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MCAT

MEDICAL COLLEGE ADMISSION TEST

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MCAT P RACTICE T EST 5R S OLUTIONS Edited, produced, typeset, and illustrated by Steven A. Leduc National Director of MCAT Research, Production & Development, The Princeton Review

Special thanks to: Jennifer Wooddell Judene Wright

Copyright © 2003, 2001 by Princeton Review, Inc. All rights reserved. MCAT is a service mark of the Association of American Medical Colleges (AAMC). TPR is not affiliated with Princeton University or with the AAMC. Version 1.0

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MCAT P RACTICE T EST 5R S OLUTIONS C ONTENTS : Physical Sciences ................. 3 Verbal Reasoning .................. 14

Biological Sciences ............ 29

030316

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PHYSICAL SCIENCES Passage I 1. B. The value of DH∞ for Reaction 1 is given as –92 kJ. Since DH∞ is negative, the reaction is exothermic, by definition. 2. C. According to the data in Table 1, the equilibrium concentration of NH3(g) (% by volume) increases as the pressure increases. This eliminates choices A and B, which show the yield decreasing as the pressure increases. To decide between choices C and D, we first notice that when the pressure increases from 1 atm to 100 atm—which is an increase of 99 atm— the % NH3(g) by volume at equilibrium jumps significantly, from 15.3 to 80.6. However, when the pressure increases from 100 atm to 200 atm—an increase of 100 atm—the % NH3(g) by volume only increases from 80.6 to 85.8. This indicates that the % yield increases sharply with pressure initially, but then increases less abruptly at high pressures. This behavior is illustrated by the graph in choice C. 3. A. The passage states that Reaction 1 is carried out “in the presence of” FeO/Al2K2O4. Furthermore, we notice from Reaction 1 that neither FeO nor Al2K2O4 is consumed in the reaction. We conclude that this mixture is a catalyst for the reaction, and the role of a catalyst is to increase the reaction rate. 4. D. Since NH4+ is a cation with a +1 charge, and SO42– is an anion with a –2 charge, we would expect that a combination of these ions would contain 2 NH4+ ions for each SO42– ion; that is, the compound would be (NH4)2SO4. The balanced acid–base reaction would be 2 NH3 + H2SO4 Æ (NH4)2SO4. 5. D. Choice A is eliminated since the product of Reaction 1, NH3, is not an ion. Because NH3 has a permanent dipole moment, we would expect dipole–dipole interactions between NH3 molecules. In addition, NH3 is capable of hydrogen bonding (the partial positive charge on an H atom of one molecule of NH3 attracted to the partial negative charge on the N atom of another NH3 molecule). 6. A. Of the ions listed, Mg2+ and H+ are cations, and being electron deficient, are acids (eliminating choices C and D). Since nitrogen is less electronegative than oxygen and since N3– has a greater negative charge than –OH, N3– is a stronger base than –OH.

Passage II 7. D. According to the passage, “there is a large electrical repulsion between these two fragments that causes them to . . . gain kinetic energy.” 8. A. According to Newton’s Third law, the magnitude of the force by Fragment 1 on Fragment 2 is equal to the magnitude of the force by Fragment 2 on Fragment 1, so choice A must be correct. Since the forces the fragments feel have the same magnitude, but the fragments have different masses, the accelerations of the fragments must be different; since a = F/m, the fragment with the greater mass experiences a smaller acceleration, eliminating choice B. Because both fragments start from rest and have different accelerations, the speeds of the fragments at any moment t will be different (because v = at); this eliminates choice C. To eliminate choice D, notice that KE = 12 mv 2 = 12 m (at ) 2 = 12 m

2

( mF t) 2 = 2Fm t 2

Since F is the same for both fragments but m is different, the fragments will have different KE values at each moment t.

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9. C. First, eliminate choice B. If all of the heavier elements were stable, then there would be naturally-occurring elements that have more protons in their nucleus than uranium does. As for choices A and D, even if we grant that they are true, they simply beg the question and do not provide an answer. Why have all of the heavier elements radioactively decayed? Why is it that heavier elements can be made only in nuclear reactors? According to the passage, the strong nuclear force is a shortrange attractive force that balances the large repulsive force between the positive charges in the nucleus. Once the nucleus get too large (that is, once the nucleus contains too many protons), we conclude that the short-range strong nuclear force becomes unable to hold the nucleus together, which is why such large nuclei do not occur naturally (they’ve spontaneous decayed). The statement in choice C provides a reasonable and direct answer as to why the strong nuclear force becomes unable to hold large nuclei together. 10. D. According to Coulomb’s law, the force between the two charged fragments, +Q and +Q, is given by the equation F = kQQ/r2 = kQ2/r2, where r is their separation. Since F is inversely proportional to r2, the graph of F vs. r must decrease nonlinearly, which is shown in the graph in choice D. 11. A. Neither choice B nor C is applicable here, since there is no mention of electrons in the fusion (or fission) of nuclei. And according to the passage, the strong nuclear force is an attractive force between the charges in a nucleus, so to fuse two nuclei together, there is no need to “overcome” an attractive force, eliminating choice D. The answer must be A. Since both nuclei are positively charged, energy must be provided to force them together and overcome their electrical repulsion. 12. C. Since each fragment (of charge +Q) experiences a decreasing force as it moves away from the other fragment (because, according to Coulomb’s law, F = kQQ/r2 = kQ2/r2, and r is increasing), each fragment will experience a decreasing acceleration (a = F/m = kQ2/mr2, where m is the mass of the fragment).

Passage III 13. D. The passage states that CFCs “can undergo photolysis in the upper atomsphere and subsequently assist in the decomposition of ozone. . . .” Therefore, we can conclude that if a compound were inert in the upper atmosphere, then it would not “significantly assist” in the depletion of ozone. 14. A. The chlorine atom in Reaction 4 is a radical, Cl•. By definition, a radical is an atom or molecule fragment with one or more unpaired electrons. 15. B. If we combine Reactions 4 and 5, crossing out the Cl• and ClO• radicals, the net reaction is given in choice B:

Reaction 4: Cl • + O 3 Æ ClO • + O 2

+ Reaction 5: ClO • + O Æ Cl • + O 2 Net reaction: O 3 + O Æ 2 O 2 16. C. First, we find the net reaction of Reactions 1 and 2: Reaction 1: O 2 Æ O + O

+ Reaction 2: O + O 2 Æ O 3 Net reaction: 2 O 2 Æ O + O 3

Now, to determine whether the overall reaction involves an increase or a decrease in free energy, we calculate DGrxn using the values for DGf given in Table 1:

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DGrxn = Â n ◊ DGf, products - Â n ◊ DGf, reactants = (1 ◊ DGf, O + 1 ◊ DGf, O 3 ) - (2 ◊ DGf, O 2 ) kJ = (230.1 mol + 163.4

kJ kJ mol ) - (2 ◊ 0 mol )

>0 Because DGrxn is positive, we can eliminate choices A and B. Choice D is eliminated since it indicates a negative activation energy (since it shows the activated complex at a lower energy level than the reactants). The answer must be C. 17. D. We calculate DS for the reaction 2 O3 Æ 3 O2 using the values for S given in Table 1: DSrxn = Â n ◊ Sproducts - Â n ◊ Sreactants = (3 ◊ SO 2 ) - (2 ◊ SO 3 ) = (3 ◊ 205.0 = 137.4

J J mol ◊ K ) - (2 ◊ 238.8 mol ◊ K )

J mol ◊ K

18. C. In Reactions 3–5, we see that the Cl• generated by the cleavage of a CFC (Reaction 3) causes the decomposition of O3 (Reaction 4). However, the Cl• is regenerated in Reaction 5, so that only a catalytic amount of CFC is needed to drive the formation of O2.

Passage IV 19. A. Apply Le Châtelier’s principle to Equation 2. “Excessively moist soil conditions” describe conditions where the amount of the H2O(l) is increased. Since H2O(l) is a reactant in Equation 2, we would thus expect that an increase in H2O(l) would shift the equilibrium toward the product side, causing a greater degree of ionization and releasing more OH–(aq). 20. D. Because the statements in choices A, B, and C are all false (since N2 accounts for more than 75% by volume of the atmosphere, the N2 molecule is nonpolar, and N2 is not a noble gas), the correct response must be D. 21. D. The species that results when an acid loses an H+ is called the conjugate base of that acid. When H2PO4– loses an H+, it becomes HPO42–. Therefore, HPO42– is the conjugate base of H2PO4–. 22. C. The equilibrium that would best account for an increase in pH would show the formation of OH– ions, so we eliminate choice A. The reactions in choices B and D are incorrect, since, for example, neither is balanced electrically nor stoichiometrically, so the best answer here is C. 23. A. Pure liquids are omitted from equilibrium expressions since their concentrations remain essentially constant (not that they’re zero), so we first eliminate choices B and D. Since Equation 2 shows the formation of OH–, we conclude that (NH4)2HPO4 is a basic salt, so A is a better response than C.

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Independent Questions 24. B. First, to balance the given equation, we need only place a coefficient of 2 in front of both the HCl and the NaCl: Na2CO3 + 2 HCl Æ CO2 + H2O + 2 NaCl Notice that this affects neither the Na2CO3 nor the CO2. Now, if we treat CO2 as an ideal gas, then 11.2 L of CO2 at STP is equivalent to 1/2 mole of CO2 (since, by Avogadro’s law, one mole of any ideal gas at STP occupies a volume of 22.4 L). According to the reaction above, we would need 1/2 mole of Na2CO3 to produce 1/2 mole of CO2. Since a 2 M solution of Na2CO3 contains 2 moles of Na2CO3 per liter, we would need 1/4 L = 250 mL of this solution to obtain 1/2 mole of Na2CO3. 25. C. The molecule MnO4– is not an exception to the rule that the oxidation number of oxygen is –2. So, if we let x denote the oxidation number of Mn in MnO4–, then x + 4(–2) = –1, so x = +7. Now, the oxidation number of Mn for the cation Mn2+ is clearly +2, so the oxidation number of Mn in MnO4– differs from its oxidation number in Mn2+ by (+7) – (+2) = 5. 26. B. First, if the waves strike the shore every 3.0 seconds, then the period of the waves is T = 3.0 seconds. Next, if the horizontal distance between adjacent crests and troughs is 1.0 m, then the wavelength is twice as much, l = 2.0 m. We now use the equation v = lf. Since f = 1/T, we have v=

l 2.0 m = = 0.67 m s T 3.0 s

27. B. The circumference of the circular path is C = 2pr = 2p(4 cm) = 8p cm. Since the particle complete 4 revolutions (or cycles) per second—that is what “moving on a circular path . . . with a frequency of 4 Hz” means—the particle travels a total distance of 4 ¥ (8p cm) = 32p cm in one second. Therefore, to travel half this distance, 16p cm, would require half the time: that is, 1/2 second.

Passage V 28. C. The buffered solution at the beginning of Experiment 1 contains 16 mmol of CDP in 1 L of aqueous solution. Since 10 mL = 1/100 L, we conclude that 10 mL of the solution contains (16/100) mmol of CDP. To find the mass of CDP in 10 mL of this solution, we multiply (16/100) mmol of CDP by its molecular mass: 16 m = ( 100 ¥ 10 -3 mol CDP) ¥

403 g 16 ª ( 100 ¥ 10 -3 ) ¥ ( 400 g) = (16 ¥ 4) ¥ 10 -3 g = 64 ¥ 10 -3 g = 6.4 ¥ 10 -2 g mol CDP

29. D. We apply Le Châtelier’s principle. To increase the yield of the product, (CP)n, in Equation 1, we could increase the concentration of the reactant (which would shift the equilibrium toward the product side). Therefore, we would expect that increasing the amount of the reactant, CDP, would increase the yield of the product, (CP)n. 30. A. The reactant, CDP, and one of the products, namely HPO42–, both have a stoichiometric coefficient of n in the balanced reaction (Equation 1 in the passage). Therefore, in the expression for the equilibrium constant for this reaction, both [CDP] and [HPO42–] must appear with an exponent of n. This eliminates choices B and C. The expression in A is a better choice than the one in D since the other product of Reaction 1 is (CP)n, not simply CP. 31. C. In Equation 1, which is balanced, the stoichiometric coefficient of the polymer, (CP)n, is 1, and the stoichiometric coefficient of HPO42– is n. Therefore, the concentration of (CP)n is 1/n times the concentration of HPO42–. 32. C. We use the Henderson–Hasselbalch equation: pH = pKa + log

[conjugate base] [weak acid]

Since the solution is buffered at pH 8.7, and pKa = 6.7, we have pH – pKa = 2, so log

[HPO 4 2 - ] [HPO 4 2 - ] = 2 fi = 10 2 = 100 [H 2 PO 4 ] [H 2 PO 4 - ] 6

Passage VI 33. A. Because 14C undergoes beta decay, it will not emit an alpha particle or neutron in the decay process; this eliminates choices B and C. Since the radioactive decay process is 146 C Æ 147 N + -01e - , we see that 14C undergoes b– decay and emits an electron, -01e - . 34. D. According to the passage, the half-life of 14C is approximately 6000 years. Therefore, a time period of 18,000 years represents approximately 3 half-lives. If the object currently contains 1000 atoms of 14C, then 1 half-life ago, it contained 2000 atoms of 14C; 2 half-lives ago, it contained 4000 atoms of 14C; and 3 half-lives ago, it contained 8000 atoms of 14C. 35. C. The passage gives the mass of a beta particle as 9 ¥ 10–31 kg. So, if its speed is 3 ¥ 107 m/s, its kinetic energy is KE = 12 mv 2 =

1 2

(9 ¥ 10 -31 kg)(3 ¥ 10 7 m s) 2

=

1 2

(9)(32 ) ¥ 10 -31 ¥ 1014 J

ª 40 ¥ 10 -17 J = 4.0 ¥ 10 -16 J 36. A. As stated in the passage, a scintillator is a substance that produces light when it absorbs the energy accompanying radioactive decay. The scintillator is attached to a photomultiplier that collects this light and converts it into electrical impulses, which are then counted. These pulses then serve to measure the rate at which decay occurs. If the scintillator were to be non-transparent to the light it emits, then it would reabsorb some of that light, which the photomultiplier would then turn into electrical pulses and add to the count (that is, in addition to the pulses that are actually due directly to the decaying object itself), thereby overestimating the radiation energy and rate of decay. This would clearly produce an inaccurate reading. To prevent (or at least to minimize) this reabsorption of light, the scintillator should therefore be (nearly) transparent to the light it emits. 37. A. The energy of a photon of frequency f is given by the equation E = hf, where h is Planck’s constant. Since f = c/l, we can rewrite the equation for photon energy as E = hc/l. For the photon described in this question, then, we have E=

hc (6.6 ¥ 10 -34 J ◊ s)(3 ¥ 10 8 m s) = l 450 ¥ 10 -9 m ª

20 ¥ 10 -26 J ◊ m 4.5 ¥ 10 -7 m

ª 4.4 ¥ 10 -19 J

Passage VII 38. C. We first eliminate choices A and B; statement A is false (after all, Cl2 is a gas at room temperature, whereas I2 is not) and although statement B is true, the relative boiling points of Cl2 and I2 are irrelevant to the determination of the oxidation state of copper in the compounds it forms in reactions with these substances. To decide between choices C and D, we look at Table 1, and notice that Cu and Cl2 form CuCl2, while Cu and I2 form CuI. In CuCl2, copper is in a +2 oxidation state, while in CuI, copper is in only a +1 oxidation state. Since Cu “gives up” two electrons to chlorine but only one to iodine in these compounds, we would select choice C over choice D. Furthermore, since Cl is higher in the periodic table than I, we would expect that each Cl atom in Cl2 would have a stronger attraction for electrons than each I atom in I2. 39. C. The nitrate ion, NO3–, has a –1 charge, so a cadmium cation would have a +2 charge in order for the molecule Cd(NO3)2 to be neutral. Since the chloride ion, Cl–, also has a –1 charge, we expect the combination of Cd and Cl to be CdCl2 and the balanced reaction between Cd(NO3)2 and NaCl to be Cd(NO3)2 + 2 NaCl Æ CdCl2 + 2 NaNO3.

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40. B. Looking to the passage for a clue about the product that would most likely form between Cd and S, we notice in Table 1 that Zn and S form the compound ZnS. Since Cd is in the same family as Zn, it is reasonable to expect that Cd and S would form the compound CdS. Now sulfur, like oxygen, is most commonly in a –2 oxidation state in its compounds with other atoms. If the oxidation state of S in CdS is –2, then the oxidation state of Cd must be +2. 41. A. First, eliminate choice B (where is the source of carbon to form CO2?). Since evolution of gas occurs only with the addition of HNO3, HNO3 must react with copper metal. Copper metal (Cu0) must be oxidized during this reaction, and HNO3 must be reduced. Of the remaining choices (A, C, and D), the only logical choice for the product of the reduction of HNO3 is NO. 42. B. Because boiling-point elevation is a colligative property, the solution whose solute dissociates into the greater number of ions will be the one with the higher boiling point. Since Zn(NO3)2 dissociates into 3 ions (Zn2+ + 2 NO3–) while AgNO3 dissociates into only 2 ions (Ag+ + NO3–), we’d expect the boiling point of Zn(NO3)2(aq) to be higher than that of AgNO3(aq). 43. D. The AgNO3(aq) solution contains Ag+ ions; as Cu atoms are oxidized, Ag+ ions are reduced to Ag, which is the “new metal [that] forms on the surface of the Cu strip.” Also, note that we can eliminate choices A, B, and C, since it is highly unlikely that the cation Ag+ would be oxidized—or that Ag or Cu would be reduced—in this situation.

Passage VIII 44. C. The temperature, T, is 673 K in Trial 5. According to the data for Trial 5 given in Table 1, the current I is 2 A, and the voltage across the wire is 28 V. Therefore, the power dissipated by the wire is P = IV = (2 A)(28 V) = 56 W. [Alternatively, since the resistance R is approximately 14 W, the power dissipated is P = I 2R = (2 A)2(14 W) = 56 W.] 45. D. In the first paragraph of the passage, the mass of the wire is given to be m = 4 ¥ 10–3 kg. Since the volume of the wire is given in the question to be V = 5 ¥ 10–7 m3, the density of the wire is

r=

m 4 ¥ 10 -3 kg = = 0.8 ¥ 10 4 kg m 3 = 8 ¥ 10 3 kg m 3 = 8, 000 kg m 3 V 5 ¥ 10 -7 m 3

46. A. According to the data in Table 1, R increases as T increases (as we can see by reading the values of R as the temperature increases from Trial 1 through Trial 5). This eliminates the graphs in B and D, which show the resistance R either constant or decreasing with temperature. Since the only choices left are the graphs in choices A and C, the question becomes, “Does R increase linearly with T?” Comparing Trials 2 and 3, then 3 and 4, then 4 and 5, we see that R increases by a steady 2.6 W for every 100 K increase in temperature. Therefore, R does increase linearly with T, and the answer is A. 47. D. From the expression given in the last paragraph of the passage, AsT4, we see that the energy radiated from the heated wire each second is proportional to T 4. So, if T increases by a factor of 2 , the energy radiated each second increases by a factor of ( 2 ) 4 = [( 2 )2 ]2 = [2]2 = 4 . 48. A. We use the equation q = mcDT, where m is the mass of the sample being heated and c is the specific heat of the sample (in this case, it is the iron wire). Since the mass of the wire is m = 4 ¥ 10–3 kg, the specific heat of iron is c = 460 J/kg·K, and DT = 573 K – 373 K = 200 K, we have q = mcDT = ( 4 ¥ 10 -3 kg)( 460

J )(200 kg ◊ K

K) = 368 J

[Note: The question uses the term “heat capacity” where it should use the term “specific heat.”]

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49. D. According to the data in Table 1 for Trial 1, the resistance of the wire was R1 = 4 W when T = 293 K. When T = 673 K (Trial 5), the resistance rose to R5 = 13.9 W ª 14 W. Therefore, if the voltage remained constant at 28 V, the current decreased, from I1 =

V 28 V = =7A R1 4W

to

I5 =

V 28 V = =2A R5 14 W

Independent Questions 50. B. Because the activity decreased to 60/240 = 1/4 its initial value, this means that 2 half-lives elapsed, since (1/2)2 is equal to 1/4. If a time period of 2 half-lives is equal to 24 minutes, then one half-life must be 12 minutes. 51. C. Since the molecules in the gas phase of a substance are much more disordered than in the solid phase, the phase change from solid to gas (sublimation) represents an increase in the entropy, S. That is, DS > 0. 52. C. The Doppler Effect implies that when the source of a sound moves away from the observer, the perceived frequency is lower than the emitted frequency. 53. A. As the diagram below shows, water at –0.1∞C and 1.0 torr is vapor, and as the pressure is increased at constant temperature, the vapor will become a solid and then a liquid:

off the diagram at P = 200 atm liquid

pressure (torr)

solid

vapor 4.6 1.0 –3.0

0.01 –0.1 temperature (∞C)

54. B. Because opposite charges attract, the negatively-charged particle will move toward the fixed positive charge Q; this eliminates choices A and C. –q

+Q

r

The negatively-charged particle experiences a force (F = kQq/r2) as it approaches +Q, so it will undergo an acceleration (a = F/m = kQq/mr2, where m is the mass of the –q particle). Since the –q charge is accelerating, its speed of approach cannot be constant. This eliminates choice D.

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Passage IX 55. D. As stated in the first paragraph of the passage, the sample XT-n contains n% Ti, where n = 0, 1, 3, or 5. Looking at the data in Table 1, we notice that the solubility of XT-n increases as n increases, so choices A and B are eliminated. Since the entry in each row is greater for toluene than for THF, we conclude that the XTs are more soluble in toluene than in THF. 56. D. Even if there were any indication in the passage that the XTs are even capable of hydrogen bonding, the formation of hydrogen bonds would not decrease the weight of a sample, so we eliminate choice A. Next, according to the data in Table 2, the masses (and therefore the weights) of the samples decrease by 20% when heated from 20∞C to 700∞C; the loss of some electrons, even if they escaped from the heating chamber, could not account for this much of a decrease in mass, so we eliminate choice B. As for choice C, the removal of protons from nuclei requires extreme conditions (like those in a nuclear reactor); it is highly unlikely that simply heating the compound to 700∞C would cause a nuclear reaction. Choice D provides the most reasonable explanation for the loss of mass by the samples as they are heated. 57. B. The transition metal titanium (Ti, atomic number 22) is in the 3d “block” of the Periodic Table, which means its valence electrons are in 3d orbitals. Titanium doesn’t contain electrons in 4p or 5f orbitals, so choices C and D are eliminated, and titanium’s 2s electrons (choice A) are not in the valence shell, so they’re unavailable to form bonds. 58. C. Of the elements listed in the choices, only zirconium (Zr, atomic number 40) is in the same family (group) as titanium (Ti, atomic number 22). The elements in each family of the Periodic Table have similar properties and have identical (or very similar) outer configurations. 59. B. Since oxygen is an element in Period 2, it has only s and p orbitals and can form no more than four hybrid orbitals, so choice C is eliminated immediately. Choice D should be eliminated immediately as well (s2p2 hybridization?). The oxygen atom in THF is bonded to two carbon atoms, so there must be four equivalent hybrid orbitals on the oxygen atom, formed by sp3 hybridization; two contain lone pairs and two contain a single electron each, which will form the s bonds with the carbons. 60. A. Since THF can participate in hydrogen bonding with H2O, but toluene cannot, we’d expect THF to be more soluble than toluene in H2O. Choice B is false (since there are no hydrogen bonds between toluene and water to compare with those between THF and water), and while the statements in choices C and D are true, they don’t answer the question.

Passage X 61. C. When the toboggan begins its slide from Point A, it has gravitational potential energy (relative to Point B), which is converted to kinetic energy as the toboggan slides down the hill. Since the passage states that the toboggan experiences friction as it slides, some of the potential energy is also converted to heat (thermal energy). Therefore, the energy conversion is best described by choice C: potential to kinetic and thermal. 62. B. According to the passage, the toboggan is opposed by a constant 60 N frictional force when it’s sliding down the hill. Since the toboggan feels this force for the entire length, l, along the hill, the work done by sliding friction on the toboggan is equal to –(60 N)(l), so the energy lost to friction is (60 N)(l). 63. A. We apply Conservation of Total Momentum to this completely inelastic collision. Since the toboggan and rider (T&R) are stationary before the collision, their momentum before the collision is zero, so the total momentum before the collision is simply MS&RvS&R, the momentum of the sled and rider (S&R). The momentum after the collision is (MS&R + MT&R)v¢. Therefore, MS& R vS& R = ( MS& R + MT& R )v ¢ fi v ¢ =

MS& R vS& R (3 kg + 47 kg) ◊ (10 m s) 50 = = m s ª 4.55 m s MS& R + MT& R (3 kg + 47 kg) + (6 kg + 54 kg) 11

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64. A. The passage states that, to the right of Point B, the sled and rider are opposed by a 50 N frictional force, so Ffrict = 50 N. Since Ffrict = mN, and N = MS&Rg = (3 kg + 47 kg)(10 m/s2) = 500 N, we have

m=

Ffrict 50 N 1 = = = 0.1 500 N 10 N

65. B. The snowball will land when it has fallen a vertical distance of y = 20 m. Let’s first figure out how long this will take. Using Big Five # 3 (with v0y = 0, since the ball is thrown horizontally), we find that y = 12 gt 2 fi 20 m = 12 (10

m 2 )t s2

fi t = 2 s

Because the snowball’s horizontal speed is vx = 25 m/s, the snowball would, in this time, travel a horizontal distance of x = vxt = (25 m/s)(2 s) = 50 m Since this equals the horizontal distance from Point A to Point B, the snowball will land at Point B. Point A snowball

y = 20 m t = 2 sec

Point B

x = 50 m

66. D. We use Conservation of Mechanical Energy. The passage states that the sled and rider slide free of friction from Point A to Point B, so the gravitational potential energy of the sled and rider at Point A is transformed into kinetic energy at Point B. Therefore, PE Æ KE fi Mgh = 12 Mv 2

fi v = 2gh

fi v µ h

If the sled and rider start at a point on the hill that is 10 m lower than Point A, then the sled and rider’s initial height above Point B is being reduced from 20 m to 10 m, a decrease by a factor of 2. Since v is proportional to h , if h is reduced by a factor of 2, then v will be reduced by factor of 2 .

Passage XI 67. D. As the diagram below shows, the distance between adjacent nodes (labeled N) on a standing wave is always equal to half the wavelength, l/2: l

N

N

N 1 2

l

11

N

N

68. A. The traveling waves whose superposition generates a standing wave travel in opposite directions, eliminating choices B and C. The oppositely-directed traveling waves must have equal amplitudes (choice A) since the resultant standing wave has displacement zero at the node positions (where the equal-amplitude traveling waves arrive exactly out of phase with each other and thus completely cancel). 69. B. That the light emitted by the laser is coherent is stated in the first sentence of the passage, so Item II is true, and we can eliminate choices A and C. If the laser has only one mode of oscillation, then it produces only one wavelength of light (since, according to the passage, “laser cavities have mode numbers that are related to the allowed cavity wavelength[s].” Because the laser produces light of only one wavelength, the light is monochromatic (“one color”), so Item I is true, and the answer must be B. (Note that we didn’t need to check Item III, but, if we did, since laser light is sharp and focused, it would not be characterized as diffuse.) 70. B. If we substitute lm = c/fm into the equation mlm = 2L (both of which are given in the passage), we find that m◊

14 c 2 Lfm 2( 13 m)(9 ¥ 10 Hz) = 2 L fi m = = = 2 ¥ 10 6 fm c 3 ¥ 10 8 m s

71. D. Because l = c/f, we have lbeat = c/fbeat. Therefore, since fbeat = fm+1 – fm, we have l beat =

c c = fbeat fm +1 - fm

Independent Questions 72. C. By Archimedes’ Principle, the buoyant force exerted by the unknown liquid on the object is given by Bunknown = runknownVg, and the buoyant force exerted by benzene is given by Bbenzene = rbenzeneVg. Thus, the ratio Bunknown/Bbenzene is Bunknown r unknown Vg r unknown = = Bbenzene r benzene Vg r benzene Since we’re given that Bunknown = 12 N and Bbenzene = 5 N, the ratio Bunknown/Bbenzene is equal to 12/5. Therefore,

r unknown 12 = 5 r benzene

fi r unknown =

12 12 7 84 168 = = 1.68r H 2 O ª 1.7r H 2 O r benzene = Ê r H 2 O ˆ = r r ¯ 50 H 2 O 100 H 2 O 5 5 Ë 10

This tells us that the specific gravity of the unknown liquid is approximately 1.7. 73. B. We apply Conservation of Total Momentum to this perfectly inelastic collision. Since the block is stationary before the collision, its momentum before the collision is zero, so the total momentum before the collision is simply mobjvobj, the momentum of the sliding object. The momentum after the collision is (mobj + mblock)v¢. Therefore, mobj vobj = ( mobj + mblock )v ¢ fi v ¢ =

mobj mobj + mblock

vobj =

1 kg vobj = 14 vobj = 14 (8 m s) = 2 m s 1 kg + 3 kg

74. B. The equilibrium is Ca(OH)2(s) Ca2+(aq) + 2 OH–(aq). Increasing the acidity of the solution has the effect of reducing the concentration of OH–(aq). By Le Châtelier’s principle, removing a product causes a shift toward the products; therefore, additional Ca(OH)2(s) will dissolve. Since the value of Ksp can be changed only by a change in temperature, the answer must be B.

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75. A. For nearsighted individuals, light from distant objects is focused in front of the retina, which occurs if the focal length of the lens of the eye is smaller than the distance to the retina. A divergent corrective lens is required to cause the light from distant objects to diverge slightly before entering the lens of the eye, so that it may be focused on the retina. diverging corrective lens retina

retina

nearsighted eye

nearsightedness corrected

76. A. The principal quantum number, n, has nothing to with the number of valence electrons of an atom (choice C) or with the mass number (choice D). And choice B is eliminated since, for example, the 3s, 3p, and 3d orbitals all have different shapes but the same principal quantum number (namely, n = 3), so the principal quantum number cannot be a “measure” of the approximate shape of an electron cloud. The best response here must be A (and the higher the value of n, the larger the radial size of the electron cloud). 77. B. The weight of the unknown solid is given by the equation wsolid = rsolidVg, and, by Archimedes’ Principle, the buoyant force exerted by the water on the solid is given by B = rH OVg. Thus, the ratio wsolid/B is 2

wsolid rsolid Vg rsolid = = B r H 2 O Vg r H 2 O Because the solid weighs 31.6 N but has an apparent weight of only 19.8 N when submerged in water, the buoyant force on the object must be 31.6 – 19.8 = 11.8 N. Since wsolid = 31.6 N and B = 11.8 N, the ratio wsolid/B is also equal to 31.6/11.8. Therefore,

rsolid 31.6 = ª 2.68 r H 2 O 11.8 This tells us that the specific gravity of the solid is approximately 2.68.

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VERBAL REASONING Passage I 78. D A: A main theme of the passage is that competition between state and local governments can be destructive. No positive benefits are mentioned. Furthermore, competition between private firms is never discussed; subnational governments compete with each other to attract private firms to their area. B: As in choice A, this statement contradicts a main idea of the passage—that competition can have negative effects. A secondary problem with this choice is that efficiency is never mentioned in the passage. C: The passage makes only two references to the national level. First, the author indicates that national economic development is not characterized by competition (lines 4-6). Second, the author cites the critics’ claim that local and state competition does not contribute to national productivity (lines 39-43). Neither reference supports a link between subnational competition and incentives for national development policy. D: Yes. The author cites the critics’ claim that cooperation would be more likely than competition to result in increased national (“overall”) productivity (lines 39-44). The passage also describes a case in which increased cooperation between jurisdictions in the Monongahela River Valley may better address those jurisdictions’ economic problems than would competition (lines 72-83). 79. B A: The primary purpose of the author is to discuss competition and cooperation on a subnational level. No role for national officials is mentioned. B: Yes. The author describes one successful attempt at cooperation, an effort organized by local leaders (lines 8083). C: Academic researchers are never discussed in the passage. D: The author does not discuss any role for media representatives. 80. C A: According to the passage, local leaders in the Monongahela region worked to increase cooperation and decrease competition between jurisdictions (lines 70-83). If these efforts were followed by prosperity, the lesson would be to cooperate more, not less (and so compete less, not more). For both choices A and B, remember that the main idea of the passage is that cooperation may be more advantageous than competition at subnational levels. B: According to the final paragraph, local jurisdictions had competed in the past, but now are managing to cooperate in the face of a regional economic downturn (lines 70-83). If this cooperation were followed by prosperity, less competition would be called for. C: Yes. If local leaders’ cooperative efforts (lines 80-83) were followed by prosperity, it is likely that continued cooperation would lead to continued prosperity for the region. D: The author does not say that local governments in the Valley region ever received federal grants, nor does the passage draw any link between federal moneys and either prosperity or cooperation. 81. C Note: To be the credited response, the answer choice must both present an assumption that can reasonably be attributed to the author and be inconsistent with the new information given in the question. A: This choice is too extreme; the author does not make this assumption. The author indicates that cooperation is difficult (lines 44-55), not that it is impossible. In fact, the passage presents a specific example of cooperation between local jurisdictions (lines 77-83). B: This choice is out of scope (“state law”) and too extreme (“only if”); the author does not make this assumption. State law is never mentioned, only pacts, policies, and agreements within and between states and localities. Furthermore, an example of cooperation is given with no suggestion that state law played any role (lines 70-83). C: Yes. In the final paragraph, the author states that “Economic conditions may ultimately serve as the catalyst for greater cooperation….” (lines 70-71). This statement is followed by the one example of a successful attempt at cooperation given in the passage; this cooperation occurred when the region was experiencing economic difficulties (lines 73-80). Thus, the author does make the assumption that cooperation is most likely in times of economic stress. The study cited in the question gives cases in which cooperation followed a period of economic improvement, and so choice C does describe an assumption that is called into question by the study. 14

D: This is not an assumption made by the author. In fact, the final paragraph presents a case in which local leaders crossed state lines to foster cooperation between jurisdictions (lines 80-83). 82. D Item I:

The passage indicates that a difference between national and state or local economic activity is that state and local economic development is competitive and that state and local governments are “awash in competition” (lines 4-9). Thus we can infer that national governments do not exhibit competitive economic behavior. Item II: Yes. This is directly stated in lines 4-9. Item III: Yes. This is directly stated in lines 4-9. 83. A A: Yes. In lines 44-46, the passage mentions the failure of Great Lakes states’ governors to keep communities within those states from pirating or stealing economic developments from each other. This example is introduced with the more general statement that cooperation is elusive (line 44). B: The experience of the Great Lakes states is one of a failed attempt to restrict competition. Furthermore, the author does not indicate that many other state governments have made similar attempts, and so the word “usually” is too extreme. C: The passage makes no such comparison between state and local governments. The fact that some state governments failed (lines 44-45) does not in and of itself indicate that local governments are or would have been more successful. Finally, while local leaders organized the successful Monongahela attempt at cooperation (lines 80-83), the author does not indicate that this can be extrapolated into a generalized comparison between the effectiveness of state and local leaders. D: This choice is both out of scope and too extreme. The power of national governments to regulate competition is never discussed. Furthermore, the Monongahela case is an example of local leaders successfully regulating or limiting local competition (lines 72-83).

Passage II 84. B A: The author explains in lines 59-64 that false memories cannot always be distinguished from correct or true memories. B: Yes. This is the main idea of the passage. In paragraphs 2, 3 and 4 the author explains how schemas, once instantiated, help us to understand new information and act upon it. In the final paragraph, the passage discusses how there may be glitches or flaws in that process that also could affect our memory and comprehension of events. C: While the author does state that certain things may be forgotten (lines 60-64), the passage does not indicate that this would necessarily be true for important things. For that reason, this choice is too extreme. Furthermore, while imperfect recall is one aspect of schema theory, it is not presented as the most important aspect of it, or as the reason why schema theory is itself important. Compare this choice to answer choice B. D: Schemas are activated and instantiated when we are confronted with situations similar to those we have experienced in the past (lines 8-15). Thus memory is activated in these cases by familiar, not unfamiliar situations. 85. D A: This choice is inconsistent with the main idea of the passage, which is in part that schemas and scripts (a special type of schema) are activated in situations that are similar to events and experiences from our past. In cases where scripts are active, then, encounters with certain events are not new learning experiences, but are affected by our past experiences. B: In lines 60-64, the passage states that we may remember information that was never part of the original event. C: As in choices A and B, this statement contradicts the passage. The very nature of a script is that once developed, it can be used to help us understand and take action in new instances (paragraphs 3 and 4). Compare this choice with choice D. Choice C is the opposite of the credited response. D: Yes. Scripts are a special form of schema (lines 22-24), and scripts not only help us understand new instances of familiar situations, but also guide our behavior in those contexts (lines 22-24). The author presents the restaurant scenario as an example of how this occurs.

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86. A Item I:

Item II: Item III:

Yes. Slot filling occurs when a script created by previous experiences helps us to understand and fill in gaps in other, similar experiences (lines 35-52). In this choice, the child has already had one year of school, presumably with significant similarities to the second year. A child first learning to ride a bicycle would not necessarily have had other similar experiences in the past. As for Item II, a child first learning the alphabet would not necessarily have had previous experience with similar situations that could have created a schema or that would lead to slot filling within that schema.

87. D A: The passage only discusses inferences, correct or incorrect, that are made within schemas (lines 35-52, 60-64). B: The author does not discuss partial activation of scripts nor suggest in any way that partial activation cannot occur. This choice may sound familiar, as the author does discuss instantiation of a script based on partial information (lines 15-19). Always be sure to go back to the passage and reread carefully. C: While it is likely that scripts are instantiated subconsciously (without our direct knowledge), the author never states it. Most importantly, the author does not make a connection between the way in which schemas are instantiated (lines 1221, 35-46) and recall error as described in the question and in the passage (lines 60-64). D: Yes. The author states that inferencing may lead to recall errors and that some information may be forgotten (lines 56-64). If different readers recalled the same text differently, this would provide evidence that recall errors do in fact occur. 88. B A: The passage does not suggest that scripts are instantiated through a deliberate or conscious act. Notice the wording of the passage. For example, in lines 14-15, the author states that a “schema is thus instantiated by the new information,” not that the person instantiates it through a deliberate act of will. B: Yes. The author explains that scripts, when activated, affect both how we process new information and how we behave based on that information (lines 10-12, 22-24). Schemas and scripts are based on memory or prior knowledge (lines 3-4, for example). C: While scripts provide general information about particular circumstances (lines 32-34), slot filling provides specific, not generalized information within that script (lines 35-52). Notice the word “however” in line 35. It is at that point that the author shifts from discussing scripts in general to describing the particular function of slot filling. D: Activation of the script influences processing of new information (lines 10-12). Inferencing occurs after activation and instantiation, when slot filling occurs (lines 35-52). Thus inferencing depends on the availability of specific pieces of information from the past, not on processing the new information itself. 89. D A: Instantiation depends on the availability of an appropriate old, pre-existing schema, not a new schema. B: Alteration of a schema may occur when slot filling and inferencing occurs (lines 56-64). This happens during or after instantiation (lines 35-46). It does not determine whether or not instantiation has occurred. C: As described by the author, instantiation depends on the quality of the information [whether or not it matches a preexisting schema (lines 12-15)], not its quantity. D: Yes. In lines 12-15, the author explains that instantiation occurs when new information is judged to be “similar enough to the content of the schema.” 90. C A: Careful reading is never suggested as a factor in recall errors, certainly not as a cause of error. B: According to the passage, errors occur when slots are inappropriately filled in a current situation with information from the schema (lines 59-64), not when the schema (prior knowledge) is incomplete. C: Yes. A person may fill in slots in a current situation [inferencing (lines 39, 49)] with information from the schema. While the situations are similar (as they must be for the schema to be activated and instantiated), they may not match exactly, and those slots may be filled with information that does not match the current reality. The person may then later remember these things as if they actually happened in that more recent situation (lines 35-45, 53-64). D: Instantiation of multiple schemas is never discussed in the passage.

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91. A A: Yes. An understanding of schema theory would most likely lead the teacher to teach reading in a way that created and/or utilized pre-existing knowledge structures. The development of background knowledge would help create such structures. B: Nothing in the passage suggests that improved pronunciation would either utilize or develop “organized knowledge structures in memory” (lines 3-4). C: While memory is involved in this choice, rote memorization would be unlikely to provide the kind of generic knowledge structures that could be applied to new situations (lines 3-7). Compare this choice to answer choice A. D: The credited response must involve the development or use of schemas or “organized knowledge structures in memory” (lines 3-4). Nothing in the passage indicates that an explanation of an unfamiliar word would help create a schema. Given that the word is unfamiliar, it is also unlikely that the explanation would draw on or utilize a schema. 92. B Item I:

Item II:

Item III:

Inferences occur when a person fills in empty spaces in the current text with details from an instantiated script. A script is not instantiated unless it is significantly similar to, and so most likely appropriate for, the current text or situation (lines 12-15). The passage does not discuss the activation of inappropriate or wrong texts. Yes. Inferencing occurs when empty slots in a text are filled in with details from a pre-existing schema (lines 35-45). When information came only from the schema and did not in fact exist in the text being read, a person may still later incorrectly remember them as part of that text (lines 59-64), essentially “rewriting” the text in their memory. Inferences are made when the reader takes information from a pre-existing schema and inserts it into gaps in a text (lines 35-45). Skimming a text to acquire specific facts does not involve this use of pre-existing memory structures, but only the current text itself.

93. B A: The passage does not indicate that the formation or use of knowledge structures requires effort or purpose, as in concentrated study. B: Yes. The author indicates that schemas are formed through previous experiences (lines 3-7), and that schemas provide a context that helps us to comprehend and utilize new information (lines 10-12, 22-24). C: Schemas help us to comprehend new situations when not all of the facts are present, as shown by the example of recognition of a face when only a few details are available (lines 15-21). D: As in choice A, the passage does not indicate that activation and instantiation of schemas requires any conscious act of will.

Passage III 94. B A: The author uses quantum mechanics as an example of a new [“in our own time” (lines 40-43)] scientific research tradition that has not managed to overcome most people’s common-sense view of “the world as being populated by substantial objects, with fixed and precise properties” (lines 40-45). B: Yes. The main idea of the third paragraph is that some world views persist despite the appearance of scientific traditions that contradict those views. The author offers quantum physics as an example of a scientific theory that has not managed to shake most people’s common-sense view of “the world as being populated by substantial objects, with fixed and precise properties” (lines 40-45). C: This is the right answer to the wrong question. It is people’s belief in indetermination (lines 62-68), not quantum mechanics, that is incompatible with the idea that we live in a universe governed by natural laws (lines 68-72). D: See answer choice A. According to the passage, most people refuse to change their world view to accommodate the insights of quantum mechanics (lines 40-45).

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95. C A: This choice takes words out of the context of the passage. Our social, political, and moral beliefs contradict the idea that all physical changes are subject to the same unchanging natural laws (lines 62-72). B: This choice also takes words and ideas from the passage out of context. Our social, political, and moral beliefs are themselves a broader system of cultural attitudes. Such systems sometimes conflict with scientific theories and traditions (lines 32-40, 62-72). C: Yes. According to the passage, the scientific tradition that holds that all physical changes are subject to natural law has been accepted since the 17th century (lines 55-59). However, most of our social, political, and moral beliefs are inconsistent with the idea that these laws could apply to human beings and perhaps higher animals (lines 68-72). D: While the author does argue that Darwinism and Marxism have been accepted by “reflective people” (lines 26-31), the author draws no direct connection between those research traditions and the social, political, and moral beliefs discussed in lines 64-72. 96. B A: Newton’s ideas were “eventually” (lines 21-22) accepted after a “process of readjustment” (line 26). Thus the words “readily” and “quickly” are too extreme. B: Yes. This choice corresponds to the author’s description of the process of readjustment that led to acceptance of Newton’s ideas (lines 14-26). C: The passage indicates that most people eventually modified their world view to bring it into line with Newton’s ideas (lines 14-26). There is no indication that the implications of Newton’s theories were ignored, or that most people’s acceptance of those theories was feigned or only superficial. D: The Newtonian view of reality was eventually accepted, not rejected (lines 21-26). 97. D A: The author refers to this claim in order to make the argument that some long-standing scientific traditions (in this case, since the 17th century) have still not been accepted by most people (lines 49-68), and that the strength of old world views may not necessarily fade over time. This choice is not consistent with the main ideas of the last two paragraphs. B: The author writes that natural laws may be either statistical or nonstatistical (lines 58-59). The passage gives no indication that the laws governing human actions must be statistical in nature, nor does the application of physical laws to humans indicate that some physical changes can only be explained by statistical laws. C: The passage states that in the 17th century, physical laws, not theories, were thought to apply to all physical objects or changes (lines 55-59). Secondly, it is the author, not (as far as we know) people in the 17th century who assume that these laws apply equally to human beings (lines 62-64). D: Yes. The view that all physical changes are completely determined would, according to the author, apply to human beings as well (lines 62-64). 98. A Item I: Item II:

Item III:

Yes. The new information in the question describes people changing their views of reality in order to accommodate new scientific findings. See the explanation for Item I. The question gives an example of adaptability and acceptance of changing world views. This choice indicates that some new scientific ideas may never find acceptance in the face of contradictory world views. See the explanations for Items I and II. This choice describes resistance to, not acceptance of new scientific ideas (lines 62-72).

99. C A: Remember the main idea of the passage. The author argues that behaviorism is not widely accepted due to the strength of the old world view in which inner mental states exist (lines 32-37, 45-47). The passage never indicates that the theory of behaviorism is itself weak. B: This is the right answer to the wrong question. The author’s discussion of the application of natural law to human behavior comes later in the passage, in a different context (lines 55-68). C: Yes. According to the author, the fact that people still believe in inner mental states indicates that the scientific tradition of behaviorism has not been able to supplant or transform an old world view that contradicts this scientific theory (lines 32-40). 18

D: While behaviorism may be a relatively new (very new is too extreme) tradition, the author does not discuss it or the contradictory belief in inner states in order to make that claim. Always keep the main idea of the paragraph in mind when answering “support” or “in order to” questions.

Passage IV 100. D A: The passage never suggests that CO2 is poisonous to fish. According to the CO2 theory, it could have been a decrease in the dissolved CO2 level in the ocean waters (and an increase in atmospheric CO2) that lead to extinctions of marine life by inhibiting the growth of algae, the base of the food chain (lines 56-62). B: Dust is never mentioned in the context of the CO2 theory. This choice illegitimately combines aspects of the two different theories presented in the passage (lines 8-11 for the first, 56-62 for the second). C: This choice contradicts the passage. Fish may in fact eat algae, or other creatures that themselves eat algae (lines 6062). However, the increase in atmospheric CO2 would cause a decrease in CO2 in ocean waters (due to increases in temperatures) (lines 54-59). Decreases in dissolved CO2 would cause algae to decline, not to flourish (lines 59-60). D: Yes. An increase in atmospheric CO2 would cause global warming (lines 46-51). Higher temperatures would cause less CO2 to dissolve in ocean waters (lines 56-59). As dissolved CO2 levels fell, so would the population of algae, which sit at the base of the oceanic food chain (lines 59-61). Disruption of the food chain could then have led to the extinction of a variety of marine species (lines 60-62), including species of fish. 101. A Note: Notice the word “would” in the question. The credited response must be something that these measurements would, not just could show, prove or indicate. A: Yes. The CO2 theory posits that increases in atmospheric CO2 would have caused decreases in levels of CO2 dissolved in ocean waters (lines 56-59). Ice cores would indicate if such a change did in fact occur. B: The new information in the question describes the possibility of measurement, not what those measurements would show. Compare this choice to answer choice A. This choice is too extreme. C: The new information in the question states only that measurements could be made, not what those measurements would be. Even if it were shown that CO2 levels had fallen in the oceans at the K–T boundary, that would not be sufficient to prove that the cause was an asteroid strike. This choice is also too extreme. D: See the explanation for choice C. Even if decreased levels of oceanic CO2 levels were found (and we don’t know that they would be), that would be insufficient evidence to either disprove the dust scenario or prove the CO2 theory. 102. D Note: Notice that the passage discusses only asteroid strikes on dry land (lines 8-11, 22) or in shallow ocean beds (line 22). Avoid a trap by going back to the passage and re-reading carefully. A: Such warming would occur if a 10 km asteroid hit in a bed of carbonate rock (lines 48-51), not in the deepest part of the ocean. B: A large enough asteroid could cause mass extinctions if it hit a layer of carbonate rock [on dry land or in ocean shallows (lines 20-27)]. There is no indication in the passage that a hit in deep ocean would release CO2 into the atmosphere and set off a chain of events leading to large-scale extinctions. C: Nowhere in the passage are extinctions limited to aquatic species discussed. Neither does the author present evidence suggesting that an asteroid hit in deep water would cause any extinction of marine species. D: Yes. The author only discusses asteroid impacts which might send up clouds of smoke and dust (lines 8-11) (and so presumably occur on dry land), or which release CO2 into the atmosphere by colliding with beds of carbonate rock on dry land or in shallow water (lines 20-23, 45-46, 49-50).

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103. A A: Yes. The CO2 hypothesis is based on the claim that a major asteroid impact on a bed of carbonate rock would release CO2 into the air, causing global warming and a decrease in oceanic CO2 levels. If a major asteroid hit on carbonate rock were not followed by climatic change, it would significantly undermine the causal claims at the heart of the CO2 theory. B: In the first paragraph, the author explains that high iridium levels at the K–T boundary inspired the idea that an asteroid struck the Earth at that time. The iridium evidence is consistent with both theories presented in the passage, and would support, not challenge the CO2 theory. C: The CO2 theory argues that increased levels of atmospheric CO2 would lead to global warming (lines 28-32, 44-53). Further correlation between CO2 levels and global warming would strengthen, not weaken the CO2 theory. D: The CO2 theory does not rest on the claim that only asteroid impact could lead to increases in global CO2. In fact, in the context of presenting the CO2 theory, the author describes other factors that increase atmospheric CO2 levels, such as burning of fossil fuels (lines 32-38). 104. C A: The possibility of a loss of carbonate rock is not addressed by the passage. Even if we were to speculate that an asteroid hit that released CO2 from carbonate rocks would entail a reduction in the rocks’ mass (and this speculation would not be sufficiently supported by the passage), there is no connection to be made to an ash-spewing volcano. B: The passage indicates just the opposite. Dust and smoke filling the atmosphere would block sunlight from reaching the Earth (lines 10-11). It is reasonable to assume that massive amounts of ash would have a similar blocking effect. C: Yes. The author states that the dust and smoke from an asteroid impact could have temporarily cooled the Earth by blocking out sunlight (lines 8-17). It is reasonable to assume that massive amounts of ash in the atmosphere would have a similar effect. D: This is the right answer to the wrong question. The passage makes no connection between dust, smoke, or ash in the atmosphere (the first theory presented, in the first paragraph) and increases in CO2 (the second theory described, in the rest of the passage). 105. D A: This choice is too extreme. According to the passage, if the CO2 theory is valid, then high levels of CO2 may have lead to large-scale extinctions, including that of the dinosaurs, by causing global warming. However, the author does not suggest that all forms of life were or would have been destroyed. Finally, the wording of this answer choice (“survive in high levels of CO2”) appears to indicate that it is the CO2 itself that kills, not its indirect effects through global warming. B: According to the passage, large amounts of dust and smoke in the atmosphere would lead to a rapid decrease in temperature and a temporary “winter” (lines 8-11). Beware of opposite answer choices. C: The passage makes no such comparison. The only reference in the passage to a species that is directly sensitive to CO2 levels is to a marine species, algae (lines 57-60). D: Yes. The author describes two theories in the passage, both of which explain the extinctions as the result of the impact of a large asteroid or comet. The author appears to prefer the CO2 theory as shown by the problems raised regarding the first theory (lines 13-17) and by the last line of the passage, “Thus, global warming could have led to the extinction of the dinosaurs.” 106. A A: Yes. The author states that low-level global warming of 2–4 ∞C due to burning of fossil fuels may lead to melting of the polar ice caps and flooding (lines 38-43). We can infer that this flooding would be caused by an increase in sea level due to the melting ice. Sea levels would have risen much more dramatically at the K–T boundary if the temperature had increased 5–20 ∞C, lasting for 10,000 years (lines 46-53). Thus, if the main theory presented in the passage, the CO2 theory, is correct, we would find increased sea levels at the K–T boundary. B: Excess iridium is mentioned in the context of the “dust and smoke” theory (lines 5-11), a theory which according to the author has serious weaknesses (lines 13-17). However, the presence of excess iridium would also be consistent with the CO2 theory, which also claims that the primary cause of extinction was an asteroid strike (lines 18-20). Thus, if the main hypothesis presented in the passage (the CO2 theory) is correct, excess iridium could well be present. C: The main theory presented in the passage, the CO2 theory, argues that major extinctions did occur at the K–T boundary (lines 60-66).

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D: As in choices B and C, this choice contradicts the passage. The CO2 theory is predicated on the claim that atmospheric CO2 levels rose (lines 23-27), not that they (and the composition of the atmosphere) remained constant. 107. B A: The author believes that asteroids may have indirectly, not directly, caused the dinosaurs’ extinction. Global warming would have affected the entire planet; there would have been no need for the dinosaurs to be located near the impact site. B: Yes. In both of the scenarios described, a consequence of the asteroid or comet impact would be slowing of plant growth (lines 11-13, 60-65). In the context of discussing the CO2 theory (the author’s preferred explanation), the author explains that reduction in plant growth disrupts the food chain, and so indicates that starvation was likely a major cause of extinction. C: The author may well believe that dust filled the air after an impact, just that the dust itself was not sufficient to cause mass extinction (lines 13-20) D: The passage states just the opposite. Decreased levels of CO2 in the oceans could have lead to a decrease in the algae population, and so disruption of the entire marine food chain (lines 56-62).

Passage V Note: This passage is particularly confusing because the author’s own position on the issue of whether or not Picasso was in fact a “cerebral structurer” as a young child, and whether or not he did in fact “begin again” as an adult artist is never made clear (see lines 8-12, 35-41). Note as well that the questions never ask for the author’s own views on those points. Don’t get distracted by things that are left unclear in the passage if they are not necessary for answering the questions. 108. D A: The author argues that Picasso was not a genius or “frankly precocious” as a child (lines 41-45), nor was his work effortless (lines 22-26, 32-35). His achievements arose from the transformation or metamorphosis that came later (lines 43-45, 52-54). B: While we know from the passage that Picasso and Braque invented cubism (line 10), the passage does not indicate that Picasso rejected the art movements of his time (he could have built upon them, for example), or that such rejection would itself be a cause of his artistic accomplishments. Be careful not to use outside knowledge. C: Only one example of revision is given (lines 61-64). We do not know from the passage that Picasso had a practice of making many revisions, or that revision was a reason for his level of artistic achievement. D: Yes. The primary purpose of the passage is to assess Picasso’s work along two related lines. First, what is the relationship between his mature work and his artistic creation as a child. Second, what is the relationship between his work as a child and as an adult, and a childlike or “unreflective” and spontaneous approach to art. According to Apollinaire and to the author, Picasso’s “spectacular progress” came when he transformed himself into an “unreflective virtuoso, who relies on nature” (lines 46-54). Picasso’s later apparent spontaneity is contrasted with his academic, literal, and precise work as a child around the age of nine (lines 24-29). 109. C A: This choice is too literal. The passage never suggests that Picasso [an example of an artist who relies on nature (lines 50-52)] painted naturalistic scenes. In fact, the “fragmented and distorted” works from his childhood, the “harbingers [forerunners] of his later genius” (lines 3-7) indicate that he did not paint naturalistic scenes or objects. His unrealistic portrait of Gertrude Stein provides additional evidence along these lines (lines 59-64). B: It is unlikely that the “unreflective virtuoso who relies on nature” would consciously refer to or employ defined aesthetic principles, even elementary ones. This is more likely true of an academic painter or “cerebral structurer.” C: Yes. The author centers the passage on a basic contrast between childhood or childlike art on one hand, and academic, literal, and precise art on the other. The “cerebral structurer” represents the latter, while the “unreflective virtuoso” represents the former category. An adult artist who draws like a child works not from rules and tradition, but with intuitive spontaneity and adventurousness (lines 22-29, 59-64). D: This choice is too extreme. According to the passage, the “unreflective virtuoso” Picasso still displayed mastery of his art. Compare this choice to answer choice C. 21

110. D A: This choice is partially correct, and so entirely wrong. Picasso’s early drawings are described as unusually skilled and his progress as spectacular (lines 41-45). However, the author rejects the standard story of Picasso’s precocious childhood talent (lines 13-15, 41-45). B: The author explicitly criticizes the standard story of Picasso’s precocious talent (lines 13-15, 41-45). Thus the author is not neutral. C: This directly contradicts the author’s rejection of the standard claim that Picasso displayed a precocious genius from early childhood (lines 13-15, 41-45). D: Yes. See the explanations above (and lines 13-15 for the standard story, lines 41-45 for the author’s opinion). 111. A A: Yes. Picasso himself claimed that his first drawings were academic, literal, and precise, and that he later transformed himself into an entirely different kind of painter (lines 22-29). He also argued that he had to entirely begin again as an adult painter (lines 15-21). This is contradicted by the author’s assertion that Picasso’s early works contained aspects of distortion, tricks, puns, playfulness, etc. that would appear in his adult creations. This indicates that Picasso was not a purely academic and realistic painter in his youth, and that there is some continuity between his youthful creations and his mature artistic style. B: Picasso’s later work was characterized by distortion (lines 61-64). Thus this choice is consistent, not inconsistent with the claim that we can see the seeds of Picasso’s adult style in his work as a child. C: The assertion made in the question is not directly relevant to the lack of drawings from Picasso’s first eight years. Therefore, it is not inconsistent. D: See the explanation for choice B. The author claims that Picasso transformed himself into an “unreflective virtuoso” (lines 47-54) who learned to “draw like a child” as an adult (lines 27-29). Thus the playfulness of his youthful graphic experimentation (lines 8-12) is entirely consistent with the claim that these experiments were precursors of his mature artistic style. 112. C A: Picasso indicates his goals through his statement that “it has taken me a whole lifetime to learn to draw like a child” (lines 27-29). Nothing in the passage indicates that piecing a work together out of found objects is a childlike approach to art. B: There is nothing particularly childlike about muted or subtle colors and elongated figures. C: Yes. Simple figures and bright colors represent a childlike style. D: Rendering “hard-edged” multifaceted perspectives would require a level of skill and purposefulness beyond the capacity of an imprecise and childlike painter. 113. B Note: The statement made in the question is new information. It is inconsistent with the claims made in the passage by Picasso and Apollinaire. Both claim that Picasso became an “unreflective virtuoso” later in life. A: This choice is inconsistent with the new claim made in the question. Raphael drew in an academic, precise, and realistic manner (lines 24-29), that is, as a “cerebral structurer.” B: Yes. Exuberant experimentation would more likely characterize an “unreflective virtuoso who relies on nature” than a “cerebral structurer who relies on understanding.” Picasso and Apollinaire claim that Picasso became an artist of the first type later in life, but “childlike” early drawings would indicate that he was such an artist from the beginning. C: As in choice A, this statement is consistent with Picasso and Apollinaire’s analyses, not with the new claim made in the question. D: Revision of Stein’s portrait from memory has no direct relevance to the argument about the nature of Picasso’s art in his childhood. This choice is out of the scope of the question. 114. C A: We must take the retort in the context of the passage and the main idea of the paragraph in which it appears. This choice is too literal. The author uses Stein’s portrait to illustrate Apollinaire’s claim that Picasso left behind realism as he metamorphosed into an artist of the first type (lines 50-64). This choice inappropriately indicates that the portrait was intended to be realistic at some time in the future. B: See the explanation for choice A. The portrait was not intended to be realistic or accurate. Secondly, the author 22

indicates that the portrait was finished at the time Picasso made his remark (line 63). C: Yes. The main idea of this (and the previous) paragraph is that Picasso left behind realism and became a more intuitive, spontaneous, and adventurous (i.e., childlike) painter in this stage of his life. Thus the example of the portrait is intended to illustrate how Picasso was now guided by concerns other than accuracy of a likeness or realism in his work. D: The author never suggests that Picasso intended to teach viewers of his paintings how to see art in a different way. Furthermore, the main idea of this part of the passage is that Picasso had left behind the academic or analytic approach to art for a more “unreflective” and childlike style. 115. A A: Yes. A cerebral structurer “relies on understanding” (line 48), and understanding would be necessary to both create and appreciate a novel written with many layers of meaning. B: Emotion is never directly discussed in the passage. There is nothing in the passage to suggest even indirectly that a novel written with academic precision [qualities which characterized Picasso in his cerebral phase (lines 24-26, 50-51)] would have great emotional power. C: While Picasso in his cerebral phase did paint in a realistic manner (lines 24-26), this choice does not match up with the contrast presented in the passage between the “unreflective virtuoso who relies on nature” and the “cerebral structurer who relies on understanding.” There is nothing uniquely cerebral, academic, or intellectual about terse (succinct) and realistic dialogue. D: Symbolism is never suggested by the passage as an aspect of a cerebral style. 116. B Note: The credited response must both accurately represent the author’s views in the passage and be relevant to the new information in the question. A: This choice contradicts the author’s argument that Picasso made striking progress after his early years, and that his first drawings as a child were not “frankly precocious” (lines 41-45). B: Yes. The author claims that Picasso’s childhood drawings were very skilled, but not “frankly precocious” (lines 41-45). Thus the new information in the question is consistent with the author’s views that the nine-year-old Picasso was not yet a uniquely skilled artist. C: If Picasso’s early work was in fact childlike [i.e., similar to the work of other children, in contradiction to Picasso’s own claims (lines 21-26)], this would indicate that the transition away from a cerebral structurer into a more “childlike” artist came earlier than the age of nine. This choice is inconsistent with the new information presented in the question. Furthermore, it is unclear from the passage exactly what the author’s position is on this topic. The author indicates that Picasso may in fact not have made a radical transformation from Apollinaire’s second kind of artist to the first kind (lines 8-12, 35-41). And yet the author cites Picasso and Apollinaire’s opinions that such a radical metamorphosis did occur without directly refuting them (lines 15-29, 46-67), and at times in apparent agreement. D: The new information in the question contradicts Picasso’s claim that his first drawings were not childlike (lines 21-26). 117. B A: This choice strengthens Picasso’s claim. In fact, the argument in this choice is made by Picasso himself (lines 15-17). B: Yes. If the naïveté which characterized children’s art does not in fact disappear, but is available on some level to the adult artist, it would weaken Picasso’s claim that the adult artist must “begin...from the beginning.” (Notice that the author himself makes the assertion given in the answer choice in lines 11-12.) C: Formal artistic training is not a relevant issue in the passage. The fact that some artists, trained or untrained, produce only childlike works provides no proof that those artists carried over that quality from their own childhood and did not “begin again” as adults. D: This choice would tend to strengthen Picasso’s claim that childlike qualities do not automatically carry over from childhood into adulthood. (It doesn’t quite qualify as a “strengthen” answer choice, however, as we do not know for sure that these artists are adults, or that if they are adults that they painted in a naïve, nonacademic style as children.)

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Passage VI 118. A A: Yes. The author introduces the passage by stating that the classification of Neandertals (as direct or distant relatives to modern humans) is “One of the hottest topics in anthropology today” (lines 1-3) and “the subject of a long and contentious debate” (lines 6-8). The rest of the passage either directly or indirectly addresses this issue. B: The author states that we do not know whether they are the direct ancestors of modern human beings (lines 1-12). This choice contradicts the main idea of the passage, which is that the classification of Neandertals has not yet been determined. C: Notice the word “primary” in the question. While the use of stone tools is a significant fact about Neandertals, it does not represent the primary significance of Neandertals. D: See the explanation for choice C. Burial of the dead is one clue to Neandertals’ potential humanness (lines 26-28), but it is not presented as the reason why Neandertals are so fascinating to anthropologists. 119. D A: While advances in the technology of stone tools occurred, the passage does not suggest that those advances were driven by a need for more sophisticated technology. The invention and refinement of the Levallois technique, for example, is not described as a response to changing conditions that necessitated more refined tools. B: While the passage describes one instance of slowing of the rate of change after a technological innovation (lines 6066), the author does not indicate that this is representative of rates of technological change in general. The word “normally” makes this choice incorrect. C: Radiocarbon dating is never mentioned in the passage. Be careful not to rely on outside knowledge. D: Yes. The author states that “Stone tools clearly signal the pace of change in human prehistory” (lines 55-56), indicating that the tools made and used by a species tell us something about that species. The passage as a whole describes aspects and characteristics of the Neandertal as a way of exploring their true nature. 120. A A: Yes. Use process of elimination! We know from the passage that Homo erectus predated the existence of modern humans (lines 3-5). The new information in the question tells us that they existed long before the appearance of the Neandertal, making it more likely (although not certain) that Homo erectus was an ancestor (direct or distant) to both Homo sapiens sapiens and the Neandertal. This choice is the best supported of the four options. B: The new information about Homo erectus tells us nothing new about the direct relationship between apes, the Neandertal, and modern humans. This choice is a more extreme version of choice A, and so is less likely. C: Even if Neandertal is not a direct ancestor of modern humans, the passage indicates that they would still be related, even if distantly (lines 10-12). The new information in the question about Homo erectus provides no clue that a completely separate human family tree exists. D: As in choice B, this statement is a much more extreme version of A, and so is even less supported by the passage. 121. B A: Tools remained few and crude until the development of the Levallois technique. The Neandertals refined that technique even further (lines 56-66). Thus this choice does not accurately describe the tools used by the Neandertals. B: Yes. The author claims that the Neandertals made a wide range of artifacts without providing specific examples of different kinds of tools. Thus, support for that claim is weak. C: The author never claims that the Neandertals were highly intelligent. The passage states that if cranial capacity were the only measure, they would have impressive mental powers (lines 21-25), but never claims that quantity is in fact the only measure. Furthermore, even if the passage did claim that the Neandertals were highly intelligent, that in and of itself would not provide strong support for claims about their resourceful tool-making. D: The Neandertals refined the Levallois technique; they did not invent it (lines 60-64).

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122. D A: The author discusses the large cranial capacity of the Neandertal, but not as a criterion for humanness (lines 20-25). In fact, it would make them different from modern humans, who have a smaller brain. B: The Neandertals did use tools, and we can speculate that they used tools to make tools with the Levallois technique (tools would be needed to strike flakes of stone from a rock). However, tool use is not given in the passage as a primary criterion for humanness. C: The author never discusses weaponry. The correct answer must be supported by the passage. D: Yes. The author describes burial ritual as a “uniquely human activity” (lines 27-28). Thus a toy that recited religious verses (performed rituals) would, among the four choices, be most human on that criterion.

Passage VII 123. D A: This statement is made in the passage (lines 1-2), but it is too narrow to be the main idea. B: Thomas and Beulah is in fact divided into two, almost-equal parts (lines 2-3), but this statement is too narrow to be the main idea of the entire passage. C: This is one theme of the passage (lines 6-8). However, it does not include the other major theme; the poems also describe the “single most important events and the resulting mind-sets in the separate lives of Thomas and Beulah” (lines 4-8). Notice the words “not only” and “but also” in the cited sentences, which indicate that there are two main themes in the passage. The correct answer must include both. D: Yes. The “psychological and emotional lives of two individuals” would include both their separate and shared experiences. 124. C A: Events connected to beautiful objects (a silk handkerchief, the Eiffel Tower, a flower, a music box) are linked to Beulah (lines 15-20, 44-47), while pain and guilt are associated with Thomas (lines 20-27, 41-43). B: Beulah desires a sensuous life (lines 44-47), while events linked to ships and water characterize Thomas’s existence (lines 41-43). C: Yes. The author introduces the passage with the statement that the collection of poems “presents not only the single most important events and the resultant mind-sets” in each separate life, but also “the significant events of their shared lives” (lines 4-8). The events described in the passage are both ordinary [parents quarreling, sweeping the floor (lines 47-53)] and extraordinary [the death of a friend (lines 9-12)]. D: This choice is too extreme. Events linked to water and ships are connected in the passage to a gap in Thomas’s life between desire and fulfillment (lines 31-41). However, the link as described is limited to these poems and these characters. The author does not suggest that this association always exists. 125. D A: This choice is only partially correct. Beulah’s preference for a pianola does represent her mixed feeling about the marriage (lines 56-66). However, the “one pierced cry” refers to Thomas’s own individual association between music and guilt (lines 20-27). Notice where each reference appears in the passage. Only the last paragraph discusses the pair’s mixed feelings about their marriage. B: Thomas’s “always jiving” represents his own individual life and experiences (lines 27-30). The passage never mentions Beulah’s desire “to hear wine pouring.” C: This is the right answer to the wrong question. These quotes illustrate Thomas’s (lines 36-37) and Beulah’s (lines 5053) individual lives and thoughts, not their mixed feelings about their marriage. D: Yes. Both quotes come from the final paragraph (lines 63-64, 66-69). The purpose of that paragraph is to present the two characters’ perspectives on their marriage, and to portray their mixed feelings. Thomas experiences longing as he wraps the yellow scarf around Beulah’s shoulders, and yet he wonders, “How did I get here?” Beulah turns her back on her father reluctantly, as she walks towards her new life with Thomas.

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126. A A: Yes. Dove describes a series of events and experiences in Thomas and Beulah’s shared and individual lives that show a mixture of positive and negative emotions, joys and uncertainties (see for example the last paragraph). If Dove’s insights into her characters’ thoughts and emotions are accurate, they should apply to other couples as well. B: This choice is too extreme. The passage describes unhappiness in the lives of both characters, but does not go so far as to suggest that people as a whole are rarely happy. C: This choice is too extreme. While Thomas and Beulah both have ambivalent feelings about their marriage, the author does not describe them as fundamentally mismatched emotionally. Furthermore, the author does not generalize from Thomas and Beulah’s experiences to all couples. D: The passage indicates that Thomas, when he marries, feels he has “raised a mast and tied himself to it” (lines 31-35, 37-39), suggesting a desire for freedom. However, the author does not say that this is true of all married men. This choice is also too extreme. 127. A Note: Neither quote given in the question is from the passage. A: Yes. The first quote in the question corresponds to Thomas’s pain and helplessness as described in such lines as “one pierced cry” (line 22), “sound quivered like a rope stretched clear to land…a man gurgling air,” and “too frail for combat” (lines 40-41). The second quote corresponds to Beulah’s desires for a beautiful and sensuous life, as shown in such lines as “she would make it to Paris one day” (line 50), her thoughts of China as she dies (lines 53-55), and her preference for “a pianola and scent in a sky-colored flask” (line 65). She is also associated with images of flowers (lines 19, 52, and the title “Canary in Bloom”). B: See the explanation for choice A. Here, the appropriate order is reversed. C: “Canary in Bloom” presents Beulah (paragraph 5), and “Mandolin” Thomas (paragraph 2). See the explanation for choice A for discussion of the quotes given in the question. D: No images associated with Lem are given in the passage; his death is described only in order to show how it affects Thomas (lines 9-12, 20-30). The second quote in the question corresponds to Beulah, not to Thomas (see the explanation for choice A).

Passage VIII 128. A Item I:

Item II:

Item III:

Yes. In the second paragraph, the author explains that ten percent of the sunlight reaching the Earth is fixed by photosynthesis. Ten percent of that energy is passed on to animals that eat the plants, ten percent of what is left goes to the animals that eat the plant eaters, and so on up to the top of the energy pyramid (lines 7-23). While the biomass pyramid on land is broad at the bottom and narrow at the top (lines 28-40) (and inverted in the sea), the author does not indicate that each successive layer weighs ten percent of the layer below it (or above it, in the ocean). There is no specific proportional relationship given in the passage. The ten percent rule relates only to the energy pyramid (lines 7-23). The author does not suggest a relationship between the energy available at each level and the number of species at that same level.

129. B Note: Eliminate the items that are consistent with the passage. Item I: The food web as described in the passage has for the most part larger animals feeding on smaller and more numerous animals and plants (lines 13-23, 52-60). Copepods are small animals, thus it is consistent with the passage that they feed on algae; in fact the passage itself states that they do so (lines 52-54). Item II: As in Item I, killer whales would be expected to feed on seals, according to the author’s description of the food chain. It is also directly stated in the passage (lines 59-60). Item III: Yes. Jackals are not top predators, and according to the passage, except for the top predators (lines 17-20), each level feeds on smaller and more numerous species below and is fed upon by those above (lines 13-20). Ticks feeding on jackals would invert this relationship, and so violate the hierarchical relationship posited by the author. Be careful not to use outside knowledge to choose or eliminate choices. 26

130. D A: The correct answer will be inconsistent with the statement cited in the question. The passage already states that algae are better than land plants at capturing solar energy (lines 49-51); if they were to capture almost all of it, it would have no effect on the author’s claim that the oceanic biomass pyramid is inverted. B: The passage states that the top carnivores on land skirt the edge of extinction (not the same, in context, as being on the brink of extinction). First of all, this does not imply that top sea carnivores are not close to that edge. Secondly, even if it did, not all whales are top carnivores. Finally and most importantly, the author mentions extinction in the context of the energy pyramid, which has no direct relevance to the shape of the biomass pyramid. Only the latter is relevant to this question. Thus the statement that whales are close to extinction has no effect on the author’s argument about inversion of the biomass pyramid. C: Zooplankton feed on phytoplankton (lines 49-54). Thus in the inverted pyramid, we would expect zooplankton to have a greater bulk. This choice is consistent, not inconsistent with the inverted pyramid. D: Yes. Marine mammals feed on invertebrate fish (lines 56-59). According to the inverted biomass pyramid, we would expect marine mammals to have a greater bulk than invertebrate fish. Thus this choice is inconsistent with the author’s statement that the pyramid is inverted. 131. C A: If some carnivores could utilize the sun’s energy directly, the percentage of the sun’s total energy utilized at the top of the pyramid would increase. However, plants would not be utilizing less (there is plenty to go around), and so the pyramid would not invert; the base would stay the same. B: The pyramid would change at least to some extent; see the explanation for choice A. C: Yes. If some top carnivores could consume a greater percentage of the sun’s energy by utilizing it directly, the top of the pyramid would broaden to include that increased amount. D: For the pyramid to flatten, a level would have to be removed (e.g., if photosynthesizing top carnivores were reclassified as plants and included in that level). Neither the passage nor the question gives us any reason to believe reclassification or collapsing of categories would occur. 132. A A: Yes. Ten percent of the sunlight reaching the Earth serves life directly through photosynthesis. Ten percent of that energy is converted to food energy (lines 7-14), or one percent of the total sunlight reaching the Earth’s surface. B: Both land plants and algae are described as “green plants” (lines 10, 63). Plants use ten percent of the sun’s energy that reaches the Earth. Animals that eat the plants get ten percent of the total available energy, and so on down the line (lines 7-20). Thus each group is able to use or absorb ten percent of the total energy from the sun available at that level— equivalent levels of efficiency. C: The passage does not discuss sources of energy available to plants other than the sun. D: The passage never indicates that top carnivores get any energy directly from the sun.

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Passage IX 133. A A: Yes. The author argues that giving may not be selfless in some cases. Beneficent acts may be performed because a subordinate beneficiary will in turn “flatter the ego” of the one in power (lines 17-21) and confirm his or her dominance. In this example, the judge is granting leniency to those who acknowledge his or her authority and flatter his or her ego through respectful forms of address. B: The boss is not giving anything to the workers, but simply commanding or requesting certain behavior on their part. C: This is, as described, a purely altruistic act, with no expectation of subservience on the part of homeless people in return. D: Both the husband and the wife can benefit from the vacation; the husband is not granting the vacation to his wife in the context of an unequal power relationship. This example appears to be more of a case of shared interests than of an expression of power. 134. D A: The author explains this assertion in lines 5-9, and gives specific examples in the second paragraph. B: In lines 29-31, the author lists three additional aspects of friendship other than caring for each other. C: The author describes the case of the selfish man and his wife in order to support this assertion (lines 15-25). D: Yes. The author makes this assertion in lines 34-35 without supporting it with explanation or example. 135. B A: The author argues that friendship is a moral activity because it entails caring about and giving to someone else for their own sake (lines 5-8, 38-41). Disagreement over shared activities does not necessarily indicate that these friends do not truly care about each other in this way. If, however, they are not truly friends in the highest moral sense, this does not undermine the author’s definition of moral friendship. This choice is not inconsistent with the passage. B: Yes. Frequent arguments over how they will spend time together indicates that a division between self- and other-interest may still exist. Thus this choice would weaken the author’s argument in lines 52-54. C: Dave and the author’s inability to perceive the interests of the other as their own would be consistent, not inconsistent with the claim that not everyone is capable of true and full moral friendship. D: This choice is consistent with the new information in the question. Through arguing with Dave, the author may be exploring and defining his own real interests and desires. 136. A A: Yes. If one identifies one’s own interests with those of another, it is reasonable to expect that those two people will naturally tend to cooperate to serve those interests, and that grounds for conflict would be lessened. Notice that the theme of this question is the same as that in Question 135 (cooperation and conflict). Compare your answers on similar questions to be sure that they are consistent with each other. B: The author does not suggest that selfishness must always exist or be expressed. C: According to the passage, friendship exists when people like each other, care about each other, and act in each other’s interests. The author never indicates that this is possible only when people know each other very well. D: Coming to see the interests of a friend as your own does not necessarily entail becoming similar to that friend in all or most ways. 137. D A: Transitivity of friendship in this mathematical sense is never discussed in the passage. The author gives us no reason to believe that two people who have a friend in common will themselves have and share all of the qualities necessary for friendship. B: If Mary and Bill are friends, they satisfy each other’s needs (lines 5-8, 42-47), but those needs may be different for each person (see also Question 136, explanation D). C: True friendship cannot exist without equality and reciprocity, as shown in the example of the power-hungry husband (lines 11-25). D: Yes. In lines 42-49, the author indicates through personal example that friendship cannot be one-sided; the trust, caring and affection must go both ways.

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BIOLOGICAL SCIENCES Passage I 138. B. Hypothesis 2 describes effector cells as those that attack and destroy antigens, and suppressor cells as those that limit the action of effector cells. It further states that the loss or inactivation of suppressor cells may lead to autoimmune disease. Therefore, individuals with both autoimmune thyroid and autoimmune liver disease must have lost or inactivated suppressor cells that respond to both of these tissues. Note that the loss or inactivation of effector cells (choices C and D) would reduce the possibility of autoimmune disease. 139. A. Hypothesis 1 states that circulating lymphocytes are activated after encountering their specific antigens, and that clones activated by the body’s own tissues (antigens) are deleted. Thus antigens that do not circulate could never encounter lymphocytes specific for them; these self-reactive lymphocytes would not be activated and would not be deleted (selftolerance would not develop). Corneal tissue is a living tissue (choice B can be eliminated). Choice C describes how the cornea might be protected against infection, but does not address the issue of the failure to develop self-tolerance, and choice D suggests that self-tolerance does develop, just through a different mechanism. 140. D. This is an ambiguous question because there is no “normal mechanism of self-tolerance” described in the passage. Judging by the answer choices available, it would appear that the question is really asking about the body’s normal mechanism of responding to foreign antigen in general, i.e., immune system function. Essentially, this is described in Hypothesis 1: “Identical groups (clones) of circulating lymphocytes remain inactive until they encounter their specific antigens, after which they proliferate.” Clones that do not encounter their specific antigen remain inactive (choice C can be eliminated), and this includes both B and T lymphocytes (choice A can be eliminated). If the antigen is from the body’s own tissues, and lymphocytes specific for that antigen were activated, this would be a mechanism for autoimmunity, not for self-tolerance (choice B can be eliminated). 141. C. Hypothesis 2 states that if the balance between effector and suppressor cells for a particular tissue is disturbed, an autoimmune reaction to that tissue may result. A disturbance to that balance could be a loss of suppressor cells (given in the passage), or it could be an increase in effector cells. Cells obtained from an identical twin would be identical to the original cells, and would be virtually ignored by the body (choice A can be eliminated). Cells from a new tissue may increase the number of effector cells for the new tissue, but would not increase the number of effector cells for the original tissue, nor would they decrease the number of suppressor cells for the original tissue; autoimmunity would not result (choice B can be eliminated). To decide between the remaining choices is difficult because an explanation of “cross-react” is not provided in the passage, and this is a vague term at best. Assuming that “cross-react” means to elicit a similar immunological effect, the injection of a foreign substance that cross-reacts with the original tissue would lead to an increase in effector cells for that tissue, thereby disturbing the balance between effectors and suppressors and leading to autoimmunity (choice C is correct and D is eliminated). 142. D. Based on the descriptions of the hypotheses given in the passage, the two do not seem mutually exclusive, rather, they seem mutually supportive. Hypothesis 1 occurs very early in life, and Hypothesis 2 seems to occur throughout life; thus, Hypothesis 2 may correct for errors made with Hypothesis 1; that is, failure to delete auto-reactive clones early on can be compensated for by creating suppressor cells to prevent the activity of auto-reactive effector cells. Note that although choice A is tempting, the passage suggests that this does not occur (“if clonal deletion...is hindered, these lymphocytes will incorrectly recognize a specific body tissue as foreign...”). 143. A. Hypothesis 1 states that self-tolerance is generated when the body is processing T and B lymphocytes. T-cell processing occurs in the thymus (remember: “T” is for “thymus”), and B-cell processing occurs in the bone marrow (remember: “B” is for “bone marrow”).

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Passage II 144. D. First, compare Compound Y to Scheme A and determine which compound it most closely resembles. Compound IV is identical to Compound Y except that it bears a methyl substituent on C-1 rather than an ethyl substituent (note their stereochemistries). Tracing the reaction pathway backwards, we see that the methyl substituent is introduced using CH3MgBr in Step 2. By analogy, Compound Y could be produced using CH3CH2MgBr in Step 2. 145. C. Nucleophilic addition reactions (Step 2) and olefination reactions (Step 4) occur at carbonyls. The ketal group protects the carbonyl at C-8 from undergoing these reactions. While ketals are not inert (eliminating choice A), they are generally less reactive than carbonyls (eliminating choice B) and are used as carbonyl-protecting groups. Ketals are converted back into their corresponding ketones by treatment with acid (eliminating choice D). 146. B. A broad absorption at 3300 cm–1 in an IR spectrum is indicative of an alcohol (O–H stretching), while a sharp absorption at 1700 cm–1 is indicative of a carbonyl (C=O stretching). 147. A. In the second half of Step 6, H2O is used to neutralize the alkoxide anion generated in the first half of the reaction. Replacing H2O with D2O will result in the deuterated alcohol:

OCH3 H O

1) LiAlH4

OCH3

OCH3

2) D2O

OCH3

OCH3

H +Li–O

H

OCH3

DO

Passage III 148. C. The binding of ACh to receptors on muscle cells initiates an action potential that results in muscle contraction. Since Drug Y blocks the ACh receptors, no action potential would be initiated and no contraction would result (choice C is correct and D is wrong). Drug Y has nothing to do with the synthesis of acetylcholinesterase (choice A can be eliminated), and the T-tubules of muscle cells are simply invaginations of the plasma membrane deep into the cell interior, they are not sites of Ca2+ influx (choice B is false). 149. A. Since Drug X is an acetylcholinesterase inhibitor, it prevents the destruction of ACh, and it therefore seems likely that the concentration of ACh in the synapse would increase. There is nothing to suggest that choice B, C, or D is true. 150. D. First, the action of Drug Y is to block ACh receptors. It does not increase ACh production (choice B is false), nor does it degrade ACh (choice C is false). The action of Drug X is to increase ACh by preventing its degradation; however, the excess ACh cannot overcome the effects of Drug Y because, according to the passage, Drug Y blocks the receptors noncompetitively (noncompetitive blockers bind somewhere other than the ligand-binding site and render the receptor unable to bind ligand; choice A is false). 151. B. A competitive inhibitor binds at the active site of an enzyme and prevents its activity. However, if the concentration of substrate for that enzyme is increased, it is more probable that the enzyme will bind a substrate molecule instead of an inhibitor molecule, and the effect of the inhibitor will be counteracted. Since in this case an increase in ACh (the substrate) decreases the effect of the inhibitor (Drug X), it must be a case of competitive inhibition. Increasing substrate concentration would have no effect on an irreversible or noncompetitive inhibitor (choices A and C are false), and there is no such thing as a “mixed competitive/noncompetitive inhibitor” (choice D is false).

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152. B. In the experiment described in the passage, the rate of learning was determined by the number of trials necessary for rats to learn to avoid an unpleasant stimulus. If the number of trials decreased, learning was assumed to have taken place more quickly. Based on that assumption, Drug Y prevented learning (the number of trials necessary increased at every tested dose), and Drug X improved learning at low to moderate doses and prevented it at higher doses. Since Drug Y blocks ACh receptors, choices C and D are false. Drug X increases ACh concentration; choice B is true and A is false. 153. A. Drug Y is a noncompetitive ACh-receptor blocker, meaning that it binds somewhere other than the ACh-binding site and prevents the receptor from binding ACh (probably through a conformational change in the receptor). Therefore, regardless of any increases in the amount of ACh present (as might be achieved by increasing the amount of Drug X), the ACh receptors will remain blocked (choice B can be eliminated). Administering a competitive-receptor stimulator would not counteract Drug Y, since the competitive stimulator would bind at the ACh-binding site, but the binding site is not available due to the action of Drug Y (choice C can be eliminated). Administering another noncompetitive inhibitor that acts at the same site as Drug Y might displace Drug Y, but only with another inhibitor! The receptors would still remain blocked; this would not counteract the effects of Drug Y (choice D can be eliminated). The only remaining choice is choice A; increase the rate of metabolism of Drug Y and its elimination from the body. 154. D. The ACh receptor is found in the plasma membrane, so the deletion of this gene would result in plasma membranes that lacked ACh receptors and that could be considered “abnormal.” The question is somewhat ambiguous since the plasma membrane is not typically considered to be an organelle; however, the ACh receptor is not associated with the organelles in choices A, B, or C, thus making D the best answer choice available.

Passage IV 155. A. The primary reason for sexual reproduction is to increase genetic variability within the population (choice A is true and C is false). The micronucleus is described in the passage as “germ-line,” not somatic (choice B is false), and sexual reproduction is not a means of rapidly increasing population size (that’s the function of binary fission, choice D is false). 156. C. The primary characteristic of extrachromosomal pieces of DNA is that they are able to undergo replication independently of the chromosomal DNA. This does not imply they are nonfunctional (choice B) nor rearranged (choice D). Some are nonlinear (e.g., plasmids), but they do not have to be (choice C is better than choice A). 157. A. Since centromeres are always present, it seems unlikely that they would trigger additional S phases. If that were the case, there would always be additional S phases, regardless of the size of the macronucleus (choice B is false). Furthermore, the presence of high concentrations of DNA or enzymes in the micronucleus would not affect the macronucleus (choices C and D are false). It seems much more likely that this system operates on a sort of negativefeedback basis, whereby a low concentration of DNA in the macronucleus triggers extra S phases to increase DNA content, and that high DNA content inhibits the S phase. 158. A. The cytoplasm of ova-producing cells (oogonia) does not contain any nuclear material (choice B is false), they do not add or skip S phases (choice C is false), and they do not undergo mitosis (oogonia undergo meiosis, choice D is false). During the meiotic divisions of oogenesis, the cytoplasm is divided unequally between daughter cells, with one cell receiving almost all the cytoplasm (the developing oocyte) and one cell receiving very little (the polar body). In this respect they are similar to the macronuclei of Tetrahymena. 159. D. Since the macronucleus does not undergo meiosis, it does not undergo crossing over (choice B is false). Furthermore, macronuclei are derived from micronuclei in Step 6 of conjugation, so they do not have distinct genetic origins (choice C is false). During binary fission the chromosomes are replicated and apportioned; macronuclear chromosomes are not lost, but they might be distributed differently between daughter cells (choice D is better than A). 160. C. The passage states that the macronucleus is the site of gene expression (i.e., protein synthesis) during the vegetative state, so it must retain DNA sequences involved in that process, such as those for transcription, translation, and ribosome production (choices A, B, and D are false). However, since the macronucleus does not under meiosis, it seems likely that these genes are eliminated during macronuclear differentiation. 31

161. B. Since the old macronuclei are destroyed during sexual reproduction, their genetic make-up cannot contribute to the genotype distribution in the F1 generation; this is due solely to the interaction of the micronuclei. A cross between two heterozygotes results in a 3:1 phenotype distribution in the F1 generation, so 75% of the offspring will express the dominant trait and 25% will express the recessive trait.

Independent Questions 162. A. In a solution at neutral pH, glycine (like all amino acids) will have a positively-charged amino group and a negatively-charged carboxylate group. This dipolar nature gives it a high dipole moment and high water solubility. 163. A. a-Carbohydrates differ from corresponding b-carbohydrates in the stereochemistry about the anomeric carbon (C-1). 6

6

COOH

H 4

OH

5

COOH

O

H OH

H

OR

H

1

4

H 3

H

OH

2

5

O

H OH

b -D-glucuronide

1

OR 3

H

OH

H

H

2

OH

a -D-glucuronide

164. A. The bigger (or higher) the taxonomic group, the less alike the members are. This is true for everything from basic appearance, to behavior, to protein amino acid sequence, etc. For example, kingdom Animalia includes insects, birds, and humans (organisms with dramatic differences), but order Primates (within kingdom Animalia) includes gorillas and humans (organisms that are fairly similar to one another). Of the choices given, “phyla” is the highest taxonomic group. Therefore, the greatest number of amino acid differences will most likely be found between members of different phyla. 165. C. The ability to initiate conjugation is a trait encoded by the F (fertility) plasmid. Transformation (choice A) is the uptake of naked DNA from the environment by bacteria and occurs when the bacteria are subjected to unfavorable conditions. Transduction (choice B) is the transfer of genetic material from one host to a new host via viral infection, and translocation (choice D) is the movement of a ribosome along a strand of mRNA during protein translation.

Passage V 166. D. If the maximum pH for the reaction of N-methylmethanesulfonamide (pKa = 11.79) with alkyl bromide is 12.2, then from the equation pH max = 12 ( pH i + pKa ) , we have pH i = 2 ◊ pH max - pKa = 2(12.2) - 11.79 = 12.61

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167. B. Replacing the central oxygen atom of acetic anhydride with a nitrogen atom produces an imide. The other nitrogen-containing functional groups are also shown below: O R

O O

O R

O

O

R

N

anhydride

R

R

NH2 amide

H imide H N R

NH2 CH3

R

imine

CH2

eneamine

168. D. Carboxylic acid anhydrides can be formed by heating the corresponding carboxylic acids. The reversal of Equation 3 requires the neutralization of the carboxylates followed by heating to produce the anhydride: O

O H3O

2 H3C

2

O

H3C

O

O OH

D

+ H2O H3C

O

CH3

169. C. The addition–elimination of an amine with acetic anhydride begins with addition of the nitrogen atom of the amine to the carbon atom of the carbonyl to form a tetrahedral intermediate. Elimination of acetate then produces the corresponding amide: O

O

O

O

O

O + H3C

O H2N

CH3 R

H3C H

O N

CH3

H3C

NHR

H3C

O

R

H H2O

170. A. Amines bonded to only one alkyl group (RNH2) are called primary amines. Amines bonded to two alkyl groups (R2NH) are called secondary amines. Note from Table 1 that aniline and benzylamine are both primary amines. [Since the nitrogen atom of aniline is bonded directly to an aromatic ring, aniline is considered an aromatic amine (choice C).]

Passage VI 171. C. The pulmonary artery carries deoxygenated blood from the right side of the heart to the lungs to pick up oxygen and to eliminate carbon dioxide. So compared to aortic blood (left side of the heart to the body), the pulmonary arterial blood has less oxygen (choices A and B are false) and more carbon dioxide (choice D is false). Not that it matters for this particular question, but carbon dioxide is converted to carbonic acid, which dissociates into bicarbonate ion and hydrogen ion, thus a higher carbon dioxide concentration is always associated with a lower pH.

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172. A. When the diaphragm contracts, it flattens (moves downward). This increases the size of the thoracic cavity and decreases the pressures there (choices C and D are false). The subsequent decrease in alveolar pressure causes air to move into the lungs (inspiration, choice B is false). 173. B. Remember that substances always flow from high-pressure areas to low-pressure areas. Thus, air flows out of the lungs when the alveolar pressure (PA) is higher than atmospheric pressure (P). If PA were lower than P, air would flow into the lungs (choice A is false), if IPP (intrapleural pressure) were lower than PA, the lungs would expand until their outer surface contacted the inner surface of the thoracic cavity (note that this is the normal situation, choice C is eliminated), and if PA were equal to P, there would be no net movement of air at all (choice D is false). 174. C. The passage states that aortic blood pressure is 100 mmHg, while pulmonary arterial pressure is only 20 mmHg. To move blood against the higher pressure of the aorta, the left ventricle must generate more force and a higher blood pressure than the right ventricle (choices B and D are eliminated). In order to do this it must have a thicker (more muscular) wall (choice A is eliminated). 175. C. As blood passes by the alveoli of the lungs, oxygen diffuses from the alveoli to the blood (alveolar PO2 decreases) and carbon dioxide diffuses from the blood to the alveoli (alveolar PCO2 increases). If blood flow to a particular region of the lung were blocked, no oxygen would be removed from the alveoli (choices B and D can be eliminated) and no carbon dioxide would be added (choice A can be eliminated).

Passage VII 176. D. Judging by its name, a hepatopancreas is a combination of a liver and a pancreas. These organs belong to the digestive system. 177. C. The passage states that insects have respiratory systems that provide gaseous oxygen directly to tissues, thus lungs would not be seen (choice A is eliminated). Insects have exoskeletons, not endoskeletons (choice B is eliminated), and malphigian tubules instead of kidneys (choice D is eliminated). Blood-filled sinuses are compatible with an open circulatory system. 178. C. The function of the malpighian tubules of insects is described as the collection of waste fluids and is similar to the mammalian kidney. The posterior intestines of insects is analogous to the bladder (choice A) since this is where the waste fluids are deposited, and the colon and the liver are not involved in collecting or processing waste fluids (choice B and D are false). 179. A. Striated (skeletal) muscle would allow for increased voluntary activity. The muscle of the digestive system is smooth, not striated (choice B is false), and pumping in the circulatory system is accomplished by means of a heart made out of cardiac muscle tissue (choice C is false). The development of a coelom has nothing to do with muscle tissue (choice D is irrelevant and eliminated). Note that even though cardiac muscle tissue is striated, the term “striated muscle” more properly refers to skeletal muscle. 180. C. The passage states that the arthropods are considered successful because of their diversity (Statement I is true) and their variety of occupied niches (Statement III is true). The passage also mentions that the species longevity contributes to the success of arthropods, but not the longevity of individuals (Statement II is false). 181. B. Nervous tissue is not derived from the coelom (choice C is false), nor are both the coelom and nervous tissue derived from mesoderm. The coelom is derived from mesoderm, but nervous tissue is derived from ectoderm (choice D is false). They are completely independent processes (choice B is true). Choice A is also true, but choice B is better since it more directly answers the question.

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Passage VIII 182. D. According to the passage, aspirin is an anti-inflammatory agent because it blocks the synthesis of prostaglandins. Therefore, prostaglandins must enhance the effects of inflammation. 183. B. Reaction 3 demonstrates the ammonolysis (cleavage by an amine) of aspirin (an ester) by active enzyme (an amine), eliminating choice A. The products of the reaction are inactive acetylated enzyme (an amide) and aspirin (a phenol), eliminating choices C and D. 184. A. The esterification (formation of an ester) reaction between the carboxylic acid of aspirin and the hydroxy group of salicylic acid produces choice A and water, as shown below: O O

OH

OH

O

O HO OH

O O

+

O

C O

+ H2 O

C CH3

O

CH3

salicylic acid

aspirin

185. C. The passage states that aspirin is purified by dissolving it in basic solution to produce its water-soluble sodium salt, eliminating choices B and D. The polymer impurities are not soluble and are filtered away, again eliminating choices B and D. Salicylic acid, having a carboxylic acid, will also be converted to a water-soluble sodium salt (eliminating choice A) and is therefore the most likely impurity in the final product. 186. D. Reaction 1 is a reversible reaction, as depicted, eliminating choice A. Early termination of Reaction 1 could lead to unused salicylic acid as a contaminant, eliminating choice B. Hydrolysis of the aspirin ester during purification would produce salicylic acid, eliminating choice C. However, the unfavorable decarboxylation (loss of carboxylate) of aspirin would not lead to salicylic acid, as shown below: O OH

O O

C O

– CO2

CH3

O

CH3

Independent Questions 187. C. The function of the gallbladder is to store and concentrate bile produced by the liver. Bile is secreted into the small intestine in response to the presence of fats; its function is to emulsify fats so that they are more easily digested by pancreatic lipases. Therefore, a person whose gallbladder has been removed should restrict the consumption of fats (triglycerides). 188. C. Of the choices given, the only one that corresponds to intestines is choice C. There is nothing similar to a shark’s spiral valve in the liver or pancreas (choice A is false). The given description of the spiral valve most closely matches the description and function of the mammalian intestinal villi and convolutions (folds). The stomach mucosa is also folded, but it’s in the stomach, not the intestines (choice C is better than B), and the pyloric valve is a sphincter and is neither similar in function nor structure to the description of the spiral valve (choice D is false).

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189. B. The prefix “tert-butyl” denotes a tertiary carbon bonded to three methyl groups, while the “amine” suffix indicates the presence of an –NH2 group: CH3 H3C

C

NH2

=

(CH3)3CNH2

CH3

190. D. Inbreeding tends to decrease the genetic diversity of a population, not increase it (choice A is false), and the rate of spontaneous mutation has to do with the error rate of DNA polymerase, not the type of breeding occurring (choice C is false). There is no reason to assume or infer that levels of aggression would increase (choice B is false). However, inbreeding does increase the incidence of expression of recessive traits since there tend to be more heterozygous carriers mating with one another. 191. B. It is precisely because the skin is relatively impermeable to water that the need for sweat ducts exists. Sweat glands are exocrine glands; they secrete their product through a duct onto the surface of the body (choice B is true). Salt would limit the diffusion of sweat since it would restrict its passage across cell membranes (choice A is false). For osmosis to occur, the barrier (skin in this case) must be water-permeable (which it isn’t, choice C is false), and sweating occurs virtually all over the body (and all areas are impermeable to water). Consider the palms of the hands and the soles of the feet; these areas are covered with extremely thick skin to prevent damage by abrasion, yet they also sweat freely (choice D is false).

Passage IX 192. B. Since nothing was boiled in Experiment 1a, nothing was denatured (choice A is eliminated), and since mitochondria are not involved in glucose production, the relative lack of mitochondria should not be an issue (choice D is eliminated). Experiment 1 (overall) shows us that membranes, cytosol, adrenaline, and glycogen are all required for glucose production. Experiment 1a is missing only the membranes, and the passage states that the membrane fraction is the site of the adrenaline receptors. In the absence of receptors, adrenaline cannot stimulate glucose production, regardless of the quantity of adrenaline present (choice C can be eliminated). This leaves only choice B. Since no part of Experiment 1 involves an intact cell, there cannot be a direct physical connection between glucose production and the binding of adrenaline. Thus there must be something produced as a result of adrenaline binding to its receptor—some chemical—that can stimulate glucose production; this chemical is the second messenger. Since Experiment 1a lacks the adrenaline receptor and no glucose was produced, it is likely that the second messenger was not formed. 193. B. This question is really just a fancy way of asking where glycolysis takes place. For the purposes of review, no portion of glucose metabolism occurs in the nucleus or on the cell membrane (choices A and D are false); glycolysis occurs in the cytosol, pyruvate dehydrogenase and the Krebs cycle take place in the mitochondrial matrix, and electron transport and oxidative phosphorylation take place along the inner membrane of the mitochondria. 194. C. The question states that adenylate cyclase is the enzyme that catalyzes the production of cAMP, not the enzyme that breaks down glycogen (choice D is false). Furthermore, it is described as a protein complex, so it would be denatured at high temperatures (it is not heat stable, choice B is false). Choice C seems correct based on the results of Experiment 2; the combination of membrane fractions, adrenaline, and ATP resulted in the production of cAMP; thus, adenylate cyclase must have been activated. There is nothing in the passage to support choice A; in fact, adenylate cyclase can be stimulated through a number of different pathways and by a number of different substances, but 5¢-AMP is not one of them. 195. A. In the diagram, Structure X is found outside the cell. The only one of the choices listed that would be found outside the cell is adrenaline. Glucose (choice B), ATP (choice C), and cAMP (choice D) would all be located in the cytosol.

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196. A. Since adrenaline is present in all four parts of Experiment 1, the necessity of its presence is not being tested. The necessity of the cytosol is tested for in Experiment 1b (choice B is false), Experiment 1d tests whether boiling the membranes has an effect (choice C is false), and Experiment 1c supports the hypothesis that membrane and cytosol must both be present for glycogen breakdown to occur (choice D is false). If we wanted to test for the necessity of adrenaline, we would have to run an experiment to measure the amount of glucose produced when only membranes, cytosol, and glycogen were present (i.e., in the absence of adrenaline). 197. D. In Experiment 2, constituents of the boiled extract were combined with cytosol and glycogen to determine their effects on glucose production. Experiment 2a shows that cAMP results in glucose production; note that this cAMP is boiled cAMP and that Item III is identical to Experiment 2a. If Item III is true, then choices A and B can be eliminated, and clearly Item II is false (without the enzymes present in the cytosol glucose cannot be produced). Item I is essentially the same thing as Item III since the boiled extract contains cAMP, so Item I is also true, and choice C can be eliminated. 198. C. The passage states that primary messengers are hormones. Of the four choices, only adrenaline (choice C) is a hormone, so this must be the correct answer. [Note that the passage states clearly that cAMP (choice D) is a secondary messenger.]

Passage X 199. A. The membrane is described as a lipid bilayer, which is the same description given to eukaryotic cell membranes. Prokaryotic cell walls (choice B) are made of peptidoglycan, bacterial spore coats (choice C) are also made of peptidoglycan, and bacterial capsules (choice D) are just a layer of sticky carbohydrates outside the cell wall that aid the bacterium in adhesion. 200. D. Figure 1 shows that the virion contains both RNA (genetic material) and reverse transcriptase (an enzyme), so the conclusions presented in choices A and C are clearly not supported by the passage. Choice B is tempting because proteins (reverse transcriptase, the core protein, the lipid bilayer proteins) are shown associated with the virion, but nowhere does the passage suggest that these proteins were synthesized by the virion. Choice D is supported by the passage; these organisms could not be grown in sterilized, noncellular media, implying that they need a cellular host to grow and reproduce. This is the definition of a parasite. 201. A. All known eukaryotic cells can be visualized with a light microscope; since this object could not be seen in this manner, it must be smaller than all known eukaryotic cells (choice A is correct and C is wrong). Cocci bacteria can be seen with a light microscope; again, if this object cannot, then it cannot be approximately the size of a typical coccus (choice B is wrong). Choice D presents a problem: we know that this organism must be smaller than bacteria, since bacteria can be seen with a light microscope and this organism can’t. However we cannot say with certainty that this organism is larger than all known bacteriophage, since bacteriophage are also not visible with a light microscope, and we are not given any information regarding the relative size of bacteriophage to this organism. This makes choice A better than choice D. 202. D. The passage states that attempts to grow the virions in sterilized, noncellular growth media were unsuccessful; therefore, growth media for this organism must contain cells (choices A, B, and C are eliminated). 203. B. Viral proteins are the result of the expression of viral genes (nucleic acid). Transcription and translation of host genes (nucleic acid) result in the production of host proteins (choices A and C are eliminated). While both transcription and translation are required for protein synthesis, translation is the more direct process (transcription is the synthesis of RNA, choice D is eliminated).

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204. B. The hallmark characteristic of diseases caused by pathogenic microbes is that they are infectious (able to be passed from organism to organism). For example, consider some diseases of known microbial origin: influenza, streptococcal pharyngitis (strep throat), AIDS, herpes, chlamydia, etc. These are all infectious diseases. Consider also some noninfectious diseases: diabetes, cystic fibrosis, multiple sclerosis, alcoholic cirrhosis of the liver; none of these are due to a microbe. Thus if a disease is infectious, it is most likely caused by a microbe pathogen (Statement III is true). Statement I is vague at best; without a better description of “suspicious objects,” it is difficult to characterize the origin of this disease. Sickle-cell anemia, a disease of genetic origin, results in the production of “suspicious objects” in the blood (misshapen, or “sickled” red blood cells). Statement I does not support the hypothesis that this disease is of microbial origin. Statement II seems to support an opposite hypothesis: that the causative agent is not a pathogenic microbe. After all, if it’s difficult to grow the pathogen, perhaps there is no pathogen.

Passage XI 205. C. The diionic mechanism shown in Figure 3 places the charges at the two carbons which will bond to complete the formation of the six-membered ring. For the diene in Entry 2 of Table 1, these positions correspond to choice C. 206. B. The enantiomer of the product in Entry 4 of Table 1 has the opposite stereochemistry at all stereocenters, corresponding to choice B. 207. C. Choices A and D do not have the same atomic connectivity as the diionic intermediate and therefore are not resonance structures. Choice B is eliminated since the molecule shown does not follow the octet rule. 208. A. According to the passage, the reaction is more favorable when dienophiles with electron-withdrawing groups are present. The dienophile in the question bears a methoxy group, which is electron donating, making the Diels–Alder reaction unfavorable.

Independent Questions 209. C. Polynucleotide sequences are nucleic acids (DNA or RNA), not peptides (amino acid sequences, choice D is wrong). Any nucleotide sequence containing U (uracil) must be RNA and not DNA (choices A and B are wrong). 210. B. Methylation of DNA involves the addition of a methyl group (–CH3) to the DNA strand. The presence of the methyl group makes it more difficult for RNA polymerase to bind and transcribe that region of DNA (choice B is correct). One can therefore expect methylation to occur more in an inactivated X chromosome than in other chromosomes (choice D is wrong). Methylation typically occurs on cytosine residues, not thymine (choice A is wrong); furthermore, it occurs on cytosines found in a 5¢C G3¢ combination. Since a particular sequence is involved, the methylation pattern of daughter strands must be very similar to the methylation pattern of template strands (choice C is wrong).

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211. C. Compound X has two peptide (main-chain amide) bonds—highlighted in the structure below—and is therefore a tripeptide.

O

O H2N

CH

C

NH

CH2

C

CH

CH

COOH

CH3

CH H3C

NH

CH3

In fact, Compound X is the tripeptide Phe–Val–Ala, and the two peptide bonds are Phe–Val and Val–Ala. 212. D. This question looks scary because it is long and complex-sounding, but it is very straightforward. Any daughter DNA molecules produced must be identical to the parent DNA molecule. If the parent DNA molecule has an (A + T) : (G + C) ratio of 3:1, then the daughter DNA molecules must have the exact same ratio, regardless of the initial molar quantities of dNTPs used to synthesize the DNA. 213. D. Based on the formula given, the structure of benzoin is H

O

C

C

OH O

O

It does not contain a carboxylic acid (RCO2H), an ether (ROR), or an aldehyde (RCH). It does contain a ketone (RCR). 214. C. If the tidal volume is 800 mL/breath, and 150 mL of that is dead space volume, then the net volume of inspired fresh air is 650 mL/breath. This gives 650 mL/breath ¥ 10 breaths/min = 6500 mL/min of inspired fresh air.

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