Full length 9 answers

MCAT Full-Length Tests

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PHYSICAL SCIENCES ANSWER KEY

1. B

2. A

3. C

4. C

5. B

6. C

7. D

8. C

9. B

10. B

11. A

12. D

13. B

14. D

15. B

16. A

17. B

18. D

19. A

20. D

21. B

22. D

23. C

24. B

25. B

26. B

27. C

28. A

29. C

30. A

31. D

32. C

33. D

34. B

35. C

36. D

37. A

38. A

39. C

40. B

41. B

42. C

43. D

44. A

45. B

46. C

47. C

48. A

49. B

50. C

51. B

52. D

53. D

54. C

55. A

56. C

57. B

58. B

59. A

60. D

61. A

62. D

63. C

64. B

65. D

66. D

67. A

68. C

69. D

70. B

71. C

72. C

73. B

74. D

75. A

76. D

77. D

Passage I (Questions 1–6)

1. BSince the two component forces are in serial in the Maxwell model, the elastic force is no

longer applicable after time zero; and according to the equation for the viscous component, agiven amount of force will continue to stretch the material as time progress. The Voight model,on the other hand, has a given length shared by both elasticity and viscosity, such that as the vis-cous component is stretched over time, it is also limited by the resistant elastic force thatincreases as the material is elongated. The correct answer is therefore choice (B).

2. AThis requires you to know the definition of Young’s modulus, which represents the stiffness

of a material. It is a property of solids and therefore, does not involve viscosity. Eliminatechoices (B) and (D). In order to increase stiffness, you need to increase the value of k. Based onthe equation F = kx, a stiff material will require a large force to produce a small amount of defor-mation. Given a large F and a small x, k must be increased to increase Young’s modulus; choice(A) is correct.

3. CAccording to the passage, the viscosity of the material can be attained from the equation

F=�v, where � is the viscosity. Since the force F is given and is constant, the remaining infor-mation required is v, which is the speed of deformation—choice (C) is correct. Both choices (B)and (D) combined can allow one to evaluate the speed of elongation, but neither by itself is asufficient.

4. CFirst, work W = Fd. Given the constant force of 25 N, you need to calculate the total dis-

tance d that is stretched over the 1 second time period. First, calculate the instantaneous stretchof the elastic component, which is x = F/k after a little rearrangement of the equation providedin the passage. We then add the stretch of the viscous component in one second, which wouldbe x=Ft/�, if we substitute x/t for v (the definition for velocity). The total calculated elongationshould equal 5 m for the elastic stretch plus 5 m for the viscous stretch; a total of 10 m. Substi-tuting back to our equation for work, we get W=(25N)(10m) = 250 J. Choice (C) is correct.

5. BAt time t = 0, the viscous component has yet to come into play, since it is time-dependent.

Hence, the initial elongation would simply be the elongation of the spring component, which isdetermined by the equation F = kx. To determine the elongation x, shuffle the equation, usingF=F0 to get x=F0/k .

6. CThis model is a combination of both the Maxwell and Voight model. Though it may seem

rather complex, it can be analyzed by its parts. First, it has an elastic spring in series with aVoight material. Since from the Maxwell model, you know that an elastic spring will deforminstantaneously, you can narrow the answer choices to (A) and (C), where there is a short verti-cal component of the graph showing an instant displacement. After t = 0, the Voight materialshould take over with the asymptotic curve as shown in choice (C). Choice (A) is a simplespring, while choice (B) is a simple damper. Choice (D) is incorrect since it does not display anasymptotic limit , which we expect when no damper is present in series. Choice (D) representsa damper in series with a Voight material.

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Passage II (Questions 7–12)

7. DIn the passage, you learned that the color of a coordination complex is the result of a promo-

tion of an electron from a low-energy d orbital to a higher energy orbital. Therefore, any transi-tion metal complex that is colorless must not be able to effect this transition. This leaves you withtwo possibilities: either the complex has no electrons in d orbitals or all the d orbitals are filledso no electrons can be promoted. The first possibility makes no sense–the definition of a transi-tion metal is that it has electrons in d orbitals. Therefore, the second option is your best bet. Thismatches what you already know about zinc ions; they are d10, meaning that there are 10 electronsin the d orbitals—all full. Now take a look at the answer choices. (D) matches this prediction; theeg orbitals are the higher-energy orbitals in a crystal field, so if they are full, all the d orbitals arefilled. This will make a promotion impossible. The complex will be colorless.

Choice (A) makes no sense—when a metal loses electrons, its properties change, so even acolorless metal could still have colored complexes once it lost electrons to become an ion.Choices (B) and (C) might seem tempting, but there is no evidence in the passage or the ques-tion stem that the crystal field splitting energy is either small or large, so there is no way to saywhat the frequency of light would be if an electron could be promoted. In any case, all orbitalsare filled for this complex.

8. CThis question might seem daunting when you first read the stem—there are so many possi-

bilities! Don’t panic; just keep the passage information in mind as you read the answer choices.The answer choices are limited to information about color, so you know that the question is ask-ing you to say something about the color of two solutions: a solution of K3[CrBr6] and a solu-tion of K3[Cr(CN)6]. Now things are a lot simpler. What is the difference between these twosolutions? This, too, might seem complicated, but as is often the case on the MCAT, you reallyonly need one piece of information to answer this question correctly. That piece of informationis going to come from the spectrochemical series given at the end of the passage. Take a look,and you will see that bromide and cyanide ions lie at opposite ends of that spectrum. Whateverdifference there is between these two solutions, it will be related to high and low spins. Ascyanide ions replace bromide ions, the solution will be changing from high spin to low spin.According to the passage, this is going to mean a larger crystal field splitting energy—a largertransition for an electron to make. In addition to the enlargement of the splitting, you also needto consider the fact that as the complex changes from high spin to low spin, there will be moreopportunities for promotion. This is because there will be more electrons in the lower energylevels and more available orbitals into which these electrons can be promoted. There will bemore promotions, so the intensity of the color will increase. Therefore, the answer is (C).

(A) makes a false comparison between bromide and cyanide—in CFT, most of the ligandswill be negatively charged, so a negative charge will not be enough to draw parallels betweenpairs of ligands. (B) assumes that the ions are all in solution together—you know this isn’t truebecause a complex is one unit. (D) tells you that the �0 decreases, but you know that it increases,so (D) is incorrect.

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9. BMany MCAT questions follow the pattern of this one: ask a yes or no question (or a ques-

tion with two possible answers) and give two possible explanations for each. So, you can usesome strategy to eliminate half the answer choices right off the bat. In this case, you can chooseto eliminate both “no” answers, both “yes” answers or you can choose to eliminate wronganswers based on their explanations, then decide whether “yes” or “no” corresponds to the cor-rect explanation. All the answer choices mention the paired or unpaired nature of electrons in amanganese complex. That’s your clue to take a look at the electron configuration of that metalion. Mn2+ has five d electrons, so it will have unpaired electrons regardless of whether it is highor low spin. (If it is high spin, there will be five unpaired electrons—one in each d orbital. If itis low spin, there will be one unpaired electron in a t2g orbital and 4 paired electrons in t2gorbitals.) Knowing that there are unpaired electrons, you can eliminate (A) and (C). Now, youneed to refer to the passage or to your outside knowledge to determine whether unpaired elec-trons lead to paramagnetism. The passage states that a paramagnetic substance is one which con-tains unpaired electrons, so the answer must be (B).

10. B The first thing to do here is to take a look at the electron configuration of the Fe2+ ion.

Iron(II) has six electrons. This means that when it is high spin, it will have four unpaired elec-trons. When it is low spin, it will have no unpaired electrons, because all the electrons will be inthe three t2g orbitals. Once all the electrons are paired, the complex will be diamagnetic; it willhave lost its paramagnetic properties, so (B) is correct.

11. ABeware of extraneous information in question stems when you’re taking the MCAT! It does-

n’t matter whether a complex is low spin or high spin when it comes to the total number of elec-trons that can fill all the d orbitals. Every transition metal has five d orbitals, each of which canhold two electrons, so ten electrons would fill the d orbitals of a coordination complex. There-fore, (A) is correct.

12. DAt first, it might seem that this question makes no sense because the first compound,

Mn(NO3)2, isn’t a coordination complex. Don’t be fooled, though; the question stem tells youthat you’re dealing with an aqueous solution. Therefore, it is a coordination complex withMn(NO3)2 as the central “ion” and water molecules as ligands. Now your job is to compare H2Oand CN– with Cl–. Don’t fall into the next trap, either—just because cyanide and chloride areboth negatively charged doesn’t mean they’re the most similar. What you need to do is consultthe spectrochemical series. Water is high spin, while cyanide is low spin. Where does chloridefall on the spectrum? It is even more high spin than water, so the complex will be high spin.Therefore, choice (D) is correct.

Passage III (Questions 13–18)

13. BThe question gives a speed, a distance, and wants you to find time. Here the rate is Mach 3,

which equals the ratio of the speed of the jet to the speed of sound. Since the jet is going 3 timesas fast as the speed of sound, the speed of the jet is 900 m/s. The question asks the time requiredto fly around the circumference of the Earth.

4

Circum. = �d � (3)(1.28 � 107 m) � 4 � 107 m

→ t = � 0.5 � 105 s � 13 hours, choice (B).

14. DThe passage mentions that the Mach number of a jet can increase as it ascends, without

increasing its velocity. Since, Mach number is the speed of the jet over the speed of sound, andthe speed of the jet is constant, the larger Mach number must mean a smaller speed of sound.The speed of sound is decreasing as the altitude increases. Remember that the speed of sound isinversely proportional to the density of the medium through which it is propagated. If speeddecreases, density must have decreased: choice (D) is correct. Choice (A) is wrong, since thespeed of sound in air does change as you change altitude. Choice (B) is incorrect since the fre-quency of a wave passing from one medium to another is constant. Finally, choice (C) is neverimplied, since the intensity of the sound wave may be different depending on whether the noseor the back is emitting the wave.

15. BThe figure indicates that the waves are building up in front of the jet. The passage states this

is exactly what happens when the jet flies at the speed of sound. If choice (A) were true, thewaves would be ahead of the jet. Choice (C) is illustrated in the passage and choice (D) wouldinvolve a change in the jet’s speed that is not illustrated by the figure.

16. ARemember that beat frequency is the difference between frequencies of a combined sound

wave. Initially, the jets start at the same place and emit identical sound waves, so there are nobeats. As one jet speeds up, the Doppler effect indicates that the detector will record a smallerfrequency for the sound emitted by that jet. Therefore, the frequencies of sound emitted by thetwo jets are now different, and a beat will be detected. So, the beat frequency increases from zeroto some larger frequency.

17. BIf the two people are underwater, then the sound waves have to travel through air and then

through water to reach them. Since sound travels faster in denser substances, it travels faster inwater than in air. Therefore, you would expect the two people swimming to hear it first. How-ever, we do not know how deep the people are underwater. In other words, the swimmers couldbe so deep underwater that the wave has so much longer to travel before it can reach them, thuseven though it travels faster it may take more time to reach them then their friend. The only thingwe can be sure of is that the frequency of sound is constant as it passes from one medium toanother: choice (B). Finally, if frequency is constant, but speed changes between two media,wavelength must also change: choice (D) is incorrect.

(4 × 107 m)��9.0 × 102 m/s

5

18. DAs the jet decelerates, the chair experiences a force equal to its mass times acceleration. So,

the spring will elongate under that force according to Hooke’s law: F = kx. These forces areequal, so one can say that the force on the spring is proportional to the change in speed of thechair: since, the final velocity is zero, the force is proportional to the original velocity of the jet,choice (D).

Passage IV (Questions 19–24)

19. AIn Reaction 1, a proton is transferred from a carbonic acid molecule to a water molecule.

Once this transfer has occurred, the bicarbonate ion has a charge of –1. Therefore, any positivelycharged species will be difficult to remove; charge separation is not favored by Coulomb’s Law.Choice (B) is incorrect because as protons are lost from carbonic acid, the concentration ofbicarbonate ion increases. Choice (C) is incorrect because the molecular geometry of bicarbon-ate ion is not changed by losing a hydrogen ion. Choice (D) is incorrect because, while hydro-gen bonds form, they do not have a significant effect on the Ka values.

20. DCarbonate, CO3

2–, has 24 valence electrons: 4 from carbon, 6 from each of the three oxygenatoms (18) and 2 from the negative charge on the whole ion. These will be arranged accordingto the following Lewis structure:

Using VSEPR Theory, you know that the three oxygen atoms will repel each other as muchas possible; this will result in atoms that are in a plane with 120º separations. This is a trigonalplanar geometry.

21. BIn the section on Experiment 2, the passage states that a dilute solution of potassium hydrox-

ide is used. This will be basic, so the solution starts out with a blue color. From this, you caneliminate choices (C) and (D). Then, recognize that as solid carbon dioxide dissolves in theaqueous solution, it will combine with water to form carbonic acid. This will neutralize thehydroxide ions and cause the solution to become acidic. Since bromothymol blue is colorless inacidic solution, the solution will lose its blue color to become clear. The answer is (B).

22. D The key to answering this question correctly is to write equilibrium expressions for Reac-

tions 1 and 2. Start out by writing:

Ka1� and Ka2

�[H3O+][CO2-

3]��

[HCO-3]

[H3O+][HCO-3]

��[H2CO3]

O O

O

C

2-

6

7

Then, solve for hydronium ion concentration in each equation to get:

[H3O+] � and [H3O+] �

Now it is clear that only choice (D) makes sense.

23. CTo calculate the equilibrium concentration of the bicarbonate ion, set up a table of values:

H2CO3(aq) +H2O(l) H3O+(aq) + HCO3–(aq)

Initial 0.034 �0 0concentration (M)

Change in –x +x +xconcentration (M)

Equilibrium 0.034 – x +x +xconcentration (M)

Now set up the following equation:

Ka1�

4.3�10–7 �

The next step is to solve for x; this gives x = 1.2�10–4 M.

All the other answer choices are traps: (A) is simply the Ka2value, (B) is the Ka1

value and(D) is the initial concentration of carbonic acid. Any of these could have been eliminated with alittle thinking!

24. BTake a look at the Lewis structure you drew for question 2. Each oxygen atom should be

identical, but in the one shown, two of the oxygen atoms have single bonds to carbon, while oneof them is double-bonded. Each of the oxygen atoms is equivalent, so there must be enough res-onance structures to account for that. The following three resonance structures are relevant:

2-

CO

O

2-

CO

O

2-OO

C

C

O O

x2

�0.034 – x

[H3O+][HCO-3]

��[H2CO3]

Ka2[HCO-3]

��[CO2-

3]

Ka1[H2CO3]��

[HCO-3]

Discrete Questions

25. BWhen light enters a medium with an index of refraction n, its speed is reduced to:

v � �n

c� � � 2.25�108 m/s.

answer choice (B).

Notice that answer choice (D) is impossible, as nothing can travel faster than the speed oflight in a vacuum.

26. BIn a galvanic or voltaic cell, the oxidation and reduction reactions are carried out in separate

compartments. These cells are used to do work, so their reactions must be spontaneous. Sincethey operate spontaneously, �G must be negative, so the answer is (B).

27. CWhen a substance sublimes, it is converted from solid to gas using the heat of its surround-

ings. Carbon dioxide sublimes at room temperature because the energy of the air surrounding itis sufficient to effect a phase transformation. Since the CO2 molecules are absorbing heat,entropy increases. Gases have greater entropy than solids, so the entropy change for sublimationmust be positive. The answer is (C).

28. AA gas absorbs heat and does work—this should remind you of the First Law of Thermody-

namics:

�U � Q–W.

So the gas in this case absorbs 900 J of heat energy, and does 600 J of work. The change ininternal energy is therefore 900 – 600 = 300 J. Whenever an ideal gas undergoes a change ininternal energy, its temperature changes as well:

�U � �3

2�n�T→�T � � � 100 K,

answer choice (A).

Passage V (Questions 29–33)

29. CThe first things we should take notice of upon reading the question are:

This is a Roman numeral question. We should test the most popular statement first, elimi-nating answer choices as we go along.

The question is asking about which forces do work, which means we need to understandwhat it means for forces to do work.

2(300)�

3(2)

2�U�

3n

3�108 m/s��

4/3

8

When an object is displaced a certain distance d through the actions of a force F, pointing inthe direction of the displacement, then the work done on the object by the force is W = Fd. Whenthe force points in the opposite direction to the displacement, the work done is negative: thisforce isn’t adding energy to the system, it’s removing energy from the system; the force is resist-ing attempts to displace the object. When an applied force is perpendicular to the displacement,then the work done by that force on the object is zero.

It might be tempting to go ahead and work out on your own which forces are doing work onthe lower box, but remember: this is an MCAT question, not a homework question. Jump imme-diately to the statements and evaluate them.

Statement I appears in the most answer choices, so start there. Does the applied force F dowork on the lower box? Yes—the lower box is displaced to the right (in Figure 1), and force Fpoints to the right. Eliminate choice (D).

What about the weight of the upper box, in statement II? Strictly speaking, the weight of theupper box acts on the upper box, not the lower box. So statement II is false, leaving choice (C)as the correct answer.

For completeness, let’s check that statement III is true. The contact force C is the force dueto friction on the upper box due to the lower box. It points to the right (think about it: it’s theforce that causes the upper box to accelerate to the right, so C had better point to the right.) Sincethe lower box exerts contact force C on the upper box to the right, by Newton’s third law, theupper box must exert contact force C to the left on the lower box. Since this force points oppo-site to the direction of displacement, it does work on the lower box.

30. AA moment’s reflection on this problem should lead you to realize that whether or not the

upper box slides is a function of friction. Static friction exists between the two boxes as theyaccelerate; the passage gives you the coefficient of static friction as m. In general, when doesone object slide with respect to the other? When the static frictional force needed to counterbal-ance the applied force reaches and then passes its maximum value, �N.

What is the normal force N in this situation? The vertical force that the lower box exerts onthe upper box. Since the upper box is not accelerating in the vertical direction, the normal forceis balanced by its weight, and N = m2g. So the maximum value of the static frictional force is�m2g, and the correct answer is choice (A).

31. DWe need to determine what additional information would be necessary to determine the

speed of the boxes at some time t. This is a kinematics problem—in all kinematics problems,begin by figuring out what you already know. The passage indicates that the boxes start fromrest, so v0 = 0. From kinematics:

vf = v0 + at.

Since we’re given the time elapsed in the problem, and the acceleration a is given in the pas-sage, we have all we need. The correct answer is choice (D).

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32. CNow that the top box is sliding, the motion of both boxes is no longer identical. This prob-

lem can be analyzed in great detail, but that is not necessary to answer this question correctly.The structure of the answer choices indicates to us that we need to figure out two things:

Does the acceleration of the bottom box increase or decrease?

Does the acceleration of the top box increase of decrease?

Your gut instinct may be to think “well, once the top box starts sliding, it should fall off theback, so the top box must be slowing down.” Go with that instinct—the physical scenario in thispassage is in line with everday experiences, so you can trust your gut.

Thinking physically about the scenario in this problem will back up our gut reaction. Thetop box begins to slide when the contact force between the boxes switches over from static fric-tion to kinetic friction. Since the coefficient of kinetic friction is generally less than the coeffi-cient of static friction, the force of kinetic friction is less than the maximum force due to staticfriction. As the contact force is now smaller, the impediment to the forward motion of the lowerbox is lessened, while simultaneously, the force that pushes the upper box forward with respectto the ground is lessened as well. The acceleration of the upper box decreases, and the acceler-ation of the lower box increases, answer choice (C).

33. DWe are told in the problem that the mass of the lower box is three times the mass of the upper

box, and then asked to compare the current acceleration of the system to the acceleration of thelower box if the top box were removed.

Let’s start with the motion of the system of boxes. The only external force on the system ofboxes is F, so the acceleration a must satisfy F = (m1+m2)a. Since m1 = 3m2,

a = �34

�.

After the top box is removed, the force F is applied only to the box with mass m1. The newacceleration of the system, a’, is F/m1, which is 4/3 times the old acceleration. Choice (D) is cor-rect.

Note that you could have eliminated answer choices (A) and (B) right away by realizing thatremoving mass from the system while keeping the force constant was going to increase theacceleration by some amount.

F�m1

10

Passage VI (Questions 34–39)

34. BWhen you’re asked about the spontaneity of a reaction, you should immediately think of the

Gibbs free energy, DG. This eliminates choices (A) and (D). You know that a reaction with anegative Gibbs free energy is spontaneous, so the answer is (B).

35. C

Titanium is bonded to two oxygen atoms in TiO2, as shown in the Lewis dot structure above.Each oxygen has a charge of –2, so titanium must have a charge of +4.

36. DYou are asked to characterize the type of reaction you are given. Choices (A) and (B) both

involve redox chemistry. You know these can’t be correct because the oxidation state of Ti doesnot change from reactants to products; it is +4 in both cases. This leaves you with (C) and (D).The transfer of chloride ions from titanium ion to the aqueous acid form might lead you to labelthis an exchange reaction, but (D) is a better choice. The fundamental aspect of this process isthe reaction between liquid water and titanium tetrachloride. Whenever a substance reacts withwater, it is a hydrolysis reaction.

37. AOnce titanium has lost all four of its valence electrons, it has the electronic configuration of

argon, [Ar]. (C) or (D) would apply if titanium had not yet lost its electrons by ionization.

38. A Titanium dioxide forms spontaneously according to Reaction 1. It is very unlikely that this

stable compound would undergo a corrosion reaction to form a gaseous “tetraoxide”. The otherthree reactions, however, all involve the formation of one of two stable substances: solid tita-nium or titanium dioxide. The reaction in choice (B) is extremely similar to Reaction 2 as shownin the passage; just substitute chlorine for bromine. The reaction in (C) forms solid titanium,which will drive the reaction because of the stability of titanium. The reaction in (D) is very sim-ilar to Reaction 1, substituting bromine for chlorine, so it is not unlikely.

39. CConsider statement (I) first since it appears most often in the answer choices. Is humidity

going to affect Reaction 1? Most certainly—the reaction is between titanium tetrachloride andwater, so humid conditions will make this reaction more favorable. Therefore, it will not inhibitthe reaction. Any choice that contains statement (I) should be eliminated. (Notice that just bydoing this, you are already done because only choice (C) does not contain statement (I).) Nowexamine statement (II). Will performing the reaction in the presence of a base affect the reac-tion? Since acid is produced, the presence of a base will drive the equilibrium to the right. There-fore, it will not inhibit production of titanium dioxide. Finally, consider statement (III). Willperforming the reaction in the presence of hydrogen gas affect the production of titanium diox-ide? A side reaction similar to Reaction 2 is very likely to occur here:

TiCl4(g) 2H2(g)→Ti(s) 4HCl(g)

O Ti O

11

This will reduce the amount of titanium dioxide produced, so statement (III) is the only onethat applies.

Passage VII (Questions 40–44)

40. BThere are two bright maxima, at 400 nm and 600 nm. To employ the formula given in the

passage, we need to know what the wavelengths of those two sources are inside the film:

� � 233 nm, � � 400 nm.

So, inside the film,

d � �m

2

� →

d � m1� (400)/2 � 200m1,

d � m2� ( ) � 400/3 m2

The question is, which integer pair m1, m2 yield the same d?

By trial and error, you can see that m1 = 2 and m2 = 3 yields a depth of 400 nm, choice (B).

41. BAnswer choice (D) is certainly wrong, as nothing prevents light from entering a medium

with a higher index of refraction at any angle. Similarly, choice (C) is wrong, because the phase-shift that wave undergo on reaching a boundary with a higher index of refraction doesn’t dependon the angle of incidence.

Choice (B) is correct. The formula assumes that the rays are coming in close to the vertical.If they were coming in shallow, the path length through the film would no longer be 2d, butsomething much longer.

42. CWhat determines the degree to which light is refracted? Snell’s Law, n1 sin�1 � n2 sin�2.

The angle of refraction has no dependence at all upon the wavelength of the light, so choices (A)and (B) are incorrect. In fact, we’ve already stated the correct answer: choice (C). Choice (D) isincorrect: it’s because the indices of refraction are different that refraction occurs at all.

43. DAre all the conditions for the soap bubble the same as those for the thin film? No. The thin

film is layered between a medium of lower index of refraction, and a layer with a higher indexof refraction. A soap bubble is surrounded on air on both sides. One of the rays isn’t going toundergo a phase shift, and so the equation no longer applies to constructive interference. Answerchoice (D) is correct—it’s ray 2 that meets the second soap-air boundary.

800/3�2

600 nm�

1.5

400 nm�

1.5

12

44. AThere are two fringes between 400 nm and 700 nm—the visible spectrum, when white light

is shone upon a film with d = 900 nm. Using the equation from the passage,

� �2md� = �

180m0 nm�.

Only two values of m yield wavelengths in the visible spectrum: m=3, and m=4. Conse-quently, � 600 nm and 400 nm can be seen. The correct answer is choice (A).

Passage VIII (Questions 45–50)

45. BRemember ROYGBIV (red, orange, yellow, green, blue, indigo, violet). Red is the lowest

energy and violet is the highest energy. As energy increases, the wavelength gets shorter. Sopotassium, which emits violet light, emits a higher energy, shorter wavelength of light than doessodium (yellow light).

46. CThe passage states that the yellow produced by sodium generally washes out all other colors

in the firework. So the presence of sodium (in the form of sodium perchlorate) will eliminate thedesired red color. Consequently, sodium perchlorate would not be a good oxidizer. (A) and (B)are wrong because the reactivity of sodium and potassium are not at issue, the reaction isbetween the perchlorate and the fuel, not the cation and the fuel. (D) is the exact opposite ofwhat is stated in the passage.

47. CThe reaction is a redox reaction, indicating that at least two atoms are going to change oxi-

dation states. Al produces a 3+ ion (see periodic table), indicating that either (A) or (C) is cor-rect (oxygen is a 2– ion). (B) and (D) have Al with a +4 charge and can therefore be eliminated.(A) is not balanced (too many Al atoms, not enough O atoms), leaving only (C).

48. A The passage states that the composition melts upon ignition, which allows them to react. So

the act of melting from a solid to a liquid allows the reaction to begin. Choice (A) is true—solidreactants do not mix and therefore cannot react. When the reactants are melted, they mix andbegin to react. We also know from common sense that (B) is false—you can go to the store andbuy fireworks, right? (C) is tempting, but we are told that dextrin is used to bind the reactantstogether, not keep them in separate compartments. (D) is completely false; dextrin is not used asan inhibitor.

49. BRed is a longer wavelength (ROYGBIV) and therefore less energetic form of light than green

light. So the difference between energy levels in the strontium atom is less than in the bariumatom. The electrons are excited from the s to the d orbital (level 5 in strontium, level 6 in bar-ium), not from s to s. So the correct answer is (B).

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50. CThe passage states that ignition melts the reactants. So (C) makes sense: melting the oxidizer

and fuel allows them to mix and begin reacting. (A) is false because the dextrin is not used toseparate the oxidizer and fuel. (B) is false because we know that the reaction generates heat andis therefore exothermic. (D) is false because the redox reaction between the oxidizer and fuelcauses the oxidizer to release its oxygen, not the heat from the fuse.

Discrete Questions

51. BThe magnetic force on a current-carrying wire is:

F�ILB�(1 A)(3 m)(0.5 T)�1.5 N.

We can eliminate choices (C) and (D). Now you can use the right-hand rule to determine thedirection of the force: your fingers point with the moving charges, up the page. Curl your fin-gers in the direction of the field, out of the page. Your thumb now points to the right—that is thedirection of the force. Choice (B) is correct.

52. DThe magnetic quantum number indicates how an orbital behaves in an electric field. This

should immediately point you to (D). The shape of the electron cloud is determined by the mag-netic quantum number. The radial size of an electron cloud is determined by the principal quan-tum number (A). The number of valence electrons about a nucleus and the number of protons inthe nucleus are determined by the location of the element on the periodic table.

53. DWhen the weight of the mass is in balance with the restoring force from the spring, then:

kx � mg → k � �mxg� � � 333 N/m,

choice (D).

54. CThe oxidation number of chromium in Cr2O7

2– is 3. This may be calculated as follows: theoxidation number of oxygen in Cr2O7

2– is –2. There are seven oxygen atoms, resulting in acharge of –14. The overall charge of the ion is –2, leaving +12 for the two chromium atoms.Each has a charge of +6. The difference between +6 and +3 is 3, choice (C).

55. AAs the two sounds are 3 Hz apart, that is the beat frequency (fb �f1 – f2). So the beats

occur three times per second, or once every one-third of a second: 0.33 s. The correct answer ischoice (A).

(2 kg)(10 m/s2)��0.06 m

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56. CIn order to answer this question, ask yourself what is meant by “expanded octet”. An

expanded octet is an atom that is surrounded by more than the usual eight electrons. In order todetermine which compound has an expanded octet, draw the Lewis structures for each answerchoice:

Clearly, POCl3 has 10 electrons around the phosphorus atom, so (C) is the correct answer.

Passage IX (Questions 57–62)

57. BThe current is running up and down the antenna. When current runs up the antenna, point

your thumb upwards and curl your fingers in the direction of the field. In this case, in a circlearound the antenna. When the current points down, point your thumb down: now the magneticfield circles the antenna in the other direction. The correct answer is choice (B).

58. BA negatively charged particle will, inside an electric field, experience a force opposite to the

field direction. Since the field points upwards, the electron will be forced downwards: theanswer is choice (B).

59. AThis is a classic MCAT Physics question, asking you to derive a property of an electromag-

netic wave. In this case, you are given the wavelength and asked to produce the period. From thewavelength, we can derive the frequency:

f � �c

� � � 6 � 107 Hz.

The period is the reciprocal of the frequency: T ��16� � 10–7 �1.7 � 10–8 s, answer choice (A).

60. DThe intensity of an electromagnetic wave is a function of the amplitude, which is increased

by boosting the power. There is no connection to the length of the antenna rods (choice (C)), noris there a connection between power and frequency (A) or period (B). However, increasing thevoltage of the AC generator will increase the power of the signal (P = IV), resulting in a higherintensity. The correct answer is choice (D).

3�108�5

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61. AThe change in potential energy in moving a charge q across a potential �V is:

�U�q�V.

In this case, to move the charge +ne across potential V will require an amount of work equalto the change in potential energy of the charge:

W � �U � (ne)(V) �neV.

answer choice (A). Notice that the length of travel has no effect on the work done—like anyconservative force, it’s only the initial and final states that count.

62. DAnother classic MCAT Physics problem: the MCAT loves power. Always remember that

power is energy per unit time. In this case, we are given the power and the time, and asked tocompute the energy:

P � �Et� → E � Pt � (50000 W)(60 s) � 3 �106 J � 3 MJ,

answer choice (D).

If the magnetic field now points from right to left, then the magnetic force on the current isgoing to be out of the page (fingers point up the page (v); fingers curl to the left (B); thumbpoints out of the page (F)). The charges are going to separate across the depth of the strip insteadof the width. In this case, the electric field that forms will also run along the depth, instead ofthe width.

Since the behavior of the charge carriers does change, choice A is incorrect.

It is true that the electric field no longer runs along the width of the strip, so answer choiceB is correct.

The electric field doesn’t point in the opposite direction (C), and we know that a new Hallvoltage is created, in this case along the depth of the strip instead of the width (D).

Passage X (Questions 63–67)

63. CThis question asks whether a reaction will go to completion. By inspection of the answer

choices, you know that the key to answering correctly will be to determine whether the reactantsare present in stoichiometric amounts or whether a limiting reagent issue will arise. So, the firstthing to do is recognize that choice (A) may be eliminated immediately; Reaction 2 has a 6:1ratio of sodium to iron oxide, so equimolar amounts of the two will not cause the reaction to goto completion. The next thing to do is begin your treatment of the limiting reagent problem. Cal-culate the molar amounts of Na(l) and Fe2O3(s) by performing dimensional analysis, using thevalues you’re given along with the stoichiometry of the reaction:

106 g NaN3� � 1.63 mol NaN3mol NaN3��65 g NaN3

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Now that you know the number of moles of sodium azide, you’ll want to calculate the result-ing number of moles of elemental sodium:

1.63 mol NaN3� � 1.63 mol Na

Now you’re ready to take a look at the iron oxide present in the reaction chamber:

35 g Fe2O3� � 0.2 molFe2O3

The next step is to compare the stoichiometry of Reaction 2 with the molar amounts you justcalculated. Do this by dividing 1.63 by 0.22. If the quotient is greater than 6 (the molar ratio ofsodium to iron oxide in Reaction 2) then iron oxide is the limiting reagent. If the quotient is lessthan 6, sodium is the limiting reagent. If the quotient is equal to 6, then stoichiometric amountsof reactant are present, and the reaction will go to completion.

� 7.4

This is greater than 6; excess sodium is present. Therefore, choice (C) is correct.

64. BIn this case, the meaning of root mean square speed is not important; this is a pure calcula-

tion question. However, you should know that root-mean-square speed is a measure of the typ-ical speed of molecules in a gas at thermal equilibrium. To calculate the root-mean-square speed,use the following formula:

urms � ��3�M

R�T��,

where R is the ideal gas constant, T is the temperature of the system in Kelvins, and M isthe molar mass of the gas in kilograms per mole. Since the answer choices are convenientlygiven in terms of RT, you only need to determine the molar mass of nitrogen to answer thisquestion correctly. The molar mass of nitrogen is approximately 28 grams per mole, so choice(B) is correct.

65. DThis is a conceptual question, designed to test your knowledge of the ideal gas law and its

applications. At a high altitude such as that of Mt. Shasta, the pressure will be lower than it is atsea level. However, the temperature is slightly lower at high altitudes, so you have to take thatinto account as well. Take a look at the Ideal Gas Law to see how these two factors will affectthe volume of the experimental balloon:

PV�nRT

Manipulate the equation to examine the effects of pressure and temperature:

V��nR

PT

�

Notice that a decrease in pressure and a decrease in temperature will cause competing effects—low-ering the pressure causes an increase in volume, while lowering the temperature causes a decrease involume. Only (D) conveys this appropriately.

1.63 mol Na��0.22 mol Fe2O3

mol Fe2O3��159 g Fe2O3

2 mol Na��2 mol NaN3

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66. DThis question indirectly tests your knowledge of reaction mechanisms. The key to answer-

ing it correctly is knowing how a catalyst works. A catalyst is a substance that takes part in achemical reaction and speeds it up, but undergoes no permanent change itself. How does a cat-alyst speed up a reaction? It does this by lowering the activation energy of the reaction. This willbe manifest in the height of the peak between products and reactants in the energy profile. There-fore, choice (D) is correct.

Choice (A) is wrong because the energy of the products will not be higher than that of thereactants; the reaction is spontaneous, so products are lower in energy than reactants. Choice (B)is also wrong, because the activation energy will be lower with a catalyst, not higher. Choice (C)is wrong because the final potential energy of the products is not equal to the potential energyof the reactants.

67. AThe MCAT expects you to know that substances in their elemental states have oxidation

numbers equal to zero. Therefore, the product, elemental iron (Fe) will have an oxidation num-ber of zero. This allows you to eliminate choices (C) and (D). Now, you just need to determinethe oxidation number of iron in iron oxide. Since oxygen has a charge of –2 and there are threeof them, the iron atoms must balance the charge with +6. There are two iron atoms in iron oxide,so each must have a charge of +3.

Passage XI (Questions 68–72)

68. CThe amount of work done by a force over a distance is W = Fd. In this case, however, there

are no forces or distances available. But since you have both the starting (10 m/s) and ending (0m/s) speed, you can apply the work-energy theorem to this problem.

The amount of work done by friction is equal to the change in kinetic energy of the object:

KEi � �12� (0.5)(10)2 � 25 J

KEf � 0 J.

Therefore, the total work done by friction is choice (C), 25 J.

69. CThe disk that started at 4 m/s decelerated to 0 m/s in 1.2 seconds, according to the table in

experiment 1. Using kinematics,

a � � � –3.3 m/s2,

answer choice (C).

0– 4�1.2

vf –v0�t

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70. BThe surface with the largest coefficient of kinetic friction will be the one who put a stop to

the disk the quickest. However, you must compare apples to apples: each of the pucks must startwith the same initial speed. You must use the data from experiment 1 for a starting speed of 4m/s, since experiment 2 uses only 4 m/s as a starting speed. Surface A actually stops the disk thequickest among the four; choice (B) is the correct answer.

71. CIt might be tempting to reason that the heavier the mass, the higher the acceleration will be.

But while it’s certainly true that the frictional force goes up with mass, the acceleration is massindependent. For example, the normal force on a disk equals the disk’s weight: N = mg. Conse-quently, the force of friction is �mg. Since this is the force that provides the acceleration, New-ton’s second law indicates that:

ma � �mg → a � �g

The acceleration is mass-independent, and the correct answer choice is (C).

72. CThe disk started at 4 m/s and came to a stop in about one second, so the deceleration was

about 4 m/s2. Kinematics tell us:

d � v0t + �12�at2 � (4)(1)–(0.5)(4)(1)2 � 2 m,

answer choice (C).

Discrete Questions

73. BThe MCAT loves to ask about systems being studied; these are considered in relation to the

surroundings. A closed system is one for which energy can be exchanged with the surroundingswhile matter is kept separate. Therefore, choice (B) is correct. Don’t get thrown off by choice(C); it is similar to the correct answer but gives just the opposite conclusion. Choice (A) is incor-rect because this is the definition of an isolated system. Choice (D) describes an open system,so it is also incorrect. Again, choice (B) is correct.

74. DThe excess pressure that affects a submerged object is hydrostatic gauge pressure, �gh,

where � is the density of the fluid. At a depth of 2 m, in a fluid with a density of 1.2 g/cm3, thehydrostatic pressure is:

�P � �gh � (1200 )(10 m/s2)(2 m) � 24000 Pa,

which is close to answer choice (D).

kg�m3

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75. ASolubility is affected by a number of factors, including temperature, the solvent, the addition

of other substances and in the gas phase, pressure. Therefore, only choice (A) does not affectsolubility. Think of a glass of water as the solvent and table salt as the substance being dissolved.If you heat the water, more salt will dissolve, so choice (C) must be incorrect. If you used a dif-ferent solvent with a lower polarity than water, less salt would dissolve, so choice (B) must beincorrect. If you add another solute, the solubility of the salt will change, so choice (D) must beincorrect. However, once the solution is saturated with salt, adding more salt will not change itssolubility, so (A) is the correct answer.

76. DIt might be easiest to handle this question by using conservation of energy. The initial energy

is all kinetic, and the final energy is all potential:

KEi � PEf

�12�(100)(v)2 � (100)(10)(20)

v2 � 400 → v � 20 m/s.

The correct answer is choice (D).

77. DIce melts at 0ºC. This means that as it is heated, it undergoes a phase transition. During the

phase transition, the temperature of a substance does not change; eliminate (C). Since thechemist only heats the ice for a moment, not all of the ice can melt. Therefore, you can elimi-nate choices (A) and (B). Hence, the answer is (D).

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21

VERBAL REASONING ANSWER KEY

78. B

79. B

80. D

81. B

82. D

83. B

84. B

85. C

86. A

87. A

88. C

89. A

90. C

91. B

92. B

93. B

94. C

95. A

96. D

97. C

98. A

99. A

100. A

101. C

102. C

103. A

104. C

105. A

106. C

107. B

108. D

109. A

110. D

111. A

112. C

113. C

114. A

115. B

116. D

117. C

118. B

119. B

120. D

121. A

122. B

123. B

124. D

125. D

126. C

127. D

128. A

129. C

130. A

131. C

132. C

133. B

134. A

135. D

136. B

137. A


Topic and Scope: The ocean’s thermohaline circulation and how disruptions in the circula-tion patterns might cause dramatic climate changes

Paragraph 1 explains the importance of the ocean in climate patterns and introduces the con-cept of thermohaline circulation.

Paragraph 2 describes the factors involved in the sinking of the water.

Paragraph 3 discusses some possible environmental changes that could alter the delicate bal-ance that keeps the thermohaline circulation going.

78. BScan for the phrase and study its context to answer this logic question. Don’t get bogged

down in the science. At the beginning of paragraph 2, the author describes how prevailing windpatterns increase evaporation rates of surface waters in the North Atlantic. Note that the ques-tion proposes a decrease in prevailing winds in the Northern Hemisphere. To answer this ques-tion you need to apply this information to the answer set. Choice (A) restates the informationcontained in the passage while (B) contradicts it. It is logical that if the winds decrease then theevaporation rate will also decrease. This makes (B) the correct answer. (C) is incorrect becausea change in the winds would cause some kind of change in evaporation rates. (D) refers to con-tinental runoff and precipitation which are mentioned in the next sentence but not relevant to thequestion. These factors do not relate to winds.

79. BReread the section around line 55 to prepare to answer this question. The question stem asks

you why the author chose to provide an example of a change in the salinity of part of the ocean.You know that the passage’s purpose is to explain the creation of deep-water and stress itsimportance in global climate. Choice (B) addresses the passage’s purpose and allows the authorto stress the fragility of thermohaline circulation. (A) is too general and strays from the pas-sage’s purpose. (C) is a faulty use of detail. Continental run-off decreases the salinity of theocean by adding freshwater. (D) is a Distortion since a drop in salinity would occur in the ocean,not in the atmosphere.

80. DTo tackle this question, evaluate how the author views climate change. In both paragraphs 1

and 3 the author mentions how a disruption in the thermohaline circulation could have a detri-mental effect on climate. Choice (D) paraphrases the main purpose of the passage and is the cor-rect answer. (A) is a faulty use of a detail. The author mentions water cascading off thecontinental shelf to illustrate how deep water is created. (B) is a detail contained within the pas-sage and may look like a tempting answer. Refer to the question stem, which asks what scientificanalysis would be most relevant for climate change. You are looking for a big picture answer. Pre-cipitation totals may be interesting but won’t tell scientists much by themselves. (C) is way out-side the scope of the passage. Human evolution has not even been mentioned in passing.

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81. BThis is a detail question. Identify and reread the section of the passage concerning the den-

sity of North Atlantic waters and remember that you want the answer that the passage contra-dicts or does not support. Density is a function of salinity and temperature. Paragraph 2 statesthat atmospheric absorption of heat, choice (C), increases density. Prevailing wind patterns (A)increase evaporation rates which increase density. Continental runoff (D) affects water densityby lowering salinity. The only answer that is not mentioned is (B), which is a Distortion of adetail mentioned in paragraph 3. Water in the North Atlantic doesn’t receive warm water fromthe North. In any case, any water coming from the North would be coming from the North Poleand be cold.

82. DThis inference question hits you with a lot of details so remember not to be intimidated by

the answer choices. Choice (A) is too extreme. The author stresses the importance of the ther-mohaline circulation but discusses both the North Atlantic and the Southern Ocean. The authorwould not endorse research focused on only one part of the ocean. (B) is a Distortion of a detailin the passage. The last sentences of paragraph 2 discuss the thermocline and how tidal actioncause the mixing of oceanic waters. (B) contradicts the information contained in the passage.(C) is opposite; while the passage doesn’t provide how disruption of the thermohaline circula-tion would change climate, the author makes it clear that it would alter world climate. (D) is thecorrect answer. It touches on the passage’s purpose, which is to discuss deep-water formationand how its interruption could alter climate.

83. BStudy the context of the section mentioned in the question. This sentence appears in para-

graph 2. The scope of this paragraph concerns the factors involved in creating the ocean’s dense,deep water, and this question asks you to evaluate why the author mentions the Atlantic’s salin-ity. Notice that the author uses a superlative to describe the Atlantic’s salinity. It’s the saltiest ofall high-latitude waters. Choice (B) is correct since the author’s use of a superlative reinforcesthe importance of salinity, and the passage describes how salinity and temperature cause waterto sink. (A) is FUD, misusing a detail found near line 21. The atmosphere’s absorption of heatdecreases the temperature of the water and does not relate to its salinity. (C) is never mentionedin the passage. While the passage mentions wind patterns and evaporation, (D) states that theauthor mentions salinity in order to discuss the factors involved in wind patterns and evapora-tion rates. That’s wrong and does not address the question.


Topic and Scope: The need to implement national standards in teaching and the means to do so.

Paragraph 1 discusses the need for guidelines and the difference between setting them andimplementing them. Paragraphs 2 and 3 discuss why schools can’t currently implement stan-dards. Paragraphs 4, 5 and 6 list the author’s recommendations: paragraph 4 teacher education,paragraph 5 teacher salaries, and paragraph 6 continuing education.

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84. BAlthough choice (D) may seem tempting, it is in fact a Distortion of the recommendation

made in the final paragraph; the purpose of the comparison to doctors and lawyers is not to showthat it is important for teachers to remain current. The author does not address anywhere in thepassage whether teachers should be accorded the same respect as doctors or lawyers (C) orwhether they should all be on the same pay scale (A). Therefore, (B) is the correct answer.

85. CThe author’s assertion throughout the passage—in fact, the main idea in this passage—is that

national science standards are useful only when teachers have the strong content knowledge andnecessary supplies to use them effectively. Choice (C) is the only answer here that works. While(A) and (B) are probably true, the fact that students did better on their tests strengthens neitherof these assertions. There is no evidence for (D) in the passage.

86. AIt would be a stretch to assume that older, more experienced teachers would be an obstacle

to instituting national standards, so choice (A) is correct here. All of the other choices involveissues that are specifically mentioned as impediments to the success of national standards—teachers avoiding topics out of ignorance, lack of proper equipment, and not holding teachersresponsible for their continuing education.

87. AThe author discusses how important continuing education is for teachers and how these

teachers need to be held accountable for their knowledge base. There is no evidence in the pas-sage to suggest that graduate schools can force higher payscales for teachers (B) or that they canimprove equipment (D). (C) would be correct only if those with science backgrounds are choos-ing to enter other fields despite the fact that they have been enrolled in teacher graduate schools.

88. CAccording to the passage, national standards can do more harm than good if they require

teachers to teach things that either the teachers don’t know or for which there are not adequateclassroom resources. Hence, choice (C) is correct. Nowhere in the passage is it suggested thatteachers can determine their own curriculum regardless of national standards (D), nor does theterm “national standards” imply that the standards would not be implemented across the wholecountry (B).

89. AIn order for national standards to be implemented, teachers need training, maintains the author,

so choice (A) is correct. The author would certainly not support the immediate implementation ofnational standards (D), nor allowing teachers to omit teaching what they did not understand (C).In addition, it is probably not a reasonable assumption that just because teachers should beretrained like doctors and lawyers, they should get paid like doctors and lawyers as well (B).

90. CThe author would agree that what is tested has a lot to do with what is taught in the class-

room, and it is possible that the testing will drive the classroom instruction. Yet new tests can-not improve teaching unless teachers are in place who can live up to the new expectations—thus,choice (C) is correct. It does not stand to reason that the teachers should take the same tests asthe students (D), even if the author would probably agree that some sort of teacher assessmentwould be important.

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Topic and Scope: Folk art and fine art; specifically, what is similar and what is different inpractice and appreciation of these two types of art.

Paragraph 1 introduces the notions “fine art” and “folk art,” and suggests that the distinctionbe recognized as a convention of language, rather than as intrinsic to each of two essentially dis-tinct forms of art.

Paragraph 2 asserts that fine artists had a tendency, at least at first, to consider more tradi-tional art as inferior to their own creations.

Paragraph 3 defines primitive art, affirms it gained perceived value sooner than naive art, andexpresses the author’s hypothesis regarding why this was so (primitive art was more easily val-ued, for its historical significance).

Paragraph 4 defines “folk art” and introduces the notion that one of the key principles forappreciation of folk art—that appropriate standards are not the same as those appropriate forevaluation of fine art—is also applied to the appreciation of modern art.

Paragraph 5 introduces the notion that a characteristic normally identifying folk art can alsobe a characteristic of fine art—different styles exist, and indicates how that characteristic istreated differently by those who are oriented toward the fine arts than by those who are orientedtoward folk arts.

Paragraph 6 questions the importance of a distinction proposed by some: that folk art maybe distinguished from fine art in that it has a practical purpose.

Paragraph 7 explains how fine art and folk art differ, in terms of their appeal to the viewer.

91. BParagraph 4 defines folk art as the art of a people, springing from their customary way of

life. Choice (B) is correct because it is a paraphrase of this definition, and thus provides a basisto infer that works of folk art would be more likely than works of fine art to meet this criteria.(C) is opposite to what would be correct. (D) is an incorrect inference. The passage does not pro-vide a basis to infer that several works of folk art would be likely to be of less aesthetic valuethan several works of fine art. (A) is a Distortion. Although the passage indicates that that folkart was disparaged by early fine artists, it does not indicate that folk art is likely to be misun-derstood by viewers today. (A) is also inconsistent with the statement in the last paragraph thatfolk art tends to have direct appeal and simple charm.

92. BThe first cave paintings can be defined as primitive art (statement II) based on the definition

in the first sentence of paragraph 3. This sentence also indicates that primitive art is a type of folkart (statement I) and that it contrasts with naive art. Thus, the passage indicates that cave paint-ings cannot be considered both primitive and naive, and choice (B), I and II only, is correct.

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93. BThe passage does not explicitly state why the exhibition at the National Gallery was greeted

with disapproval, questioning, and cynicism. However, the reason can be inferred from the con-text of the statement. The earlier part of the paragraph refers to “maturation,” and the MarkTwain quote is an ironic statement. Clearly, the change in Twain’s assessment of his father wasnot actually due to what his father had learned in seven years, but due to Twain’s maturation.Choices (A) and (B) are possible reasons why those who attended the exhibit were notimpressed. Which is more likely in this context? Is it the curators of the exhibit or the attendeeswho are most likely to have lacked a “mature” sense of the artistic value of folk art? If it wasprimarily the curators, then it is necessary to infer that the public would have had the capacityto appreciate the artistic work of particularly valuable folk art (had it been displayed), which isnot supported by anything in the passage. Those who attended the exhibit are analogous to theyoung Mark Twain; it is their assessment that the passage refers to. Thus, the best inference ischoice (B). (C) is not supported by the passage. If fine artists had exercised undue influence onthe gallery, it would be more likely that the exhibit would not have been shown at all. (D) is aDistortion of the passage. Although the sentence in question refers to “folk painting,” it does notindicate that the paintings were on canvas rather than, for example, on furniture.

94. CThis question is best answered by an elimination strategy. Choice (A) cannot be correct

because naive art and modern art are defined in a mutually exclusive way in lines 32–35 of para-graph 4. (B) cannot be correct because of how primitive art is defined in the first sentence ofparagraph 3. (D) cannot be correct because of how folk art is defined in the first sentence ofparagraph 4. This leaves (C). The passage indicates that the fine arts have “conventions” and thatmodern artists choose to violate “contemporary rules” of the fine arts. This implies that the rulesof fine art are not static. Thus, it is not inconsistent with the passage to conclude that the rulesof modern art could evolve, and a piece of modern art that violates today’s rules could beaccepted as fine art in fifty years.

95. ABy contrasting the close relationship of folk art with the traditional culture of an area to the

close association of the fine arts with sophisticated urban centers, the passage focuses on thesocial context (statement I) in which the artist’s skill was developed. The intention of the artist(statement II) is not relevant to distinguishing folk art from fine art. Rather, the intention of theartist is related to distinguishing modern art from folk art and modern art from the rest of fineart. While the country of origin of the masks that served as a model for Picasso’s Les Demoi-selles d’Avignon is mentioned in the paragraph, country of origin is not focused on as a way ofdistinguishing fine art from folk art.

96. DSince the author mentions that the a distinction between “fine art” and “folk art” postdated

the first cave paintings and that primitive art is part of the history of art, the author can beinferred to agree that folk art and fine art share a common origin, choice (D). The passage men-tions both similarities and differences between folk art and fine art, but provides no support foran overall judgment that they are more different than they are similar (A). (B) is a Distortion ofthe passage. While the second paragraph of the passage makes reference to the maturation offine art, it also likens the early stages of fine art to adolescence. (C) is inconsistent with the pas-sage’s focus on specific criteria which distinguish folk art and fine art.

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97. C Since the passage indicates that folk art “springs naturally from the customary way of life

and beliefs of the people of a region,” artists living in the same area would be likely to be stylis-tically similar. In contrast, the passage indicates that styles of fine art are likely to diffuse toother areas. Thus, choice (C) is correct. While the tendency of younger artists to learn fromskilled adults (A) is certainly characteristic of folk art, the passage provides no basis for thisbeing less true in the fine arts. (B) is inconsistent with the statement in the passage that folk artsprings naturally from customary way of life and beliefs. (D) is only partly true, since “sophis-ticated techniques” are not associated with folk art.

98. AThe referenced statement follows, and questions, a sentence in which the fact that folk art is

often executed as a decoration to objects of practical value is presumed to represent a funda-mental distinguishing characteristic of folk art. Since the referenced statement challenges thispresumption on the basis of “economic constraints,” the implication is that economic constraintsare a consideration more external to the artist than his other artistic decisions. Why would thisbe so? Choice (A) is most directly related to this point. It focuses directly on how economicsmay force the artist to choose objects that also have practical value. The remaining choices relatein some way to the economics of artistic life, but are not as directly related to the artist’s choiceof working surface.


Topic and Scope: The passage proposes that there is a problem with the plurality system ofvoting in multi-candidate contests and describes two alternative systems.

Paragraph 1 suggests that the type of system used for casting and tabulating votes can affectthe outcome of an election and influence what candidates will run for office.

Paragraph 2 describes the plurality system of voting.

Paragraph 3 describes the Borda count, an alternative to the plurality system of voting.

Paragraph 4 describes approval voting, another alternative to the plurality system.

Paragraph 5 expresses both the author’s hope that an alternative system might be adoptedand also the author’s belief that this would be beneficial.

99. AThe argument assumes that voters considered Taft and Roosevelt (former and incumbent

Republican presidents) similar, and that Taft voters would not have switched to Wilson and Roo-sevelt voters would not have switched to Wilson. However, the passage does not state that bothwere Republicans at the time they ran for election in 1912. Furthermore, “similarity” betweencandidates can be assessed on many different parameters. Absent evidence that actual Taft orRoosevelt voters would have voted (in a two candidate race) for the other of the two, rather thanfor Wilson (statement III), must be considered an assumption. Statement I is not an assumptionbecause it is not clear from the passage what role party affiliation played in the election, partic-ularly since the passage does not identify Roosevelt’s affiliation at the time of the 1912 election.Statement II contradicts the author’s explicit statement, i.e., if Taft and Roosevelt “split a major-ity of the popular vote,” by definition they collectively received more votes than Wilson, whomust have received a minority of the popular vote. The correct answer is choice (A).

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100. AThe passage mentions the 1912 Presidential election as an example of choice (A). There is

no example of an election in which a different voting system produced a different outcome (B),a sophisticated voter voted for his second choice (C), or a change in voting system had an impacton what candidates ran for office or what viewpoints they represented (D).

101. CIn using the 1912 election as an example of the points made in the preceding two sentences,

the author implies that Roosevelt and Taft were similar to each other and that they split the votethat either one, if opposed only by Wilson, would have received. The passage explicitly statesthat Roosevelt and Taft split “a majority of the popular vote.” Thus, under the premises of thepassage Roosevelt would have received Taft’s votes, had Taft not run, and this would have con-stituted a majority of the votes cast. Thus, statement I is implied. Had a Borda count been used,the author’s premises indicate that Wilson would have been third choice of both Roosevelt andTaft voters, while Roosevelt and Taft would have been ranked second choice by each other’s vot-ers. If there were only the three candidates, Roosevelt and Taft would have received two pointsfrom each voter who ranked them first and one point from those who ranked them second. Wil-son would have received two points from each voter who ranked him first, but would have hadno voters who ranked him second. The implication is that Wilson would have lost under a Bordacount. Statement II is not implied because the passage does not take any position on whether ornot the voters in the 1912 election were sophisticated; the passage does not provide a basis toinfer that voters viewed either Roosevelt or Taft as “unable to win.”

102. CTo weaken an argument, a piece of information often attacks an assumption made in the

argument. In this case, the argument assumes that Roosevelt and Taft were considered similar,and would thus have been ranked second choice by each other’s voters. Evidence that Rooseveltwas more similar to Wilson than to Taft would undermine this assumption. Thus, choice (A)weakens the argument by providing evidence of a measure by which Roosevelt could be con-sidered more like Wilson than like Taft. (B) weakens the argument by providing evidence of away in which Roosevelt and Taft could be considered quite different from one another. (D) pro-vides evidence to infer that Wilson might have been able to produce a majority for the Demo-crat party in 1912 if opposed only by either Roosevelt or Taft, since Wilson was subsequentlyable to do so in 1916, when opposed by only one other major candidate. (C) is outside the scopeof the argument and neither strengthens nor weakens it.

103. AThe passage uses the phrase “those in serious contention” to make a distinction among can-

didates in an election. The passage as a whole focuses on voting systems, rather than on char-acteristics of candidates. The only answer that makes sense in the context of the passage is thatthe distinction is between candidates who are more likely to win (“those in serious contention”)and those less likely to win (not in serious contention). While, in a different context, the phrasecould reflect the attitude, energy, or moral commitment of the candidate, none of these alterna-tive meanings makes sense in the context of the distinctions on which the passage focuses.

104. CSince approval voting provides the opportunity to vote for more than one candidate, it is rea-

sonable to assume that those who favor the system in a particular election probably would berelatively happy with more than one candidate winning the election. Conversely, those who haveone strongly preferred candidate who is markedly different from the rest might hope that plu-

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rality voting would give them an advantage, since the similar candidates might “split” the votesof voters who were opposed to their preferred candidate. Thus, the correct inference is choice(C). Since we don’t know what system the students are already used to, there is no basis toassume that preference for plurality reflects a resistance to change (A). There is nothing in thepassage to indicate that students who are more mathematically inclined (B) would prefer theplurality system of voting.


Topic and Scope: The New Historicism approach to history, followed by a complaint aboutthe way history constantly changes its approaches.

Paragraph 1 introduces New Historicism. According to the author, this historical approach usesseveral different academic disciplines to create something called “thick description”, where schol-ars study entire cultures, and compare the works of major authors to nearly unknown writers.

Paragraph 2 tells us about some of the positive results of the New Historicism, as well assome of the initial criticisms.

Paragraph 3 tells us about the main backlash against the New Historicists. The New His-toricists, critics claim, have overemphasized the cultural unity of past cultures, and thus overem-phasized the importance.

Paragraph 4 seems to go a little off track, as the author criticizes the tendency of historiansto switch their approaches to writing history every few years ago. The result, says the author, isthat this casts doubt on earlier historical approaches—if we got rid of them so quickly, how goodwere they?

105. AWe’re told that New Historicism was first used by scholars studying Renaissance Italy, but

it doesn’t follow that New Historicism necessarily clarified any problems in the field. Choice (C)is a major trap answer, since it does appear later in the passage. However, (C) is what some crit-ics say about New Historicism. We don’t know if this is an actual fault, or even if our authoragrees with that point. Since the passage later criticizes the critics (confusing, isn’t it?), wemight even guess that the author is not in complete agreement with New Historicism’s critics.(D) appears to be true of historians in general, not simply the New Historicists. (A), the correctanswer, can be found in the second paragraph, where we’re told that the work of the New His-toricists heightened both popular and academic interest.

106. CChoice (A) certainly sounds good, since the author complains that only history appears to

change its techniques so frequently. But the main argument here isn’t so much that historychanges techniques more frequently than other disciplines, but that history needs to change atall. The author seems to be arguing for a complete halt, or at least a slowdown, in the swiftchange of historical approaches. (C), which argues that these changes are necessary in order toform historical judgements, is the best answer to that argument. Of the other choices, (B) istempting, but the author is complaining about historians in general, not merely the New His-toricists, when she complains that historians no longer make historical judgements. (D) is off themark. The issue isn’t so much whether New Historicism is going strong or not, but why histori-ans feel the need to constantly change their academic approaches.

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107. BThe author tells us that the New Historicists study marginalized and understudied groups,

including pornographers, so (D) is clearly out. The New Historicists also try to study works inthe context of other works from the era, so (A) and (C), where two contemporary works are com-pared, is also out. But because the New Historicists focus on marginalized groups, they’re prob-ably not too likely to study the ruling group of any area.

108. DBecause history isn’t an actual science, it lacks a single disciplinary approach, the point

expanded in the rest of the paragraph. The author never contrasts history and hard sciences suchas physics directly, so choice (A) is out. The author also never suggests that economists, anthro-pologists and historians can even use the same methods used in the hard sciences, so she can’tbe criticizing scholars for not using these methods. The same thing goes for (C), especially sincethe paragraph never urges any of these scholars to try a more straightforward, objectiveapproach.

109. AThe passage tells us that the new scholars (not the New Historicists, but the scholars who

followed them) study the same groups as their predecessors. Since the New Historicists studiedmarginalized groups, we can infer that the new scholars are probably studying them too. Sostatement I is correct, and we can toss out choice (C). Statement II is definitely tricky, since thepassage notes that, like the New Historicists, these new scholars study the same marginalizedgroups. But that’s not necessarily the same flaw. The flaw for the New Historicists was that theyassumed all cultures had some sort of cohesion; the flaw of the backlash scholars is that theyaren’t really studying anything new. As far as statement III goes, we don’t know whether or notthe backlash provided any new historical insights, so we can’t infer that, either.

110. D The answer can’t be choice (B), since the passage never claims that the New Historicists

unfairly concealed anything, let alone the uniqueness of Dante, only that their approach oftenclouded the uniqueness of canonical texts—not quite the same thing. The author never says, asin (A), that the backlash against the New Historicism lacks all claim to originality, only that it’sflawed. (C) wouldn’t necessarily be inconsistent with the author’s admiration. Since the author,however, seems to be against the need for this historical backlash against the New Historicists,admiring a book from this backlash seems pretty inconsistent.


Topic and Scope: How animals “map” their environments.

Paragraph 1 discusses the basic dichotomy between moving and unmoving objects. Para-graph 2 discusses why animals can’t use their own body movements to “map”. Paragraph 3 givesan example, a desert ant.

Paragraph 4 Explains why that any isn’t really “mapping”. Paragraph 5 gives other examplesof animals that store food to relocate it later. Paragraph 6 describes experiments being done tolearn more about animal mapping.

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111. AThe passage clearly states that navigational errors would multiply rapidly in an kinesthetic

system because small movements would throw the system off; this supports choice (A). Thereis no evidence to support (B) or (C). (D) is contradicted by other information in the passage.

112. CThis question requires you to consider what events are more likely or easier than others.

There is no reason to suspect that it is easier to remember many objects than one object, andcommon sense suggests that the reverse is likely true; this goes against choices (A) and (B).Also, common sense suggests that one object could more easily be moved than several objectscould, contradicting (D). (C) is the correct answer.

113. CIn an experiment, you must alter variables one at a time to see what each one’s individual

contribution is; this is choice (C). If you did them all at once, that would actually be a new exper-iment, lending no support to (A). There is no evidence for (B) and (D).

114. AIf in every instance animals operate similarly, it is reasonable to assume that something

about animals requires them to act that way. This would eliminate choice (C). (D) is contradictedby information in the passage. There is no reason to think that the muscles, themselves, performany computations, so (B) is out. (A) is the best choice.

115. BChoice (B) is the best answer because it takes into account the fact that animals need to use

these cues from a distance. (C) is directly contradicted by information in the passage. There isno evidence for (A) and (D).

116. DIf we find that every example of something operates similarly, then it is reasonable to assume

that a newly found example will operate in the same way. This would eliminate choice (A). So,we would assume that this animal will find food the way other animals do, by a configurationof landmarks, (D).


Topic and Scope: The benefits of fairy tales in comparison to modern “safe” children’s stories

Paragraph 1 introduces the underlying psychoanalytic aspects present in the simplistic plotsof fairy tales, and presents the thesis that such polemical tales actually are more effective forhelping children to deal with basic issues of life and death than are modern stories.

Paragraph 2 develops this thesis by discussing the polarization of the existential dilemma asit is present in fairy tales. By presenting good and evil so clearly, fairy tales allow children tobetter comprehend the differences than if the issues were more ambiguous or “true to life.”

Paragraph 3 continues this discussion and addresses parent’s concerns about children’s fearsand emotions, claiming that fairy tales provide children with an outlet for dealing with their anx-ieties and even gives them solutions to such struggles in ways that children can understand.

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117. CThe final sentences in paragraph 3 attest to the fact that the fairy tale “addresses itself

directly to anxieties and fears,” just like choice (C). (A) is the opposite of the third sentence inparagraph 2; (B) is the opposite of the last few sentences of paragraph 1, which claim that “safe”stories do not discuss death, aging, limits of existence, or the wish for eternal life; and (D) is aDistortion of part of paragraph 3 where parents’ fears, and their relation to different types of sto-ries, are discussed.

118. BParagraph 1 lays out the argument that fairy tales present children with conflicts between

good and evil, and that Freud’s recommendation to struggle against odds to find meaning in lifeis encapsulated by these types of stories. Choice (A) is outside the scope, since the validity ofpsychoanalytic examination of literature is not at issue. (C) is a Distortion of the differencesbetween modern stories and fairy tales, as described in paragraphs 1 and 2. (D) contradicts thefirst sentence of the paragraph, which clearly implies that the “dark side” of humanity does existand should be acknowledged.

119. BThe author professes that children gain comfort from the typical fairy tale endings, but that

they can also discern that “living happily ever after” should not be taken literally, suggesting thatchildren are able to discriminate between fiction and reality in the meaning of the fairy tale (B).Choice (A) is a Distortion of the final sentence of the passage, and a FUD, misusing the state-ment made in lines 7–8 that “struggling against what seem like overwhelming odds” can givemeaning to life. (D) is too extreme, with the word “always,” and is a Distortion of the first sen-tence of Paragraph 3. (C) is a Distortion of the middle part of the final paragraph, which dis-cusses parents’ discomfort with their children’s fears or negative emotions.

120. DMoral lessons and values are never mentioned directly by the author as a benefit of the polar-

ities in fairy tale plots. Choices (A), (B), and (C) are all discussed in paragraph 2 as support ofthe main idea expressed in the first sentence of the paragraph that fairy tales “state an existen-tial dilemma briefly and pointedly.”

121. ABy stating that children need the symbolic suggestions of fairy tales in order to understand

life and death issues and mature properly, the author suggests that parents may not be able toprovide this type of instruction to their own children. Choice (B) does not give any indication asto whether or not the fictional representations lead to a deeper understanding of good and bad.(C) is outside the scope, since children’s own instinctive responses are never mentioned. (D) isopposite and would undermine the author’s assertion, since it would in fact disprove the centralthesis of the passage.

122. BThe central thesis of the passage is that traditional fairy tales are more beneficial to chil-

dren’s growth and development than modern stories that attempt to provide “safe” plots andcharacters, implying that such modernization is in fact detrimental to children’s maturation.Choices (A) and (D) are outside the scope, since the passage never discusses the intentions ofthe authors of children’s stories, nor the utopic lives presented in many fairy tales and the effectsof such perfection on child readers. (C) is the opposite of the author’s belief that the inclusionof evil characters in fairy tales is a useful and necessary element of fairy tales.

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123. B “Safe” stories paint an overly rosy picture of life, and choice (B), a paraphrase of the author’s

point, smacks the nail on the head. The other choices all sound positive, whereas the author saysnothing good about modern stories. “Safe” stories don’t deal with problems (A) and they fail todepict struggles between good and evil (C). (D) describes a theme that, the author suggests, ischaracteristic of fairy tales.

124. DThis is a common type of Evaluation question. Why does the author mention Freud and his

“prescription”? The author is using the reference to state an idea (“major struggles are funda-mental to living”) which is central to his own argument about fairy tales. (C) may have beentempting, but Freud’s prescription is not an analogy; it’s simply a view that accords with theauthor’s. If you were hesitant about (C), that should have been a signal that something was notquite right. By contrast, (D) is clear-cut. (A) is inconsistent with the passage; the author’s pur-pose is to counterpose Freud’s idea to the dominant cultural attitude. (B), like (C), soundsmurky—it just doesn’t fit. No contradiction is being exposed.


Topic and Scope: Threats to wetlands, current protection methods, and problems with theireffectiveness.

Paragraph 1 relates the history of wetlands conversion in the mid-19th and mid-20th cen-turies.

Paragraph 2 provides a detailed list of wetlands functions and benefits which have been lost,and which could have been achieved with less wetland loss through “regional planning, strongerregulation, and greater public understanding of wetland values.”

Paragraph 3 discusses the current obstacles to wetland protection laws and programs, includ-ing both governmental and private efforts to prevent the continued erosion and loss of these envi-ronmentally important areas.

Paragraph 4 explains why the current approach, using stringent permit guidelines that do notdistinguish by wetland types or values, is ineffective and offers an alternative strategy, based onregional management and prioritization of the most severely-threatened areas. The authorimplies that this second method would be more effective.

125. DThe third sentence of paragraph 2 declares that budget concerns cause limits on the acquisi-

tion of new wetlands for preservation, and the Emergency Wetlands Resources Acts is men-tioned as an example of a stop-gap acquisition measure that is constrained by the small amountof federal funds allocated for wetlands protection. Choice (B) is opposite, since the author’s con-tention is that federal funds are insufficient for wetlands preservation. (C) is outside the scope,and (B) is a FUD from the second sentence of paragraph 2.

126. CChoices (A) and (D) are directly mentioned in Paragraph 1, in the second and last sentences

of the opening paragraph. (B) is implied in the final sentence of paragraph 2, which contendsthat “…public health benefits have been lost to agricultural forestry and development enterprises

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of all kinds.” (C) is a Distortion of this same sentence since no mention is made of either anincrease in rezoning of wetlands nor of the impact of industry on the conversion of wetlands.

127. DThe author describes only two strategies for protecting wetlands, the “present approach” and

an “alternative strategy,” in the first and third sentences of paragraph 4. For this reason, choice(A) is a Distortion of the information in these sentences. (B) is a Distortion of the author’s crit-icism of the present approach, and FUD of the “stringent permit guidelines” mentioned in thefirst sentence of the paragraph. (C) is a Distortion of the phrase “Cooperating federal, state, andlocal interests” and also uses extreme language by claiming that such collaboration is “the onlyway” to reach the goal of wetlands preservation.

128. AThe author suggests that federal, state, and local interests need to be considered, and that

although no “general federal authority” currently exists for the protection of wetlands systems,“several authorities” do exist which could develop this conservation plan (last three sentences ofparagraph 4), which infers that these organizations would cooperate for the purpose of wetlandsprotection. Choices (B) and (C) both contain extreme language (“never” and “only”). (D) is half-right and half-wrong, since the author never mentions that any areas of wetlands should beexempt from protective regulations.

129. CStatement I is implied in the middle of paragraph 3, with the phrase “qualitatively important

but quantitatively limited,” which suggests that an increase in wetlands purchases by privategroups and trusts would benefit wetlands protection. Statement II is implied in the final sentenceof the paragraph, which declares that “the marketplace does not generally recognize the publicbenefits of wetlands,” indicating that increased recognition of these benefits might bring abouta change in business decisions and behaviors. Statement III is a Distortion of the first sentenceof paragraph 4.

130. AGovernment and private acquisitions are mentioned in paragraph 3. The author says the first

“will always be limited by severe budget constraints” and will save only “a small percentage”of remaining wetlands. The second offers “quantitatively limited” protection. Protection by pri-vate developers, in paragraph 4, is also treated pessimistically—only “some” developers havetried to combine protection and profit, and the marketplace generally does not favor such anapproach. The most straightforwardly pessimistic choice (A) is the one you want. (B) makes adistinction among the three strategies that the passage does not support. (C) makes a recom-mendation—further study—that doesn’t appear in the passage. (D) sounds okay until the end—the author never suggests that any policy will actually reverse the trend. All these choices arestraightforwardly wrong and your only problem should be with their wordiness. Try not to getbogged down in minutiae, and to go straight to the key idea—like “reverse” in (D)—that makeseach one wrong.

131. CJust after mentioning “severe budget constraints” that limit government acquisition, the

author states that the Emergency Wetlands Resources Act (EWRA) allocates “only $40 millionper year” in federal funding for wetland purchases. So, the EWRA is an example illustrating thepreceding generalization, which is rephrased in choice (C). (A) incorrectly focuses on the needfor legislation to save American wetland areas. Although the EWRA is undeniably a piece of

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legislation, it is clearly meant to illustrate the problem of inadequate federal funding, not thelack of helpful environmental legislation per se. (B) is way off the mark; the author neveraccuses the federal government of interfering with state government efforts. Choice (D) is sim-ilarly misleading. Although paragraph 4 says the marketplace generally fails to recognize thebenefits of wetlands, the author never implies that there’s a widespread perception that wetlandsare not worth saving, or that the EWRA would be evidence of it.


Topic and Scope: Binary star systems and the discovery, composition and development ofsymbiotic stars.

Paragraph 1 introduces a scientific curiosity—binary star systems whose radiation emissiondoes not fit that pattern of that of other binary stars.

Paragraph 2 provides a potential explanation for this unusual radiation pattern, and supportsthis explanation with scientific evidence about the discovery and composition of symbiotic starpairs.

Paragraph 3 continues to describe the structure of symbiotic stars, and concludes with thepossibility that symbiotic stars may mark a brief phase in the evolution of certain binary sys-tems.

Paragraph 4 presents some theories about the development of symbiotic stars, and concludeswith suggestions about the evolutionary course of symbiotic pairs, based on conjecture aboutother binary systems.

132. CParagraph 2 clearly states that photographic plates only revealed the giant star in a symbi-

otic pair, but that newer satellite-borne instruments can detect ultraviolet radiation that “cannotbe detected by instruments on the ground.” Without these types of instruments, scientists wereunable to study symbiotic stars, since they didn’t really understand what the star formationswere. Choice (A) is FUD of information in paragraph 4, which tells us that “the phase must beextremely brief, perhaps as short as a million years.” (B) is a Distortion of the information men-tioned in paragraph 2. (D) is FUD of the sentence in paragraph 4 that tells us there is a “com-paratively small number of known symbiotics in our galaxy.”

133. BThe final sentence of the passage contains the reference to red giants and Mira variables, and

finishes with a conclusion about the mass of “the original cloud from which a symbiotic systemis formed.” Choice (A) is FUD of the last sentence in paragraph 2—although symbiotic stars doexist outside of our galaxy, this question is only concerned with a specific reference found inparagraph 4, so anything from other parts of the passage is incorrect, even outside the scope, forthis particular question. (C) is outside the scope because there is no size comparison made inparagraph 4. And (D) is a Distortion of the previous sentence in paragraph 4, which declares thatthe “evolutionary course of a binary system is predetermined”—this sentence refers to binarysystems in general, and not just to giant stars, as the answer choice does.

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134. AParagraph 2 described the structure of symbiotic stars, which together “travel around a com-

mon center,” not around each other. Choice (B) is stated at the end of paragraph 1; (C) is statedin the second sentence of paragraph 3; and (D) is stated at the first sentence of paragraph 4.

135. DParagraph 3 ends with a concluding sentence indicating that “Symbiotic stars may therefore

represent a transitory phase in the evolution of certain types of binary systems.” Choice (A) isoutside the scope, since the usefulness of research methods is never discussed. (B) is outside thescope as well, for dangers of radiation are not mentioned in the passage. (C) is a Distortion ofparts of paragraphs 3 and 4, and uses extreme language with the terms “all” and “must.”

136. BThe “recurrent outburst of a nova” are used as a comparison to the outbursts that occur with

symbiotic stars as a way of explaining a certain process observed in the latter that is similar toa well-documented and understood process that occurs with a nova. Choice (A) is outside thescope for this question, since the contrast made between symbiotic stars and other binary starsoccurs earlier in paragraphs 1 and 2. (C) is FUD, since the mention of the nova does not provethe existence of matter transfer, which is described in the final sentence of paragraph 3. (D) is aDistortion of the previous sentence in paragraph 3, which claims only that symbiotic stars emitflares, not that all binary systems do so.

137. AParagraph 4 tells us that the course of a binary system is “predetermined by the initial mass

and angular momentum of the gas cloud within which binary stars are born,” so that binary starsmust begin inside this cloud, which has both mass and motion. Choice (B) is a Distortion of thefinal sentence of paragraph 3, which mentions “transfer of matter from the larger partner to thesmaller.” (C) is a Distortion of the second sentence of paragraph 4, for this sentence actually isa hypothetical—“if all binaries of modest mass normally pass through a symbiotic phase in theirevolution…”—and the answer choice does not contain this all-important word “if” but insteadturns this hypothetical statement into a definitive fact. Likewise, (D) is a Distortion of the firstsentence of the passage, which tells us only that there are “several hundred million binary sys-tems estimated…and thus theoretically detectable on sky-survey photographs…”. Again, theanswer choice here is formulated as a statement of fact, whereas the original sentence in the pas-sage is much less certain.

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37

BIOLOGICAL SCIENCES ANSWER KEY

138. C

139. B

140. B

141. D

142. A

143. D

144. D

145. A

146. B

147. A

148. D

149. B

150. B

151. A

152. A

153. D

154. B

155. B

156. C

157. B

158. C

159. C

160. D

161. D

162. B

163. D

164. A

165. C

166. B

167. D

168. C

169. A

170. A

171. C

172. A

173. D

174. C

175. A

176. C

177. D

178. C

179. A

180. C

181. A

182. B

183. D

184. B

185. C

186. B

187. A

188. D

189. C

190. B

191. C

192. A

193. D

194. B

195. D

196. A

197. B

198. D

199. B

200. D

201. C

202. B

203. C

204. C

205. B

206. D

207. C

208. A

209. A

210. C

211. C

212. D

213. B

214. B


138. CTheoretically, medication that inhibits the production of gastric acid, choice (A), would not

allow for the proper environment for both the release of cobalamin from its complex with foodcomponents and the binding of cobalamin with R-protein. Remember that every enzyme hasoptimal conditions (pH, temperature) under which it operates. So choice (B) would also affectthe absorption of cobalamin since the pancreas is a major source of the bicarbonate that allowsfor an alkaline environment in the duodenum, without which the enzymatic dissociation of thecobalamin-R-binder complex would not be possible. Bile, however, is not a necessary compo-nent of this process, though it is indispensable in the absorption of lipids and other nutrients.Choice (D) would result in a person without an ileum, and the ileum is where the cobalamin-IFcomplex is actually absorbed. Recall that the small intestine is composed of three segments—first the duodenum, second is the jejunum, and third is the ileum, which then connects to the firstpart of the colon.

139. BA person given this supplement would only depend on the latter steps discussed in the first

paragraph of this passage; one would still require both an absorptive surface (ileum) with thespecific receptor and transcobalamin to transport the absorbed vitamin to the liver and red bloodcells. Thus, a person with a transcobalamin deficiency would still suffer from a lack of cobal-amin despite such a supplementation, and choice (B) is correct. Pernicious anemia is brieflymentioned at the end of the passage and is characterized by absent or deficient IF. So a supple-ment such as the proposed one would treat cobalamin deficiency due to this disorder; thus,choice (A) is incorrect. The same holds for choices (C) and (D), which would both affect earliersteps and which could both be bypassed by this supplement.

140. BRecall that myelin is a major factor in the speed of neuronal transmission. Myelin acts to

insulate the axon and allow increased speed of propagation of an action potential down a givenaxon. (It does not, however, affect the other major factor—synaptic transmission of a messagefrom neuron-to-neuron.) Also, recall that myelin is composed of one type of cell in the periph-eral nervous system—the Schwann cell, and a different cell type in the central nervous system—the oligodendrocyte. The question describes an experiment being done at a site of the peripheralnervous system, which is composed of all parts of the nervous system outside the brain andspinal cord proper. The passage states that cobalamin deficiency causes demyelination, so youcan assume that slower transmission in this case is due to a defect of the myelinating cells of theperipheral nervous system; choice (B), not choice (A), is correct. The nodes of Ranvier are theareas of the axon between the myelinated segments which are devoid of myelin; this unmyeli-nated space is obviously not a site of demyelination; thus choice (C) is incorrect. Lastly, the den-drites are the receivers of inter-neuronal transmission; there is no mention of a defect of synaptictransmission in cobalamin deficiency, and choice (D) is incorrect.

141. DControl of involuntary muscle groups falls within the autonomic division of the peripheral

nervous system, choice (B), as opposed to the somatic division, choice (D) which controls onlyvoluntary muscle action. Both the autonomic and somatic divisions are subsets of the peripheralnervous system, choice (A). The nervous system can also be divided into those cells that relatesensory information from the internal or external environment (afferent neurons) and those thatbring reflexive or processed information to the effectors (such as muscle) in response to the sen-

38

sory information gathered (efferent neurons). Think of the alphabet: “a before e” to help remem-ber this sequence. Communication to the sphincter muscles described in the question stem is viaefferent neurons by definition; thus, efferent neurons could be involved in this effect of cobal-amin deficiency.

142. ARecall that the purines are adenine (A) and guanine (G)—“PURe As Gold.” The transcrip-

tion process that will be most affected will be the one that must produce mRNA from the mostA’s and G’s, or the DNA segment with the most T’s and C’s, respectively. Thus, choice A is cor-rect. Choices B and D also have some T’s and C’s, but much fewer than the segment given aschoice A. Choice C is incorrect because DNA is made of the four bases A, T, C, G; uracil (U) isonly found in RNA.

143. DLipid digestion begins once food reaches the small intestine, choice (B); the stomach has no

role in this process. Bile produced by the liver, choice (A), is secreted into the first part of thesmall intestine (the duodenum) where it begins to emulsify and break down lipids into smaller,more soluble units. These emulsified units are then acted on by lipases from the pancreas, choice(C), which break down the lipids further into free fatty acids. Fatty acids can then be absorbedby the cells lining the small intestine.


144. DA sharp peak at 1735 cm–1 is indicative of a carbonyl, and a peak at 2800 cm–1 indicates a

C-H stretch. These peaks would be present both in isoamyl acetate and in acetic acid. However,acetic acid would not be present in the final solution because the student extracted the reactionmixture with sodium bicarbonate. Therefore, the compound giving rise to the carbonyl and C-Hstretches is most likely isoamyl acetate. The wide boiling point range and the presence of an O-H stretch (broad band at 3200 cm–1) indicate the presence of some impurity. The passage statesthat the student forgot to dry the compound with anyhdrous sodium sulfate. This suggests thatwater is present which agrees with the O-H stretch.

145. ALe Châtelier’s Principle states that if reactant is added, the reaction will be driven to prod-

uct. Similarly, if product is removed, the reaction will be driven to product. The boiling point ofisoamyl acetate is ~140°C according to the passage. The boiling point of water is 100°C. Forthis reason, if one product is to be removed by distillation, the product removed must be water.Removing water by another method, such as by adding a dehydrating material, will also work.Isoamyl acetate cannot be distilled from the solution because isopentyl alcohol and acetic acidwould distill with it. (i.e. replace “water” with “isopentyl alcohol and acetic acid)

146. BSodium bicarbonate is a base and will consequently deprotonate an acid. In this case, acetic

acid will be deprotonated because it is the only acid in solution. The resulting salt (acetate) iswater-soluble and so will be separated from the hydrophobic product. Addition of water would,in fact, drive the reaction toward formation of products. Isopentyl alcohol is completely used inthe reaction because it is the limiting reagent (remember acetic acid is used in excess!). Isoamylacetate is hydrophobic and will not dissolve in aqueous solution.

39

40

147. AThe carbonyl carbon is electrophilic (eliminate choice (B)). The electron-withdrawing

effects of the chloride ion make the carbonyl more electrophilic than acetic acid making it moresusceptible to nucleophilic attack by the alcohol. Esterification is an addition-elimination reac-tion so no carbocation is formed (eliminate choice (C)) and an SN1 reaction does not take place(eliminate choice (D)).

We recognize that the reaction is similar to the one depicted in Equation 4, except with a dif-ferent alkene. First, we need to determine the correct structure, then we apply the reaction to thecompound.

148. DAn extreme shift such as 11.2 ppm is likely due to a carboxylic acid such as acetic acid. Also

suspect acetic acid because it is used in excess in the reaction. Again, isopentyl alcohol isunlikely, because it is the limiting reagent in the reaction and because the hydroxyl hydrogen inmost alcohols is usually found around 5-6 ppm. Water usually does not appear in nmr spectra,and if it does it is a very broad peak due to extensive hydrogen bonding. Acetyl chloride wouldnot be found in the reaction mixture because it was not used in this reaction.

149. B This question is essentially asking which compound has the lowest boiling point and which

the highest. Compounds that exhibit the most hydrogen bonding have the highest boiling points,those with the least hydrogen bonding have the lowest boiling points. Octane clearly has theleast hydrogen bonding and so has the lowest boiling point. It will therefore distill first. Alco-hols have more hydrogen bonding than esters (alcohols have an O-H group, esters do not). N-octyl alcohol has the highest boiling point and will distill last.


150. B Glucagon is the only hormone that is not lipid-soluble like steroid hormones, and therefore

glucagon will not cross the lipid bilayer. Water-soluble hormones require membrane receptors inorder to conduct the signal. Lipid-soluble hormones, on the other hand, can cross the lipid bilayerand bind to cytoplasmic or nuclear receptors. Of the answer choices given, aldosterone, testos-terone, and estrogen are all lipid-soluble steroid hormones, which are derived from cholesterol.

151. A The net effect of the Na+/K+ pump is to transport one ion out of the cell, thereby lowering

intracellular osmolarity. Placement of an animal cell into a hypertonic environment creates a ten-dency for water to leak out of the cell because of high extracellular osmolarity. To prevent waterloss from the cell, the Na+/K+ pump will slow down the rate of ion exchange, thereby main-taining cellular osmolarity. Note that this process has nothing to do with “need”. It occurs auto-matically, since a hypertonic solution around the cell would decrease the concentration gradientsuch that there would be a significantly lower draw on ions to leave the cell. Choice (B) is incor-rect because a hypotonic medium outside the cell will increase the rate of ATP entry. Choices(C) and (D) are incorrect because both channels dissipate the function of the pump, causing itto consume more ATP.

152. AThe passage states that the small bleached region regains its fluorescence after the fluores-

cently labeled molecules exchange places with bleached molecules. The rate at which this

41

exchange occurs is directly dependent on the diffusion rate of the fluorescently labeledmolecules. Since the lipids diffuse faster than proteins, the rate of lipid exchange is greater thanthe rate of protein exchange. It will thus take longer for the bleached region to recover if pro-teins are labeled instead of lipids. The answer is choice (A).

153. DTo answer this question one must be familiar with the mechanism of enzyme action. An

enzyme catalyzes a reaction by lowering its activation energy, which essentially means lower-ing the potential energy of the transition state intermediate. During their transfer from one layerof the bilayer to the other, lipid molecules represent the transition state from reactants (lipids onone side) to products (lipids on the other side). Incorrect answer choices can be eliminated usingthe knowledge that enzymes do not modify the free energy or the enthalpy of reactants and prod-ucts. Choice (A) is incorrect because if lipid molecules were stabilized after their transfer thenthe free energy or enthalpy of products would be lowered. Choice (B) is incorrect because desta-bilization of lipid molecules before their transfer means that the free energy or the enthalpy ofthe reactants is raised. Choice (D) is correct.

154. BSince valinomycin permits passive transport of K+ across a membrane, K+ moves down its

concentration gradient and out of the vesicle. When K+ exits the vesicle, it carries a positivecharge, lowering the overall charge of the vesicle’s interior. The correct answer is (B).


155. BThe key to answering this question is realization that G protein-coupled receptors are trans-

membrane protein molecules and thus must be synthesized by ribosomes associated with theendoplasmic reticulum. Ribosomes that are associated with ER synthesize all proteins that haveto cross the membrane, e.g., transmembrane, lysosomal or excreted proteins. This occursbecause it is energetically unfavorable to insert a finished protein molecule into the membraneand the cell meets the challenge by inserting the protein into the membrane as it is being syn-thesized. Choice (B) is correct.

156. CSince selective bradykinin antagonists block pain mediated only by bradykinin, one must pre-

sume that pain mediated by other molecules such as ATP and protons must be intact. The ele-ments of the pain transduction pathway that are shared by bradykinin and other pain producingstimuli must thus be operating, while the elements that are used only by bradykinin can be inhib-ited. Among the given answer choices, choice (C) presents a model where only pain mediated bybradykinin will be inhibited. Choices (A) and (D) can be eliminated because in both scenariospain will be augmented rather than inhibited. Choice (B) presents a model where pain mediatedby a broad range of stimuli, not just bradykinin, will be diminished. Choice (C) is correct.

157. BThe passage states that activation of the G protein-coupled receptors leads to an activation

of a signaling pathway that dampens nociceptive response. Since the question stem indicates thatadenylyl cyclase is activated in response to binding of endogenous stimuli to the G protein-cou-pled receptors, one can deduce that adenylyl cyclase is one of the enzymes that mediates damp-ening of the nociceptive response. This means that if adenylyl cyclase is constitutively activethen nociceptive response will be diminished; i.e., the response to the endogenous pain stimuliwill be reduced. Choice (B) is correct.

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158. CThe passage states that below pH 7, protons augment nociceptor activation by thermal,

chemical or mechanical stimuli. This indicates that the threshold for activating nociceptors inresponse to various stimuli is reduced under acidic conditions. Thus, the thermal stimulus gen-erated by an actively respiring muscle can be augmented by acidity and become sufficientlystrong to activate nociceptive neurons. Choice (A) is incorrect because the passage states thatprotons can directly activate nociceptive neurons only at pH values lower than 6. Choices (B)and (D) are unsupported by the passage. Choice (C) is correct.

159. CThe action of opiates must account for the inhibition of postsynaptic neurons in the spinal

cord. Among the listed answer choices all but choice (C) present a scenario that will actuallylead to activation of the postsynaptic neurons. Choices (A) and (B), for example, will lead to anincrease in the number of excitatory neurotransmitters in the synaptic cleft, thereby stimulatingpostsynaptic neurons. In choice (D), opiates activate sodium channels, which inevitably leads todepolarization of the postsynaptic neurons. Choice (C), on the other hand, describes a mecha-nism that can explain the inhibitory effect of opiates. Inhibition of cation channels in postsy-naptic neurons can prevent activation of those neurons in response to excitatory stimuli. Thismechanism is also described in the passage for the dampening of nociceptive signaling inresponse to pain producing stimuli. Choice (C) is correct.

160. DThe passage states that G protein-coupled receptors are involved in dampening of the noci-

ceptive signaling in response to endogenous pain stimuli. Thus if the G protein-coupled recep-tors are blocked the sensation of pain will be increased rather than diminished. Choice (D) iscorrect.

Discrete Questions (Questions 161–164)

161. DOxaloacetate is the molecule regenerated at the end of the citric acid cycle. Oxaloacetate

combines with acetyl CoA, choice (A), produced during pyruvate decarboxylation to form cit-rate, choice (B). Malate, choice (C), is the molecule that precedes oxaloacetate in the citric acidcycle. Since oxaloacetate is the molecule regenerated at the end of the citric acid cycle, choice(D) is correct.

162. BDuring the S phase, each chromosome is replicated to produce two identical sister chro-

matids joined by a centromere. Although the cell contains twice as much genetic materials as itdid prior to the S phase, because the chromatids have not yet separated, the cells are still 2N,even at the end of the S phase. Thus choice (B), 2N, is correct.

163. DHydrogen bromide adds to alkenes in the presence of peroxides with a regioselectivity oppo-

site to that predicted by Markovnikov’s rule; anti-Markovnikov addition is observed and theless-substituted alkyl bromide is the major product. The reaction is a free-radical chain processin which a bromine atom adds to the double bond in the first propagation step. The intermedi-ate formed has the structure shown in choice (D). Choice (C) is incorrect because the more sub-stituted (Markovnikov) addition product would be formed. Choice (A) is incorrect because thefree radical is on the same carbon as the bromine. Choice (B) is incorrect because the bromineis missing from the intermediate.

43

164. AUracil is a nucleotide found only in RNA. Transcription is the process whereby information

coded in the base sequence of DNA is transcribed into a strand of mRNA. RNA contains thenucleotide uracil (U) rather than thymine (T), so the process of transcription requires the pres-ence of uracil. Translation is the process whereby mRNA codons are translated into a sequenceof amino acids. Mitosis is the process of cellular division that results in the formation of twodaughter cells that are genetically identical to each other and to the parent cell. Conjugation isthe temporary joining of two organisms via a tube called a pilus, through which genetic mate-rial is exchanged. Choice (A) is the correct answer.


165. CThe isomer of the product that is formed is controlled by the position of attack of the elec-

trophile on the benzene ring. Two major factors influence this. As in all other mechanisms, amechanism step proceeds to give the most stable product if there are no other overriding fac-tors. (An example of an overriding factor would be the C-H/C-LG periplanar requirement for theE2 reaction.) Since the electrophilic attack affords a carbocation, the attack leading to the moststable carbocation is favored.

Attack at any position leads to a carbocation with at least three resonance structures. In thespecific example of this question, X = CH3, is an electron-donating group. The best carbocationin the ortho and para cases is tertiary, while the best carbocation in the meta case is sec-ondary. Based on carbocation stability, we predict attack at the ortho and para positions to befavored over meta attack. (If the X group can share lone pairs through resonance then the orthoand para carbocations can have a fourth resonance structure, further improving their stabilityover the meta carbocation.) The X group disfavors attack at the ortho position because it is morecrowded than the meta or para positions. Resonance generally dominates, so the carbocationstability issue is more important than the steric effect. Thus the order of attack when X = CH3or any other electron-donating group is: para (most) > ortho > meta (least). Electron-donatinggroups are therefore termed ortho/para directors.

Ortho attack:

Meta attack:

Para attack:

E+E

HX

+ +

XHE +

XHE

X

E+

X X

E

H

+ +

X

EH +

X

EH

E+

X X

+

E H

X

+

E H

X

E H

+

166. BElectrophilic attack on the benzene ring disrupts aromaticity and forms a carbocation, and

thus requires a powerful electrophile. Molecular bromine is not sufficiently electrophilic to over-come aromaticity; thus, choice (A) is incorrect. A more powerful electrophile is needed. TheAlBr3 serves to increase the electrophilicity of the bromine by strongly polarizing the Al-Brbond, thus increasing the amount of positive charge on the bromine by decreasing its electrondensity. This is the first step of the mechanism shown. (In cases where an arenium ion with afull octet resonance structure is produced, the bromine may be sufficiently electrophilic byitself.) Choice (D) is incorrect because AlBr3 is regenerated in the reaction. Rather, HBr is thebyproduct.

When X is a electron-withdrawing group, it destabilizes an adjacent carbocation by increas-ing the net positive charge on that carbon. This occurs in the carbocations resulting from orthoand para attack, but not from meta attack. In this case the order of attack is meta > para >ortho. Electron-withdrawing groups are termed meta directors.

167. DReaction rate is controlled by the rate-determining step (rds) of the mechanism. For EAS,

this step is attack of the electrophile on the aromatic ring. This is the rds because aromaticity issacrificed. Since the rds involves electrophilic attack, factors which make the benzene ring a bet-ter nucleophile will increase the rate of reaction (e-donating groups also produce more stablecations). The OCH3 group is a strong e-donating group which increases the e-density of the ben-zene ring making it a better nucleophile. CH3 is a weaker e-donating group than OCH3. NO2 isan e-withdrawing group and thus nitrobenzene (III) is a weaker nucleophile than benzene (IV).

168. CSide-chain halogenation takes place under conditions that favor the formation of radicals

(e.g. UV light). The halogen dissociates to produce halogen atoms and then the halogen atomsinitiate chains by abstracting hydrogens of the methyl group:

+ HBr + AlBr3

Br3Al Br Br

CH3 CH3

Br H AlBr3Br

CH3

Br

CH3

Br H

CH3

Br H

44

Abstraction of a halogen from the methyl group of toluene produces a benzyl radical. Thebenzyl radical then reacts with a halogen molecule to produce a benzyl halide and a halogenatom. Steps 2 and 3 repeat. Choices (A) and (D) are incorrect because, whereas many arylhalides are UV-sensitive, exposure to UV light over short periods of time wouldn’t lead to sig-nificant decomposition. If Choice (C) were correct, the carbocation intermediate must capture anucleophilic bromide ion instead of losing a proton. Capture of a proton does not restore aro-maticity whereas deprotonation does. This provides a significant driving force for deprotonationover capture of chloride ion.

169. AIn an EAS reaction, the rate-determining step is electrophilic attack on the benzene ring (not

deprotonation of the aromatic ring) because it results in loss of aromaticity. The purpose ofAlBr3 is to increase the electrophilic nature of Br2 by polarizing the Br-Br bond. H2OBr+ is astronger electrophile (better able to accept electrons) because the bromine bears a positivecharge (it is electron deficient). Also, H2O is an excellent leaving group. Thus, Answer B is cor-rect. H2OBr+ is an ion and is therefore highly soluble in H2O compared with Br2 (Answer Dincorrect).

170. AThe mechanism of an EAS reaction involves two steps. This automatically eliminates

choices (B) and (C) which are potential energy diagrams for one-step reactions. The rate deter-mining step is the first step which involves electrophic attack and carbocation formation. Step 2involves deprotonation and restoration of aromaticity which produces a product much more sta-ble than the carbocation intermediate, and more stable than the starting materials since the reac-tion proceeds spontaneously. Thus choice (A) is correct.

Br2UV light

2 Br

Step 1:

Step 2: Chain Propagation

CH3

+ Br

CH2

Benzyl radical

+ HBr

Step 3:

CH2

Benzyl radical

+ Br2

CH2Br

Benzyl halide

+ Br

45


171. CTo answer this question correctly you must know the relative intracellular and plasma con-

centration of various components in the blood. Since fluoride causes hemolysis, it allows intra-cellular components to mix with plasma, thereby raising plasma concentration of intracellularcomponents. Potassium is found in high concentration inside cells and in low concentration out-side cells due to the action of Na+/K+ ATPase. Thus during hemolysis, the plasma concentrationof K+ will be raised significantly. Choice (A) is incorrect because Na+ is present mostly in theplasma. Choice (B) is incorrect because albumin is a plasma protein. Choice (D) is incorrectbecause after bicarbonate ion is made inside red blood cells, it diffuses into plasma.

172. AFrom Figure 1, you can see that glucose concentration changes from approximately 4.8 to 3

mmol/L in 8 hours. The net change of glucose concentration was thus 4.8 – 3 = 1.8 mmol. Theaverage rate of disappearance is simply the slope of the line and is equal to (1.8 mmol/L)/8 hrwhich is approximately 1/4. This corresponds with answer choice (A).

173. DThe beginning of the third paragraph states that D,L-glyceraldehyde is an antiglycolytic

agent and one can infer that it will inhibit a glycolytic enzyme. Choice (A) is incorrect becausethe passage is discussing the effects of D,L-glyceraldehyde on blood after it was collected frompatients. D,L-glyceraldehyde will thus have no influence on insulin secretion by the pancreas.Choice (B) is incorrect because if D,L-glyceraldehyde is an antiglycolytic agent the last thing itwill do is increase the rate of glycolysis. Choice (C) is not only unsupported by the passage, butalso incorrect because if D,L-glyceraldehyde increased the rate of glucose synthesis then ananalysis would provide an overestimate of plasma glucose concentration thus rendering D,L-glyceraldehyde unsuitable as preserving agent.

174. CTable 1 shows that creatinine concentration values for blood treated with D,L-glyceralde-

hyde are significantly different from those obtained from blood with no additives. D,L-glycer-aldehyde will thus interfere with measurements of creatinine. Choices (A), (B) and (D) areincorrect because for urea, calcium and potassium there is almost no difference in concentrationvalues obtained with and without D,L-glyceraldehyde.

175. AThe last sentence of the passage states that the L stereoisomer was responsible for all or most

of the antiglycolytic activity of the racemic mixture. Since 30 mmol/L of racemic D,L-glycer-aldehyde will have 50% or 15 mmol/L of the L isomer, 15 mmol/L of pure isomer will berequired to have the same effect. Choice (A) is correct.


176. CSuccinylcholine competes with acetylcholine for the receptor (paragraph 2) and causes a

depolarization (activator). In contrast, vecuronium competes but does not cause a depolarizationand is classified as a competitive inhibitor. The correct answer is choice (C).

46

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177. DDecreased heart rate and increased salivation are signs of parasympathetic activity; edro-

phonium increases parasympathetic activity (paragraph 4). Choices (A) and (C) are signs ofincreases in sympathetic activity. Choice (B) is incorrect because atropine (a parasympatheticblocker) would not be expected to interfere with edrophonium’s neuromuscular effects.

178. CLoss of the acetylcholine receptors, choice (A), would not allow the impulse to be transmit-

ted to the skeletal muscle cells. As in spinal cord injury, severing motor neural axons, choice (B),would also prevent impulse transmission. Preventing calcium ion release, choice (C), within themuscle cells would prevent calcium binding to troponin; this will cause the actin-binding site toremain covered. Downregulating the acetylcholinesterase gene would result in hyperexcitationrather than paralysis. The correct answer is choice (C).

179. ASuccinylcholine binds to the receptor and causes a depolarization (paragraph 2). Succinyl-

choline depolarizes muscle cells individually as the succinylcholine reaches them. It is the chan-nels remaining open that causes the subsequent paralysis. Choice (B) describes allostericbinding (induced fit type). Choice (C) is incorrect because although succinylcholine does bindand prevent ACh from binding, the very binding of succinylcholine causes a depolarization.Choice (D) is false because succinylcholine cannot cross the blood-brain barrier (paragraph 1)and would not suppress brain activity. The correct answer is choice (A).

180. CThis experiment suggests that curare interferes with transmission of the signal to contract

rather than contraction itself. Because the muscle is able to contract with an external electricaldepolarization, the muscle cell contraction apparatus must be intact. This rules out any elementsof this apparatus (statement III), leaving only elements related to signal transmission (statementsI and II). Differentiation between statements I and II is impossible without further experimenta-tion. The correct answer is I and II only, or choice (C).


181. AThe alkene in choice (A) is favored over that of choice (B) since it has more substituents on

the double bond. Alkene A has three substituents while alkene B has only two. The stability ofan alkene is directly related to the number of substituents. Choices (B) and (C) are incorrectsince, in each case, one alkene has fewer substituents than the other. Choice (C) is incorrectbecause the alkenes have different substituents and consequently are formed in differentamounts. Choice (D) is incorrect because relative product distribution can be predicted accord-ing to the alkene substitution pattern.

182. B2-Methyl-2-butene and 2-methyl-1-butene are constitutional isomers. The two alkenes have

the same molecular formulas but the atoms are bonded in different arrangements. Choice (A) isincorrect because configurational isomers are stereoisomers (enantiomers or diastereomers).The atoms are bonded in the same manner but have different arrangements in space. However,molecules cannot be transformed into one another by rotation around single bonds. Choice (D)is incorrect since conformers have different arrangements of atoms in a molecule according torotation around single bonds. Choice (D) is incorrect because diastereomers are configurationalisomers or stereoisomers.

183. DThe reaction is a second-order reaction that proceeds in one step. 1-Bromooctadecyl bro-

mide is a primary halide that reacts with a good nucleophile (ethoxide ion) through a transitionstate involving a pentacoordinted carbon atom. The substitution product (the ether) is formed viaan SN2 mechanism (substitution, nucleophilic, bimolecular), and the elimination product (thealkene) is formed via an E2 mechanism (elimination, bimolecular). Choice (A) is incorrect sincethe reaction does not proceed through a carbocation intermediate as SN1 substitution (substitu-tion, nucleophilic, unimolecular) and E1 (elimination, unimolecular) mechanisms indicate.Choices (B) and (C) are incorrect because substitution and elimination reactions are either bothunimolecular or both bimolecular, never a combination of the two.

184. BThe tert-Butoxide ion is very bulky when compared to the methoxide ion. Steric considera-

tions reveal that removal of a proton to form the elimination product is achieved more easily thanattack of the electrophilic carbon center to give the substitution product. Choice (A) is incorrectsince tert-butoxide ion is actually much more basic than methoxide ion. This higher (not lower)basicity also significantly contributes to the product outcome. Choice (C) is incorrect becausethe time and temperature differences are minimal when compared to the steric and basicity dif-ferences of the nucleophiles.

185. CThe best solvent choice is dimethylformamide since it is a polar, aprotic solvent. The solvent

has no protons to participate in hydrogen bonding and, consequently, the nucleophile (cyanideion) is not solvated and is therefore a much more powerful nucleophile. Choices (A), (B) and(D) are incorrect because water, methanol, and formamide have hydrogens that participate inhydrogen bonding and solvate the nucleophile. The nucleophilic power of the cyanide ion wouldbe reduced thus slowing the reaction.

186. BAn anion is a better nucleophile than the corresponding neutral compound since the elec-

trons on the anion are not as tightly held. Therefore, the electrons of the anion are more easilyshared than those of the neutral compound. Choice (A) is incorrect because a more electroneg-ative species is less nucleophilic. The more electronegative species has a tighter hold on its elec-trons and does not share them as easily. Choice (C) is incorrect since larger atoms are betternucleophiles. In larger atoms, the outer electrons are further from the positively-charged nucleusand so are less tightly held. The electrons are more easily shared and so the atom is a betternucleophile. Choice (D) is incorrect since the more polarizable a species is, the more nucle-ophilic it is. Polarizability is directly related to the magnitude of the force with which the elec-trons are attracted to the nucleus. When the electrons are more loosely held the atom is morepolarizable and its electrons are more easily shared.


187. AChondrocytes are the specialized cells that secrete chondrin, which forms a firm but elastic

matrix to form cartilage. Epiphyses aren’t actually cells. Rather, they are the rounded, dilatedends of the bone. The diaphysis is the long, cylindrical shaft in between the epiphyses.Osteoblasts are the cells that help build bone and function antagonistically to osteoclasts, whichare the cells that help break down bone. Thus, choice (A) is correct.

48

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188. DHelper T cells activate B cells and other T cells through the secretion of lymphokines. Cyto-

toxic T cells destroy foreign antigens. Suppressor T cells downregulate B and T cell activityagainst antigens. T cells are not, however, responsible for recognizing antigen upon secondaryexposure to it. Rather, that is the job of memory B cells. There is a class of memory T cells, buttheir job is to proliferate and produce a large number of cytotoxic T cells upon secondary expo-sure. Choice (D) is correct.

189. CIf the normal X chromosome is denoted by X+, and the X chromosome with the gene for

hemophilia is denoted by Xh, a female who is a carrier for hemophilia has a genotype of X+Xh,while a male who is a hemophiliac has a genotype of XhY. Utilizing a Punnett square, it may bedetermined that the genotypes of their offspring are 1 X+Xh, 1 XhXh, 1 X+Y, 1 XhY. Thus, ofthe female children, 1/2 are carriers of the gene for hemophilia, and 1/2 are hemophiliacs. Thus,50% of the female children of a hemophilia carrier female and a hemophiliac male will behemophiliacs. This corresponds with answer choice (C).

190. BImines are formed from the reaction of an aldehyde or ketone and a primary amine. The

imine in the correct answer choice (B) is formed by the reaction of propanal and methylamine.

Choice (A) is incorrect because it is an amine, not an imine. Choice (C) is an enamine,formed by the reaction of an aldehyde or ketone with a secondary amine. Choice (D) is acarbinolamine. Carbinolamines are intermediates in the formation of imines; the carbinolaminein choice D would spontaneously lose water to form the imine in the correct answer (B).

191. CElectron withdrawal stabilizes the conjugate base of a carboxylic acid, the carboxylate ion,

and thus the acidity of a carboxylic acid is increased by the presence of an electron-withdraw-ing group such as chlorine. The proximity of the electron-withdrawing group to the carboxylgroup determines the degree of stabilization of the carboxylate ion. Thus the conjugate base ofacid I is stabilized to the greatest extent, and I is the strongest acid. Acid III, having a chlorineon the carbon, is stronger than acid II where there is no chlorine present.

CH3CH2CHNHCH3 CH3CH2CH NCH3 + H2O

OH

CH3CH2CH + CH3NH2 CH3CH2CH NCH3

O

50


192. ABased on paragraph 2 of the passage, we know that dark adaptation involves an increase in

sensitivity. The results of the experiment, however, show a descending threshold curve. Basedon the definition of threshold intensity, one should deduce that the threshold is inversely pro-portional to sensitivity. Hence, the curve demonstrating the greatest increase in sensitivity is alsothe curve showing the greatest drop in threshold. The answer is curve A, which shows the great-est reduction in threshold intensity between 0 and 40 minutes after the onset of darkness.

193. DBased on the first paragraph of the passage, we learned that cones are more active in brighter

conditions and rods are more active in darker conditions. We are also told in the final paragraphthat there is a dominance of cone sensitivity above the rod-cone break (the first phase) and adominance of rod sensitivity at intensities below the break (the second phase). The passage doesnot suggest that only one kind of photoreceptor is responsive at different light intensities. There-fore, cones are more sensitive than rods in bright conditions. The correct answer is choice (D).

194. BAs explained above, rod sensitivity is dominant during the second phase of the adaptation.

As the curves proceed from A to D (with lower pre-adapting intensities), one can see that thesecond phase occurs earlier and earlier in time until curve D, at which point there is no longer avisible cone adaptation phase. However, note that this does not mean that cones do not adapt atlow pre-adapting intensities (or that rods do not adapt at high pre-adapting intensities), as sug-gested by choices (C) and (D). The graphs show a total sensitivity that is defined by the mostsensitive photoreceptor. In other words, at low pre-adapting intensities, the rods are more sensi-tive so the curve follows rod sensitivity. At high pre-adapting intensities, the cones are at firstmore sensitive than cones but become less after the cone-rod break, thus showing a rod-conebreak at the inflection point.

195. DSuch a subject lacks functional cones. If you understand that the first phase of adaptation is

the cone-dominant phase, you should realize that a person without cones will only have the rod-dominant phase of adaptation. Another way to tackle this question is to see that all the curveswith the exception of curve D have an inflection point.

196. AChoices (B), (C), and (D) are all correct in this question. The first part of the adaptation

curve demonstrates a predominance of cone adaptation, which is faster than rod adaptation. Rodadaptation occurs more slowly, but takes over after 10 minutes and achieves a higher sensitivity,as demonstrated by the lower threshold. An absolute threshold is indeed reached such that anydarker light is simply imperceptible by the human eye under any lighting conditions. Choice (A)is incorrect because sensitivity of the photoreceptor depends on the lighting condition (cones aremore sensitive in bright light) or time in the dark during dark adaptation (cones are more sensi-tive at first but becomes less sensitive over time). Though the absolute sensitivity of rods maybe higher, choice (A) is too much of a generalization to be a correct statement.

51

Passage X (Questions 197–202)

197. BThe narrow melting point range obtained for the recrystallized sample suggests the unknown

is a pure compound (Choice (A) incorrect). The presence of excess H2O in the sample wouldcause a lowering of the melting point, therefore choice (D) is possible. Qualitative tests rule outacetaminophen as the identity of the unknown, however (negative Hinsberg Test for 2º amine)so choice (D) must be incorrect. The only explanation consistent with the qualitative test resultsis choice (B). Salicylic acid (mp 156-158°C) is the hydrolysis product of acetylsalicylic acid.

Choice (C) is incorrect because esters are more prone to hydrolysis than amines.

198. DThe mechanism for the Williamson ether synthesis is an SN2 reaction involving two steps:

Initially, the acetaminophen reacts with carbonate to form the corresponding phenoxide ion:

HO–C6H4–NH–COCH3 + CO32– –O–C6H4–NH–COCH3 + HCO3

–

The phenoxide then acts as the nucleophile in the SN2 reaction with ethyl iodide to formphenacetin:

–O–C6H4–NH–COCH3 + CH3CH2I CH3CH2–O–C6H4–NH–COCH3 + I–

199. BA chiral carbon has four different groups attached to it. Of the compounds in Table 1, Ibupro-

fen is the only one that contains a chiral center. The IUPAC name for Ibuprofen is (±)-2-(�-isobutyl phenyl) propionic acid.

CH3

CH *

HO

O

H C CH2

CH3

CH3

Ibuprofen

CH3

OH

O

Salicylic acid

O

200. DThis question is asking if the Hinsberg method is useful for identifying an amine as either

1°, 2° or 3°. In other words, does the Hinsberg test give different VISIBLE results depending onwhether an amine is 1°, 2° or 3°? The best way to tackle this question is to replace the 2° aminein the Hinsberg equation from the passage with 1° and 3° amines. Recall the chemistry ofamines and decide when (and if) precipitates will form.

The passage states the Hinsberg Test is conducted in aqueous base. The amine immediatelyreacts if it is 1° or 2° with the resulting HCl being neutralized by the base. The passage tells youthat sulfonamide derivatives of 2° amines are usually insoluble solids (neutral compounds!).You should recognize that the sulfonamide derivative of a 1° amine will be acidic and will dis-solve in aqueous base. Acidification of this solution will then precipitate the sulfonamide of a 1°amine. Because of the heterogeneous nature of this system, the rate at which the sulfonyl chlo-ride reagent is hydrolyzed to the sulfonic acid is relatively slow.

So the correct answer is choice (D). The Hinsberg Test is useful for distinguishing 1°, 2° and3° amines. As shown in the following equations, 1° and 2° amines react to give sulfonamidederivatives with loss of HCl, whereas 3° amines do not give any isolable products other than thestarting amine. In the latter case a quaternary “onium” salt may be formed as an intermediate,but this rapidly breaks down to liberate the original 3° amine.

N H

H

R

1o amine

S

O

Cl

O

NaOH

HClS

O

N

O

H

R

acidic hydrogen

NaOHH2O

S

O

N

O

R

Na

water-soluble salt (clear solution)

NaCl

HClS

O

N

O

H

R

water-insoluble (precipitate)

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201. CThe Ferric chloride test does not involve oxidation or reduction (Choices (A) and (B) incor-

rect) but relies on acid-base chemistry. Iron remains in the +3 oxidation state. Pyridine is a baseand reacts with the HCl which is generated in the reaction to form pyridine(HCl). The neutral-ization provides the driving force for the ligand exchange since the reverse reaction is prevented.

202. BThe IUPAC name for acetaminophen is N-(4-hydroxyphenyl) acetamide. 4-hydroxy benzoic

acid is the IUPAC name for salicylic acid which is the hydrolysis product of acetylsalicylic acid(IUPAC name is 2-acetoxy benzoic acid). Choice (C) is therefore incorrect.

Passage XI (Questions 203–208)

203. CThe first experiment showed that male mice deficient in -galactosyltransferase are infertile,

while the second experiment showed that sperm derived from such mice don’t bind eggs aseffectively as sperm from the � or the wild type strains. Both of these experiments thus supportHypothesis 2 and suggest that -galactosyltransferase is required for effective sperm-egg bind-ing. Choice (A) is incorrect because such a conclusion cannot be drawn from the experimentsdescribed in the passage. Strain �, for example, can have a number of defects that would renderthis strain incapable of reproducing outside the laboratory conditions. Choice (B) is incorrectbecause the second experiment shows that some strain sperm do attach to eggs. Choice (D) isincorrect because such a conclusion can not be drawn based on the information in the passage.Choice (C) is correct.

N R

R

R

3o amine insoluble

S

O

Cl

O

2 NaOHS

O

O

NaN R

R

R

starting amine insoluble

NaCl

2 H2O

HCl

N R

R

R

quaternary ammonium salt (soluble)

H

Cl

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204. CStrain a lacks �-galactosyl residues because it is deficient in functional �-galactosyltrans-

ferase, which attaches �-galactosyl residues to oligosaccharides.

According to Hypothesis 1 those �-galactosyl residues must be present on the egg surfacein order for sperm-egg binding to occur. Thus if �-galactosyl residues were essential for sperm-egg attachment (i.e. Hypothesis 1 were true), then eggs from strain a mice would be defectiveand strain � females would be infertile. The reason sperm will be unaffected by �-galactosyl-transferase mutation is because sperm surface doesn’t contain �-galactosyl residues, but(according to Hypothesis 1) contains protein that binds those residues. Hence strain � spermwould be capable of attaching to normal eggs that express �-galactosyl residues. Choice (C) iscorrect.

205. BIf reagents that change the specificity of -galactosyltransferase for its substrate also inhibit

sperm-egg interaction, then this would support the essence of Hypothesis 2 which states thatsperm-egg binding is mediated by -galactosyltransferase and its substrate. Choice (A) wouldnot support nor contradict Hypothesis 2, which discusses sperm-egg attachment and not fertil-ization. Choice (C) would contradict Hypothesis 2, according to which N-acetyl-glucosaminylis a substrate for -galactosyltransferases and its removal must interfere with sperm-egg attach-ment. Choice (D) would also contradict Hypothesis 2 because if -galactosyltransferase is thetrue sperm binding protein then excess of this enzyme would bind to the zona pellucida and pre-vent the sperm from attaching to an egg. Choice (B) is correct.

206. DAccording to Hypothesis 1, �-galactosyl residues are important in sperm-egg binding. Con-

sequently the observation that the removal of �-galactosyl residues from ZP3 does not effectsperm-egg binding would undermine Hypothesis 1. Choice (A) would support Hypothesis 1because strain � eggs don’t express �-galactosyl residues on their surface and according toHypothesis 1 shouldn’t be fertile. Choices (B) and (C) would neither support nor contradictHypothesis 1. Choice (D) is correct.

207. CThe experiment described in the question stem is designed to isolate sperm proteins that bind

to ZP-3. The isolated peptide turned out to be a 56-kDa peptide, the same peptide that was sug-gested by Hypothesis 1 to be the protein on the sperm surface that binds ZP3. Hence the exper-imental observation will support Hypothesis 1, and choice (C) is correct.

208. AA competitive inhibitor is a substance that binds the active site of a protein thereby prevent-

ing it from interacting with its normal substrate. One important characteristic of a competitiveinhibitor is its resemblance to the actual substrate. Among the given answer choices onlyoligosaccharides with N-acetyl-glucosaminyl and �-galactosyl residues resemble the substratesthat bind β-galactosyltransferase and Sp56 respectively. To decide between choices (A) and (B),one must look at the experimental evidence. The experiments described in the passage provideevidence in support of Hypothesis 2. Therefore the best candidate for a competitive inhibitormust be a molecule that interferes with the receptor-ligand interaction of molecules described inHypothesis 2, namely the β-galactosyltransferase and an oligosaccharide with N-acetyl-glu-cosaminyl residues. Since an exogenous supply of oligosaccharides with N-acetyl-glucosaminylresidues will bind β-galactosyltransferase on the sperm surface and thereby prevent the attach-

54

ment of the sperm to the ZP3, choice (A) must be the correct answer. Choices (C) and (D) areincorrect because antibodies typically inhibit the action of proteins in a noncompetitive fashion,i.e. they often do not bind the active site.


209. AThe reaction is an example of acid-catalyzed esterification, also known as Fischer esterifi-

cation. The mechanism involves nucleophilic addition to the carbonyl group, followed by elim-ination of the leaving group. The nucleophile is the oxygen atom of methyl alcohol.

As the reaction shows, the oxygen atom of the alcohol (18O in this problem) becomes thesingly-bonded oxygen atom of the ester product. Choice (A) is therefore the correct answer tothe problem. Choice (B) is incorrect because the isotope label in on the carbonyl oxygen of theester. Choice (C) is incorrect because it is not an ester. Choice (D) is incorrect because the iso-tope label is on the oxygen atom of water, not the ester.

210. CThe compound shown in the problem is an aldopentose. This can be seen readily be exam-

ining the structure in its open-chain form.

Sugars belong to the D series when the hydroxyl group on the highest-numbered stereogeniccenter is in the same configuration as D-glyceraldehyde, that is, on the right. In a pentose it isthe configuration of the C-4 hydroxyl group (answer choice (C)) that determines the D or L con-figuration of the sugar. Choice (A) is incorrect because the direction of optical rotation does notdetermine absolute configuration. A D sugar can rotate plane-polarized light either in the (+) orthe (–) direction. Choice (B) is incorrect because the configuration at the anomeric carbon deter-mines whether the cyclic hemiacetal form of the sugar is + or –, not D or L. Choice (D) is incor-rect because C-5 is not a stereogenic center in a five-carbon sugar (pentose).

HO H

HO H

H OH

CH2OH

CH O

CH318OH + C6H5COH C6H5COH C6H5C18OCH3 + H2O

O OH

18OCH3 O

55

211. CBe sure to expand parentheses and correctly identify the longest chain in order to assign a

systematic (IUPAC) name to a compound. Remember that the longest chain of carbon atomsneed not just be from left to right across the page, but can go “around corners.” Begin number-ing at the end of the chain closest to the first substituent.

This reveals methyl groups on carbons 2, 4, and 7 of a 9-carbon chain. Thus the stem of thename is nonane and the correct name is 2,4,7-trimethylnonane. Choices (A), (B), and (D) incor-rectly identify the longest chain; it is 9 carbons, not 6 (hexane) or 8 (octane).

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

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

213. BIncreasing concentrations of hydrogen ion in the blood result in a low pH, which decreases

hemoglobin’s affinity for oxygen. Increasing concentrations of carbon dioxide in the blood alsodecrease hemoglobin’s affinity for oxygen. The increased concentration of oxygen in the alveo-lar capillaries promotes oxygen uptake in the lungs. Thus, statements I and II decreasehemoglobin’s affinity for oxygen, while statement III increases hemoglobin’s affinity for oxy-gen. The correct answer is therefore III only, or choice (B).

214. BThe three primary germ layers—ectoderm, endoderm, and mesoderm—are formed during

gastrulation. Muscles and the heart are derived from the mesoderm. The pancreas is formed fromthe endoderm. The spinal cord and skin are derived from the ectoderm. Thus the correct answeris choice (B).

CH3CHCH2CH2CHCH3

CH2CH3 CH2CHCH3

7 6 5 4

2 1

CH38 9

3

56

Health & Medicine

Full length 9 answers