MCAT Full-Length Tests

Dear Future Doctor, The following Full-Length Test and explanations are an opportunity to bring it all together in simulation. Do not engage in Full-Length practice until you have adequately prepared your knowledge and critical thinking skills in Subject, Topical, and Section tests. Simply g the tests is inadequate; a solid understanding of your performance through your Score Reports and the explanations is necessary to diagnose your specific weaknesses and address them before Test Day. All rights are reserved pursuant to the copyright laws and the contract clause in your enrollment agreement and as printed below. Misdemeanor and felony infractions can severely limit your ability to be accepted to a medical program and a conviction can result in the removal of a medical license. We offer this material for your practice in your own home as a courtesy and privilege. Practice today so that you can perform on test day; this material was designed to give you every advantage on the MCAT and we wish you the best of luck in your preparation. Sincerely,

Albert Chen Executive Director, Pre-Health Research and Development Kaplan Test Prep © 2003 Kaplan, Inc. All rights reserved. No part of this book may be reproduced in any form, by Photostat, microfilm, xerography or any other means, or incorporated into any information retrieval system, electronic or mechanical without the written permission of Kaplan, Inc. This book may not be duplicated, distributed or resold, pursuant to the terms of your Kaplan Enrollment Agreement.

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PHYSICAL SCIENCES ANSWER KEY

1. A

2. D

3. B

4. C

5. D

6. D

7. A

8. C

9. C

10. A

11. B

12. D

13. B

14. D

15. C

16. C

17. C

18. D

19. D

20. C

21. C

22. D

23. A

24. D

25. C

26. A

27. D

28. B

28. A

30. B

31. D

32. C

33. A

34. B

35. D

36. C

37. C

38. C

39. A

40. D

41. B

42. B

43. C

44. C

45. C

46. B

47. A

48. D

49. A

50. A

51. A

52. B

53. D

54. C

55. B

56. B

57. B

58. C

59. B

60. C

61. D

62. A

63. D

64. A

65. D

66. D

67. C

68. A

69. A

70. B

71. A

72. C

73. B

74. B

75. A

76. D

77. C

Passage I (Questions 1–6)

1. AThis question is asking us to apply the definition of momentum, p = mv. Since each bullet

has a mass m = 5.0 � 10-3 kg (remember to convert to kilograms) and leaves its gun at 250 m/s,its momentum is p = (5.0 � 10-3 kg) (250 m/s) = 1.25 kg � m/s The correct answer is choice A.

2. DThis question asks about the conservation of the total momentum of the system and the con-

servation of kinetic energy in Stunt 2. The total momentum of the system is not conserved, sincegravity acts on both bullets. The total momentum of a system of objects is conserved only whenthere is no net force on the system. The kinetic energy of the system is also not conserved, sincethe collision is inelastic. Remember that the kinetic energy of a system is only conserved in acollision that is perfectly elastic. The correct answer is choice D.

3. BLet’s apply Roman Numeral strategy by evaluating the statement that appears in the most

answer choices: statement II. The target’s translational kinetic energy is zero throughout; thetarget’s motion is all rotation, it never changes position. Statement II must appear in the correctanswer; eliminate answer choice A.

What about statement I? The target’s potential energy is a function of how high it sits abovethe ground. In fact, the zero level of gravitational potential energy is arbitrary. We usually set thepotential energy at ground level equal to zero; but we could give it any number at all. So thepotential energy of the target doesn’t have to be zero; any answer containing statement I iswrong. Only answer choice B remains.

We don’t have to check statement III, but let’s check it just in case. The target’s rotationalkinetic energy is not zero; according to the passage, the target rotates at constant angular veloc-ity between the two shots.

4. CThe first bullet causes the target to rotate because it applies a torque. The second bullet will

stop the target’s rotation by applying an equal torque in the opposite direction. Since the bulletsstrike on opposite sides of the center of the target, we know the torques will be in opposite direc-tions. We need to know the magnitude of the torque from the first bullet. Applying the defini-tion of torque (� = rF sin �, but � = 90º, so � = rF), we see that this torque is �1 = (0.5m)(F0) =0.5F0, using the fact that the radius of the target is 0.5 m. The second bullet strikes the target0.25 m away from the center with a force we’ll call F. Setting the two torques equal, we see that0.5F0 = 0.25F; F = 2F0.

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5. DThe two bullets travel the same horizontal and the same vertical distance before they collide:

In this projectile motion problem, it would be easier for us to deal with motion in the hori-zontal direction, since there is no acceleration horizontally. The horizontal component of thevelocity of each bullet is (250 m/s) cos�. Since there is no acceleration horizontally, we canapply d = vt for each bullet. How long does it take each bullet to travel the 2.5 horizontal metersbefore the collision?

(2.5 m) = (250 m/s cos�) t; t = = s.

The correct answer is choice D.

6. DHow should the shooter compensate for the wind? The wind blows to the northeast. The east-

erly part of the wind will act to speed up the bullet a little bit, but won’t affect its aim. However,the northerly component of the wind will blow the bullet off course. To compensate for that, theshooter will have to aim a bit south of the target. This eliminates answer choices A and C.Answer choice D offers a different wrinkle: does the shooter have to aim high to compensate forgravity? Yes, she does. The acceleration due to gravity will start pulling the bullet downwardonce it leaves the gun. Over a long distance, the bullet could fall far enough to pass under thetarget, so the shooter should also aim a bit high, so that the bullet will arc upwards, and strikethe target on the way down.

Passage II (Questions 7–12)

7. AThis question asks us to know the distinction between molecular geometry and electronic

geometry. In XeF2, the central xenon atom is bonded to two fluorine atoms and carries threeelectron lone pairs. The three lone pairs inhabit sp2 hybridized orbitals (technically, Xe is sp3dhybridized). These three orbitals fall into a trigonal planar configuration, with the two fluorineatoms positioned on opposite sides of the xenon atom, perpendicular to the planar electron con-figuration, as shown below:

F

Xe

F

0.01�cos�

2.5 m��(250 m/s) cos�

250 m/s 250 m/s

A B250 cos � 250 cos �

� �

3

We could look at this electronic geometry and mistakenly choose answer choice D, trigonalbipyrimidal. But why would this be wrong? Because though the electronic geometry is indeedtrigonal bipyrimidal, the molecular geometry is simply linear. Lone pairs are not considered partof molecular geometry. Choice A is correct.

8. CVSEPR Theory allows us to generate the three-dimensional structure of XeF4. Xenon

tetrafluoride has 36 valence electrons: 7 from each of the four fluorides and 8 from the centralxenon atom. Xenon therefore has 4 fluoride single bonds and two lone pairs. Six regions of elec-tron density (4 single bonds + 2 lone pairs) correspond to octahedral electronic geometry andsquare planar molecular geometry, as shown below:

However, the information relevant to answering the question is simply the number of lonepairs on xenon, which again is 2. Choice C is the correct response.

9. CTo determine the intermolecular forces in a sample of XeF6, we must again look to the

molecular structure. Xenon has 8 valence electrons: 6 fluoride single bonds and 1 lone pair.Since there is but one lone pair on xenon, there can be no electronic symmetry in the molecule.XeF6 must have a dipole moment. Therefore the predominant type of intermolecular interactionis dipole-dipole interaction, choice C. Answer choice D can be immediately discounted sincethere are no hydrogen atoms in XeF6. Choices A and B are wrong since XeF6 is not ionic.

10. A According to the kinetic theory of gases, the average kinetic energy of a gas particle is pro-

portional to the absolute temperature of the gas, KE=3/2kT, where k is the Boltzmann constant.Thus the average kinetic energy of a xenon gas particle can be calculated as follows:

KEXe = (3/2)(1.38 � 10-23 J/K)(400K � 273K)

� (3/2)(673)(1.4 � 10-23 J)

� (1000)(1.4 � 10-23 J)

� 1.4 � 10-20 J

The correct answer is therefore choice A.

11. BTo produce XeF4, xenon reacts with fluorine as shown:

Xe(g) � 2F2(g) → XeF4(g)

F

Xe

F

F F

4

In Experiment 2, the chemist uses 19 g, or 0.5 mole, of F2(g). If she adds xenon and fluo-rine in a 1:5 molar ratio (as stated in the passage) , she must then add 0.1 mol Xe(g). Based onthe stoichiometry of the above reaction, xenon is the limiting reagent in Experiment 2. If all 0.1mol of Xe get consumed, then twice as much fluorine, or 0.2 mol to be exact, get consumed aswell. This leaves 0.5 mol – 0.2 mol = 0.3 mol of fluorine gas left after the reaction has gone tocompletion. At STP, the volume of 0.3 mol of F2(g) is , choice B.

12. DAlthough this question asks about a topic (lasers) with which we may not be familiar, the

question can be answered using the information provided in the question stem, as well as ourfundamental understanding of light and wave properties. The question stem tells us that xenon-chloride and argon-fluoride lasers can be used for medical procedures. Why? Because they havesufficient energy (actually, this is in addition to other desirable properties of rare gas-halidelasers – but we do not need to know this to answer the question!) So why might xenon-fluoridelasers not be a good choice for medical lasers? Probably because they do not have sufficientenergy. But since we do not know everything about these lasers, we should use this predictionas a good (but not limiting) starting point as we look through the answers.

Choice A tells us that the instability of gas-halide molecules precludes the use of xenon-flu-oride lasers. While gas-halide molecules are indeed unstable (again, we do not need to knowthis), we can discount this choice for the simple reason that the passage stem tells us that twodifferent types of gas-halide molecules are already used for medical purposes. We cannot gen-eralize the flaws of xenon-fluoride lasers to all types of gas-halide lasers. Furthermore, intensityis a property of light sources, not individual photons.

Choice B can be eliminated since the longer wavelength of the xenon-fluoride laser implieslower, not higher, energy. As the question stem states, xenon-fluoride lasers have a longer wave-length (351 nm) than either xenon-chloride (308 nm) or argon-fluoride (193 nm) lasers. Energyis related to frequency and wavelength by the equation E = h�= hc/.

Choice C is somewhat of a nonsensical response – we already know that xenon-fluoridelasers exist. Choice D must then be the correct response. Indeed, lower output frequencies docorrespond to lower energies.

Passage III (Questions 13–18)

13. BFigure 2 gives the distance traveled by Puck A in the x-direction; but, since the puck’s speed

in the y-direction is the same as the puck’s speed in the x-direction, the distance traveled in they-direction is the same as it is in the x-direction. Looking at Figure 2, we see the pucks collideat about 35 cm, choice B. A similar way of seeing this is to notice in Figure 1 that the pucks areinitially separated by 0.70 m. Since the pucks travel at the same speed and angle, they will meetin the middle, at 35 cm. Having traveled the same distance in both the x- and y-directions, puckA travels 35 cm up the table before the collision.

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14. DRemember that an inelastic collision means that the kinetic energy of the system is not con-

served. Momentum is conserved in both elastic and inelastic collisions! We can immediatelyeliminate choices A and B, because choice B will not distinguish between elastic and inelasticcollisions, while choice A indicates that something more complicated has occurred (i.e. an opensystem). In addition, the work-energy theorem tells us that KE = Work. Therefore, if kineticenergy is lost via an inelastic collision, the work done on the pucks cannot be zero: eliminateanswer choice C. The correct answer must be choice D. In an inelastic collision, kinetic energyis lost; the energy is lost as sound, heat, or deformation. The heat energy will raise the puck’stemperature.

15. CThis question requires a little calculation. Think about what the question asks, before just

plugging in numbers. The change in momentum, or impulse, is given by average force � thetime of impact. Looking at Figure 3, we see that the force rises and falls at a steady rate (i.e.symmetrically over the time interval), so the average force is half the maximum force (1000 N),and

p = (500 N) � (3 � 10-3 sec) = 1.5 N � s, choice C.

16. CThe work-kinetic energy theorem tells us that if 6 J of kinetic energy were lost, then 6 J of

work was done on the puck. Since work equals force times distance, we have what we need todetermine the force due to friction:

W = Fd→F = �Wd� = �0.

65

Jm� = 12 N.

The correct answer is choice C.

17. CLook at the initial conditions of the pucks before they collide. Puck A and puck B are trav-

eling with the same velocity. They are traveling in the y-direction with the same speed. They alsohave the same speed in the x-direction, but in opposite directions. So when they collide, therewon’t be any forces in the y-direction, which eliminates answer choices B and D. Since puck Bis coming from the right, the force it exerts on puck A will be from right to left, in the negativex-direction; since the net acceleration of puck A is proportional to the net force, the correctanswer is choice C. Answer A is the acceleration puck B experiences due to colliding with puckA.

18. DWe want the total momentum of the system to be zero. The total momentum of the system

doesn’t change during a collision, so the nature of this collision doesn’t matter at all! Since weknow the speed of puck A before the collision, all we need to do is to determine what speed puckB should have before the collision so that the momentum of the system is zero.

mAVA � mBVB = 0

The above equation sets the total momentum of the system equal to zero. Since mA = 2 kg,mB = 3 kg, and VA = 2.6 m/s, we can solve for VB:

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VB = = = -1.73 m/s.

If the initial speed of puck B is 1.73 m/s, the total momentum of the two-puck system is zero; thecorrect answer is choice D.

Passage IV (Questions 19–24)

19. DAtomic iron is a transition metal with an electronic configuration of [Ar] 4s2 3d6. The iron (III)

ion (ferric) has a formal charge of +3 and loses three electrons from atomic iron. The first two elec-trons are stripped away from the outermost orbital, the s orbital. The third electron comes from thed orbital, the next outermost orbital, leaving the ion with the configuration, [Ar] 3d5. Choice D is cor-rect.

20. CThe first step is to write out the dissociation reaction:

Fe(OH)3(s) Fe3+(aq) � 3OH–(aq)

Ksp = [Fe3+] [OH–]3

Let x = the concentration of Fe3+. Then 3x is the concentration of OH–.

Ksp = (x) (3x)3 = 27x4

= 2.64 � 10-39 (from question stem)

Set these two value equal to each other and solve for x:

27x4 = 2.64 � 10-39

x4 = (2.65/27) � 10-39

x4 � 0.1 � 10-39 = 10-40

x � 10-10 � choice C

21. CEach one of the reactions listed in the passage is an oxidation-reduction reaction, except for

Reaction 3. Oxidation-reduction reactions are distinguished by a change in the oxidation state ofsome participating species.

Reaction 1: sulfur is oxidized from a –2 in iron disulfide to a +6 in the sulfate ion. The iron iniron disulfide is a +4 undergoing reduction to the ferrous state of +2. Diatomicoxygen, oxidation state 0, undergoes reduction to an oxidation state of –2 whenrecombined to make the sulfate ion.

-(2)(2.6)�(3)

� mAVA�mB

7

Reaction 2: the ferrous iron ion (+2) undergoes oxidation to become the ferric iron ion(+3). Again, diatomic oxygen undergoes reduction to form the oxide ion ofwater with an oxidation number of –2.

Reaction 3: the oxidation states of each atom remain unchanged, and therefore Reaction3 is not an oxidation-reduction reaction. The correct answer is thereforechoice C.

Reaction 4: fourteen ferric iron ions are reduced to fourteen ferrous iron ions; the iron ofiron disulfide is oxidized to produce one additional ferric iron ion. The (-2)sulfur in iron disulfide is oxidized to (+6) sulfur in the sulfate ion.

22. D Assuming Reaction 3 – which produces 3 moles of H+ for every mole of Fe3+ consumed –

is the only reaction contributing to the pH, the H+ concentration would be:

� � � 8.14 � 10-4 mol H+/L

Using the concentration of H+, the pH can then be calculated:

pH = –log[H+]

= –log (8.1410-4) mol/L)

� –log (10-3 mol/L)

� 3

This approximate pH best matches the actual pH of 3.09. Therefore the correct answer ischoice D.

23. AThe key to this question is understanding the physical set-up described in the question stem.

If pyrite were found under a water table, there would be access to plenty of water, but therewould be little or no access to oxygen since there is no exposure to the atmosphere. We knowfrom the passage that rock needs to be exposed to atmospheric oxygen, which is why AD is asso-ciated with mining and deep excavation. In the absence of oxygen, there is no oxidizing agentand therefore Reaction 1 would not proceed toward the acidic products. Also note that paragraph3 states that removal of air and/or water can slow or stop pyrite from being oxidized. The answermust be choice A. Choice C is false because disulfide does not get hydrolyzed – water getshydrolyzed.

24. DHenry’s Law states that the solubility of a gas in a liquid is proportional to its partial pres-

sure above that liquid:

Solubility = kH � partial pressure

where kH is the Henry’s Law constant. With the Henry’s Law constant given in the passage stemand the atmospheric partial pressure of oxygen given in the passage stem, the solubility is simply:

3mol H+

��1mol Fe3+

1mol Fe3+

��56 g Fe3+

0.38 g Fe3+

��25 L

8

Solubility = ( 1.3 � 10-3 mol/(L·atm) ) � (0.21 atm)

= 2.7 � 10-4 mol/L

The answer is therefore choice D. Some of you may have incorrectly selected choice C,thinking that the question has asked you about the lowest oxygen concentration that can sustainlife. However, all you have been asked to compute is simply the solubility of oxygen, not theconcentration.

Discrete Questions

25. CThe first step in solving a percentage yield question is to determine the limiting reagent. The

balanced chemical equation shows H2S reacting with SO2 in a 2:1 molar ratio.

204 kg of H2S corresponds to = 6 � 103 mol H2S.

256 kg of SO2 corresponds to = 4 � 103 mol SO2.

The molar ratio of H2S to SO2 is 6 � 103: 4 � 103 = 6:4 = 1.5:1. Since there are only 1.5moles of H2S for every mole of SO2 (instead of 2:1), H2S is the limiting reagent.

The chemical equation shows that the Claus process produces 3 moles of S2 for every 2moles of H2S. Hence, if we start with 6 � 103 mol of H2S, the theoretical maximum of the reac-tion would be:

theoretical yield � � (6 � 103 moles H2S) � 9 � 103 moles S

However, the question stem tells us that we only end up with 224 kg of S. The actual yieldis then:

actual yield � � 7 � 103 moles S

Therefore the percentage yield of sulfur is:

percentage yield � � 100% � � 100% � 77%, choice C.

26. ABy putting the object in space, the question is telling you that there are no other forces on

the object other than those caused by the ropes. What the question doesn’t tell you is in whatdirection each of the forces pull on the object. What’s the largest possible net force on theobject? If both ropes pulled in the same direction, the net force would be 30 N + 10 N = 40 N.Answer choice D is possible, and so it isn’t the right answer. What’s the smallest possible netforce on the object? If both ropes pull in opposite directions, the net force will be 30 N – 10 N= 20 N. Since 10 N is smaller than the smallest possible net force on the object, choice A mustbe the correct answer.

7 � 103 moles S��9 � 103 moles S

actual yield��theoretical yield

224 � 103 g S��

32g/mol S

3moles S��2moles H2S

256 � 103 g���(32.0 � 2 � 16.0) g/mol

204 � 103 g���(2 � 1.0 � 32.0) g/mol

9

27. DThe gist of this question is that triple bonds are shorter than double bonds, which themselves

are shorter than single bonds. That is, the more electrons there are between two nuclei, the closerthose two nuclei get. The question stem tells us that the carbon-carbon double bond in ethene(C2H4) is longer than the carbon-carbon triple bond in ethyne (C2H2), 134 pm versus 120 pm.We should therefore expect the carbon-carbon single bond in ethane (C2H6) to be longer thanthe carbon-carbon double bond, i.e. >134 pm. The only answer choice that meets this criterionis choice D, 150 pm.

28. BSince the object travels on a circular path with uniform speed, its acceleration must be cen-

tripetal: a � � � 18 � 104 m/s2. The force causing this circular motion is

due to the magnetic field: F � qvB(6 � 10-9 C)(300 m/s)B � (18 � 10-7 �N

T�)B. Now we can use

Newton’s Second Law to equate force and acceleration, and solve for B (don’t forget to convertthe mass to kg):

F � ma→ (18 � 10-7 �N

T�)B � (7 � 10-11 kg)(18 � 104 �

s

m2�)

B � 7 T

The correct answer is choice B.

Passage V (Questions 29–34)

29. AIn the passage, we are told that the diving bell is descending below the surface of the ocean,

and that the outside water pressure is increasing. The bottom of the diving bell is open to thewater, and so the water must exert an increasing pressure on the air inside the bell. The volumeof the trapped air decreases as the water volume inside the bell increases, so the air pressuremust increase in direct proportion to the increase in water pressure. Answer choice A is correct.The air pressure remains constant (B) only when the bell’s depth is constant, and the air pres-sure decreases (C) only when the bell is rising.

30. BSince the air pressure in the bell equals the water pressure at the air-water boundary, we need

to find the hydrostatic gauge pressure at that boundary. Since the boundary exists at depth D-L(not depth D+L, as in answer choices A and C), the gauge pressure is g(D-L). The total pres-sure is:

P � P0 � g(D-L).

The correct answer is choice B. Choice D gives a total pressure that is lower than atmo-spheric pressure, which makes no sense since the bell is below the surface of the ocean.

(300 m/s)2

��0.5 m

v2

�r

10

31. DThe total pressure of a fluid at a particular depth is the sum of atmospheric pressure and the

hydrostatic gauge pressure. In this question, we are given the gauge pressure and asked for thetotal pressure. Since the gauge pressure is 2.46 atm, the total pressure at the air-water boundarymust be 2.46 atm + 1.00 atm (atmospheric pressure) = 3.46 atm, answer choice D.

32. CThe pressure of the air inside the bell has doubled after the bell has been lowered so far into

the ocean. The beginning of the passage gives us the link between pressure and volume: Boyle’sLaw. It says that the volume of a gas varies inversely with the pressure of a gas. Since the pres-sure of the gas has doubled, it must now be taking up half its original volume. That happenswhen the bell is half-full: L = H/2, answer choice C.

33. AThe difference between this scenario and the scenario in the passage is the density of the

fluid in which the bell is submerged. The question behind the question here is: how does thepressure of a fluid depend on its density. Since the gauge pressure of a fluid at depth z is gz,the pressure must increase as density increases. Therefore, the air pressure in the bell submergedin pure water must be less than the air pressure in the bell when submerged to the same depthin ocean water, since the density of pure water (1000 kg/m3) is less than the density of oceanwater (1024 kg/m3). The correct answer is choice A.

34. BIn order to lift the object toward the surface, we need to overcome its “apparent” weight in

the water. That is to say, we need to counteract the weight of the object, but it’s not so bad, sincewe have the buoyant force lightening our load. The weight of the object is W � mg �(2.2 � 105)(10) � 2.2 � 106 N. The buoyant force on the object (which points upwards) is the

weight of the displaced ocean water. FB � watergV � (1024 )(10 )(25 m3) � 2.5 �105 N.

Consequently, the net force downward on the object is W �FB � 1.9�106 N. The minimumforce needed to raise the object must counteract this force, so it has to be at least 1.9�106 N.The correct answer is choice B.

Passage VI (Questions 35–39)

35. DTo begin attacking this problem, first figure out exactly what is going on in the passage.

When does the cloudiness disappear? The cloudiness disappears when the homeowner adds HClto a swimming pool that has been disinfected using calcium hypochlorite, which produces HClOby the reaction shown in Reaction 2. What was causing the cloudiness? Cloudiness in generalis caused by some species exceeding its solubility. In this case, calcium hypochlorite reacts(completely) to form calcium hydroxide and a weak acid. It is thus the Ca(OH)2 which is satu-rated. The addition of HCl must then function to increase the solubility of Ca(OH)2. How canHCl do this? It must decrease the concentration of either OH- or Ca2+. Indeed, hydrochloric acidneutralizes the hydroxide ions, forming water. With a lower hydroxide ion concentration, morecalcium hydroxide can be solvated.

m�s2

kg�m3

11

Now go through the answer choices, one-by-one, looking for a false or unlikely statement.Choice A is a true statement. At equilibrium, the calcium and hydroxide ion concentrations areinversely related to each other (actually, inversely-squared) through the solubility product con-stant. To “fill in the gaps” made by hydroxide ions that have “left” to neutralize the acid, cal-cium hydroxide dissociates into both calcium and hydroxide ions, and thus increases calciumion concentration.

Choice B is a true statement as well. Once the homeowner has added HCl to the pool, all thespecies on the product side of Reaction 1 will then be present. Naturally in an equilibrium, someof these products will proceed toward the reactant side, thereby forming chlorine gas. Choice Cis as well a true statement. Adding HCl lowers the hydroxide ion concentration, thereforeincreasing the pOH.

Choice D, however, is a false statement. A saturated species is one whose ion product isequal to its Ksp. As the homeowner adds HCl, the ratio of [Ca+] to [OH-] may change, but the

ion product, [Ca2+][OH-]2, does not. As the pool owner keeps dumping acid into the pool, cal-cium hydroxide molecules enter solution to keep the ion product the same. The owner adds acidonly until the precipitate disappears, at which point the ion product is still equal to the solubil-ity constant.

36. CBoth Hypotheses 1 and 2 are true, making choice C the correct answer. In aqueous solution,

calcium hydroxide dissociates into Ca2+ and OH-, making it an Arrhenius base (a species thatproduces hydroxide ions). The existence of hydroxide ions raises the pH of solution. Hydrochlo-ric acid is then added to neutralize the hydroxide ions and dissolve the precipitated calciumhydroxide. This explanation agrees with Hypothesis 1. But Hypothesis 2 is true as well. Calciumhypochlorite does release a Lewis base, OH-. The hydroxide ion then exchanges an electron pairwith the acid, forming water in a neutralization reaction.

37. CThis question tests your understanding of the relationship between pKa and pH. Generally

speaking, when an acid is in an aqueous solution whose pH is lower than the acid’s pKa, the acidwill be found in its protonated form. If the pH is greater than pKa the acid will give up its protonand will be found in its deprotonated conjugate base form. Applying this knowledge to the ques-tion, when the swimming pool is cloudy with precipitate, there is an excess quantity of calciumhydroxide present. Therefore the pool water is alkaline. In solution where the pH is higher than 7,the indicator - which has a pKa of 5.0 - will be found in its deprotonated, conjugate base form.Matching this form with the appropriate color on the table, yellow, the correct answer is choice C.

38. C A buffer system consists of a mixture of a weak acid and its salt, or a mixture of a weak base

and its salt. Therefore we can answer this question by reading through the answers, eliminatingchoices which are weak acid/base-salt systems, and choosing the one that is not.

Choice A is an example of a weak base-salt system. NH3 grabs a proton to become NH4+

and ionically binds to Cl-, forming NH4Cl. Choice B is an example of a weak acid-salt system.To form NaClO2, HClO2 donates a proton to solution (becoming ClO2

-) and ionically binds toNa+. Choice D is another example of a weak acid-salt system. C6H4(CO2H)2 loses a proton,becoming C6H4(CO2H)(CO2)- and then complexes with a free sodium ion.

12

Choice C, on the other hand, is a salt-salt system, not a buffer solution. One alkali metal ion,K+, simply replaces another, Na+. There is no proton movement.

39. AThe passage tells us that the optimal pH is 7.5. In aqueous solution, pH � pOH = 14. There-

fore, when a pool is at optimal pH, the pOH is 14 – 7.5 = 6.5. The relationship between pOHand [OH-] is [OH-] = 10-pOH. The optimal pOH is then 10-6.5. You can rewrite this as 10-6.5 =10-7 � 100.5 = �1�0� � 10-7. Though not an integer, you should be able to recognize that thesquare root of ten is slightly larger than 3. Thus, pOH. Choice A is the correct answer.

Passage VII (Questions 40–44)

40. DIn the second paragraph, we learn that the electric field near the Earth’s surface points

downward. Since the question has a positive charge moving against the lines of electric field(which they don’t want to do!), the charge will be gaining electric potential energy, just as aboulder will gain gravitational potential energy by being rolled up a hill. Eliminate answerchoices A and C. The formula for the change in potential energy of charge q is U = qV. Theelectric field strength near the surface is 100 V/m; for every meter of travel, the change in elec-tric potential is 100 V. Since the positive charge in the question moves up a half-meter, the poten-tial change is 50 V. Since the net charge of the molecule is e = 1.6 � 10-19 C, the change in thecharge’s potential energy is U = (1.6 � 10-19 C)(50V) = 8 � 10-18 J answer choice D.

41. BThis question is asking us to plug in numbers from the passage into our expression for the

capacitance of the Earth/electrosphere system. The radius of the Earth is given in the passage; theradius of the electrosphere is the radius of the Earth plus the extra 60 km, or 6.46 � 106 m. Noticethat the difference in the radius of the electrosphere and the radius of the Earth that appears in thedenominator is simply the 60 km gap between the capacitor’s “plates.” Applying the formula,

C =(1 � 10-10 �

N

C

�

2

m2�) (6.5 � 106 m)(6.5 � 106 m)

= 7 � 10-2 F.

The only close answer choice is B. The other answer choices are traps waiting for you tomake calculation mistakes.

42. BThis question asks you to again consider the expression for the capacitance of the Earth/elec-

trosphere system. This time, the Earth’s radius is halved, but the distance between the Earth andthe electrosphere is still the same. Looking at the equation, we need to figure out the effects onthe radius of the electrosphere. Since the Earth’s radius makes up the vast majority of the radiusof the electrosphere ( rE = 6.4 � 106 m, while the distance from the surface to the electrosphereis only 60 km), we can say that the radius of the electrosphere has been halved as well. In thecapacitance formula, two factors in the numerator have halved, while the denominator stays thesame. Consequently, the capacitance has fallen by a factor of four, answer choice B.

4��0rErS�

rS-rE

13

6 � 104 m

43. C What do we know electrically about the gap between the ground and the electrosphere? We

know that the distance is 60 km, and that the potential difference is 300 kV. The work requiredto raise a positive charge through a potential is the change in potential energy of the charge. (Thework you would need to do in carrying a boulder up a hill is the change in potential energy ofthe boulder between top and bottom.) Therefore,

W = U = qV = (5 � 10-6 C)(3 � 105 V) = 1.5 J.

The correct answer is choice C.

44. CWhat do we know about the leakage current? The leakage current density is 10-12 A/m2; 10-

12 coulombs of charge leave every square meter of the Earth’s surface every second. But we needto know about every minute; there are 60 seconds in a minute, so leaves every minute. To figureout how much charge leaves the entire surface of the Earth every minute, we need to multiplythis by the surface area of the Earth:

charge = (60 � 10-12 C/m2)(5.1 � 1014 m2) = 306 � 102 C = 3.1 � 104 C

The correct answer is choice C.

Passage VIII (Questions 45–50)

45. CWe are given a specific temperature and pressure, so we should first look at the phase dia-

grams to see what we can learn. Both Figure 1 and Figure 2 indicate that Compound A is a liq-uid at 3.5 atm and 92ºC (365 K). The smallest G (i.e. the most negative) will be associated withthe process that is the most spontaneous or favored. Therefore, any process that favors the pro-duction of a liquid will be the most spontaneous: G for that process will be smaller (more neg-ative) than for any other. Perusing the answers, we can immediately eliminate choice B, sincethis inhibits the formation of liquid. In addition, we see choices A and D favor the production ofa gas and solid, respectively. While either deposition or sublimation may have a G < 0, it cer-tainly won’t be as negative as condensation. Therefore, condensation will be the most favoredand G for the condensation reaction will be the most negative (i.e. smallest). The correctanswer is choice C.

46. BRemember that compressibility is a property of gases, while solids and liquids are consid-

ered incompressible. Therefore the question essentially just asks us to pick the temperature andpressure at which Compound A is a gas. Chemist 2’s phase diagram indicates that a gas isfavored at 310 K and 0.1 atm. Choice B is correct.

47. AAt first, this question may seem unrelated to the passage; however, a bit of scientific rea-

soning will elucidate the choices. Consider a liquid substance at a given temperature. If theambient pressure is increased, then Le Châtelier’s principle intimates that the denser phase willbe favored (notice that Le Châtelier’s principle has applications outside of solution chemistry!).Normally, we expect the solid-liquid boundary line on a phase diagram to be positive, indicat-ing that as pressure increases, a solid will be favored. So Compound A, having a positive slope,

14

is denser as a solid than as a liquid. Therefore, I is not correct: eliminate choices C and D. Next,remember that as water freezes, a lattice structure is favored, resulting in the same mass pergreater volume, i.e. lower density. Therefore, water is, in fact, less dense as a solid. (Thisexplains why ice floats in water). We know II is correct: looking at the choices, this leaves onlychoice A, the correct choice. Notice, we do not even need to consider carbon dioxide, though wemay remember that CO2(l) is less dense than CO2(s).

48. D Sublimation is the process by which a solid immediately becomes a gas, without entering

the liquid phase first. The question asks for a possible value of Hsub. However, to what can wecompare it? Well, the passage indicates that Hvap = 25.76 kJ mol-1. Since sublimation includesvaporization and fusion, we expect the enthalpy of sublimation to be higher than the enthalpy ofvaporization. (This exemplifies Hess’s Law.) Therefore, the only answer choice that is greaterthan 25.76 is choice D, our answer.

49. ABefore answering this question, let us stop, think, and predict. What are some noticeable dif-

ferences in the two phase diagrams? It appears that Chemist 2’s diagram is wider. Carefulinspection indicates that boiling point is higher and melting point is lower. Sound familiar?Seems as though the colligative properties of the compound have been altered; clearly, an impu-rity has been introduced. Now, attack the answer choices. Choice A mentions that the chemist’shands had salt on them (it does not matter what salt), this sounds like a possible contaminant.Check the other choices as well. Choices B, C, and D assume that the quantity of sample usedor the specific temperatures and pressures used to determine state properties influence the result– but they do not. Boiling points and melting points are affected by colligative properties, notpathways and not quantity. Therefore, the answer is choice A.

50. ADon’t over-think this one. O2 at standard conditions (25ºC, 1 atm) is obviously a gas.

Gaseous oxygen will have a very low density; Compound A will have the most similar densitywhen it too is a gas. So we need to look for the pressure and temperature that makes CompoundA a gas. Again, examining Figure 1 we see that at 0.5 atm and 450 K, Compound A is a gas:choice A is correct. Notice, choices B and C are the pressures and temperatures at STP and stan-dard conditions, respectively. Finally, choice D shows Compound A as a solid.

Discrete Questions

51. AThe reaction rate of a multi-step reaction is determined by the rate-limiting step. In this reac-

tion, it is Step 2. The rate for this step is k[B][BC]. However, the rate for this step cannot be therate for the overall reaction, since Compound B is neither a product nor a reactant, but an inter-mediate. (Therefore choice B is incorrect.) [B] is a function of the first step, which is first-orderwith respect to BC and first-order with respect to A2. Therefore if we can replace [B] with[BC][A2], making the reaction third-order overall, with a rate expression k[BC][A2][BC], ork[A2][BC]2. Choice A is the correct answer.

15

52. BHow does the period of oscillation of a mass-spring system depend on the mass and the

spring, in general? The angular frequency � = 2�f = ��m

k�� is the easiest formula to remember.

Since the period is the reciprocal of the frequency, T = �1

f� = �

2

�

�� = 2� ��

m

k��. Now that we have

the formula, we can deduce what happens. Doubling the mass and the spring constant has no neteffect on the fraction inside the square root; the extra factors of two cancel each other out. Theperiod remains unaffected; the correct answer is choice B.

53. DAn equilibrium constant of 0.05 means that at equilibrium the ratio of product [A-][H3O+]

to the amount [HA] is 5:100. This eliminates choices A and B. If we forget that there are twoproducts in the formula for K (K = [H3O+][A-]/[HA], not K = [A-]/[HA]), choice C wouldappear to be correct. With an equilibrium constant K << 1, the reactant HA undergoes little ion-ization (also called hydrolysis or dissociation) in water. Choice D is therefore the correct answer.Choice A would also be arrived at by looking at the balanced equation (a 1:1 mole ratio) andignoring the value of K. Balanced equations only tell us how many individual product specieswill be present at equilibrium relative to one reactant species, not how many of those reactantsare actually converted to products by the time the equilibrium is established.

54. CTo determine how many neutrons the daughter nucleus has, we’ll need to conserve charge

and mass number for this decay. Let’s start with charge. Consulting the periodic table, a cobaltnucleus has 27 protons. The electron on the other side of the decay has a proton number of –1.Consequently, the proton number for the daughter nucleus must be 28: 27 = Z –1, Z = 28. Nowwe’re ready to move on to conservation of mass. The mass of the cobalt nucleus is 60, and themass of an electron is zero; so the mass number for the daughter nucleus must be A = 60. Since28 of the 60 nucleons are protons, the remaining 32 must be neutrons. The correct answer is C.

55. B An amphoteric species is one that can act as either an acid or a base, depending on the con-

ditions. Amino acids, for example, are amphoteric species because they have both an acid groupand an amino group. The plan of attack for this question is to find, by process of elimination,the answer choice that can act only as a base or only as an acid. Finding the correct choice doesrequire a small but fundamental knowledge of the species listed. Choice A, ammonia, can eitheraccept a proton to become the ammonium ion, NH4

+, or lose a proton to become NH2-. We

should recognize that choice B, nitric acid, is a very strong acid and will only dissociate – it istherefore not amphoteric. Choice C, partially dissociated sulfuric acid, can either regain a pro-ton to return to acid form or continue dissociating into the SO4

- ion. Choice D, water, canbecome both a hydronium ion, H3O+, or the hydroxide ion, OH-. Therefore choice B is the cor-rect answer.

16

Passage IX (Questions 56–61)

56. BWhat criterion does the passage establish for the usefulness of materials as target materials?

In the final paragraph, we are given an empirical expression for the efficiency of x-ray produc-tion. This efficiency depends on two factors; the proton number of the target material and thepotential difference between the cathode and the anode. Since the question asks about the anodematerial, we need to figure out the proton number of carbon. A quick check of the periodic tabletells us that Z = 6. That’s low compared to tungsten and rhenium. This is what must make car-bon a bad target material; answer choice B is correct.

Nothing in the passage indicates that the anode has to be a metal (choice A), or that a car-bon target will react with the anode material (choice C), or that the target material is heated(choice D).

57. BThe amount of energy that is wasted will be defined by the efficiency of the x-ray produc-

tion. The periodic table tells us that Z=74 for tungsten, so the efficiency is � = (10-7)(74)(8 � 104) = 600 � 10-3 = 0.6 = 60%. The fraction of energy that’s wasted isaround 40%; answer choice B is correct. If you forgot to subtract the efficiency from 100%, youchose choice C.

58. CThis is a discrete question in the middle of a passage. Given the wavelength of a photon, you

can calculate its energy using the formula:

E = hf = �hc� = = 10 � 10-15 J = 1 � 10-14 J.

which is very close to answer choice C.

59. BWe need to read carefully and understand how the apparatus works in order to answer this

question. Let’s try it without looking at the answers. The filament current heats the filament,which glows white-hot and boils off electrons. These electrons pick up speed due to the poten-tial difference between the filament (cathode) and the anode. The electrons slam into the anodetarget, releasing x-rays. So apparently, the energy of the x-rays comes straight from the incom-ing electrons. This matches with answer choice B. We don’t know anything else about the heatemitted by the cathode (A), we aren’t told anything about the heat produced at the anode (C),and the metal housing doesn’t come into the picture until the x-rays are already created (D).

60. CFrom Figure 1, we can see that the circuit containing the tube voltage contains the cathode

filament and the anode. What role do these play in this setup? The filament is heated until itbecomes so hot, that electrons peel off of it. These electrons could travel in any direction, justlike the light from a light bulb flies off in all directions. But because of the circuit with the tubevoltage, there is a potential difference between the cathode and anode, pulling these electronstoward the anode. So the purpose of the tube voltage is to provide the potential differencebetween the cathode and anode with directs where the electrons go after they leave the filament.The correct answer is choice C. The filament is heated by the filament current (A), the anodeisn’t heated at all (B), and x-rays are emitted from the anode; they don’t strike the anode (D).

(6.6 � 10-34 J�s)(3 � 108 m/s)����

(2 � 10-11 m)

17

61. DThe word “efficiently” in the question stem should be a clue that this question is calling on

you to use the efficiency equation from the passage. Consulting the answer choices, we see thatwe have a choice between a tungsten anode and a rhenium anode, and a choice between 10000Vand 300000 V (not 300 V, read carefully!) We know that the higher the proton number Z, thehigher the efficiency. A quick check of the periodic table shows us that rhenium (Z = 75) has thehigher proton number; eliminate answer choices A and B. Since the efficiency is also propor-tional to the potential difference between the cathode and anode, we must choose the higher ofthe two voltages, 300 kV. The correct answer is D.

Passage X (Questions 62–66)

62. AA catalyst increases the velocity of a reaction by lowering the activation energy of that reac-

tion. However, at high temperatures the catalyst – often a protein – may begin to denature. Oncethe catalyst begins to denature, the reaction’s velocity will cease to increase. Looking at Figure1, we can see that the velocity of the catalyzed reaction stops increasing at around 55ºC, indi-cating that the catalyst is beginning to denature. Notice that at 30ºC the catalyst is still func-tioning properly since that velocity of the reaction continues to increase. However, at 85ºC thegraphs of the catalyzed and uncatalyzed reactions have already crossed, indicating that the cat-alyst has already lost its catalytic abilities, and the reaction proceeds as though it were uncat-alyzed. The correct answer is then 55 + 273 = 328 K, choice A.

63. DWe can answer this question by inspection of Figure 2. Looking at the reaction profile of the

catalyzed reaction we see a local energy minimum in the center of the pathway. This minimumindicates the presence of some relatively stable state – i.e., an intermediate. The single energymaximum of the uncatalyzed reaction indicates that this is a single step reaction with one high-energy transition state. Some catalysts may boost the reaction rate by stabilizing this intermedi-ate state, thereby lowering the activation energy. The corresponding reaction profile would looklike that of the uncatalyzed reaction, except with a smaller “hump.” For this reason, choices Band C are incorrect. Choice A is wrong because although the dissociation of Product A mayaffect the position of the Equation 1 equilibrium, it does not change the rate at which the reac-tion proceeds to equilibrium.

64. AAt 55ºC the rate of the Experiment 2 reaction begins to drop. Presumably this is because the

catalyst has ceased to function properly. Up until this temperature the catalyzed reaction hasbeen proceeding at a much faster rate than the catalyzed reaction. Consequently, more reactantsare consumed in the catalyzed reaction. Indeed, the passage states that 28% of the reactants inExperiment 2 are consumed, while only 2% are lost in Experiment 1. By 75ºC, the catalyst hascompletely denatured. Above this temperature, only the uncatalyzed reaction is observed isExperiment 2. This reaction, as the passage states, is rate-dependent on the concentration of bothreactants. Therefore, the Experiment 2 reaction will proceed more slowly than the Experiment1 reaction because there are fewer reactants present – these “missing” reactants have been con-sumed by the catalyzed reaction that occurred at lower temperatures. Choice A is correct.

18

Choice B is alluring and probably true. Dissociation reaction rates do depend on the con-centration of the dissociating compound, and more Product A will have accumulated by 75ºC.However, the passage tells us that Product A dissociates very slowly, so this does not give a goodexplanation as to why the rate of formation of Product A would lessen. After all, if the dissoci-ation happened very quickly relative to formation, it would be difficult to actually measure anysort of formation rate. Although possibly true, choice B does not give the best explanation.

Choice C is incorrect, since catalysts are not consumed by a reaction. If a catalyst does react,it will always show up as a product. Choice D is incorrect because the reaction quotient is belowthe equilibrium constant at the start of the reaction – this is why the reaction proceeds.

65. D First, set-up the expression for K2:

K2 = �[B

[A

][C

]

]�

Before Product A begins to dissociate, we start with some unknown quantity of Product A.Let’s call this N. We also have X moles of Product B and no moles of Product C. Summarizingthe initial conditions (in 3 liters of water),

[A] = N/3 [B] = X/3 [C] = 0

Then Y moles of Product A dissociate, and the system reaches equilibrium. Since every moleof Product A dissociates into one mole of Product B and one more of Product C, at equilibrium:

[A] = (N – Y)/3 [B] = (X + Y)/3 [C] = Y/3

The equilibrium constant may then be expressed as:

K2 = �X�

3Y

� � �Y3� � �N�

3Y�

= …manipulating to solve for N:

N�Y = �Y(

3XK�

2

Y)�

N = �Y(

3XK�

2

Y)��Y= �

Y(3XK�

2

Y)� � �

33KK

2

2

Y�

This is the number of moles of Product A that were initially formed. Choice D is thereforecorrect.

66. DThis is more than just a math question. The question requires us to understand the kinetics

of each reaction. Notice that the passage tells us that the uncatalyzed reaction is dependent ontwo species, while the catalyzed reaction is dependent on just one. We can conclude that theuncatalyzed reaction is second-order, while the catalyzed reaction is first-order.

rateuncat = kuncat[X][Y]

ratecat = kcat [Z]

Y(X�Y)�3(N�Y)

19

Through algebraic manipulation we see that:

kuncat = rateuncat [X]-1 [Y]-1

kcat = ratecat [Z]-1

Therefore, dividing the second equation by the first:

kcat / kuncat = (ratecat / rateuncat) · [X] · [Y] / [Z]

The rate ratio is dimensionless, as is the ratio of [Y] to [Z]. This leaves the dimensions of[X] as the only dimensions left. Hence kcat / kuncat has concentration dimensions. Choice D istherefore correct.

Passage XI (Questions 67–71)

67. CThe question stem tells us that QH = 900000 J and that QC = 80000 J. Since the question is

asking for the work done by the engine, answer choices B and D, which simply repeat these twonumbers, are probably traps. The conservation of energy (and the first law of thermodynamics)applied to this refrigerator implies that the energy poured into the hot reservoir must all comefrom the heat leaving the cold reservoir and the work done on the engine:

QH = W � Qc

In this case, W = QH -QC = 900000 J – 80000 J = 820000 J; the correct answer is choice C.

68. AThe final paragraph explains what must be true for an ideal engine: �

QTH

H� = �QTC

C� . We need to

use this relationship to rewrite the coefficient of performance, �. Start with what we know:

� = �QH

Q-CQC� = �1

(-Q(Q

C

H

/Q/Q

C

C

))� = �1 � (Q

1

H/QC)�. We can rearrange our expression from the ideal

engine: �QQ

H

C� = �

TT

H

C�. Substituting into our expression for �: � = �1-(QH

1/QC)� = �1-(TH

1/TC)� =

�TH

T-CTC�, answer choice A. The other answer choices are traps you can fall into by making

algebra mistakes.

69. AWe’ll need to evaluate each answer choice individually. We have an immediate winner in

choice A. Heat from the cold reservoir can’t spontaneously leave for someplace warmer; workhas to be done to transport that heat energy. Let’s have a look at the wrong answers:

Choice B: This is the second law of thermodynamics in a nutshell; the entropy of a systemand its surroundings must increase, or at worst stay constant.

20

Choice C: This is the conservation of energy – because work energy is added to the systemto pull out heat energy from the cold reservoir, the heat energy dumped into the hot reservoirmust greater than the heat energy pulled out of the cold reservoir.

Choice D: The definition of the coefficient of performance is the heat energy extracted fromthe cold reservoir over the amount of work done. The more efficient the refrigerator, the lesswork is needed to pull the same amount of heat out of the reservoir. Another way of seeing thisis to say, “OK, I fix the numerator (QC). How do I change the denominator (W) to increase theefficiency? I decrease it.”

70. BThis is a discrete question calling on your knowledge of thermal expansion. The formula for

the change in volume of a material as its temperature changes is:

V = �VT

Since you know the original volume, the volumetric coefficient of expansion, and the tem-perature change for the water in question, we can calculate the change in volume directly:

V = (3.67 � 10-3 Cº-1)(3 m3)(100 Cº) � 1.1 m3.

answer choice B. Making a calculation mistake will lead you to the other answers.

71. ATo calculate the power needed by the refrigerator, we need to know how much work the

refrigerator does. Energy must be conserved; since the refrigerator pulls 2200 J of heat energyfrom the cold reservoir (QC = 2200 J) , and puts 4000 J of heat energy in the hot reservoir (QH = 4000 J ), the work done by the refrigerator must make up the difference: W = 4000 J –2200 J = 1800 J. If the refrigerator take 5 minutes (or 300 seconds) to do this, then the powerused is:

P = �W

t� = �

1

3

8

0

0

0

0

s

J� = 6 W.

The correct answer is choice A.

Discrete Questions

72. CThe aluminum cube will experience forces due to gravity (weight) and due to the surround-

ing water (buoyant force). Since we don’t know the mass of the cube, we’ll have to calculate itusing our knowledge of its density and volume:

mcube = cube Vcube = (2.7 � 103 kg/m3)(0.01 m)3 = 2.7 � 10-3 kg.

So the weight of the cube is W = mcubeg = 2.7 � 10-2 N. The buoyant force is the weight ofthe displaced fluid. We can calculate the mass of the displaced fluid in the same way we calcu-lated the mass of the cube, since both take up the same volume:

mwater = water Vwater = (1.0 � 103 kg/m3)(0.01 m)3 = 1.0 � 10-3 kg.

21

And so the buoyant force on the cube is Fb = (1.0 � 10-3 kg)(10 m/s2) = 1.0 � 10-2 N. Thenet force on the cube will be downward, since the magnitude of the cube’s weight is greater thanthe buoyant force. The magnitude of the net force downward is 2.7 � 10-2 N � 1.0 � 10-2 N =1.7 � 10-2 N. This is answer choice C.

73. BThe best way to attack this question is to carefully examine each answer choice and see how

it fits in with the five assumptions of the kinetic theory of gases:

1. Gases are made of particles whose volumes are negligible compared to the container vol-ume.

2. Gas particles exhibit no intermolecular attraction or repulsion.

3. Gas particles are in continuous, random motion, undergoing collisions with other parti-cles and the container walls.

4. Collisions between any two gas particles are elastic, meaning that there is no overall gainor loss of energy.

5. The absolute kinetic energy of an individual gas particle is proportional to the averagetemperature of the gas, and is the same for all gases at a given temperature.

While we need not have these assumptions exactly memorized, we should know that kinetictheory assumes gases to be something like a dilute system of perfectly colliding balls: the ballsdon’t lose energy during collisions and don’t see each other unless they are colliding – i.e., theydon’t interact with each other and have radii much smaller than the distances between them.

Choice A says that molecules that are close to one another exert forces on one another. Thisis a direct violation of assumption 2. Choice B maintains that no energy is lost when a moleculecollides with the container. Assumption 4 tells us that these types of collisions are elastic (i.e.,no energy is lost). Choice B must be correct. Finally, let’s eliminate the last 2 to be sure. ChoiceC talks about “high pressures.” High pressure means that the gas particles are constantly bump-ing into one another – i.e., the separation between them has decreased. This goes against the firstassumption. Choice D may be tempting, but assumption 2 clearly states that no forces areexerted between collisions, it says nothing about during collisions.

74. BIt is certainly possible to work out this problem using variables, but it’s probably going to be

easier just to pick a few numbers and work out the consequences. Let’s pick 10 V for the bat-tery voltage, and 2 � and 3 � for the resistances. Since the two resistors are in parallel, we’llneed to find their equivalent resistance to determine the total current. Using the formula for par-allel resistors, we find:

�R

1

eq

� = � �2

1�� + �

31��; Req = �

6

5� �.

Using Ohm’s law, we find the total current in the circuit: I = �R

V

eq� = = �

2

3

5� A.

10 V�6/5 �

22

Now we remove one of the resistors: let’s take out the 3 � resistor. Since the total resistance

of the circuit is now 2 �, the total current becomes I = �1

2

0

�

V� = 5 A. The current has decreased,

so the correct answer is B. The total current doesn’t increase (A); the equivalent resistance ofthe circuit doesn’t decrease, it increases (C); the total voltage drop across a circuit is always thesame as the total voltage gained, which is constant in this problem (D).

75. AThe dipole moment of molecule is determined by the strength and orientation of all the elec-

tronegative groups. Ammonia has an electron lone pair in addition to three covalently bondedhydrogens, giving it tetrahedral electronic geometry. The lone pair, not drawn in any of theanswer choices, repulses the three bonding electron pairs. The nitrogen-hydrogen bonds arepolar covalent, with a partial negative charge on the electronegative nitrogen and partial positivecharge on each hydrogen. With the partial positive charges on the hydrogens and the partial neg-ative charge on the nitrogen, as well as the electron lone pair, ammonia has a dipole moment asshown in answer choice A.

76. DThe easiest way to answer this question is to remember the equation that gives the relation-

ship between the wavelength of a standing wave and the length of the pipe, when one end is

open and the other is closed: = �4

n

L�; n = 1,3,5,... From this we can see than when

n = 1, = 4L; answer choice (D) is correct.

If you didn’t remember the formula, perhaps making a sketch of the situation will help.Here’s a sketch of a pipe of length L, with the fundamental (n=1) wave drawn inside:

Since only one quarter-wavelength fits inside the pipe, for this case, = 4L, matching choiceD. Fortunately, we don’t need to consider n = 3, 5, etc. Notice that you can’t arrive at any of theother answer choices by choosing a legal value for n.

77. CThis type of question cannot be answered by prediction – we must read through the answer

choices until we find a true statement. Scanning through the answer choices, we can quickly seethat they are all statements regarding the directional flow of anions and electrons in galvaniccells. Now, recall that galvanic cells are spontaneous electrochemical cells. That is, in a galvaniccell a spontaneous chemical reaction is used to generate an electric current. In all electrochem-ical cells, electrons will flow from the anode to cathode. Therefore we can eliminate choices Band D. Now, we must determine the direction of anion movement in a spontaneous versus a non-spontaneous electrochemical cell. In all electrochemical cells, oxidation will occur at the anode(mnemonic: AN OX), and reduction will occur at the cathode (RED CAT). In a spontaneous cell,the anode will be negative since the electrochemical potential of the cell drives electrons to be

L

L = λ/4

23

move from anode to cathode. Since positive charges attract negative charges and vice versa,cations (positive always) will move toward the anode (negative in a galvanic cell), and anions(negative always) will move toward the cathode (positive in a galvanic cell). Choice C is there-fore correct.

24

25

VERBAL REASONING ANSWER KEY

78. D

79. A

80. D

81. D

82. D

83. D

84. C

85. C

86. D

87. B

88. B

89. B

90. A

91. B

92. A

93. B

94. B

95. C

96. B

97. C

98. C

99. B

100. A

101. C

102. B

103. D

104. A

105. A

106. D

107. C

108. C

109. D

110. A

111. B

112. B

113. C

114. D

115. A

116. C

117. C

118. D

119. A

120. B

121. C

122. A

123. C

124. D

125. B

126. A

127. B

128. D

129. D

130. B

131. D

132. B

133. D

134. B

135. D

136. C

137. D

Passage I (Questions 78–83)

Topic and Scope: The basic factors that shaped the U.S. political system explain both the fre-quency and the weakness of third parties.

Paragraph Structure: Paragraph 1: US third parties don’t tend to achieve national status.Paragraph 2: Contrast between the major parties and the third parties, and the reasons for the riseof the third parties and for their weakness. The rest of paragraph two and paragraph three gothrough the factors that shape US politics: first, the size and heterogeneity of the country; thefederal structure; and finally, in paragraph three, the winner-take-all voting system. Paragraph 4:the nondoctrinal character of US politics results and gives third parties their impetus. New issuestend to be ignored by the big parties and are taken up by third parties. Finally, paragraph 4 saysthird parties don’t achieve lasting power, they tend to remain isolated or fade, or they get co-opted by the major parties.

78. DThis question asks for a major reason for the rise of third parties. The answer focuses on the

biggest point made about the major parties and third parties, the one stated in the first sentenceof paragraph 4: new issues tend to be ignored by the major parties. (D) paraphrases that state-ment. That’s really the key to the situation: Issues like opposition to immigration or slavery, orthe rights of labor, which are all mentioned in the next sentence: (these are all nineteenth-cen-tury illustrations, by the way) aren’t taken up at first by the major parties. So they spark the riseof third parties. Choice (A) is never mentioned. It’s an example of an idea familiar in the out-side world that you should not choose as your answer, because the author of the passage hasnever mentioned it. (B) is also never discussed in terms of third parties. Regionalism is men-tioned up in paragraph 2 as a factor that major parties overcome by the diversity of their appeals.You might infer that third parties, with their more specific platforms, might not overcome it soeasily. But in any case, it’s never mentioned in relation to them. Choice (C) is implicitly contra-dicted. In lines 32-5, we’re told that mainstream voters have usually viewed certain issues asdivisive or threatening. Implicitly, these are issues raised by third parties. So the statement inchoice (C) is not true.

79. AQuestion 13 asks for a choice that did not contribute to the weakness of third parties. Your

answer here is (A). Paragraph 4 suggests repeatedly that third parties tend not to avoid but toembrace sharply defined programs. That same list of issues I was just referring to—oppositionto immigration, abolitionism, the rights of farmers and workers—they are all examples of thistendency. This choice just might be a little tricky because it stressed that the major parties dotend to avoid sharply defined programs, but not third parties–just the opposite. All the incorrectchoices are details mentioned about the major parties or about third parties. (B) is mentioned inparagraph 3; the winner-take-all system, instead of being a system of proportional representa-tion, rewards broad-based political strategies—those of the major parties, with their non-ideo-logical approach—and undercuts parties with more restricted support. This is referring to thirdparties, with their appeal to specific issues and constituencies. (C) is the flip side of (B). The ten-dency of at least some third parties to adopt programs with a narrow appeal is mentioned a cou-ple of times toward the end of paragraph 4 and in paragraph 5. Choice (D) is stressed at the endof the passage. When a third party does gain some prominence, like the Populists or the Social-ists, the major parties take over their programs, or parts of their programs.

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80. DThis question asks about the causes of the nondoctrinal character of U.S. politics. It’s a detail

question, but it focuses on big issues in the passage, not obscure points. Option I refers to a pointmentioned first in lines 11-13 and then elaborated a little in the next sentence. Option II refersto the next big factor, the federal system brought up in line 14. If you think of this as referringto the winner-take-all voting system, that’s less precise, but it doesn’t matter. That factor cuts thesame way. Finally, option III paraphrases the discussion at the end of paragraph 3 and the startof paragraph 4. It’s also virtually a restatement of the phrase “nondoctrinal character” in thequestion stem. All three have helped bring about the situation the question is asking about.

81. DThe question is asking for an inference: how would voters react to ideas put forward by a

third party? And the answer choices characterize their reactions in broad, descriptive terms—thesame pairs of adjectives and nouns that you get when you’re asked for the tone of the passageor the author’s attitude. According to the passage most voters don’t immediately support theissues raised by third parties. You get this most specifically around line 32. Mainstream votershave usually viewed certain issues as divisive or threatening, so that a dedicated minority—thatis, the third party—has been the one to put them on the agenda. So mainstream voters don’t likethese issues at first, and you want a negative choice. That narrows it down right away to (A) or(D), and (D) is the one that best reflects the tone of that phrase, “divisive or threatening.” Itworks much better than (A), which could be an initial response to new political ideas, but itdoesn’t fit the tone of that phrase. Think about it. If you view something as divisive or threaten-ing, your response will be “suspicio[n] and disapprov[al],” not “shock and disbelie[f].” So (D)will be our choice. The other three don’t fit at all. The idea that voters see these new issues asdivisive or threatening implies a definite and negative view. They are not “curious and open-minded,” (C), because they do have a sharp response. Finally, they are not “confused and inde-cisive,” (B), because they are negative, initially at least. (B) might be a better choice to apply toa later period when some of the third- party ideas have sunk in a bit, despite the initial negativeresponse.

82. DThis is a genuine detail question. It’s not dealing with one of the major points in the passage,

but one that comes up pretty much in passing. The Republican party is mentioned only once, upin paragraph 1, and the only thing we learn about it there is that it gained prominence when theWhigs were declining in the l850’s. So you should go right to choice (D). The Republicans wereable to become a major party because “a more established party was...in decline.” This is theonly thing you can infer from the one fact that’s given about the Republicans. The other choicespick up on irrelevant parts of the discussion or are never mentioned at all. (A) refers to a pointthe author makes about major parties in general. But here he or she never tells us anything aboutthe Republican’s program, so it won’t fit. (B) steals a bit from the discussion of co-optation inthe last paragraph, but we’re never told that the “Whigs...[tried] to steal from the Republicanplatform.” Anyway, we are told that they were already declining. (C) is entirely invented; noth-ing is mentioned about any of this. It also happens not to be true.

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83. DThe answer doesn’t jump out at you in this one, but it is there, if you look carefully at para-

graph 3, where the electoral system is discussed. The very end of the paragraph says that voterswho support minor parties may be left without representation because of the nature of the win-ner-take-all voting system. Earlier in the paragraph, that system has been contrasted with thesystems of unnamed other countries that provide for proportional representation. Presumably, inone of these countries, some minority party or parties would be represented in Congress or Par-liament on the basis of their proportion of the vote. That is the underlying, unstated idea, and itjustifies choice (D). Those systems would give representation to some voters that don’t get it inthe U.S. system. Of the wrong choices, (A) is a somewhat implausible idea. Why would a moredecentralized system, like the U.S.’s, be less flexible? Normally, one associates centralism withrigidity, but in any case, this issue–flexibility–is never raised at all. (B) also makes comparisonsthat never come up in the passage. The relative importance of the three levels of government isnever mentioned. (C), finally, is one of those just-the-opposite choices. The main point theauthor is making about the U.S. system is that it discourages this kind of polarization.

Finally, in the last paragraph, we get the reasons why third parties don’t achieve lastingpower. The ones that raise marginal issues or issues with short-term appeal tend to remain iso-lated or fade. The ones that do raise important issues get co-opted by the major parties, whichhave much greater resources—sort of a catch-22 situation.

The major thing to keep your eye on here is that opening generalization that everything thatgives U.S. politics its basic character also explains why third parties keep rising, but never getanywhere. Time after time, the questions will be focusing on that same idea. If you keep that inmind, and the basic structure of the passage--first, the basic factors about U.S. politics, then thethird-party stuff in paragraphs four and five--you shouldn’t have real trouble with the questions.They all focus on those big points about the passage.

Passage II (Questions 84–90)

Topic and scope: Health insurance for seniors in the United States, specifically, the problemsin providing prescription drug benefits to seniors.

Paragraph structure: Paragraph 1 introduces the Medicare program, identifies its historicalsimilarity to and current differences from typical private insurance programs, and its omissionof outpatient prescription drug coverage. Paragraph 2 mentions two possible approaches to rem-edying the omission and introduces the Medicare+Choice (M+C) program as a basis for com-parison. Paragraph 3 and 4 describe the M+C program and its problems. Paragraph 5 warns thatone of the alternatives mentioned in paragraph 2 may be subject to the same problems as the pre-vious M+C program.

84. CThe passage identifies that there is already consensus that an OPD should be added to Medi-

care. Thus, (D) may be eliminated. While the passage discusses the history of the Medicare pro-gram and contrasts it with private medical insurance, it does so in order to accomplish itsbroader purpose of criticizing a particular approach to expanding coverage. Thus, (C) is correct,and (A) and (B) may be eliminated.

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85. CThe third paragraph states that few plans have chosen to participate, and many beneficiaries

lack access to the plans. Therefore, (C) does not describe a success of the M+C program and isthe correct answer. (A), (B) and (D) describe successes of the program: SOME seniors had morechoices (choice A); 84% of seniors who chose participating plans had coverage in 1997 and thisincreased to 87% in 1998 (choice D); most participating seniors had the benefit (choice B).

86. DThe passage strikes a cautionary note about the author_s expectations regarding success of

the proposal, if adopted, but provides no basis for determining whether the proposal will or willnot be adopted. Thus, (A) and (B) can be eliminated. (C) is too extreme. (D) is correct becausethe passage indicates that Medicare+Choice had some success, and this provides a basis to inferthat the current proposal might also have some success, but that this will be limited.

87. B(C) can be eliminated because the passage does not question the motives of those who advo-

cate providing the OPD benefit through government subsidies to private plans. Rather, the pas-sage questions whether the method will be effective in practice. Thus, (B) is correct. The passageindicates that there may be lack of agreement on which method should be used, but not that thiswill cause lack of action (A). The passage suggests that (D) is a reason that Medicare did nottraditionally offer an OPD benefit, but not that (D) is a reason Medicare will offer it in the future.

88. BThe second paragraph describes reasons that alternative plans end up either limiting OPD

benefits, charging beneficiaries higher premiums, or both. (B) is an additional reason this couldoccur, would thus further support for the argument, and is the correct answer. (C) cannot rein-force the argument as it is inconsistent with information in the passage. The passage does notstate that beneficiaries fail to choose alternative plans. (D) is plausible, but off the mark in thecontext of the paragraph.

89. BThis question is best answered by an elimination strategy. Choice A is eliminated because it

is supported by the second sentence of paragraph 1. (D) is eliminated as it is supported by thelast two sentences of paragraph 1, which indicate BOTH plans have undergone change. (C) iseliminated because it is supported by second sentence of paragraph 2. This leaves (B). Althoughboth plans have changed, there is no specific evidence or assertion to establish that the changesin the Medicare program increased plan efficiency.

90. AAlthough the passage begins with the premise that an OPD benefit is necessary, its primary

focus is on identifying an effective mechanism for providing the benefit. Thus (D) can be elim-inated as it is not most central to the author’s probable reaction. (A) is correct because it focuseson the specific problems that have been encountered in the Medicare+Choice program and thatcan, by logical extension, be applied to the PBM proposal. (B) and (C) are incorrect because,while these are plausible critiques of the proposal, they are not critiques that are consistent with,or predictable from, the direction of the critiques expressed in the passage.

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Passage III (Questions 91–97)

Topic and Scope: The passage presents an overview of clinical allotransplantation, then abriefer discussion of clinical xenotransplantation.

Paragraph Structure: Paragraph 1 gives Fox & Swazey’s three questions for experimentalmedical treatments. Paragraphs 2 and 3 cover the history of allotransplantation, and paragraph4 is the history of xenotransplantation.

91. BThe first question receives the most attention, and is most central to the author_s conclusion

that comparison of the level of laboratory success in allotransplantation prior to 1964 with thelevel of laboratory success which currently exists in xenotransplantation suggests that xeno-transplantation will soon be evaluated further in the clinical arena. The second half of the sec-ond paragraph addresses the issue of whether the Barnard procedure, an initial clinicalexperiment, was justified in terms of prior laboratory success. Most of the third paragraphfocuses on the lack of laboratory success sufficient to support the 1964 xenotransplantation andthe subsequent laboratory success. The remaining second question is relevant to the criteria bywhich Barnard selected his first patient (the third sentence of the second paragraph). The thirdparagraph provides a basis to infer that expansion of the scope of clinical evaluation hasoccurred since Barnard’s initial procedure, but does not directly address the issue of what levelof initial success warranted this expansion, which is the focus of the second question.

92. AThe second paragraph indicates that the 1967 procedure was highly controversial; the fourth

paragraph indicates that the 1964 procedure was also criticized. Paragraph 4 explains that the1964 was not preceded by laboratory success, but was followed by laboratory success. Labora-tory success is not mentioned in connection with the period following the 1967 procedure, ratherthe focus is on clinical progress.

93. BThe xenotransplantation in 1964 was the first clinical experiment, and was not preceded by

success in the laboratory. The premise presented in the first paragraph is that some level of lab-oratory success is required before a clinical experiment should be attempted. Thus, the passageindicates the author would consider that procedure not a legitimate clinical experiment. There isnothing in the author_s argument to suggest that he would consider a procedure an appropriatetreatment that had achieved neither laboratory nor experimental clinical success. In the secondparagraph, the passage indicates that the author agrees with editorial suggesting that the allo-transplant performed by Barnard in 1967 was a legitimate experiment but not a treatment. If theauthor does not consider the 1967 procedure a legitimate treatment, there is no basis to believethat he/she would consider the 1964 procedure a legitimate treatment.

94. BThe increase in survival rate is mentioned in the third paragraph. While it is plausible that

changes in immunosuppressive medications and availability of donor organs could affect accep-tance of the procedure, the passage itself provides no basis information on the degree of or direc-tion of change in these areas. The number of procedures performed is not discussed.

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95. C(A) defines one type of xenotransplantation, but the passage also refers to transplantation

between two different species other than human beings as xenotransplantation in paragraph 4.

96. BEach time the passage refers to “clinical” experiments, humans are involved; “laboratory”

experiments, on the other hand, can be part of the basis for determining that “clinical” experi-ments are warranted. So “laboratory” experiments are those that do not involve humans. There-fore, “clinical” procedures include both I and II, but not III.

97. CFox and Swazey are indicated to have identified three questions. The first transplant into a

human being would involve the first question as it involves movement from the laboratory exper-iment to a clinical experiment, and as the conditions under which the initial xenotransplant tookplace are not identified. The passage references lack of any laboratory success as the basis formoving into the clinical arena (paragraph 4), and compares it to the record of laboratory successin allotransplantation prior to the first allotransplant. Thus, the justification was differentbecause it was nonexistent.

Passage IV (Questions 98–104)

Topic and Scope: History and causes of “breeches” roles in English Restoration theatre.

Paragraph Structure: Paragraph 1 is the history of the breeches role. Paragraph 2 gives detailsof the nature of the roles. Paragraph 3 is the analysis by “most” modern critics, focusing on howthese roles exploited women. Paragraphs 4 and 5 state the author’s opinion about how these rolesbenefited women.

98. C

The fact that female playwrights created breeches roles helps to reinforce the author’s claimthat 18th century women did not object to this type of role. (A) is a Distortion, for there is nomention of how these playwrights created these roles or what their intentions were in doing so.(B) and (D) are Outside the Scope of the passage since no mention is made of protest againstbreeches roles nor does the author ever criticize the playwrights.

99. BThe author begins with a brief sketch of the socio-historical context in which breeches roles

emerged, followed by details about this role, presents a discussion of contemporary criticism ofthese roles, and ends by contradicting this criticism by justifying advantages of these roles. Asfor the wrong choices, (A): the author doesn’t support the modern interpretation of breechesroles. (C): Although the other offers some criticism to modern objections to this role, the major-ity of the final paragraphs is dedicated to illustrating the positive aspects of the breeches role.(D): The third paragraph does not focus on the impact of the breeches role on women.

100. AParagraphs 3 and 4 mention several reasons why women benefited from the royal permis-

sion to act in public (control, new career option, etc.). (B) is never mentioned in the passage. (C)is Outside the Scope, as is (D).

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101. CIn the first paragraph, we are told that women were forbidden to act in public because of fear

that they would provoke illicit behavior in the audience or corrupt the morals of actors. Theassumption, then, is that men acting on the public stage would not produce similar negativeeffects. (B) and (D) are OS and never mentioned in the passage, and (C) is an irrelevant com-parison.

102. BThe amount of compensation women received is never mentioned. (A) appears in Paragraph

3, and (C) and (D) are found in Paragraph 4.

103. DThis is a paraphrase of the final sentence of Paragraph 2. (A), (B), and (C) are all half-wrong:

(A) The author declares that some parts of a modern interpretation may be valid, but goes on toillustrate how factoring historical context into the equation results in a much different interpre-tation; (B) The author does not discount all of contemporary criticism, but rather cautionsagainst ignoring differences in socio-historical contexts; (C) Finally, the author does validateparts of the standard modern interpretation of the breeches role, but then goes on to point outhistorical influences that counter such a critical interpretation.

104. ANo mention is made of the production values like scenery, lighting, costumes, or script qual-

ity in terms of how they impacted audience response to actresses on stage. (B) is implied in thefirst sentence of Paragraph 3; (C) the appeal of actresses’ showing off their legs and being ableto act and speak well are mentioned in Paragraph 4; (D) Paragraph 3 describes new situationscreated by the breeches role.

Passage V (Questions 105–111)

Topic and Scope: An exploration of the validity of the “Mozart Effect” in education.

Paragraph Structure: Paragraph 1 gives the origin and history of the “effect.” Paragraph 2describes the results of research on it. Paragraph 3 contrasts the benefits of musical training.Paragraphs 4 and 5 cite the real issues raised by the “effect” and paragraph 6 states the author’sconclusion.

105. A(B) does not make any sense, since there is nothing in the passage about meteorology. There

is a list of types of intelligence, but Mozart intelligence is not on it, so (C) is wrong. Heritabil-ity better describes how much something can be explained by heredity, so (D) is wrong.

106. DThe answer to this is directly stated in the passage in paragraph 4. (A) contradicts what is

stated in this paragraph. (B) contradicts the point of the entire passage. There is no support forchoice (C). The passage clearly states that intelligence is composed of multiple abilities.

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107. CThe author seems to think that I is true, which is why he thinks that starting something ear-

lier rather than later is generally better. By saying that it makes sense that children should startformal instruction earlier, the author implies that play serves no instructional purpose, so theauthor assumes II. Of course, if children could not benefit from instruction, it would make littlesense to offer it to them, so the author assumes III.

108. CStarting on line 24, the author explains that the Mozart Effect has had only a small and brief

impact on any population. The author then cites the results of several studies, claiming that theimpact of the Mozart Effect is indistinguishable from the variability in results expected whenpeople take the same test more than once. There is no evidence for (A). (B) and (D) are bothcontradicted by information in the passage.

109. DHistory is not once mentioned in the passage, so (A) is incorrect. While (B) and (C) are gen-

erally considered a part of intelligence, they are not facilitated by the Mozart effect. (D) is theonly thing specifically mentioned in the passage.

110. AThe passage does not discuss whether twins show similar effects from listening to Mozart,

so (B) is incorrect. Identical twins are discussed prior to the discussion of kinds of intelligence,so (C) is incorrect, but they are discussed in the context of heritability, so (A) is correct. Thereis no mention about twins and playing instruments, so (D) is incorrect.

111. BThe author clearly states that the effect, however small that it is, is not limited to Mozart.

But, since the effect is small, listening to others composers would offer no better benefit, so (A)is incorrect. The author never discusses listening to music while studying, so (C) is incorrect.The author never discusses nutrition, so (D) is incorrect because it is Outside the Scope of thepassage. (B) is clearly implied by the author noting that individuals who learn to play an instru-ment have shown some improvements on intelligence tests.

Passage VI (Questions 112–118)

Topic and Scope: Problems of determining a definition of “realism” in literature

Paragraph Structure: Paragraph 1 introduces the idea of realism and its fundamental basis, andthen highlights an important problem with this term. Paragraph 2 describes one of the firstattempts to arrive at a definition of realism and of how it relates to mimesis; Auerbach’s work andits contributions, weaknesses and contradictions. Paragraph 3 presents a contemporary definitionof realism and a more recent theory of language, and an example that upholds Kristeva’s semi-otic theory of language. Paragraph 4 links common aspect of Auerbach’s and Kristeva’s studies,and shows the consequences of a semiotic system that becomes unstable when removed from itsproper historical context. Paragraph 5 summarizes the important factors that influence realism,and concludes that a satisfactory fixed definition of the term has not yet been formulated.

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112. BThe author provides varying definitions of realism in Paragraphs 2 and 3, and also poses

leading questions – implying doubt about being able to define this term – at the end of paragraph1. (A) is a distortion, since the author tells us in paragraph 4 that there are certain similaritiesbetween Auerbach’s work and the study of semiotics. (C) is opposite from what paragraph 3explains, and (D) is opposite from what is mentioned in paragraph 2.

113. CThis is a paraphrase of the first sentence of the passage. (A) is FUD – although this common

understanding is mentioned several times, it is also questioned in paragraph 2 and does not pro-vide a foundation for the theory of realism. (B) is opposite from what paragraph 2 tells us aboutmimesis. (D) is a Distortion and contradicts the first sentence of paragraph 4.

114. DThe currency is the arbitrary sign, the value assigned to it is a mental signifier, and the goods

and services purchased represent the signified objects. (A) and (B) are simple bartering systemsand are not based on arbitrary signs. (C) uses signs—colored chips—but there is no transfor-mation from the chips to the cash, since they represent the exact same value.

115. AThe author gives no indication of downplaying the importance of Auerbach’s work, and

would not mention it if it did not have some relevance to the topic and scope of the passage.However, the author also points out problems with Auerbach’s study at the end of the paragraph.(B) goes too far, and the second part of it is a Distortion of the author’s criticism of Auerbach’swork. (C) begins with a Distortion of the author’s initial evaluation of Auerbach’s work. (D)begins with an adjective that is a bit too strong, just like (B), and becomes outside the scope atthe end with the mention of “other literary genres.”

116. CThe author’s intent is completely outside the scope. (A), (B), and (D) are all mentioned in

the first sentence of paragraph 5 as being important to the theory of realism.

117. CThis is a summary of the main idea of paragraph 3. (A) and (B) are Distortions of the rela-

tionship between Kristeva’s and Auerbach’s theories as explained in paragraphs 2 and 3 and thenintegrated in paragraph 4. (D) is Opposite from one of the consequences of semiotics mentionednear the end of paragraph 3.

118. DThis a paraphrase of the last part of the first sentence in paragraph 5, showing how the author

and reader must share a certain common understanding to achieve effective communication. (A)is FUD since literary techniques are only one of the elements mentioned in paragraph 2. (B) and(C) are opposite from what we are told in the first sentence of paragraph 3.

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Passage VII (Questions 119–124)

Topic and Scope: A study of the behavior of young children in nursery school and at home.

Paragraph Structure: Paragraph 1 outlines the study and paragraph 2 gives the categories andthe procedures. Paragraph 3 starts analysis of the data. Paragraph 4 discusses correlations foundbetween sibling relationships and behavior in school, which paragraph 5 discusses only correla-tion found between behavior at home and behavior at school, and the reason for the general lackof such correlations.

119. ATotal observation time was sixty-eight hours; thirty-four subjects were observed for two

hours each. (B) is half the correct amount; you counted either one hour of observation time persubject or seventeen subjects. (C) represents a misreading; in each setting for each subject anhour of observation time was collected in twenty periods of three minutes' duration each. Thereis no discussion of how long collecting the data took overall (D).

120. BAccording to paragraph 3, the subject's age (I) and sex (III) demonstrate the greatest num-

ber of correlations (B). According to paragraph 4, “having a same-sex sibling” (II) revealed cor-relations with both “onlooker behavior” and adult-child interaction in the nursery school, butthere is no indication that these correlations extend to other types of behavior, and the questionasks which variables correlate “most frequently” with “behavior” in general. Therefore, choice(B), I and III only, is our answer.

121. CAccording to the study results given in paragraph 3, solitary play varies inversely with the

age of the preschool child. Paragraph 2 discussed solitary play in the context of Parton's socialplay hierarchy, placing this activity early on the scale and implying that Parton would expect tofind this type of play more among younger preschoolers and less among older preschoolers.Thus the study under discussion tends to confirm Parton's hierarchy insofar as the two studiescan be compared, and (C) is correct. The study reveals nothing about the specificity or general-ness of activities, but the passage defines these categories in paragraph 2, where solitary play issaid to be a general class of behavior (A). (B) is true of solitary play in the home, not in the nurs-ery school. The study does not suggest that solitary play is rare among preschoolers (D). All weknow from the study's findings is that solitary play tended to vary inversely with the age ofpreschoolers. The findings tell us nothing any more specific about the incidence of solitary play.

122. AThe last paragraph suggests one problem with the study might be that differences between

the various home environments were too great in relation to the number of subjects in the study;a larger subject pool drawn from more homes (A) would remedy this situation, since you wouldpresumably then get groups of children with similar homes—making it possible to statisticallycorrect for selected home variables and treat home environment as a constant. The author neversuggests that observation time per subject is insufficient; nor would increasing such observationtime even indirectly remedy the other possible problems discussed in the last paragraph (B). (C)runs contrary to paragraph 2, since breaking the observation time into three minute periodsapparently yields more reliable results. The second part of (D) would solve the same problem aschoice (A), but reduction of the subject pool would impair the results; in any case, preselectingfor similar home environments might distort the study in other ways.

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123. CAccording to paragraph 4, “onlooker behavior” most closely related to both “having a sis-

ter” and “having a same-sex sibling.” Since a girl with an older sister fulfills both these qualifi-cations, (C) is correct. Choices (A) and (B) each involve only one of these characteristics.Solitary play at home (D) appears unrelated to onlooker behavior, but if anything one wouldexpect these two behaviors to be inversely, not directly, related, since solitary play at home isinversely related to “having a sister” (paragraph 3) while onlooker behavior is directly correlatedwith the same variable.

124. DWhen sibling-related variables revealed correlations with school behavior (paragraph 4), the

passage took this as evidence for the influence of the home environment on interactions outsidethe home; using analogous reasoning, this question requires an example in which some descrip-tive variable at school correlates with some behavior at home. The only answer choice whichdoes this is (D). In (A), since sibling-unrelated variables (specified in the passage as the age andsex of the subject) are also school-unrelated, the nursery school environment does not influencehome behavior. (B) presents an example of the home environment influencing home behavior.(C) offers no indication of causality; school could be influencing home, or vice-versa, or someother variable could be influencing both (analogous to “demonstrates aggression” in the lastparagraph and creating the same problems of interpretation).

Passage VIII (Questions 125–131)

Topic and Scope: The author argues against a party system in a government elected by thepopulace.

Paragraph Structure: Paragraph 1 argues parties are “factions” dividing the populace. Para-graph 2 says parties operate through rivalry and revenge, at worst this leads the people to seeka despot for stability. Paragraph 3 says even if this extreme isn’t reached, constant rivalry of par-ties is distracting from the work of governing. Paragraph 4, some think parties serve as a checkon the administration, but the author believes there is sufficient tendency to disunity in humannature to serve this purpose without institutionalizing it.

125. BThe term as used in the passage refers to feeling and passion aroused by an organization

directed toward a political goal. Thus (B) is the correct answer. (A) is too general. The passageidentifies the “spirit of party” as “having its root in the strongest passions of the human mind.”It is, thus, not part of the mind, but an impulse that springs from the mind. (C) is wrong becauseit omits the important characteristic of parties that distinguishes them from factions: they areorganized. (D) is incorrect because it refers to organizations in general rather than to the specifictype of organization discussed in the passage--an organization directed at political goals.

126. AThe passage states a belief that the tendency to have strong passions is inseparable to our

nature, and that the “spirit of party” is a natural tendency, which cannot be quenched, but onlycontrolled. Thus, the author would probably dismiss as impossible to achieve the goal of indi-viduals being free of selfish interests and advocate, instead, that these individual tendencies bemitigated and assuaged.

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127. B(B) is correct. The author refers to (B) as an “extremity” occurring “at length,” i.e., not

immediately, and contrasts it to the “common and continual” mischiefs of the “spirit of party”.The remaining choices all paraphrase these other mischiefs: (A) paraphrases the second sen-tence of paragraph 3, (C) the last sentence of paragraph 3, and (D) the first sentence of para-graph 1. A “faction” is by its dictionary definition a “contentious minority.”

128. DThe metaphor has two parts: The first part is that “spirit of party” is not to be “quenched”

(destroyed or eliminated). This reflects the first aspect of the author_s argument, suggested inthe last two sentences of the first paragraph: “. . . the ‘spirit of party’ is inseparable from ournature . . .” and it “. . . exists everywhere . . .” This aspect of the argument is taken further in thelast paragraph: “parties . . . serve to keep alive the spirit of Liberty.” but there will always exist“. . . enough of that spirit [in elective governments] for every salutary purpose. . . .” This part ofthe metaphor could be applicable to all four answer choices. The second part of the metaphor isthat the “spirit of the party” must be managed so that it does not “consume.” What might thespirit consume? “Liberty,” which exists under elective, popular governments (democracies).Thus, (D) is the correct choice. Although it could be argued that the “spirit of party” could also“consume” monarchies or dictatorships (A and C), the passage is not concerned with that pos-sibility. Patriotic movements (B) may be the basis for formation of a party, but the author doesnot warn that these movements may be “consumed” (i.e., destroyed) by the “spirit of party”.Rather, the argument suggests that patriotic movements may be strengthened and stimulated bythe “spirit of party”--to the detriment of an established elective government.

129. DD is correct since the passage speaks of the “delegated” will of the nation. What one dele-

gates one authorizes to act in one’s behalf. (A) and (C) are incorrect as there is no indication inthe passage that a government that is more liked would foster greater “spirit of party.” Indeed,since this “spirit” attempts to “counteract, control, or awe the regular deliberation and action ofthe constituted authorities,” it is logical to infer that the “spirit” would flourish most readilywhen the government was acting in such a manner as to make it LESS popular with the people.(B) is incorrect because delegation means the authorization to act for, not the tendency to actwith. Furthermore, the focus in the passage on the detrimental effects of factions, associations,and other combinations and on the desirability of discouraging them indicates that maximal par-ticipation of the people at large in the functions of the government is not a necessary to the typeof government being discussed.

130. BThe passage accepts the “spirit of party” as inevitable, but counsels against it being orga-

nized and encouraged. What the passage does encourage is the use of popular opinion to “miti-gate and assuage” the effects of the “spirit of party.” Thus, the writer would probably encourage(A), (C), and (D). Rebutting any statements made by the dissenters (B) could have the effect ofinflaming the disagreement and producing more polarization of opinion.

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131. DIn the last sentence of paragraph 1 the passage states that the “spirit of party” exists in all

governments. Thus elective government (D) cannot be fundamental to the “spirit of party.” Thepreceding sentence specifies that the “spirit of party” is inseparable from our nature. Accordingto the first sentence of the passage, parties organize (C) faction (B) toward political action.

Passage IX (Questions 132–137)

Topic and Scope: The Federal Reserve System’s purpose and operations.

Paragraph Structure: Paragraph 1: Typical explanations of Fed operations are purposely mis-leading; some key concepts that few people fully appreciate. Paragraph 2: the standard explana-tion of the Fed as “lender of last resort;” various different terms often used to describe thisprocess are identical and can be best understood as: The Fed buys assets from a bank to increasethe bank’s “cash.” Paragraph 3 explains two tools the Fed uses to achieve its goal of increasingor decreasing the money supply. The discount rate tool and reserve requirement changes. Para-graph 4: Open Market operations are the primary monetary tool, though they were hardly men-tioned or used in the early years. The purpose is to manipulate the money supply in a mannerthat accommodates the government’s desire for deficit financing. Paragraph 5: Control of moneysupply is sole purpose of the Fed.

132. B Selling foreign currencies is not mentioned in the passage and not part of monetary policy.

All the other choices are specifically referenced in the passage: purchasing government securi-ties (A) on line 56, lowering the required reserve ratio (C) on line 45, and selling governmentsecurities (D) on line 59.

133. D It is inferable that most people are unfamiliar with the essence of the Fed’s operation; most

of paragraph 1 provides the evidence. The first five lines state the author’s contention that mostdescriptions of the role of the Fed are confusing. (A) is not correct; the purpose of the passageis to provide an explanation of what the Fed does. The author doesn’t express a strong opinionabout how the banking system should operate (C), and the author also doesn’t express an opin-ion about how the Fed handles open market operations (B). The only strong opinion the authorexpresses is that most people don’t understand what the Fed does because most descriptions ofit serve to obscure instead of illuminate.

134. B Choice (B) is stated explicitly in paragraph 1. (A) is a Distortion; banks don’t create debt.

(C) is outside the scope; the relationship between interest rates and inflation is never mentioned.(D) is not supported by the passage; reserve requirements are discussed in paragraph 3, but nothow often they are changed.

135. D There is no discussion whatsoever of open market operations being done in secret in paragraph

4. Each of the other choices are stated explicitly (A) or can be reasonably inferred (B and C).

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136. C (B) and (D) are Opposite. Furthermore, the generous wording used to talk about the oppo-

nents on fractional reserve banking is a tip-off. On the other hand, the term “fractional reservebanking” does not appear anywhere in the text, and can therefore make the question harder.There is no support for (A) in the passage.

137. D All three statements are supported. I by inference; II, a main theme at beginning and end but

may be difficult to ferret out in view of the discussion of lender of last resort as, perhaps, some-thing significantly different and substantial in degree; III explicitly stated at beginning.

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MCAT FULL-LENGTH 6 WRITING SAMPLE REPRESENTATIVE RESPONSES

Essay 1 Statement:

Prisoners should be granted the same rights as the other members of free society.

Sample Response: (Level 5–6)

As the prison population continues to rise in America, the issue of treatment of inmates has come into thelimelight. Many champions of reform question whether the harsh conditions found in some prisons canreally accomplish the goal of rehabilitating criminals. Furthermore, apart from pragmatic concerns, onemay object to the treating of prison inmates as less than fully integrated citizens as a violation of the fun-damental principle upon which our society is founded. The Constitution and its Amendments guaranteeto the American populace certain fundamental, inalienable rights, including among them the right to dueprocess of law and freedom of religion. These rights, according to some, should not be removed or dimin-ished because of an individual’s status as an inmate. The statement that “prisoners should be granted thesame rights as the other members of free society” arises from this sentiment.

However, it is obviously absurd to pretend that prison inmates are or should be equal to members of freesociety in every respect. As members of free society, for example, we enjoy the right to enter into arrange-ments of our own choosing of buying and selling. Granting this right to perpetrators of crimes wouldenable them to benefit from their wrongdoings if they were to sell their stories to members of the mediaof dubious ethics. This practice, if it were allowed to occur, would constitute disastrous public policy byproviding an incentive to commit heinous crimes and would be a flagrant offense to public sensibility.

In trying to delineate the distinction between rights of which people should never be deprived and rightsthat could justifiably be taken away from inmates, it is necessary to keep in mind the goals of incarcera-tion. First and foremost, we would like to see the prison term serve a rehabilitative purpose. Accom-plishing this goal will necessarily involve depriving inmates of access to some comforts and rights, inorder to induce them to appreciate the benefits of being free, law-abiding citizens. At the same time,human nature simply dictates that there will always be in addition a retributive aspect to sentencing,designed solely for the purpose of inflicting some degree of misery upon the inmates. As long as the dis-comfort does not venture into the territory of cruel and unusual punishment, this vindictive sentiment,albeit hardly noble, is a reasonable one. The recent “victims’ rights” movement has reinforced this notion.In short, therefore, a prison sentence serves both a rehabilitative and a punitive goal. Any right that, ifgranted, would directly interfere with either of these goals should be denied.

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Essay 2 Statement:

The scientific pursuit of truth is flawed by economic and personal interests.

Sample Response: (Level 5–6)

The statement suggests that the practical realities of scientific study fall short of the scientific ideal ofobjectivity. Ideally, scientists study the physical and natural world using objective, reproducible experi-mentation and honestly disclosed results. Towards this goal, scientific advancement should be driven bya pursuit of truth and a progression of logic. However, both those who perform research and those whofund it may not be driven by truth, and may instead be controlled by personal interests, both economicand otherwise. Frequently, the bottom line can take priority over the objective pursuit of truth. For exam-ple, biotechnology and pharmaceutical companies – home to some of the world’s largest collaborative sci-entific research efforts – have no interest in discovering the fundamental causes of and solutions todisease. Rather, these companies are interested in producing drugs or products that are economically prof-itable. Research that cannot demonstrate significant market potential will be dropped, even if it is objec-tively worthy of study. Indeed, the great technological advancement of the last 100 years is closely tiedto the confluence of science and economics.

Yet, scientific study is always constrained by the laws of the physical and natural universe. Advancementsnot based in firm science do not hold up to the strict scrutiny of the scientific community. Technologiesand theories are perpetually being put to the test by those with no vested interest in their success. Forexample, despite a considerable economic imperative to push a new drug onto the market as fast as pos-sible, a pharmaceutical company must pass through a series of scientific hurdles – clinical trials and FDAapproval – before it can sell its product. And once on the market, clinical studies will continue to test theeffectiveness of the drug. The economic interests of one company cannot bend the rules of science.

Though considerations both economic and otherwise may select and direct the advancing frontiers ofhuman knowledge, the scientific pursuit of truth retains its objectivity so long as experiments are repro-ducible, theories are testable, and outcomes are not misrepresented. If a drug’s effectiveness were to beevaluated solely by the drug’s maker, then yes, scientific objectivity may become compromised. But solong as others can confirm these results, then the pursuit of truth has not been flawed. That some packetsof human knowledge are deemed more attainable or more useful makes them no less factual and no lessscientific. Since experimental science must be performed in the real, physical world, it is unreasonable toassume that practical interests cannot become an influence. And even so, these influences are kept incheck by the laws of the physical and natural universe, as well as the objectivity of human scientific dis-course.

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43

BIOLOGICAL SCIENCES ANSWER KEY

138. B

139. A

140. B

141. B

142. B

143. A

144. C

145. B

146. D

147. A

148. B

149. C

150. A

151. B

152. D

153. B

154. B

155. C

156. B

157. D

158. B

159. C

160. A

161. D

162. C

163. A

164. A

165. A

166. A

167. D

168. C

169. A

170. B

171. D

172. D

173. A

174. C

175. A

176. D

177. B

178. A

179. C

180. A

181. A

182. C

183. C

184. A

185. B

186. A

187. B

188. C

189. A

190. B

191. C

192. D

193. D

194. D

195. A

196. B

197. D

198. D

199. B

200. B

201. A

202. B

203. B

204. A

205. C

206. D

207. D

208. C

209. A

210. B

211. D

212. B

213. C

214. B

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Passage I (Questions 138–143)

138. BThe passage states that the mutant Hb-S has a valine residue instead of a glutamic acid

residue at position 6 in the b chain of the hemoglobin. For this substitution to occur in every pro-tein and be heritable, a mutation must occur at the DNA level. Choice A is wrong because her-itable mutations do not occur at the protein level. Choice C can be eliminated as potentiallylethal since a ribosomal translation error implies that every protein made in the cell, not justhemoglobin, would have improper additions of valine in place of glutamic acid. Choice D iswrong because mRNA is a transient and renewable molecule, and mutations in the mRNA arenot heritable.

139. ASince both parents are heterozygous, their genotype for sickle-cell would be “Ss”. For their

child to have the full-blown sickle cell disease, he or she must be homozygous. To find out thechance of their child’s being homozygous, one can set up the appropriate Punnett square dia-gram. The Punnett square diagram shows that 1/4 or 25% of the possible gametic combinationswill result in the homozygous SS genotype, as shown below:

S s

S SS Ss

s Ss Ss

140. BAccording to the Hardy-Weinberg equation, the sum of the frequencies of the dominant and

recessive alleles for a particular trait, p and q, is equal to 1. In other words, p + q = 1. When onetakes the square of both sides of the equation, the result is: p2 + 2pq + q2 = 1, where p2 is thefrequency of the homozygous dominant genotype, 2pq is the frequency of heterozygous geno-type, and q2 is the frequency of the homozygous recessive genotype. Since we know that the fre-quency of the sickle cell disorder is 4% or 0.04, we know that this number represents q2 becausesimply one p or one q does not give the disease – an individual must have TWO alleles to haveany trait – hence, the chance of having one sickle-cell allele is q, and the chance of having twoalleles is q x q or q2. The question asks, however, about the heterozygotes or carriers, which arerepresented by 2pq in the equation. The mathematics for determining the frequency of the het-erozygous (carrier) genotype is as follows:

q2 = .04q = 0.2

Since p + q = 1 then, p = 1 - qp = 1 - 0.2

p = 0.8

Then, the frequency of the heterozygous genotype is equal to 2pq or 2(0.8)(0.2) = 0.32, andthe percentage of these individuals in the population can be found by multiplying this frequencyby 100 → 0.32 � 100 = 32%. The closest answer to this is Choice B, 34%.

...

141. B This question requires knowledge of different types of mutations. A missense mutation

occurs when a codon for one amino acid becomes a codon for another amino acid as is the casein sickle cell anemia. A nonsense mutation results when a codon for one amino acid becomes astop codon. The outcome is typically a mutant protein which is shorter than the wild type pro-tein and functionally useless. A deletion mutation, as the name suggests, occurs when one ormore nucleotides are deleted from the DNA sequence. This is often a serious mutation resultingin loss of a protein’s function. A frameshift mutation, which can result from deleting or addinga base into the DNA sequence, is another example of a serious mutation. The number ofnucleotides deleted or inserted is a non-multiple of 3, which causes a shift in the codon readingframe and results in a sequence of amino acids that is completely different from the originalsequence after the point at which the mutation occurred. Thus, Choice B is correct since sickle-cell involves a single amino acid substitution.

142. BAnti-clotting agents prevent blood clot formation caused by plasma proteins and platelets;

yet, sickled red blood cells clot because they jam in blood vessels due to their sickled, inflexi-ble state. Their clotting has nothing to do with the normal clotting response. Even if you didn’tunderstand this technicality of blood clotting, which may seem somewhat obscure, you couldeliminate all the other answer choices based on information readily available in the passageitself. Choice A is wrong because, according the last paragraph, polymerization of Hb-S isreduced as the concentration of Hb-F is increased. A drug that raises Hb-F and lowers Hb-S will,thus, be an effective treatment for sickle cell anemia. Choice C is wrong because a drug that pre-vents interactions between hemoglobin molecules will prevent the polymerization ofhemoglobin molecules that occurs in sickle cell anemia, and the drug may well be an effectivetreatment for this disorder. Choice D is wrong because ADH will increase the permeability ofthe collecting duct to water thereby absorbing more water and elevating blood volume. As statedin the passage, dehydration is one of the factors that can initiate sickling, so the ADH wouldkeep someone well-hydrated and help prevent the sickling.

143. AThis question tests your knowledge of the function of various organs/systems in the human

body. The spleen is an organ of the immune system and functions in clearing infectious agentsfrom the bloodstream. In addition, the spleen is one of the primary organs that function in thebreakdown and elimination of worn out, used red blood cells. The spleen is a repository of whiteblood cells that “scan” the blood as it filters through the organ – somewhat like a large lymphnode. Choices B, C and D can be eliminated because those are the functions of kidneys, thy-roid/parathyroid and pancreas/small intestine respectively.

Passage II (Questions 144–149)

144. CThe carbon atom in a carbene does not have a full octet: It only has six valence electrons

(two pairs of bonding electrons and one pair of nonbonding electrons). We would thereforeexpect it to react as an electrophile. This expectation is borne out by its reaction with the alkene,which is electron rich because of its pi (�) electron “cloud.” (In general, you may recall thatelectrophiles undergo addition reactions with alkenes and alkynes.) The pi electrons are used toform a sigma bond between the alkene and the carbene.

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46

Choice A is incorrect because Grignard reagent is an alkylmagnesium halide, with the gen-eral formula RMgX.

Choice B is incorrect because ylides are species with the following typical structure:

145. BEach of the products shown contains two stereogenic (or chiral) centers. The compound

shown on top is a meso compound: It possesses a plane of symmetry that cuts vertically downthe middle of the molecule. Such a plane of symmetry, however, does not exist for the bottommolecule, as one of the methyl groups points “up” (towards us out of the page) and the otherpoints down (into the page). The two molecules therefore differ in the configuration about onestereogenic center only. They are diastereomers. Notice that we do NOT need to actually assignany absolute (R/S) configuration.

Choice A is incorrect because the two molecules need to differ in the configuration of everystereogenic center in order to be enantioners (nonsuperimposable mirror images). One can alsoeliminate this answer choice because a meso compound does not have an enantiomer.

Conformational isomers (choice C) are isomers that can be interconverted through the rota-tion of single bonds. This is not the case here.

Choice D is incorrect because only one of the products is a meso compound.

146. DExamination of Equation 4 tells us that the Simmons-Smith reaction will lead to the creation

of a cyclopropane ring where a double bond used to be. (Very crudely, the double bond can bethought of as the location of the “base” of the triangle.) Only the compound in choice D willlead to the desired product. The other choices, if they were to react via this path, would lead tothe following products. (However, notice that some of the compounds shown are highly strainedand the desired reaction may not result.)

=

(C6H5)3P—CR2

+ –

147. Atert-Butoxide and ethene are reactants that we recognize from the reaction scheme described

under Method 1. So we know that this question is asking us to apply a given reaction scheme toa new set of reactants. In this reaction, instead of using chloroform, HCCl3, we are using 1-chloro-2,2-dimethylpropane. The carbene generated will be different. The abstraction of a pro-ton followed by the elimination of a chloride ion leads to the following carbene:

The proton that is abstracted is the most acidic because it is closest to the electronegativechlorine atom, which stabilizes the negative charge on the conjugate base. Since the chlorine isbonded to the same carbon atom (and not an adjacent one), elimination of the chloride ion doesnot lead to the formation of a double bond (as with E1 and E2 reactions) but instead leads to thecarbene.

The carbene then reacts with ethene as follows.

This is the compound shown in choice A.

148. BWe recognize that the reaction is similar to the one depicted in Equation 4, except with a dif-

ferent alkene. First, we need to determine the correct structure, then we apply the reaction to thecompound.

This is the compound shown in choice B, 1,1,2-trimethylcyclopropane.

149. CThe role of the tert-butoxide is to abstract a proton from the chloroform. In other words, the

ion acts as a base. However, we know that very often a base can also act as a nucleophile becauseof its lone pair of electrons, and in this case an SN2 reaction can occur: Chloride is a good leav-ing group, so we can obtain a substitution product Nu-CHCl2. To prevent this from happening,we employ a bulky base so that steric factors may favor elimination. Therefore, hydroxide is notas good a choice.

CH3 CH3

H3C

H3C

H3CH3C

CH3

CH3

CH3C

CH3H3C

CH3CHHC

H2C CH2

CH3t-butoxide

CH3

CH3CCl H2C

CH3

CH3

CH3CCl HC–

CH3

CH3

CH3C– Cl– HC

47

Choice A is incorrect because the two ions are comparable in basicity. (tert-Butoxide is infact expected to be a slightly stronger base because of the electron-donating alkyl groups.)

Choice B is incorrect because the tert-butoxide ion actually is nucleophilic, although it maynot be as effective as hydroxide as a nucleophile because of its bulkiness. (We do NOT want theion to act as a nucleophile anyway, but as a base.)

Choice D is incorrect because we DO want elimination to occur. The reaction leads to thenet elimination of HCl from chloroform. However, hydroxide ion may favor substitution instead.

Passage III (Questions 150–154)

150. AThe passage states that the transfer of electrons in the respiratory chain results in the trans-

port of H+ ions from the mitochondrial matrix to the intermembrane space. When an alternativeenergy source such as fatty acids is introduced, it can enter the respiratory chain by transferringits electrons to ubiquinone. Ubiquinone can then proceed to transfer the electrons to other inter-mediates of the respiratory chain, causing the transfer of H+ across the inner mitochondrialmembrane. Choice B is incorrect because taking electrons from succinate does not reduceNADH. However, succinate’s electron transfer WILL have an effect on ATP synthesis – it willincrease – so choice C is also incorrect, as is choice D.

151. BGlycolysis requires the oxidized electron carrier NAD+ that is regenerated through fermen-

tation and cellular respiration. When both fermentation and cellular respiration are blocked, cellswill continue generating ATP via glycolysis until the supply of NAD+ is exhausted. Choice B isthe correct answer because glycolysis generates a NET (overall) 2 ATPs for every glucosemolecule broken down; thus, it is not possible for cells to produce 4 molecules of ATP for everyglucose molecule if fermentation and cellular respiration pathways are blocked.

152. DSince nigericin dissipates the H+ concentration gradient needed to generate ATP, the amount

of ATP produced will be decreased. Choice A would be true if nigericin transported H+ ionsagainst the concentration gradient. Choice B is wrong because an equal exchange of ions willnot result in increased osmolarity in the cytosol. Choice C is also incorrect because the drugaffects the hydrogen ion gradient across the inner mitochondrial membrane and, thus, will affectATP production in some way.

153. BMaintaining the mitochondria in a solution at pH 8 until saturation will result in a slightly

basic environment throughout the mitochondria. Basic solution – think: decrease in hydrogenion concentration! When mitochondria are then transferred to an acidic solution, the mediumoutside the inner membrane will immediately become acidic while the matrix will still be basic.Thereby the conditions that are normally present in mitochondrion and are necessary for ATPsynthesis (i.e. high [H+] outside and low [H+] inside the mitochondrial matrix) are reproduced.The pH gradient across the inner mitochondrial membrane will then drive the synthesis of ATPin the presence of ADP and Pi.

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154. BThe fact that the Krebs cycle does indeed break down larger molecules to reduce NAD+ and

FAD does not, in itself, lend support to the chemiosmotic model. All it says is that electrons areproduced when bonds are broken. Choices A, C and D are wrong because all three would sup-port the chemiosmotic model by illustrating that an electrochemical gradient is necessary forATP synthesis. Decoupling agents do not permit ion exchange across ATP-synthase membranepumps; the physical continuity of the membrane is necessary for the constant flow of hydrogenions and for the interaction of proteins involved in the ETC; and, ATP synthesis depends upona hydrogen ion gradient across the inner mitochondrial membrane. These are the reasons whychoices A, C, and D support the chemiosmotic model, which invokes all of these ideas toaccount for mitochondrial ATP synthesis.

Passage IV (Questions 155–160)

155. CIn the synapse, acetylcholinesterase inactivates acetylcholine by hydrolyzing the neuro-

transmitter. Thus, acetylcholine is not available to activate its receptors. Anticholinesterase drugsinhibit the activity of this enzyme causing the amount of acetylcholine to significantly increasewithin the synapse and increasing the probability of activating its receptors. This phenomenonis important because receptor number is reduced in autoimmune MG.

156. BAntibody production against oneself (autoimmunity) is an example of abnormal immune func-

tion (I) as well as a distinguishing characteristic of autoimmune MG. Altered acetylcholinesteraseactivity is a characteristic of the inheritable form of MG (II) only. Skeletal muscle weakness is acharacteristic of MG and will be present in both forms (III), so skeletal muscle weakness could notbe used to distinguish autoimmune MG from congenital MG, and choice B is correct.

157. DIf 90% of MG patients have autoantibodies, they have autoimmune MG. Eliminate choice

A. According to the passage altered acetylcholinesterase is seen only in the inheritable form ofMG. Eliminate choice B. In autoimmune MG, receptor number is normal but antibodies preventthe interaction of the neurotransmitter with these receptors. If MG is an autosomal dominant dis-ease, it is inheritable. Everyone who gets at least one copy of the MG gene will contract the dis-ease. The autoimmune form is not inheritable.

158. B The parasympathetic system is the “rest and digest” system. Overstimulation of the

parasympathetic system would result in a decreased heart rate, increased blood flow to the diges-tive tract, and decreased blood flow to the skeletal muscles. Choice B, dilation of the pupils,would not be seen as an effect of overstimulation of the parasympathetic system. Overstimula-tion of the parasympathetic system would result in constriction of the pupils.

159. CIf a form of myasthenia gravis were discovered in which acetylcholine receptors were

impaired in their ability to bind neurotransmitter, a drug used to treat this form of MG wouldneed to increase the probability that neurotransmitter will bind to acetylcholine receptors ormaintain depolarization of muscle cells. Choice C would not alleviate the symptoms of this formof MG, since anti-ACh receptor antibodies would attach to the ACh receptors and prevent neu-rotransmitter from binding. Choice A is a plausible treatment, since blocking acetyl-

49

cholinesterase activity means that more neurotransmitter is available, increasing the probabilitythat acetylcholine receptors will bind neurotransmitter. Choice B is a plausible treatment, sinceif repolarization is blocked, the muscle cell remains depolarized for a longer period of time with-out stimulation from additional neurotransmitter. Choice D is a plausible treatment, since ifadditional acetylcholine is released, the probability that acetylcholine receptors will bind neu-rotransmitter increases.

160. ABoth MS and autoimmune MG result from the body's autoimmune response against a "self"

antigen. In MS, the body mounts an autoimmune response against myelin. In MG, the bodymounts an autoimmune response against acetylcholine receptors on the surface of muscle cells.Choices B and C are incorrect because they refer to other forms of MG, not autoimmune MG.Furthermore, Choice B is incorrect because the information given in the question stem does notmention anything about enzyme inhibition as a cause of MS. Similarly, choice C is incorrectbecause the information given in the question stem does not mention desensitization or down-regulation of postsynaptic neurotransmitter receptors as a cause of MS. Choice D is incorrectbecause the passage mentions nothing about defects during protein synthesis as a possible causeof MG, and the information given in the question stem does not mention defects during proteinsynthesis as a cause of MS.

Discrete Questions (Questions 161–164)

161. DThe pancreas secretes amylase, which digests starch, lipase, which digests fats, as well as

trypsin and chymotrypsin, which digest proteins; thus choice D is correct. The salivary glandssecrete saliva, which contains salivary amylase. Salivary amylase breaks down starch into mal-tose. The liver is responsible for storing certain nutrients, detoxifying chemicals and drugs,forming urea, regulating the storage/breakdown of glycogen, and producing bile, which emulsi-fies fats. The gallbladder stores bile and secretes it into the small intestine.

162. CA sarcomere is a unit of layered actin and myosin filaments. These sarcomeres shorten when

the muscle cell is signaled to contract and they lengthen (relax) upon muscle cell relaxation. Theshortening is regulated by a nerve cell impulse outside the muscle fiber which results in calciumion release from a specialized ER called the sarcoplasmic reticulum (SR) found in all musclecells. Calcium allows conformational change in the sarcomere such that a protein (tropomyosin)that covers the binding site (troponin) of myosin on the actin filaments is displaced. This allowsmyosin to bind to actin and, with the help of ATP, cause sarcomere shortening. Choice A isincorrect because the SR releases calcium, not sodium, and choice B is incorrect because the T-tubules themselves do not shorten – the T-tubules are part of the SR through which the nervousimpulse is transmitted. Choice D is also wrong because the SR does not release ATP.

163. ADNA is composed of 4 nucleotides: cytosine (C), thymine (T), adenine (A), and guanine (G).

Cytosine base pairs with guanine via three hydrogen bonds, and thymine base pairs with ade-nine via two hydrogen bonds. Thus, a strand of DNA that contains fewer (C) and (G) would havea lower melting point than one with more (C) and (G), since there are fewer hydrogen bonds tobe broken. Looking at the answer choices, choice A contains 2 G-C (or C-G) pairs, choices Band C contain 3 G-C (or C-G) pairs, and choice D contains 4 G-C (or C-G) pairs. Thus the strandwith the lowest melting point is the strand shown in choice A.

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164. AThe rate of an electrophilic aromatic substitution (EAS) reaction is affected by substituents

already present on the ring. Substituents can be classified as activators or deactivators. Activa-tors include (neutral) amine groups, the hydroxy group, alkoxy groups and alkyl groups. Theirpresence makes the compound more reactive (i.e., react more rapidly) in EAS reactions. Deac-tivators include halogens, the nitro group, the cyano group, and carbonyl groups. Their presenceslows down the rate at which the compound reacts in EAS.

Choices A and B are both compounds with activating substituents. Methoxybenzene has themethoxy substituent –OCH3, while toluene has the methyl group substituent. The methoxygroup is more strongly activating than the methyl group, so the rate of EAS reaction will be mostrapid for choice A.

Passage V (Questions 165–170)

165. AA compound that is hygroscopic is one that readily absorbs moisture from the atmosphere.

Choice A is an anhydride. You may recall (or you may deduce from its name) that an anhydrideis derived from the loss of water. As a consequence, anhydrides generally react rapidly withwater to form the corresponding acids.

166. AWe are told in the passage that compound X was refluxed in aqueous acid to yield two new

compounds, one of which was an amine salt. We are also given information on the other com-pound, but at this point we can examine the choices and see if any of them would be consistentwith what we know. An amide can undergo acid-catalyzed hydrolysis to yield an amine (in theform of a quaternary ammonium salt) and a carboxylic acid. Indeed, of the four choices, this isthe only one to involve a nitrogen-containing functional group. Therefore choice A is correct.

167. DAccording to the structure given for 6-azathymine, there are three nitrogen atoms in the

molecule. All three are sp2 hybridized. The nitrogen atom that forms a double bond to carbonhas a nonbonding pair of electrons in an sp2-hybridized orbital that is in the plane of the ring.

Each of the other two nitrogen atoms also possesses a lone pair of electrons, but these pairsare in unhybridized p orbitals and are delocalized both throughout the ring and with the carbonylpi electrons.

CH3

H

H O

ON

N

-120°

-120°H3C

51

Notice that all the carbon atoms in the ring are also sp2 hybridized. Since we have six pi elec-trons that can be delocalized (two from the C=N bond and four from the nitrogen atoms), thesystem is aromatic.

168. CThe solubility data included in paragraph 2 of the passage indicate that compound Z was

slightly soluble in water and more soluble in dilute aqueous base. This tells us that the com-pound most likely contains an acidic functional group. In the presence of a base, the groupdeprotonates and thus becomes ionized. Because of the net charge, the compound’s solubility inthe aqueous medium increases since water molecules solvate ions very well. When no base ispresent, the carboxylic acid functional group is still expected to be somewhat soluble because ofhydrogen bonding, but if the organic portion of the molecule is large, the solubility will be com-promised. This is exactly consistent with the solubility data. So choice C is correct.

Choice A is incorrect because amines are expected to increase their solubility in acidic con-ditions. In the presence of an acid, the nitrogen atom may acquire a proton and become a posi-tively charged quaternary salt.

Choices B and D are incorrect because neither an amide nor a ketone is expected to partic-ularly soluble in either an acid or a base.

169. AThe proton labeled (1) is a carboxylic acid proton and is therefore most readily abstracted.

The conjugate base that results from deprotonation at this site is stabilized by the carbonylgroup.

The proton labeled (2) is a methoxy proton and is not appreciably acidic. The negativecharge on the conjugate base –OCH2

- is not well stabilized.

The proton labeled (3) is a phenolic proton. It is more acidic than typical alcoholic (hydroxy)protons because the negative charge can be delocalized throughout the ring, but is not moreacidic than a carboxylic acid proton.

The proton labeled (4) is an aryl proton and is not very acidic.

170. BYou are expected to know the characteristic IR frequencies of common functional groups.

The O-H stretch of alcohols usually occurs in the range 3200-3600 cm-1. (The O-H stretch of acarboxylic acid occurs at a lower frequency.) The carbonyl stretch occurs anywhere from about1650 to about 1800 cm-1 depending on the specific compound. The N-H stretch occurs in the3000’s. The aromatic C-C stretch occurs somewhere between 1500 and 1600 cm-1. Choice B istherefore correct.

Passage VI (Questions 171–175)

171. DThe passage states that lithium is used to treat individuals with mood disorders. If the mech-

anism proposed by Hypothesis 1 is correct, one would expect that individuals with these disor-ders have an overactive GSK-3 which phosphorylates and degrades catenin proteins. The levelsof catenin proteins would, thus, be reduced rather then elevated. Choice A would support

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Hypothesis 1 because it would provide genetic evidence to link abnormal GSK-3 to psychiatricdisorders. Choice B also supports Hypothesis 1 because it would give additional evidence show-ing that lithium targets GSK-3. Choice C would support Hypothesis 1 by showing that byinhibiting GSK-3, lithium compensates for the defective protein (the disheveled protein that nor-mally inhibits GSK-3) thereby providing therapeutic effect.

172. DThe disheveled protein inhibits GSK-3, whose function is to phosphorylate b-catenin and g-

catenin. Therefore if disheveled protein is overactive, phosphorylation of b-catenin and g-catenin is reduced. Choice A is wrong because the levels of catenin proteins fall due tophosphorylation. Choice B is not relevant to the first hypothesis, because it is the second hypoth-esis which deals with inositol. Choice C is an incorrect answer because an overactive disheveledprotein would decrease the activity of GSK-3, not increase it.

173. AHere, you need to recall that sodium ions flood into a nerve cell axon when sodium chan-

nels open. Also, the inside of the neuron rapidly depolarizes (becomes more positive in relationto the outside of neuron) when these positive ions flood in. If lithium behaved like sodium, theneuron would rapidly depolarize as lithium flowed into the neuron.

174. C This is essentially a reading comprehension question. According to Hypothesis 2, lithium’s

inhibition of two enzymes leads to the reduced level of inositol in cells. Choice C would showthe opposite and, thus, would challenge the hypothesis. Choice A would support Hypothesis 2by showing that lithium targets a protein whose function is overactive in psychiatric patients.Choice D would support Hypothesis 2 by giving evidence that cells do, in fact, become unre-sponsive to signals when lithium is introduced.

175. AIn Hypothesis 1, lithium is thought to inhibit overactive GSK-3, while in Hypothesis 2,

lithium is believed to target IPPase and IMPase. Choice B is wrong because both hypothesesincorporate findings from animal models. Choice C is true for Hypothesis 2 but not for Hypoth-esis 1. Choice D is directly contradicted by the first sentences of the third and the fifth para-graphs.

Passage VII (Questions 176–180)

176. DThe term “lysosomal storage disease” implies that an excess of material gets stored in lyso-

somes and does not get broken down as usual. This makes sense because, in I-cell disease, cer-tain lysosomal enzymes are not present in the lysosomes, causing a buildup of material normallydigested there. While Choice A is a correct statement, it does not explain the term “storage dis-ease,” since there are many other types of mechanical problems in many different organelles thatresult from the absence of enzymes that are supposed to be in certain places. There is no sug-gestion in the passage that I-cell or any related disease results from the lysosomes themselvesbeing transported out of the cells (choice C), or that lysosomes are not being properly formed(choice B), so B and C should be eliminated.

53

177. BIf proteins meant for the lysosomes are actually found outside the cell, that means these pro-

teins have been incorrectly targeted after translation. Proteins are “labeled” for transport aroundthe cell in both the ER and the Golgi, where various enzymes cleave and shape the newly-formed proteins and also add sugar chains to these proteins. Yet, if certain sugar chains do notget added, the proteins will not be targeted to proper places – i.e. to the cell membrane, out ofthe cell, or to intracellular compartments like lysosomes. It is suggested in the last paragraph ofthe passage that a gene defect in the enzyme N-acetylglucosamine phosphotransferase preventsthe enzyme from accurately adding certain suagrs to certain proteins destined for the lysosomes– hence, choice B is the most likely reason that the enzymes get exported rather than sent to thelysosomes. Ribosomal enzymes (choice A) could be implicated as a cause if they did not directtheir protein synthesis to the ER membrane, which would cause the protein to be released free-floating in the cytoplasm, but that is not what choice A says. Choice C is incorrect because thereare no sugars to interact with – they were never added onto the proteins properly. Choice D isincorrect because lysosomes are not vesicles and do not transport proteins from the ER or Golgito the plasma membrane. Once the properly glycosylated proteins are inside lysosomes, theywill stay there.

178. AThe passage describes skeletal abnormalities and joint movement restriction as signs and

symptoms of I-cell disease. This alone suggests that choice A is correct. While choice B cor-rectly describes the overloaded lysosomes that occur in I-cell, the symptoms in the passage areexplained by A. There is no evidence in the passage whatsoever for choice C or choice D. I-cellis a very specific disease caused by a very specific enzyme deficiency due to a particular genedefect.

179. C This question tests your knowledge of the function of various cellular organelles. The

smooth ER is responsible both for manufacturing lipids, which end up in the cell and organellemembranes, and for enzymatically detoxifying poisons such as alcohol and other toxins. This isoften done using an enzyme like cytochrome p450, which changes toxins so that they becomewater-soluble and can diffuse out of cells and be excreted in the urine, rather than remain insol-uble within cells.

180. AAlthough the Golgi is extremely important for continued modifications to proteins destined

for export or for other sub-cellular compartments, the ER is the first stop for all proteins that willeventually make it to the Golgi. As seen here in the passage, the Golgi cannot target lysosomalenzymes at all if they do not first receive the proper sugar residues in the ER. Therefore, ChoiceA is the most reasonable answer to this question given the information found in the passage.

Passage VIII (Questions 181–185)

181. AThe first step of the reaction is similar to Equation 2 in the passage, except that we are using

n-heptane instead of n-octane. The alkylborane intermediate will therefore only have a seven-carbon alkyl chain. The second step of the reaction, however, uses hydroxylamine instead ofalkaline hydrogen peroxide. According to the last paragraph, the intermediate reacts withhydroxylamine to yield an amine. The amine functionality will occur at the end of the alkylchain. Hence the product is 1-heptanamine. The structures for the other choices are as follows.

54

Choice B is what we would have obtained if we had used alkaline hydrogen peroxide instead ofhydroxylamine.

182. CThe starting material is 2-methyldecane, a saturated hydrocarbon. Therefore none of the

schemes in choice A, choice B or choice D would work as these are all ways to synthesize alco-hols from alkenes. Choice C is exactly the reaction described in the passage. Even though weare told that borylation is slowed with branching of the alkane, this method remains the only fea-sible synthetic route given. The preference for non-branching is manifested in the site of bory-lation, leading to the formation of 9-methyl-1-decanol instead of 2-methyl-1-decanol.

183. CBoron has the special characteristic that it often violates the octet rule when forming com-

pounds. It usually has six valence electrons in compounds, and can act as a Lewis acid. Becauseit only has three pairs of valence electrons, the molecule adopts a trigonal planar structurearound the boron atom. The boron atom is therefore sp2 hybridized.

Upon reacting with a Lewis base (such as ammonia), the boron atom has eight valence elec-trons and acquires a negative formal charge. In the initial state, the boron has no formal charge.

In summary, statements I and II are correct, but statement III is not.

184. AAs mentioned in the first paragraph, hydroboration-oxidation of an alkene leads to the addi-

tion of a water molecule across the double bond. The hydroxy group adds preferentially to theless substituted carbon. Since the question asks us to determine the alkene that would lead to 3-methyl-2-butanol, we should first have a grasp on its structure:

We can proceed in one of two ways. We can either work backwards from the structure of thisalcohol, or deduce the product of each of the four choices. If we employ the first method, weknow that the carbon bearing the hydroxy group is one of the double-bonded carbon atoms inthe alkene. But to which carbon is it double-bonded: carbon 1 or carbon 3 (the one with the

OH

CH3

NH

OH

1-heptanol

1-amino-1-heptanol

N-heptyl-1-heptanamine

OH

NH2

55

methyl group)? If the former, then the product would actually have been different, since carbon1 is less substituted than carbon 2, so the hydroxy group should have added to carbon 1 instead.Therefore, the double bond must have been between C-2 and C-3. The hydroboration-oxidationproduct of each of the choices is as follows.

185. BMarkovnikov addition of water to an alkene occurs in acid-catalyzed hydration. The hydroxy

group adds to the more substituted carbon. This is most easily remembered by understanding themechanism of the reaction. In the first step, a proton adds to the less substituted carbon, so thatthe more substituted carbon bears the positive charge, resulting in the more stable carbocationintermediate. A water molecule then adds to this more substituted carbon, and upon deprotona-tion (to regenerate the acid catalyst) an alcohol is formed. With hydroboration-oxidation, how-ever, the hydroxy group ends up at the less substituted carbon. The addition is thereforecharacterized as anti-Markovnikov.

Less Substituted

Less Substituted

Less Substituted

Less Substituted

OH

OH

OH

HO

56

57

Choice C, free radical addition, also leads to the anti-Markovnikov product. However, freeradical addition is usually carried out with hydrogen halides rather than with water.

Choice D, hydrolysis, refers to the breaking of a compound with water: for example, hydrol-ysis of esters and amides.

Discrete Questions (Questions 186–190)

186. AThis question calls on your knowledge of where specific hormones are released. Although

this hormone has the word “adrenal” in it, it is in fact released by the pituitary gland. Very fewhormones are released by the posterior pituitary gland (only ADH and oxytocin are), so anteriorpituitary is a safe guess for all hormone release if you know the hormone is released by the pitu-itary. ACTH, adrenocorticotropic hormone, causes the adrenal cortex to release glucocorticoids,steroid hormones which affect glucose metabolism and which can suppress the immune system.

187. B Passive immunity results when antibodies are transferred from one individual to another.

Often, this occurs when a pregnant woman passes her own antibodies on to her fetus through theplacenta or to her baby through her breast milk. Since antibodies are proteins, they are eventuallydegraded, and this immunity will last for only a short time. Passive immunity can also be trans-ferred by giving someone antibodies from another person or animal already exposed to a certainpathogen. Choices A, C, and D are all examples of active immunity, whereby one’s immune sys-tem, after being exposed to a particular pathogen, mounts a response and creates memory B andT cells that can be activated very quickly in the event of later exposure to the same pathogen.

188. CThe chorionic villi are the main site of gas and nutrient exchange between the mother’s

blood and the fetus’s blood. Keep in mind that maternal and fetal blood do NOT mix, yet doexchange gases and nutrients through thin walls in tree-like projections call villi that the pla-centa forms within the uterine wall. The umbilical cord, choice A, is simply a set of vessels forthe conveyance of blood to and from the fetus – although the maternal end of the cord spreadsout into the chorionic villi. The fallopian tube, choice B, is the usual site of fertilization, and theamniotic fluid, choice D, is the fluid that cushions the developing fetus, not a site of gas or nutri-ent exchange.

189. AHemophilia is a sex-linked recessive trait, so the genotype of a man with hemophilia would

be XhY. A woman whose father was a hemophiliac would be a carrier of hemophilia with geno-type XXh. A cross between these two individuals would produce the following results:

X Xh

Xh XXh XhXh

Y XY XhY

Since the question asks about their sons only, let’s find the probability that a son born tothem will be a hemophiliac. From the Punnett square, 1/2 of their sons will be hemophiliacs(genotype XhY). The probability that three of their sons will be hemophiliacs is therefore(1/2)(1/2)(1/2) = 1/8. This is equal to 12.5%, choice (A).

190. BEven though you may automatically recognize choice A as a carboxylic acid, don’t leap to

that as the answer immediately! Keep in mind that organic acids are typically weak acids whencompared to inorganic acids such as HCl. Acetic acid, for example, has an acid dissociation con-stant much less than one (about 10-5), or a pKa of 4.75. (However, you do not need to have mem-orized this number to obtain the correct answer.) Carboxylic acids are only acidic relative toother organic compounds.

Are any of the other choices a stronger acid than the carboxylic acid of choice A? We knowthat ammonia (choice D) is considered a weak base, so it cannot be a stronger acid than the car-boxylic acid. Consider the conjugate base of each: NH2

- is a very strong base while CF3COO-

is stabilized both by resonance and by the fluorine atoms. The stronger the conjugate base, theweaker the acid. It would take very harsh conditions to cause ammonia to deprotonate and gen-erate the strong conjugate base.

Choice C, the quaternary ammonium ion, looks more promising. Its conjugate base isammonia, which, as a neutral compound, is not expected to be a strong base. However, ammo-nia is still recognized as being capable of reacting as a base, whereas CF3COO- is not. In otherwords, since NH3 is still a stronger base than CF3COO-, NH4

+ is a weaker acid than CF3COOH.(You can also arrive at the same conclusion if you recall how an amino acid deprotonates: Thecarboxylic acid functionality –COOH deprotonates first before the –NH3

+ group, implying thatthe former is more acidic.)

Choice B, however, is a different story. Its conjugate base is the neutral alcohol (methanol),which is not considered basic. In neutral conditions, the equilibrium between CH3OH2

+ andCH3OH favors the neutral molecule. (Only if we add an acid such as H2SO4 can we protonatethe alcohol.) In other words, the compound shown in choice B is generally “eager” to deproto-nate, rendering it a strong acid. Compare this to the carboxylic acid: Because it is again a weakacid “in the grand scheme,” equilibrium favors CF3COOH over the conjugate base carboxylateanion. Therefore, choice B is the stronger acid and the correct answer.

Passage IX (Questions 191–195)

191. CThe passage states that graft rejection is due to the activity of the specific immune system –

lymphocytes, B and T cells, are the main responders to these grafts. If macrophages (compo-nents of innate or nonspecific immunity) were found in the mixture with the transferred lym-phocytes, this would cast doubt on that statement. The passage already states that transplantrejection is a result of an immune response, so choice A is not correct. Second set rejection arisesbecause an animal has already been exposed to a transplant from a particular strain and, whenexposed again to that same strain, mounts a much quicker rejection. This is equivalent to beingimmunized, and this statement would not be challenged by the situation set up in the questionstem, so choice B is incorrect. Choice D is wrong because second set rejection, as it is similarto the secondary immune response (i.e. after exposure to a pathogen or vaccination), will alwaysbe quicker than the first set (or primary) response – this statement would not be challenged bya finding of macrophages along with lymphocytes.

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192. DThe question is analogous to the scenario represented in last row of table 1. In both cases a

particular strain is sensitized to another strain, but is then given a graft from a third strain. Sincethat strain has never been exposed to this third strain, it will mount a slow, first set rejection tothe graft. In other words, one can only mount a second-set rejection if this is one’s second timebeing exposed to the same tissue – in the case of this question, strain C has never been exposedto strain A before, and it will mount a slow (standard), first set rejection to the strain A graft.Therefore, Choice D is correct.

193. DHumoral immunity is a form of immunity that is mediated by substances present in body flu-

ids (humors). These substances are the serum proteins termed antibodies that can be transferredin a cell-free serum. In order to demonstrate that humoral immunity plays a role in transplantrejection, one must show that transfer of serum from a sensitized animal into one never beforeexposed will transfer immunity against a particular graft. Choice D clearly supports the humoralmodel. Notice that the question does not expect one to know whether humoral immunity playsa role in graft rejection. It simply asks what form of evidence would implicate humoral immu-nity in response against foreign grafts. Choice A is wrong because it states that unexposed ani-mals do not mount a faster response against strain A after having received antibodies from strainB already exposed to A. If this were so, this result would actually provide an argument againstthe role of antibodies in graft rejection. Choice B is wrong because it supports the cell-mediatedmodel of graft rejection, not the humoral model. Choice C supports neither the cell mediated northe humoral model.

194. D This question draws on your knowledge of the difference between the nonspecific and spe-

cific defenses of the body. Nonspecific defenses include physical and chemical barriers, theinflammatory response, and widely-released chemicals such as interferons (cytokines). Physicalbarriers include the intact skin and mucous membranes. These barriers are aided by variousmicrobe catching fluids such as mucous, Choice D. Other components of non-specific immunityinclude phagocytic cells such as neutrophils and macrophages. Choices A, B, and C, however,are all examples of specific immunity – T cells, B cells, and antibodies released by B cells areall very specific for the pathogens (antigens) they will recognize and bind to.

195. AThe last paragraph of the passage outlines the mechanism by which killer T-cells mediate

rejection of foreign grafts. The mechanism involves T-cell recognition of foreign MHCmolecules followed by the lysis of infected cells. If T-cells bind self MHC molecules, this canresult in lysis of the individual’s own cell. In fact, failure to eliminate self-reactive lymphocytesis the basis for many autoimmune diseases. Choices A, B and D are not supported by the pas-sage. There is no evidence for T-cells’ being less able to recognize foreign cell surface proteins,nor evidence that more T-cells would be produced than B cells. Although killer T-cells are notinvolved in defense against bacterial invaders, one could eliminate Choice D based on the factthat if fewer T-cells are killed off in the thymus, one should have more circulating T-cells forinfection fighting.

59

Passage X (Questions 196–202)

196. BThe product of Step 3 includes an alkene functional group. One of the sp2-hybridized car-

bon atoms is bonded to a nitrogen atom, which in this case is protonated (hence it is the conju-gate acid of some compound). The question, then, is asking us to identify the name of thefollowing functional group:

This species is an enamine.The structures of the incorrect answer choices are as follows:

197. DA �-hydroxyketone, when heated, will undergo dehydration in which a water molecule is

eliminated. The driving force for this reaction is the formation of a carbon-carbon pi bond thatcan be conjugated with the carbonyl pi bond. The double bond, therefore, has to be between thealpha and beta carbons.

Choices A and B are incorrect because heating alone is not sufficient to cause the compoundto decompose into two carbonyl compounds. This process would involve breaking a carbon-car-bon single bond, which is relatively strong.

Choice C is incorrect because the hydroxy group should no longer be a part of the moleculeafter dehydration.

198. DStep 1 involves the nucleophilic attack of the nitrogen in proline on the carbonyl carbon of

acetone. The conditions must NOT be acidic. If it were, the nitrogen atom would be protonatedand there would not be a lone pair of electrons available for nucleophilic attack. The mediummust therefore be neutral or slightly basic to ensure that the nitrogen does not form a quaternarysalt. However, immediately after the addition has taken place, we obtain an alkoxide ion. Inorder to get the dehydration in the next step to occur, protonation must occur. This can only beaccomplished in acidic conditions. Choice D is therefore correct.

OO

C NNH

ImideImine

O

N

Amide

N

60

199. BThe nitrogen atom in proline is bonded to two alkyl groups (which connect to each other to

form the five-membered ring). It is therefore a secondary amine. Choice D is incorrect becausethe ring is not aromatic: The carbon atoms are all saturated.

200. BThe stereogenic center in the �-hydroxyketone is created in Step 4 when the aldehyde adds to

the proline-acetone complex. The transition state of Step 4 is shown in the passage. As mentionedin the passage, the transition state is a tricyclic complex, and one of the three rings is a six-mem-bered ring involving the aldehyde. Notice that the R group is in the favored, equatorial position. Inother words, the aldehyde adopts the more stable orientation relative to the complex, and it is thisorientation preference that leads to the predominant creation of one enantiomer. In order to createthe other enantiomer, the R and H will have to be swapped in the aldehyde. (I.e., the aldehyde willhave to be rotated 180º about an axis going through the carbonyl bond.) That, however, will leadto higher strain in the transition state, as the R group will be in the axial position.

Choice A is incorrect because we are not adding different halves of a molecule across thedouble bond, so the Markovnikov versus anti-Markovnikov designation is irrelevant.

Choice C is incorrect. A general principle in organic synthesis is that we will never be ableto create net optical activity using only optically inactive reagents. In other words, in order tocreate an enantiomer without its mirror image, there must be some chiral environment orreagents introduced at some point. If we started with a racemic mixture of proline, the finalproduct would be a racemic mixture of the R and S isomers of the �-hydroxyketone.

Choice D is incorrect because the aldehyde is achiral. It does not have an enantiomer.

201. AThe two carbonyl compounds involved in the reaction are a ketone and an aldehyde. Neither

possesses a good leaving group. So no nucleophilic acyl substitution reaction is possible. Thegeneral scheme for a nucleophilic acyl substitution reaction is shown below:

The last step of the reaction is similar to the second step of a nucleophilic acyl substitutionreaction under acidic conditions. The carbon-oxygen pi bond is formed as the proline acts as aleaving group and departs. However, the new carbonyl oxygen has never been part of a carbonylcompound. Instead, it comes from the water molecule that adds to the compound in Step 5, sothe reaction is not a nucleophilic acyl substitution reaction.

Choice B is incorrect because it does occur in Step 1: The nitrogen acts as a nucleophile andattacks the carbonyl carbon.

O

Y

X–

Y–

O

X

O

YX

–

+

61

62

Choice C is incorrect because addition to an alkene does occur in Step 4. The pi electrons ofthe C=C bond in the enamine are used to form the sigma bond to the aldehyde.

Choice D is incorrect because Steps 5 and 6 together constitute a hydrolysis reaction.

202. BWater adds to the substrate-catalyst complex in Step 5, with the hydroxy group adding to the

enamine carbon. The oxygen of this hydroxy group becomes the carbonyl oxygen in Step 6. If18O water were used, therefore, the carbonyl oxygen of the �-hydroxyketone would be 18Olabeled.

Passage XI (Questions 203–208)

203. BFrom the information presented in the figures one can see that daily injections of leptin will

reduce the amount of food consumed by the wild type mice. We also learn from the last sentenceof the passage that leptin, among other things, decreases appetite. We would thus expect elimi-nation of leptin from the blood by antibodies to increase appetite and food intake in wild typemice. Of all the answer choices, only choice B represents increased food intake after adminis-tration of antibodies to leptin and is thus the correct answer.

204. AThis mating represents a cross between a homozygous recessive strain Ob1 and a homozy-

gous dominant strain. The progeny of the cross will all be heterozygous at the leptin locus as canbe seen from the following Punnett square diagram.

LL x ll

L L

l Ll Ll

l Ll Ll

Since all the progeny are heterozygous, and only homozygous recessive mice are expectedto develop obesity, none of the offspring are expected to become obese.

205. CThe answer to this question must be a process that does not occur when insulin levels are

low. Insulin levels are reduced at times of low blood glucose concentration in order to conserveglucose for the use by the brain. Thus when insulin levels fall, tissues such as muscle and liverdecrease glucose uptake and utilization (choice A) and start using fatty acids instead (choice D).When insulin levels are low, glucagon levels rise by default and promote conversion of glyco-gen into glucose (choice B) to maintain blood glucose levels. The only process that doesn’toccur, in response low insulin levels, is thus choice C, increased utilization of glucose as fuel.

206. DBoth figures in the passage show that the Ob2 strain of mice did not respond to daily admin-

istration of leptin. This could be explained if the Ob2 strain had a defect in gene coding for theleptin receptor. Choice A is wrong because if this were the case then the Ob2 strain would haveresponded to leptin administration just like the wild type. Since the Ob2 response notably devi-ated from that of the wild type, there is clearly a mutation in the leptin pathway. Choices B and

C can be eliminated because exogenous administrations of functional leptin would have restorednormal feeding behavior of the mutations involved the production of leptin.

207. DIn order for levels of neuropeptide Y to be reduced, mice must produce functional leptin and

be cable of responding to it. Since the Ob1 strain is defective in the former and the Ob2 strainis defective in the latter, both strains will have elevated levels of neuropeptide Y.

208. CFrom the figures in the passage one can see that leptin promoted weight loss in mice with-

out functional leptin in the serum and in wild type mice. This suggests that individuals with nor-mal body weight, with mutations in the leptin gene and without defects in the leptin pathway,will respond to leptin administration. The leptin receptor, on the other hand, represents a com-ponent of the leptin pathway downstream of leptin. Thus if leptin is administered to individualswithout functional leptin receptors, it will not be able to exert its effects and those individualswould be least likely to lose weight.

Discrete Questions (Questions 209–214)

209. AThe enthalpy of formation of a molecule is an indication of its stability. The more unstable

(the more strained) the compound, the higher the enthalpy of formation. However, direct com-parison of heats of formation of compounds that are not isomers is misleading, so often whatorganic chemists do (for hydrocarbons) is refer to the heat of formation per CH2 group, whichoffers a common basis for comparison. Among the compounds listed as answer choices, choiceA, cyclopropane, has the highest ring strain, arising from both the angle strain imposed by itsgeometry and the torsional strain among its CH bonds. As we increase the size of the ring, thering strain is alleviated (until we reach cyclohexane which has zero ring strain in its chair con-formation). Therefore, when we normalize the value for the heat of formation to that per CH2group, cyclopropane is expected to have the highest value.

210. BThe alkyl halide given in the question can lead to two different elimination products, depend-

ing on which proton is abstracted. The formulas for the products are given, but we should alsobe able to predict them for ourselves. We can see that the product described as predominant ismore substituted: It only has one vinylic hydrogen (hydrogen attached to a double-bonded car-bon atom) whereas the other product has two. More substituted alkenes are more thermody-namically stable, a fact that explains the observed product distribution. Choice B is the correctanswer.

Note that in order to form the more stable (more substituted) alkene, the base needs toabstract a proton from the more sterically hindered secondary carbon. Under some conditions,therefore, we may expect the less substituted alkene to form predominantly, since it is formedby abstracting a proton from the more accessible methyl groups. In the question, however, weare told which product is formed predominantly, so choice A is incorrect. Besides, steric con-siderations apply to the haloalkane substrate, not the product alkene.

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It is true that the less substituted alkene has a larger dipole moment, but this is irrelevant tothe product distribution. It will only affect the physical properties of the alkene. So choice C isincorrect.

A compound is kinetically stable if, for example, it faces a high activation energy of reac-tion, or if the temperature is low so that even though the reaction can lead to more thermody-namically stable products, it does not have the energy to surmount the activation energy barrier.This characteristic of the less substituted alkene, even if true, cannot account for its small pro-portion among the products.

211. DWhen we see the starting aromatic compound and the reagent bromine, we may conclude

that this is an electrophilic aromatic substitution (EAS) reaction. However, such a reactionrequires a Lewis acid catalyst (FeBr3). Simply heating the mixture would not lead to an EASproduct. Instead, what we would obtain are bromine free radicals, which can substitute for alkylhydrogens. Free radical reactions involving bromine tend to be relatively selective, proceedingthrough radical intermediates that are the most stable. The alkylbenzene substrate can lose ahydrogen atom from either its methylene (CH2) group or its methyl group. Losing a hydrogenfrom the methylene group leads to the more stable secondary (benzylic) carbon radical. This iswhere the bromine will substitute. Some relevant steps in the chain reaction include:

CH2CH3

+ Br

Br

+ HBr

CH2CH3

CH2CH3

+ Br

Br2 Br + Br

+ Br

CH2CH3

CH3

Leaving Group

Abstract Proton from Secondary Carbonto Form More Stable Alkene

Abstract Proton from Primary Carbon to Form Less Stable Alkene

Cl

CH3CH3 CCH2

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The first step shown is an initiation step, while the next two steps are chain propagationsteps. The benzyl radical can, of course, also participate in a chain termination reaction with abromine radical to yield the same substitution product.

Choices A and B are incorrect because they are EAS products, which will not form in thiscase. Choice C is incorrect because it involves the less stable primary carbon radical intermedi-ate, and is therefore not expected to be the major product.

212. BThis question asks you to remember the distinction between a virus’s lytic cycle, which

causes the host cell to burst rapidly from quickly manufactured “baby” viruses, and the lyso-genic cycle, in which viral DNA integrates for a period of time within the host cell’s DNA,becoming “active” only after some sort of environmental cue. Choice C, integrative cycle, ismeant as a distractor answer choice for those who are unfamiliar with the terms lysogenic, andchoice D might be confusing since HIV, the virus causing acquired immune deficiency syn-drome, can act in a lysogenic fashion.

213. COne of the five tenets of the Hardy-Weinberg Equilibrium law is that, in order to keep the

frequency of alleles for traits constant over several generations – i.e. in other words to be a non-evolving population – members of the population must mate in a random fashion with eachother. While this rarely, if ever, occurs in nature due to certain characteristics always attractingcertain mates, random mating would allow alleles to be randomly shuffled from generation togeneration in order that no allele is particularly selected to be passed on over another allele. Ifrandom mating were coupled to the other Hardy-Weinberg tenets, it would result in the stabi-lization of a particular allele’s frequency. Therefore, choices A, B, and D are all incorrect.

214. BCompared to inhaled air, the pulmonary capillaries have a higher partial pressure of CO2 and

a lower partial pressure of O2. Thus, in the alveoli, gas exchange occurs when CO2 flows downits concentration gradient from the capillaries into the alveoli, and O2 flows down its concen-tration gradient from the alveoli into the pulmonary capillaries. The presence of surfactant low-ers the surface tension of the alveoli and facilitates gas exchange. Thus, I and III contribute togas exchange in the alveoli, and B is the correct answer. (Note that II is a false statement, sincethe partial pressure of CO2 in the pulmonary capillaries is high compared to inhaled air.)

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