AAMC MCAT test 5r a

MCAT Practice Test 5R Solutions ® Solutions 5R www.PrincetonReview.com MCAT MEDICAL COLLEGE ADMISSION TEST 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 www.PrincetonReview.com MCAT P RACTICE T EST 5R S OLUTIONS C ONTENTS : Physical Sciences Verbal Reasoning 14 Biological Sciences 29 030316 PHYSICAL SCIENCES Passage I B The value of DH∞ for Reaction is given as –92 kJ Since DH∞ is negative, the reaction is exothermic, by definition 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 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 A The passage states that Reaction is carried out “in the presence of” FeO/Al2K2O4 Furthermore, we notice from Reaction 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 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 NH4+ ions for each SO42– ion; that is, the compound would be (NH4)2SO4 The balanced acid–base reaction would be NH3 + H2SO4 Æ (NH4)2SO4 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) 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 D According to the passage, “there is a large electrical repulsion between these two fragments that causes them to gain kinetic energy.” A According to Newton’s Third law, the magnitude of the force by Fragment on Fragment is equal to the magnitude of the force by Fragment 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 = 12 m (at ) = 12 m ( mF t) = 2Fm t Since F is the same for both fragments but m is different, the fragments will have different KE values at each moment t 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 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 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 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 and 5, crossing out the Cl• and ClO• radicals, the net reaction is given in choice B: Reaction 4: Cl • + O Æ ClO • + O + Reaction 5: ClO • + O Æ Cl • + O Net reaction: O + O Æ O 16 C First, we find the net reaction of Reactions and 2: Reaction 1: O Æ O + O + Reaction 2: O + O Æ O Net reaction: O Æ O + O 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: DGrxn = Â n ◊ DGf, products - Â n ◊ DGf, reactants = (1 ◊ DGf, O + ◊ DGf, O ) - (2 ◊ DGf, O ) kJ = (230.1 mol + 163.4 kJ kJ mol ) - (2 ◊ 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 O3 Æ O2 using the values for S given in Table 1: DSrxn = Â n ◊ Sproducts - Â n ◊ Sreactants = (3 ◊ SO ) - (2 ◊ SO ) = (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 “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 shows the formation of OH–, we conclude that (NH4)2HPO4 is a basic salt, so A is a better response than C Independent Questions 24 B First, to balance the given equation, we need only place a coefficient of in front of both the HCl and the NaCl: Na2CO3 + HCl Æ CO2 + H2O + 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 M solution of Na2CO3 contains 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) = 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 revolutions (or cycles) per second—that is what “moving on a circular path with a frequency of Hz” means—the particle travels a total distance of ¥ (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 contains 16 mmol of CDP in 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 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 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 - ] [HPO - ] = fi = 10 = 100 [H PO ] [H PO - ] 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 half-lives If the object currently contains 1000 atoms of 14C, then half-life ago, it contained 2000 atoms of 14C; half-lives ago, it contained 4000 atoms of 14C; and half-lives ago, it contained 8000 atoms of 14C 35 C The passage gives the mass of a beta particle as ¥ 10–31 kg So, if its speed is ¥ 107 m/s, its kinetic energy is KE = 12 mv = (9 ¥ 10 -31 kg)(3 ¥ 10 m s) = (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 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 + NaCl Æ CdCl2 + NaNO3 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 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 ions (Zn2+ + NO3–) while AgNO3 dissociates into only 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 According to the data for Trial given in Table 1, the current I is 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 = ¥ 10–3 kg Since the volume of the wire is given in the question to be V = ¥ 10–7 m3, the density of the wire is r= m ¥ 10 -3 kg = = 0.8 ¥ 10 kg m = ¥ 10 kg m = 8, 000 kg m V ¥ 10 -7 m 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 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 and 3, then and 4, then 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 So, if T increases by a factor of , the energy radiated each second increases by a factor of ( ) = [( )2 ]2 = [2]2 = 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 = ¥ 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 = ( ¥ 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.”] 49 D According to the data in Table for Trial 1, the resistance of the wire was R1 = 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 half-lives elapsed, since (1/2)2 is equal to 1/4 If a time period of 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 > 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 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 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 10 122 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 25 126 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 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 27 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 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 28 BIOLOGICAL SCIENCES Passage I 138 B Hypothesis 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 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 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 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 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 not seem mutually exclusive, rather, they seem mutually supportive Hypothesis occurs very early in life, and Hypothesis seems to occur throughout life; thus, Hypothesis 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 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”) 29 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 By analogy, Compound Y could be produced using CH3CH2MgBr in Step 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 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) 30 152 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 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 not add or skip S phases (choice C is false), and they 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 of conjugation, so they 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 COOH H OH COOH O H OH H OR H H H OH O H OH b -D-glucuronide OR H OH H H 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 = ◊ pH max - pKa = 2(12.2) - 11.79 = 12.61 32 167 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 requires the neutralization of the carboxylates followed by heating to produce the anhydride: O O H3O H3C 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 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 33 172 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 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 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 34 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 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 is a reversible reaction, as depicted, eliminating choice A Early termination of Reaction 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) 35 189 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 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 (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 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 36 196 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 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) 37 204 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 places the charges at the two carbons which will bond to complete the formation of the six-membered ring For the diene in Entry of Table 1, these positions correspond to choice C 206 B The enantiomer of the product in Entry of Table has the opposite stereochemistry at all stereocenters, corresponding to choice B 207 C Choices A and D 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) 38 211 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 39 ... 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... is a catalyst for the reaction, and the role of a catalyst is to increase the reaction rate D Since NH4+ is a cation with a +1 charge, and SO42 is an anion with a charge, we would expect that a. .. main idea of the passage In paragraphs 2, and the author explains how schemas, once instantiated, help us to understand new information and act upon it In the final paragraph, the passage discusses