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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1. D

2. B

3. C

4. C

5. B

6. D

7. B

8. A

9. A

10. D

11. A

12. D

13. C

14. A

15. B

16. C

17. A

18. C

19. A

20. B

21. A

22. B

23. D

24. D

25. C

26. B

27. A

28. B

29. C

30. D

31. C

32. B

33. C

34. A

35. D

36. D

37. B

38. D

39. D

40. B

41. A

42. A

43. C

44. A

45. C

46. C

47. D

48. A

49. B

50. B

51. D

52. C

53. C

54. B

55. A

56. B

57. D

58. C

59. A

60. D

61. C

62. B

63. B

64. A

65. B

66. C

67. B

68. C

69. A

70. C

71. A

72. D

73. D

74. D

75. C

76. B

77. B

PHYSICAL SCIENCES ANSWER KEY

Passage I (Questions 1–5)

This passage concerns the physics of fluids. The first paragraph discusses the two forces thatact on an object when it is submerged in water: gravity and the buoyant force. As you are read-ing, you might want to write Archimedes’ principle in symbols as F = mwg, where F is the buoy-ant force, mw is the mass of the water displaced and g is the acceleration of gravity. It is easierto use as a reference in this form. The second paragraph discusses an apparatus shown in the fig-ure. It is helpful to look at the figure as you are reading the description of it in the text. In thisparagraph you are told that the gas in the balloon is an ideal gas. The final paragraph, althoughshort, also provides some very useful information. You are told that the pressure is not constantat all points in the pool. You might want to ask yourself why the pressure varies. The pressureat a given point in the pool is dependent on the depth of the point below the surface of the fluid.

1. DIn the passage you are given Archimedes’ principle, which states that the buoyant force is

equal to the weight of the water displaced. Mathematically, the buoyant force is given by F = mwg, where mw is the mass of the water displaced and g is the acceleration due to gravity.Since you are not given information about the mass of the water, you can express it as mw = ρwVw,where ρw is the density of water and Vw is the volume of the water displaced. Since the volumeof the water displaced is identical to the volume of the submerged balloon, Vb, the mass of thewater displaced mw = ρwVb and the buoyant force F = mwg = ρwVbg. Since the buoyant forceis proportional to the volume of the balloon, doubling the volume of the balloon will double thebuoyant force. Thus, choice D is correct.

Choices A, B, and C are wrong for the following reasons. The weight of the balloon is givenby W = mbg, where mb is the mass of the balloon. Since the mass of the balloon is unchanged,so is the weight, and choice A is incorrect. The average density of the balloon is given by ρb =mb/Vb. Since the volume of the balloon is inversely proportional to the density, doubling the vol-ume will halve the density, making choice B wrong. Choice C is wrong because the density ofwater remains constant.

2. BAccording to the passage, the balloon experiences no net force when the buoyant force and

weight are exactly equal. This condition is called equilibrium and occurs at depth d1.

If the water is replaced by a denser fluid, the pressure at depth d1 will increase since the pres-sure is given by P = P0 + ρgd, where P0 is the pressure at the surface, ρ is the density of thefluid, and g is the acceleration due to gravity. In response to this increase in external pressure,the volume of the balloon will shrink. Since the mass of the balloon is constant, a decrease inthe volume means that the density of the balloon will increase. Choice B is thus an incorrectstatement and is the answer we are looking for.

Choice A is incorrect because the pressure does increase. Choice C is incorrect because thevolume of the balloon does decrease. Choice D is incorrect because the weight of the balloon isits mass times gravity, neither of which changes.

3. CAs discussed in the explanation to question 1, the buoyant force on the balloon is given by

F = ρwVbg, where ρw is the density of water, Vb is the volume of the object, and g is the accel-eration due to gravity. Therefore, the buoyant force on the second object can be expressed as

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F = ρwVog, where Vo is the volume of the object. Moreover, the weight of the object is given byW = mog = ρoVog, where mo is the object’s mass and ρo is the object’s density. The object ismotionless when the buoyant force exactly equals the weight; mathematically expressed, ρwVog= ρoVog. By canceling Vog on both sides, we obtain ρw = ρo. In other words, to be in equilib-rium, the object must have the same density as water. Since the balloon and the second objectare both in equilibrium at depth d1, their densities must both be equal to ρw. Their densities arethus equal to each other, and choice C is correct.

We can only determine the density, which is an object’s mass to volume ratio, from the infor-mation given. We cannot infer anything about the mass or volume separately, so choices A andB are wrong.

The buoyant force is dependent on an object’s volume and no other characteristic of theobject. Since the relative volumes of the balloon and the second object are unknown, the rela-tive buoyant forces on the balloon and the second object cannot be determined, and choice D isincorrect.

4. CTo answer this question, you have to figure out what physical conditions change as the bal-

loon is moved to depth d2 (which is deeper than depth d1).

As discussed in the explanation to question 2, the pressure increases with increasing depth.Hence, the balloon’s volume will decrease when it is moved to depth d2, and thus the buoyantforce, F = ρwVbg, will decrease as well. The balloon’s weight, however, will not change becausethe mass of the balloon will remain unchanged. At depth d1, the balloon experiences no netforce, so the weight must equal the buoyant force. At depth d2, then, the unchanged weight isgreater than the decreased buoyant force. Since the weight is directed downwards and the buoy-ant force is directed upwards, the net force on the balloon at depth d2 is directed downwards.The balloon is thus accelerated downwards, and choice C is correct.

Choice A is incorrect because the balloon does not move upwards. Choices B and D areincorrect because the buoyant force is less than the weight at depth d2.

5. BThis question requires information outside of that given in the passage. For questions about

the motion of an object, your first step should be to identify the forces acting on the object soyou can determine the net force on the object. Once you figure out the net force, you can deter-mine the net acceleration, and then use the appropriate kinematic equation to find the time. Inthis case there are two forces: the force due to gravity and the buoyant force. The buoyant forceis directed upward, and the gravitational force is directed downward. Therefore, the net down-ward force on the object is F = mba = mbg – ρwVbg, where mb is the mass of the ball, a is thenet acceleration of the ball, g is the acceleration due to gravity, ρw is the density of the water,and Vb is the volume of the water. Solving for the acceleration gives a = g(1 – ρwVb/mb). Sincemb = ρbVb, where ρb is the density of the ball, a = g(1 – ρw/ρb). To find the time it takes theball to reach the ground, use the equation d = at2/2, where d is the distance traveled and t is thetime. Solving for time, you find t = 2d/a. Substituting in the equation for acceleration gives

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t = g(1 –

2

d

w /b)

.

Plugging in the numbers, you get

t = (9.8(2)()

1(0–.3

15/7)

.9) .

Since the answer choices differ only in their order of magnitude, you don’t have to do anexact calculation. Rounding off the numbers, you get

t 10(10

–.71/8)

7

7

/

0

1

/

0

8 8/100 0.2.

This is closest to answer choice B, the correct answer.

Passage II (Questions 6–12)

Passage II presents you with information about two solvents: ammonia and water. The firstparagraph discusses ammonia and its properties. The passage then points out a few differencesbetween water and liquid ammonia, and how these differences affect their solvating ability.

In this passage, you are given two reactions. Always pay close attention to these reactions;you may be asked questions about them later. You are also given some experimental data, so thechances are you will be asked to work with this data in some way. In fact, the first question asksyou to compare the freezing points of Solutions 3 and 4, so you need to extract data from thepassage to get the right answer.

6. DThe passage states that compounds dissolved in liquid ammonia and water behave differ-

ently. One of these differences is that weak acids completely ionize in ammonia and partiallyionize in water. Therefore, Solution 4 will have a greater amount of dissociated ions.

The freezing point depression is proportional to the concentration of solute particles in solu-tion and to the Kf of the solvent. (∆Tf = Kfm, where m is the molality of the solute particles.)Notice that the value of Kf is larger for water than it is for ammonia. We therefore have two fac-tors that oppose each other. The ammoniacal solution has a higher concentration of solute par-ticles but a smaller Kf. Which effect is more important? To decide, we need to have a rough ideaof the extent of dissociation of acetic acid in water. Even though you do not need to memorizethe acid dissociation constant for acetic acid, you should be aware that its value is orders of mag-nitude smaller than full dissociation. (The acid dissociation constant for acetic acid in water ison the order of 10–5.) The values of the two Kf’s, however, differ just by a factor of two. Thedissociation of acetic acid in the two solvents is therefore a much more significant effect. Thefreezing point depression is larger for solution 4, the ammoniacal solution, even though ammo-nia has a smaller Kf. Choice D is the correct response.

Choice A is incorrect because there is more undissociated acid when the solvent is water.Choice C is contradictory to the information given in the passage. The freezing point depression

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of Solution 3 would only be larger than that of Solution 4 if acetic acid ionized to a greaterdegree in water. Choice B is incorrect since the ionizing ability of acetic acid differs in each sol-vent, which affects the freezing point depression.

7. BSilver nitrate, just like calcium chloride, will dissociate into its constituent ions. Since Solu-

tion 2 contains 4 moles of chloride ions, you would expect 3 of those moles to react with 3 molesof silver ions to form insoluble silver chloride. Now, because 3 moles of chloride ions have beenremoved from the solution as a precipitate, you would expect the effect upon the freezing pointand boiling point to diminish—the freezing point would increase and the boiling point woulddecrease. However, 3 moles of nitrate ions are still in the solution, which counteracts the precip-itation of 3 moles of chloride ions. Therefore, there are the same number of ions in the solutionas there were previously, so the boiling point and freezing point will be unaffected—choice B.

Choice C is incorrect because this effect would only be seen if the number of solutemolecules decreases. Choice A is incorrect because this effect would only be seen if the numberof solute molecules increases. Finally, choice D is wrong because both the boiling point and thefreezing point cannot decrease—if the boiling point increases, the freezing point decreases andvice versa.

8. AThe question tests you understanding of Le Châtelier’s principle, which states that when a

system in equilibrium is placed under stress, the equilibrium will move in such a way as to coun-teract that stress. Therefore, if ammonium chloride is added to a liquid ammonia solution, thereis an increase in the number of NH4

+ ions. To re-establish the equilibrium, the extra NH4+ ions

have to be ‘used up,’ and this is done by shifting the equilibrium to the left. Choice A is, there-fore, the correct response.

Choice B is incorrect because the equilibrium would only shift to the right if NH2– or NH4

+

were removed from the solution. Choice C is incorrect because a reagent has been added to thesystem which upsets the equilibrium. Therefore, there has to be a shift in the equilibrium to cor-rect for the excess of reagent. Choice D is incorrect because it is the addition of a catalyst, notthe addition of reactants or products, that affects the speed at which an equilibrium is reached.

9. ATo work out the volume of gas formed, the mass of liquid ammonia has to be calculated. This

is done by using the equation, mass equals density multiplied by volume. In other words, themass of liquid ammonia equals 0.692 g/cm3 multiplied by 11 cm3. Straight away, you can dis-card choice C and choice D, since the density is divided by the volume, not multiplied. The nextstep is to calculate the number of moles of liquid ammonia. This is given by the equation: num-ber of moles equals the actual mass divided by the molar mass. Hence, the number of moles ofliquid ammonia is (0.682 × 11)/17. Now we have the number of moles of gas, and we need toconvert this into the volume of gas. Another equation you should be familiar with is: number ofmoles of gas equals the volume of gas divided by the volume occupied by one mole of gas (22.4liters at STP). Therefore, the volume of gas is the number of moles of gas multiplied by 22.4.So, our final calculation will be: the volume of gas formed equals (0.682 × 11/17) × 22.4, orchoice A.

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10. DYou are told in the question stem that the boiling point of the solution increases to 101.56°C.

Since pure water boils at 100°C, the boiling point elevation is 1.56°C. Since the boiling pointelevation equals the boiling-point constant—Kb—multiplied by the molality of the solution, themolality of the solution is equal to 1.56 divided by the Kb for water, 0.52°C/m. This works outto be 3 molal. Remember, this observed molality is due to the contribution of one Ca2+ ion and2 Cl– ions. Therefore, the true molality of calcium chloride is 1 molal. If there is 1 mole of cal-cium chloride in 1000 g of solvent, there will be 1.5 moles in 1500 g of solvent. Therefore, theactual mass of calcium chloride is the number of moles—1.5—multiplied by the molar mass—111 g/mol. This works out to be 167 g, and the only answer choice near to this value is choice D.

11. AThe equilibrium constant, K, tells us the relative composition of the equilibrium mixture.

The value of K is equal to the product of the concentrations of reaction products divided by theproduct of the concentrations of reactants. Large values of K indicate that the equilibrium favorsthe products, whereas small values of K indicate that the reactants are favored in the equilibrium.If any pure solid or liquid is involved in the equilibrium, it can be discarded because its activityis equal to one. In addition, the concentrations of the reactants and products have to be raised toa power equal to their stoichiometric coefficients in the balanced equation. Looking at Reaction2, you should see that the silver halide will not be involved in the equation (since it is a solidand has an activity of 1), and ammonia will not be involved in the equation (since it is a liquidand has an activity of 1). Therefore, the equilibrium constant is equal to the product of the con-centration of the silver-ammonia complex—Ag(NH3)x

+ and the concentration of the halideion—X–. This corresponds to choice A.

Choices B, C, and D all take into account the concentration of ammonia, which we alreadyestablished has an activity equal to 1, and is not involved in the equilibrium expression. In addi-tion, choice B is incorrect because it does not include the concentration of the halide ion, andchoice C is incorrect because it not only includes the concentration of solid silver halide, but alsorepresents the equilibrium constant as being equal to the product of the concentrations of reac-tants, divided by the product of the concentrations of reaction products.

12. DAlthough ammonia is a neutral ligand, it possesses a lone pair of electrons on its nitrogen

atom, allowing it to act as a Lewis base. In coordination complexes, the central ion is usuallypositively charged, and it is the attraction of the ligand’s lone pairs to this positive charge thatlowers the energy of the complex, making it more stable. Therefore, in the complex shown inReaction 2, silver has a positive charge and the ammonia ligands each have a lone pair which isattracted to this positive charge—choice D is correct.

If ammonia acted as a Lewis acid by accepting electrons, the coordination complex wouldbe very unstable since there would not be any electrons to coordinate with the central ion. Any-way, you should know that the valence of nitrogen in ammonia is 8, so it cannot accept anymoreelectrons since this would exceed its octet. Therefore, choice C is wrong. Choices A and B arewrong since Brønsted acids and Brønsted bases are species that donate and accept protons,respectively. Since there are no protons involved in the coordination complex, these answerchoices can be discarded.

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Passage III (Questions 13–17)

This passage discusses two experiments involving a circuit designed to generate a magneticfield. The first paragraph presents a formula for the magnetic field generated by a current loopand then goes on to describe the experimental apparatus shown in Figure 1. It is useful to referto the figure as you are reading its description. The second and third paragraphs describe theexperiments. Note the difference between the two experiments. The voltage source used inExperiment 1 provides a constant voltage, but the voltage source used in Experiment 2 providesa voltage that varies with time. Figure 2 is a graph of the voltage supplied to the circuit in Exper-iment 2 as a function of time.

13. CFrom the information in the passage you can infer that the magnitude of the magnetic field

at point P is given by B = µ0i/2r, where µ0 is a constant, i is the current through the loop, andr is the loop’s radius. After the switch is closed, the current i through the loop can be deter-mined from Ohm’s law V = iReq, where V is the voltage and Req is the equivalent resistance ofR1, R2, and R3 combined. Solving for i, we obtain i = V/Req. For the two resistors in parallel,R2 and R3, the equivalent resistance R23 is given by 1/R23 = 1/R2 + 1/R3. Therefore, increasingthe resistance of either R2 or R3 will decrease 1/R23 and thus increase R23. Since R1 is in serieswith R23, the equivalent resistance of the three resistors combined is given by Req = R1 + R23.In this way, you see that increasing R1, R2, or R3 will increase Req, which in turn decreases ibecause V is constant.

If i decreases, then B decreases as well. So statements I, II and III are correct. Because theradius is inversely proportional to the magnitude of the magnetic field, decreasing the radius willincrease the magnitude of the magnetic field. Hence, statement IV is incorrect, and the correctanswer is choice C.

14. AUse the right-hand rule to determine the direction of the magnetic field. Point your right

thumb in the direction of the current. Your remaining right fingers then mimic the magnetic fieldlines, which circle around the wire. Your fingers, curling around the wire, point in the directionof the magnetic field. Note that this right-hand rule works for determining the direction of themagnetic field resulting from current flowing through a straight wire as well as a circular loop.

Since the current runs from the positive to the negative terminal of the voltage source, yourright thumb should point downward if you place it along the wire between points P and Q. Yourcurling fingers then indicate that the magnetic field is directed into the page at point P and outof the page at point Q, so choice A is correct.

15. BTo answer this question, first find the voltage drop across the equivalent resistance of R2 and

R3, and then use Ohm’s law to calculate the current through R2. Since R1 and the equivalentresistance of R2 and R3 are in series, the current that passes through R1 also, in effect, passesthrough the equivalent resistance of R2 and R3. To calculate the equivalent resistance R23, usethe formula 1/R23 = 1/R2 + 1/R3. Rearranging algebraically, we obtain R23 = R2R3/(R2 + R3) =(1)(2)/(1 + 2) = 2/3 Ω. Again, since R1 and the equivalent resistance R23 are in series with eachother, the 2.7 A that pass through R1 must also pass through R23. By Ohm’s law, the voltageacross R23 is thus V23 = i23R23 = (2.7 A)(2/3 Ω) = 1.8 V. Since R2 and R3 are in parallel, the volt-

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age V2 across R2 and the voltage V3 across R3 both equal the voltage V23 across the equivalentresistance R23 which equals 1.8 V. You can then calculate the current through R2 using i2 =V2/R2. Plugging in the numbers gives i2 = 1.8 /1 = 1.8 A, which is choice B.

16. CThe magnetic field at point P is given by B = µ0i/2r, where i is the current through the wire,

r is the radius of the loop, and µ0 is a constant. After the switch is closed in Experiment 2, thecurrent i through the loop is given by Ohm’s law i = V/Req, where V is the time-varying volt-age produced by the voltage source and Req is the equivalent resistance of R1, R2, and R3 com-bined. Substituting this relation for i into the formula for B gives B = µ0V/2rReq. Since only thevoltage varies in time, to determine how B varies in time we can rewrite the last equation as B= αV, where α = µ0/2rReq is a time-independent constant. Since B is directly proportional to V,and V is a linear function of time, B will also be a linear function of time (but with a differentslope). Mathematically, if V = mt + b, where m is the slope, t is the time, and b is the y-inter-cept, then B = αV = α(mt + b) = (αm)t + αb, which is simply the equation for another line withslope αm and y-intercept αb. Choice C is thus correct, and choices A and D are incorrectbecause they are not linear. Since the slope of V is non-zero, the slope of B is non-zero as well,and thus choice B is incorrect even though it is linear.

17. AConsider a graph where the area under the curve is a rectangle. The magnitude of this area

is equal to the product of the length and the width. Similarly, the units of the area under the curvewill be equal to the product of the units on the y-axis and the units on the x-axis. Although thearea under the curve in the question stem is not rectangular, the principle is the same. Therefore,the area under the given curve has the units of power × time. Since power can also be expressedas energy/time, the units of the area under the curve are power × time = (energy/time) × time =energy, and choice A is correct.

Passage IV (Questions 18–23)

Passage IV is about lead-acid batteries. The first paragraph gives you some general infor-mation about these batteries, and you are told that these cells are classed as secondary cells.Although the term secondary cell may not be familiar, the terms galvanic cell and electrolyticcell—mentioned in the last sentence of the first paragraph—should be familiar. Think aboutwhat these terms mean; for example, recall that a galvanic cell utilizes a spontaneous chemicalreaction to generate an electrical current, whereas in an electrolytic cell, an electric current hasto be passed through the cell in order to drive a nonspontaneous reaction.

The second paragraph then gives you some information about the set-up of the lead-acidcells, and the third paragraph describes the reaction that occurs when the lead-acid cell ischarged and discharged. The discharging reactions are shown in Reactions 1 and 2. Always payclose attention to reactions; identify which species, if any, are being oxidized and reduced;notice if the equation is balanced; and, look at the sign of the standard reduction potentials(remember, a spontaneous reaction is accompanied by a positive reduction potential).

The final paragraphs then look at problems associated with lead-acid batteries, and howover-charging or over-discharging can result in decreased efficiency in the battery.

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18. CIn this question, you need to use Faraday’s constant to calculate the right answer. You are

told in the question that the car engine was started for 3 seconds, and a current of 333 amps wasgenerated. This means that there are 3 × 333 coulombs that flow, or approximately 1000coulombs. Faraday’s constant states that there is 9.6485 × 104 coulombs of charge per mole of elec-trons. Therefore, the number of moles of electrons generated during this time is 1000/9.6485 × 104.By approximating 9.6485 × 104 C/mol to 100,000 C/mol, the number of moles of electronsworks out to 0.01. Looking at the discharge reaction at the anode (Reaction 1), we can see thateach lead generates two electrons. Therefore, the number of moles of lead is half the number ofmoles of electrons, or 0.005. Since the atomic mass of lead is 207 g/mol, the actual mass of leadused up in the reaction is 0.005 × 207, which is approximately 1 g. Choice C, therefore, is thecorrect response.

You would have arrived at choices A, B, or D if you had misplaced decimal points at vari-ous stages of the calculation. Remember, always watch your units.

19. AThe discharge reactions for the lead-acid cell are shown in Reactions 1 and 2. After charging,

both of these reactions favor the reactants; that is, lead is deposited at the cathode, and lead(IV)oxide is deposited at the anode. When the reaction moves in the forward direction, electrons aregenerated and the cell gives out a certain voltage. So, what would be the effect on this forwardreaction and, hence, the output voltage if lead sulfate was added to the cell after charging? Sincelead sulfate is a solid, it is not appreciably soluble in the electrolyte. As a result, the concentra-tion of lead sulfate—and the position of the equilibrium—does not change. Since the value of thecell potential is affected by the equilibrium constant, it follows that no change in the position ofthe equilibrium results in no change in cell potential—choice A is the correct response.

Choices B and C would only be likely answer choices if lead sulfate was soluble in the elec-trolyte. Only then could the position of the equilibria in Reaction 1 and Reaction 2 be affected.This would result in the generation of more or fewer electrons and, therefore, a different outputvoltage. Choice D is incorrect because the discharge voltage (2V) does not change, and can beeasily measured.

20. BYou are told in the first paragraph that automobile batteries usually consist of six lead-acid

cells—each generating 2V—connected in series to give a total output voltage of 12V. If the bat-tery only generates 6V, then only three of the six cells must be working properly to generate avoltage. Therefore, choice B is the correct response. Choice A is incorrect because the passagestates that the cell should not be discharged to the point where there is more than 30% conver-sion of lead and lead(IV) oxide to lead sulfate. Therefore, less than 30% conversion of lead andlead(IV) oxide to lead sulfate, as stated in choice A, will not be detrimental to the output volt-age of the cell. Choice C is incorrect because as long as the electrolyte level is sufficient toobtain a reading, it will not affect the magnitude of the voltage reading. Choice D is incorrectsince 100% conversion of lead and lead(IV) oxide to lead(II) sulfate corresponds to total dis-charge of the lead-acid battery (i.e., Reactions 1 and 2 proceed to the right to completion).Therefore, the voltage generated at this point would be 0 V, not 6 V, and choice D can be elim-inated.

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21. ARecall that charging the lead-acid cell involves driving Reactions 1 and 2 to the left (i.e.,

forcing a nonspontaneous reaction). An electrolytic cell is defined as an electrochemical cell inwhich a nonspontaneous reaction is forced by the input of electricity. You are told that chargingthe cell results in: 1) the reduction of lead(II) sulfate at the cathode to form lead, and 2) the oxi-dation of lead(II) sulfate at the anode to form lead(IV) oxide. You are also told that charging thecell beyond the point where all the lead(II) sulfate is used up results in the electrolysis of waterto form hydrogen and oxygen.

The question stem asks you to identify the reaction at the anode when the cell is over-charged. Remember, oxidation always occurs at the anode, and reduction always occurs at thecathode. In addition, oxidation is loss of electrons and reduction is gain of electrons. Rememberthese two points! Choice A is correct since this represents the half-reaction at the anode: the oxi-dation number of oxygen changes from –2 in H2O, to 0 in O2, hence oxygen has been oxidized.

Choice B is incorrect since this reaction is a reduction reaction: protons gain electrons toform 2 molecules of hydrogen. This reaction occurs at the cathode, not the anode, when the lead-acid cell is over-charged. Choice C is incorrect because this is the overall redox reaction for theelectrolysis of water; it is a combination of the two half-reactions in choices A and B. Since thisreaction involves reduction, choice C can be eliminated. Choice D is also a reduction reaction.In addition, it is an incorrect reduction reaction since we are told in the passage that oxygen andhydrogen are generated when water undergoes electrolysis, not hydrogen and hydroxide ions.

22. BThe question asks which of the measures would be most useful in preventing loss of capacity

in the battery. To answer this question, you need to understand the definition of Le Chatelier’s principle, which states that when an equilibrium is subjected to stress, it will move to alleviate that stress.

The battery loses capacity when Reactions 1 and 2 move to the right. In essence, we arelooking for a procedure that will drive Reactions 1 and 2 to the left, or at least inhibit the for-ward reaction. In Reaction 2, we can see that water is produced, which in turn can generatehydrogen and oxygen. If the battery is sealed, hydrogen and oxygen will not be lost from thebattery, so the forward reaction—and hence the loss of capacity in the cells—will be inhibited;choice B is the correct answer.

Choice A is incorrect since lead(II) sulfate is a solid, and we established in question 84 that theaddition of lead(II) sulfate will not affect the position of the equilibrium. Choice C is incorrectbecause the addition of sulfuric acid will result in a shift to the right in the equilibrium of Reactions1 and 2. Choice D is incorrect for the same reason as choice A: lead(IV) oxide and lead are solids,so their concentrations will remain the same and will not affect the position of the equilibrium.

23. DReaction 1 is the discharge half-reaction at the anode, and Reaction 2 is the discharge half-reac-

tion at the cathode. So, to answer this question, you need to add the two discharge half-reactions:

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The balanced equation is represented in choice D—the correct response. Choice A is incor-rect since this is the opposite of this reaction: it is the charging reaction. Choice C is incorrectbecause this reaction is the half-reaction for the discharge reaction, not the balanced reactionequation. Choice B is correct in that it is a balanced reaction, but it is not the correct balancedreaction. This reaction fails to include protons on the left-hand side of the equation, and wateron the right-hand side.

Independent Questions

24. DBefore trying to solve this problem quantitatively, look at the answer choices. You can rule

out two choices immediately by remembering that metals expand when heated and contractwhen cooled. Therefore, when the temperature drops, the length of the flagpole will decrease,so choices A and B can be ruled out. To choose between C and D, you have to use the formulafor linear expansion: ∆L = αL∆T, where ∆L is the change in length, α is the linear expansioncoefficient, L is the original length, and ∆T is the change in temperature. Plugging in the num-bers gives ∆L = (11 × 10–6)(20)(30) = 6.6 × 10–3 m = 6.6 mm, and choice D is correct.

25. CThe sum of the oxidation states of the elements in a polyatomic ion is equal to the net charge

on that ion. In this case, the ion in question is an anion, and the net charge on the ion is –1. Eachoxygen has an oxidation number of –2, so 4 oxygens contribute a total charge of –8. Therefore,for the net charge to equal –1, chlorine must have an oxidation number of +7. By the way, inmost cases, the oxidation number of oxygen is equal to –2, but there are exceptions; for exam-ple, in hydrogen peroxide, the oxidation number of oxygen is equal to –1, and in compoundslike OF2, where the element that combines with oxygen is more electronegative, the oxidationnumber of oxygen is equal to +2. Choice B is incorrect since this is equal to the formal chargeof chlorine, not the oxidation number. Let’s briefly look at how formal charge is calculated.Chlorine has 7 valence electrons, each oxygen has 6 valence electrons, and the molecule gainsan extra electron since it has a negative charge. Therefore, the number of electrons that have tobe accounted for is 7 + 24 + 1: 32 electrons, or 16 electron pairs. These 16 electron pairs arrangeaccording to the following Lewis structure:

Cl O

O

O

O

—

PbSO4(s) + 2e

–Pb(s) + SO4

2–(aq)

PbO2(s) + SO42–(aq) + 4H

+(aq) + 2e

–PbSO4(s) + 2H2O(l)

PbO2(s) + 2SO42–(aq) + 4H

+(aq) + Pb(s) 2PbSO4(s) + 2H2O(l)

11

Three of the chlorine-oxygen bonds involve two bonding electron pairs; in other words, adouble bond forms. One of the chlorine–oxygen bonds involves one pair of bonding electrons;this constitutes a single bond. In the bonds that involve two electrons pairs, we assign one elec-tron pair to chlorine, and one electron pair to oxygen. In the bond that involves one electron pair,we assign one electron to oxygen and one electron to chlorine. Therefore, the number of elec-trons that chlorine possesses in the Lewis structure is (3 × 2) + 1, or 7. The number of valenceelectrons chlorine possesses in the free state is 7, so the formal charge is the number of electronsin the free state (7) minus the number of electrons in the Lewis structure (7), which equals zero.Choice A is incorrect because this value corresponds to the charge on the anion, not the chargeon the chlorine. Finally, choice D is a “red herring”—it doesn’t correspond to oxidation state,formal charge, or the charge on the anion.

26. BIn solution, barium sulfate will exist in the following equilibrium:

BaSO4(s) Ba2+(aq) + SO42–(aq)

The extent to which barium sulfate ionizes depends on the Ksp—the solubility product con-stant. According to the equilibrium reaction, a large Ksp favors the products, whereas a small Kspfavors the reactants. Since the Ksp of barium sulfate is small, the equilibrium will favor the reac-tants (very little barium sulfate will ionize in solution). The Ksp of barium sulfate is calculatedas follows:

Ksp = [Ba2+][SO42–]

Since the Ksp—which is equal to 1.1 × 10–10—is the product of the concentration of Ba2+

and SO42–, the concentration of Ba2+ is equal to the square root of 1.1 × 10–10, or 10–5 M. If you

stopped at this point, you would have ended up with choice C. However, you are being asked tocalculate the number of moles of barium, not the concentration. Since the volume of the solu-tion is 500 mL, the number of moles of barium is 10–5 × 0.5, or 5 × 10–6; choice B. A simpleerror in your calculation could have led to choices A and D, so always be careful when usingdecimals.

27. AA compound that can react with an acid or a base is classed as amphoteric. In the first reac-

tion, aluminum hydroxide acts as a basic hydroxide oxide by reacting with hydrogen ions—donated by acid—to form Al3+ and water; in the second reaction, aluminum hydroxide acts asan acidic hydroxide by reacting with hydroxide ions—donated by a base—to form Al(OH)4

–.Therefore, choice A is the correct response. Choices B and C can be eliminated because, as wecan see in these reactions, aluminum hydroxide acts as a base and an acid. Since aluminumhydroxide acts as a neutralizing agent (neutralizing acid in the first reaction, and base in the sec-ond reaction), it cannot be a neutral compound—it must be acidic and basic, so choice D can beeliminated.

28. BTo answer this question, you need to understand Dalton’s Law of partial pressure, which

states that the partial pressure of a gas is equal to the mole fraction of gas multiplied by the totalpressure. The first step, therefore, is to calculate the number of moles of nitrogen and oxygen,and the mole fraction of nitrogen. The number of moles of nitrogen in the flask is equal to the

12

Ksp

mass, 0.28 g, divided by the molar mass of nitrogen, 28 g/mole—0.01 moles. The number ofmoles of oxygen is equal to the mass of oxygen, 0.64 g, divided by the molar mass of oxygen,32 g/mole—0.02 moles. Since the total number of moles of gas in the flask is 0.03, the molefraction of nitrogen is 0.01/0.03, or 0.33. You are told that the total pressure in the flask is 3atmospheres, so according to Dalton’s Law, the partial pressure of nitrogen is 0.33 × 3, or 1atmosphere: choice B. (By the way, a short cut would have been to establish that the molar ratioof nitrogen to oxygen is 1:3, so the partial pressure of nitrogen is a third of the total pressure, or1 atmosphere.)

Since the partial pressure of nitrogen is 1 atm, and the total pressure in the flask is 3 atm,oxygen must contribute a pressure of 2 atm, so choice C can be eliminated. Choice D is incor-rect because there are two gases in the flask, both of which exert individual pressures; therefore,no one gas can contribute solely to the pressure in the flask. You may have picked choice A ifyou incorrectly calculated the number of moles of nitrogen in the flask.

Passage V (Questions 29–33)

Passage V is about collision theory. This theory states that molecules with sufficient kineticenergy must collide with each other before they react to form a product. The minimum kineticenergy these reactant molecules possess is called the activation energy, and the reaction speciesthat is formed is called the activated complex. When the activated complex is formed, it caneither go on to form product molecules, or fall apart into reactant molecules.

The topic of collision theory should be familiar to you. What is probably not familiar is thetwo reaction mechanisms that are proposed for the formation of hydrogen iodide. These mecha-nisms follow the collision theory; the reactant molecules collide to form an activated complex,which then goes on to form a product. In the first mechanism, a hydrogen and iodine moleculecollide to form an activated complex; in the second mechanism, two iodine atoms collide with ahydrogen molecule to form an activated complex. Each reaction mechanism is drawn and the rateequations for each are given, so pay special attention. The passage then gets more detailed in itsdiscussion of the second mechanism. This mechanism is more complex than the first since therate of the reaction is dependent on the concentration of an unstable intermediate—atomic iodine.The steady-state approximation is another concept that you may be familiar with. It assumes thatduring a reaction, the concentration of the intermediate remains constant and small; this greatlysimplifies kinetic calculations for reactions that involve a number of steps. Finally, you are givendata for the rate of reaction when the concentrations of iodine and hydrogen are varied. When-ever you are presented with a table, chances are you will be asked to interpret the data.

29. CThe rate equation for mechanism 1 is: rate = k [H2][I2]; the rate equation for mechanism 2

is: Rate of reaction = k1k3/k2[H2][I2]. Therefore, the rate of reaction for both mechanisms is sim-ilar in that it is first order with respect to hydrogen, and first order with respect to iodine. Theorder of the reaction is confirmed by looking at the data in Table 1: As the concentration ofhydrogen increases from 0.25 mol/L to 0.5 mol/L, the rate of formation of hydrogen iodideincreases from 1.5 × 10–5 mol L–1s–1 to 3.0 × 10–5 mol L–1s–1, (i.e., both the concentration ofhydrogen and the rate of formation of hydrogen iodide double). Therefore, the rate of thereaction is first order with respect to hydrogen. Similarly, as the concentration of iodine dou-bles (from 0.25 mol/L to 0.5 mol/L), the rate of hydrogen iodide formation also doubles(from 7.5 × 10–6 mol L–1s–1 to 1.5 × 10–5 mol L–1s–1), and the rate of the reaction is also first

13

14

order with respect to iodine. Therefore, both mechanisms are supported by the table and choiceC is correct.

Choice D would be correct if the results gave a reaction order different to that observed (firstorder with respect to hydrogen and iodine). The results, however, support both mechanisms, sochoice D is incorrect. The only way we could establish whether mechanism 1 or mechanism 2has occurred would be by performing experiments that give conclusive results supporting onemechanism. For example, an experiment in which the effect of iodine radicals upon the rate ofreaction would allow us to decipher the mechanism. Since we are not given any data to distin-guish between the mechanisms, choices A and B are incorrect.

30. DThe change in enthalpy for this step is positive, (it is an endothermic reaction), and the

change in entropy is also positive. For a reaction to be spontaneous, the Gibbs free energychange, ∆G, must be negative. ∆G is equal to ∆H – T∆S. Since the enthalpy change and theentropy change for this step is positive, the term T∆S must be larger than the term ∆H, so that∆G is negative and the step is spontaneous. Since we are not given the temperature of the reac-tion, we cannot determine whether this step is spontaneous—choice D is correct.

Choice A would only be correct if this step were carried out at low temperatures, and choiceC would only be correct if the reaction were carried out at high temperatures. Again, we are notgiven the temperature of the reaction, so choices A and C can be discarded. Choice B is incor-rect since the stability of hydrogen iodide is irrelevant in determining whether the step is spon-taneous. In addition, hydrogen iodide is a relatively stable molecule, so choice B is an incorrectstatement.

31. CAccording to Mechanism 1, the rate of the reaction is equal k[H2][I2]. The rate constant,

therefore, is equal to the rate of the reaction divided by the product of the concentration ofhydrogen and iodine. If we take a set of values for any one experiment from Table 1, the rateconstant works out to be 1.2 × 10–4 L mol–1s–1, choice C.

32. BThis question asks us about the rate of the reaction in Mechanism 2, so we have to look at

Mechanism 2 and Equation 2. In Mechanism 2, we are told that the rate of the reaction is depen-dent on the product of the rate constant, k3, the concentration of atomic iodine raised to the sec-ond power, and the concentration of hydrogen raised to the first power. Therefore, the rate of thereaction is second order with respect to atomic iodine—choice B. Recall that the overall orderof a reaction is the sum of the exponents in the rate law, so, according to Mechanism 2, the rateof reaction is third order with respect to atomic iodine and hydrogen, not second order: choiceA is incorrect. In Equation 2, we can see that the rate of the reaction is first order with respectto molecular iodine, so choice C can be eliminated. Finally, Mechanism 2 and Equation 2 showthat the rate of the reaction is first order with respect to hydrogen, so choice D can be eliminated.

33. CThe energy needed to break the bonds of 1 mole of a gaseous molecule is called the bond

dissociation energy. Since chemical reactions involve bond breakage and bond formation, bonddissociation energies can be used to calculate the enthalpy of a reaction. Recall that energy hasto be provided to break chemical bonds, and energy is always released when chemical bonds areformed. Mechanisms 1 and 2 are similar in that the covalent bond in hydrogen is broken and the

covalent bond in iodine is broken, and two hydrogen iodide bonds are formed. Therefore, wehave to put energy in to break the hydrogen and iodine bonds, while energy is given back whenthe hydrogen iodide bonds are formed. If the energy input is greater than the energy given out,the reaction is endothermic; if the energy input is less than the energy given out, the reaction isexothermic.

The standard enthalpy of formation of hydrogen iodide is the energy change when onemole of hydrogen iodide molecules is formed. Since the balanced equation for the reaction isH2(g) + I2(g) → 2HI(g), the bond dissociation energies of hydrogen and iodine must be halvedin order to calculate the standard enthalpy of formation of HI. The equation often used to cal-culate the enthalpy of a reaction is:

∆H° = Total bond dissociation energy(reactants) – Total bond dissociation energy(products)

or

∆H° = Total energy input – Total energy released

Inserting values into our equation, we get:

∆H° = (218 + 75.5) – (298)

∆H° = 293.5 – 298

∆H° = –4.5 kJ/mol

If you incorrectly calculated the enthalpy of the reaction to be the energy released minus theenergy input, you would have ended up with incorrect choice B. If you assumed that 2 moles ofhydrogen iodide were formed in the reaction, you would have ended up with choice D. Finally,choice A is incorrect for the same reasons as choices B and D: the enthalpy of the reaction isincorrectly calculated as the energy released minus the energy input, and the bond energy of 2moles of hydrogen iodide is used erroneously in this calculation.

Passage VI (Questions 34–38)

34. AFrom the equation given in the question stem, we know that the speed of sound increases

with pressure and decreases with density. Air is composed primarily of nitrogen and oxygen,both diatomic gases. They are also both heavier (have a higher molecular weight) than hydro-gen. Recall that one mole of an ideal gas, regardless of its chemical identity, occupies the samevolume as a mole of another ideal gas under the same conditions. Therefore, the gas with thehigher molecular weight is more dense. Hydrogen is therefore less dense than air, and sound willtravel at a higher speed in hydrogen than in air. We then need to decide whether sound will travelfaster at a higher or lower temperature. We can apply the ideal gas law: pV = NkBT, where p isthe pressure, V is the volume, N is the number of particles, kB is the Boltzmann constant, and Tis the temperature. Dividing both sides by N, you get pV/N ∝ T. N/V is the number of particlesin a given volume which is the volume density ρ, so V/N is just 1/ρ. Thus, we have p/ρ ∝ T.Inserting this into the equation for velocity v, we have that v ∝ T. Thus when the temperatureincreases, then so does the speed of sound. In short, sound travels faster when the temperatureis higher. Choice A is correct.

15

35. DWe are not really given much information about the wave set up in the aluminum rod. How-

ever, one characteristic of waves that we should know is that as it travels from one medium toanother, its frequency remains constant even though its speed and wavelength may change. If wecan determine the frequency of the sound wave in the glass tube, therefore, we know the fre-quency of the sound wave in aluminum rod.

We can find the wavelength of the standing wave set up in the Kundt’s tube from the factthat the dust clusters occur at nodes, and that these clusters are 5.5 cm apart.

As can be seen from the diagram above, the wavelength must be 2 x 5.5 cm, or 11 cm. Sincewe are given the speed of sound in air, we can determine the frequency of the sound wave:

f = λv

= 3

0

4

.1

4

1

m

m

/s ≈ 340 10 s–1 ≈ 3400 Hz

A more exact calculation would yield 3.1 103 Hz. Since we have determined the fre-quency in air to be about 3400 Hz, we therefore also know that this is the frequency in alu-minum, and hence choice D is correct.

36. DWe are told that the clamp in the middle of the rod forces there to be a node, while the ends

are antinodes. We should be able to picture the following for the fundamental:

The wavelength is therefore twice the length of the rod.

37. BThe standing wave sustained in the glass tube is a sound wave. Sound waves are mechani-

cal waves, which require a material medium in which to travel, in this case air. This is opposed

nodeantinode antinode

L

λ

nodes

5.5 cm

16

to electromagnetic waves, which can travel throughh vacuum. Furthermore, sound waves arelongitudinal waves because the compression and rarefaction of air molecules occur in a direc-tion parallel to that of the wave itself.

38. DAs described in the passage, a step in the Kundt’s tube experiment consists of moving the

glass tube back and forth so that the length can sustain the standing waves generated by thestroking of the aluminum rod. Choice A can be eliminated since sound waves cannot travel invacuum.

The distance between the dust clusters is then used to determine the wavelength (see theexplanation to #35 above). It is then stated in the passage that this value can be used to deter-mine the frequency of the wave when we know the speed of sound in air. Obviously, the valueof the speed of sound must be applicable to the situation in the glass tube. If we seal the tubewith a certain amount of air inside and prohibit any exchange, then as we move the tube backand forth, the density and pressure of the air inside will change. There is then no way of know-ing exactly what value of the speed of sound is valid. Choices B and C are therefore both incor-rect. Instead, we must somehow allow for the free exchange of air with the environment (usuallywith some form of an inlet on the top) so that the air in the tube is at atmospheric pressure.

Passage VII (Questions 39–44)

This passage discusses X-rays. The process used to make X-ray images is briefly describedin the first paragraph. In this paragraph you are given an equation for the intensity of an X-raybeam transmitted through a medium. Note the relationship between the variables: the intensitytransmitted decreases exponentially as the distance the X-ray beam travels through the materialincreases. The second paragraph discusses what happens as a result of the collision between anX-ray and an atomic electron. Long wavelength photons are emitted when X-rays collide withan atom. So when X-rays, short wavelength photons, collide with low level atomic electrons,long wavelength photons are emitted. The final paragraph discusses how Geiger counters work.

39. DAccording to the passage, the probability of a collision between an X-ray and an electron in

a low atomic energy level only depends upon the charge of the nucleus. Remember, the isotopesof a given element all have the same number of protons, and hence the same atomic number, butthey have different numbers of neutrons, and hence different atomic masses. Since all isotopesof tin have the same atomic number, the same number of protons, and the same nuclear charge,their interactions with X-rays are the same. Choices A, B, and C are all isotopes of tin and so allwould scatter X-rays equally well.

40. BIn this question you are given wavelengths and asked to find energies. Therefore the two

relations involving wavelength and energy given in the first paragraph are useful. Combiningthem and solving for energy in terms of wavelength, you get E = hc/λ. Inserting the constantgiven in the passage, hc = 1240 nm·eV, you get E = 1240/λ, where E is measured in eV and λis measured in nm. Now it remains to choose the correct energy. This can be done by convert-ing the boundaries of the X-ray spectrum into energies. 0.01 nm corresponds to 1240/(.01) =124000 eV or 124 keV and 1 nm corresponds to 1240 eV. So any photons between 1240 eV and124 keV are X-rays.

17

41. AThe passage and question both refer to the picture as a “shadow.” This means that the pic-

ture is a picture of the incident X-rays which passed through the body without interacting at all.This X-ray shadow is similar to a visible light shadow; a dark shadow indicates that very few X-rays passed through the body. Because the picture is a negative of a shadow, the light areas indi-cate the absence of X-rays. The X-rays are absent because they have been scattered. Thus bonesand connective tissue scatter X-rays more efficiently. The passage states that elements withhigher nuclear charge scatter the X-rays more efficiently, but the answers are all in terms of“heavy” and “light” elements. Since elements with greater atomic number are also heavier ele-ments, heavier elements will scatter the X-rays more efficiently. Thus heavier elements scatterX-rays more efficiently and bones and connective tissue scatter X-rays more efficiently, so theanswer must be choice A. Choices B and D are incorrect because they imply that the lightnessor darkness of the X-ray picture is due to the behavior of cascade photons. Choice C is incor-rect because the picture is a negative.

42. AThe passage states that the cascade photons are long wavelength photons and that X-rays are

short wavelength photons. Therefore, the cascade photons must have a longer wavelength thanX-rays. Since energy is proportional to frequency, and frequency is inversely proportional towavelength, energy is inversely proportional to wavelength. Therefore, long wavelength photonshave lower energies than short wavelength photons. Choices B and D are incorrect because bothimply that the cascade photons have shorter wavelengths. Choice C is wrong because all pho-tons travel at the same speed through the same medium.

43. CTo answer this question, which asks for a ratio, it is necessary to set-up the ratio as a frac-

tion. The numerator is the intensity of the X-ray beam after it has passed through 1 cm of plasmaand the denominator is the intensity of the X-ray beam after it has passed through 6 cm ofplasma. In the first paragraph of the passage, you are told that the intensity of an X-ray beamtransmitted through a medium is given by I = I0e–µx, where I0 is the intensity of the beam whenit is incident on the material, µ is the absorption coefficient of the medium, and x is the distancetraveled through the medium. Using the fact that the absorption coefficient of plasma is 0.2cm–1, you can determine that the numerator of the fraction is I0e–(0.2)(1) and the denominator isI0

–(0.2)(6). Therefore, the ratio is equal to I0e–(0.2)(1)/I0e–(0.2)(6) = e(0.2)(5) = e, making choice Cthe correct answer. You could have ruled out choice A immediately because it is less than one,and common sense and the formula in the first paragraph indicate that the intensity of the beamdecreases as the distance it travels in a given medium increases.

44. AIn the last paragraph of the passage, you are told that within a Geiger counter there is a high

voltage difference between the wire and the wall of the gas chamber. This implies that a poten-tial difference exists between the wall and the wire, and choice C can be ruled out. Recall fromyour knowledge of electric circuits that current flows from high to low potential. Since electronsflow in the direction opposite to that of the current, electrons will flow from low to high poten-tial. Therefore, the wire must be at a higher potential than the wall of the gas chamber.

18

Passage VIII (Questions 45–50)

Passage VIII, like so many passages found on the MCAT, describes an experiment. In addi-tion, the apparatus used to carry out this experiment is shown. When you are presented with pas-sages such as these, you can be sure that you will be asked questions about the experiment,whether it be about the apparatus, the experimental conditions, or the results of the experiment.

Gravimetric analysis is an important analytical tool. This technique is often employed todetermine the composition of an analyte (the substance being analyzed), whether it is isolatedor part of a compound. The composition of the analyte is exclusively determined by changes inweight, which can be monitored using the apparatus shown in Figure 1. So, let’s move on to thequestions.

45. CIn this question, you are being asked to interpret the results of the experiment, and to deci-

pher the reaction that occurs when nickel is heated in the presence of oxygen. When nickel isoxidized, it will readily form an Ni2+ ion. Therefore, nickel(II) oxide, or NiO forms. Choice Calso possesses the correct stoichiometry for this reaction, so it is the correct response. Choice Ais incorrect because although nickel combines with gaseous oxygen, the oxidation state of nickelin the product is +3. To form Ni2+, two electrons are lost from the 4s subshell. To form Ni3+, asin choice A, an additional electron would have to be lost from the 3d orbital. This is energeti-cally unfavorable, so nickel prefers the +2 oxidation state. Choice D is incorrect because nickelwould have to be oxidized to Ni4+ to form NiO2. This would involve the loss of two electronsfrom the 3d orbital; even more difficult than the loss of one electron to form Ni3+. Choice B isincorrect since this represents the reaction of nickel and oxygen in aqueous solution; therefore,the reaction conditions are incorrect.

46. CYou could have gotten the answer here by two methods. First, you could have used the num-

ber of moles of nickel that reacted with oxygen. The number of moles of nickel is the actualmass of nickel in grams (0.05 g) divided by its atomic mass (58 g/mol). This works out to be8.6 × 10–4 moles; choice C. On the other hand, if you knew that nickel(II) oxide was formed inthe reaction, you could have calculated the number of moles of product by dividing the mass ofthe product (0.064 g) by the molar mass of nickel(II) oxide (74.7 g/mol). Again, this works outto be 8.6 × 10–4 moles. By the way, don’t be worried by the fact that you don’t have a calcula-tor. Learn to approximate. You could have gotten the right answer by approximating 0.05/58, or0.064/74.7 to be slightly less than 10. Adding in the decimal point, you would have gotten thenumber of moles of product to be slightly less than 10 × 10–4, and the only answer choice nearto that is C.

Now let’s look at the wrong answers. If you forgot to convert milligrams into grams of prod-uct, you would have ended up with choice A. This leads us to another point: always watch yourunits! A simple mistake in your calculation could easily lead you to the wrong answer. If youassumed that 1 mole of nickel went on to form 1/2 a mole of product, you would have ended upwith choice D. If you forgot to convert milligrams into grams, and you assumed 1/2 the amountof product was formed, you would have ended up with choice B.

19

47. DA phase change is defined as the point where a substance changes into a different state of

matter; for example, when ice melts to form water. When a phase change occurs, there is alwaysa transitional period in which the different phases co-exist in a dynamic equilibrium; this isknown as the phase boundary. Phase changes are not accompanied by a change in weight, andsince thermogravimetric analysis is used to detect changes in weight, choice D is the correctresponse.

Since a phase change is not accompanied by a change in weight, choices A and B can be dis-carded. Choice C states that TGA is an unsuitable method to detect the occurrence of a phasechange because the structure of the sample does not change. Molecular structure is irrelevant inexplaining whether or not thermogravimetric analysis is useful in determining phase changes, sochoice C is incorrect.

48. ATo answer this question, you need to be familiar with the electronic structure of the transi-

tion metals. The transition metals have partially filled d orbitals. Nickel possesses the electronicconfiguration, 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d8. Since nickel likes to lose two electrons from its4s subshell, the most stable oxidation state is +2. Iron has the same configuration as nickel,except that it possesses six 3d electrons. Iron can also lose two electrons from its 4s subshell toform Fe2+.

Unlike nickel, iron will readily lose an electron from its 3d subshell, resulting in the forma-tion of a stable Fe3+ ion. So, why will iron readily form a +3 ion, whereas nickel will not? Toanswer this question, we need to review Hund’s rule. Hund’s rule states that when orbitals ofidentical energy are available to electrons, the electrons will occupy these orbitals singly insteadof in pairs. Hence the d5 configuration is very stable, since the electrons occupy differentorbitals, each with parallel spins. Fe3+ has the configuration 1s2, 2s2, 2p6, 3s2, 3p6, 3d5, so it isa very stable ion. If nickel loses a 3d electron to form Ni3+, the stable 3d5 configuration is notformed. Instead, a 3d7 configuration arises. In this subshell, 2 out of the 5 orbitals contain pairedelectrons. According to Hund’s rule, this configuration is not as stable.

Choice C states that the d subshell in the Ni2+ ion is half-filled which imparts stability. How-ever, we have just seen that it is the Fe3+ ion that possesses a half-filled d subshell, so choice Cis incorrect. Choice B is irrelevant since the question focuses on why iron(II) oxide is easily oxi-dized to iron(III) oxide. In addition, nickel can possess an oxidation state higher than +2, so thisanswer choice is an incorrect statement. Finally, choice D is incorrect because iron(II) oxide isrelatively stable.

49. BIn the reaction of nickel and oxygen, nickel acts as a reducing agent: donates electrons and

becomes a positive ion. However, when nickel oxide reacts with hydrogen, nickel acts as an oxi-dizing agent (it gains electrons), and hydrogen acts as a reducing agent. Reduction is defined asthe gain of electrons, and Ni2+ gains two electrons to form Ni0. Therefore, one of the productsis solid nickel, and choices A and C can be eliminated. Notice that there was a big clue in thepassage which may have led you to nickel formation; the weight of the product drops to 50 mgwhen it is reacted with hydrogen, insinuating that solid nickel was re-formed. When oxygen islost from nickel oxide, it combines with hydrogen to form water, not hydrogen peroxide. There-fore, choice D can be eliminated and choice B is the correct answer.

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50. BWhen Group II metal oxides react with water, their corresponding hydroxides are formed.

These metal hydroxides—with the exception of beryllium hydroxide—are very strong bases.Since the question stem tells you that a strong base is formed, you could have figured out thatthe product was calcium hydroxide and the reactant was calcium oxide.

Another way to get to the right answer, which is a little more difficult, is to look at Graph 1.As the temperature increases, hydrated calcium oxalate breaks down into lower molecularweight compounds. Initially, water is removed from hydrated calcium oxalate, resulting in theformation of anhydrous calcium oxalate (Compound A). Calcium oxalate then breaks down intocalcium carbonate (Compound B) which in turn breaks down to form calcium oxide (CompoundC). Therefore, the correct answer is choice B.

The question states that a strong base is formed when water is added to Compound C. How-ever, calcium carbonate and calcium oxalate are both insoluble in water and do not form calciumhydroxide, so choices A and C are incorrect. Finally, choice D is incorrect because calciumhydroxide is the product formed when compound C dissolves in water. Even if calcium hydrox-ide were Compound C, it would be sparingly soluble in water.

Independent Questions

51. DBoron belongs to the group III elements, so it possesses three electrons in its outer valence

shell. When BF3 forms, all these electrons take part in covalent bonding. According to VSEPRtheory, a molecule which possesses 3 bonding electron pairs such as BF3 will have a trigonalplanar geometry:

Since fluorine is more electronegative than boron, each individual B–F bond will have adipole moment in the direction of fluorine. However, the BF3 molecule is symmetrical, so theindividual bond dipoles will cancel out, and the molecule will have a zero net dipole moment,i.e., it will be nonpolar.

All the other molecules have a geometry whereby the dipole moments of the individualbonds do not cancel out: SO2 has a bent geometry, and the dipole moment lies toward the oxy-gens; CH3Cl has a tetrahedral geometry, with the dipole moment in the direction of chlorine (theC–H bond is essentially nonpolar); and PCl3, due to a nonbonding pair of electrons on phos-phorus, has a trigonal pyramidal geometry, with the dipole moment in the direction of chlorine.Therefore, all of these molecules are polar and again, choice D is correct.

52. CTo determine the initial velocity vector of charge P, consider the forces on charge P. The

magnitude of the electrostatic force between two charges is given by F = kq1q2/r2. Since thecharges Q and R are equidistant from P, the magnitude of the force on P due to Q is equal to themagnitude of the force on P due to R. Since P and Q have the same sign, P will be repelled by

B

F

FF120°

BF

F

F

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Q. Since P and R have opposite signs, P will be attracted to R. The vector representing the forceon P due to charge Q will be oriented along the line connecting P and Q and it will point awayfrom Q. In other words, it will point in the same direction as D. The vector representing the forceon P due to R will be oriented along the line connecting P and R and it will point toward R. Inother words, it will point in the same direction as B. To determine the initial force that P expe-riences when it is placed on the corner of the triangle, you have to sum the forces acting on P.The resultant force vector will point in the same direction as C. Since the initial acceleration vec-tor points in the same direction as the initial force vector and the initial velocity vector points inthe same direction as the initial acceleration vector, choice C is the correct answer.

53. CTo answer this question, use the equation for the period of oscillation of a spring T = 2m/k,

where m is the mass of the block and k is the spring constant of the spring. You might have theequation for the frequency memorized. If so, you can get the period by using the fact that theperiod equals one over the frequency. The best strategy for Roman numeral questions is to eval-uate the statements one by one and rule out answer choices after you’ve evaluated each state-ment. Since the period is proportional to the square root of the mass, increasing the mass of theblock will increase the period, and statement I is correct. At this point you can rule out choiceD because it does not include statement I. The amplitude of the oscillation doesn’t appear in theformula for the period. Furthermore, the mass of the block and the spring constant of the springare properties of the block and the spring, respectively, and are not dependent upon the ampli-tude of the motion. Therefore, the amplitude does not affect the period and statement II is incor-rect. So choice B can now be ruled out. Finally, since the period is inversely proportional to thesquare root of the spring constant, decreasing the spring constant increases the period, and state-ment III is correct. Thus, choice C is the correct answer.

54. BTo determine how long it takes the helicopter to travel 3 miles, first you have to determine

the magnitude of the helicopter’s resultant velocity. There are three components of the heli-copter’s velocity. The helicopter is moving west at a speed of 10 mph and upward at a speed of5 mph, and the air current is moving it south at a speed of 10 mph. You can think of the north-south axis as the x-axis, the east-west axis as the y-axis, and the altitude axis as the z-axis.According to the Pythagorean theorem, the magnitude of the resultant velocity is given by

v = v2x + v2

y + v2z .

Plugging in the numbers gives v = 225 mph = 15 mph. Since the velocity is constant, youcan use the following equation to find the time: distance equals velocity times time. Solving fortime, you get time equals distance over velocity. Therefore, the time it takes to travel 3 milesequals 3/15 or 0.2 hr. This is equal to 12 minutes which is answer choice B.

55. AA reaction where an ion or atom in a compound is replaced by another atom or ion is defined

as a single displacement reaction. In this case, the ion Cu2+ is replaced by the atom Zn to formzinc sulfate and solid copper. More specifically, this reaction is an example of a metal displace-ment reaction, and it occurs because zinc is higher than copper in the metal activity series (there-fore, copper will not displace zinc from zinc sulfate). A metathesis, or double displacementreaction is one in which exchange occurs between two reactants to from two new compounds,so choice B is incorrect. Choice C is incorrect because a decomposition reaction is defined as areaction in which a single substance breaks down into two or more products. This does not occur

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in this reaction, so choice C can be eliminated. Finally, choice D can be eliminated because adisproportionation reaction is a type of redox reaction where the oxidation state of a substanceincreases and decreases. In this reaction, the oxidation state of copper decreases only, and theoxidation number of zinc increases only, so choice D can be eliminated.

Passage IX (Questions 56–60)

This passage addresses several important concepts in optics, including polarization, refrac-tion, dispersion, and reflection and transmission. The first paragraph describes Figure 1, whichdepicts polarization by reflection of an initially unpolarized light beam when it is incident on adielectric surface at Brewster’s angle. The first paragraph also mentions Snell’s law for therefraction of the beam upon entering a material of a different refractive index, but it doesn’t givethe formula. The second paragraph is concerned with the second figure, which consists of twopolarizers. You are told that a polarizer only transmits the electric field component of light thatis parallel to its transmission axis. Malus’ law for the transmittance through two polarizers isalso introduced.

56. BThe second paragraph of the passage states that a polarizer transmits 100% of the electric

field component parallel to its transmission axis and none of the component perpendicular tothat axis. Therefore, for minimum transmission, the polarizer must be oriented with its trans-mission axis perpendicular to the electric field of the light beam. According to the first para-graph, the electric field vector of the reflected light is parallel to the glass surface. Since theelectric field vector is parallel to the surface of the glass, transmittance through the polarizer willbe at a minimum when its transmission axis is perpendicular to the surface of the glass, makingchoice B the correct answer.

57. DThe polarization of a wave is the resolution of its vibration into a plane perpendicular to its

propagation direction. Longitudinal waves cannot be polarized because their vibration is paral-lel to their propagation direction. Therefore, choice D is correct. Only transverse waves can bepolarized because their vibration is perpendicular to the direction of propagation. Choices A andB are wrong because the polarizability of wave is independent of whether or not it is reflectedor attenuated when it passes from one medium to another. Choice C is wrong because any trans-verse wave can be polarized. Light, which has an electric field component, is just one type oftransverse wave. For example, transverse waves on a string that are traveling in the z-directioncan be polarized so that the displacement of the string is in a plane perpendicular to the motion.

58. CTo answer this question, use the fact (given in the second paragraph) that unpolarized light

incident on a polarizer can be represented as the superposition of any two waves with electricfield components that are equal in magnitude and perpendicular to each other. You can chooseone of the waves to have an electric field component parallel to the transmission axis of polar-izer P1. Therefore, the other wave must have an electric field component perpendicular to thetransmission axis of P1. Since these two waves’ electric field components are equal in magni-tude, the intensity of each wave must be equal to one-half the intensity of the unpolarized light.You are told in the passage that a polarizer transmits 100% of the electric field component par-allel to its transmission axis and none of the electric field component perpendicular to its axis.Therefore, only the wave with the electric field component parallel to the transmission axis of

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P1 will be transmitted through P1, and the intensity at point A will be one-half the intensity ofthe unpolarized light or I0/2, and choice C is correct.

59. ATo answer this question, use the fact given in the passage that the intensity of light trans-

mitted through P2 is equal to I1cos2φ First consider the case before the third polarizer isinserted. When the transmission axes of P1 and P2 are perpendicular to each other, the anglebetween the axes φ = 90°. Since the cosine of 90° is zero, the intensity at point B would be zero.Since the light intensity cannot be less than zero, choice B is incorrect.

If a third linear polarizer is placed between P1 and P2, with a transmission axis at an anglegreater than 0° but less than 90°, with respect to the transmission axis of P1, the light thatpasses through it will have an electric field component parallel to the transmission axis of P2.This is a result of the cos2φ dependence of the transmission through a linear polarizer. Sincethe third polarizer has a transmission axis at an angle of 45° with respect to that of P1, theintensity of the light that passes through the third polarizer is I1(cos 45°)2 which is non-zero.This light will be incident on P2. Since the third polarizer’s transmission axis is also at anangle of 45° with respect to that of P2. The light that passes through P2 will have an intensityequal to [I1 (cos 45°)2](cos 45°)2 which equals I1/4. Therefore, the intensity of the light at pointB increases when the third polarizer is inserted. You don’t have to do the math to answer thisquestion. Once you set up the equations, you will realize that the intensity at point B will benon-zero. Thus, the effect of the third polarizer would be to increase the amount of light pass-ing through P2, and choice A is the correct answer. Choices C and D are wrong because theintensity at point B would change. Choice C is incorrect for another reason: the polarizationat B is defined by P2, and therefore is not affected by placing another linear polarizer in frontof P2.

60. DYou are told in the passage that when light is incident at the Brewster angle, the angle

between the reflected ray and the refracted ray is 90°. Looking at Figure 1, you can see that thesum of the angle of reflection θr, the angle of refraction θt, and the angle between the reflectedand refracted rays is equal to 180°. Since the angle between the refracted and the reflected raysis 90°, the sum of the angles of reflection and refraction must equal 90°. Therefore, θt = 90° – θr.Since θr equals the angle of incidence θi, θt is also equal to 90° – θi. Therefore, θi = 90° – θt. Sincethe light is incident at the Brewster angle, the Brewster angle equals 90° – θt, and choice D isthe correct answer.

Passage X (Questions 61-66)

This passage discusses the effect of gravity on planetary gases. The first paragraph presentstwo equations which express the surface gravity g and the magnitude of the escape velocity vescin terms of the mass and radius of a planet. Understand the general structure of the relation-ships—surface gravity increases as the mass increases, but decreases as the radius increases.Escape velocity also increases as mass increases, and it decreases as radius increases. Under-stand the meaning of escape velocity—the minimum velocity needed to escape. Any velocity ofequal or greater magnitude is sufficient.

The second paragraph describes a physical process—atmospheric leaking. Any particles ofa gas with speeds greater than the magnitude of the escape velocity may escape the confininggravity of the planet. If the fraction of such particles of a gas is large, the rate of escape can be

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also. If the fraction is small, the rate of escape will be small. The process of escape is an imbal-ance between thermal kinetic energy and gravity.

The third paragraph describes a probability distribution called the Maxwell distribution. Aprobability distribution is a formula giving the relative probability of various values of a param-eter. In this case the parameter is speed v, so the probability represents the probability that a par-ticle has a certain speed at a given temperature. The paragraph also gives the formula for theroot-mean-square speed per particle and the average translational kinetic energy. You shouldnote that for a given temperature, the root-mean-square speed increases as the mass decreases.

61. CThis question does not refer to temperature or gases, so you may correctly guess that only

information in the first paragraph is required to solve it. To answer this question, you have todetermine the ratio of the magnitude of the escape velocity on Jupiter to that on Earth. In thequestion stem you are given the ratio of the two planets’ masses and radii, therefore, you candirectly calculate the ratio of the magnitude of the escape velocity on Jupiter to that on Earth.The ratio equals

= MJRE/MERJ

Since the mass of Jupiter is 360 time Earth’s mass and its radius is 11 times Earth’s radius,the ratio of the magnitude of the escape velocity on Jupiter to that on Earth equals 363/11. Byestimating, you can determine that this is approximately the square root of 30 which is closestto choice C, the correct answer.

62. BThe minimum velocity a particle needs to escape from the surface of the Earth is the escape

velocity. Therefore, the minimum kinetic energy it must have to escape is equal to 1/2 mvesc2.

From the passage, you can see that the magnitude of the minimum velocity required to escapefrom the surface of the Earth is

vesc = 2GME/RE.

Therefore, the minimum kinetic energy required to escape the Earth’s surface equalsGmME/RE, and choice B is correct. Choice A can’t be right because it has units of force, notenergy. Choices C and D are wrong because they both represent the average translational kineticenergy per particle of ideal gas, which is not a measure of how much kinetic energy a particleneeds to escape the Earth’s surface.

63. BThe best way to answer this question is to evaluate each of the answer choices in the order that

they appear to determine which one is true. Choice A states that the maximum speed of oxygenmolecules is greater than the maximum speed of the nitrogen molecules. You should know that allof the molecules in a gas do not travel at the same speed. Each molecule’s motion is random. Some-times a molecule moves at a speed faster than the root-mean-square speed, while at times it movesat a slower speed. This implies that the kinetic energy of the individual molecules also variesbecause kinetic energy = 1/2 mv2. The probability of finding a molecule of speed v is plotted in Fig-ure 1. From this graph, it is clear that very high speed molecules are very rare. However, there is noupper limit to the speed indicated in the graph. If you were not sure about this from looking at thegraph, the statement in the third paragraph that the probability goes to zero as the speed goes to

2GMJ/RJ2GME/RE

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infinity is another indication that there is no upper limit to the speed of a gas molecule. Therefore,we have no way of comparing the maximum speed of oxygen to the maximum speed of nitrogenmolecules, and choice A is incorrect.

Choice B states that the average translational kinetic energy of the oxygen molecules is equalto that of the nitrogen molecules. According to the passage, the average translational kineticenergy per particle of gas at temperature T is equal to 3/2 kBT. Since the oxygen gas and thenitrogen gas are at the same temperature, the average translational kinetic energy of the oxygenmolecules will be the same as the average translational kinetic energy of the nitrogen molecules,and choice B is most likely the correct answer. To check this, let’s look at the remaining choices.If each oxygen molecule had 4/3 more translational kinetic energy than each nitrogen molecule,as choice C suggests, the average translational kinetic energy per oxygen molecule would begreater than the average translational kinetic energy per nitrogen molecule, and this is not thecase. So choice C is incorrect. Choice D is incorrect because the question states that themolecules are all at the same temperature. This is an equilibrium condition, and according to thesecond law of thermodynamics the temperature can not spontaneously rise.

64. ASince you are asked to consider shifts in the composition of the Earth’s atmosphere and you

are told to neglect the effects of changes in the climate and ecology, it is reasonable to assumethat these shifts are due to the fact that some gases leak away from the Earth more rapidly thanother gases. The second paragraph states that at a given temperature particles with greater speedsleak away more rapidly. In the third paragraph of the passage, you are told that the root-mean-square speed of a molecule is given by 3 kBT/m. Therefore, if two gases are at the same tem-perature, the molecules of gas with smaller mass will have a greater root-mean-square speed andwill leak away more rapidly. So the ratio of a lighter gas to a heavier gas will decrease over timebecause the lighter gas will leak away more rapidly. Looking over the answer choices, itbecomes clear that choice A is correct because it states that the ratio of a lighter gas, hydrogen,to a heavier gas, oxygen, will decrease over time. Choice B is wrong because it states that theratio of a heavier gas, water vapor, to a lighter gas, ammonia, will decrease over time. Choice Cis wrong because it states that the ratio of a lighter gas, nitrogen, to a heavier gas, oxygen, willincrease over time. Finally, choice D is wrong because it states that the ratio of a lighter gas,methane, to a heavier gas, carbon dioxide will increase over time.

65. BThis question requires that you use the fact that at a given temperature particles with greater

speeds leak away from planets more rapidly. Without looking at the answer choices, you should tryto figure out why the Moon might have no atmosphere based on the information in the passage. Itis likely that the gases on the Moon’s surface had great enough speeds to escape from the Moon’sgravitational field long ago. Look at the answer choices. Choice A is unlikely considering the largeamount of gaseous elements on the Earth and elsewhere in the Solar System. Choice B states thatthe temperature on the Moon is too high for its size and mass. First, you must consider temperature.A relation given in the passage states that the root-mean-square speed of a gas is proportional to itstemperature. Therefore, gases near the surface of the Moon will have root-mean-square speeds thatare proportional to the temperature of the Moon. Now consider the escape velocity. From the infor-mation in the first paragraph, you can deduce that the escape velocity associated with a planetdecreases as the planet’s mass decreases. Therefore, relatively low-mass planets will have relativelylow escape velocities. From this you can infer that the statement in choice B means that the tem-perature of the Moon is too high for it to retain an atmosphere because all of the gases around it

26

would have high enough temperatures to escape its gravitational field. So choice B presents a rea-sonable explanation for why the Moon has no atmosphere.

At this point it is a good idea to look over the remaining answer choices to make sure thereis not a better explanation. Choice C is incorrect because it directly contradicts choice B. If thetemperature of the Moon were cooler, the Moon would be more likely to maintain an atmo-sphere. Choice D suggests that the atmosphere is restrained by the magnetic field rather than thegravitational field. This statement is not supported by the information in the passage.

66. CTo answer this question consider how the Earth’s rotation affects the force on a particle on

the Earth’s surface. Since the Earth is rotating, a particle on its surface exhibits uniform circu-lar motion at a radius of r, which is equal to the distance of the particle from the axis of rotationor the polar axis. The centripetal force is therefore F = mv2/r , where m is the mass of the parti-cle and v is the speed of the particle. The speed of the particle is equal to the angular frequencywith which the Earth rotates ω times by the distance between the particle and the axis of rota-tion r. Substituting v = rω, gives F = mω2r. Since m and ω are constants of the motion, the cen-tripetal force increases as r increases. Since r is the distance between a particle on the surfaceof the Earth and the polar axis, r is a maximum at the equator, and therefore the centripetal forceis a maximum at the equator.

Remember, the centripetal force is the force required to keep a particle traveling in a circu-lar orbit of radius r. In this case, the force of gravity is the centripetal force. Since the force ofgravity is approximately the same for all particles on the surface of the Earth and since the par-ticles at the poles require a smaller attractive force to keep them at the surface of the Earth thanthe particles at the equator, the velocity required to escape from the surface of the Earth will begreater at the poles than at the equator, and choice C is correct.

Passage XI (Questions 67–72)

The topic of Passage XI is hard water. The passage discusses the problems associated withthe use of hard water and then goes on to describe how the concentration of carbon dioxide inwater can affect the precipitation of salts from hard water. In this case, the topic of the passageis probably familiar to you. You may see these on the MCAT, so don’t be surprised.

Again, notice how the information is presented to you—the second paragraph gives youmuch information as do the reactions and the table. Try to “weed out” the relevant information—there is no point trying to remember everything in the passage. Absorbing the important infor-mation may also help you decipher what sort of concepts the questions will contain. Forexample, Table 1 gives you a number of Ka, Kb, and Ksp values. Straight away, you should startto think about acids and bases, and how your understanding of these concepts will help you toanswer probable questions on this topic.

67. BThe first point to establish is that hydrogen carbonate is acting as an acid and a base: in Reac-

tion 3, hydrogen carbonate acts as a Brønsted acid since it loses a proton; in Reaction 2 hydro-gen carbonate can act as Brønsted base in the reverse equilibrium reaction by accepting a proton.In addition, hydrogen carbonate possesses an acid and a base dissociation constant, which fur-ther confirms that it is amphiprotic. So, we have established that hydrogen carbonate is an acidand a base, but now we have to decide what type of acid and base. The answer to this lies in

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Table 1. From Table 1 we can see that the Ka and Kb for hydrogen carbonate is quite small. SinceK is an equilibrium constant, a low Ka means that HCO3

– does not readily dissociate into H+ andCO3

2–. If there are few protons around, the solution will not be very acidic—HCO3– is, there-

fore, a weak acid and choices A and C can be eliminated. A low Kb means that in the equilib-rium, HCO3

– + H+ H2CO3, the equilibrium favors the reactants because HCO3– does not

like to accept a proton to form H2CO3. In other words, hydrogen carbonate is a weak base.Choice D can be eliminated and choice B is the correct response.

68. CTo answer this, you need to know how to calculate the hydrogen ion concentration of a solu-

tion and how this is related to the pH. In the question stem, you are told to assume that carbonicacid dissociates into H+ and HCO3

– only; therefore, you do not have to worry about the seconddissociation of HCO3

– into H+ and CO32– when calculating the pH of the solution.

Remember that the Ka is simply the equilibrium constant when an acid, HA, dissociates intoits constituent ions, H+ and A–. Since the equilibrium constant is equal to the concentration ofproducts divided by the concentration of reactants, the Ka is equal to the product of the concen-trations of H+ and A–, divided by the concentration of the undissociated acid, HA. Therefore, inthe case of carbonic acid, the Ka is equal to the product of the concentrations of H+ and HCO3

–

divided by the concentration of H2CO3. Now, to work out the hydrogen ion concentration, wehave to re-arrange this equation. Let us call the product of the hydrogen ion and hydrogen car-bonate concentration x2. The equation can now be re-arranged to read: x2 = Ka[H2CO3]. Sincethe acid dissociation constant of carbonic acid is 4.3 × 10–7 and the concentration of carbonicacid is 0.023, x2 works out to be roughly 1 × 10–8. The hydrogen ion concentration, therefore,is the square root of 10–8 10–4 M. You should know how to work out the pH from the hydro-gen ion concentration. The equation is as follows: pH = –log [H+]. The minus log of 10–4 worksout to be 4, so choice C is the correct answer.

If you forgot to take the square root of 10–8, you may have ended up with choice D. A pHof 8 indicates a basic solution, and carbonic acid is a weak acid, so choice D can be eliminated.Again, be careful in the steps of your calculation—a simple mistake can lead to the wronganswer. Choices A and B are wrong because these pHs are characteristic of a strong acid, not aweak acid.

69. AThe solubility product constant, Ksp, is defined as the product of the concentrations of ions

in solution, divided by the concentration of the parent compound. In the case of calcium car-bonate and magnesium carbonate, these compounds are insoluble in solution. Therefore, theiractivities are equal to 1 and the denominator in the Ksp expression can be discarded. So, the Kspof calcium carbonate is equal to the product of the concentrations of Ca2+ and CO3

2–, whereasthe Ksp of magnesium carbonate is equal to the product of the concentrations of Mg2+ andCO3

2–. Therefore, if a compound possesses a small Ksp, it will not readily ionize in solution; inother words, the compound has a low molar solubility. The Ksp of CaCO3 is around 350 timessmaller than the Ksp for MgCO3, so calcium carbonate is much less soluble than magnesium car-bonate, and choice A is the correct answer.

Choice B is incorrect because the Ksp of MgCO3 is much larger than the Ksp of CaCO3, soMgCO3 will be more soluble in solution, and will make up less of the precipitate. There would

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only be an equal distribution of CaCO3 and MgCO3 if the Ksp’s of each compound were thesame. Since they are different, choice C is incorrect. Finally, choice D is incorrect because thesecond sentence in the passage tells you that Mg2+ and Ca2+ form slightly soluble carbonatesalts.

70. CA buffer is a solution that resists a drastic change in pH when an acid or base is added to it.

An acid buffer is a mixture of a weak acid and its conjugate base. Let’s consider a system inwhich a conjugate base, A–, and a weak acid, HA, are present. The equilibrium for this systemis as follows: HA + H2O H3O+ + A–. You should be able to see that when an acid is addedto this system, the hydrogen ions are consumed by the conjugate base, A–, to form the undisso-ciated acid, HA. Therefore, there is not a drastic decrease in the pH. Conversely, if a base, OH–,is added to the system, protons are removed from the weak acid to form H2O, so the concentra-tion of base remains nearly unchanged and the pH doesn’t rise too drastically. With this in mind,let’s use a real example—carbonic acid. Carbonic acid is a weak acid, so in solution the follow-ing equilibrium exists: H2CO3 + H2O H3O+ + HCO3

–. The conjugate base in this systemis HCO3

–, so the addition of NaHCO3, which dissociates into Na+ and HCO3–, will result in the

formation of an acidic buffer, and choice C is the correct response.

All of the other answer choices are wrong because they will not form conjugate bases of car-bonic acid. Sulfuric acid—choice A—dissociates into hydrogen ions and sulfate ions, and will,when added to carbonic acid, cause a large drop in the pH. Since H2CO3 and H2SO4 are bothacids, the solution will not be buffered against large pH changes, and choice A can be elimi-nated. Sodium sulfate—choice B—will dissociate into Na+ and SO4

2– ions. SO42– is the conju-

gate base of HSO4–, so sodium sulfate is a good buffering agent for bisulfate not carbonic acid,

making choice B incorrect. Finally, choice D is incorrect because NaCl dissociates into Na+ andCl– ions, and Cl– is not the conjugate base of H2CO3.

71. AThis question tests your understanding of Le Châtelier’s principle, which states that when a

system in equilibrium is placed under stress, the system will move in such a way as to alleviatethat stress. An increase in the pH means that there is a depletion of protons. Looking at the reac-tions, you can see that protons are involved in reactions 2 and 3. Removing protons in these reac-tions is, in essence, the same as removing one of the products. In compliance with Le Châtelier’sprinciple, if a product is removed, the system will move to counteract this loss by forming moreproduct. In other words, the equilibria in reactions 2 and 3 will shift to the right. When both equi-libria shift to the right, the concentration of CO3

2– will increase. Since there are more carbon-ate ions present, there is increased precipitation of CaCO3. Choice A is, therefore, the correctresponse.

Choice B states that an increase in pH results in an increased proton concentration. Anincrease in pH results in a decreased proton concentration since the solution is more basic, sochoice B is wrong. Choice C states that the proton concentration does not affect the equilibriaof Reactions 2 and 3. This would only be true if the protons were not part of the reaction. Sincethey are part of the reaction, however, their absence or presence can influence which directionand to what extent the equilibria move. Choice D is contradictory to Le Châtelier’s principle.The depletion of protons results in the movement of the equilibria to the right. If the pHdecreased, resulting in an increase in the number of protons, then the system would move infavor of H2CO3 production to “get rid” of the excess protons.

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72. DFirst, let’s look at the effect of pH. An increase in the pH results in the depletion of protons;

as a result, the equilibria in reactions 2 and 3 shift to the right. Since the equilibria in Reaction 2shifts to the right, the carbon dioxide dissolved in the solution will be used up. To replenish thesupply of dissolved carbon dioxide, more gaseous carbon dioxide dissolves; Reaction 1 is alsoforced to the right. Therefore, increasing the pH results in an increased amount of gaseous car-bon dioxide that dissolves; statement I is correct. Thus, choice B can be eliminated immediately.

The temperature of the water—statement II—also affects the solubility of gaseous carbondioxide. Reaction 1 is an exothermic reaction: heat is given out to the surroundings. If the tem-perature of the surroundings is increased, the system will move to counteract that increase byabsorbing heat. In other words, increasing the temperature favors an endothermic reaction, andthe equilibrium in Reaction 1 shifts to the left. On the other hand, if the temperature of the sur-roundings is decreased, the system will move to increase the temperature of the surroundings bygiving off heat. Therefore, decreasing the temperature favors an endothermic reaction, and theequilibrium in Reaction 1 shifts to the right. So, the temperature will affect the amount ofgaseous carbon dioxide that dissolves, and statement II is correct. Notice that the only feasibleanswer is choice D, even without considering statement III. Let’s confirm that statement III iscorrect, anyway. If the partial pressure of carbon dioxide is high, there will be many carbon diox-ide molecules in contact with the surface of the reservoir. As a result, these molecules are morelikely to dissolve in solution. If the partial pressure of carbon dioxide is low, fewer moleculeswill be in contact with the reservoir and fewer will dissolve in solution.

Independent Questions

73. DIn order to find the acceleration of an object, you need to first determine the net force on the

object since force equals mass time acceleration. In this case, there are two forces acting on theobject, the force due to gravity and the force of friction. Since the object is on an incline, youhave to find the component of the force of gravity along the incline. It is given by mg sin 30°,where m is the mass of the object, g is the acceleration due to gravity, and 30° is the angle of theincline. This force is directed down the incline. The force of friction is always in the oppositedirection to that of the motion. So in this case it is directed up the incline. Therefore, the netforce on the object along the plane of the incline is (mg sin 30° – 2) N. By Newton’s second lawthis is equal to ma, where a is the net acceleration of the object. Therefore, a = (g sin 30° – 2/m)m/s2 = (4.9 – 1) or 3.9 m/s2 which is choice D.

74. DSince chromium has an atomic number of 24, the outer valence electrons are associated with

the n = 3 and n = 4 levels. Electrons always fill lower energy orbitals first, and the energy of theorbitals with which we are concerned increases in the order, 4s, then 3d, then 4p. Since the 3dorbital is lower in energy than the 4p orbital, electrons will enter the 3d orbital first, so straightaway, choice A can be eliminated. Since the 4s orbital is lower in energy than the 3d orbital, wewould expect the 6 outer valence electrons to fill the orbitals in the order 4s2, 3d4; however, it isenergetically more favorable for 5 of the 6 valence electrons to enter the 3d orbital, each elec-tron aligning singly in each of the 5 degenerate orbitals (thus avoiding electron repulsion whichoccurs when the electrons align in pairs in each degenerate orbital). Hence, one valence electronenters the 4s orbital, and 5 electrons enter the 3d orbital; the most stable outer electron config-uration of chromium is 4s1 3d5, and choice D is the correct response.

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Although choice B is a possible outer electron configuration for chromium, it is not the moststable since the electrons are paired in the 4s shell, resulting in electron repulsion and a higherenergy configuration than that shown in choice D. Choice C is incorrect since this also resultsin the pairing of electrons in one of the five 3d orbitals which decreases the stability ofchromium.

75. CIn aqueous solution, ammonia undergoes the following reaction:

NH3(aq) + H2O(l) → NH4+(aq) + OH–(aq)

If 1% of 0.1 M hydrolyzes in solution, it follows that the concentration of hydroxide ionswill be 1 percent of 0.1 M, or 10–3 M. Just as the pH of a solution is equal to –log[H+], so thepOH of a solution is equal to –log[OH–]. In addition, the sum of pH and pOH must always beequal to 14. Since the concentration of OH– ions is 10–3 M, the pOH is equal to –log[10–3], or3. Therefore, the pH of the solution is equal to 14 – 3, or 11—choice C. Choice A is incorrectsince this is the pOH of the solution, not the pH. Choice B is incorrect because this pH is neu-tral, and solutions of ammonia are very basic. Choice D corresponds to a pH of 13 and a pOHof 1. A pOH of 1 corresponds to a hydroxide ion concentration of 0.1 M, which would result if100% of ammonia hydrolyzed, not 1%.

76. BIonic crystals consist of tightly-packed anions and cations which are not free to move

around. Magnesium fluoride, sodium chloride, and potassium nitrate all form these crystals.Sodium chloride and potassium nitrate form an ionic crystal known as the rock-salt structure,which consists of an overlapping arrangement of face-centered cubic anions and cations. Mag-nesium fluoride forms an ionic crystal known as the fluorite structure. On the other hand, car-bon dioxide forms a molecular solid: a crystal lattice in which the CO2 molecules are heldtogether by weak intermolecular forces, namely dispersion forces. Since they are held togetherby weak intermolecular forces, molecular solids tend to have low melting points and many,including carbon dioxide, undergo sublimation at low temperatures.

77. BThe half-life of a radioactive element is the time that it takes one-half of the initial amount

of the element to decay. To determine how much of the initial amount remains after 16 days, firstdetermine how many half-lives are in 16 days. Since the half-life is 4 days and 16/4 = 4, thereare 4 half-lives in 16 days. After one half-life has passed 1/2 of the initial sample remains. Afteranother half-life passes 1/2 times 1/2 of the initial sample remains. Following this trend, afterfour half-lives have passed, (1/2)4 or 1/16 of the sample remains. This is equal to 0.0625 or6.25%, which is answer choice B.

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78. C

79. B

80. C

81. C

82. C

83. D

84. B

85. C

86. B

87. D

88. A

89. A

90. D

91. C

92. C

93. D

94. A

95. B

96. D

97. B

98. C

99. C

100. C

101. B

102. A

103. D

104. C

105. A

106. B

107. C

108. B

109. C

110. A

111. C

112. C

113. D

114. C

115. C

116. C

117. C

118. C

119. A

120. A

121. C

122. A

123. D

124. C

125. B

126. C

127. D

128. B

129. D

130. C

131. A

132. C

133. C

134. C

135. A

136. D

137. A

VERBAL REASONING ANSWER KEY

Passage I (Questions 78-83)

TOPIC AND SCOPE: The origins of modern society’s troubles; the author writes a complexrebuttal to the assertion that the liberalism of the 1960s is the source of present-day societalwoes. The author contends that the decay of western society has been occurring for hundreds ofyears, and that the 1960s “unmasked” rather than created our moral collapse. Even more impor-tant, the author argues that the 1960s were only the manifestation of the centuries-old rejectionof traditional beliefs. Such philosophies gained an audience in the 1960s due to timing and cir-cumstances.

In the first paragraph, the author introduces the reader to “neoconservative” opinions that statethe 1960s are the root of moral decline.

The second paragraph introduces the central argument—“the foundations of civilized lifewere crumbling long before Chicago.” Here, the author introduces the theory that many mod-ern-day criminals and derelicts were raised in the 1950s, “supposedly an era of wholesomenuclear families, raised by the right values.”

The third paragraph begins to present the crux of the author’s argument: Societal collapsebegan centuries ago, not decades ago. The author cites several examples.

The author bolsters his argument in the fourth paragraph by asserting that, based on flawedhuman nature, the populace was eager to embrace nontraditional ideas—some even advocatingimmoral or amoral behavior. Further, the author places blame on “authority.”

In paragraphs five and six, the author continues to unravel the “family values” of the 1950s,which set the stage for aberrant behavior to take root and blossom in the 1960s.

The last paragraph restates that the 1960s did not “cause our moral collapse [but] onlyunmasked it.”

78. CThe question asks at what time the author might state that nontraditional, free-thinking

began. The correct answer can be inferred from the multiple examples of nontraditional and“risqué” happenings listed in the third paragraph. Here, the author does cite the Enlightenmentand determinism as rationalization of misconduct and spontaneity. Thus, free-thinking, nontra-ditional behavior dated back centuries ago to “movements such as the Enlightenment and deter-minism,” as stated in choice C.

Choice A is incorrect. Although the author mentions nontraditional behavior in the 1950s inparagraphs five and six, these references come after the discussion of late-nineteenth centurysociety in paragraph three. The passage mentions neither choice B nor choice D. The authornever mentions the Great Depression or the post-Civil War era in the passage. Also, the authornever implies that misconduct and free-thinking evolve from economic hardships, as indicatedby these two choices. Rather, the author’s examples of the infancy of such thoughts occur dur-ing times of economic stability to prosperity—Victorian England and the American 1950s. So,choices B and D are incorrect, and choice C is the credited response.

79. BThis question asks where this passage might be found. It is best to consider what type of pas-

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sage this really is: it is opinionated yet well written, probably by an author who is an expert insocial or historical science. Thus, choice B, “the editorial page of a scholarly journal,” is the bestanswer choice and is correct.

Choice A is very misleading, but not quite correct. If this passage was part of a graduate soci-ology thesis, it would probably have a different form. Most significantly, this is a very opinion-ated passage; a graduate thesis would express findings, conclusions, and perhaps some opinionsbut not at this level of vehemence. Further, this passage lacks any references to other works orpapers—an essential element to any research thesis. Thus, choice A is incorrect. Choice C issimilarly wrong. This passage is far too opinionated for a history text. Also, a sidebar from a his-tory text would be “historical” in nature—citing events for factual purposes, not to confirm anopinion. Finally, choice D is misguided. If this were a speech, its tone would be different. Itwould address the audience in some manner, which this passage does not. Also, a speech wouldnot use parentheses (as seen in paragraph three) since the author speaks every word. Thus,choice D is incorrect, and choice B is right.

80. CIn paragraph four, the author states, “Ultimately, the blame for our lack of ‘guardrails’ lies with

flawed human nature.” Therefore, the author would consider founding the government in a deeplyfelt religious faith as too simplistic. He would much rather analyze and maybe even reassess whatconstitutes appropriate behavior.

Choice A is incorrect because one would not typically consider democracy as a form of gov-ernment that is “founded in a deeply felt religious faith,” but rather a society whose memberspractice the religion of their choice. Choice B is incorrect because the statement in the questionstem actually agrees with, not alienates, neoconservative beliefs. Choice D is incorrect becausethe author uses the term “Progressive educators” as an example of leftist intellectuals, whowould characteristically reject government founded in a deep religious faith.

81. C“Violence” (choice C) is often a result of societal collapse, not a cause of it. Choices A, B,

and D all appear in the passage. In paragraph three, the author mentions that, “Darwinism [was]undermining religion” and “Decadence [was] inspired by ... Marquis de Sade.” In paragraphfive, the author states that, “Authority flopped at defending the civilization ... as [did] the fam-ily.” The author never refers to “violence” as a contributor to societal “moral collapse.”

82. CChoice A would not weaken the author’s argument; it would support the point he made

regarding the increase in the illegitimacy rate between 1940 and 1960. Even if choice A did con-tradict one of the author’s points, it would not be encompassing enough of a statement to weakenthe author’s main argument. Choice B is incorrect because the theme of the passage is not finan-cial wealth; it is traditional morality versus the reality the author presents. The second bestchoice is D. Choice D would lead the reader to believe that people in higher income levels andpeople who hold a strong religious faith are less likely to commit crime, and therefore maintaina higher level of morality. This weakens the author’s point that citizens raised in “wholesomenuclear families informed by the right values” could be “murderers, rapists, muggers and riot-ers.” However, choice C is best because it attacks the central argument and focus of the essay bystating that traditional morality can exist in a society despite the influence of “pornography, cap-italistic mass consumption and free thought.” The author appears to be resigned to the idea thattraditional morality is ineffective.

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83. DThe author’s tone is ironic throughout much of the essay. Hence, the reader should realize

that when the author writes “all was right with the world,” he is mocking a viewpoint. This view-point happens to be held by conservatives who feel that traditional values, such as those at theturn of the century, kept the world in an ideal equilibrium. The author implies choice D, that,“conservatives were naive in believing a utopian society [ever] existed.” Choice C is incorrectbecause the author argues throughout the essay that the structure of western society did and doesnot prevent people from going against traditional morality. Choice B indicates that the authorfeels that “global society was at complete peace”; however, the author traces the roots of vio-lence and society’s problems to long before the end of the nineteenth century. Therefore, it isillogical to choose an answer that indicates that the author believes that the world was ever “atcomplete peace.” Regardless of the author’s implications, choice A is incorrect. The sentencethat contains the quote from the question stem also states, “socialism, anarchism and pragma-tism [dissenting philosophies] were gaining popularity.”

Passage II (Questions 84–90)

The passage tracks the development and much-needed recognition of African-Americanentertainers from the late nineteenth century into the 1920s. As the passage refers to these indi-viduals as “black entertainers,” that term shall be used in these explanations. The first paragraphsets the stage for the passage by noting that the tradition of the black entertainer truly began afterthe Civil War, and by the 1890s, black entertainers were part of the entertainment industry.

The author highlights the expansion of black entertainers throughout the early part of the20th century in the second and third paragraphs. The author notes, however, that black enter-tainers still were forced to overcome hardships and second-class treatment during this time. Inthe fourth paragraph, however, the author mentions that a musical in 1921 called Shuffle Alongcreated a significant interest in black entertainers. Shuffle Along was a huge success andincreased the recognition of black performers in the entertainment industry. Despite the ShuffleAlong’s successes for all black entertainers, the author points out how far black entertainers hadyet to go through a quotation from a black journalist, J.A. Jackson, at the end of this paragraph.Jackson notes that “things were picking up...for the colored group in this industry...Directors aredesirous of naturalness and have...eliminated the made-up, white actor in Negro characteriza-tions.” Although this quote highlights the monumental achievements of Shuffle Along, as well aspioneering black entertainers such as, Josephine Baker and Florence Mills, it indicates howblack performers were still not considered completely equal to their white counterparts.

In the final paragraph, however, the author ends the passage on an upbeat note. The authorends by citing that the achievement from Shuffle Along lead to successes by scores of other blackentertainers. Specifically, the author notes that after Shuffle Along, many acts by black enter-tainers were booked into “white locations” and top theaters, several of which were located onBroadway.

84. BChoices A, C, and D are all stated within the passage. The author states, in the fourth para-

graph, that Shuffle Along became a smash hit and fueled a renewed interest in black theater—choice A. The author states, in the third paragraph, that black acts sometimes accepted bookingsin lesser theaters, as choice C indicates. The entire first paragraph describes the rise of blackentertainers and their gradual acceptance during the nineteenth century, per choice D. Accord-ing to the passage (2nd paragraph), the men who tutored Duke Ellington wrote and directed

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black shows. These men may have been musicians in their own right, yet this is not indicated bythe passage. Although Duke Ellington was most likely taught by some of the most talented musi-cians of the early 1920s, the passage does not contain this information. Therefore, according tothe passage B is untrue, and the correct response.

85. CTo choose the correct answer, one must understand that the author sees Shuffle Along as an

impetus for the black entertainment renaissance. C is the correct answer. The author states thatShuffle Along “started a renewed interest in black entertainment” and quotes a black entertain-ment columnist from Billboard as saying the “show began the renaissance of the Negro in musi-cal comedy.” Choice A is incorrect because the second paragraph states that by the end of thecentury (nineteenth), black artists played to mixed audiences. Therefore, this is weak support forthe author’s assessment of the play because black artists had already been performing for mixedaudiences since the 1890’s. B is incorrect because the author does not focus on the popularity ofthe music in his assessment of Shuffle Along’s impact. Choice D, in saying that white and blackaudiences enjoyed the show, does not offer strong support to the idea of the black entertainmentrenaissance.

86. BEach of the decades listed as choices is mentioned in the passage. However, A, the 1880s, is

noted only as the decade in which minstrel shows gave way to variety shows. C, the 1890s, isused as an endpoint to illustrate the gradual integration of black performers over a period ofmore than fifty years. D, the 1830s, is incorrect, mentioned only as a date when the first blackminstrels formed their owns shows. B, the 1920s, is correct because it was in that decade, theauthor asserts, that white performers were no longer used to portray blacks, and black acts wereallowed to play in white clubs.

87. DThe question requires the reader to disregard the author’s opinion and interpret that of

another author he has quoted. Jackson states (lines 36–43) that the increase in filmmaking inNew York had increased black artists’ employment, as in D. Choice A is incorrect, in that it isthe author’s assessment of a decade not in question. While B might be true, and mentioned inthe same lines as the proper choice D, it is treated as an aside, of secondary importance to thesurge of filmmaking in New York. C is also as an aside.

88. ABaker’s name is included as a detail, not a point of argument, but her mention is important

to his assertion that Shuffle Along spawned many new careers. For B to be true, the author wouldhave had to make the assertion that talents like Baker’s were critical to the show’s success, some-thing he does not do. C is not true, as Baker was not the only entertainer mentioned who gainedfame from Shuffle Along. Also, presumably all the artists in the show were black—the authordrew attention to many others as well. Baker’s future successes are not directly relevant here,and the author does not link her future successes to Shuffle Along. Nowhere does he mentionBaker’s popularity with white audiences, so D is also incorrect. Choice A is the proper responsebecause Baker is presented as one of many black performers that contributed to the show’s suc-cess and gained notoriety from it.

89. CChoice C is the only incorrect statement, and thus the correct response. The entire passage

is concerned with A, as evidenced strongly by the passage’s last paragraph. B is discussed

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specifically in lines 5–9, as is D in line 15. C is the correct answer because the author mentionsthat full-dress musicals were not being created by black writers and composers until the lastyears of the century, not the decade after the Civil War (1865–1875).

90. DThe key word here is “discuss.” D is the correct response, the only piece of information not

discussed in the passage. A is examined in the second paragraph, B in the first, and C in the finalparagraph. The role of the black entertainer in the 18th century, choice D, is only noted in pass-ing, with no attempt at description or argument.

Passage III (Questions 91–96)

TOPIC AND SCOPE: This is a passage about the end of the First World War and the conse-quences of the peace settlement.

The author begins her discussion in the first paragraph by noting that major wars are often“punctuation marks” of history, as they result in major realignment of “relationships amongstates,” and that World War I was no exception. Here the author comments that the First WorldWar ended Europe’s world dominance, as European countries were exhausted and could notlonger sustain such dominance. In fact, the author notes that Japan and The United States werethe only victors of this conflict who were not so affected.

In the second paragraph, the author notes that the end of World War I came sooner thanexpected by the major Allied powers (USA, Great Britain, and France). At the War’s end, theAllies had placed most of their recent energies into military accomplishments and had madeonly cursory preparations for peace.

91. CThe author states clearly in lines 22–26 that the victors were unprepared for Germany’s

quick defeat, thus negating answer A. B is incorrect; the author says in line 14 that Europe neverreally recovered from defeat, but the author is referring to World War I, not the 1815 Final ActOf Vienna. The need for a new international organization as in D is noted in line 7, but the authorassures us in lines 34–36 that the Allied powers were much more concerned with winning thewar than outlining the peace. C is nearly identical to the author’s own statement in lines 9–11,and thus the correct response.

92. CChoice C is the proper definition, both by the dictionary and by the author’s own contextual

references. Thus choice C is correct. Since the “European victors ... suffered ... and none everreally recovered”, this indicates that they paid a price for their victory or, in other words, expe-rienced a “costly triumph.” A does not apply because the sentence refers directly to suffering anda difficult recovery on the victor’s part which does not relate to “total defeat.” B is wrongbecause the author suggests that the European nations would never recover from such a victory,a concept at odds with any notion of complete triumph. D, a disastrous truce, is incorrect—dis-astrous is the proper word for the author’s tone, but the end of the war certainly was not a truce,by her or any other historical definition.

93. DThe author’s central argument is that World War I marked a drastic change in global politics;

to challenge it, there must come a direct refutation. Thus D is the correct answer. Even if A were

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true, it would only discredit the author’s assertion that the European nations never would recoverfrom some unnamed after-effect of World War I, and not that they ceased to play a role as pri-mary arbiters of world issues. The same holds true for B; if it were true, and the economies ofthe world had rapidly stabilized after the war, there would still remain the issue of Europe’sdominance. C is not germane to the central argument. Only D would pose a direct challenge tothe author’s assertion that Europe lost its top-drawer international stature after World War I.

94. AThe author’s thesis is that Britain, France and the United States were charged, by default,

with the peace process that ended the war. A, therefore, is the correct answer. Nowhere does theauthor mention the possibility of B, the allies “could have ended the war much sooner.” Simi-larly, the author asserts the Allied powers suspected the peace process would be a lengthy ordeal,not a short one. Thus, choice C is incorrect. As for D, the author notes that the small states madetheir claims but that the larger Allied powers gave only cautious approval, not close attention, totheir claims.

95. BChoice A is discounted where the author judges that some—not most—of the early planning

for peace took place in less effectual quarters. C is untrue: the author states that the Alliesexpected Germany to last until mid-1919, not into the 1920s. As for D, the author leads his sec-ond paragraph with the suggestion that most wars end quickly and unexpectedly, exactly theopposite logic of the choice. Thus B is the only remaining choice, and can be reaffirmed by re-reading lines 23–26.

96. DChoice A is stated in the second paragraph, as is B. C is answered, somewhat more vaguely,

in the end of the first paragraph. Only D is neatly refuted, in line 6, and is thus the right answer.

Passage IV (Questions 97–103)

TOPIC AND SCOPE: In this passage, the author criticizes a work called Prisoners Of Men’sDreams by Suzanne Gordon. In this way, the author also indirectly criticizes one approach ofmodern feminist thought.

The author begins in the first two paragraphs by outlining Gordon’s argument and evidence inthe book. The author notes that Gordon seems to advocate “cold, ruthless, ‘equal opportunity’ fem-inism,” and Gordon seeks “caregiving” as the ultimate American value instead of competition.

The author begins her pointed criticism of Gordon’s work in the third paragraph. The authorstates that Gordon’s entire approach is ineffective and inappropriate. The author proceeds in thefourth paragraph to underscore that Gordon’s argument is not well constructed by noting her useof “awkward, unintegrated quotes,” which “disguise [Gordon’s] lack of familiarity with eco-nomics, history, and political science.”

Starting in the sixth paragraph, the author begins to express her own opinions. The authorstates that modern America provides more freedom to women than in any time in history. Theauthor adds, in paragraphs seven and eight, that Gordon’s portrayal of women as “nurturant andcompassionate” is misguided, as the author notes that those values were prevalent in the 1950swhen women were more repressed.

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97. BThis question asks the reader to assume the author’s side of the argument in order to delin-

eate Gordon’s opposing argument. Answer B would be the author’s most likely description ofsomeone opposed to her own idea of feminism. Answer A would be inappropriate because theauthor identifies with the “modern independent woman.” The author declares herself rebelliousin line 43, thus negating answer B. And while Gordon claims to attack a “patriarchy,” this doesnot mean that Gordon, or the author, advocate a matriarchy; thus D is an incorrect choice. The“sentimental” woman the author derides in line 16 would be her description of her opposition,and thus B is the proper choice.

98. CAnswers A, B, and D are all true. Choice A is true because the author asserts in the third para-

graph that Prisoners Of Men’s Dreams is a work that “most men rightly ignore.” Similarly forchoice B, since the author describes Gordon as “well-meaning” in line 15; the reader can rightlyassume that any plan of Gordon’s would likely take on her personal characteristics. According tothe author in lines 38–39, Gordon is guilty of stereotyping womanly values, as stated in choiceD. C is the right answer, a topic to which the author devotes the fourth paragraph.

99. CAlthough the author apparently agrees with all the choices, C is the most significant point of

her essay. The author notes in passing, as in choice A, that Gordon is an intelligent woman, butit is not central to her dismissal of Gordon’s work. Again, the author dismisses Gordon’s argu-ment concerning B, this time in the fifth paragraph. D is obviously her opinion, as inferred fromthe end of the passage. Only C fully conveys her frustration with Gordon’s line of questioning.

100. CThe author gives the name “transformative feminism” to Gordon’s theory of womanhood as

“nurturant and compassionate.” Answer A is wrong because Gordon’s National Care Agenda isportrayed as a political outgrowth of this theory. The author states that Gordon herself is againstthe cold “equal-opportunity” feminism; thus answer B is incorrect. Gordon already sees the fem-inine role as nurturant, so D is also incorrect. C is central to the author’s interpretation of Gor-don’s argument—that we must re-center our social values around compassion instead ofcompetition.

101. BThe author argues against Gordon’s definition of women as caring and compassionate

throughout the passage. The question asks what the author does NOT discuss. Therefore answerA is incorrect, because the central argument is fundamentally anchored to her belief that Gor-don’s idea of womanly virtue is false. The author also argues vehemently for C and D. Thoughone might infer that the author is in favor of equal-opportunity feminism, she does not state orimply an opinion on the matter. Thus B is the proper choice.

102. AThe author refers to Gordon’s work as “unlearned” (line 17), questions her writing and inter-

pretive ability (paragraph 4), and associates her opinions with “know-nothings” (line 29). B isfalse, because the author does not broach the issue of morality in the passage. Although theauthor may agree with C, it is not central to the argument; the same holds true for D. These areimportant facts in the passage’s contexts, but not germane to the whole thesis.

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103. DThe author is clearly interested and engaged in Gordon’s novel, though she may not approve

of its logic, so answer A is incorrect. She also goes to great lengths to deride the current phaseof feminism, which eliminates B. Answer C is incorrect: although the reader is not apprised ofthe author’s credentials, the author can be taken as a valid critic because she seems to have readthe book and has personal experiences to share. D is correct—the author even states in lines16–20 that such works are rightly ignored, then proceeds to challenge and negate the importantaspects of Gordon’s argument.

Passage V (Questions 104–111)

TOPIC AND SCOPE: This passage concerns the actor/director, Orson Welles, and the diffi-culties he had in directing himself in films, as opposed to his ease in self-direction on stage.

The first paragraph introduces this topic and notes some fundamental differences that Wellesencountered with film.

In the second paragraph, the author compares Welles to legendary comedic film actor, Char-lie Chaplin.

In the third paragraph, the author outlines some of the techniques Welles used to improve hisfilm acting.

104. CThe passage is concerned with the difficulty that Welles encountered during the filming of

Citizen Kane, and of translating his stage direction skills to film. A is incorrect because theauthor offers no evidence that Welles used a double for his stage roles, nor does he thoroughlyexplain Welles’ stage directing technique—only his stage acting technique and his occasionaluse of a double in film production. While B is true, it is only a detail that demonstrates howWelles approached filmmaking, not his central argument. The same holds true for D, which ismerely a detail that suggests Welles’ predecessors shared his early mistaken notions. C is cor-rect because nearly every paragraph is oriented around the idea that Welles had to learn film-making through experience and mistakes.

105. AThis question asks the reader to determine the author’s goal in the second paragraph. A is

the correct answer. B and D are touched upon, but only as support for the main idea of the para-graph. C is not discussed fully enough for the reader to pass judgment. A is the right answer,because the author cites Chaplin as an example of another great actor who directed himself, andthen examines the added burden Welles assumed with pacing and dialogue.

106. BIn this question, the reader has to infer the author’s central theme, which is Welles’ uneasy

transition from a stage director to a film director. You must also decide which choice would mostdebilitate the author’s argument. B is the proper choice, for if stage and film acting did indeeduse the same techniques, the author’s continual referrals to Welles’ difficulty in choreographyand timing would be patently false. A is incorrect because this statement is more general anddoes not directly attack the author’s principal argument as does B. C is not the proper choicebecause even if Chaplin had been involved in “talkies”, this does not weaken the author’s argu-ment that Welles had a difficult transition. D is incorrect because the interaction of the cast is

40

not used as a point regarding the difficulty in transition. (The interaction between the actor andthe audience is used instead.) Also, this point does not dominate the passage.

107. CThe context of the paragraph has Welles directing actors on film and examining their prod-

uct—logically, film. A would apply to Welles’ work if it were on the stage, and if the author inti-mated that Welles made sketches of his progress. The author does not. B is wrong—the authordiscusses Welles’ inability to hit his chalk marks on film is discussed in another paragraph.Choice D is never mentioned in the passage.

108. BAgain, the question asks the reader to arrive at the author’s central theme. A is false as stated

in the second paragraph, “it is true that drama did not contain the demands of pacing found incomedy.” The author does suggest that Welles’ task may have been tougher than Chaplin’s, butdoes not draw a firm enough conclusion to make C a correct choice. D is incorrect, according tothe author’s summation in the first paragraph. However, the author does suggest that spoken dia-logue combined with manic pacing, as found in most comedies, would make a difficult task forany director. B is correct because the second paragraph says “the dimension of spoken dialogue”in drama would cause “enormous problems” and on top of this Welles’ paced the film in themore difficult style of comedy.

109. CWelles’ stage acting technique is fully described in the first paragraph as one that changed

and responded to the audience. All answers except C accurately describe this style. Also, C doesnot fit in with any of the other choices.

110. ANowhere in this passage does the author support the claim that Beerbohm Tree or Duse were

screen disappointments—the author merely takes it as fact. Answer A, then, is the correctresponse. The majority of paragraph three discusses answer B and gives the other actor’s name.One can infer from the first paragraph that choice C is accurate. Line 35 mentions that WilliamAlland was slighter than Welles, discounting choice D.

111. CWelles’ stage acting is discussed in the first paragraph, and within the passage his technique

is taken by the author as emblematic of all stage acting. Welles’ technique allowed him to alterhis movements, speak directly to the audience, or change his approach to the role nightly—thusA and B are incorrect. D is proven over the passage to be incorrect. C must be inferred by thereader, from the last three paragraphs, but it is the only proper choice.

Passage VI (Questions 112–118)

TOPIC AND SCOPE: The subject of this passage becomes clear right in the first paragraph; theauthor is discussing the mystery of why we sleep.

The following three paragraphs critically analyze the theory that we sleep in order to rest.On the one hand, the author says in paragraph 2 that it makes sense to think of sleep as “restful”and “restorative,” because that’s how we experience it—even though there is no physiologicalevidence to support the idea. On the other hand, he points out in paragraph 3 that there is prob-ably more to sleep than rest. After all, you don’t need to go to sleep to get rest. In paragraph 4,

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the author speculates that higher animals may restore their attention and memory during sleep,and the “brain-mind” may review and reorganize acquired data at the same time.

The final three paragraphs are devoted to consideration of the problems sleep researchersencounter when trying to investigate sleep function. Using the sleep-deprivation approach,which would seem to make the most immediate sense, is fraught with difficulties, as explainedin paragraph 5. In paragraph 6, the author outlines the ideal alternative to the sleep-deprivationapproach, but notes that the alternative has never been used successfully. Returning to his cri-tique of the sleep-deprivation approach in the final paragraph, the author argues that even ifsleep-deprivation were somewhat successful, this wouldn’t tell us anything about what’s reallyhappening on a cellular or neurobiological level.

112. CThe author says that the researchers’ inability to establish functional hypotheses has been

due to methodological problems, which we learn about in the last three paragraphs. Sleep-depri-vation experiments are difficult to carry out and give only limited information, and the idealalternative approach has not proved successful. Choice C is correct.

The author never suggests that the failure to establish functional hypotheses has any impli-cations regarding Freud’s hypotheses (choice A) or the passive rest theory of sleep, whether itbe for the theory (choice D) or against it (choice B).

113. DThe author offers several reasons in the first four paragraphs that passive rest is probably not

the sole function of sleep: 1) sleep is too complex and has different component phases, 2) youdon’t need sleep to rest, and 3) sleep is probably associated with the active processes of restor-ing efficiency to attention and memory and of reviewing acquired data.

The question is: which one of the answer choices contradicts one or more of these reasonsand thus weakens the author’s argument? A doesn’t. The fact that folk wisdom and science sel-dom agree (but do in supposing that the sole function of sleep is rest) has nothing to do with theauthor’s argument. B and C are wrong because they both support the author’s argument. If selec-tive deprivation of REM sleep has a different effect from that of selective deprivation of non-REM sleep, then this implies the component phases have different functions—just what theauthor says. And the author also says that some animals get plenty of rest when they’re awake,implying that sleep must provide more than just rest; therefore, C strengthens the author’s argu-ment as well. D is the one that weakens the author’s argument. If measures of attention andmemory are stable regardless of sleep-time, then the author’s idea that attention and memory isrestored during sleep seems to be wrong. This strengthens the possibility that rest actually is thesole function of sleep.

114. CThe author’s point in the second paragraph is that our subjective experience of sleep as rest-

ful and restorative should “more powerfully motivate us to seek physiological or behavioralexplanations” of this phenomenon, even though there is no physiological evidence. From thisyou can infer that the author thinks that a theory—in this case, that sleep is physiologicallyrestorative—may be true even if there isn’t any scientific evidence for it (choice C).

The author never says that subjective experience can prove a theory correct all by itself (A);he thinks that subjective experience should prompt scientists to move in a specific direction. Nor

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does the author link dreaming to physiological restoration (B) or state that subjective experienceis always accurate (D).

115. CThe best way to approach this question is to run through the choices quickly to see whether

one of them jumps out as a claim the author makes but doesn’t support with evidence. A iswrong because the author spends all of paragraphs 5 and 7 arguing that the sleep-deprivationapproach is flawed. B is labeled a “naive notion” by the author, so it’s out as well. C, on the otherhand, is a statement made by the author at the end of the third paragraph for which he never givesany supporting evidence or statistics—C is the correct answer. As for D, the author mentions theresults of experiments with sleep-deprived rats to back up this statement.

116. CThis is essentially a sleep-deprivation experiment in which the resident goes without sleep

and then is tested for signs of a memory deficit. Remember what the author says in the fifth para-graph about such experiments: there is no way to know whether the memory deficit is the resultof sleep deprivation or other factors, such as the things the subject did to keep herself awake. Cis therefore correct.

A is tricky because the author does say that sleep-deprivation doesn’t provide informationon a molecular level; however, the experiment does NOT establish that sleep deprivation affectsmemory function. B is wrong because the author never directly states that if you don’t reviewand reorganize information during sleep, your memory will decrease. Finally, D, testing the res-ident’s memory during the work shift may have made it possible to get a better look at the res-ident’s memory capacity over the course of the shift, but the author would still object by sayingit’s not clear that her memory is being affected only by sleep deprivation.

117. CReview the sixth paragraph to understand what a “positive functional model” is: a measure

of brain function or behavioral capability or psychological process that could be tested aroundthe clock and show deterioration as the subject stayed awake longer yet recover after sleep.

A doesn’t fit the model because the subject’s mood doesn’t show deterioration. B doesn’t fiteither, because it doesn’t show the subject’s brain electrical activity deteriorating prior to sleepand then recovering. D is incorrect for similar reasons: reading speed does not deteriorate orrecover, and is not tested around the clock. C is the best example of the successful application ofthe author’s model, since the concentration of neurotransmitter decreases the longer the subjectis awake but recovers after sleep (unfortunately, such a neurotransmitter has not been discovered).

118. CThe author’s point in the final paragraph is that even if you could carry out a sleep-depriva-

tion experiment successfully, you wouldn’t learn anything on the cellular or neurobiologicallevel. The example he gives is the sleep-deprived rats experiment; even though we can concludefrom this that sleep deprivation can be fatal, we cannot answer the question, “How are sucheffects mediated?”, on the cellular level. The question drives home the limitations of the sleep-deprivation approach (choice C).

The author says nothing in the final paragraph about knowing which question to ask (choiceA), the staggering lack of knowledge about sleep on the molecular level (choice B), or the natureof productive scientific research (choice D).

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Passage VII (Questions 119–124)

TOPIC AND SCOPE: The premise of this passage is that racist speech on campuses is not per-missible. This type of speech can not be restricted by government because of the first amendment.The University can, however restrict this speech on the basis of appropriateness as illustrated bythe rules of etiquette.

Paragraph 1 introduces the topic in the form of a question. It acknowledges the right of freespeech and asks what kind of a frill etiquette is for those in pursuit of truth. In paragraph 2 anexample is given of an expulsion which took place at Brown University. People argued that allrestrictions of free speech are intolerable at the university. In response the president of the uni-versity reclassified the speech as behavior.

Paragraph 3 is one sentence: The premise is wrong: Not all restrictions of free speech areintolerable at the university. In paragraph 4 the authors respond to the premise by defining therole of the university. There is a contrast between unlimited speech and unlimited inquiry.

Paragraph 5 states that no one would argue that offensive speech will advance knowledge.The course of action a university should take comprises paragraph 6—to ban speech that inter-feres with free exchange of ideas. Offensive topics may be discussed but offensive mannersshould be prohibited. Attack ideas not people.

Some ideas on education and the purpose of structure in education are discussed in para-graph 7. Paragraph 8 shows how the law prohibits free speech in the pursuit of juridical truth inthe courtroom.

Paragraph 10 provides more proof in the form of speech and behavior required by parlia-ment. But even in the U.S., as demonstrated, we maintain order in the court; and hopefully, wedo when our government bodies debate amongst themselves as well.

Paragraph 11 restates that improper speech prohibits substantive argument and it wasteseveryone’s time. Paragraph 12 adds that responding to insulting speech with insults only pro-longs a verbal interaction that is going nowhere.

Finally, the authors advocate that impolite speech should be prohibited throughout the uni-versity, not just in the classroom. There is a reiteration of definition of university—a forum ded-icated to rational debate.

119. AIt is implied by the fact that the Brown University president was able to reclassify the stu-

dent’s offensive speech as offensive behavior and then expel him. According to the essay, stan-dards of comportment are regulated only in the courtroom. Offensive speech is not something astudent can be expelled for, otherwise the president of Brown would not have had to reclassifythe student’s speech.

120. AThe strongest argument in favor of the use of etiquette at the university is the use of etiquette

to govern the courtroom. Parliament (choice B) is not such a good example, because, as statedin the passage explanation, Free Speech is considered a strength of the American constitution asopposed to the regulation of speech in other countries. It may be true that racist remarks are not

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likely to lead to advances in knowledge (choice C), and this is a good argument, but it is not agood argument to pose against the first amendment, which after all is the strongest argumentagainst the regulation of speech at the university. Choice D is insufficient for the same reasonsas choice C.

121. CYou need to backtrack by asking questions. 1) Where is the word premise or the definition

of the word used previously? In the previous sentence Brown’s president “agreed with thatpremise.” In the sentence before that “many argued that all restrictions of free speech are intol-erable in the university.” That is the premise. There is a logical progression. Offensive behavior(choice A) is the grounds on which the student was expelled. The premise is that there can beno restrictions of free speech, so the behavior was reclassified to conform with grounds for dis-missal. The expulsion of a student on the basis of offensive speech (choice B) never took place.It is not the premise. The word “premise” refers to a precedent, a conclusion on which later con-clusions or actions are based. So we want to find the first conclusion on which all others arebased. The first conclusion is that free speech must not be restricted. The following conclusionsare that a student cannot be expelled on the basis of offensive speech, but can be on the basis ofoffensive behavior. Regarding choice D, the concept that the university can be hampered by therestrictions of civility is part of the opening statement of the essay. It is a rhetorical question andhas a different function which is more universal to the essay than the third and fourth paragraphwhich deal with a more specific situation.

122. AThis is correct because it is closest to the main idea of the authors. You can tell this because

it is contained in the last sentence of the paragraph. It is also supported by statements like “It isto foster unlimited inquiry” and “it must protect the discussion of offensive topics.” B is true,but less so than choice A. Choice A is a general definition. Choice B is more specific, and that’swhy it merely supports the main definition. Since the main definition is in the last paragraph,that also indicates that it’s the main definition. Choice C may be true, but it is not mentioned inthe passage. Choice D is definitely not the authors’ definition of the university. In fact, in the lastparagraph the authors say that invective is detrimental to rational debate. Detrimental meansdamaging.

123. DThis answer is actually mentioned word-for-word in the ninth paragraph. Choice A, “In par-

liament,” is obviously not true after the whole discussion in paragraph 11. It is implied that thisanswer B is true, but since answer D is actually stated, it is a better answer. Answer choice C isnot mentioned at all in the passage.

124. CThe reason this answer is correct is because it is closest to the main idea of the passage, which

is that etiquette should be used to guide speech in the university. A clue is the word “main” usedin the question. For choice A, the restriction of free speech being intolerable is contrary to every-thing the authors argue for in the passage. This answer is actually a counter argument. The authorsdo actually use the statement in choice B. It is, however, only in support of the main idea, whichis that the rules of etiquette should restrict speech instead. Neatly reclassifying the offensivespeech as offensive behavior, as seen in choice D, is actually a complaint the authors make againstthe president of Brown University. This is not a response the authors would condone.

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Passage VIII (Questions 125–132)

TOPIC AND SCOPE: This social sciences passage is about George F. Kennan’s strategy forinfluencing the international outlook of the Soviet leaders and how it was undermined by theformation of NATO.

The first two paragraphs explain Kennan’s idea of using behavior modification on the Sovi-ets. The formation of NATO, we find out at the end of the second paragraph, made effectivebehavior modification more difficult. Paragraph 3 explains that NATO got started because theWest Europeans were worried that the Russians might attack with their superior forces. Para-graphs 4 and 5 discuss the three reservations that Kennan voiced about NATO and the adminis-tration’s position on each of them. Finally, paragraphs 6 and 7 relate that NATO was formeddespite all reservations. Kennan reluctantly agreed that the Western Europeans needed NATO, butNATO definitely made it more difficult to change the Soviet concept of international relations.

125. BSeveral things are stated or suggested about Kennan’s “behavior modification” approach in

the passage, particularly in the first paragraph, so it’s best to run through the answer choices tosee which ones can be eliminated. A is wrong: Kennan’s approach easily could have been suc-cessful even if the Americans were never forced to employ negative reinforcement. C is incor-rect because the passage never says that Kennan had any kind of background in behavioral psychology. Nor does the passage ever say that “behavior modification” was used on the Rus-sians during WWII (D).

B is correct because war and appeasement, the approaches Kennan rejected, are the twoextremes on the spectrum of possibility. Responding positively to the Soviets when they’re con-ciliatory and countering them when they’re not is a moderate strategy in comparison to all-outfighting or constant attempts to make peace.

126. CWe learn in the second paragraph that Kennan regarded the formation of NATO as “certain

to reinforce Soviet feelings of suspicion and insecurity, and, hence, to narrow opportunities fornegotiations.” Kennan didn’t want opportunities for negotiations to narrow; America had to beprepared to negotiate in order to give positive reinforcement of conciliatory Soviet behavior. Theformation of NATO, then, was going to make Kennan’s strategy difficult to carry out, and heclearly disapproved, choice C.

Choices A, B and D all imply that heightening Soviet insecurity and suspicion wouldimprove Western/Soviet relations. Kennan never thought that heightening Soviet suspicion andinsecurity would be a positive thing, regardless if it meant negative reinforcement (choice A),forcing the Soviets to be nicer (choice B), or making negotiations easier (choice D). Thesechoices may have made sense to you, but the correct choice here is the one that reflects Ken-nan’s attitude.

127. DThe third paragraph explains that the West Europeans started to push the idea of a military

alliance with the U.S. and Canada because the Russians had three times as many divisions inEurope as the combined U.S., British and French forces. They were afraid that the Russians mightsweep to the English Channel etc., as they were reported capable of doing. D paraphrases this.

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Choice A is a misreading of the sentence about the intelligence estimates, which only saysthat the Russians could easily attack, not that they would attack. Choice B is wrong because it’snever stated that none of the nations in NATO could withstand a Soviet attack without assis-tance. This may have been true of some of the European nations, but what about the U.S.? Asfor C, it’s true that America wanted to aid the European economic recovery, but this isn’t whyNATO was formed. Both Kennan and the politicians in Washington believed that focusing on themilitary would hinder economic recovery. Again, choice D is correct because the primary rea-son for NATO’s formation was that the Western Union countries were afraid of a Soviet attack.

128. BKennan’s first point about the formation of NATO was that the Europeans “had mistaken

what was essentially a political threat for a military one, and that they risked, as a result, ‘a gen-eral preoccupation with military affairs, to the detriment of economic recovery.’” If Kennanthinks that the Soviet threat is political and not military, then he’s assuming the Soviets aren’treally going to attack with their thirty divisions (choice B).

A contradicts the statement that a preoccupation with military affairs would be a detrimentto economic recovery. There is no evidence in the passage to support answer C. Choice D iswrong because though Kennan thinks putting military affairs above economic affairs is wrongin this case, he’s not saying that it’s always a mistake.

129. DIn his second reservation, Kennan says that the countries not included in the alliance would

be rendered more vulnerable. You can infer that he would be concerned about the vulnerability ofGreece, Turkey and Iran, since they were excluded from the alliance. D correctly paraphrases this.

Kennan does mention that the Western European countries had common defense interestsrooted in geography, but he never says that countries outside this region should be excluded fromthe alliance A. Choice B is wrong because Kennan never says anything, so far as we know, abouthow encompassing the concept of “North Atlantic” should be. Nothing is ever said about Ken-nan wanting to use particular countries in an anti-Soviet strategy, choice C.

130. CThe last sentence of the fifth paragraph explains that nobody was worried about “a final mil-

itarization of the present dividing line” because most people figured that it was too late; the divi-sion of Europe was already complete. Choice C is correct.

Choice C is the only reason given in the passage for why people weren’t concerned aboutKennan’s third point, so the notions that the administration thought it possible to alter the linepeacefully (choice A) or that they wanted a strong American military presence in Europe (choiceB) can be thrown out. Answer D is incorrect because the administration’s agreement with Ken-nan’s first two points does not explain their lack of concern for his third point.

131. AThe explanation for the previous question points out that the Truman administration agreed

with Kennan on two of his three points. They decided, however, to go ahead with the treatydespite their (and his) reservations. Therefore, the administration was “often in agreement” withKennan but “ultimately deterred” by the reservations he had (choice A).

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Kennan never made proposals for the improvement of NATO (at least not in the passage) sochoice B can be eliminated. Choice C is wrong because the administration didn’t try to counterhis criticisms—they agreed with him on some of them. Their overall objectives (choice D) werenever explicitly discussed nor re-examined.

132. CLet’s take each of the choices one-by-one. Choice A is silly; your common sense should tell

you that military alliances can’t always be flawed and ineffective. Choice B is more tempting,but it’s wrong too. The author of the passage says that Kennan thought “behavior modification”was the best way to deal with the Soviets—this is not the same thing as saying “behavior mod-ification” is the only way to change a government’s concept of international relations. Youshould always be careful when choosing a response that contains “always” or “only” becausethese statements are often too definitive. Answer choice C is correct because Kennan’s coherentstrategy of behavior modification was hampered by the formation of NATO, which came aboutas a result of the military imbalance in Europe. Finally, choice D accords with what the futuremembers of NATO thought but not with what Kennan believed. Therefore, D cannot be consid-ered to accord with a “theme of the passage.”

Passage IX (Questions 133–137)

TOPIC AND SCOPE: This passage is about the timing of the ape-human split in human andape evolution.

Paragraph 1 describes the dispute between the two sides. The paleontologists believe fossilsare the only direct evidence and molecules are not relevant to extinct species versus molecularanthropologists who have introduced new data, specifically molecular biological data.

Paragraph 2 describes the paleontologists’ approach.

Paragraph 3 describes the molecular anthropologists’ approach—they study proteins. Theirmain goal is to date the divergence (separation) of two species from a common ancestor. This isdone by comparing the amino acid sequences. The more similar the sequence, the more recentthe divergence.

In paragraph 4, the author takes a position that molecular data is more valid because resultscan be scientifically duplicated in different laboratories. She sees paleontology as less conclu-sive because the analysis is too subjective and researchers often disagree.

Paragraph 5 discusses the paleontologists’ position on the ape-human split. Paleontologistsbelieve that humans and apes have had separate ancestry for an extremely long time. They placeRamapithecus in the human ancestral line.

Paragraph 6 discusses the molecular position on the ape-human split. The molecular anthro-pologists believe that there was a recent divergence only 5 mya. Ramapithecus existed beforethe split, so they consider Ramapithecus a common ancestor to both humans and apes.

Paragraph 7 contains additional evidence supporting the molecular approach. This paragraphrefutes a claim made by the opposition, and then presents arguments to support the author’scause. It uses the hour hand, minute hand and second hand analogy.

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Paragraph 8 concludes with a reiteration of the molecular position that Ramapithecus is acommon ancestor to both humans and apes.

133. CThis question requires attention to detail. Paragraph 6 states that molecular phylogeny rules

out the possibility that Pliopithecus is an ancestral gibbon, or that Dryopithicus or Proconsul areearly apes, therefore I. and II. are false. However, the three-way divergence about 5 mya (mil-lion years ago) does not preclude (prohibit, exclude or prevent) Australopithecus as an early rep-resentative of the human family, so only III. is true. Although the findings of the paleontologistsare different from the findings of the molecular anthropologists, the author later defends themolecular methods and their truth by saying that the findings are an observation based on anenormous amount of data. The findings that are true according to the passage are the ones thatthe author is arguing to be true.

134. CThis is really a main idea question because of the words “most significant.” The passage is

a straightforward explanation of the author’s view that Ramapithecus has an equal place inhuman history as well as in ape history. The main point is reiterated in the last paragraph.Answers A and B are true, but not the best answers because they are less significant in the pas-sage and therefore less significant to the main idea. Answer D is nonsense because the “molec-ular clock” is a metaphor, not an actual object that was invented.

135. AChoice A is the best choice because it augments the point in the first paragraph that paleon-

tologist evidence is not as accurate as the evidence of molecular anthropologists. The origin ofthe name, Ramapithecus, is not significantly relevant to the concept of the first paragraph. So,choice B is wrong. Answer choice C is not relevant because the similarity of humans to apes isnot dealt with in this paragraph. “Ontogeny recapitulates phylogeny” (choice D) refers to theconcept that the development of the individual (ontogeny) reflects the stages of it’s ancestors inevolution (phylogeny). This concept is not dealt with in the passage, although both paleonto-logical and molecular phylogeny are. Paleontological phylogeny refers to the stages in the evo-lution of a species that can be observed through fossil bones and teeth. Molecular phylogenyrefers to the stages in the evolution of a species that can be observed on a molecular level.

136. DChoice D is true because Ramapithecus represents one of the greatest differences in the find-

ings of paleontologists and molecular anthropologists. The former contend that Ramapithecus isa human ancestor only. If Ramapithecus were proven to be as much ape-like as human-like thiswould support the molecular findings that Ramapithecus is a common ancestor to both humansand apes. Choice B would be evidence against the accuracy of molecular anthropologists, sinceit is their theory (Paragraph 6) that the Asian apes split off earlier than the rest. Choice C wouldnot strengthen the claim of molecular anthropology because the paleontologists also believe thisto be true. The strengthening of molecular anthropology would be most effective when using afinding the paleontologists disagreed with, as in choice D.

137. AChoice A is obviously correct because the author makes this claim yet does not offer any

supporting evidence. Choice B is not a claim that the author makes, so we don’t need to lookany further for supporting evidence. Choice C is a claim which the author makes and supports.Evidence supporting the statistical constancy of the molecular clock is found in the hour, minute

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and second hand analogy. Choice D is also a claim that the author makes and supports. Thisstatement is part of the author’s conclusion. Paragraph 7 is devoted to proving it.

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I.

Essay 1 Statement:

The future of scientific research should not be decided by public opinion.

Sample Response: (Level 5–6)

The statement suggests that future directions of science and technology should be dictated by forces otherthan mass appeal. If policy is left in the hands of an all-too-often short-sighted and ill-informed populace,science and technology will be relegated to the status of cultural follower rather than be where it should—on the vanguard of invention, innovation, and imagination. If science is forced to follow the whims ofpublic opinion, we’ll stagnate progress and compromise possibility. Governments or governing bodiesmust nurture and support scientific research, the fruition of which may not come for generations.

On the other hand, the dollars which feed pure science are part of a societal budget—one which mustoften meet conflicting needs and demands. Suppose a small country were to suffer a devastating naturaldisaster like a flood or an earthquake. In such a case—where the wellness of its people is at risk—a nationcould hardly justify spending money on speculative scientific inquiry, whatever the ultimate payoff downthe road. Also, one could argue that a populace should control the distribution of its communal funds, andif that means limited commitment to science then so be it. In such a case, scientific research would bemore and more privatized so that “public opinion” would be reflected in success or failure of businesses.

There is a certain momentum in the growth of science which should not be halted by changing the pub-lic sentiment. A present-thinking populace can’t always make long-term decisions if the decision requiresshort-term sacrifice, so it seems risky to trust the future of science to anything less than an expert panel,group, or board. However, a culture’s growth and change should be reflective of its people. One couldargue that science has no more of a manifest destiny than other cultural endeavors and should be subjectto the same constraints—financial and political—which bind other aspects of a society. If dedicatingmoney and effort to future scientific projects would damage the well-being of a society, then scienceshould yield to public need. Under other circumstances, it is important that scientific progress not bethreatened by naive opinions and attitudes.

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II.

Essay 2 Statement:

Structure stifles creativity.

Sample Response: (Level 5–6)

The statement suggests that a structured environment—one in which processes are spelled out and pre-planned—inhibits the kinds of thinking necessary for creativity. Insofar as creativity relies on unconven-tionalism, novelty, and abstraction out of specific circumstances, it seems to be incompatible with thetight, rigid structures of, say, a military or political hierarchy.

However, considering that creative leaps require a certain freedom within the mind, it stands to reason thatstructure can actually help liberate creativity by holding other concerns to a stable structure. Take, forexample, a writer who rigidly compartmentalizes her day so she won’t have to use any “mental overhead”to make decisions about how to spend her time. By freeing up her cognitive load, she can focus more onher writing, giving herself the chance to reach that state of immersion and flow necessary for creativeinsight. Another example of when structure does not stifle creativity would be a musician who, throughpractice and diligence, acquires great technical skill so that he has more musical “tools” at his disposalwith which to create new music.

Structure can be a noose around the neck of creativity, or it can be a platform which supports creativity.The distinction is determined by the relationship between the specific type of structure and the creativeprocess. In cases where the structure directly blocks the activities or mental states necessary for creativework, then it is a hindrance. In situations where structure acts to release the mind so it can focus on higher-order endeavors, it is not a hindrance to creativity—but rather is a necessary component.

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138. A

139. B

140. B

141. D

142. C

143. D

144. C

145. C

146. D

147. B

148. C

149. A

150. C

151. C

152. A

153. B

154. D

155. C

156. D

157. B

158. A

159. D

160. B

161. D

162. C

163. C

164. B

165. D

166. A

167. B

168. B

169. A

170. B

171. D

172. C

173. B

174. D

175. A

176. D

177. B

178. C

179. C

180. D

181. A

182. B

183. D

184. A

185. C

186. A

187. D

188. B

189. B

190. D

191. C

192. C

193. A

194. B

195. D

196. A

197. C

198. B

199. A

200. C

201. C

202. B

203. D

204. C

205. A

206. A

207. A

208. B

209. D

210. A

211. B

212. A

213. A

214. C

BIOLOGICAL SCIENCES ANSWER KEY

Passage I (Questions 138–143)

138. AAccording to the passage, RB results when both copies of the wild-type (normal) RB gene

are inactivated in a retinal cell, regardless of whether it is sporadic RB or familial RB. In otherwords, an RB cancer cell is homozygous recessive at the RB gene locus. Therefore, any RB can-cerous cell is going to have the genotype rr, so choice A is correct. Normal retinal cells arehomozygous dominant at the RB gene locus and therefore have the genotype RR. Rr is the geno-type that results when only one copy of the RB gene in a retinal cell has been inactivated, as ina child with familial RB, prior to the somatic mutation that triggers the onset of RB.

139. BAccording to the passage, people with familial RB are born with one mutant RB gene and

one normal RB gene. Also, the gene is located on chromosome 13, which means that this traitis autosomal, not sex-linked. So while the retinal cells of this man have the genotype rr, hisgerm-line cells—the cells that are the progenitors of spermatogenesis (meiosis)—carry thegenotype he was born with: Rr. Thus, spermatogenesis will result in 50% of the sperm carryingthe R gene and the other 50% carrying the r gene. In other words, any child that this man has—boy or girl—has a 50% chance of inheriting the mutant RB gene (r). Therefore, choice B is cor-rect. Remember, the gender of the child is irrelevant because the gene is not sex-linked.

140. BThe cell cycle represents the 4 stages in the life of a cell: G1, S, G2, and M. The G1 phase is

a period of cell growth and biochemical activity; new organelles are produced and the cell dou-bles in size. The S (synthesis) phase is the period of DNA replication. During the G2 phase, thecell continues to grow and assemble more new organelles. The M (mitotic) phase is the time dur-ing which mitosis and cytokinesis occurs. Since p105-RB inhibits cell progression from G1 toS, it must inhibit DNA synthesis and therefore inhibit mitosis. Thus, choice B is correct. Thismakes sense if you approach this question looking at p105-RB as a tumor suppressor gene. It’sonly when tumor suppressor genes are inactivated that uncontrolled cell growth and tumorigen-esis can occur. Choice A is incorrect because microtubule formation occurs during G1, and alsobecause the inhibition of cell movement would not prevent tumor growth. Choice C is wrong forthe same reason: protein synthesis occurs during G1. Moreover, p105-RB is described asrestraining cell growth, not promoting it. Choice D is wrong because p105-RB prevents, not pro-motes, RB development.

141. DThe products of tumor suppressor genes function in ways that limit cell proliferation,

thereby suppressing uncontrollable cell growth and tumor growth. Therefore, a protein that func-tions as an activator of cellular proliferation is most likely not the product of a tumor suppres-sor gene. So, choice D is correct. Since the main components of cell growth are an increase incell mass and cell number, a protein that functions as an inhibitor of cell growth could be theproduct of a tumor suppressor gene. Therefore, choice A is incorrect. Apoptosis, or programmedcell death, is a cell’s self-destruct button. The product of a tumor suppressor gene could func-tion as an activator of programmed cell death, because a dead cell certainly can’t proliferateuncontrollably, as cancerous tumor cells do. So, choice B is wrong. For example, some lym-phomas result from the failure of a subpopulation of lymphocytes to enter programmed celldeath. Finally, the product of a tumor suppressor gene could function as a regulator of cellulardifferentiation, because specialized cells lose the ability to rapidly proliferate. So, choice D is

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incorrect. For example, a teratoma is a germ cell tumor that arises because of a lack of differen-tiation control.

142. CAccording to the passage, cells that are homozygous mutant at the RB locus have effectively

lost both copies of the wild-type RB gene. This implies that these cells are unable to make thewild-type RB gene product, p105-RB. Yet according to the question stem, some retinal cells thatare homozygous mutant at the RB locus do synthesize p105-RB. You don’t know if the p105-RB is normal or functional (it isn’t); you just know that it is being produced. At the very least,in order for p105-RB to be synthesized, the DNA coding for it must still be transcribed andtranslated. In other words, the mutation that triggered the RB cancer did not affect the portionof the RB gene responsible for promoting transcription. So, choice C must be correct, since it isthe only one that explains how it is that the protein is still produced by cancerous retinal cells.

Choice A is incorrect because incomplete penetrance would not account for the synthesis ofp105-RB. Penetrance is the percentage of people with a given genotype who actually expressthe corresponding phenotype. Given the nature of RB cancer, one can infer that the penetranceof RB gene inactivation is about 100%, meaning that if you lose both copies of the wild-typeRB gene, you will get an RB tumor. Choice B is tempting because it is a true statement. Ahomozygous RB mutant cell still capable of producing p105-RB would produce a nonfunctionalversion of the protein. In other words, transcription can occur, even though somewhere withinthe DNA a mutation exists that results in the translation of a nonfunctional protein. But, you areasked to explain how it is that the cell can still produce the protein, not what level of function-ality the synthesized protein possesses. So, choice B is incorrect. Choice D is wrong because,according to the passage, cells that are homozygous mutant at the RB locus develop RB tumors,and are therefore not phenotypically normal.

143. DAfter entering a host cell, a DNA virus can either replicate in the cytoplasm using the cell’s

“machinery,” or it can integrate into the host cell’s genome, remaining there through several cellcycles before re-emerging to replicate and infect other cells. Since the RB genes of the SV40-infected retinal cells are still intact and functional, this implies that normal p105-RB is stillbeing synthesized. This, in turn, implies that SV40 must be integrated in the retinal cell genome,but in a site other than the RB gene locus. Remember, however, that these retinal cells are can-cerous, and so SV40 must somehow be inactivating p105-RB. Since choice D is the only plau-sible explanation for this phenomenon, it is correct.

Choice A is incorrect because p105-RB is synthesized despite infection with SV40. If SV40had inserted in an exon in the RB gene, then the genes would not be normal and functional. Sochoice B is wrong. Choice C is wrong because even if it is assumed true that the paternalgermline is more susceptible to viral infection (which it isn’t), this does not explain how anSV40-infected retinal cell can be cancerous and still have two functional RB genes.

Passage II (Questions 144-149)

144. CUnfortunately for Fred Griffith, subsequent research proved him wrong. In 1943, O.T. Avery

and his colleagues discovered that it is DNA, not protein, that is the transforming agent. Trans-formation occurs when a fragment of DNA from one bacterial cell enters another bacterial celland is incorporated into the host cell’s DNA. This new piece of bacterial DNA is transcribed and

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translated along with the host DNA. Foreign genes are thereby incorporated into previously purebacterial cultures, endowing them with the characteristics coded for by the foreign DNA, e.g.,the ability to produce polysaccharide capsules. However, this question must be answered fromGriffith’s point of view, which was that protein was the transforming agent. Since DNA is not aprotein, but rather a nucleic acid, the observation that DNA transformed bacteria would mostcontradict Griffith’s interpretation. Thus, choice C is correct.

Choice A is incorrect because it supports Griffith’s hypothesis. It makes sense that if heat-killed smooth-strain bacteria can transform live rough-strain, then live smooth-strain bacteriawould be able to do the same. So, choice A is wrong. Choice B is wrong because it neither con-tradicts nor supports Griffith’s interpretation. While the polysaccharide capsule of the smooth-strain bacteria does play a role in making the bacteria virulent, it is not the transforming agentnor is it a protein. Choice D is incorrect because it supports Griffith’s theory. Since amino acidsare the biochemical building blocks of protein, discovering that the transforming agent was com-prised of amino acids lends credence to Griffith’s interpretation.

145. CExposing this strain of bacteria to heat will essentially kill the bacteria and thus destroy its

ability to cause infection. Choices A and D are incorrect because they are not the best answers.It is true that heat-killing the bacteria will destroy the polysaccharide capsules (A) and that deadbacteria do not divide (D); however, the reason that the mice did not die is only an indirect resultof these two answers. The direct reason that the mice survive is that the bacteria were no longerinfectious. Choice B is wrong because the DNA was not denatured in this experiment. Heat-killing of bacteria is normally done at 42 degrees Celsius while DNA denatures around 96degrees Celsius. Moreover, if the DNA had been denatured by the heat-killing in this experi-ment, then the denatured DNA would have been unable to transform the rough strain and pro-duce dead mice in the Group 4 trial. Therefore, the heat-killing could not have denatured theDNA from the smooth strain. Choice C is the best answer.

146. DAs a result of his experimental results, Griffith correctly concluded that some component of

the heat-killed smooth strain-bacteria had been transferred to the live rough-strain bacteria(though he incorrectly concluded that this component was a protein). The only way that nonvir-ulent unencapsulated bacteria can be “transformed” into virulent encapsulated bacteria is if thenonvirulent bacteria acquire bacterial genes that allow them to manufacture their own capsules.In the Group 4 experiment, the only bacterial DNA coded with the messages needed to producepolysaccharide capsules was the DNA of the heat-killed smooth strain. Since the heat-killedsmooth strain was incapable of killing the mice on their own (as evidenced by the survival of theGroup 3 mice), the smooth-strain DNA must have been incorporated into the rough-strain DNA,giving the latter the ability to manufacture the capsules, and thereby “transforming” them fromrough to smooth. So, choice D is correct.

Choice A is incorrect because it is the smooth strain that transfers material to the roughstrain, not vice versa. Choice B is wrong because the transfer of tRNA would not transform bac-teria. tRNA functions as a carrier of amino acids during protein synthesis. Choice C is incorrectbecause it is not the polysaccharide capsules themselves that are transferred from one strain tothe other, but rather the DNA coding for the production of the capsules.

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147. BOne of the ways that RNA differs from DNA is in the use of the nucleotide uracil. In RNA,

uracil is substituted for thymine. In other words, RNA has uracil and DNA doesn’t. During DNAreplication, adenine pairs with thymine. During transcription (mRNA synthesis from a DNAtemplate) and translation (translation of mRNA into a peptide chain), adenine pairs with uracil.Thus, if bacteria were exposed to a drug that cleaves adenine from uracil, transcription and trans-lation would be disrupted, whereas transformation, which is the transfer of DNA from one bac-teria to another, would not be directly affected by this drug. Therefore, I and II would be directlyaffected and III would not. Choice B is correct.

148. CThe shape of individual prokaryotes is a function of their cell wall. There are three major

shape-groups that most of the major bacteria groups fall into. Spiral-shaped bacteria are calledspirilla; spherical bacteria are called cocci; straight, rod-shaped bacteria are called bacilli. Thepneumococcus bacteria is a member of the spherical-shaped family of bacteria.

149. A The only reason that the combination of heat-killed smooth strain plus live rough strain

killed the Group 4 mice was because smooth-strain DNA coding for the polysaccharide capsulewas incorporated into the genome of the rough strain. When the rough strain bacteria repro-duced, all of the progeny inherited the ability to manufacture the capsule. In other words, therough strain had been transformed into the smooth strain, and could now infect the mice and killthem. The key to the transformation was that the rough strain reproduced. Heat-killed bacteria,whether they are encapsulated or not, cannot reproduce and cause infection because their DNAhas been denatured. Therefore, if Griffith had injected a fifth group of mice with a combinationof heat-killed smooth strain and heat-killed rough strain the mice would have lived. So, choiceA is correct.

Choice B is incorrect because even though the mice would live, it’s not because the heat-killed smooth strain cannot transform mice cells. Bacterial cells can transform other bacterialcells; they cannot transform animal cells or plant cells. Choice C and choice D are wrongbecause the mice would survive. FYI: Acquired immunity is the kind of immunity one gets froma vaccine; it is usually not permanent.

Passage III (Questions 150–154)

The first organic chemistry passage is about carboxylic acids and their derivatives. TheMCAT loves to test oxygen–containing compounds, so make sure you are familiar with alco-hols, aldehydes, ketones, carboxylic acids, and acyl halides, to name a few.

Notice that you are given information in the form of a diagram and text. Learn to absorb theinformation given, but don’t try to memorize it—the passage isn’t going anywhere! A good tech-nique when tackling these passages is to read to understand, not to memorize. Also, try to filterout the information that doesn’t seem so important; for example, the first paragraph may giveyou some general information on the subject, but you may not be tested on this. When you aregiven a reaction scheme such as that in Figure 1, look at the reactants and products closely, andtry to identify the types of compounds, and the reactions they undergo. Since you are given anexperimental procedure with unidentified compounds, there is a good chance that the reactionsthese unknown compounds undergo are based on the reaction scheme in Figure 1.

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150. CAs you can see from the NMR spectrum of Compound D in Figure 2, this molecule contains

aromatic protons. Aromatic protons generally give a signal in the region 7–8 ppm, a value thatyou should know. Straight away, you could have eliminated choice A since cyclohexanecar-boxylic acid does not contain any aromatic protons. In addition, cyclohexanecarboxylic aciddoes not have a molecular formula of C8H8O2, so it definitely can’t be Compound D.

Choices B, C, and D, all have a molecular formula of C8H8O2, so you have to look a bit moreclosely at the information in the passage. In the procedure carried out by the student, CompoundA is converted to Compound B by reaction with silver oxide. If you look at Figure 1, you cansee that silver oxide is used in the conversion of an aldehyde (Compound II) to a carboxylic acid(Compound V). So, Compound A is an aldehyde and Compound B is a carboxylic acid. You arealso told that Compound B is insoluble in water, but is soluble in the aqueous layer upon theaddition of sodium hydroxide. At this point, you may have identified Compound B as benzoicacid; this molecule is insoluble in water, but is soluble after the addition of sodium hydroxidesince it is converted to its sodium salt—sodium benzoate. Compound B is then heated withCompound C in the presence of mineral acid to form Compound D. Again, there is another cluein Figure 1. The only reaction that involves the use of mineral acid and heat is the conversion ofCompound V (a carboxylic acid) to Compound VII (an ester). An ester is formed when a car-boxylic acid reacts with an alcohol in the presence of a mineral acid; this is another fact that youshould have committed to memory. If Compound B is a carboxylic acid, Compound C must bean alcohol and Compound D must be an ester. The only alcohol with a molecular formula ofCH4O is methanol, and the only aromatic acid with a formula of C7H6O2 is benzoic acid, so theend product must be methyl benzoate, choice C. Choice B and Choice D are incorrect becausethey are not esters.

151. CWe have already established that Compound A—an aldehyde—reacts with silver oxide to

form Compound B—a carboxylic acid. This immediately eliminates choice D. Compound B notonly forms an ester that possesses aromatic protons, but dissolves in the aqueous layer upon theaddition of sodium hydroxide, so it must be benzoic acid—choice C. Choice B would be a pos-sibility if Compound D did not possess aromatic protons. However, the source of the aromaticprotons must come from Compound B since Compound C is not aromatic. Therefore, Com-pound B cannot contain a cyclohexane ring. In addition, the molecular formula of choice B isC7H12O2, which doesn’t match the molecular formula of Compound B, so choice B is wrong.Choice A is wrong for the same reasons as choice B; it does not contain any aromatic protonsand its molecular formula does not match that of Compound B.

152. AIn this reaction, Compound V, a carboxylic acid, is converted to Compound VII, an ester. So,

why is a mineral acid needed for this reaction to occur? To answer this, we need to briefly reviewthe process of esterification. Like so many reactions of carboxylic acids and their derivatives,esterification occurs through a mechanism known as nucleophilic acyl substitution. Remember,a nucleophile is a species that loves a positive charge—like the oxygen in an alcohol, forinstance. In esterification, it is the oxygen of the alcohol that attacks and adds to the positivelypolarized carbonyl carbon of the carboxylic acid to form a tetrahedral intermediate. Water is thenlost from the tetrahedral intermediate, resulting in the formation of an ester. Mineral acid isadded to the reaction vessel because it can donate protons that attach to the carbonyl oxygen ofthe carboxylic acid, thus encouraging electron density to be pulled away from the carbonyl

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carbon. Pulling electron density away from the carbon further increases its positive charge, so itis more susceptible to attack by the nucleophile. Therefore, choice A is the correct response.

Choice B is wrong because the acid donates a proton to the carbonyl oxygen, not thehydroxyl group. The hydroxyl group does eventually become protonated, but the proton isdonated by the attacking alcohol. The protonated hydroxyl group is then expelled from the tetra-hedral intermediate in the form of water. Choice C is wrong because an acid does not donatehydroxide ions. A species that donates hydroxide ions is a base. Finally, choice D is wrongbecause the catalyst does not accept hydroxide ions—it donates protons. Again, the correctanswer is choice A.

153. BYou need to be familiar with the reactions in Figure 1 to answer this question. One method

used to synthesize a carboxylic acid is the reaction of a Grignard reagent (Compound IV) withcarbon dioxide, followed by hydrolysis with aqueous acid. Grignard reagents are prepared bythe reaction of an organic halide (Compound III) with magnesium metal in the presence of ether.When carbon dioxide is bubbled through an ether solution containing the Grignard reagent, thealkyl portion of the Grignard reagent, which is negatively polarized, can add to carbon dioxideto generate the magnesium salt of the carboxylic acid, which has the general formula ofRCOO–MgX+. Upon treatment with aqueous acid, the carboxylic acid is generated. An impor-tant point to note in this reaction is that the carbon chain always extends by one. Therefore, ifthe alkyl group in the Grignard reagent is a butyl group, the resulting carboxylic acid will pos-sess 5 carbons. Therefore, choice B is the correct response since butylmagnesium chloride willform pentanoic acid, not butanoic acid. Propylmagnesium chloride, on the other hand, will formbutanoic acid, so choice D can be discarded.

Another common method used to synthesize carboxylic acids is the oxidation of primaryalcohols with potassium permanganate, shown in the conversion of Compound I to CompoundV. Butanol is a primary alcohol and will, upon reaction with potassium permanganate followedby acidification, form butanoic acid. Therefore, choice A can also be eliminated.

Carboxylic acid salts are formed by the addition of a base, such as sodium hydroxide, to thecarboxylic acid. This is shown in the conversion of Compound V (a carboxylic acid) to Com-pound VI (its corresponding sodium salt). Similarly, the addition of acid to the sodium saltregenerates the carboxylic acid. Therefore, the reaction of sodium butanoate with aqueous min-eral acid will result in the formation of butanoic acid, and choice C can be eliminated.

154. DIn this question, you are being asked to compare the boiling point of a carboxylic acid (Com-

pound V) to its corresponding alcohol (Compound I). Alcohols and carboxylic acids both havehigher boiling points than their corresponding alkanes because they possess a proton attached toan electronegative oxygen. This allows the molecules to undergo intermolecular hydrogen bond-ing, i.e., bonding with each other. However, carboxylic acids have higher boiling points thantheir corresponding alcohols since they can undergo intermolecular hydrogen bonding to agreater degree; this is possible because carboxylic acids possess a hydroxyl group and a car-bonyl group, so they are held together by two hydrogen bonds rather than one. Choice D, there-fore, is the correct answer.

Choice B is wrong. Compound I will undergo hydrogen bonding to a lesser degree becauseit can only form one hydrogen bond to another molecule. Choice A and choice C are wrong

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because intramolecular hydrogen bonding, which is the bonding that occurs within a molecule,has nothing to do with the boiling point of a molecule. In addition, Compound V and CompoundI will not undergo intramolecular hydrogen bonding in solution.

Passage IV (Questions 155-159)

155. CThe cerebellum is a region of the brain that sits below the rearmost portion of the cerebral

cortex. The cerebellum continuously processes information received from the motor cortex andperipheral position sensors to allow unconscious coordinated movement of skeletal muscle.Damage to the cerebellum would therefore impair fine motor coordination, so choice C is cor-rect. Choice A is incorrect because visual image processing is a function of the occipital region,the rearmost region of the cerebral cortex. Choice B is incorrect because speech is a function ofthe region of the brain known as Broca’s Area, which is part of the cerebral cortex. Choice D iswrong because respiration is governed by the part of the brainstem known as the medulla.

156. D Hypothesis 1 is clearly about the immune system. According to Hypothesis 1, the demyeli-

nation associated with MS is the result of an autoimmune response. Normally, the body recog-nizes its own tissues through a system of cell markers and doesn’t damage them. Occasionally,however, the immune system turns on the body’s own tissues and begins to destroy them as ifthey were foreign substances. In this case, macrophages present myelin as a foreign antigen tothe body’s T cells and B cells, resulting in the destruction of oligodendrocytes. Therefore, choiceB and choice C can be ruled out. Hypothesis 2 involves lymphocytes, which are another immunesystem cell. Lymphocytes are cells that recognize one specific antigen and react to it by pro-ducing immunoglobulins, or antibodies. Lymphokines are chemicals released by sensitized T-cells that help turn on and off the immune response, as well as direct cells from other parts ofthe body to travel to the site of an infection. In Hypothesis 2, an unknown signal causes chemi-cals that damage the oligodendrocytes to be released by the lymphocytes. Based on that infor-mation alone, it would be impossible to determine which of the two is the dominant mechanismin MS, since both hypotheses involve the immune system. Thus, choice A is incorrect and choiceD is correct.

157. BSince the passage states that the axons of demyelinated nerve cells sustain little damage, a

decreased ability to regenerate axons would NOT explain findings of reduced nervous systemfunction in MS patients. Thus, choice B is correct. Myelin acts in a similar way to the insulationon an electrical wire - its main task is to speed up the propagation of signals along the axon. So,choice A is incorrect. Signals moving down the axon jump from node to node in a process calledsaltatory conduction. For this reason, cells with myelin sheaths conduct much faster than thosewithout. A cell that loses its myelin will have slow conduction or lose conduction altogether. Forinstance, in MS, signals moving between the brain and the leg will arrive at the wrong time whenmoving through the demyelinated regions in the brain. So, choice C is incorrect. Choice D isincorrect because according to the passage, the nerve cells of the central nervous system aredemyelinated with MS.

158. AOne of the hallmarks of Hypothesis 2 is the proposal that lymphocytes release chemical

agents that damage the outer layers of the oligodendrocytes first. If you remember cell anatomy,the interface between the axon and myelin sheath is the innermost part of the oligodendrocyte.

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If choice A were true, then Hypothesis 2 would be wrong. Therefore, choice A is correct. ChoiceB actually supports Hypothesis 2, since the passage states that the triggering factor may be aviral infection somewhere else in the body. It may take 15 to 20 years before the damage pro-gresses to the point where one can observe the symptoms. Choice C is outside the scope of thehypothesis presented here. Although T cells and B cells can release lymphocyte-stimulating sub-stances, why they do, or even if it is important to the initiation of the process, is not mentionedin Hypothesis 2. So, choice C neither supports nor contradicts the hypothesis. Choice D is a truestatement, although you don’t know if this is important to the mechanism proposed in thehypothesis. Since it doesn’t contradict Hypothesis 2, choice D is also incorrect.

159. DThe knee jerk reflex does not involve the brain. It is a very simple nervous loop that goes

from the patellar tendon to the spinal cord and back again with only one synapse in the spinalcord. Simultaneously, the opposing muscle is inhibited to prevent resisting the reflex motion.This is accomplished with only two synapses in the spinal cord. Therefore, plaques limited tothe brain should NOT have any effect on the reflex motions such as the knee jerk. Thus, choiceD is correct. Eye muscle control could be affected if plaques developed along the pathways thatconnect eye motion control areas in the brain stem with the eye muscles. Therefore, choice A isincorrect. Choice B is incorrect because plaques in the regions of the cortex governing reason-ing would lead to deficits in these abilities. Although often a late developing symptom, MS suf-ferers have cognitive difficulty when the cortex is involved. Choice C is incorrect because senseof touch is handled by the neurons that travel from the sensory receptors in the arm through thespinal cord to the sensory area of the cortex. A lesion in the peripheral or central system couldlead to loss of this ability.

Discrete Questions

160. BEach molecule contains two chiral centers: the two carbons attached to the methyl groups.

Since these molecules do not possess an internal plane of symmetry, and they each containtwo chiral centers, there will be 22 or 4 different stereoisomers of this molecule. The fourstereoisomers are made up of two pairs of enantiomers, and the molecules in this question areone of those pairs. An enantiomer is a molecule that is nonsuperimposable on its mirrorimage. You can see that the molecules are mirror images of each other, yet they cannot besuperimposed on each other, hence they are enantiomers.

Diastereomers are stereoisomers that are not enantiomers. A feature of enantiomers is thatthey have an opposite configuration at every chiral carbon, whereas diastereomers have the sameconfiguration at one or more carbons. The configuration about both chiral carbons in thesemolecules is opposite, so they cannot be classed as diastereomers, and choice C can be elimi-nated. Choice A is incorrect since both of these molecules are cis isomers; both methyl groupsare arranged equatorially on the ring, and therefore lie on the ‘same side’ of the ring. If one ofthe methyl groups in either molecule were arranged axially, then this molecule would be thetrans isomer of the molecule in which the two methyl groups were arranged equatorially.Finally, choice D is incorrect since meso compounds are defined as molecules that possess chi-ral centers, but also possess an internal plane of symmetry, so they are superimposable on theirmirror images, and are optically inactive. Due to the puckered conformation of the molecules inquestion, we cannot draw a plane of symmetry through the molecule, so they cannot be classedas meso compounds.

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161. DEven if you didn’t have a clue about the differences between fetal hemoglobin and maternal

(adult) hemoglobin, you should have realized that choice D is the only one that does NOT plau-sibly explain how a fetus gets enough oxygen. While it is true that fetal lungs are not functionaluntil birth, this fact does not explain how fetal blood gets enough oxygen to keep the fetus alive.

A fetus does not breathe on its own, getting a fresh supply of oxygen from its lungs; rather, itobtains oxygen from its mother, across the placenta. The placenta serves as the fetus’ lungs and isits only link to the external world. Oxygen dissolved in the placental sinuses diffuses into the fetalblood because the maternal blood has a mean PO2 of 50 mmHg (for comparison, blood leaving theaorta has a mean PO2 of 95 mmHg) while the mean PO2 in the fetal blood returning to the placentais 30 mmHg. Given the low PO2 of oxygen entering the placental sinuses, it is a wonder that thefetus doesn’t suffer from hypoxia. What prevents this from happening is fetal hemoglobin, whichdiffers from maternal hemoglobin in that it has a greater affinity for oxygen. So, choice A can beeliminated. Thus, for a given partial pressure of oxygen, fetal hemoglobin will be able to carrymore oxygen than the maternal hemoglobin. Another way of saying this is that at low PO2s fetalblood will be more saturated with oxygen than will maternal blood. In fact, at low PO2s, fetalhemoglobin can carry between 20 to 30% more oxygen than can maternal hemoglobin. So, choiceB is incorrect. Furthermore, another way that fetal blood obtains enough oxygen is by having itsconcentration of hemoglobin 50% higher than that of maternal blood. This further enhances theability of fetal blood to transport oxygen. Thus, choice C is also incorrect.

162. CSN1 reactions initially involve the loss of a good leaving group, and the subsequent forma-

tion of a carbocation. Since all of these compounds are alkyl halides, they have to lose a bro-mide in order to form a carbocation. Since all the carbons attached to the bromine are tertiary,the compounds could all theoretically form tertiary carbocations. However, when the tertiarycarbon loses a bromide, and hence acquires a positive charge, its hybridization changes from sp3

to sp2. sp2 hybridized carbons have a planar geometry: this accounts for why nucleophiles canattack and add to the tertiary carbon from above and below the plane of the molecule. Thisallows for the formation of a racemic mixture of products. The ability to form a planar carbo-cation decreases as we go from Compound I to Compound III. This is because the bondsattached to the tertiary carbon become increasingly more rigid, hence they cannot orient them-selves in a planar arrangement. The molecule that will be least likely to form a planar carboca-tion is Compound III. Due to the rigid ring system, the bridgehead carbon that links one side ofthe ring to the other and the two carbons next to the tertiary carbon will be pulled downwards,and will not be able to form a planar arrangement around the tertiary carbon.

Steric hindrance is not a factor in SN1 reactions since the slowest step, and hence the rate ofSN1 reactions, is dependent upon the loss of a leaving group to form a carbocation, and does notinvolve the nucleophile. When the group attached to the leaving group sterically hinders aspecies involved in the slowest step (such as the nucleophile in SN2 reactions), only then cansteric factors be considered. Choice A can be eliminated. Choice B is incorrect because we can-not assume that these molecules differ in their reactivity. All of these compounds are tertiaryalkyl halides, so they are all susceptible to the same type of reaction. Choice D is incorrectbecause backside attack by a nucleophile occurs in SN2 reactions. Since this question comparesthe ability of the molecules to undergo SN1 reactions, the features of SN2 reactions do not haveto be considered, and choice D can be eliminated.

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163. CThis question is essentially testing your knowledge of the difference in ATP output between

aerobic respiration and anaerobic respiration. Aerobic respiration (glycolysis + Krebs cycle +electron transport) requires oxygen, and yields a net of 36 molecules of ATP per molecule ofglucose. Anaerobic respiration (glycolysis) occurs in the absence of oxygen and yields a net of2 molecules of ATP. Eukaryotic unicellular organisms such as a Paramecium obtain energy viaaerobic respiration, while E. coli are facultative anaerobes that can undergo aerobic or anaero-bic respiration. An E. coli inhabiting the human intestinal tract which contains no supply of oxy-gen would certainly undergo anaerobic respiration. Thus, from the degradation of one moleculeof glucose, a Paramecium would produce 18 times as much ATP than would an E. coli withinthe human intestinal tract.

164. BBecause of the incredible number of similarities between modern day mitochondria and bac-

teria, scientists have hypothesized that mitochondria evolved from primitive aerobic bacteriathat entered and established relationships with primitive anaerobic eukaryotes. The primitiveaerobic bacteria possessed enzymes and a transport system on their plasma membranes that gavethem the ability to respire aerobically, using oxygen as the final electron receptor in the synthe-sis of ATP. This relationship between aerobic bacteria and anaerobic eukaryotes gave the eukary-otes a new energy source, and thus the ability to reproduce at a faster rate, and gave the bacterianutrients and protection. Mitochondria are believed to be the descendants of these bacteriabecause they possess circular, non-histone bound DNA, bacteria-like rRNA, enzymes, and ribo-somes, and they reproduce by similar means. Mutually beneficial relationships such as the onedescribed above are referred to as being symbiotic, and this particular theory is referred to as theendosymbiotic hypothesis. Another example of a symbiotic relationship involving bacteria andeukaryotes is found in the plant world. There are certain modern-day plants that have nitrogen-fixing bacteria living symbiotically inside of their cells. Just knowing the meaning of the termsymbiotic would have allowed you to eliminate all of the wrong choices.

Choice A, the chemiosmotic hypothesis, refers to the mechanism by which ATP is formedfrom ADP in mitochondria (and chloroplasts). The oxidation of NADH releases protons andelectrons. The electrons release energy as they move from a higher to a lower energy state viaan electron transport system in the inner membrane. The energy is used to pump the protonsfrom the inner mitochondrial matrix into the intermembrane space of the mitochondrion via anATP synthetase complex. This creates both a pH gradient and an electric potential across theinner mitochondrial membrane, since the inner mitochondrial membrane is impermeable to pro-tons. When the concentration of protons reaches a certain critical mass, however, they flow backinto the matrix along their concentration gradient, through the ATP synthetase complex. Themovement of the protons creates a proton-motive force that drives ATP synthesis. Choice C, thecoevolutionary hypothesis, and choice D, the endocytotic hypothesis, are not even actualhypotheses. They are fabricated terms designed to tempt you with their scientific-soundingnames. Thus, choices A, C, and D, are all incorrect.

Passage V (Questions 165-169)

165. DAccording to the passage, the primary function of cortisol is to stimulate the formation of

glucose from the body’s stores of protein and fat during stressful periods. This implies that theprotein stores in the muscles of a patient with Cushing’s syndrome are going to be depleted,since an excess of cortisol results in increased protein catabolism (breakdown) and decreased

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protein synthesis. The observed muscle weakness is due to this increased protein catabolism,and so choice D is correct. Choice A is incorrect because the high blood glucose associated withCushing’s syndrome would not cause muscle weakness, although it might put a strain on thebeta-islet cells of the pancreas that produce insulin. Choice B is wrong because individuals withCushing’s syndrome can still respond to stress. Choice C is wrong because fluctuations in theconcentrations of CRH and ACTH wouldn’t cause muscle weakness. Furthermore, patients withCushing’s syndrome caused by an adrenal or ectopic tumor exhibit consistently low levels ofCRH and ACTH (see next explanation), while individuals with Cushing’s syndrome caused bya pituitary or hypothalamic abnormality will exhibit consistently high levels of CRH or ACTH,or both.

166. AAccording to the passage, cortisol secretion is regulated by a negative feedback mechanism

involving the hypothalamus and the pituitary. The hypothalamus secretes CRH, which stimu-lates the pituitary to secrete ACTH. ACTH then stimulates the adrenal glands to secrete cortisol.In turn, ACTH and CRH secretion is inhibited by a high concentration of cortisol in the blood-stream. The excessive cortisol secretion found in patients with Cushing’s syndrome caused byan adrenal tumor, then, would result in the suppression of both CRH and ACTH secretion. Thus,cortisol concentration in these patients would be high, while CRH and ACTH concentrationswould be low, and so choice A is correct. Choice B would only be correct if the patient hadCushing’s syndrome caused by overproduction of ACTH by the pituitary, while choice C wouldonly be correct if the Cushing’s syndrome were the result of overproduction of CRH by thehypothalamus. Choice D is incorrect because if cortisol were high because of an adrenal tumor,then CRH would most likely be low, as a result of the negative feedback effect that cortisol hason the hypothalamus.

167. BAccording to the question stem, cortisol decreases glucose utilization by inhibiting glucose

uptake at the cell level. This is one of the means by which cortisol raises blood glucose. There-fore, cortisol acts in direct opposition to insulin, which decreases blood glucose by promotingglucose entry into most cells of the body. A patient with Cushing’s syndrome has excess corti-sol in the bloodstream, which means that glucose uptake will be even more suppressed andblood glucose levels will be higher than in a normal individual. The normal balance betweencortisol and glucose in stabilizing blood glucose levels is upset. The end result of an excessiveand consistent hyperglycemia is a decreased sensitivity to insulin, along with the developmentof diabetes mellitus. Thus, choice B is correct, and choices A and D are wrong. Choice C isincorrect because the consistently elevated cortisol secretion seen in patients with Cushing’ssyndrome ensures that recovery from low blood glucose would never be an issue.

168. BIn addition to stimulating cortisol secretion, ACTH regulates the secretion of the adrenal

androgens. The most active androgen is testosterone, the hormone responsible for secondarymale characteristics, among other things. 80% of testosterone activity is derived from testes-secreted testosterone; the adrenal-produced testosterone accounts for the remaining 20%. InCushing’s syndrome caused by an excess of ACTH, the concentration of both cortisol and theadrenal androgens is abnormally high. In males, this excess adrenal androgen merely accentu-ates their existing secondary sex characteristics. However, in females with this type of Cushing’ssyndrome, the excess androgens result in the appearance of masculine traits, such as increasedfacial hair. Thus, choice B is correct.

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Choice A is incorrect because estrogen secretion is regulated by a complicated feedbackmechanism involving LH, FSH, and GnRH. Choice C is wrong because ACTH stimulates theadrenals, not the testes, to produce testosterone. Testes-produced testosterone, like estrogen, isunder the control of LH and FSH. Finally, even though cortisol and testosterone are producedby the same metabolic pathway that begins with cholesterol as its precursor, ACTH cannot con-vert cortisol into testosterone.

169. AThe dexamethasone suppression test is a diagnostic tool used by physicians to distinguish

patients with Cushing’s syndrome caused by a malfunctioning hypothalamic-pituitary axis frompatients with other types of Cushing’s syndrome. It is based on the ability of exogenous gluco-corticoids to suppress pituitary release of ACTH. When dexamethasone is injected into a healthyindividual, it exerts the same effect as a very high dose of cortisol: it suppresses the secretion ofCRH by the hypothalamus and the secretion of ACTH by the pituitary, resulting in a decrease incortisol secretion. When dexamethasone is administered to a patient with Cushing’s syndromecaused by hypothalamic-pituitary oversecretion of ACTH, ACTH secretion will decrease, as willcortisol secretion. By virtue of its cortisol-likeness and its extreme suppressive activity, dexam-ethasone can induce the negative feedback mechanism that is malfunctioning in patients withCushing’s syndrome caused by excessive CRH/ACTH. However, a similar effect will NOT beseen in patients with other types of Cushing’s syndrome because adding exogenous cortisol(dexamethasone) to an already elevated plasma cortisol not due to excessive ACTH will notdecrease cortisol output.

Since the patient in the question stem exhibits a decrease in cortisol levels after injectionwith dexamethasone, the Cushing’s syndrome is most likely the result of pituitary-dependentoversecretion of ACTH, and so choice A is correct. Choice B is wrong because an undersecre-tion of CRH could never lead to the development of Cushing’s syndrome. Choices C and D areincorrect because Cushing’s syndrome caused by adrenal-dependent oversecretion of cortisol orover-ingestion of exogenous steroids would not respond to a dexamethasone suppression testwith a decrease in cortisol.

Passage VI (Questions 170–174)

Passage VI is about lithium aluminum hydride and sodium borohydride, which are bothreducing agents. LiAlH4 and NaBH4 are very important carbonyl reducing agents, so you prob-ably encountered these when studying aldehydes and ketones. Figure 1 shows the mechanismby which these reagents reduce carbonyl compounds.

After the introductory paragraph, the passage gets a little more detailed and tells you aboutthe selectivity of these reagents toward electrophilically polarized carbons other than carbonylcarbons. In general, sodium borohydride is more selective than lithium aluminum hydride,except in the reduction of conjugated carbon-carbon double bonds where the opposite trendoccurs. Whenever you see a statement about a trend, always take note: you may be asked to pre-dict the outcome of a reaction according to this trend. Always look closely at the reactions youare given; they may reiterate what is said in the passage and help you understand the passagecontent.

170. BThis is a reading comprehension question. The passage states that in conjugated systems,

lithium aluminum hydride is the more selective reducing agent: it will only reduce the carbonyl

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bond. This is demonstrated in Reactions 1 and 2—lithium aluminum hydride will only reducethe carbonyl bond in cinnamaldehyde, whereas sodium borohydride will indiscriminately reducethe carbonyl bond and the carbon-carbon double bond (although some unsaturated product isobtained). To arrive at choice B, you need to apply both the information from the text and theinformation from Reactions 1 and 2 to the reaction in the question.

Choice A is incorrect because this is contradictory to the information in the passage. Lithiumaluminum hydride and sodium borohydride both reduce aldehyde and ketone functionalities toalcohols, so the carbonyl group will not stay intact, as choice A insinuates. In the molecule inchoice C, the carbon-carbon double bond and the carbonyl group are not reduced, thus choice Cis incorrect. In addition, the ether functionality would not be converted to an alcohol group uponthe addition of lithium aluminum hydride. Finally, choice D is incorrect since the reaction ofketone functionalities with LiAlH4 always results in their conversion to alcohols, not theirremoval.

171. DTo answer this question, you have to extract information from Figure 1, and use outside

knowledge. Figure 1 shows the mechanism of attack by LiAlH4 and NaBH4 upon aldehydes andketones. In the first step, a hydride ion—H–—attacks and adds to the electrophilic carbon toform a tetrahedral alkoxide intermediate. Since the carbonyl carbon is sp2 hybridized and has aplanar geometry, the hydride ion can attack from above or below the plane of the carbonyl bond.The alkoxide ion that is formed when 1-deuteropropanal is attacked by the hydride ion is chi-ral—the central carbon is attached to four different groups. Since the hydride ion can attack fromabove or below the plane, 50% of the alkoxide ion formed will be an R-enantiomer and 50% ofthe product formed will be an S-enantiomer. An equal mixture of R and S enantiomers consti-tutes a racemic mixture. Therefore, choice D is the correct answer.

Choice A and choice B would only be correct if the attack of the hydride ion upon the alde-hyde was stereospecific; that is, the aldehyde yields one particular isomer. To be a stereospecificreaction, the hydride ion would have to attack the aldehyde from one side of the plane of the car-bonyl bond. Since there is nothing to prevent the hydride from attacking from either side of thecarbonyl bond, a specific enantiomer will not form, and choices A and B can be eliminated.Choice C is incorrect because 1-deutero-1-propanol contains a chiral carbon; carbon 1 isattached to a hydroxyl group, a hydrogen atom, a deuterium atom, and an ethyl group. There-fore, this molecule is chiral and choice C can be discarded.

172. CYou are told in the first sentence of the passage that “two of the most common carbonyl

reducing agents in use today are lithium aluminum hydride and sodium borohydride.” Therefore,the answer must include roman numerals I and III, eliminating choices A and B.

By the way, this question may be slightly confusing because 9-fluorenone is a conjugatedsystem; that is, a system in which double or triple bonds are separated by alternating singlebonds. Therefore, you may have been considering the action of the reducing agents upon theconjugated carbon-carbon double bonds. This is where your knowledge of aromatics comes intoplay. Of course, fluorenone is a conjugated system, but the pi electrons in the ring system delo-calize to form a stable system. As a result, the carbons in the ring do not behave like isolateddouble bonds, and this accounts for why benzene is more susceptible to substitution, not addi-tion. Because this ring system is so stable, it will not be susceptible to reduction by lithium alu-

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minum hydride or sodium borohydride. This is confirmed by Reactions 1 and 2: lithium alu-minum hydride and sodium borohydride do not reduce the aromatic ring of cinnamaldehyde.

Finally, we have to consider statement II—a mixture of chlorine and iron (III) chloride.Again, from your knowledge of aromatics, you should know that a mixture of chlorine and iron(III) chloride will result in halogenation of the aromatic ring. Therefore, 9-fluorenone will notform 9-hydroxyfluorene when reacted with chlorine and iron (III) chloride, so roman numeralIII and choice D can be discarded. This leaves us with choice C, the correct response.

173. BThis question is pure reading comprehension—if you understand what is going on in the pas-

sage and in Reactions 1 and 2, you will probably be able to answer this question. Again, the partof the passage that you need to focus on is the information about conjugated carbon-carbonbonds and the ability of LiAlH4 and NaBH4 to reduce them. Choices A, C, and D are all conju-gated molecules. Sodium borohydride, as shown in Reaction 2, will reduce conjugated carbon-carbon double bonds (although it will not reduce them 100%—notice that Reaction 2 shows theformation of cinnamyl alcohol, an unsaturated alcohol). Since choice B is the only choice thatdoes not contain a carbon-carbon double bond in conjugation with the carbonyl bond, the reduc-tion of its carbonyl group by sodium borohydride will result in the formation of a fully saturatedmolecule—choice B is the correct answer.

In choices C and D, some of the sodium borohydride will reduce the carbonyl bond and thecarbon-carbon double bond, and some of the sodium borohydride will reduce the carbonyl bondonly. Therefore, in these molecules, there will be a mixture of unsaturated and saturated alco-hol. In choice A, the presence of the two phenyl groups ensures that there will always be anunsaturated product—even if the carbonyl bond and the adjacent carbon-carbon double bond isreduced. Therefore, the order of increasing saturation will be choice A, then choices C and D,followed by choice B, the correct response.

174. DIn order to form 1-butanol, hydrogen has to be added to the carbon-carbon double bond in

cis-2-butenal, and the carbonyl group has to be reduced. When cis-2-butenal reacts with waterin the presence of acid, water adds across the carbon-carbon double bond to form an alcohol. Inaddition, water will add to the carbonyl group to form a 1,1-diol, or gem-diol. The resultingmolecule will be a triol; therefore, choice D is the correct response.

From your knowledge of alkene and aldehyde reactions, you should know that hydrogen—in the presence of a catalyst such as nickel, palladium, or platinum—will add to the double bondof alkenes to form a carbon-carbon single bond, and will reduce the aldehyde group to an alco-hol. Therefore, choices B and C can be eliminated since the reaction of cis-2-butenal with hydro-gen over palladium or nickel will result in the formation of 1-butanol. We have established inthe passage that sodium borohydride will indiscriminately reduce any conjugated bond, so 1-butanol also forms when cis-2-butenal reacts with NaBH4; therefore, choice A can be elimi-nated.

Passage VII (Questions 175-181)

175. AAccording to the passage, erythropoietin production is stimulated by a deficiency in oxygen

delivery to the tissues. The decrease in oxygen delivery can be the result of conditions such as

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anemia, hemorrhaging, high altitudes, or a genetic disease such as hereditary spherocytosis.Anemias, hereditary spherocytosis, and hemorrhage are all characterized by a decreased RBCnumber; fewer RBCs means that less oxygen is delivered to the tissues. High altitude, on theother hand, does not affect RBC number. Oxygen delivery is decreased at high altitudes becausethe partial pressure of oxygen in the air is lower at higher altitudes—the higher up you go, theless oxygen there is. Thus, at high altitudes, erythropoietin production is increased, even thoughRBC number is not decreased. The body acclimates to high altitudes by increasing its produc-tion of RBCs and hemoglobin. So, choice A is correct. Choices B, C, and D are incorrectbecause although all three statements are true, they support the conclusion that erythropoietinproduction is stimulated only by a decreased RBC number.

176. DAccording the passage, human bone marrow contains pluripotential hemopoietic stem cells,

which are the precursor cells of RBCs, white blood cells, and platelets. According to the ques-tion stem, aplastic anemia results in a hypocellular bone marrow. The prefix “hypo” meansbelow normal; therefore, a hypocellular bone marrow sample is one that contains fewer cellsthan normal bone marrow, including fewer pluripotential hemopoietic stem cells. Therefore, apatient with aplastic anemia will have a decreased number of platelets, white blood cells, andRBCs. So, choice D is correct and choice C is incorrect. Choice A is wrong because aplastic ane-mia would not increase the number of reticulocytes. Choice B is incorrect because iron defi-ciency is not a result of anemia, it is a cause of it. Furthermore, although iron deficiency canresult in certain types of anemia, it does not cause aplastic anemia, which affects the bone mar-row, not iron concentration.

177. BSickle cell anemia is an autosomal recessive disease that causes RBCs to assume a sickle-like

shape rather than their normal biconcave disk shape. The sickle-shaped RBCs are very fragile andsusceptible to destruction, especially when traveling through tight spots, such as capillaries. Thus,patients with sickle cell anemia have fewer RBCs than normal. According to the passage, adecreased RBC count means that both the concentration of hemoglobin and the percentage ofblood constituted by RBCs are below normal. Furthermore, there is an increase in the number ofcirculating reticulocytes, or immature RBCs, in patients with decreased RBCs. Thus, Romannumerals I and II must be a part of the correct answer, and so choices A and C can be eliminated.Unlike aplastic anemia, sickle cell anemia affects only the production of RBCs. The gene forsickle cell anemia results in the production of an abnormal hemoglobin, HbS. Thus, a patient withsickle cell anemia is not expected to have any white blood cell abnormalities. So, Roman numeralIII can be eliminated, which means that choice D is wrong and choice B is correct.

178. CAnemia is defined as a deficiency in the number of RBCs. Fewer RBCs means that there will

be a decrease in the overall oxygen-carrying capacity of the blood, since RBCs are the oxygen-carrying component of the body. A decrease in RBC number necessarily means that there willbe a decrease in the amount of oxygen getting delivered to the tissues. Thus, choice C is correct,since an anemic patient would not have an increased oxygen-carrying capacity. Choices A, B,and D are incorrect because they are all symptoms that are expected in an anemic patient. ChoiceA is incorrect because a decrease in RBC number will result in an increase in erythropoietinsecretion, as the body tries to compensate for the loss of RBCs by stepping up their synthesis.Choice B is wrong because blood viscosity will be decreased due to the decrease in the numberof RBCs found in the blood. Choice D is wrong, because fewer RBCs means that there is lesshemoglobin available to carry oxygen.

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179. CAn autosomal dominant disease is one that is expressed whether the individual carries one

or two copies of the gene. According to the question stem, the mother of the man with heredi-tary spherocytosis was unaffected. This means that her genotype was homozygous recessive,which in this case, is the normal genotype. If she were heterozygous, she would have had thedisease herself, since the disease is inherited as an autosomal dominant. The fact that the man’smother was homozygous recessive means that the man must have inherited the disease from hisfather, and that the man’s genotype is heterozygous. The man therefore has one copy of thehereditary spherocytosis gene. If this man has a child with a normal woman (i.e., she’s homozy-gous recessive), there is a 50% chance that his sperm will have the hereditary spherocytosisgene. And since the gene is dominant, there is a 50% chance that their child will inherit the dis-ease, and so choice C is correct.

180. DAccording to the passage, erythropoietin is a glycoprotein hormone produced primarily by

the kidneys. It is the primary factor controlling RBC production. Since RBCs are produced inthe stem cells of bone marrow, it can be inferred that these cells are the target cells of erythro-poietin. While steroid hormones diffuse across the cell membranes and directly into the nucleiof their target cells, effecting change at the nuclear level, peptide hormones alter cellular activ-ity extracellularly. Typically, peptide hormones bind to extracellular receptors located on theexterior of the cell membranes of their target cells. The binding of hormone to receptor initiatesa series of reactions inside the cell cytoplasm that accomplish the job for the hormone. Sinceerythropoietin is a glycoprotein hormone, it acts on its target cells—the stem cells—by bindingto erythropoietin-specific receptors found on the cell membranes of the stem cells. Therefore,choice D is correct and choice C is wrong. Choices A and B are incorrect because the RBCsthemselves are the end product of erythropoietin’s action. By binding to the receptors on stemcells, erythropoietin stimulates a series of reactions that results in the differentiation of the stemcell into an immature RBC (reticulocyte).

181. ARBCs are not typical of the rest of the cells found in the body. They are produced in the bone

marrow, and during their maturation, they lose their nuclei, mitochondria, and all membranousorganelles. RBCs have an average life span of 120 days, after which they are removed from theblood by the spleen. Since RBCs do not have nuclei, they cannot divide. And since they do nothave mitochondria, they cannot produce ATP via aerobic respiration (i.e., the Krebs cycle andoxidative phosphorylation). However, RBCs, like all cells, need ATP. And even though theydon’t have mitochondria, they can still synthesize ATP anaerobically, via glycolysis. Glycolysisoccurs in the cytoplasm and results in the production of 2 ATP molecules for every one moleculeof glucose. Thus, choice A is correct, and choices B and C are incorrect. Choice D is wrongbecause according to the passage, platelets, white blood cells, and red blood cells are all derivedfrom the same type of stem cell—pluripotential hemopoietic stem cells. The type of cell that astem cell differentiates into is dependent on the actions of specific growth inducers and differ-entiation inducers. Erythropoietin is the main inducer responsible for the differentiation of astem cell into a RBC.

Passage VIII (Questions 182-186)

182. BThe results of the researcher’s experiment reveal that the neuronal receptors have almost

twice the affinity for THC that white blood cell (WBC) receptors have for THC. According to

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Theory B, this is because the WBC contains a membrane protein near the THC receptor thatinterferes with the receptor’s binding of THC. You’re not told this in the passage, but perhapsthis other WBC membrane protein is a receptor that also binds THC, thereby decreasing the fre-quency with which THC binds to its receptor. This is essentially the definition of noncompeti-tive inhibition. Noncompetitive inhibition results from interference at a site other than the activesite, which in this case, is the THC receptor. So, if Theory B were correct, choice B would bestaccount for the experimental results. Choice A is wrong because competitive inhibition occurswhen a substance directly competes with the substrate for access to the active site; e.g., if a sub-stance competes with THC for access to the THC receptor. Choice C is incorrect because it sup-ports Theory A. Choice D is wrong because even if it were true, it would not affect the affinitybetween THC and its receptor.

183. DThe results of the researcher’s experiment reveal that the neuronal receptors have almost

twice the affinity that white blood cell (WBC) receptors have for THC. According to Theory A,this is because the two types of receptors differ in their affinity for THC. Choice D is correctbecause it essentially describes why two receptors for the same substance would have differentaffinities for that substance. If THC has a better fit with the neuronal THC receptor than withthe WBC THC receptor, then the experimental results make perfect sense. Choice A is incorrectsince the experiment was conducted in vitro, making any in vivo variables irrelevant. Choice Bis wrong because cell size differentials would not make any difference at the receptor level,which is the level at which affinity is determined. Choice C, which has to do with the possiblemechanism of action of THC, is wrong because these effects happen after binding occurs. There-fore, they cannot affect receptor binding.

184. AAccording to the question stem, THC induces the synthesis of a neuropolypeptide called

ASF. Inducing protein synthesis involves the transcription of the DNA sequence that codes forthe peptide into a strand of mRNA, followed by the translation of the mRNA into a peptide andthe modification of the peptide into the finished product. Transcription occurs in the nucleus,while translation occurs in the cytoplasm. Thus, choice A is correct. Choices B, C, and D areincorrect, since mitochondria, lysosomes, and cilia are subcellular organelles that are notdirectly involved in protein synthesis.

185. CTheory A, which purports that the neuronal THC receptors have a different affinity for THC

than WBC THC receptors, would be favored by the discovery that there are significant structuraldifferences between the two types of receptors. A known example of this phenomenon are thedifferent dopamine receptors in the human body. The D1, D2, and D3 dopamine receptors havedifferent binding affinities for dopamine. So, choice C is correct. Choice A is wrong because itwould account for differences in THC binding affinities among different types of white bloodcells, but not for differences between white blood cells as a group and neurons. Choice B isincorrect because although it would be consistent with the experimental results, it does notaccount for the results, nor does it favor one theory over the other. Choice D is wrong becauseit doesn’t support one theory over the other, and because it requires a leap from in vitro testingto in vivo functional differences—an area which the passages does not even address.

186. AIf Theory A is assumed correct, then it can most likely be concluded that the DNA sequences

that code for the two types of THC receptors are different. The differing affinities for THC can

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most likely be attributed to a difference in the physical structure of the receptors themselves, anda structural difference can be attributed to a difference in the DNA that codes for the receptors.The THC receptors are proteins, and protein synthesis begins with the transcription of a specificsequence of DNA into a corresponding strand of mRNA. The mRNA is than translated into apeptide, which is later modified into its final form, which in this case, is a THC receptor. If theDNA sequences coding for the two THC receptors were identical, then the receptors themselveswould be identical and would have the same affinity for THC. Given that the experimentalresults contradict this, and assuming that Theory A is correct, choice B must be wrong andchoice A must be correct. Choice C is incorrect because it would only help to assess the recep-tor’s mechanism of action; it would have no bearing on the validity of Theory A over Theory B.Choice D is completely out in left field. By definition, the control cultures had the lowest con-centration of either kind of THC receptors - ideally, they had few if any. The fact that the con-trol cultures exhibited the lowest binding affinity for THC supports this. So, choice D is wrong.

Discrete Questions

187. DWithout knowing anything about the skeletal structure of these four animals, you should

have been able to get the right answer. Basically, birds are reptiles specialized for flight. Theirbones are hollow, and their bodies contain air sacs. Even the female reproductive system hasbeen reduced to a single ovary in female birds, thereby decreasing their weight even more.Therefore, a sparrow would be expected to have the least dense bones in a group consisting ofa sparrow, an iguana, a squirrel, and a whale. Again, choice D is correct.

Even though they are small, both the iguana and the squirrel do not have hollow bones, sincethey have to support their bodies on land. Active terrestrial lifestyles require the skeletal supportthat only solid bones can provide. So, choices A and B can be eliminated. Furthermore, eventhough whales live in the ocean and have less stress on their bones than non-aquatic animals,whales are very large mammals. Therefore, they possess large, solid bones that are most defi-nitely denser than the hollow bones of a sparrow, so choice C is also incorrect.

188. BIn order to answer this question, you need to understand the hormonal intricacies of the men-

strual cycle. During the first half of the menstrual cycle (the follicular phase), the hypothalamussecretes gonadotropin-releasing hormone (GnRH), which stimulates the anterior pituitary tosecrete both luteinizing hormone (LH) and follicle-stimulating hormone (FSH). LH and FSH acton the ovary to promote follicle maturation and estrogen production, which in turn, further pro-mote LH secretion. After ovulation, which occurs midway through the cycle, LH and FSH pro-mote the formation and maintenance of the corpus luteum from the ruptured follicle. During theluteal phase, the corpus luteum secretes estrogen and progesterone, which together develop theendometrial lining. Towards the end of this phase, the high levels of estrogen and progesteroneinhibit GnRH secretion via a negative feedback mechanism. As a result, LH and FSH are inhib-ited, and if fertilization does not occur, the corpus luteum atrophies. Consequently, estrogen andprogesterone levels drop, the endometrial lining degrades, and menstruation occurs. The cyclethen begins again because GnRH is no longer inhibited.

Oral contraceptives suppress ovulation by introducing a constant level of estrogens and pro-gesterone into a woman’s body. This inhibits GnRH secretion via the negative feedback mech-anism. This means that LH and FSH, which induce ovulation, are not produced, so ovulationcannot occur. Thus, choice B is correct. Choice C is incorrect because menstruation is induced

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by a drop in estrogen and progesterone levels. Oral contraceptives are only administered forthree-fourths of the cycle, so during the last part of the cycle, the sudden drop in estrogen andprogesterone induces menstruation. (FYI: Oral contraceptives come in either 28-day packets, or21-day packets. The last 7 pills of the 28-day packets are placebos.) Choice A is incorrectbecause the high levels of estrogen and progesterone in oral contraceptives inhibit LH and FSHby inhibiting, not promoting, secretion of GnRH. Choice D is incorrect because HCG is not pro-duced by high levels of estrogen and progesterone. HCG is produced only by the developingembryo and placenta, so it would only be present if fertilization had occurred. HCG is respon-sible for maintaining progesterone and estrogen secretion by the corpus luteum during the firsttrimester of pregnancy. After the first trimester, the placenta secretes estrogen and progesterone;this means that GnRH is inhibited throughout pregnancy and ovulation cannot occur.

189. BWhen alcohols are heated in the presence of mineral acid, unimolecular elimination (E1)

occurs. E1 reactions proceed through the formation of a carbocation. The first step of the reac-tion involves protonation of the –OH group by the acid to form an –OH2

+ group; this group is agood leaving group, and its loss from the molecule results in the formation of a secondary car-bocation:

However, the secondary carbocation can undergo intramolecular rearrangement to form amore stable tertiary carbocation:

The loss of a proton adjacent to the positive charge results in the formation of 2,3-dimethyl-butene, choice B. Choice A is incorrect since this is the minor product in the reaction. Thisalkene forms when a proton adjacent to the positive charge is lost from the secondary carboca-tion, not the tertiary carbocation. Recall that a tertiary carbocation is more stable than a sec-ondary carbocation. Therefore, more product will originate from the tertiary carbocation thanthe secondary carbocation. Choice C is incorrect since a ketone will only be produced when thealcohol is treated with an oxidizing agent. Heating the alcohol in the presence of mineral acid,as is the case here, will not result in oxidation, and choice C can be eliminated. Finally, choice

CH3 C

CH3

CH3

CH CH3+

+CH3CH

CH3

CCH3

CH3

CH3 C

CH3

CH3

CH CH3

OH

H+

CH3 C

CH3

CH3

CH CH3

OH2+

CH3 C

CH3

CH3

CH CH3+

–H2O

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D can be eliminated since this molecule still possesses a hydroxyl group, and we have alreadyestablished that the hydroxyl group is protonated and then lost from the molecule.

190. DAlthough this reaction may look complex, it is not; all the alcohol functionalities in the reac-

tant molecule are converted to ester functionalities. Esterification occurs because the oxygens inthe alcohol functionalities act as a nucleophiles, by attacking and adding to one of the ester car-bons in the acid anhydride (CH3CO)2O. When the alcohol attacks and adds to the acid anhy-dride, a tetrahedral intermediate is formed; the loss of CH3CO2

– ion then results in formation ofthe ester functionality. Since an acyl group is lost from the anhydride, the reaction is an exam-ple of nucleophilic acyl substitution, not nucleophilic substitution; therefore, choice B can beeliminated. Choice A is incorrect since the alcohol functionalities act as nucleophiles, and sub-stitute for an acyl group in the acid anhydride. Electrophilic addition is usually observed inspecies that contain multiple bonds. Finally, choice C is incorrect since this reaction is actuallyan example of acetal formation. If we look at the anomeric carbon in the reactant molecule, wecan see that it is a hemiacetal group. By adding the acid anhydride in the presence of pyridine,this hemiacetal group is actually converted to an actual group. Therefore, choice C can be elim-inated, leaving choice D as the correct response.

191. CSkeletal muscle fibers are comprised of smaller units of contraction known as sarcomeres.

Each sarcomere is made up of thick and thin filaments. The thick filaments consist of myosinprotein molecules wrapped around each other, with the head of each individual myosin moleculeturning outward from the filament. The thin filaments are composed of actin filaments, whichare associated with two other proteins—tropomyosin and troponin. The actin filaments havemyosin-binding sites. When the muscle is in a resting state, the tropomyosin strands cover themyosin-binding sites on the actin. During contraction, Ca2+ is released into the cytoplasm of themuscle cell and binds to the troponin molecules on the thin filaments. This causes thetropomyosin molecules to shift, exposing the myosin-binding sites on the actin. The myosinheads now bind to the actin, and the sarcomere contracts. Thus, tropomyosin physically inhibitsmuscle contraction until Ca2+ binds to troponin. Therefore, the role played by tropomyosin inskeletal muscle contraction is most similar to that of the role played by a repressor in proteinsynthesis. A repressor molecule binds to DNA and physically inhibits mRNA transcription, andits removal allows transcription to occur. Thus, choice C is correct. In inducible systems, tran-scription is repressed until a molecule called an inducer binds to the repressor and inactivates it.In repressible systems, the repressor must first bind to a co-repressor molecule before it can bindto the DNA, so in the absence of the co-repressor, transcription occurs.

Choice A is incorrect because as just discussed, an inducer is the molecule responsible forremoving the repressor in inducible systems. (FYI: In embryology, an inducer is a tissue, bodypart, or substance that causes the differentiation of another tissue or body part.) The action ofCa2+, not tropomyosin, in muscle contraction is most similar to that of an inducer. Choice B isincorrect because an immunoglobulin (antibody) is a globular protein synthesized by B lym-phocytes. An immunoglobulin is complementary to a specific antigen, and binds to it specifi-cally. Immunoglobulins are secreted in response to the presence of the antigen in the body. Thisis known as the immune response. Choice D is incorrect because a catalyst is any substancewhich speeds up a chemical reaction by lowering the activation energy of the reaction. Enzymesare organic catalysts that speed up biochemical reactions. ATPase is one of the enzymes involvedin sarcomere contraction.

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Passage IX (Questions 192-198)

192. CSince Roman numeral I is a part of all of the answers, this means that glucose can be gen-

erated from glycogen. In fact, glucose is stored as glycogen in liver and muscle tissues, essen-tially providing a metabolic energy source for between meals. According to the passage,glycerol, derived from triacylglycerol, can be used to synthesize glucose via gluconeogenesis.Therefore, the right answer must include Roman numeral II; this eliminates choice A. You’realso told that fatty acids are converted into ketone bodies, which go on to supply fuel to braintissue, which the brain consumes instead of glucose during the latter stages of starvation. Thisimplies that fatty acids cannot be converted into glucose. Since Roman numeral III is not cor-rect, choices B and D can be eliminated, which means that the correct answer is choice C andthat Roman numeral IV is part of the answer. The passage says that muscle protein is degradedto provide precursors for glucose synthesis. Since protein degradation yields amino acids, it canbe inferred that amino acids can be utilized as substrates for gluconeogenesis.

193. AGlucagon is produced by the islet cells of the pancreas and functions to control blood glu-

cose concentration. Glucagon causes blood glucose concentration to increase by promotingglycogen breakdown and glucose synthesis and by promoting fat and protein degradation. So,choice A is correct and choice C is wrong, since it is insulin that decreases blood glucose con-centration. Choice B is wrong because glucagon does not regulate the concentration of triacyl-glycerol in the blood. Choice D is wrong because it is a low blood glucose level, not glucagon,that stimulates the feedback inhibition of insulin secretion.

194. BAccording to the passage, triacylglycerol holds six times more energy than glycogen. Thus,

if an imaginary migratory bird used glycogen rather than triacylglycerol as its stored energysource, it would have to lug around six times the weight in fuel than a real migratory bird car-ries. This bird would be so weighed down by the amount of glycogen necessary to make theflight that it wouldn’t even be able to get off the ground to fly. So, choice B is correct. ChoiceA would be correct if the bird didn’t have to carry so much glycogen in order to get same amountof energy stored in triacylglycerol. Since glycogen is a stored form of glucose, the bird wouldhave burned glucose as its main metabolic fuel during migratory flight if it wasn’t weighed downby the glycogen. But since the bird wouldn’t be able to fly, choice A is incorrect. Choice C iswrong because acetoacetate is synthesized from the acetyl CoA generated by the beta-oxidationof fatty acids. Fatty acids come from the hydrolysis of triacylglycerol; but our imaginary bird’sfuel source is glycogen, not triacylglycerol. This bird would be burning glucose derived fromglycogen breakdown; it would not use any of the products of triacylglycerol breakdown. ChoiceD is incorrect because while it is true that the bird wouldn’t be able to complete the voyage, itwould have a ready supply of glucose in its stored form, glycogen.

195. DAccording to the passage, muscle protein is degraded during periods of starvation so that

amino acids can be used to synthesize glucose. Therefore, the likelihood of protein synthesisoccurring during starvation is slim, so choice D is correct. The passage also tells you thatglucagon secretion is increased during the early stages of starvation, resulting in the promotionof glucose synthesis. Since gluconeogenesis is the conversion of non-glucose molecules to glu-cose, choice A is wrong. In addition, the first paragraph informs you that during the first day of

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starvation insulin secretion is inhibited, and triacylglycerols are hydrolyzed to fatty acids. So,choice B and choice C are incorrect.

196. AIn order to enter the Krebs Cycle (citric acid cycle), acetyl CoA is joined to oxaloacetate by

the enzyme citrate synthetase, yielding citrate. In subsequent steps, citrate loses two carbons toCO2 and ends up as oxaloacetate, now ready to join with a new acetyl CoA molecule to com-plete another round of the cycle. However, during starvation, large amounts of acetyl CoA aregenerated by the degradation of fatty acids. In addition, the oxaloacetate concentration isdepleted, because it is converted into phosphoenolpyruvate and then used to synthesize glucose.So although excess acetyl CoA is made during starvation, very little of it can be incorporatedinto the Krebs cycle because of low oxaloacetate concentrations. Instead, the acetyl CoA is usedto produce ketone bodies. So, choice A is correct. Choice B is way off course; no special enzymeis mentioned in the passage and, even so, the existence of an enzyme that catalyzes the forma-tion of ketone bodies would not explain why acetyl CoA is necessarily used only to produceketone bodies. So, choice B is incorrect. Both choice C and choice D are true statements—acetylCoA can’t be converted into pyruvate and triacylglycerols are mobilized from adipose tissueduring starvation. However, neither choice answers the question; neither accounts for why acetylCoA can’t just enter the Krebs cycle like it does under normal, non-starvation conditions. So,choices C and D are incorrect.

197. CThe capillaries of the brain are unique because they do not allow substances in the blood to

have free access to the neurons of central nervous system. These capillaries have no fenestra-tions, or gaps, between capillary epithelial cells; therefore, a physical barrier between the circu-latory system and the brain exists, which is called the blood-brain barrier. To enter brain tissue,a compound must pass through the epithelial cells. The only substances that can enter the brainare those that: (1) are hydrophobic, because they can dissolve through the plasma membrane, or(2) have a specific carrier molecule on the epithelial cell surface that will transport the (typicallyhydrophilic) substance across the plasma membrane. Glucose, which is normally the brain’smetabolic fuel, and continues to be so during the early stages of starvation, enters the brain viathis second method. Fatty acids, on the other hand, do not have a carrier molecule that allowsthem to cross the blood-brain barrier, and so they cannot be used by the brain as fuel when bloodglucose is low. Thus, choice C is correct. Choices A, B, and D are wrong because they all implythat fatty acids are normally used by the brain as its metabolic fuel.

198. BImmediately following a meal, the concentration of glucose in the blood rises from the

recently digested food. Processes that function to lower blood glucose, such as insulin secretion,glycogen synthesis, and glycolysis, will occur following a meal. On the other hand, processesthat function to raise blood glucose, including processes that occur during starvation, such asglucagon secretion, gluconeogenesis, and hydrolysis of triacylglycerol, will not occur followinga meal. Therefore, choice B is correct, since glucagon secretion would be inhibited following ameal. Choices A and D would occur only when blood glucose was low, so they are both incor-rect. And choice C is incorrect because cellular fermentation would only increase under anaer-obic conditions, such as those experienced when exercising muscle, which isn’t even mentionedin the question.

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Passage X (Questions 199–203)

Passage X is another passage in which the subject matter is probably familiar to you. SN2reactions are an important part of organic chemistry. By knowing the conditions for this type ofreaction, you are able to predict whether other reactions will occur by this mechanism.

The passage discusses the disadvantages of carrying out SN2 reactions in polar aprotic sol-vents, and then goes on to describe phase-transfer catalysis, an alternative method by which SN2reactions can proceed. If phase-transfer catalysis isn’t familiar to you, don’t worry—the passagenot only describes the process in detail, but also provides you with a drawing of the apparatus.Finally, you are given a number of reactions that can be carried out by phase-transfer catalysis.Look closely at these reactions—try to identify the type of reaction, the mechanism by which itoccurs, and the reaction conditions. You may need this information later when you are answer-ing the questions.

199. AThis is another question that tests your nomenclature skills. In this case, you are given a

compound and asked to name it. You may also be required to identify the structure of a com-pound from its IUPAC name.

The fact that the compound contains a C6H5 group confirms that the compound is a substi-tuted benzene. Now, if we lost the bromo group, the substituent on the ring would be an ethylgroup and the compound would be called ethylbenzene. However, since a bromo substituent isattached to the ethyl chain, we have to establish where the bromo substituent is positioned onthis chain. Since the compound is named as a substituted benzene, the carbon of the bromoethylgroup that is attached directly to the ring is carbon 1. Therefore, the carbon that is attached tothe bromine atom is carbon 2, so the functionality attached to the benzene ring is 2-bromoethyl.This functionality is in parentheses so that there is no confusion about where the bromo groupis positioned. Since all the carbons on the benzene ring are equal, we do not need to distinguishto which carbon the 2-bromoethyl group is attached. Therefore, the name of the compound is (2-bromoethyl)benzene.

Choices B and D are incorrect because the bromo and ethyl groups are not part of the samefunctionality. In these answer choices, the bromo and ethyl groups are attached separately to thering—this is not the case with the compound shown in Reaction 3. Finally, choice C is incorrectbecause this choice names the compound as a substituted ethane with a benzyl group attachedto it. A benzyl group consists of a CH2 group attached to a phenyl group, so this molecule wouldhave too many carbons and hydrogens to qualify as the reactant in Reaction 3.

200. CThe passage tells you that phase-transfer catalysts are usually quaternary ammonium salts.

Quaternary ammonium salts contain a central nitrogen attached to four groups, none of whichare hydrogens. Because the nitrogen is attached to four groups, it possesses a positive charge.To counteract this positive charge, there is always an anion that coordinates with the ammoniumportion of the molecule. The ammonium portion and the anion are what constitute a quaternaryammonium salt. The only answer choice that does not contain a nitrogen attached to four func-tional groups that are NOT hydrogens is choice C. This compound contains a central nitrogenattached to three hydrogens, and is called an aminium salt. Aminium salts are compounds inwhich the nitrogen is attached to at least one hydrogen, and they are formed when primary, sec-ondary, and tertiary amines react with acids. All the other answer choices are quaternary ammo-

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nium salts since the central nitrogen in each is attached to four alkyl groups, and the ammoniumcation coordinates with a halide ion. Again, the correct answer is choice C.

201. CIn Reaction 1, 1-chlorooctane—a primary alkyl halide—reacts with cyanide—a strong

nucleophile—to form nonanenitrile. Therefore, a substitution reaction has taken place. Primaryalkyl halides will only undergo SN2 and E2 reactions, so the mechanism is SN2. In this reaction,the nucleophile, CN–, attacks the unhindered, positively polarized carbon 180° to the leavinggroup (in this case, Cl–). Then, a transition state forms in which the incoming nucleophile andthe leaving group are partially bonded to the nucleophilic carbon. As the bond between thenucleophile and the carbon starts to form, the bond between the carbon and the leaving groupbegins to weaken and eventually break, resulting in the loss of an anion. The answer choice thatcorrectly represents this transition state is choice C. Notice that the chlorine leaving group pos-sesses a slight negative charge and the cyanide nucleophile possesses a slight negative charge(since cyanide brings electrons to the substrate and chlorine leaves the molecule as an anion, soit “pulls” electrons away from the carbon).

Choice B is incorrect because the chlorine is positively polarized and the cyanide is nega-tively polarized. Remember, chlorine is leaving the molecule with a negative charge, and startsto acquire a slight negative charge in the transition state. Choices A and D are incorrect sincethey show the alkyl chain partially bonded to the electrophilic carbon. In addition, chlorine isfully bonded to the carbon and appears as though it will not leave the molecule; however, chlo-rine is a good leaving group and should be partially bonded to the electrophilic carbon.

202. BThe product obtained in Reaction 2 is benzoic acid. Carboxylic acids with five carbons or

more are fairly insoluble in water; benzoic acid contains 7 carbons, so it is insoluble in aqueoussolution. However, the acidity of benzoic acid—like other carboxylic acids—ensures that it canlose a proton and react with sodium hydroxide or sodium bicarbonate to form a water-solublesodium carboxylate salt. Mixing sodium bicarbonate and the organic layer obtained in Reaction2 will result in the formation of sodium benzoate, which moves from the organic layer to theaqueous layer. By the way, this process is demonstrative of extraction—a process often carriedout in the laboratory to isolate a pure compound from its contaminants. This is an importanttechnique in organic chemistry: make sure you are familiar with it.

Choice A is wrong because sodium salts are ionic compounds, so they cannot dissolve in theorganic layer. Choices C and D are wrong because benzoic acid will form a sodium salt.Remember, carboxylic acids contain an acidic proton which can be lost to form a carboxylateion. This ion is stable because the negative charge—acquired by the loss of the proton—candelocalize across the two oxygens of the carboxylate ion. The presence of sodium ensures thata cation is present to balance out the negative charge, thereby forming a stable sodium salt.

203. DThe passage states that the most favored solvents in SN2 reactions are polar aprotic solvents.

Let’s briefly discuss why. Polar aprotic solvents and polar protic solvents will both dissolveionic compounds. However, polar protic solvents possess acidic hydrogens—that is, hydrogensattached to an electronegative atom such as oxygen. These hydrogens can form very stronghydrogen bonds to the nucleophile or anion and form a ‘solvation sphere.’ As a result, it is dif-ficult for the nucleophile to ‘shake off’ the solvent molecules and effectively attack the substrate.Aprotic solvents also solvate the nucleophile, but the absence of acidic hydrogens means that

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they do so less strongly (the nucleophile is solvated by electrostatic interaction with the posi-tively-polarized portion of the solvent). As a result, nucleophiles are more effective in polaraprotic solvents than in polar protic solvents. The only compound that is a polar aprotic solventis choice D—sulfolane. This molecule is polar because the electronegative oxygens pull elec-tron density away from the sulfur, but is aprotic since the molecule does not contain any acidichydrogens.

Choice B,acetic acid, and choice C, ethylene glycol, are both incorrect because they containacidic hydrogens, hence they are polar protic solvents. Choice D, isobutane, is a nonpolar sol-vent. SN2 reactions, in the absence of a phase-transfer catalyst, could not be carried out in thissolvent, since it does not dissolve ionic compounds.

Passage XI (Questions 204–209)

Passage XI is a short passage about isomers. You are given the specific isomer of three com-pounds: A, B, and C. Try to identify what types of isomers these compounds represent. Forexample, Compound A does not contain any double bonds, which should hint at conformationalisomers. Compound B is a disubstituted cyclohexane, so you should be thinking about configu-rational isomers. Finally, Compound C is a disubstituted alkene, so start to think about geomet-ric isomers. You are then given a reaction scheme for the conversion of Compound C toCompound D. Try to identify what types of reactions are occurring here, and establish the rela-tionship between Compound C and Compound D.

204. CSince Compound C contains a double bond, which is a region of high electron density, it will

be susceptible to electrophilic attack. More specifically, it will be susceptible to electrophilicaddition. In the first step of the reaction, hydrogen adds to the double bond, obeyingMarkovnikov’s rule. This rule states that hydrogen adds to the least substituted carbon—ensur-ing that the most stable carbocation is formed. Since both double-bonded carbons in CompoundC are monosubstituted, it does not matter which carbon forms the carbocation:

Straight away, we can see that choice C is the correct response since this is the only choicethat contains electrophilic addition. Let’s just finish off the mechanism to confirm that choice Cis the correct answer. In order to form Compound D, HBr has to be eliminated from the bro-

CH3 CH2 H

CH2 CH3H

HBr H

CH3 CH2

H

CH2 CH3

BrH

KOH/ethanol

Compound C

H

CH2 CH3

H

CH3 CH2

Compound D

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moalkane (recall that elimination is defined as the formation of a multiple bond by the loss of asmall molecule). Elimination occurs in the presence of a strong base such as potassium hydrox-ide; the base abstracts an alpha proton, while electron density is given to the leaving group.Compound D differs from Compound C because the carbon-carbon bond in the bromoalkanecan freely rotate, so elimination results in the formation of the alkene where the substituents arearranged differently. Compound D is thermodynamically favored over Compound C since theethyl groups are farther apart; hence, the molecule is more stable. Therefore, when the bro-moalkane is heated in the presence of potassium hydroxide and ethanol, the activation energybarrier to form the thermodynamic product is overcome, and Compound D is formed.

Choices A and D are incorrect since neither nucleophilic substitution nor nucleophilic addi-tion occurs. Since Compound C contains a region of high electron density, it will be susceptibleto electrophilic attack, not nucleophilic attack. In addition, the presence of a double bond meansthat addition will occur, not substitution. Choice B is wrong because free radical addition wouldonly occur in the presence of radical initiators, such as UV light or peroxide. Since neither ofthese are present in Reaction 1, choice B can be eliminated.

205. ACompound B and Compound E are similar in that they are both disubstituted cyclohexanes,

and the substituents attached to the ring are trans. The key difference, however, is that the –Cland –CH3 substituents in Compound B are arranged equatorially (in the plane of the ring),whereas in Compound E, the –Cl and –CH3 substituents are arranged axially (perpendicular tothe ring). In Compound E, there is more nonbonded strain since there is 1,3-diaxial interactionbetween –Cl and –CH3 groups, and the hydrogens in the ring:

Compound E

Since Compound B has less nonbonded strain, it is the favored isomer in the equilibrium;choice A is the correct response. Choice C is incorrect since we have just established that Com-pound E has more nonbonded strain than Compound B. Choices B and D are incorrect becauseCompound B and Compound E are both chair conformations of cyclohexane, with minimalangle strain (since the sp3 carbon bond angle is 109.5°).

206. ACompound C and Compound D are geometric isomers, or more specifically, cis/trans iso-

mers. In Compound C, the top priority groups attached to each double-bonded carbon (the ethylgroups) lie on the same side of the double bond, so the compound is a cis-isomer. In CompoundD, the top priority substituents lie on opposite sides of the double bond, so the compound is atrans-isomer.

H

H

H

H

H

H

HH

H

H

CH3

Cl

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Meso compounds are species that contain chiral centers, but contain an internal plane ofsymmetry; therefore, they are optically inactive. Compounds C and D do not contain chiral cen-ters, so choice C can be eliminated. Choice B is incorrect because configurational isomers ariseby rotation about a carbon-carbon single bond (like Compounds A and B, for example). Sincewe are dealing with the arrangement of substituents about a carbon-carbon double bond, choiceB can be eliminated. Choice D is incorrect because enantiomers are stereoisomers that haveopposite configurations at every chiral center, and are nonsuperimposable mirror images of eachother. Compounds C and D do not contain chiral centers, and they are not nonsuperimposablemirror images, so they cannot be classed as enantiomers.

207. AThe Newman projection of Compound A is as follows:

Newman projection of Compound A

In this conformer, there is a lot of nonbonding interaction between the halogen groups. Thereis also nonbonding interaction between the methyl groups since they are only 60° apart; how-ever, the methyl groups are smaller than the halogen atoms, so the magnitude of nonbondinginteraction is lower.

In the compound in choice A, the methyl groups are anti to each other (180° apart). Thisensures that there is less nonbonding interaction between the methyl groups and the halogengroups. For example, in the Newman projection of Compound A, we can see that the rear chlo-rine interacts with the bulky chlorine and bromine, both of which are attached to the front car-bon. However, in the Newman projection of the compound in choice A, the rear chlorine isclosest to a methyl and a chlorine group, and since the methyl group is less bulky, there is lessnonbonded interaction.

Newman projection of Choice A

Since there are more repulsive van der Waals forces between the groups in Compound A, themolecule suffers from more nonbonded strain, and is higher in energy than the conformer inchoice A. Therefore, choice A is lower in energy, and is the correct response.

CH3

Cl Br

Cl Br

CH3

Cl

Br CH3

Cl Br

CH3

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Choices B, C, and D are incorrect since these conformers are higher in energy than Com-pound A. All these conformers are eclipsed: the groups attached to each carbon are in directalignment with each other. In the eclipsed conformation, the molecules are very high energysince a lot of van der Waals interaction exists between the groups, and there is also a great dealof torsional strain (brought about when the carbon-carbon single bond rotates to form theeclipsed conformation).

208. Bsp2 hybridization is associated with double-bonded carbons. Two electrons in the 2s orbital

of carbon hybridize with two of the three 2p orbitals of carbon. As a result, three sp2 orbitals areformed. Since one of the 2p orbitals remains unhybridized, it can overlap with an adjacent porbital to form a pi bond, while the three sp2 orbitals contribute to sigma bonding (which, in thecase of Compound C, is a sigma bond to hydrogen atom, a sigma bond to the carbon of the ethylgroup, and a sigma bond to the double-bonded carbon).

Choice A is incorrect since sp3 hybridization involves the hybridization of the 2s orbitalswith all the 2p orbitals. As a result, there are no unhybridized p orbitals available for overlap;therefore, sp3 hybridization is always found on saturated carbons, such as those found in alka-nes. The hybridization of the 2s orbital and one 2p orbital in carbon results in sp hybridization—choice C. In this case, two 2p orbitals remain unhybridized, so there is the possibility offormation of two pi bonds; therefore, sp hybridization is observed in triple-bonded species suchas alkynes. Finally, choice D is incorrect since hybridization between the 2s and 2p electrons incarbon always occurs when a chemical bond is formed.

209. DHBr reacts with alkenes according to Markovnikov’s rule; that is, the hydrogen adds to the

least substituted double-bonded carbon, so that the most stable carbocation is formed. Sinceboth double-bonded carbons in Compound C are attached to one hydrogen, they will both formsecondary carbocations upon reaction with HBr. When alkenes react with hydrogen halides inthe presence of UV light, anti-Markovnikov addition occurs: hydrogen adds to the most substi-tuted double-bonded carbon. In the case of Compound C, it does not matter whetherMarkovnikov or anti-Markovnikov addition occurs, since each double-bonded carbon is equallysubstituted. Therefore, the outcome of the reaction will be the same—the bromoalkane willform, there will be rotation about the carbon-carbon single bond, and HBr will be eliminatedfrom the molecule to form Compound D. Therefore, choice D is the correct response.

Since Compound C reacts with HBr in the presence or absence of UV light, choice C can beeliminated. Reaction 1 shows that the bromoalkane undergoes elimination to form CompoundD only. Since the reaction of Compound C with HBr in the presence of UV light results in thesame haloalkane that was obtained by the reaction of Compound C with HBr, it follows thatelimination of the haloalkane will also result in the formation of the same thermodynamicallyfavored elimination product: choice B can be eliminated. The chain propagation step in the anti-Markovnikov addition of HBr to alkenes generates bromine radicals. If there is an excess ofalkene, then it will undergo self-polymerization. However, we are told that Compound C doesnot form a polymer, but instead forms Compound D, so choice A can be eliminated.

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Discrete Questions

210. ABreathing is the means by which our bodies obtain a fresh supply of oxygen. When a person

inhales, the diaphragm contracts, drawing in air from the mouth or nasal cavity through the tra-chea, and into the lungs via the two bronchi. The bronchi divide into the bronchioles, which ter-minate in clusters of small air sacs called alveoli. Inhalation fills the alveoli with oxygen-richair. The alveoli are surrounded by a dense network of blood vessels known as the pulmonarycapillaries. The pulmonary capillaries contain deoxygenated, carbon dioxide-rich blood deliv-ered from the right side of the heart via the pulmonary arteries. It is across the thin walls of thealveoli and the pulmonary capillaries that gas exchange occurs—oxygen is exchanged for car-bon dioxide via diffusion. Following their respective concentration gradients, oxygen diffusesfrom the alveoli into the pulmonary capillaries, and carbon dioxide diffuses from the pulmonarycapillaries into the alveoli. Once inside the capillaries, the oxygen binds to hemoglobin (theoxygen-carrying molecule in mammals) and the oxygenated blood returns to the left side of theheart via the pulmonary veins. During exhalation, when the diaphragm relaxes, the carbon diox-ide exits the body via the same route through which oxygen enters. Thus, choice A is correct andchoice B is wrong.

Choice C is wrong because oxygen does not dissociate from hemoglobin until it reaches thetissues, via systemic circulation. The dissociation allows the oxygen to enter the oxygen-poorcells and the waste product, carbon dioxide, to leave them. Choice D is wrong becausehemoglobin, a component of red blood cells, is confined to the vessels of the circulatory system.The alveoli never actually come into direct contact with the blood. The circulatory system isreferred to as a closed system because the blood it carries is confined to a closed circuit of bloodvessels.

211. BEven if you couldn’t recall everything that bile does do, you might have been able to recall

what it DOESN’T. Bile is secreted by the gallbladder in response to the secretion of the hormonecholecystokinin (CCK), which is secreted in response to the entrance of fatty foods into the duo-denum. Bile is comprised of bile salts, bile pigments, cholesterol, and other substances dissolvedin an alkaline electrolyte solution. While you might not have known that bile, along with intesti-nal juices and pancreatic juice, neutralizes the gastric acid entering the duodenum from thestomach, you should have known that this neutralization occurs in the small intestine, not thelarge intestine. Acidic chyme entering the duodenum is raised to an almost neutral pH of6.0–7.0. Therefore, choice B is the correct answer.

Choices A, C, and D all describe actual functions of bile. Bile, in conjunction with mono-glycerides and phospholipids, emulsifies fats entering the small intestine, thereby decreasing thesurface tension of the fat particles. Emulsification allows pancreatic lipase, the most importantenzyme in fat digestion, to break down the particles into free fatty acids and monoglycerides.Bile interacts with the end-products of fat digestion to form complexes called micelles. Thehighly-soluble micelles ferry the lipids into the intestinal mucosa, where they are then absorbed.Thus, choices A and C are wrong. In addition, bile is the means of excretion for several wasteproducts from the blood, such as bilirubin and cholesterol. The majority of bilirubin is formedin the tissues from the degradation of hemoglobin. The bilirubin enters the circulation and bindsto albumin until it reaches the liver and dissociates. The bilirubin undergoes a reaction withinthe liver and is then actively transported into the bile canaliculi, and ultimately, into the gallbladder. Thus, choice B is correct.

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212. AAll of the methyl hydrogens in this compound are equivalent; therefore, a signal will be

observed at only one δ value. Since all of the methyl hydrogens are attached to a carbon that isadjacent to a central carbon that has no hydrogens attached to it, none of the hydrogens willundergo spin-spin coupling. As a result, a single peak will be observed. When integrated, the sin-gle peak corresponds to 12 methyl protons.

213. A Parathyroid hormone and calcitonin are antagonistic to each other, which means that they

produce opposite physiological effects. Parathyroid hormone, which is secreted by the parathy-roid glands, is responsible for increasing the Ca2+ level in the blood, while calcitonin, which issecreted by the thyroid gland, decreases the Ca2+ level in the blood. An easy way to distinguishbetween the two hormones is to remember that calcitonin tones down calcium. Parathyroid hor-mone increases the Ca2+ level in the blood by reducing Ca2+ excretion by the kidneys, increas-ing calcium uptake in the small intestine, and increasing bone degradation. Bone degradationresults in the release Ca2+ into the blood as calcium is a constituent of bone matrix. The bonecells responsible for dissolving bone are the osteoclasts, so choice A is correct.

Choice B is incorrect because osteoblasts are stimulated by calcitonin, not parathyroid hor-mone. Osteoblasts, which build bone, absorb Ca2+ from the blood, so they decrease the level ofcalcium ions in the blood. Choice C is incorrect because parathyroid hormone has no effect onchondrocytes, the cartilage-forming cells. Cartilage does not contain calcium, so neitherparathyroid hormone nor calcitonin affects chondrocyte activity. You may have been tempted bychoice D because parathyroid hormone does act on the kidney to increase the calcium ion levelin the blood. However, parathyroid hormone causes the kidneys to reduce Ca2+ excretion, notincrease vitamin D excretion. Vitamin D actually works with parathyroid hormone by aiding cal-cium uptake in the small intestine. That’s why milk is often fortified with vitamin D! So, choiceD is incorrect.

214. CThe hormone aldosterone is secreted by the adrenal cortex and is responsible for regulating

blood pressure and potassium ion concentration in the blood. In a normal individual, aldosteroneis secreted in response to low blood pressure. Its action causes potassium ions and hydrogen ionsto be secreted into the nephron’s distal tubule and collecting ducts and causes sodium ions to bereabsorbed from the nephron into the extracellular fluid. The tubular reabsorption of sodiumcauses water reabsorption, since the reabsorbed sodium causes osmosis of water through thetubual epithelium. Thus, aldosterone maintains extracellular fluid volume, and therefore main-tains blood volume, as well. If the adrenal cortex is not producing enough aldosterone, as inAddison’s disease, tubular reabsorption of sodium ions and water will decrease, causing extra-cellular fluid volume and blood pressure to drop. Also, there will be a decrease in the amount ofpotassium and hydrogen ions secreted into the kidneys (and excreted in the urine), which meansthat there will be an increase in the concentration of potassium ions in the extracellular fluid.Thus, choice C is correct.

Choice A is wrong because these are the effects observed when aldosterone secretion is inexcess. Choice B is wrong because potassium concentration in the extracellular fluid increases,not decreases, in Addison’s disease. Choice D is wrong because aldosterone insufficiency wouldclearly affect both blood pressure and potassium ion concentration in the extracellular fluid,since aldosterone is the primary hormone responsible for regulating these two variables.

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