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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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Page 1: Full length 8 answers

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

1. D

2. D

3. A

4. B

5. D

6. A

7. A

8. A

9. C

10. C

11. C

12. B

13. B

14. A

15. D

16. A

17. B

18. A

19. D

20. D

21. D

22. B

23. B

24. D

25. D

26. D

27. B

28. A

29. C

30. B

31. D

32. D

33. A

34. C

35. B

36. D

37. C

38. C

39. D

40. D

41. A

42. B

43. C

44. C

45. D

46. C

47. C

48. D

49. A

50. A

51. D

52. D

53. C

54. A

55. D

56. C

57. A

58. B

59. D

60. D

61. C

62. C

63. A

64. D

65. D

66. B

67. A

68. B

69. A

70. D

71. C

72. B

73. C

74. B

75. B

76. B

77. C

PHYSICAL SCIENCES ANSWER KEY

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Passage I (Questions 1-6)

1. DAt first glance, it might seem that only three hydrogen bonds are possible for each boric acid

molecule, since there are three hydroxyl groups. However, each hydroxyl group has a hydrogenatom with one possible H-bond and an oxygen atom with two lone pairs; two more possible H-bonds. Therefore, each hydroxyl group is capable of forming three hydrogen bonds; there arenine possible hydrogen bonds for each boric acid molecule.

2. DAccording to the passage, solubility is determined by the intermolecular reactions both in

solvent-solute interactions and solute-solvent interactions. In order to arrive at the correctanswer, consider each Roman numeral separately. Statement I is true; acetylene molecules arenonpolar because of their symmetry, while water molecules are very polar. Since the polarity ofacetylene molecules differs greatly from that of water molecules, acetylene will interfere withthe hydrogen bonding between water molecules, thereby lowering their miscibility. Since state-ment I is true, you can eliminate choice (B). Now consider statement II. The intermolecularinteraction between acetylene and water is weak because of the different polarities of the twomolecules, so this statement is true. Eliminate choice (A), since it does not contain statement II.Finally, consider statement III. Acetylene is not capable of hydrogen bonding because it does notcontain an electronegative element, so the statement is true. Since all three statements are true,choose (D).

3. ABefore you consider the explanations provided in the answer choices, ask yourself how the

information in the passage might be relevant to the question. The passage discusses hydrogenbonding, so the answer to the question will probably have something to do with hydrogen bond-ing. Ammonia is capable of hydrogen bonding because it contains nitrogen, a highly elec-tronegative element, bonded to hydrogen. X-ray studies suggesting more hydrogen bonding thanpredicted would indicate higher attractive intermolecular forces and raise the melting point, sochoice (A) makes sense. Choice (B) would decrease the melting point, since the moleculeswould have decreased intermolecular interactions. (C) and (D) are irrelevant to the meltingpoint.

4. BThe passage states that the structure of ammonium chloride is a cubic arrangement of 8 Cl-

around a single NH4+. This is a simple description of a body centered cubic crystal as shown

below. Therefore, the answer is B. The MCAT also expects you to be able to distinguish betweensimple cubic and face centered cubic structures, so keep those in mind as well and you’ll be pre-pared for any structure that appears on the exam.

5. DThe passage states that the heat of vaporization is affected by both the number of hydrogen

bonds and inhibition of rotation. The molecules in choices (A), (B), and (C) are all capable of

2

Page 4: Full length 8 answers

multiple H-bonds and have three-dimensional geometry. HCN is not capable of H-bondsbecause the hydrogen is not bonded to an electronegative element in HCN. In addition, its lin-ear geometry allows rotation to occur even when the molecules are bonded in series:

Therefore, the answer is (D).

6. AWhile many H-bonding molecules are soluble in water this is not a requirement for H-bonding.

Large organic molecules containing a single hydroxyl group can form hydrogen bonds and yet beinsoluble in water. Therefore, choice (A) is correct. (B) is incorrect because an available lone elec-tron pair is necessary for hydrogen bonding to occur; if the electronegative element doesn’t have alone pair of electrons, it won’t be able to bond to a hydrogen atom. Choice (C) is incorrect becausethe hydrogen atom must be slightly acidic in order for the H-bonding to occur. This is a consequenceof the fact that a polar bond will lead to a more acidic hydrogen atom than a nonpolar bond. Choice(D) is incorrect because the two molecules must be in close proximity for an intermolecular bondsuch as hydrogen bonding to occur. Again, the answer is (A).

Passage II (Questions 7-12)

7. AAccording to Equation 1, the speed of a seismic wave in water equals the square root of its bulk

modulus B divided by its density ρ. Therefore, we set up the expression,

v 21

2.

.

0

40

0

g

1/

0

c8

mP

3a 1.4 103 m/s.

This is closest to choice (A). Be sure to watch the units on the table. The density is given asg/cm3 and must be changed to kg/m3 in an expression necessarily involving Pa.

8. ASolid 1 will store more potential energy. This question requires integrating information drawn

from the passage with information regarding compressibility, restoring forces, and potential energyin springs: A large bulk modulus is essentially equivalent to a large spring constant, k, in the mediumthrough which sound is traveling. We know this because the passage tells us that a large B indicatesa large restoring force in response to compression (see paragraph 3). Potential energy is proportionalto k (recall U=1/2 kx2 for springs). Thus, the atoms in solid 1 will be able to hold more potentialenergy; choice (A) is correct.

9. CThe wavelength of any wave is given by the expression v=fλ, where f is the frequency and λ is

the wavelength. This formula is an MCAT favorite. The speed v of a seismic wave in a rock near theEarth’s surface = 6000 m/s (paragraph 2). Thus, λ = v/f = (6000 m/s) / (5000 Hz) = 1.2 m, choice(C).

10. C Recall that the speed of the wave depends on ρ, µ, and B. Reading the question stem carefully

reveals that we want to find which answer choice will independently explain the difference in wavespeed. The surface wave has a smaller speed than the core wave (see paragraph 2): A smaller shear

H C N: H C N:H C N:

3

Page 5: Full length 8 answers

or bulk modulus or a larger density could explain a decreased wave speed. Likewise, a larger shearor bulk modulus or a smaller density would explain an increased wave speed in the core wave.Choice (C) is the correct answer. Temperature has not been discussed in the passage – eliminatechoice (D).

11. CAgain, Equation 1 tells us what we need to know. Now consider that vA/vB = 3/1 =

So ρB/ρA = 9/1, and the ratio of the density of Soil A to Soil B is 1:9;

the correct answer is choice (C).

12. BSeismic waves are longitudinal waves. If a seismic wave is traveling through a solid, then its

molecules are moving parallel to the movement of the wave, in this case horizontally. Since the solidis immobile, there is no vertical movement of the molecules, and so their kinetic energy in the ver-tical direction is zero – choice (B) is correct. Remember that the MCAT is primarily a critical think-ing test; knowing your concepts will get you a long way.

Passage III (Questions 13-18)

13. BAs heat is transferred to the bath, the ice begins to melt. Since ice has a smaller density thatwater, its melting will decrease the volume of the bath. Here the height of the bath increases, sovolume increases: therefore, we are not melting ice, but freezing water. Heat must be transferredfrom the bath! The only way to spontaneously transfer heat from the bath is to add a substancewith a lower temperature. Choice (B) is the only choice with a substance at a lower than zerotemperature. Therefore, choice (B) is correct. All the other substances listed would either trans-fer heat to the bath or not at all.

14. AThe passage states that heat transfer between two substances is directly proportional to the

temperature difference between those two substances. So, before thermal equilibrium is reached,the bath with a temperature farthest away from 68°C will experience the fastest (so after 10 sec-onds, the most) heat transfer. Eliminate choice (D), since ice will not exist at 32°C and 1 atm.Don’t be tempted by this choice; it is designed to test whether you are confusing the Celsius andFahrenheit scales. Choice (A) is correct, since it sets up the largest temperature gradient betweenthe bath and the substance.

15. DStraightforward calculation question. We know that all the heat lost from the unknown sub-

stance is gained by the bath: Qsub = Qbath. In addition, both the substance and water ended up at15.8°C.

Q = mC∆T

Qsub = Qbath

ρBρA

ρB (43

µ Β )ρA(

43

µ Β )

4

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(150 g)(C)(75° 16) = (300 g)(4.18)(16°− 10°)

Now make an approximation:

C = (

(

1

3

.5

10

1

2

0

g2

)

g

(4

)(

)

6

(6

0

°

°

)

) ≈ 8 × 10−1, choice (D).

16. AThis question tests your knowledge of colligative properties. Once the CaF2 is added, the

water is no longer pure solvent, but rather a solution. We know that it will have a boiling pointelevation and a freezing point depression. So, the water is cooled to −30°C ice and then heatedby the copper at 75°C. The ice will melt, but it will melt at a lower temperature (don’t forget,freezing point depression also implies melting point depression!). A lower temperature corre-sponds to less heat required for melting, or fusion. This is because the crystal lattice of the iceis not as strong if there are impurities (CaF2), and will require less heat to break the lattice.Notice, choice (D) makes no sense since deposition is the process of going from a gas to a solid:the opposite of sublimation.

17. BBe careful. If there is a hole in the insulating top, then the bath is no longer thermally iso-

lated. However, it will still reach thermal equilibrium, but now it will do so with the surround-ings. Since the room is large compared to the bath or substance, thermal equilibrium will bereached at 27°C, the temperature of the room. This is analogous to leaving hot coffee open in aroom: a few hours later, the room will be at approximately the same temperature, and the cof-fee will have reached the temperature of the room.

18. AHere the substance lost heat, or transferred it to the bath. In transferring heat to the bath, the

ice will melt and any left over heat will be used to increase the temperature of the water. First,figure out how much heat is required to melt 100 g of ice. We know that −6.007 kJ are requiredto freeze 1 mole of water, so 6.007 kJ are required to melt 1 mole of ice.

Heat required: 100 g of ice × 118m.0

o2leg

iiccee

× 1 m

6.o0l0e7okfJice

≈ 33.3 kJ of heat.

Therefore, after we melt all the ice, there was no heat left over to raise the temperature ofwater: the final temperature must be 0°C, choice (A).

Passage IV (Questions 19-24)

19. DThe passage states that in order to stop the heart and bring it back to normalcy, a current in

excess of 1 A must be applied. The resistance of human skin is, at most, so the minimum volt-age needed can be calculated using Ohm’s law: V= IR = (1A)(104Ω) 104 V. Choice (D) is thecorrect answer. If you used 70 mA as your current, or made a mistake with scientific notation,you might have arrived at one of the other answer choices.

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20. DIn order to minimize the damage, we want less current passing through the person. Looking

at statement I, we see that insulating rubber shoes would inhibit the flow of current, so the per-son would experience less current and thus, the damage would be minimal. We can eliminatechoices (C) and (B). Now, we need only check statement III. The passage states that in general,a current below 10 mA will cause less damage. Choice (D) is correct. Finally, note that a volt-age of 120 V, could still produce a high current if the resistance is very low.

21. DUse the definition of power: P=I2R. The total resistance is the resistance of the wire plus the

resistance of the person with dry skin: 15 Ω + 104 Ω ≈ 104 Ω. So, in this case P= (12)(104) =104 W. The question asks how much energy is dissipated in 45 seconds, since 1 Watt equals 1J/second the total amount of energy dissipated is 104 W× 45 seconds = 4.5 × 105 J, choice (D).

22. BThe first person acts as a resistor with resistance 104 Ω. If a second person latches on to the

first, he too would be a resistor, one that has been added in parallel. Moreover, the second per-son has wet skin, so will have an even lower resistance. Adding resistors in parallel acts todecrease the overall resistance of the circuit. Therefore, if resistance decreases, total currentpassing through the circuit will increase. Since current is ∆charge/∆time, an increase in currentimplies that less time will be required to pass the same amount of charge. The correct answer ischoice (B).

23. BThe passage states that cells have low resistance since there are dissolved ions in the cells.

If there is a high cellular ion concentration current will pass even more readily. Do not forgetthat ions are also called electrolytes, since they conduct electricity. Therefore, more damage isexpected since more current flows – choice (B) is correct.

24. DRecall that a 70 mA current causes more damage to cardiac tissue than a current in excess

of 1 A: eliminate Choice (B). Notice, the passage says nothing about a 70 mA current versus a65 mA current, so we can eliminate choice (A). Finally, we can eliminate choice (C), since thepassage states that current in excess of 1 A can stop the heart, unlike current of 70 mA. The cor-rect answer is choice (D), which is implied by the passage: a current of 70 mA will cause theheart to contract irregularly, resulting in improper blood flow.

Discrete Questions

25. DA species has a full octet of electrons when it resembles a noble gas in electron configura-

tion. When fluorine gains one electron, it resembles neon, so it has a full octet. When rubidiumloses an electron, it resembles krypton, so it also has a full octet. When magnesium loses twoelectrons, it resembles argon, giving it a full octet. When phosphorus gains two electron, it isisoelectronic with chlorine, so it is one electron short of a full octet.

26. DAll that matters in this problem is the relative velocity of the car with respect to the truck. If

you think of the truck as stationary, then the car is moving forward at 40 – 30 = 10 m/s. So the

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real question is, how long does it take the back end of the car to travel the 15 meters it needs forthe back end to come level with the front end of the truck, if the car is moving at 10 m/s?

t dv

1

1

0

5

m

m

/s 1.5s,

answer choice (D).

27. BWhenever you see resistor in parallel, the first thought that should come to mind is, “the volt-

age drop across each resistor in parallel is the same.” In this case, there is a 12 V drop acrosseach resistor. Use that to determine the current through the 5-W resistor:

V IR → I A,

answer choice (B).

28. AOnce a species has gained an electron, its atomic radius will increase. This is because the num-

ber of protons in the nucleus stays the same, but the number of negative charges the nucleus has topull on has increased; the pull is not as great, so the radius expands. This means that chloride andbromide will have larger atomic radii than chlorine or bromine. Br- will have the largest atomicradius, followed by Cl-. Cl- will have a slightly larger radius than Ar, which will be slightly largerthan K+, so the answer is (A).

Passage V (Questions 29-33)

29. CRecall that the rate expression differs from the equilibrium expression. The rate expression

is equal to the rate constant times the concentrations of the reactants; products don’t enter intothe equation at all. Therefore, the answer is k1[CH3OH][CH3OH], which equals k1[CH3OH]2

The answer is (C).

30. BThe first step is to balance the equation, then compare the coefficients with the answer

choices provided. Always balance the atom that appears the least on the right- in this case, car-bon. You’ll need two molecules of carbon dioxide to balance the two on the left. Therefore, c=2.Once you know that, you’ll need to balance the hydrogen atoms. There are six on the left, soyou’ll need three water molecules (each has two hydrogen atoms.) That means that d=3. Nowthat you know the coefficients of the products, you can take a look at the oxygen molecule onthe left. How many are necessary? There are seven oxygen atoms on the right (four from carbondioxide, three from water) and one from the dimethyl ether, so six more are needed from dioxy-gen. This corresponds to three oxygen molecules, so b=3. a=1, so you’re done. The answer is(B).

31. DReaction 1 is an equilibrium reaction, so you’ll want to examine the specific elements of eachside in order to determine how reaction conditions will affect the equilibrium. There are equalnumbers of moles of gas on each side, so altering the pressure at which the reaction is performedshould not affect the equilibrium. However, the reactant side contains two of the same molecule,while the product is comprised of two different gas molecules. Is there a way to alter the prod-

125

12 V5 Ω

VR

7

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ucts? Yes: the products contain a water molecule, so any method of water removal will favor theproducts by Le Châtelier’s principle. Performing the reaction in a moist environment will nothelp, but using a desiccant to remove water will drive the reaction. The answer is (D).

32. DBoiling points depend on a number of factors: molecular weight, structural differences,

hydrogen bonding and other intermolecular forces. Therefore, it is helpful to examine the struc-ture of formaldehyde in order to better answer this question. Drawing yourself a quick sketch ofthe structure of a molecule will only take a few moments, and could save you valuable timewhen answering the question. Take a look at the structure of formaldehyde:

Formaldehyde does not have the correct structure to exhibit hydrogen bonding (no elec-tronegative elements are bonded to hydrogen in formaldehyde.) While it is true that formalde-hyde has a lower boiling point and molecular weight than the CFCs in Table 1, its lower boilingpoint has more to do with the size and composition of the CFCs: they contain many atoms andeach have a halogen atom, which adds electron density. The added electron density contributesto the dispersion forces that lead to a high boiling point. Overall, the key to this question is thatformaldehyde has ketone functionality: it contains a carbon-oxygen double bond which is keyin making its boiling point high compared to other compounds.

33. AWhen two competing reactions are possible, it is helpful to consider exactly what drives each

separate reaction. Reaction 2 is a combustion reaction; products and reactants are in gas form,so performing the reaction at low pressures will favor these products relative to those of Reac-tion 3. Therefore, (B) is incorrect. Removing water from the reaction mixture won’t help eitherbecause it is a product of Reaction 2- removing it will drive that reaction. So, choice (C) is incor-rect. Methanol is a product of the desired reaction, so having an excess of it to begin with willbe detrimental to the equilibrium- excess methanol will cause the reverse of Reaction 3 to pro-ceed more readily, so choice (D) is incorrect. Choice (A) works because less oxygen is requireto perform Reaction 3 than is required for Reaction 2. Choice (A) is correct.

Passage VI (Questions 34-39)

34. CThe passage states that the volumes of the test pieces are equal. Therefore, the increase in

mass is the same as the increase in density. So, what is the percent increase in density going from0.4 × 103 kg/m3 to 7.8 × 103 kg/m3? The density, or equivalently mass, increases by 7.4 × 103

kg/m3.

7

0

.

.

4

4

× 1

1

0

03

3

k

k

g

g

/

/

m

m3

3 100 % ≈

70..24 × 100% =

742 × 100% = 1800 %

O

H H

8

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35. BThe data show that wood essentially does not move in the fluid, which iron and concrete

sink. The density rule implies that iron and concrete have a larger density than the liquid com-pound, while wood has a smaller density: Choice (B) says exactly that. Notice that Choice (A)implies that the unknown compound has a larger density and Choice (C) is completely off thetopic. Be careful with choice (D), the work-kinetic energy theorem states that the net work doneon an object equals the change in the kinetic energy of that object. Here the work done by theunknown object is only part of the net work done: gravity is also doing work on the test piece.

36. D The passage states that the engineer compares the experimental data with the data predicted

by the equation. Clearly, the engineer is looking for deviations in the experimental data. The rea-son deviations may appear is that Bernoulli’s equation does not hold for viscous fluids. Viscos-ity is a measure of internal friction, while Bernoulli’s equation is a conservation of energyequation. Thus, Choice (D) is correct. Choice (A) is certainly untrue, since Pascal’s principlestates this for a non-viscous fluid. Choice (B) is a 180, since Bernoulli’s equation only holds fora non-viscous fluid. Viscosity is a measure of the internal friction of a fluid; it is not proportionalto the average kinetic energy of the fluid. In anything, you would expect a more viscous fluid todissipate more energy and be inversely proportional.

37. CA higher altitude has lower pressure, so the pressure exerted on the top of the fluid is smaller.

However, the air pressure exerted on the hole is smaller as well. The air pressure is the same forthe top of the fluid and the hole: this experiment could be held at any air pressure, and the effecton the speed of the departing fluid would be unchanged. The correct answer is choice (C).

38. CThe continuity equation guarantees that AA × vA = AB × vB = AC × vC = AD × vD. In other

words, as cross sectional area increases, velocity decreases. Here we are looking for the largestvelocity, and so the smallest cross-section area. Choice (C) corresponds to the narrowest regionof the funnel or test tube. The smallest velocity would be a point A.

39. DWhen a substance is heated, a phenomenon known as thermal expansion occurs. If the iron

piece is heated to 100°C, the volume of the cube will increase, though the mass will not; con-sequently, the density of the cube will decrease. Looking at the answer choices, we see thatChoice (A) is incorrect since the heating has negligible effect on the liquid itself. While Choices(B) and (C) may seem appealing, but remember that weight is mg and here neither mass norgravity is changing. Finally, Choice (D) is correct, since Fb = ρgV: if the volume of the cubeincreases, so will the volume of the fluid displaced and subsequently, the buoyancy force.

Passage VII (Questions 40-45)

40. DThe passage tells you that sap contains mostly water but that it also contains sucrose, malic

acid, dissolved salts and phenolic compounds. All of these solutes will raise the boiling point ofa solution. The normal boiling point of water (the solvent in a sap solution) is 100°C, so the tem-perature at which sap boils will be slightly higher than that. Therefore, (D) is correct.

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41. AThe starred atom is a carbon that has two single bonds and a double bond. This means that

it will have sp2 hybridization- three substituents. On questions like this one, don’t get intimi-dated by answer choices like (D)- they are just there to distract you and make you think that youdon’t know your stuff. You’ve put in a lot of time studying, so when you know your generalchemistry, stick to it.

42. BThis is a situation that appears frequently on the MCAT: you are asked to choose from two

possible outcomes, each of which has two possible explanations. In this case, you need to deter-mine whether the purity shows an inverse or a direct relationship to the pressure at which thesyrup is obtained. Once you have done that, you need to decide why it is the case. The first partof this problem is simply math: notice that as the pressure increases, the value for the purity ofthe syrup decreases. Therefore, the purity and pressure are inversely related. This means that youcan eliminate (C) and (D). Now, you need to decide why the inverse relationship exists. Theexplanation portion of the question is a bit more conceptual: does boiling point increase ordecrease with decreased pressure? The simplest way to think about this is to remember the def-inition of boiling point: the temperature at which the vapor pressure of a liquid equals the exter-nal pressure. This will definitely decrease with decreased pressure, so (B) is correct.

43. CBefore you even confront the answer choices, you should recall from the passage that the

purpose of Reaction 1 is to determine how much sucrose is in the syrup. This means that sucroseis the unknown, while dinitrogen tetroxide will be in excess. Vanadium oxide is the catalyst, soit will not be consumed and therefore isn’t a limiting reactant. Water is one of the products, soit is not a limiting reactant, either. Therefore, the answer is (C).

44. CThe passage tells you that 3.36 moles of NO gas are collected from the oxidation of sucrose

in the syrup. This might seem like an unknown volume, but the reaction was performed at STP,so you know that each mole of gas corresponds to 22.4 liters. Just multiply the two values to get75.4 L.

45. DThe passage gave you a table of various bonds and their frequency range for IR spec-

troscopy. Your job is to determine what changes when malic acid is converted to oxaloaceticacid. To do this, you’ll need to compare the two structures. The difference is that one of themcontains an alcohol functionality, while the other has ketone functionality at that site. Thismeans that oxaloacetic acid will not have an alcohol signal in the IR spectrum. Two answerchoices can now be eliminated- (B) and (C). (A) and (D) both deal with the loss of alcohol func-tionality. How are you going to choose between the two? A tells you that losing alcohol func-tionality will cause a signal to disappear in the 1050-1300 cm-1 range. While it is true that thealcohol is lost, the signal in that range won’t go away, because the molecule still contains car-boxylic acid functionality, so the signal will remain. The signal that will disappear is the one thatcorresponds to a monomeric alcohol: 3590-3650 cm-1. Therefore, the answer is (D).

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Passage VIII (Questions 46-50)

46. CThis is a very conceptual question, calling on your knowledge of first principles. Heat energy

flows between to substances when they differ in temperature. The laws of thermodynamics, con-versely, tell us that when two systems are in contact and are both at the same temperature, thatno energy flows. Choice (C) is correct. Notice that answer choice (B) is necessary for heat toflow through matter, but not sufficient.

47. CRadiation can travel through a hard vacuum (the Sun’s energy reaches the Earth), so state-

ment III will be in the correct answer; eliminate choices (A) and (B).

Since there is no matter to either conduct or convect heat, answer choice (C) must be correct.

48. D You are being asked to convert a photon wavelength into a frequency:

ƒ cλ 1.5 1015 Hz,

answer choice (D).

49. ASince the gas lies in a fixed volume, it can’t do work or have work done on it. Consequently,

all the heat energy entering the gas is converted into internal energy. Eliminate answer choices(B) and (D).

Now the question is, does the entropy go up or down? Since heat is being added to the gas,and its temperature is going up, the entropy must be increasing as well. Answer choice (A) iscorrect.

50. AIn the absence of a formula, we’ll have to use common sense. The longer the material, the

longer it will take for the heat to get through. So we expect the rate of conduction to be inverselyproportional to L. Eliminate (B) and (D).

When the temperature difference ∆T is greater, we expect heat to flow faster. Temperaturedifferences are what causes heat energy to flow, after all. The rate of conduction should be pro-portional to ∆T; eliminate choice (C).

Choice (A) is left as the correct answer. The bigger the cross-sectional area, the more “room”there is for heat to flow through.

Discrete Questions

51. DBased on the wording of the questions, your first instinct should be to balance the equation.

When dealing with stoichiometry, it is always worth your time to obtain a balanced chemicalequation. Once you have balanced the equation, you’ll see that the coefficient of hydrogen gasis 6, while that of diborane is 1. 13.4 L of hydrogen represents 0.6 mol of gas at STP, so you’ll

3 108200 10-9

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need to divide that by 6 to obtain the number of moles of diborane. 0.1 mol diborane correspondsto 2.76 g, since the molar mass of diborane is 27.6 g/mol. Choice (A) is designed to trap the testtaker who neglects to account for the fact that there is 0.1 mol of diborane as opposed to 1 mole,so it is incorrect. Choices (B) and (C) are incorrect, and choice (D) is correct.

52. DThe bare facts: an electron (charge = -e) crosses a distance of 30 meters because the poten-

tial difference is 5000 V (be careful of units). You are asked to find the electric field present inthe region. You know a field must be present, otherwise the electron wouldn’t accelerate any-where. Eliminate choice (A).

One of the units for electric field is volts per meter; use the handy formula:

V Ed → E Vd

530000

mV

1.7 102 V/m

choice (D).

53. CRecall that ionic character increases when the electronegativity of the two bonded species

differs greatly. Lithium and hydrogen don’t differ appreciably in electronegativity, nor dosodium and iodine when compared with rubidium and chlorine. This is because electronegativ-ity decreases down a column and increases across a period. Eliminate choice (D); carbon israrely involved in ionic bonding. The answer is (C).

54. AAlpha-decay is the loss of a helium nucleus from a parent, radioactive nuclei. Since helium

has Z = 2 and A = 4, ejecting a helium nucleus must mean losing 2 protons and 4 – 2 = 2 neu-trons; answer choice (A).

55. DThe MCAT expects you to know the various ways of expressing the Gibbs Free Energy. One

of these is in terms of R, T and the equilibrium constant. (D) is correct. The other involves ∆H°and ∆S°: ∆G° ∆H° T∆S°. Commit these to memory so you won’t be thrown off on test day.

56. CTo calculate the change in a quantity, you should figure out the before and after quantities:

KEi 21

(3 kg)(5 m/s)2 37.5 J

KEf 21

(3 kg)(3 m/s)2 13.5 J.

So, |∆ΚΕ| 37.5 13.5 24 J

answer choice (C).

Passage IX (Questions 57-62)

57. AThe passage gives the anode reaction for the battery as it discharges: Cd (s) + 2 OH− (aq) →

Cd(OH)2 (s). In order to recharge the battery, the reaction must be reversed: from this, you can

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eliminate choices (C) and (D). Between choices (A) and (B), choice (B) is not mass-balanced;the nickel reaction is missing from the right hand side. Therefore, choice (A) is correct.

58. BIn order to recharge an exhausted battery, it must be connected to an external battery: in this

case, a Zn-HgO battery is being used. The external battery voltage must be larger than the volt-age of the Ni-Cd battery in its original state. The passage tells you that the Ni-Cd battery has avoltage of 1.4 V. Therefore, you need the Zn-HgO battery to have a voltage greater than 1.4 V.Since the question tells you that E°

anode = −0.762:

E°Zn-HgO = E°

cathode − E°anode > 1.4 V

= E°cathode + 0.762 > 1.4 V

So, E°cathode > 0.638 V, Choice (B).

59. DAs the concentration of aqueous species changes, the reaction quotient, Q, gets larger and

larger. According to the Nernst equation:

∆E = ∆E – RT

nlFn Q

as Q increases, ∆E will decrease. Qualitatively this makes sense, since as the battery dis-charges, the redox reaction is approaching equilibrium and will not “want” to proceed any fur-ther. Using the same logic (and ∆G = -nF∆E° as proof), we expect ∆G to get less negative (i.e.increase) and the reaction to become less spontaneous. Choice (D) is correct.

60. DThe Leclanché cell and alkaline cell are galvanic batteries, while the Ni-Cd is an electrolytic

battery. Remember that the oxidation reaction always occurs at the anode and the reduction reac-tion always occurs at the cathode. However, in an electrolytic battery, the anode is positive andthe cathode is negative. Therefore, for the Leclanché cell and alkaline cell, the cathode is posi-tive, but for the Ni-Cd cell, this is not true. Therefore, choice (D) is correct.

61. CThe passage states that the ZnCl2 and NH4Cl allow ion movement to neutralize the overall

charge moving to the inner core. Here the mixture serves a function exactly analogous to the saltbridge in any other galvanic cell. A salt bridge allows the reaction to occur without causing sep-aration of charge from the moving electrons by allowing ions to flow back and forth. If this mix-ture were missing, the reaction would not occur, since it would result in a too large a separationof charge. Choice (C) is correct. Notice, choices (A) and (B) are incorrect since zinc and man-ganese are the reducing and oxidizing agents.

62. CIf the metal X is being oxidized, it is losing electrons. If ∆G is given, use an equation that

relates ∆G° to amount of electrons lost:

∆G° = −nF∆E°

−347 × 103 J = − n (105) (1.2 V)

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14

n ≈ (31..52

×(1

1005)

5) ≈ 3

Therefore, 3 moles are lost and the net charge on X is +3, Choice (C).

Passage X (Questions 63-67)

63. ASince all four bows shoot arrows at the same angle and reaches the same target when the

string is drawn back at different distances, they must have different bow strengths, which is rep-resented by the spring constant k. Intuitively, one can think of bow #1 as a very “stiff” bow sinceit only needs to be retracted half as much as bow #4, which is a lot looser, with a smaller k value.Hence, bow #1 must have the greatest k value, and when drawn back the same distance as theothers, will have the greatest potential energy stored via the equation E=1/2kx2. The correctanswer is choice (A).

64. DSince all four bows in the experiment shoot arrows at the same angle and reaches the same

target, the velocity of all four arrows must be the same. The only differences between all the tri-als lie in the k value of the bow and the distance drawn back in order to attain the same kineticenergy when the arrow leaves the bow. Choice (D) is correct.

65. DIf the bow needs to be retracted twice as far, then x is doubled. From the equation E=1/2kx2:

in order for E to be the same (not accounting for gravitational potential) when x is doubled, kmust be reduced four fold. Choice (D) is correct.

66. BWhen the archer draws the string back at an angle, the arrow is lowered in height but the

string is stretched. Therefore, gravitational energy is lost while the elastic potential energy isgained, choice (B). Notice this question doesn’t ask about which is bigger, the gravitational PEchange or the change in bow potential energy.

67. AWind resistance always opposes the direction of motion. Since the object is fired upwards

and outwards, the air resistance force would act downwards and towards the archer. This wouldreduce the highest height attainable and the maximum distance attainable. The resistant forcemay also slow down the pellet, but which would increase the time it takes to fall. Therefore onlystatements I and II are correct, and choice (A) is the correct answer.

Passage XI (Questions 68-72)

68. BThe question stem asks for a net reaction. When dealing with half-reactions, you should

always be sure to balance the number of electrons, the charge and the number of each species.In this case, simply adding the two half-reactions gives the correct net reaction since there aretwo electrons on each side. Therefore, (B) is correct.

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69. AIn a battery, the oxidation reaction will always occur at the anode and the reduction reaction

will always occur at the cathode. To remember this, use a tried-and-true mnemonic: red cat; anox. The more difficult part of this question is determining which of the two reaction is the oxi-dation reaction and which is the reduction reaction. Always remember that “leo the lion goesger” or “lose electrons = oxidation; gain electrons = reduction.” Since sodium loses electrons, itmust be the oxidation reaction, while sulfur gains electrons in a reduction reaction. Therefore,(A) is correct.

70. DThe passage states that ions with large, polarizable electron clouds are considered “soft”. It

also tells you that soft anions prefer to bond with soft cations. Sulfur is quite soft, so it will bondstrongly soft cations. Mercury appears two rows below zinc in the periodic table, so it has sub-stantially greater electron density. Hence, its polarizability, or softness, will be greater than thatof zinc. Since Hg is softer, it will bond more strongly to a sulfide ion, making the dissociationof HgS less favored. Hence, its Ksp will be smaller than that of ZnS.

(A) is incorrect because a mercury ion will have a larger electron cloud than a zinc ion. (B)is true (mercury is a liquid at room temperature and zinc is a solid) but this is irrelevant to thequestions. (C) is incorrect because both zinc and mercury have filled s-orbitals. Again, (D) iscorrect.

71. C The key to answering this conceptual question is to compare the reactions upon which Ksp

and Kspa are based. The equilibrium in Equation 3 represents Kspa. Write an equilibrium equa-tion that corresponds to Ksp:

MS(s) H2O(l) M2+(aq) HS-(aq) OH-(aq)

The left-hand side of this equilibrium is favored because of the strongly basic nature of HS-

and OH-; both of these compounds prefer to be protonated, so the metal sulfide cannot dissoci-ate appreciably. For the acidic solution, however, both of the basic compounds appear in proto-nated form as H2S and H2O, so the metal sulfide can better dissociate. This causes Kspa to belarger than Ksp for metal sulfides.

(A) is incorrect because some ionic compounds dissolve better in aqueous or basic solutionsthan in acidic solutions. (B) is also incorrect; it says the same thing as (A), but uses the K val-ues to illustrate it. (D) is incorrect because the charges on the ions are irrelevant to the question.Again, (C) is correct.

72. BBefore you jump into the answer choices for this question, think about what a small Kspa

means: the two ions are tightly bound and do not want to separate, even in acidic solution. So,as the Kspa values decrease, the ions are less inclined to dissociate. The passage gives informa-tion about why this might be the case: hard-hard and soft-soft interactions are favored. This mustmean that as you move from left to right within a period, the electron density increases andcations become softer. As the cations become softer, they bond more tightly with sulfide and theKspa value for the compound decreases.

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(A) is incorrect because cations become softer as you progress from left to right. (C) is incor-rect because larger electron clouds are more polarizable (and hence softer) than small one, solarger cations are softer than small ones. (D) is incorrect because as you move down a column,cations become softer. The choice is also irrelevant because the question asks about movingfrom left to right within a period, not down a column. Again, choice (B) is correct.

Discrete Questions

73. CThe rate law is determined by the slow step in a proposed mechanism. The given mechanism

shows that the second step is slow, so the rate expression should be equal to k[A][B2]. However,the rate expression can only contain species that are in the overall reaction. The overall reactionis not stated in the question, but can be inferred from the question stem. It is A + B → AB. Basedon this reaction, the rate expression can only contain concentrations of A and B. To get rid ofB2, use the first step of the mechanism. The rate of formation of B2 is equal to some rate con-stant times [B]2. The overall rate expression is therefore equal to k[A][B]2, choice (C).

74. BLet’s evaluate the answer choices, one by one. The entropy of the block of ice has not

increased, it has decreased – the ice gets colder. However, the universe must pay the price; sincethe process isn’t reversible, the entropy of the universe must increase. So choice (B) is correct.

The temperature of the air surrounding the block may have increased, or it may havedecreased – we don’t know exactly how the block was cooled. Choice (C) is wrong. The aver-age kinetic energy per molecule, or temperature of the water hasn’t increased, it’s decreased...choice (D) is wrong, too.

75. BIn collisions, the first thing you should think of is: momentum is conserved. In fact, apply-

ing this fact is the quickest way to arrive at the final speed of the combined final mass.

pi (2m)(v) (m)(0) 2mv;

pf (5m)(vf) 5mvf ;

Conserving momentum,

5mvf 2mv → vf 25

v

choice (B). The question tries to fool you by asking for the final speed of the original mass(3m) piece, but since it moves together with the other object, you can proceed normally.

76. BA solution with a high pH has a large number of hydroxide ions in solution. When a strong

base is mixed with a weak acid, some of the hydroxide ions will be neutralized by the acid pro-tons, but some basicity will remain. Choice (B) is therefore the correct answer.

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77. CA beam of light with wavelength 400 nm has

E hλc

1244000eVnm

.nm 3eV

of energy per photon. Since 3100 eV of energy strikes the detector, you would expect a lit-tle over 1000 photons. This jibes with answer choice (C).

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18

78. B

79. C

80. C

81. B

82. B

83. D

84. D

85. B

86. C

87. B

88. A

89. C

90. C

91. C

92. C

93. A

94. C

95. C

96. B

97. B

98. C

99. B

100. D

101. A

102. D

103. A

104. C

105. A

106. B

107. D

108. C

109. C

110. B

111. B

112. B

113. A

114. C

115. D

116. D

117. B

118. D

119. D

120. C

121. C

122. D

123. C

124. A

125. D

126. B

127. A

128. B

129. C

130. D

131. A

132. C

133. B

134. D

135. B

136. C

137. C

VERBAL REASONING ANSWER KEY

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Passage I (Questions 78-83)

Topic and Scope: The character of Dalila in Milton’s Samson Agonistes and the effects thather decisions have on Samson.

Paragraph 1 introduces the poem and discusses the hero/antihero dichotomy present betweenSamson and Dalila, although it is not yet clear that Dalila will be the focus of the passage. Para-graph 2 introduces the chorus of visiting friends, who try to convince Samson that he is blame-less; then, Samson explains his culpability. Paragraph 3 compares Samson Agonistes andParadise Lost, but delineates the differences between the characters of Eve and Dalila. Para-graph 4 examines Dalila’s motives and explains Milton’s comparison of her character to the ser-pent in the Garden of Eden. Paragraph 5 discusses Dalila’s employment of antifeminist rhetoricin order to make her argument. Paragraph 6 reemphasizes the strength of Dalila’s character.

78. BIn paragraph 1, Dalila is introduced as the character who is to be most closely studied choice

(B). Paradise Lost is discussed, but only for the span of one paragraph—the essay is not a thor-ough comparison of Paradise Lost and Samson Agonistes, as choice (A) would imply. We aretold in paragraph 5 that Dalila employs sexual essentialism and antifeminist rhetoric, ruling outchoice (C). While the biblical book of Judges is mentioned, it is included merely as a referencepoint for the original Samson and Dalila story, and is not discussed at length nor compared toMilton’s version, as choice (D) would imply.

79. CDalila employs all of these tactics except for choice (C), arguing the superiority of the

female intellect to the male. In Paragraph 5, the author discusses Dalila’s appropriation of patri-archal stereotypes in choice (A). In paragraph 1, we learn that Samson felt he had been ensnaredby his lust for Dalila, choice (B). Finally, we see Dalila apologize to Samson in paragraph 4, rul-ing out choice (D).

80. CIn paragraph 4, the author references the serpent in order to emphasize how strongly Milton

felt about Dalila’s deceptive nature; thus, choice (C) is correct. While the author mentions theCleopatra and the asp, this is only to give a nod to a common association, and not necessarilythe association that Milton intended—making choice (A) incorrect. In mentioning the serpent insuch close affiliation with Dalila, the author is making the point that Milton associated Dalilamore with the serpent than with Eve, so choice (B) is incorrect. Finally, while the asp in the storyof Cleopatra is mentioned, there is no real comparison of it to the serpent of biblical tradition,rendering choice (D) incorrect.

81. BAll of the answer choices are valid comparisons except for choice choice (B). As discussed

in paragraph 3, Eve beseeches Adam to taste the fruit not out of deception, but out of an honestdesire for him to gain the knowledge that she believes she has gained. Both Adam and Samsonexperience a “fall” because of a decision made by a female character, choice (A). To differingdegrees, both Adam and Samson accept the blame for their situation, choice (C). And, both sto-ries are expansions upon biblical passages, choice (D).

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82. BNone of the answer choices would adequately challenge the author’s claim that Milton views

Eve and Dalila as two clearly different “types” of women except for choice (B). If Eve madeAdam taste the fruit so that she could gain power over him, she would fit the same mold of thedeceptive and treacherous woman that Dalila does. Even if Samson’s punishment was equalAdam’s, this does not have any bearing on the personalities of Eve and Dalila, making choice(A) incorrect. Similarly, even if Milton has given his Samson a few more positive characteris-tics than are granted to the biblical Samson, this does not have an effect on the way Dalila treatshim or on the way Eve treats Adam, thus choice (C) is incorrect. Finally, even if Dalila was oncegenuinely in love with Samson, as (D) suggests, this does not have any real bearing on her laterdecision to betray and deceive him—and it does not bring her closer in personality-type to Eve.

83. DThe best inference is that the author of this passage is choice (D), a literary scholar. The pas-

sage does not include enough comparisons between Samson Agonistes and the Old Testamentstory to conclude that the author is a biblical scholar, making choice (A) incorrect. BecauseDalila uses antifeminist rhetoric, she would most likely not be included in research of feministcharacters, making choice (B) incorrect. Finally, choice (C) is too broad and is not entirely on-topic; nowhere in the passage does the author make mention of general societal attitudes towardfigures such as Dalila.

Passage II (Questions 84-90)

Topic and Scope: the paleogeology of southern Chilean islands. the formation of a group oflimestone islands found off the coast of southern Chile.

Paragraph 1 explains the present terrain of the islands and introduces the scope of the pas-sage. Paragraph 2 explains how and when the limestone formed. Paragraph 3 explains how thetheory of the plate tectonics can be applied to the formation of these islands. Paragraph 4describes the supercontinent Pangaea and discusses how the islands came to be in their presentlocations. The passage concludes by giving information about other parts of the world wheresimilar islands can be found.

84. DThis question requires that you determine why the author discusses Pangaea. Paragraph 4

describes Pangaea and explains how the Earth looked when the Tarleton limestone formed. (A)is a faulty use of a detail found in paragraph 2. Pangaea has nothing to do with foraminfera for-mation. Choice (B) is not directly related to Pangaea as well as being inaccurate. Rocks wouldtravel from the lithosphere down to the athenosphere so eliminate choice (B). Choice (C) iswrong. Paragraph 1 explains karst formation, which has nothing to do with Pangaea. Choice (D)is the correct choice. It restates one of the main ideas of the passage: limestone that formed inthe Tropics is now found in a subpolar environment.

85. BParagraph 4 tells you that during the Triassic and early Jurassic periods (lasting from 230 to

160 millions years ago), the Tarleton limestone accreted to the South American tectonic plate.Pangaea would have had to have broken apart for this accretion to occur. Choice (B) is correct.You can eliminate choice (A) as incorrect since Pangaea assembled much earlier than the Trias-sic period in the early Permian. Choices (C) and (D) are incorrect since you know that during

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the Triassic the limestone was colliding with the South American plate and therefore could notbe located in either the Tropics or above the Equator.

86. CChoice (A) would support the author’s conclusion, found in paragraph 4, that the Tarleton

limestone was formed separately from the Chilean mainland. Choice (B) is irrelevant to the con-clusions in the passage. Whether or not the process that formed Pangaea was gradual, Pangaeastill formed. Choice (C) would refute the author’s conclusions. Paragraph 3 explains that vol-canic and earthquake activity is evidence of plate tectonics, and plate tectonics is necessary toexplain how the limestone moved position. Choice (D) can be eliminated since it is outside thescope of the passage and would neither strengthen nor weaken the author’s conclusions.

87. BAs stated in the first sentence of paragraph 2, scientists dated the limestone in order to date

the islands. Choice (B) restates this concept. Choice (A) confuses the purpose of the passagewith a detail contained within it. The author only uses the foraminifera to date the islands.Choice (C) is outside the scope of the passage. Choice (D) is wrong because the limestone hasnothing to do explaining plate tectonics.

88. AStatement I is correct. The passage clearly says that Tarleton Island is not made of the same

rock as the mainland. Statement II is a distortion. The island is formed from what was once acoral reef but is no longer. The island is limestone. Statement III is also incorrect. As paragraph4 states, the plate collision happened in the Triassic or early Jurassic period not the Carbonifer-ous. So the correct answer is choice (A) since only Statement I is correct.

89. CTo answer this question, apply what the passage says about plate tectonics to how paleoge-

ologists determined the origin of these Chilean islands. Choice (A) is a faulty use of a detailfound in the passage. The formation of limestone does not apply to plate tectonics. While choice(B) makes logical sense since the passage talks about these geologic events resulting from mov-ing plates, the question asks how plate tectonics relates to paleogeology and this answer does-n’t address the question. It is outside the scope of the passage. Choice (C) is correct. Continentaldrift is explained by plate tectonics. Choice (D) is another faulty use of detail.

90. CThis question asks you to analyze how the author can tell that an equatorial reef has been

dispersed. Choice (A) is a distortion of a detail in the passage. The age of the Chilean mainlandhas nothing to do with the dispersal of an equatorial reef. Choice (B) is outside the scope of thepassage. Choice (C) supports the idea that a coral reef was dispersed by plate tectonics. Choice(D) would weaken and not strengthen the argument.

Passage III (Questions 91-97)

Topic and scope: the relationship between biography (fact) and fiction illustrated by a dis-cussion of the journals of Canadian author Lucy Maude Montgomery.

Paragraph 1 is a very general paragraph, introducing the concept of the relationship betweena writer’s life and his or her fiction, using the example of Jane Austen to illustrate the difficul-ties of knowing any great details about an author’s life.

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Paragraph 2 introduces Lucy Maud Montgomery and briefly discusses her novels.

Paragraph 3 gets to the meat of the question – Montgomery’s journals – explaining thatalthough the journals might appear to be a solid, day to day record of the author’s life, thingsaren’t so simple. Montgomery may have rewritten some passages, and her work as an authorcaused her to write up her life in a neat narrative format, which may not give an accurate depic-tion of her life.

Paragraph 4 gives us a very fuzzy sum up, suggesting that we still don’t know what effectreal life has on fiction – although Montgomery’s journals do suggest that fiction writing mayhave an effect on how novelists depict their own lives

91. CStatement I is true. That lengthy third paragraph cites the fact that some of Montgomery’s

journal pages were cut out and replaced. While these alterations don’t prove anything, the authormentions them to suggest that Montgomery censored the originals. Eliminate answer choice (B).Statement II is false. According to the passage, we don’t know if art imitates life, or vice versa.Authors may put “real life” material in their novels, just as they may put novelistic elements intotheir journals or autobiographies. But the incidence of real events in the novels doesn’t mean thatthose events aren’t factual. Eliminate answer choice (D). Statement III is true. According to thepassage, it’s the novelists’ urge to write in a narrative format that instinctively causes them to telltheir own life stories in a narrative format. In other words, novelists think in novel forms, sowhenever they tell a story, it comes out sounding like a novel. Eliminate choice (A). The answeris (C).

92. CAnswer choice (A) is clearly out; there’s nothing in the passage to suggest that author’s

reject boring material, only that they reject material that doesn’t seem to fit the narrative format.In choice (B), we don’t know that the authors – or even Montgomery – are trying to make them-selves look better, only that they may, when writing up their lives, make their biographies seemlike novels. In choice (D), it’s still not clear from the passage whether novelists put real lifeevents in their novels, or put events from their novels into their autobiographies.

93. A As the passage itself says, we have only the copies of the journals – not the originals. Thus,

we have no idea of what Montgomery changed, only the physical evidence that some pages fromthe journals were replaced by other pages. The author states that the page introducing Mont-gomery’s husband was one of those pages. It’s certainly possible that Montgomery decided torewrite her initial impressions of her husband. But it’s equally possible that the removed pageoriginally contained a description of something else entirely, and Montgomery decided toremove it, and replace it with a description of her husband. Thus, the author’s comment that“later knowledge of him may well have caused her to rewrite some of her original impressions”assumes that these impressions were there. Of the other answer choices, choice (B) goes too farin the second half. The author never states or assumes that Montgomery’s impressions of herfuture husband were misleading and inaccurate, only that they may have changed. Choice (C) isa good trap answer, since it paraphrases the text, but we’re not looking for a paraphrase here,we’re looking for the flaws in the author’s reasoning. Choice (D) also simply restates theauthor’s argument, instead of pointing out the flaw in the argument itself.

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94. C Montgomery’s psychological problems are never mentioned, so choice (D) is out. The

author seems to think that the relationship between truth and fiction is still pretty unclear, so it’sdoubtful that just one more volume of Montgomery’s journals will help matters much. Choice(A) is out. Since the first four volumes didn’t tell us much about Montgomery’s editing process,only that she replaced and removed pages from her journals, we can’t infer that the fifth volumewill tell us anything. According to paragraph three, however, events in the novels mirror thosein the first four volumes of the journals, so it seems reasonable to expect that this trend will con-tinue.

95. C Choice A is opposite; we are expressly told we have comparatively little of Jane Austen’s

writings other than her novels. Austen’s family is never directly criticized, despite the possiblymisleading use of the word “infamous”. That said, the author simply mentions the destruction,and does not specifically blame the Austen family for it, so answer choice (B) must be wrong.Choice (D) is way outside the scope; we have no way of knowing from the passage how muchtime Austen scholars have spent studying the relationship between Austen’s letters and her fic-tion.

96. B The author might use choices (A) and (C) to study the relationship between fact and fiction,

but not between the author’s published journals and bestselling novel. The author might usechoice (D) to study how novelists can reshape and self-edit their autobiographies, but thatwouldn’t necessarily be the best way to study the relationship between the journals and a novel.A textual comparison, as in choice (B), could show any direct links between the journals and thenovel.

97. B This is clearly not irrelevant to the author’s argument, so choice (D) is out. We don’t know

that Montgomery claimed that she had a prodigious memory; all we know is the author says thatshe did, a claim apparently supported by evidence outside the journals. So (C) is FUD or a Dis-tortion. (A) is actually the direct opposite of what we’re looking for. The author suggests thatMontgomery did have access to all of her journals despite her claims that the earliest ones weredestroyed, a point proven by her quotations from the journals she wrote when she was ten. IfMontgomery later gave different versions of events from those in her journals, however, then it’smore likely that she was quoting from memory and didn’t have access to those journals – thatMontgomery was telling the truth all along and wasn’t editing her journals. In any case, findingthe journals that Montgomery wrote when she was ten would certainly tell us more about herwriting habits, so the best answer would be (B).

Passage IV (Questions 98-103)

Topic and scope: Fan-fiction writing, in particular the legal issues surrounding it and the dif-ficulty of regulating it.

Paragraph 1 introduces us to fan-fiction writing, defining it as works written by amateurwriters which use the characters or settings created by professional writers.

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Paragraph 2 explains that writers in the medieval and Renaissance eras also wrote storiesbased on previously created characters, as did nineteenth century writers trying to make somequick money. The trend was dying down, however, until the Internet.

Paragraph 3 explains some of the reasons why fan-fiction is so hard to regulate. Sites are dif-ficult to find and may be located in foreign countries. Plus, shutting down the fan-fiction sitesmay anger fans.

Paragraph 4 sums up the defenses fan-fiction writers give for their writing. People did it backin the Middle Ages, fan-fiction might even increase sales for professional writers, well knowncharacters might be in the public domain, and the fan-fiction authors aren’t making any money.

Paragraph 5 is a short summary paragraph, and just tells us that fan-fiction cases for the mostpart aren’t ending up in court.

98. C The author says that medieval and Renaissance authors wrote stories based on established

characters, but there’s no sense that either the author or the fan-fiction writers consider this tobe a legal precedent. That gets rid of choice (A). Choice (D) is a lot trickier, but the fan-fictionauthors also aren’t assuming that the historical precedent alone is enough to change internationallaw; it’s just one thing that they’re bringing up to make their case. We don’t know, as in choice(B), whether or not the medieval or Renaissance authors objected to other people using theirwork, but that’s also not quite the issue here. (The passage makes it pretty clear that at least someof the professional authors don’t mind other people using their characters, anyway – but thatdoesn’t change the fact that some professional authors do mind, a lot.) The main thing that thefan-fiction authors are assuming, however, is that the medieval/Renaissance situation can berelated to today’s situation. That might be true, but it’s still an assumption: as the passage makesclear, a lot of things – attitudes towards writing, payment, and legal issues have changed sincethe Renaissance period.

99. B The two paragraphs above discuss the difficulty of regulating fan-fiction, thanks to the

nature of the Internet and the fact that most fan-fiction writers don’t seem to think they’re doinganything wrong. The author seems to think that this situation will continue, so she can’t beassuming that a clear legal statement will take care of the situation. That takes care of (A).Choice (C) has two problems. First, choice (C) restates something in the passage. In an assump-tion question, you’re looking for something that the author has left out. Second, (C) contains thatoften tricky word “all”, which often – as in this case – indicates that the answer choice is tooextreme. The author’s not saying that all amateur writers can’t afford the expenses of a courtcase – just that most can’t. The passage does mention the threat of lawsuits as one way of deal-ing with fan-fiction, but never states or suggests that this is the only legal avenue available towriters – meaning that we also need to throw out choice (D). The last two sentences, however,contain a major assumption. The author tells us that these fan-fiction cases have mostly not goneto court, which has prevented professional authors from making a clear legal statement regard-ing the harm caused by fan-fiction. But why do they need to make these statements in court? It’sequally possible that the professional authors could take their case to Congress or other legisla-tive bodies, and ask for clear revisions to the law explaining the harm found in fan-fiction.

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100. D Choice (A) looks very tempting. After all, the passage says that these amateur writers are

generally breaking copyright law. But if you note, the passage also mentions one exception:cases where professional authors have given permission to amateur authors to write stories basedon their works. If the amateur author’s story fits into that category, it isn’t breaking copyrightlaw – making (A) wrong. If, however, the story doesn’t fit into that categories, it probably isbreaking copyright law, making (C) wrong. The end of the passage tells us that most amateurauthors can’t afford expensive litigation procedures, and thus apparently either close down theirsites or move them to foreign domains, and avoid facing expensive litigation, making (B) wrong.Paragraph 3 tells us that the fan-fiction sites are extremely difficult to regulate and remove,which ties in nicely with answer choice (D).

101. A We’re told that international copyright laws came into existence in the twentieth century,

which means that writers must not have been protected by them. With choice (B), we’re toldonly that medieval and Renaissance writers frequently wrote about well-established characters;we don’t know if they necessarily preferred to do this, or if they wrote about these charactersbecause they had to. Choice (C) gets its timing a bit confused – it’s the nineteenth century writ-ers and publishers that were interested in making quick money out of other people’s charactersand settings, not the medieval and Renaissance writers. With choice (D), we don’t know thatmedieval and Renaissance writers never wrote original stories with original settings, only thatthey frequently wrote about established characters and settings. That doesn’t mean that they did-n’t take the time now and again to write something brand new.

102. D Choices (A), (B), and (C) are all ways that fan-fiction writers try to get around breaking

copyright law, or claim that they aren’t breaking copyright law. But the first thing that must bedetermined is whether or not the work is based on a professionally written novel, TV series, ormovie in the first place. A story set in outer space may be perfectly legal; if it uses the Romu-lans from Star Trek, it probably isn’t. With choice (A), a story printed on a foreign site may stillbe breaking international copyright laws; it’s just going to be very difficult to take down. Withchoice (B), fan-fiction writers use the “but we aren’t getting paid!” line as an excuse – but thatdoesn’t mean that they aren’t still breaking the law. Choice (C) is another good excuse for fan-fiction writers – but even if their work ends up increasing the sales of professional authors, thatdoesn’t necessarily mean that they aren’t still breaking the law.

103. A We’re told that the chief argument made by the fan-fiction writers is that they aren’t getting

paid for their work, so where’s the harm? If the fan-fiction writers can prove that the professionalwriters are not harmed by fan-fiction, this would strengthen their argument. (The reverse, ofcourse, is equally true: if the professional writers can prove that they are being harmed by fan-fiction, it would destroy the argument of the fan-fiction writers.) Even if the fan-fiction authorsaren’t getting paid, they might still be harming the professional writers. (Readers might read thefree fiction available on the Internet instead of paying for the professional work.) Besides which,the passage makes it clear that whether or not the fan-fiction writers get paid, they’re still break-ing international copyright law, which means that (B) is wrong. An independent determinationof the harm caused by fan-fiction might help out the amateur writers, or it might actually workagainst them, which makes choice (C) wrong. A clear legal statement detailing which charactersand settings are in the public domain, and which aren’t, might help to clear things up – but itwon’t strengthen the argument of the fan-fiction writers who are writing stories about characters

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that clearly aren’t in the public domain, so (D) is wrong. The correct answer choice here has tostrengthen the argument of all the fan-fiction writers, as choice (A) does here.

Passage V (Questions 104-110)

Topic and Scope: The debate over the applicability of European languages by Africanauthors.

This passage consists of five paragraphs and is primarily expository, describing a disagree-ment over the appropriate language for use by African authors.

Paragraph 1 lays out the two sides of the debate and locates it in time (since the 1960s).

Paragraph 2 discusses the impact of European colonialism on literature in Africa, in increas-ing the volume of written literature and leading to a new identification of writing and authors as“African.”

The third and fourth paragraphs discuss the two sides of the debate through looking at theworks of two authors said to be typical of each position.

Paragraph 3 explores the position of Chinua Achebe; Achebe argues that English is anAfrican language and that it embodies and unifies the nation of Nigeria.

Paragraph 4 author discusses writer Ngugi wa Thiong’o; Ngugi argues that the use of Euro-pean languages reproduces the colonial encounter and that an African author has a responsibil-ity to the African people to use indigenous languages. In the final paragraph the author discussessome other African writers who don’t take one side or the other in the debate.

104. CIn line 52 the passage states that Ngugi’s contributions “focused the issue and increased

attention to it.” Answer choice C is a paraphrase of this. Choice (A) is unsupported. Althoughthe passage states that most literature in Africa is published in European languages, it does notimply that the percentage increased due to this debate. B is opposite. The passage states that fewwriters have switched to writing in African languages (line 54). Choice (D) is outside the scope.Literacy is not addressed directly in the passage.

105. AThe question asks which statement Ngugi would disagree with. Thus, three of them are all

either statements he might agree with, or deal with issues we don’t know his opinion on. Ngugi’sposition is characterized in paragraph four, so to answer this question you want to skim thatparagraph again. At the end of the paragraph it states that Ngugi believes there is no intermedi-ate position between oppression and resistance. We know he equates the use of European lan-guages with oppression and the use of African languages with resistance. Therefore, he woulddisagree that this difference could be bridged by the use of both languages.

Choice (B) is outside the scope. Although Ngugi argues that the author’s main responsibil-ity is to the majority of Africans, who don’t read a European language, we can’t infer that hewould disagree with this statement. People literate in a European language might be a muchbroader audience than those who aren’t; they just are not Ngugi’s audience of choice. For exam-ple, he might agree that writing in English would allow an African writer to reach readers in

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Europe or the United States, but he isn’t writing for those people. (C) is opposite. This is a state-ment Ngugi would agree with. (D) is Outside the scope. Ngugi is not concerned about theexpressive potentials of language, according to this passage.

106. B Ngugi is introduced in paragraph three as having written in Gikuyu since the 1960s. Later in

the passage the author refers to other writers who have not “switched” to writing in an Africanlanguage as Ngugi did. Thus, we can infer that Ngugi did change language in the 1960s. Thispoint is made only subtly in the passage so this might be a good candidate for answering byeliminating wrong answer choices.

Choice (A) is outside the scope. Whether they addressed each other directly is never men-tioned nor implied. (C) is also Outside the scope. Preference? We don’t know anything aboutwhat readers prefer, as the author notes in the final sentence. (D) is too extreme. We know that“many” African authors have taken sides in the debate, but not “most.” Further, the passage dis-cusses cases in which authors took a position somewhere other than “one side or the other.”

107. DSince this is primarily a descriptive passage, the author’s own point of view is not seen

clearly for most of the passage. Much of the passage consists of reporting of others’ opinions.The clearest expression of the author’s own idea is in the final sentence, which concerns the lackof attention to readers in this debate. Choice (D) reflects this idea.

It’s important in a passage like this one to be clear about whose opinion is whose. Some ofthese were reported in the passage, but attributed to others. Choice (A) was cited as a belief thatNgugi claims lingered in Africa at least until the 1980s, but it is not attributed specifically.Choice (B) is Ngugi’s point of view; we can’t infer that the author endorses it. Choice (C) is notsupported by the passage. There is no suggestion of chaos or disarray in literary circles as aresult of the debate.

108. CWe know that Achebe and Ngugi don’t agree on much – to answer this question, you would

want to re-skim the paragraphs about those two authors (paragraphs three and four) and then goto the answer choices. Both authors make their claims about language as it relates to their com-munity or communities of significance. For Achebe, this community is the post-colonial nation-state, which is unified by a European language. For Ngugi, the community is the broader cultureof African people, which is still suffering from the legacy of colonial oppression. Though theseare very different arguments, both link language to the broader socio-political context.

These are statements that reflect the beliefs of just one, or neither, author. (A): Achebe atleast would disagree with this, as his audience is his national community. Ngugi would proba-bly also disagree, as he is writing (albeit in a local language) for the whole continent. (B):Achebe would make this claim, but Ngugi wouldn’t; we don’t know his stance on contemporaryAfrican nations, except that he abhors any remnant of colonialism. (D): Achebe might agreewith this, as the passage states that he writes in Igbo as well as English; however, Ngugi appearsto see language choice as an unequivocal expression of political loyalty – “on the side of oppres-sion [or] the side of resistance” (line 42).

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109. CThis question concerns colonialism, which is addressed in most detail in paragraph two

(though also mentioned in paragraph four). To answer the question, you should review that para-graph again before going to the answer choices. Since it’s an “EXCEPT” question, three answersare supported by the passage and the correct one is not. The paragraph states that written litera-ture existed before colonialism, choice (C) is the correct answer (not supported by the passage).

Answer choices (A), (B) and (D) are all mentioned in paragraph two as effects of Europeancolonialism in Africa.

110. BSince the passage mentions that Achebe is Nigerian, to answer this question you should skim

paragraph three to recall the information about Nigeria’s language situation. It states thatEnglish, the national language, is believed (by Achebe) to unify the country despite its multitudeof local languages. Thus, choice (B) is correct.

All of the wrong answer choices distort the information given in paragraph three, or are out-side the scope.

Passage VI (Questions 111-118)

Topic and Scope: language in non-human primates.

Paragraph 1 lays out the topic and scope, and hints at the author’s purpose. Paragraph 2 dis-cusses non-human communication systems known as call systems, which are used by non-human primates in the wild. Paragraph 3 describes how language differs from call systems,focusing on the features of symbolism, productivity, displacement, and cultural transmission ascritical to true language. In Paragraph 4 the author discusses the success of various apes whohave learned language. Paragraph 5 presents a later study that attempted to prove the apes hadnot truly learned language; the author discounts this study. Paragraph 6 sums up by asserting thatthe similarities between humans and apes are greater than the differences.

111. BThe author of this passage believes that non-human primates have an innate ability to learn

and understand language, although they have not so far displayed this ability in the wild. Com-bination of calls would provide further evidence of this ability.

Choice (A): Outside the scope. The passage does not discuss the possible effects that con-tact with humans would have on wild chimpanzees. (C): The author thinks that chimpanzeesalready possess linguistic ability. (D): Call systems might display some features of true lan-guages, but the passage outlines several differences between them and language; the lack of pro-ductivity is only one of them. We cannot infer that a call system with one feature of languagewould qualify for status as a language; for example, it would still lack symbolism, which theauthor considers the most important element of language (line 21).

112. B Displacement is the ability to discuss items that are not present or do not exist. The passage

links lying with displacement specifically in reference to Lucy: she “not only was able to dis-cuss items she couldn’t see, she could also lie” (line 40). Lying may be the ultimate form of dis-placement: creating a scenario in your mind and talking about it.

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Choice (A): Productivity is the ability to combine words or symbols to create new meanings.There is no evidence that Lucy was combining signs into new meanings when she lied. (C):Though Lucy’s response could be interpreted in this light, this answer choice does not answerthe question posed. The question asks about a linguistic principle, not a judgment about the eti-quette of lying. (D): Linguistic transmission presumably refers to the ability to pass on languagethrough learning (line 30). It is unrelated to Lucy’s ability to lie.

113. ASymbolism refers to an arbitrary connection between a symbol (such as a word, sign, sound

or picture) and the thing it represents. This connection is also conventional, meaning that it isunderstood by some community of language users. There is no inherent connection between theshape of a sign and the meaning “stop”: these differences are based on regional or geographicalcustoms.

Choice (B): Although not exact, “meow” is inherently connected to the sound that a catmakes, not arbitrarily as a symbol would be. (C): The mercury in a thermometer is directlylinked to the temperature. It is not based on custom or convention that higher temperatures makemercury expand, but on natural properties of mercury. (D): The connection here may be arbi-trary, but there is no symbolism in a young child’s scribbles because the pictures are not con-ventionally understood, and there is no indication that the same scribbles represent the sameitems each time. Pictorial symbolism (such as a stick figure) would be recognizable by peopleother than the artist.

114. CThe author admits that call systems and languages are both used for communicating, but

notes that there is “a qualitative difference between human languages and primate call systems.”This suggests that the differences are more important to the author than the similarities.

Choice (A) is opposite. Paragraphs 2 and 3 delineate several important differences betweenlanguages and call systems. (B): Outside the scope. The author does not speculate on the evolu-tion of language. (D) is opposite. The passage spends a lot of time discussing the differencesbetween these types of communication.

115. DThe author points out that Terrace’s experiment was conducted to disprove the theory that

chimpanzees could learn languages. However, she asserts that the reason this particular animalfailed to learn language was lack of linguistic interaction with humans. Since humans also needobservation and personal involvement to learn language, the failure of the study supports thenotion that the apes’ acquisition of language is essentially the same as humans’.

Choice (A) is opposite. Terrace may have been trying to disprove this, but to the author ofthis passage, the experiment actually supports the theory. (B) and (C): The author does implic-itly disagree with the methodology, but this criticism is not the purpose of the example. She doesdraw an analogy, so that we can understand the meaning of the example.

116. D In this passage the author discusses what she calls “human language,” to differentiate it from

non-human communication systems. The author never refers to other kinds of language nor toother communication systems that can be called language. The fact that she uses “human lan-guage” interchangeably with “language” implies that they are one and the same. The implica-

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tion of paragraphs 2 and 3 is that language differs fundamentally from other forms of animalcommunication.

All of the wrong answer choices are statements that the author agrees with. (A): The authorstates as much in the final paragraph, and also provides examples in paragraph 4 of great apesthat have learned language. (B): The author agrees with this statement. In the final sentence shestates that the difference between apes’ use of language and humans’ use of language is a dif-ference of degree. (C): The author discusses this in the third paragraph, where linguistic learn-ing is introduced, and at the end of the fifth paragraph (lines 63-64). She states that a child mightnot learn language at all without interacting with others.

117. BChildhood contact with users of a language is mentioned as a prerequisite of language in

paragraph 5. A biological capacity for language is implied throughout the passage, particularlyin the final paragraph where the author mentions “naturally present cognitive capacities” (lines68-69). The passage specifically suggests that the anatomical features enabling speech are notnecessary for language, since apes have been taught language although they cannot speak (lines35-37).

Even if confused by the roman numeral format and the choices, students should have elim-inated choice (C) because language cannot be learned by an organism without the biologicalability to learn and use it. This passage centers on the question of whether non-human primatesare capable of language, and the author argues that they are. So the correct answer choice mustinclude statement II.

118. DCall systems are introduced in the passage as a form of communication other than language.

Both call systems and languages are used by primates: call systems by non-human primates andlanguages primarily by humans. “Natural systems” most likely refers to their use outside ofexperimental settings, which is accurate for both call systems and languages if we remember thatlanguages are used by humans.

Choice (A): Language has not been observed in wild, non-human primates, as stated in thefinal paragraph. (B): Call systems do not involve symbolism, which is a key feature of language.(C): This is true of call systems, but not of languages; we can and do combine words intophrases, sentences and compound words.

Passage VII (Questions 119-124)

Topic and Scope: The Florida Homestead law, and specifically how that law may be used bydebtors from other states seeking to avoid paying off their creditors.

Paragraph 1 describes a hypothetical case of a woman managing not to pay a large courtordered settlement, a scenario that might seem unlikely, but is apparently possible in Florida.

Paragraph 2 describes Florida’s Homestead law, explaining the unusual nature of the state’slaw. First, Florida homeowners get a substantial property tax break; second, their homes cannotbe taken by creditors, even if they run up huge credit card bills.

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Paragraph 3 explains Florida’s other advantages for debtors – no state income tax, and cred-itors aren’t allowed to take money directly from a debtor’s wages. (Sounds like a great deal,doesn’t it?) The paragraph also notes that these laws don’t just help swindlers; they protect poorand working class people as well.

Paragraph 4 suggests one solution for the hapless creditors: a proposal currently under con-sideration by the Senate, which would apparently remove some of the benefits of the HomesteadLaw. The problem? This law may be unconstitutional.

119. D The passage tells us that only one-half of one acre in a municipal locality can be considered

part of a homestead. Therefore, in this hypothetical case, the remaining one and a half acres arenot part of the homestead. Problem is, nowhere in the passage do we hear how Florida law treatsreal property that isn’t part of a homestead – only about the protections that the homestead getsunder Florida law. We’re never told what creditors can or cannot do to real property that isn’tpart of the homestead. It’s possible that creditors could force the property owner to sell, but wedon’t know that, so A is out. It’s unclear from the passage just what the proposed Senate legis-lation would cover, so B is out. Choice (C) sounds possible, but according to the passage, thedebtor would be able to find shelter from creditors on at least one half acre of the property, sothe debtor should be able to find relief from creditors on at least part of the property. What’spretty clear, however, is that the law covers only that one-half of one acre. The rest of the prop-erty would not be immune from creditors, as in answer choice (D). We don’t know what thecreditors could do with the property, but it wouldn’t be covered under the homestead law.

120. C The main financial benefit – and problem for creditors – with the Homestead law isn’t so

much the tax benefit, but the ability to shelter assets from creditors. It follows that this is prob-ably why the author thinks Florida voters will be so resistant to the idea of changing the Home-stead law. But, if the law changing the Homestead law also created new shelters for Floridaresidents to hide assets in, Florida residents might well be less interested in defending the Home-stead law. (That would still leave the creditors out in the cold, but this question is about the vot-ers, not the creditors.) This gets rid of choice (B), which suggests that homesteaders are mostlyconcerned about paying property taxes; as the rest of the passage shows, however, the dimin-ished property taxes on a homestead is not the main reason people pour money into them. As faras (A) goes, many Florida voters might well be creditors...but that doesn’t necessarily mean thatthe majority of Florida voters are creditors (or that the state legislature will listen to them.)Besides, the voters who are creditors may well want to keep the ability to shelter assets in home-steads themselves. Choice (D) is way out of scope. The issue here isn’t the state’s tax revenue,but tax laws. (It seems to be pretty clear that Florida has other ways of raising tax money, in anycase.)

121. C A detail question. The passage tells us that only three liens against homesteads are legal in

Florida: those for unpaid taxes, mortgages, and home improvements. It follows that if homeimprovement liens are legal, the people who did the home improvement work would be able toput a lien on the homestead – if they hadn’t been paid, of course. Of the other answer choices,(A) is opposite; the sentence that states the lien can be imposed also states that a sale may beforced for its payment. Choice (B) also directly contradicts the passage, since we’re told that inFlorida, wages can’t be attached. Choice (D) may sound plausible, but we don’t know whatFlorida law says about a homestead owner’s other assets. It’s probable that creditors – including

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people who did home improvement work – could demand payment from these, but we don’tknow that.

122. D We don’t know that all debtors have IRAs and pensions, only that some do, which throws

choice (A) out. (Keep an eye out for extreme-sounding answer choices like this one.) (C),income tax, is outside the scope of the passage. We know that Florida doesn’t have a stateincome tax, but we know nothing about federal income taxes, which may or may not apply topensions and IRAs. Choice (B) may be true – there’s nothing to say that a person couldn’t applyincome from a pension or an IRA towards a homestead – but it’s not mentioned in the passage,so (B) is out. The hypothetical case given in the first paragraph tells us that the woman in ques-tion is enjoying her income from pensions and IRAs – and not paying her creditors. It must fol-low that the creditors are unable to attach the assets and income from the pensions and IRAs.

123. CThe passage seems to be in doubt about which should prevail – state or federal law. More

importantly, the passage is fairly clear on one point: state law prevails in cases of real property.That doesn’t necessarily mean that state law prevails in all cases – particularly in cases that don’tdeal with real property. In the case of choice (A), the author states in the third paragraph thatFlorida’s laws do keep some poor and working class people from facing utter financial ruin, sug-gesting that these people do need the protection of Florida’s bankruptcy laws. The author alsosuggests, in the last line of the passage, that the Homestead law is unlikely to be repealedbecause it is in the best interest of Florida voters. That in turn suggests that the author believesthat voters generally do not vote against what they perceive as their best interest. That throwsout (B). Choice (D) is tricky. But the author does mention a “growing rush of high-profilecases”; the implication is that the Senate is taking a hand only because these cases are likely toincrease as the word gets out.

124. A The laws of other states aren’t mentioned, so the author can’t be contrasting them to

Florida’s, so (B) is out. With choice (C), we have no reason to think that the Florida legislaturepasses laws to protect poor and working class people, so that can’t be the reason – or at least notthe only reason – why the legislature won’t overturn the Homestead law. With choice (D), theauthor doesn’t say that poor and working class people are unfairly hiding from creditors. Quitethe opposite: the passage says that the Homestead law keeps poor and working class people fromfalling into utter poverty. This is a good thing about the law, and is probably mentioned in orderto defend the law at least in part – as in answer choice (A).

Passage VIII (Questions 125-130)

Topic and Scope: The historical writings of the Roman historian Tacitus. The passage askswhether he wrote The Annals or The Histories first, and concludes, on the basis of writing style,that The Histories, although it describes a later time period, was written first.

Paragraph 1 plunges us right into the question: did Tacitus write The Annals or the Historiesfirst? The paragraph lists the similarities between the two works, but concludes that The Annalswas written second.

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Paragraph 2 discusses the narrative format of both works, pointing out that The Historiescontains several narrative digressions, apparently a sign of immature historical writing for ourauthor.

Paragraph 3 talks about the narrative digressions that appear in The Annals, taking pains toassure us that these are quite different than the ones in the The Histories, and apparently of themore mature sort.

Paragraph 4 tells us that, despite all this, the books still remain basically similar: two histo-ries of Rome that focus on the inevitable corruption of man.

125. D Choice (A) is actually the direct opposite to what you’re looking for. If a rigid year by year

format is a more mature form of historical writing, than the author’s main argument – that TheAnnals was written after The Histories – is strengthened, not weakened. Choice (B) is slightlyoff the topic of the question here. Moral digressions may be an essential part of a historical nar-rative, but that doesn’t tell us whether The Histories or The Annals were written first. Choice (C)is definitely outside the scope of the question: we need to know which book was written first,not which book is more intriguing. That leaves us with (D). The passage tells us that The His-tories contains lengthy moral commentaries, and The Annals does not. If the ability to writelengthy moral commentaries only comes with age, then it’s more likely that The Histories waswritten second – weakening the author’s main argument.

126. B The second paragraph tells us that one mark of the maturity of The Annals is that Tactitus

presents his stories about the emperors without further comment – in direct contrast to The His-tories, where he seemingly feels the need to moralize everyplace. Thus, the author apparentlybelieves that telling stories without adding further morals to them is a mark of mature historicalwriting – as in the example given in choice (B), where Stalin’s deeds are presented without fur-ther comment. Choice (A) is definitely tricky here, since The Annals, supposedly the moremature work, was written in a rigid year by year format. But the author never says that that rigidyear by year format is one of the reasons for assuming that The Annals was written after TheHistories. The Annals also discusses other world events outside the ones occurring in the city ofRome, but again, the author doesn’t state that this is a sign of mature historical writing, whichthrows out choice (B). Trade restrictions aren’t mentioned anywhere in the passage, so choice(D) is out.

127. A The passage never tells us about what effect, good or bad, the emperor Vitellus had upon the

Roman empire, so we have to throw out answer choices (B) and (C). Actually, about all the pas-sage tells us is that Vitellus seems to have been a rather useless sort of guy, We don’t knowwhether he wrote poetry (bad or good) or not, although the contrast made with Nero might makeus suspect that he didn’t. The author does say, however, that Tacitus may have needed those longparagraphs describing moral depravity because the evil of the characters in The Histories was-n’t as obvious. We can infer, therefore, that no matter how worthless Vitellus may have been, hecould not have come across as badly as Nero did.

128. B The first paragraph tells us that Tacitus wrote histories full of colorful anecdotes – which

would suggest that this is the way Tacitus feels history should be written. This would throw out

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answer choice (A), since we’re looking for what Tacitus would disagree with. According to theauthor, Tacitus mentions Nero’s bad poetry writing to illustrate his utter uselessness as anemperor, which would suggest that Tacitus thinks that writing poetry is not something emperorsshould do, throwing out answer choice (C). According to the author, Tacitus used both Nero andTacitus to show the corruptive influence of power, throwing out choice (D). Tacitus would not,however, agree with choice (B). The rulers who followed Agrippina and Nero may not have beenquite as bad as they were, but Tacitus, according to the author, believed that all humans wereinevitably corrupted; there’s no reason to believe that he would have thought that the laterRoman emperors were any less corrupted by power than were the earlier ones.

129. C We’re told in the first paragraph that Tacitus, unlike most ancient historians, actually tried

for a little historical objectivity in his writings. Since the author points this out as unusual, wecan assume that most ancient historians didn’t follow his example. Answer choices (A), (B) and(D) do come from the text, but discuss the historical writings of Tacitus, not that of Roman his-torians in general. It’s possible that not all Roman historians chose to focus on evil historical fig-ures, or on the history of Rome. It’s equally possible that other historians avoided both the rigidyear by year narrative format and a biographical format. (The second format isn’t directly men-tioned in the passage either, giving you another reason to avoid answer choice (D).

130. D Much of the point of the passage is that the author is unable to give a precise date for the

writing of either The Annals or The Histories – and neither can anyone else. It follows that anynew writings found will probably be equally impossible to date precisely – as in choice (D). (A)might seem plausible, since the author states that the later Annals is more bitter in tone than TheHistories, but it doesn’t necessarily follow that Tacitus would become even more bitter later inlife or in writings. He might even have written the newly discovered, more bitter writingsbetween writing The Histories and The Annals – and then mellowed a bit while writing TheAnnals. Choice (B) also sounds plausible, since the author criticizes Tacitus for using all of thosemoral digressions in The Histories and uses the digressions as evidence that The Histories werewritten before The Annals. But it doesn’t follow that these newly discovered writings wouldhave been written before The Histories, just because they also contain moral digressions. In fact,the discovery of these new writings, which seem to combine aspects of both works, would seemto cloud the whole issue further – and certainly keep the author from dating The Histories andThe Annals with any more precision, as in choice (C).

Passage IX (Questions 131-137)

Topic and Scope: Analysis of William James’ resolution of a traditional philosophical dis-tinction between the material and the spiritual.

Paragraph 1 introduces the traditional philosophical views. Paragraph 2 covers Plato’s dis-tinction of our mental images of objects from the objects themselves. In Paragraphs 3 and 4James’ view is introduced, which gives no separate reality to the perception. Paragraph 5 offersthe critics’ challenges to James’ theory.

131. AThe most reliable way to answer questions of this type is to eliminate, which makes them

good candidates to save until last. Choice (B) is a summary of the second paragraph supportedby the example of the fountain. Answer choice (C) is the premise of the last paragraph and is

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elaborated on with specific examples of possible challenges. Choice (D) is a paraphrase ofJames’ argument and is supported by the example immediately following. A shortcut to answerthese rare question types is to look for answer choices that fall at the end of paragraphs or theend of the passage. Very often the item that has no support will be at the end of a thought beforethe author segues into something different. This will often allow the answer choices to be nar-rowed down quickly even if it does not immediately yield the correct answer.

132. CThis question is primarily a logic question that tests the structure rather than the content of

the passage. The author argues that two divergent and abstract philosophies compete to explainhuman consciousness. These questions are particularly vulnerable to the denial test: if an answerlooks tempting, deny it and investigate whether the author’s argument is hurt. If it is, the authormust rely on the idea behind the statement and it is a correct answer choice. In this case, if theauthor believes that abstract ideas cannot lend insight into human behavior, both philosophiesare irrelevant. Choice (A) supports the dualistic theory, which the author does not endorse.Choice (B) is an extreme answer choice mentioned nowhere in the passage, and Choice (D) isouter space. Each of these wrong answer choices, if denied, won’t devastate any part of theauthor’s argument and so are less attractive than choice (C).

133. BThe primary claim to superiority of James’ theory mentioned in the fourth paragraph is its

“efficiency” (line 47), that it is simpler, or more economic than a dualistic philosophy. If this isproven false, James’ theory is stripped of its primary advantage. Choices (C) and (D) would bothlend support to James’ theory, while choice (A), involving concepts of morality, is Outside theScope of the passage.

134. DApplication questions are common on abstract passages, and will require you to apply the

ideas in the passage to specific situations. The key is to identify what relevance the situation hasto what part of the passage. In this instance, an action is performed on an object, removing themoral sense while leaving consciousness intact. Since self-awareness is unchanged, neither phi-losophy is supported or hurt by the experiment, thereby leading to choice (D).

135. BAssumption questions are the most vulnerable to the denial test. The test should not be used

on every answer choice, as this is time consuming, but can be used to verify an answer choicethat looks tempting. Answer choice (B), involving consciousness and the spiritual sense, bothcritical to the dualistic philosophy, is tempting. Denying it by stating that beings without a spir-itual sense can have consciousness, undermines the foundation of dualism set up in the firstparagraph. This must therefore be a primary assumption. The remaining answer choices are allDistortions of the claims made by the dualistic philosophy.

136. CThe first step in this question is to determine what philosophy is being talked about. Since it

refers to a strictly material universe, it must be James’ philosophy. The correct answer will be anecessary conclusion based on what the author says about James’ philosophy. Choice (C) fol-lows from the discussion in the last paragraph of the philosophy’s shortcomings: it assumes thatone type of matter exists to prove the same thing. Choice (A) is extreme, as James’ philosophyacknowledges consciousness but not as a separate entity. The remaining answer choices areopposite, supporting the dualistic hypothesis.

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137. CLogic questions are answered easily if a good map has been created beforehand. The first

paragraph presents the dualistic view and gives examples of sub-philosophies on opposite endsof that same philosophy, which matches choice (C). Choices (A) and (D), while on oppositesides of James’ philosophy, are both wrong for that same reason: they cannot counter or supportthe materialistic philosophy as it has not yet been brought up in this part of the passage. Choice(B) is a faulty use of detail—although a disparity between a modern and ancient philosophy ispresented, Socrates is not introduced into the passage for that reason.

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37

138. A

139. A

140. D

141. A

142. B

143. C

144. B

145. A

146. D

147. A

148. D

149. B

150. A

151. C

152. B

153. D

154. C

155. A

156. C

157. B

158. B

159. A

160. C

161. C

162. A

163. A

164. D

165. D

166. C

167. D

168. B

169. B

170. D

171. D

172. C

173. D

174. B

175. C

176. D

177. C

178. B

179. C

180. B

181. D

182. D

183. A

184. B

185. D

186. D

187. A

188. D

189. D

190. B

191. A

192. D

193. D

194. B

195. A

196. B

197. B

198. C

199. D

200. C

201. A

202. C

203. B

204. D

205. A

206. B

207. A

208. C

209. A

210. A

211. C

212. B

213. C

214. D

BIOLOGICAL SCIENCES ANSWER KEY

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Passage I (Questions 138-143)

138. AThe reaction with sulfuric acid proceeds via the E1 mechanism where the formation of a car-

bocation is involved so the product ratio should be thermodynamically determined according toSaytzeff’s rule where the more favored product is the more substituted alkene. For the KOHreaction, it proceeds via the E2 mechanism and therefore the concertedness of the reaction givesmore of the less substituted product.

139. ATrans-4-methyl-2-pentene is the product of the reaction between vic-2,3-dibromo-4-methyl

pentane and Zn in ethanol. The debromination reaction takes place by an E2 mechanism:

140. DThe NMR spectrum of the ketone should appear as follows:

a should be a singlet (no neighboring H’s) and shifted downfield compared to a regularmethyl group (2.3 ppm)

b doublet (1 neighboring H) and shifted downfield compared to a regular methylene groupbecause it’s next to a carbonyl (2.1 ppm)

c multiplet (probably 9 peaks) overlapped with the singlet a

d doublet (1 neighboring H), huge because it corresponds to 6H’s or two methyl groups (1ppm)

141. AThe best synthetic route to (CH3)3COCH3 is Choice (A). In Choice (B), the difference in

basicity of hydroxide and tert-butoxide, the conjugate base of (CH3)3COH, is not sufficientlylarge to allow essentially complete deprotonation of the alcohol. When CH3I is added to thereaction mixture, there will be two alkoxides (hydroxide and tert-butoxide) competing for thereaction. In choice (C), reaction of CH3O- (a strong base) with a tertiary alkyl halide will giveE2, not SN2, products.

O

c

d

db

a

H

BrMe

iPr

H

Br

Zn

Me

HH

iPr

+ ZnBr2

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142. BPrimary and secondary alcohols can be oxidized to aldehydes or ketones; ethers cannot be

oxidized to aldehydes or ketones. For example:

Choice (A) is incorrect because hydroxide and alkoxide ions are not good leaving groups, soboth are inert to nucleophilic attack in the absence of a strong acid. Similarly, both ethers andalcohols have oxygen atoms that bear two lone pairs, so both can function as poor nucleophilesor weak bases (choice (C) incorrect). It is true that alcohols have a removable hydrogen atomwhile ethers do not. However, the hydrogen can only be removed with base rather than acid.Choice (D) is therefore incorrect.

143. BThe most stable alkoxide has the smallest magnitude of negative charge on the oxygen (and

thus must also be the weakest base). In this series of alkoxides, we can compare the effect ofreplacing a hydrogen atom (IV) with a chlorine atom (I), fluorine atom, or methyl group. Chlo-rine and fluorine are both electron withdrawing by the inductive effect, resulting in decreasedcharge density on the oxygen atom and increased stablility. Fluorine is more electronegative thanchlorine so the oxygen atom of the trifluoro alkoxide has a smaller electron density than the oxy-gen atom of the trichloro alkoxide. Alkyl groups such as methyl are weak electron donors,resulting in an increase in the charge density of the oxygen atom. Thus the order of basicity andstability is: F3CCH2O- (most stable; weakest base) > Cl3CCH2O- > CH3CH2O- >(CH3)3CCH2O- (least stable; strongest base).

CH3OH + P H2C O

CH3OCH3 + no reaction

C C

P C C

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Passage II (Questions 144-149)

144. BThis particular step in the reaction sequence is a bimolecular nucleophilic substitution reac-

tion. The amine functional group acts as the nucleophile as its lone pair of electrons forms abond with the methyl substituent of adoMet. Since the sulfur atom of adoMet is positivelycharged, it serves as a very good leaving group for SN2 displacements on the methyl carbon.Essentially, the chemistry is exactly as if the adoMet molecule were a methyl halide. The entirereaction mechanism can be seen below:

145. AThe alkyl component of alanine is directly attached to the ring allowing it to donate electron

density to the ring and thereby activating it. This stabilizes the carbocation intermediate causingit to form at a faster rate. Any alkyl substituent directs any further substitution at the ortho orpara positions because those intermediates possess the most stable resonance structures, specif-ically the formation of tertiary carbocations. Choices (B) and (C) are incorrect because, whileboth amino and hydroxyl groups are also both ortho/para directing activators, neither is a sub-stituent directly attached to the benzene ring prior to hydroxylation. Choice (D) is incorrectbecause the carboxylic acid functionality is also not attached to the benzene ring and is a metadirector.

O

H

CH2

H

N

N

N

N

NH2

H

HO

H

S

CH3

CH2CH2CH

NH2

HOOC

HO

HO CHCH2 NH2

HO

HO 2 NHCH3

SN2

O

H

CH2

H

N

N

N

N

NH2

H

HO OH

H

SCH2CH2CH

NH2

HOOC

Adrenaline

OH

CHCH

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146. DThe passage indicates that this particular step in the reaction sequence requires a reducing

agent. Choice (A) is correct because LAH is a very strong reducing agent that will act to reducealdehydes, ketones, and carboxylic acids. Choice (B) is also correct because sodium borohy-dride is specific for aldehydes and ketones. Choice (C) is incorrect because PCC is an oxidizingagent used to oxidize primary alcohols to aldehydes.

147. AThe carboxylic acid functional group is the most oxidized, and therefore has the highest pri-

ority when naming the compound. The base name is propanoic acid because the longest carbonchain containing the highest priority functional group is 3 carbons in length. The other sub-stituents are then appropriately named in alphabetical order along with the number carbon towhich they are attached.

148. DEssentially, this is a Friedel-Crafts acylation mechanism in which the reaction between the

Lewis acid catalyst (POCl3) and the chloroacetylchloride forms an acyl cation. This acyl cationis the reactive electrophile and is subsequently attacked by the aromatic ring. The reaction mech-anism can be seen below:

Choice (B) is incorrect due to the inductive effect resulting from the electronegative atom ofthe hydroxyl group. Proximity to an e-withdrawing group decreases electron density at C2thereby destabilizing the carbocation relative to that of choice (D).

Cl CH2Cl

O POCl3

CH2Cl

O

HO

HO

CH2Cl

O

+

HO

HO

H

CH2Cl

O

HO

HO

CH2Cl

O

+ HCl

Cl

+ POCl4

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149. BPriorities are assigned to the four substituents on a chiral carbon according to the atomic

number of the atom directly attached to the chiral carbon. If a determination cannot be reached,the atomic numbers of the second atom of each substituent are considered. The higher the atomicweight, the higher the priority and the lower the assigned number. The hydroxyl group has thehighest priority followed by the amine containing substituent and finally the phenyl substituent.Orienting the lowest priority substituent into the page, the priorities follow in a clockwise pat-tern. Choice (B) is epinephrine. Answer A is S-epinephrine. Choices (C) and (D) are both con-stitutional isomers of epinephrine (same molecular formula, different connectivity).

Passage III (Questions 150-154)

150. A As explained in the passage and seen in Figure 1, CheA is a cytosolic protein. The cytosol

is the liquid or aqueous environment of the cell. Therefore, proteins that are in the cytosol mustbe soluble in a liquid environment. Charged or polar surface residues make a protein water sol-uble. The reason for this is the ability of water to hydrogen bond. Water is a polar molecule.Therefore, to be water soluble, a molecule must also be polar or charged. The opposite is truefor proteins that are membrane soluble (like the transmembrane receptors described in the pas-sage). The cell membrane is a phospholipid bilayer, with a hydrophobic center. The hydropho-bic center comes from the lipid “tails” of the phospholipids that make up the membrane. To bemembrane soluble, a protein must have a large number of hydrophobic (i.e. uncharged or non-polar) residues on its surface. This is why choice (B) is incorrect. Answer choice (C) is a dis-tortion of information presented in the passage. Histidine kinases are mentioned, but not in thecontext of histidine residues predominating the surface of the protein. Choice (D) is incorrectsince it has nothing to do with the passage. Cysteine residues typically form disulfide bridgesthat covalently attach proteins together.

151. CThis question tests your knowledge of the differences between eukaryotes and prokaryotes.

You are told that E. coli are bacteria and, therefore, prokaryotes. The patient is a eukaryoticorganism (as all humans are). You would administer a drug that distinguished between the twoby exploiting one of the differences between eukaryotes and prokaryotes. Both types of cellshave a phospholipid bilayer and tRNAs (transfer RNA), however bacteria do not contain mito-chondria (or other membrane-bound organelles). This eliminates choices (A), (B), and (D),respectively. The peptidoglycan cross-links are what make up the cell wall of bacteria. Eukary-otes do not have cell walls; thus, a drug that targets the cell wall of the bacteria, but does not tar-get any eukaryotic structure would be a good one to administer. Choice (C) is correct.

152. BThis question tests your understanding of the passage and the signal transduction pathway

used in chemotaxis. CheA autophosphoylation is followed by CheY phosphorylation, whichthen induces the flagellar motor to turn clockwise, causing the cell to tumble. As the cellapproaches an increasing maltose gradient, it swims smoothly so as to swim into the maltose.Thus CheA autophosphorylation would decrease, decreasing the levels of phospho-CheY and the relative activity of CheZ, causing increased counter-clockwise rotation, and less tumbling tosmooth swimming. Choice (B) is correct.

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153. D A CheA with a greater affinity for ATP would slow down the bacterium’s ability to respond

to its environment. Greater affinity for ATP would mean tighter binding of ATP; therefore, theCheA would have a decrease ability to pass phosphates. Greater affinity should not be confusedwith a faster rate of autophosphorylation and phospho-transfer. The ability of CheA to formcomplexes with the transmembrane receptors does influence CheA’s binding of ATP. However,the ability of CheA to bind ATP does not have any influence its ability to form a complex withthe transmembrane receptors. Choice (D) is correct.

154. CThe cellular process that most resembles chemotaxis is a G-protein being activated by

epinephrine. The key to this question is understanding the two-component signaling nature ofchemotaxis. G-proteins are part of two-component regulatory systems. Similar to CheA activa-tion, when a epinephrine receptor binds its ligand, it will activate the G-protein, which will theninteract with other effector proteins. Choice (A) is a trap because regulation of phosphofructok-inase (PFK) does occur by phosphorylating the enzyme, but PFK does not signal to other pro-teins. Activation of a T-cell happens when interleukin (IL) are secreted by cells involved withimmune responses (e.g. macrophages). The IL signal to induce the T-cell to change is not simi-lar, however, to the two-component system. Similarly, the troponin/tropomyosin response to cal-cium is not analogous to the two component regulatory system. When troponin binds calciumions, it physically moves tropomyosin to expose the ATP binding sites on myosin. There is not,however, a signal transduction pathway in which some message is being sent to effector pro-teins. Choice (C) is correct.

Passage IV (Questions 155-160)

155. AThe passage states that the force pushing the fluid from the glomerular capillaries must be

balanced by the osmotic pressure in the capillaries. Since the effective osmotic pressure in thecapillaries is due to plasma proteins, increased plasma protein concentration will elevate theosmotic pressure. Greater osmotic pressure will resist the flow of fluid out of the capillaries andwill reduce the amount of filtrate entering Bowman’s capsule. Choices (B) and (D) are incorrectbecause small molecules such as glucose and ions are easily filtered across the capillary mem-brane and are thus incapable of contributing to the difference in osmolarity between the plasmaand the filtrate. Choice (C) is incorrect because an elevated arterial pressure will force more fluidthrough the glomerular capillaries and will increase the amount of fluid entering Bowman’s cap-sule.

156. CThe percent of urea reabsorbed can be found by comparing clearance of urea to the glomerular

filtration rate (GFR). Clearance of urea can be determined using the equation Cx = ([Ux] * V)/ [Px].Since urea concentration in the urine is 100 times greater than urea concentration in the blood, then[Uurea]/[Purea] is equal to 100. Clearance of urea can thus be expressed as:

Curea = 100*V where V is urine flow rate and is equal to 1L/day.

= 100*1L/day

= 100L/day.

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Since the amount of blood that is filtered in one day is 200L (GFR is 200 L/day) and urea iscompletely cleared from 100L (Curea is 100L/day), then 50% of filtered load of urea is reab-sorbed back into blood. This corresponds with answer choice (C).

157. BThe clearance of a substance is essentially the rate at which the substance is eliminated

from the body in urine. The question stem states that penicillin is eliminated through bothfiltration and secretion into the tubular fluid. When one of the ways by which penicillin isexcreted is inhibited, the rate at which penicillin is removed from the body is decreased. Onecan also arrive at this conclusion by looking at the equation for clearance. When secretion ofpenicillin is inhibited, penicillin clearance must decrease because concentration of penicillinin the plasma [Ppenicillin] is increased, while the concentration of penicillin in the urine[Upenicillin] is decreased. The correct answer is choice (B).

158. BIf the amount of creatinine excreted by each nephron is greater than the amount of creatinine

that is filtered across the glomerulus, then there must be an additional source of creatinine in thetubules. Among the presented answer choices, only choices (B) and (C) are consistent with anincrease of substance in the tubules (if either choice (A) or (D) were true there would be adecrease in the amount of creatinine in the urine). In order to decide between the remaining twoanswer choices one must look for clues presented in the passage. The passage states that one ofthe functions of the peritubular capillaries is to transport proteins that are to be secreted by thetubules. Choice (B) is thus consistent with the information presented in the passage and sincethere isn’t any evidence in support of choice (C), choice (B) is the correct answer choice.

159. AThe volume of the urine ultimately depends on the osmotic gradient between the fluid in the

tubules and the interstitium. When a substance enters the tubular fluid and cannot be reabsorbed,it will contribute to the osmolarity of the urine and will resist the flow of water out of the tubule.Thus the amount of fluid remaining in the tubules will increase and a greater volume of urinewill be produced. Choice (A) is correct.

160. CThe second paragraph of the passage discusses the forces responsible for moving fluid and

solutes out of the capillaries into either intercellular space (in systemic capillaries) or intoBowman’s capsule (in glomerular capillaries). The same forces in opposite direction are alsoresponsible for moving the fluid back into the capillaries from the intercellular space. Amongthe given answer choices only choice (C) presents a scenario where both the hydrostatic pres-sure and the osmotic pressure favors the migration of water and solutes into to capillaries.Choice (C) is correct.

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Discretes (Questions 161-164)

161. COnce the action potential reaches the interior of the muscle fiber, the sarcoplasmic reticulum

is induced to release calcium. This calcium binds to troponin molecules to cause thetropomyosin to shift and expose the myosin-binding sites, choice (C). With actin’s myosin-bind-ing sites exposed, myosin and actin can form cross bridges. Choice (C) is correct.

162. AWhen the ovum is not fertilized, the corpus luteum atrophies. This leads to a decline in the

levels of both estrogen and progesterone, which causes the sloughing off of the endometriumand the beginning of menses. Thus, choice (A) is correct.

163. AThe correct functional groups present can be readily identified.

Choice B is incorrect because neither a carboxylic acid nor an ether are present. Choice (C)is incorrect because an alkene is not present. Choice (D) includes an ether that is not present.

164. DA Lewis acid is a species that can act as an electron-pair acceptor. In order to accept a pair

of electrons the species must be electron deficient (choice (D)). Choices (A) and (B) are incor-rect because they are the definition of a Lewis base, not acid. Choice (C) is incorrect becauseBrønsted acids are proton donors, and there are no protons in FeBr3.

Passage V (Questions 165-170)

165. DOf the three hydrocarbons, only cyclohexene reacts with bromine. Cyclohexane and benzene

do not react with bromine under the conditions of the experiment, as indicated by the charac-teristic red color of bromine when a few drops of reagent were added. Cyclohexene reacts read-ily with HBr. Benzene reacts with HBr (via electrophilic substitution NOT addition) only in thepresence of a catalyst.

166. CThe ease of evaporation of a liquid is related to its vapor pressure which can, in turn, be

related to its boiling point. Of the compounds present following reaction with bromine, trans-1,2-dibromocyclohexane has by far the lowest vapor pressure so not much evaporated overnight.

EtherAlkeneCarboxylicacid

C O C C CCOH

O

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Compound b.p.

dichloromethane 42

bromine 59

cyclohexane 81

cyclohexene 83

benzene 80

Trans-1,2- dibromocyclohexane 146 at 100mm Hg

Bromine reacts with alkenes by electrophilic addition. The reaction occurs via a bromoniumion with anti stereochemistry so the trans isomer is formed.

167. DLitmus paper will be blue in basic solution, red in acidic solution. In neutral solution there

will be no color change. HBr is an acid, so if no reaction occurred, it will still be in solution. HBr was consumed in the reaction of cyclohexene with bromine and thus the litmus test indi-cates a solution which is not acidic (no color change = neutral).

168. BAnother way to state Markovnikov’s Rule is: in the addition of HX to an alkene, the hydro-

gen atom adds to the carbon atom of the double bond that already has the greater number ofhydrogen atoms. This rule extends to the ionic addition of unsymmetrical reagents. The rate-determining step is the formation of a carbocation. The mechanism for hydration of an alkene issimply the reverse of the mechanism for the dehydration of an alcohol.

169. BTwo stereogenic centers are generated in the reaction with cyclopentene and the product is

actually a racemic mixture of (R,R)-trans-1,2-dibromocyclopentane and (S,S) –trans-1,2-dibro-mocyclopentane. Another way of naming the racemic mixture is (±)-1,2-dibromocyclohexane.

Br2 Br

BrBr

BrBr

Br

R,R S,S

andand

+

+

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170. DThe relative rates of the three reactions are controlled by their rate-determining steps, all of

which involve carbocation formation. The most stable carbocation is formed more quicklybecause its formation has a lower energy of activation. All the carbocations are secondary, butonly the carbocation formed in reaction II has the extra stability afforded by resonance. There-fore choice (D) is correct.

Passage VI (Questions 171-175)

171. DAs explained in the passage, ssDNA exists after the DNA strands have separated. Looking

at the graph, ssDNA exists when UV light absorption is at its highest, and at or above Tm. At85°C here, which happens to also be Tm, the DNA has more single-stranded character than atany of the other temperatures in the answer choices. Hence, choice (D) is correct.

172. CThis question calls upon your knowledge that adenines (A) and thymines (T) only form two

hydrogen bonds with each other, while guanines (G) and cytosines (C) form three. This meansthat A-T rich areas of DNA are held together by fewer bonds and will separate more easily thanG-C rich areas. Choice A is incorrect because complementary strands with an equal composi-tion of A, C, T, and G base pairs would be held together by more hydrogen bonds than those ofchoice C, and thus be more difficult to separate or denature. Choice B is incorrect because com-plementary strands cannot be identical. Choice D is incorrect because as stated in the passage,DNA is actually more difficult to denature in a sodium rich solution.

173. DAs stated in the passage, longer strands of DNA are more stable at higher temperatures. Thus

the Tm of long DNA strands would be higher than the Tm measured in the first experiment. Theonly answer choice that gives a possible value is choice (D), since all other values are at or belowthe Tm of the DNA tested in the original experiment. Hence, choice (D) is correct.

Cl H

I:

II: Cl HOCH3 OCH3 OCH3

2

Cl HIII:

2o carbocation

o carbocationwith resonance

+

+

+

+2o carbocation

47

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48

174. BBecause double-stranded DNA (dsDNA) is made of two strands, it can be oriented in an

antiparallel fashion, with a 5’ to 3’ strand running in one direction, and a 3’ to 5’ strand com-plementary to that one. This is necessary in all DNA double helices for proper alignment andbonding; hence, choice (B) is correct, since ssDNA only has one strand. ssDNA does not haveany uracil (U) – only RNA contains uracil, so choice (A) is incorrect. In addition, RNA, notssDNA, is translated at the ribosomes in protein synthesis, so choice (C) is incorrect. Choice (D)is incorrect since both ss and ds forms of DNA would have the characteristic 5’ to 3’ orientationof phosphate-sugar-base bonding.

175. CRecall that increasing temperature is what breaks apart the strands in the first place. Thus, it

can be assumed that high temperatures would prevent renaturation from occurring as rapidly asit might otherwise occur. Only chemicals, and not a minor change in temperature, could breakapart the actual bonds holding the nucleotides together, so choice (A) is incorrect. Recall thatheating breaks only the G:C and A:T hydrogen bonds. The heat of the solution would have noeffect necessarily on the separation of future solutions of DNA, which is what choice (B) sug-gests. Thus, choice (C) is the best response here.

Passage VII (Questions 176-180)

176. DThis question requires you to be able to read the graph of experimental data and to conclude

from information in the passage that VE increases with heavier breathing. Yet, VE quicklyreaches its maximum in patients with abnormal respiratory conditions, and for the congestiveheart failure patients, this maximum is around 40 L O2 / min. Thus, choice (D) is correct.

177. CAs stated in the passage, all three abnormal groups of patients had increased resistance in

their airways due to alveolar damage. Thus, it is more difficult to move air into and out of thelungs, especially out of the lungs as can be seen in the graphed data. This happens becausepatients with obstructive lung diseases use chest muscles to try to force out air from their lungswhen they exhale, which compresses their airways and makes it more difficult to get rid of theair – this explains the rapid drop in VE seen with exercise. However, you don’t have to know thatto get this question right. If it is more difficult for these patients to breathe, they will breathemore slowly to minimize their overall work – in other words, to minimize the cost of breathing.Thus, given that reasoning and the information presented in the passage, choice (C) is the onlycorrect answer here.

178. BThis is a question about basic respiratory anatomy. Air passing into the lungs travels through

the nose or mouth into the pharynx, or throat, and then past the larynx, or voicebox, beforereaching the bronchi, bronchioles, and finally alveoli. Gases are diffused from the alveoli intothe bloodstream, but do not pass through capillaries first; thus choice (D) is incorrect. The pleuraare tissues that surround the outside of the lungs and cushion them against the chest wall – noair passes through either of the pleura. Thus, choice (B) is correct.

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179. CBoth the residual volume and total lung capacity will increase in emphysema patients. As the

passage states, the experiment shows that emphysema patients are unable to exhale all of the airthey inhale (see graph). Thus, the amount of air held in after each breath is much greater than itshould be. As RV increases, TLC must also increase since TLC includes the RV, hence choice(C) is the correct answer.

180. BThe medulla oblongata, within the brain stem at the back of the head, controls breathing rate

and depth. The cerebellum, choice (A), controls and coordinates many muscular movementsthroughout the body, while the pleura, choice (C), are the tissues surrounding the outside of thelungs, and the sinoatrial node, choice (D), controls heart rate. Choice (B) is correct.

Passage VIII (Questions 181-185)

181. DAll three statements are true; therefore, choice (D) is the correct answer. Any three-mem-

bered ring structure can only attain a planar geometry resulting in torsional strain due to theeclipsing of bonds on neighboring atoms. Three-membered rings are also unable to achievemaximum bonding because the overlapping orbitals are unable to point directly toward oneanother. Lastly, there is also considerable angle strain due to the compressed bond angles whichdeviate greatly from the normal 109.5 degrees of a sp3 hybridized atom.

182. DReaction I is simply the reaction of an alkene with a peroxyacid as seen in Figure 1. Reac-

tion II is a two step reaction involving first the formation of a halohydrin from an alkene, thenan intramolecular Williamson ether synthesis using sodium hydroxide and water. Both I and IIwill produce an epoxide. However, reactions III and IV will not. While reaction III also appearsto be an intramolecular Williamson ether synthesis, because this reaction follows a SN2 mecha-nism, the attacking alkoxide must be anti in configuration to the leaving group. In this particu-lar case, both are on the same side of the cycloalkane. Reaction IV does result in a cyclical ether;however, a 5-membered oxirane is formed instead of a three-membered epoxide.

183. AAddition reactions to alkenes often occur with either syn or anti addition. In this case, we

are seeking a trans-substituted product which would eliminate choice (B). Of the other threechoices, only choice (A) follows a similar mechanism to the acid-catalyzed epoxide ring open-ing. The final step of alkene bromination involves the opening of the cyclic bromonium ion bynucleophilic attack resulting in a trans product.

184. BAll Grignard reagents are nucleophilic in nature and methylmagnesium bromide is a Grig-

nard reagent. Therefore, this reaction would follow a path similar to that of a nucleophilic basesuch as an hydroxide or alkoxide ion. According to the passage, this would be a typical SN2mechanism.

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185. D4-membered cyclic ethers are generated much more slowly than either 3 or 5-membered

cyclic ethers. While the ring strain is about the same as in 3-membered cyclic ethers, the entropyfactor is considerably decreased because the two reacting centers are separated by an extramethylene group.

Discretes (Questions 186-190)

186. DSomatostatin is increased by high blood glucose (statement I) and high blood amino acid

levels. It also decreases both insulin (statement II) and glucagon (statement III). Thus, all threestatements are true, making choice (D), I II, and III, the correct answer.

187. AThis question involves the use of the Hardy-Weinberg equations: p2 + 2pq + q2 = 1 and p +

q = 1, where p = the frequency of the dominant allele, q = the frequency of the recessive allele.In the question stem, we are told that the frequency of individuals expressing the recessive phe-notype is 0.36, so q2 = 0.36, and q = 0.6. p = 1 - q = 1 - 0.6 = 0.4. The frequency of individualsexpressing the homozygous dominant phenotype would be p2 = (0.4) 2 = 0.16. This correspondswith choice (A).

188. DIn a fetus, the ductus arteriosus shunts blood from the pulmonary artery to the aorta; thus, if

the ductus arteriosus fails to close completely after birth, blood would flow from the aorta to thepulmonary artery (generally, more pressure is generated by the the left side of the heart as com-pared to the right side of the heart). Also, in a fetus, blood is shunted from the liver via the duc-tus venosus, and blood is routed from the right atrium to the left atrium via the foramen ovale.The tricuspid valve separates the right ventricle from the right ventricle in both fetal and adultcirculation. Since the question asks what happens if the ductus arteriosus fails to close afterbirth, the correct answer is choice (D); blood would flow from the aorta to the pulmonary artery.

189. DIn the plant species mentioned in this question, the alleles for flower color are codominant:

the phenotype of the heterozygote is a reflection of both alleles. Let’s denote the allele for redflowers R and the allele for white flowers W. A plant with pink flowers would have the geno-type RW, and a cross between two such plants would produce the following results:

R W

R RR RW

W RW WW

The question asks about resulting plants. From the Punnett square, we see that 25% of theoffspring are genotype RR (red flowers), 50% of the offspring are RW (pink flowers) and 25%of the offspring are WW (white flowers). Choice (D) is the correct answer.

190. BThe reagents given (sodium iodide in acetone) are typical for a reaction proceeding by the

SN2 mechanism. Bimolecular (SN2) substitution reactions proceed with inversion of configura-tion. In this problem, however, the leaving group is not attached to the stereogenic center.

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As a result, the product will have the same absolute configuration as the starting material (S);the correct answer choice is (B). Choice (A) is incorrect because the stereochemistry of thestereogenic center is inverted. Choice (C) is incorrect because racemization accompanies SN1reactions, not SN2. Choice (D) is incorrect because the product retains a stereogenic center andis chiral.

Passage IX (Questions 191-195)

191. AIf nest site choice is genetically encoded rather than chosen by physical and environmental

conditions, then this finding strongly supports Hypothesis 1, which maintains that the twopython groups ought to be classified as separate subspecies; thus choice (A) is correct. This find-ing does not support the idea that the pythons cannot be classified according to convention,choice (C), because they can still fit into those classifications as two different subspecies. Choice(B) is incorrect since the finding implies that the two groups have different genes – differentenough to warrant classification as separate subspecies. Choice (D) is incorrect because the find-ing does not invalidate the r-selection and K-selection modes of classification – it simply sup-ports the idea of a separate subspecies classification.

192. DBoth Hypothesis 1 and Hypothesis 2 state that maternal nest attendance varies between the

warm and cool nest pythons. A finding that nest attendance is invariable across all pythonspecies does not strengthen or lend support to either of these ideas. Thus, choice (D) is correct.

193. DThe best answer here is choice (D), that Hypothesis 1 argues for r and K grouping, even if

it means dividing the pythons into subspecies, while Hypothesis 2 argues against that. There isnot enough evidence in the passage to account for any other answer choice.

194. BThese female pythons would be exhibiting a mixture of r-selected traits, such as large num-

bers of eggs laid, and K-selected traits, such as remaining with the young until birth or maturity.Thus, this finding would strengthen Hypothesis 2, which suggests that the pythons cannot beeasily separated into r and K groups, that they exhibit a mixture of traits from both strategies.

195. ABecause Hypothesis 1 explains how warm nest pythons can leave their offspring alone while

cool nest pythons cannot, it is suggested that the temperatures force cool nest pythons to stayaround and incubate their eggs, while warm nest ones do not. This is further reinforced by theassertion that cooler nest pythons have to wait longer to reproduce, that reproduction takes moreenergy out of them than it does the warm nest pythons. Hence, choice (A) is the best answerhere. Choice (B) is contradicted in the passage, choice (C) is never mentioned specifically inHypothesis 1, and boom and bust cycles in choice (D) are never elaborated on more than just themention that they exist for the r-selected behaviors of warm nest pythons.

Cl

H OHNaI

acetone?

I

H OH

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Passage X (Questions 196-202)

196. BSince increasing meat consumption will tend to lower plasma pH, compensation mecha-

nisms by either the kidneys or the lungs must be working to increase plasma pH. Among the pre-sented answer choices, only choice (B) will increase plasma pH because reabsorbed HCO3

- willbuffer the free H+ ions. Choice (A) is incorrect because the activity of the tubular Na+/H+

antiporter is required for excretion of acid. Choice (C) is incorrect because a decrease in the rateof ventilation will tend to increase plasma pH. Choice (D) is incorrect because an increase inglomerular filtration rate without the corresponding increase in the rate of reabsorption ofHCO3

- will favor excretion of HCO3- in the urine, which will lower plasma pH even further.

197. BLow intracellular concentration of sodium is required to generate the driving force for the

Na+/H+ antiporter. If Na+/H+ ATPase is inhibited, the concentration of Na+ inside the cells willrise and there will no longer be secretion of H+ ions into the tubular lumen. Since secretion ofH+ ions is required for HCO3

- reabsorption, the amount of HCO3- reabsorbed will decrease.

Choice (C) is incorrect because if sodium is not removed from inside the cells, intracellularosmolarity and volume will increase. Choice (D) is incorrect because a decrease in H+ secretionwill lower the acidity of the blood rather than raise it.

198. CThe decline in the activity of carbonic anhydrase would be a maladaptive response because

carbonic anhydrase is needed for rapid secretion of H+ and for reabsorption of HCO3-. Choice

(A) is incorrect because proton pumps can hasten excretion of acid by actively pumping protonsinto the tubular fluid. Choice (B) is incorrect because rapid activation of brainstem chemore-ceptors can rapidly increase ventilation rate and restore plasma pH. Choice (D) is incorrectbecause an increase in the rate and the depth of breathing would be an appropriate response tometabolic acidosis.

199. D The Na+/H+ antiporter uses the energy of the Na+ gradient to transport H+ ions against the

concentration gradient. This process is termed secondary active transport because ATP energy isused indirectly. The energy of ATP is initially used to establish the Na+ gradient through theNa+/H+ ATPase. The Na+ gradient is then used for secretion of H+ out of the tubular cells.Choice (D) is correct.

200. CSince increased respiration will raise blood pH by getting rid of CO2, a renal compensatory

mechanism must act to lower plasma pH. Among the given answer choices, only choice (C) willlower plasma pH by decreasing the amount of basic HCO3

- in the blood.

201. ABlood’s buffering capacity is its ability to counteract addition of acid or base with a minimal

change in pH. Increased blood volume will increase blood’s buffering capacity because theeffect on pH will be smaller if same amount of acid or base is added to a larger volume of fluid.Choices (B) and (D) are incorrect because those will make blood more basic and acidic respec-tively. Choice (C) is incorrect because carbonic acid is important for the rapid conversion ofexcess acid or base into neural products.

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202. CFunctional carbonic anhydrase in the lumen of the tubule and in the tubular cells is required

for effective reabsorption of HCO3- and secretion of H+. When any one of the enzymes is inhib-

ited the cycle stops and both the secretion of H+ and absorption of HCO3- are compromised.

Choice (C) is correct.

Passage XI (Questions 203-209)

203. BThe Kent reaction is an example of a nucleophilic substitution in which the type of car-

boxylic acid derivative (an acid itself in this case) is replaced with another (an amide). Choice(A) is incorrect because SN2 reactions proceed through a concerted mechanism and can onlytake place on sp3-hybridized primary or secondary carbons. Transesterification choice (C) is anexample of nucleophilic substitution, but neither the reactants nor the products of the Kent reac-tion are esters. A Michael addition (choice (D)) involves the addition of a nucleophile to the β-carbon of an α,β-unsaturated carbonyl which is not present in a polypeptide. There is a transientringed intermediate formed in the Kent reaction, but it is not formed via Robinson annulation.

204. DTo answer this question, it is necessary to know some of the factors that determine whether

or not an enzyme will be active. Some of these factors include: folding into the correct shape,presence of the correct substrates, absence of an inhibitor, solubility of the enzyme and sub-strates, and bioavailability of the substrates. It is also necessary to recognize that 2-mercap-toethanol is both a thiol and an alcohol from the “mercapto” and “ethanol” parts of the IUPACname. We are told that lysozyme catalyzes hydrolysis, the splitting of chemical bonds withwater, so it is unlikely that 2-mercaptoethanol would bind in the active site since it doesn’t lookmuch like either substrate (sugar or water). Choice (A) is therefore incorrect. Choice (B) is alsoincorrect because 2-mercaptoethanol is not a strong enough reducing agent for the reduction ofamide bonds. Amide C=O bonds are very robust and are only reduced by much stronger reduc-ing agents such as LiAlH4. As well, 2-mercaptoethanol is a thiol, not a hydride. Choice (C) isincorrect because Boc groups are cleaved by trifluoroacetic acid rather than by reduction, asstated in the passage. Choice (D) is the correct answer. The correct tertiary structure of anenzyme is crucial to its proper function. Moreover, disulfide bonds play an important role informing and maintaining the structure. Reducing agents such as 2-mercaptoethanol can breakthese bonds thus causing the protein to unfold, rendering it inactive.

205. ATrifluoroacetic acid (TFA) is a carboxylic acid. It therefore follows that the reaction should

be acid-catalyzed. In fact, TFA is a strong acid (it has a COOH moiety in addition to electron-withdrawing fluorines) with a pKa of 0.23. Formation of carbocations from tertiary carbons isfavorable under acidic conditions when there is a good leaving group such as a carboxylic acid.The subsequent decarboxylation of the carbamide leaving group drives the equilibrium furtherin favor of deprotection. In choice (B), trifluoroacetate is a very weak nucleophile, and the sub-stitution of an ester for an amide is energetically unfavorable. Choice (C) is incorrect because,in an acidic solution containing trifluoroacetic acid, water would be an insufficiently strongnucleophile to hydrolyze and amide bond. Choice (D) is incorrect because a tertiary carbon is apoor substrate for an SN2 reaction and trifluoroacetate is a poor nucleophile.

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206. BThis question addresses the reaction shown in Figure 2. Boc-protection prevents coupling of

the free amino acids, shown below, so choice (B) is the correct answer.

While choice (A) is theoretically true, reactive side chains are usually protected before theiruse in synthesis. Because the amino acids are added in excess, coupling is a more likely thanreactions involving the side chains. Choice (C) is incorrect because the Boc-group does not sig-nificantly affect the molecule’s reactivity. The reaction shown in Figure 2 involves the car-boxylic acid end of the Boc-protected amino acid, which is several atoms away from theprotected amino end. The Boc-group protects the amino end of the incoming amino acid, whichis not involved in the coupling reaction. Choice (D) is also incorrect. Dichloromethane, the sol-vent used in solid-phase peptide synthesis, is a polar solvent which is expected to solvate aminoacids which are polar compounds (remember, like dissolves like). If this weren’t true, CH2Cl2wouldn’t be a very useful solvent for this reaction!

207. AThe isoelectric point, or pI, of an amino acid occurs at the pH where the overall charge on

the amino acid is zero. Aspartic acid has two acidic groups and one basic group, so the pI occurswhen the peptide carboxylic acid is deprotonated (negatively charged), the side chain carboxylicacid is protonated (neutral), and the basic amide is protonated (positively charged). This point isfound by taking the average of the two lowest pKas yielding 2.76 (choice (A)). Choice (B) (3.65)is simply the median pKa of the amino acid. Choices (C) and (D) are both the result of averag-ing the incorrect numbers: 5.04 is the average of all three pKas and 6.63 is the average of the twohighest pKas.

208. CAn important point to address is the reason why the side chains of amino acids need to be

protected in the first place, namely because certain side chains contain reactive functionalitiessuch as amine or carboxyl groups. These reactive side chains must be protected by different pro-tecting groups to prevent unwanted side reactions, as stated in choice (C). If the amine side chainis also Boc-protected, it will become deprotected simultaneously with the peptide nitrogen,allowing it to react with the next amino acid, as shown below.

OH

OH2N

RO

OH2N

R

NC

N

NH

C

NOH

O

H2N

R

NH

O

H2N

R

ROH

O

DCC

+

new amino acid new amino acid

The self-coupling reaction

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Choice (A) is a false statement; Boc groups are useful protecting groups for amines, whichis why they are used in protein synthesis. Choice (B) is also incorrect because trifluoroaceticacid will remove the Boc group from all amines by the same mechanism. Finally, choice (D) isincorrect because the E1 mechanism shown in question 2 is more favorable than substitution ona tertiary carbocation under the conditions of deprotection because the methyl groups would hin-der the attack of a nucleophile, making the E1 mechanism faster than SN1.

209. AChoice (A) is correct. The linkage between the resin and the polypeptide chain is an ester

linkage, which is weaker than the amide bond between amino acids. If a peptide bond wereweaker than an ester bond, the polypeptide chain would decompose under the conditions ofcleavage from the Merrifield resin. Therefore, we can eliminate choices (C) and (D). Choice (B)is incorrect because the correct amino acid sequence (primary structure) is insufficient for ensur-ing that a protein will be active—the secondary and tertiary structure of the protein must also becorrect, and these are dependent on the chemical environment as well as the primary structureof the protein. Therefore, choice (B) is also wrong, and we are left with A as the correct answer.Choice (C) is also incorrect. The Kent reaction requires the presence of an N-terminal aminoacid with a nucleophilic side chain such as cysteine which can attack the thioester on the C-ter-minal of the other polypeptide chain, as shown in Figure 3.

Discretes (Questions 210-214)

210. AA woman who is color-blind would have the genotype XcXc. A man with normal color

vision would have the genotype XY. A cross between these two individuals would produce thefollowing results:

Xc Xc

X XXc XXc

Y XcY XcY

The question asks about what percentage of their children will have normal color vision.From the Punnett square, 1/2 of their children will have normal color vision (genotype XXc).The probability that three of their children will have normal color vision is therefore(1/2)(1/2)(1/2) (1/8). This is equal to 12.5%, so choice (A) is the correct answer.

OH

O

BocHN

RO

O

H2N

R

N

C

N

NH

C

N

NH

O

H2N

R

H2NOH

O

H2N

O

NH3+

DCC

+

lysine

Coupling of unprotected lysine and a newly added amino acid

new amino acid

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211. CArylamines are about 106 times less basic than a typical alkylamine. The only alkylamine,

that is an amine whose nitrogen atom is directly attached only to sp3 carbon atoms, is the cor-rect answer, choice (C). Choices (A), (B), and (D) are all arylamines. In each case the nitrogenatom is directly attached to a benzene ring (two benzene rings in the case of choice (B)).

212. BDigestion of carbohydrates begins in the mouth, where salivary amylase hydrolyzes starch

to maltose, so statement I is correct. Carbohydrate digestion does not take place in the stomach,so statement II is incorrect. Carbohydrate digestion continues in the small intestine, where dis-accharides are broken down into monosaccharides, so statement III is correct. The correctanswer is I and III only; this corresponds with choice (B).

213. CThe antibody binding site is the region of an antibody that interacts chemically with an anti-

gen, and is determined by an antibody’s three-dimensional structure. There is no reason to sup-pose that the antigen binding site is nonpolar; for instance, a polar antigen would bind to anantigen binding site that had many polar amino acids; thus choice (A) is incorrect. By the samereasoning, there is no reason to exclude hydrogen bonds from the antigen binding site; thus,choice (B) is incorrect. Since the antibody binding site determines what antigen a particular anti-body can bind to, the antibody binding site varies from antibody to antibody, making choice (D)incorrect. Choice (C) is the correct answer.

214. DTranslation is the term used to describe the process of converting a sequence of mRNA

codons into a sequence of amino acids. Transformation is the uptake and incorporation of“naked” DNA (fragments found in the environment) by a recipient bacterial cell. In the lyticcycle, a bacteriophage takes control of a host’s genetic machinery, and manufactures progenyuntil the bacterial cell bursts, releasing the manufactured bacteriophages to infect new bacteria.In the lysogenic cycle, a bacteriophage becomes integrated into the host bacterium’s genome andremains dormant for one or more generations. The situation described in the question stemdescribes the lysogenic cycle; thus choice (D) is correct.

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