being agriculturally selected for. supported to being industrially supported. The steel (B) The black moths blended better with the mill released a lot of smog and soot into the air, which color of the tree bark and, thus, were collected on the bark of trees in a wooded area near the selected for. outskirts of town. Over a period of ten years the bark (C) The black moths blended better with the gradually darkened, then maintained a constant dark color of the tree bark and, thus, were color. selected against. (D) The white moths blended better with the A variety of animals and insects lived in the wooded color of the tree bark and, thus, were area. In particular, a certain species of moth served as selected against. the primary food source for a population of birds. The (E) The black moths did not blend with the color moths lay their eggs in the bark of the trees and, thus, of the tree bark and, thus, were selected must spend a fair amount of time sitting on the tree against. trunks. Table 1 presents data on the moth population. 95. If a seed from one of the trees was planted in an area far from the steel mill, what color would the Table 1 bark of the tree be? (A) Black, because the parent tree had black bark Year % white moths % black moths (B) White, because the gene causing black 1910 95 5 bark was mutated due to environmental pollution 1920 95 5 (C) Black, because the gene causing white 1930 95 5 bark was mutated due to environmental pollution 1940 50 50 (D) White, because the black bark was an 1950 20 80 acquired characteristic and is therefore not 1960 5 95 passed on to progeny (E) The color of the bark is not able to be determined. 93. The wings of the moths and the wings of the 96. Birds track their prey visually, whereas bats rely birds are both used for flight (similar functions); on sonar to locate their food. If the bird population however, their underlying structures are very were replaced with a bat population in 1940, different. Moth wings and bird wings are thus what would be the ratio of white moths to black classified as moths? (A) homologous structures (A) 95% white, 5% black (B) autologous structures (B) 80% white, 20% black (C) divergent structures (C) 50% white, 50% black (D) analogous structures (D) 20% white, 80% black (E) emergent structures (E) 5% white, 90% black 414 | 4Cracking 14 the SAT Biology E/M Subject Test Cracking_SAT_Bio_21_Ch 16_1314_111612_DAS.indd 414 11/16/2012 3:53:48 PM  BIOLOGY E/M SUBJECT TEST 1—Continued Questions 97-100 98. Why does the mass of Tube 1 remain relatively unchanged throughout the experiment? Dialysis tubing is a semipermeable membrane. It (A) The dialysis tubing in Tube 1 is defective and allows small molecules, such as water, to pass through does not allow water to cross. easily, while larger molecules, such as sucrose, are (B) There is no concentration gradient to drive restricted. Movement of molecules across the tubing the movement of sucrose. is due to concentration gradients. In an experiment (C) The dialysis tubing broke, allowing the tube designed to study osmosis, several pieces of dialysis contents to mix with the beaker contents. tubing were filled with sucrose solutions of varying (D) There is no concentration gradient to drive concentration and placed in beakers containing the movement of water. distilled water. The rate and direction of water (E) The experimenter failed to record the data movement was determined by weighing the bags properly. before and after placing them in the distilled water. The data are recorded below. 99. Which of the following graphs best illustrates the Table 1 relationship between Tube 2 and Tube 3 ? Tu b e co n te n ts Tu b e M as s (g ) M as s (g ) M as s (g ) M as s (g ) M as s (g ) Tube 2 Tube 3 Tube 3 (b e ak e r (A) (B) (C) n u m b er 0 m i n u te s 15 m i n u te s 30 m i n u te s 45 m i n u te s 60 m i n u te s co n te n ts ) D is t ille d Tube 3 Tube 2 w a t er mass mass mass Tube 2 1 (d is t ille d 22.3 g 22.4 g 22.2 g 22.3 g 22.3 g w a t er ) 10% s u cr o s e time time time 2 (d is t ille d 24.8 g 25.3 g 25.7 g 26.4 g 26.9 g w a t er ) (D) (E) Tube 2 40% s u cr o s e Tube 2 3 (d is t ille d 25.1 g 26.3 g 27.5 g 28.9 g 29.6 g w a t er ) Tube 3 mass mass D is t ille d Tube 3 4 w a t er 22.7 g 21.3 g 20.5 g 19.8 g 18.7 g (40% s u cr o s e ) time time 100. Cell membranes are also semipermeable, 97. Why does the mass of Tube 3 increase while the allowing water but not other substances to cross mass of Tube 4 decreases? easily. A red blood cell placed in a 0.9% NaCl (A) Water is moving into Tube 3, and sucrose is solution will neither swell nor shrivel. Based on moving into Tube 4. this knowledge, and the information presented in (B) Water is moving into Tube 4, and sucrose is Table 1, what would happen to a red blood cell moving into Tube 3. placed in a 20% NaCl solution? (C) Water is moving into Tube 3, and water is (A) Water would be drawn out of the cell and the moving out of Tube 4. cell would swell. (D) Sucrose is moving into Tube 3, and sucrose (B) Water would be drawn into the cell and the is moving out of Tube 4. cell would swell. (E) Sucrose is moving out of Tube 3, and water (C) Water would be drawn out of the cell and the is moving out of Tube 4. cell would shrivel. (D) Water would be drawn into the cell and the cell would shrivel. (E) No change would occur to the cell.