79
MCAT Full-Length Tests Dear Future Doctor, The following Full-Length Test and explanations are an opportunity to bring it all together in simulation. Do not engage in Full-Length practice until you have adequately prepared your knowledge and critical thinking skills in Subject, Topical, and Section tests. Simply g the tests is inadequate; a solid understanding of your performance through your Score Reports and the explanations is necessary to diagnose your specific weaknesses and address them before Test Day. All rights are reserved pursuant to the copyright laws and the contract clause in your enrollment agreement and as printed below. Misdemeanor and felony infractions can severely limit your ability to be accepted to a medical program and a conviction can result in the removal of a medical license. We offer this material for your practice in your own home as a courtesy and privilege. Practice today so that you can perform on test day; this material was designed to give you every advantage on the MCAT and we wish you the best of luck in your preparation. Sincerely, Albert Chen Executive Director, Pre-Health Research and Development Kaplan Test Prep © 2003 Kaplan, Inc. All rights reserved. No part of this book may be reproduced in any form, by Photostat, microfilm, xerography or any other means, or incorporated into any information retrieval system, electronic or mechanical without the written permission of Kaplan, Inc. This book may not be duplicated, distributed or resold, pursuant to the terms of your Kaplan Enrollment Agreement.

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MCAT Full-Length Tests

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

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

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Physical SciencesTime: 100 Minutes

Questions 1–77

DO NOT BEGIN THIS SECTION UNTIL YOU ARE TOLD TO DO SO.

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2

PHYSICAL SCIENCES

DIRECTIONS: Most of the questions in the PhysicalSciences test are organized into groups, with adescriptive passage preceding each group of ques-tions. Study the passage, then select the single bestanswer to each question in the group. Some of thequestions are not based on a descriptive passage; youmust also select the best answer to these questions. Ifyou are unsure of the best answer, eliminate thechoices that you know are incorrect, then select ananswer from the choices that remain. Indicate yourselection by blackening the corresponding circle onyour answer sheet. A periodic table is provided belowfor your use with the questions.

1

H

1.0

2

He

4.0

3

Li

6.9

4

Be

9.0

5

B

10.8

6

C

12.0

7

N

14.0

8

O

16.0

9

F

19.0

10

Ne

20.2

11

Na

23.0

12

Mg

24.3

13

Al

27.0

14

Si

28.1

15

P

31.0

16

S

32.1

17

Cl

35.5

18

Ar

39.9

19

K

39.1

20

Ca

40.1

21

Sc

45.0

22

Ti

47.9

23

V

50.9

24

Cr

52.0

25

Mn

54.9

26

Fe

55.8

27

Co

58.9

28

Ni

58.7

29

Cu

63.5

30

Zn

65.4

31

Ga

69.7

32

Ge

72.6

33

As

74.9

34

Se

79.0

35

Br

79.9

36

Kr

83.8

37

Rb

85.5

38

Sr

87.6

39

Y

88.9

40

Zr

91.2

41

Nb

92.9

42

Mo

95.9

43

Tc

(98)

44

Ru

101.1

45

Rh

102.9

46

Pd

106.4

47

Ag

107.9

48

Cd

112.4

49

In

114.8

50

Sn

118.7

51

Sb

121.8

52

Te

127.6

53

I

126.9

54

Xe

131.3

55

Cs

132.9

56

Ba

137.3

57

La *

138.9

72

Hf

178.5

73

Ta

180.9

74

W

183.9

75

Re

186.2

76

Os

190.2

77

Ir

192.2

78

Pt

195.1

79

Au

197.0

80

Hg

200.6

81

Tl

204.4

82

Pb

207.2

83

Bi

209.0

84

Po

(209)

85

At

(210)

86

Rn

(222)

87

Fr

(223)

88

Ra

226.0

89

Ac †

227.0

104

Rf

(261)

105

Ha

(262)

106

Unh

(263)

107

Uns

(262)

108

Uno

(265)

109

Une

(267)

*

58

Ce

140.1

59

Pr

140.9

60

Nd

144.2

61

Pm

(145)

62

Sm

150.4

63

Eu

152.0

64

Gd

157.3

65

Tb

158.9

66

Dy

162.5

67

Ho

164.9

68

Er

167.3

69

Tm

168.9

70

Yb

173.0

71

Lu

175.0

90

Th

232.0

91

Pa

(231)

92

U

238.0

93

Np

(237)

94

Pu

(244)

95

Am

(243)

96

Cm

(247)

97

Bk

(247)

98

Cf

(251)

99

Es

(252)

100

Fm

(257)

101

Md

(258)

102

No

(259)

103

Lr

(260)

PERIODIC TABLE OF THE ELEMENTS

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Passage I (Questions 1-6)

Hydrogen bonding is rarely considered a major com-ponent of intramolecular bonding. It is only whenmolecules like proteins approach a sufficiently large sizethat hydrogen bonding contributes significantly to molec-ular geometry. The impact of the hydrogen bond in smallermolecules, therefore, lies not in intramolecular bonding,but in intermolecular bonding. The intermolecular conse-quences of hydrogen bonding are seen in solubility, physi-cal properties, and crystal structure of molecules.

Some molecular structures do not meet the strict crite-ria for hydrogen bonding, but nevertheless they are stillcapable of unusually strong dipole interactions. This isbecause they possess a sufficiently bare proton and a suffi-ciently electronegative heteroatom.

When an uncharged organic compound dissolves to anappreciable extent in water, the solubility may beattributed to these strong dipole interactions. Thus,dimethyl ether, forming two of these strong dipole interac-tions, is completely miscible with water, whereas dimethylsulfide is only slightly soluble in water. Beyond the effectof solvent on solute, the effect of solute on solvent must bedetermined to predict solubility; solute molecules inca-pable of hydrogen bonding or even appreciably strongdipole interactions may insinuate themselves between sol-vent molecules and disrupt solvent-solvent interactions.

Hydrogen bonding also plays a role in many nonaque-ous solvents. Chloroform is a good solvent for fatty acids,since its polar C—H bond may engage in strong dipoleinteractions. Ethers can serve as solvents for hydrogenchloride, and acetylene is very soluble in acetone becauseof strong dipole interactions.

Intermolecular hydrogen bonding in pure substancesincreases the heat of vaporization in two ways: first, byincreasing the attraction between molecules. Either thisattraction must be overcome on vaporization (as in the caseof H2O) and/or small polymeric molecules of the liquidmust be vaporized – for example, (HF)x. Second, the heatof vaporization increases by the restriction of rotation ofthe molecules in the liquid. Such rotation is possible in thegas, and the energy absorbed in exciting this rotation addsto the heat of vaporization. Hydrogen bonding thusincreases the boiling point of a liquid beyond that expectedfrom comparison with comparable non-hydrogen-bondedliquids and causes a greater entropy of vaporization thanthat of unassociated liquids.

In ionic substances, hydrogen can play a major role insolid state structure formation. When hydrogen bonding ispossible in molecular crystals, the resultant structure isgenerally the one that yields the maximum number ofhydrogen bonds. Discrete hydrogen-bonded ions exist inacid salts such as KHF2. Other acid salts, such as sodiumhydrogen carbonate, form polyanion chains or three-dimensional networks, as in KH2PO4. The difference instructure between NH4F (wurtzite structure with a 4 F-

arranged tetrahedrally around NH4+) and NH4Cl (CsCl

structure with a cubic arrangement of 8 Cl- around NH4+)

stems from the greater strength of the N—HF bond thanthe N—HCl bond.

1. Given the molecular structure below, how manyhydrogen bonds is H3BO3 capable of forming?

A. 0

B. 3

C. 6

D. 9

2. Possible explanations for the low solubility of acety-lene in H2O include:

I. acetylene interferes with hydrogen bondingbetween H2O molecules

II. intermolecular attraction between acety-lene and water is weak

III. acetylene does not readily form hydrogenbonds

A. I and III only

B. II and III only

C. I and II only

D. I, II and III

H

O O

O

B H

H

3

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3. Which of the following experimental results wouldexplain the melting point of NH3 being higher thanpredicted by the trend evident in the following exper-imental data?

A. X-ray studies of solid ammonia indicate that 3hydrogen atoms from different nitrogen atomspoint toward the lone pair of electrons.

B. X-ray studies of solid ammonia indicate thateach nitrogen atom lone electron pair abuts alone pair from an adjacent molecule.

C. IR studies of ammonia show the H—N bond tohave strong basic character.

D. Ammonia is found to act as a Lewis base inwater, forming NH4

+

4. According to the passage, the crystalline structure ofNH4Cl is best described as:

A. simple cubic

B. body centered cubic

C. face centered cubic

D. hexagonal close packing

5. The passage lists two mechanisms for hydrogenbonding to increase the heat of vaporization of puresolutions. Which of the following molecules wouldbe least affected by hydrogen bonding in �Hvapexperiments?

A. H2O

B. H3BO3C. HCF3D. HCN

6. Which of the following is NOT a necessary compo-nent or condition for hydrogen bonding?

A. aqueous solubility

B. an available lone electron pair

C. a hydrogen atom with some degree of acidity

D. spatial proximity between the two interactingmolecules

4

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5

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Passage II (Questions 7-12)

Refraction seismology uses high frequency, longitudi-nal waves known as seismic waves to map geologic struc-tures below the Earth’s surface. The seismic wave may begenerated by an explosion, a dropped weight, a mechani-cal vibrator, or other sources and travels through variousmedia below the Earth’s surface. The seismic wave isdetected by a device called a seismometer, which generatesa voltage when a wave produces relative motion of a wirecoil in the field of a magnet on the surface of the Earth.Data are usually recorded on magnetic tape for subsequentprocessing and display.

Variations in the speed at which seismic waves propa-gate through the Earth can cause variations in seismicwaves recorded at the Earth's surface. By measuring thesevariations, geologists can gain information about the com-position of underground structures. Longitudinal seismicwaves travel at speeds from about 6 km/sec in surface rockto about 10.4 km/sec near the Earth's core. The speed ofthese waves (�) in solids, liquids, or gases depends on sev-eral factors as shown in Equation 1, including the densityof the medium (�), its shear modulus (�), and the bulkmodulus of the media (B). Any change in rock or soil prop-erty that causes �, �, or B to change will cause seismicwave speed to change.

� � ��Equation 1

For example, going from a dry soil to one saturatedwith water will cause both the density and the bulk mod-ulus to change. A material’s density describes its inertia.A material’s bulk modulus describes its compressibility.B is the ratio of the change in pressure to the fractionalchange in volume, as given by Equation 2. A positivepressure increase corresponds to a decrease in volume ofthe medium. The bulk modulus is a measure of a mate-rial’s elasticity, with a large value of B describing a“stiff” material, one that is hard to compress. If B issmall, then a small pressure can compress the materialby large amounts. The stiffer the material, the strongerthe restoring forces that develop in response to com-pression and the faster a longitudinal wave can travelthrough it. In moving from dry to moist soil, the bulkmodulus changes because air-filled pores become filledwith water, and water is much more difficult to com-press than air.

B � – ���VP/V�

Equation 2

Table 1 shows the bulk modulus of elasticity and den-sity of several compounds.

Compound Bulk Modulus Density (g/cm3)of Elasticity (Pa)

Benzene 14.7 � 108 0.879Water 22.4 � 108 1.000

Mercury 28.5 � 109 13.546

Table 1

7. What is the speed of a seismic wave in water?(Assume the shear modulus of water is 3/4 Pa).

A. 1.49 � 103 m/s

B. 2.24 � 103 m/s

C. 4.73 � 104 m/s

D. 22.4 � 108 m/s

8. Solid 1 has a large bulk modulus, while solid 2 has asmall bulk modulus. If a seismic wave travelsthrough them, which will be able to store morepotential energy in its atoms as they are displacedlongitudinally from their original positions?

A. Solid 1

B. Solid 2

C. The bulk modulus of the solids will not affecttheir ability to store potential energy.

D. The solid’s ability to store potential energy can-not be determined from the information given.

9. A seismic wave with a frequency of 5000 Hz enters arock in the Earth’s surface. What is the wavelength ofthe wave in this rock?

A. 5.0 mB. 2.0 mC. 1.2 mD. 0.8 m

�43�� B��

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10. A seismometer detects a wave that passes through theEarth’s surface and another that passes through theEarth’s core. Which of the following is LEAST likelyto be the only explanation for the difference detectedin the speeds of the waves?

A. The surface rock may have a smaller bulk mod-ulus.

B. The surface rock may have a larger density.

C. The core rock may have a smaller shear modu-lus.

D. The core rock may have a lower temperature.

11. The speed of a seismic wave in Soil A is 3 times thatin Soil B. Assume that each has the same shear mod-ulus and bulk modulus, what is the ratio of densitiesof Soil A to Soil B?

A. 3:1

B. 9:1

C. 1:9

D. 1:1.7

12. The vertical kinetic energy of an atom within a solidthrough which a seismic wave is traveling from leftto right is:

A. equal to 1/2mv2, where m is the mass of thesolid and v is the speed of the wave travelingthrough it.

B. zero.

C. dependent on the density of the solid.

D. dependent on the frequency of the sound wave.

6

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Passage III (Questions 13-18)

Calorimetry allows for an accurate measure of heatflow between a system and its surroundings. An icecalorimeter, for example, uses a bath containing a mixtureof ice and water which is thermally insulated. This meansthat the inner contents of the calorimeter do not transfer orabsorb any heat from the outside environment. When aheated substance is placed inside the bath, heat will flowinto the ice-water mixture at a rate directly proportional tothe difference in temperature between the substance andthe bath. As heat flows into the bath, some of the ice melts,and the change in volume of the bath can be used to mea-sure the amount of heat transferred. Though this type ofcalorimeter is common, a bath of pure water may alsoserve the same purpose. However, instead of measuringchanges in volume, in this case the student measureschanges in temperature.

In an experiment to determine the specific heat of anunknown substance, a student fills a calorimeter with 300 gof water at 10ºC. The student then heats 150 g of theunknown substance to 75ºC, and plunges it into the ice bathwithout loss of heat to the surroundings. After the substanceand the bath reach thermal equilibrium, the student mea-sures a temperature of 15.8ºC for the bath. She comparesthe specific heat calculated for the substance to a table ofspecific heats. (Note: all experiments are done in a 27ºCroom; the specific heat of water is 4.18 JK-1g-1).

13. A cylinder is filled with an ice-water mixture to aheight of 15 cm. After a substance is added to thebath, a transfer of heat causes the water-ice mixtureto rise H cm in height. Which of the following sub-stances when added could have caused this rise?

A. Cu(s) at 0ºC

B. Hg(l) at –2.5ºC

C. Fe(s) at 75ºC

D. O2(g) at 15ºC

14. A sample of the unknown substance is heated to 68ºCat and plunged into an ice bath at 1 atm. After 10 sec-onds, before thermal equilibrium is reached, theunknown substance will have lost the most heat if theice is at:

A. -15ºC

B. -10ºC

C. 0ºC

D. 32ºC

15. According to the experimental data, the unknownsubstance is most likely:

A. Cu (specific heat = 0.385 JK-1g-1)

B. Fe (specific heat = 0.449 JK-1g-1)

C. SiO2 (specific heat = 0.739 JK-1g-1)

D. CaCo3 (specific heat = 0.818 JK-1g-1)

16. In another experiment, the student adds 3 moles ofCaF2 to a water bath at 15ºC. She then cools the bathto -30ºC ice and plunges a piece of Cu(s) at 75ºC intothe bath. Compared to a bath of pure ice, this ice bathwill require:

A. less heat to be transferred for fusion to occur.

B. more heat to be transferred for fusion to occur.

C. the same amount of heat to be transferred forfusion to occur.

D. the same amount of heat to be transferred fordeposition to occur.

7

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17. As the student is constructing the calorimeter, sheaccidentally punctures the insulating top, exposingthe bath to outside air. Consequently, the approxi-mate final temperature of the bath at thermal equilib-rium is most likely:

A. 8.2ºC.

B. 27ºC.

C. 75ºC.

D. There is not enough information to determinethe final temperature.

18. Suppose the student used a bath with 100 g of ice at0ºC in which to conduct her experiment. What wouldbe the final temperature of the bath if the unknownsubstance lost 33.3 kJ of heat. (Note: for H2O(l)→H2O(s) , �H = -6.007 kJ)

A. 0ºCB. 15ºCC. 32ºCD. 78ºC

8

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Passage IV (Questions 19-24)

When the human body completes an electrical circuit,the result may be an extremely painful shock. The proba-bility that the shock will prove fatal increases as both themagnitudes of the current and the contact time increase.However, not all parts of the body pose the same dangerwhen shocked: current passing through electricallyexcitable tissue like muscle and nerves often inflicts themost pain and damage.

Current between 1 and 10 mA from a live wire ispainful, though usually harmless to a healthy individual.However, once current above 10 mA passes through mus-cle tissue, the damage inflicted may be more severe sincethe individual may not be able to release the wire. Ironi-cally, currents passing through cardiac tissue pose less of athreat if the magnitude of the current exceeds 1000 mAthan if the magnitude were 70 mA. A lower current dis-rupts the electrical coupling of the heart, causing irregularcontraction, or ventricular fibrillation. A current in excessof 1 A passing through cardiac tissue is so large that itcauses the heart to stop completely; once the current isremoved, the heart usually regains its normal rhythm.

Ohm’s law implies the resistance of the body influ-ences the severity of damage resulting from a shock.Human tissue has relatively low electrical resistivity sincethere is an abundance of dissolved ions in the cells. How-ever, the waterproof outer layer of skin on the body has ahigh resistance when dry, approximately 104 �; this resis-tance substantially decreases when the skin is moistened.The total electrical resistance of the body can be approxi-mated using the electrical resistance of the skin.

19. A defibrillator is a device used to cause complete ces-sation of heartbeat followed by resumption of normalrhythm. What is the minimum voltage neededbetween the leads of the defibrillator in order for it tobe effective against dry skin?

A. 7 V

B. 102 V

C. 7 � 102 V

D. 104 V

20. A person who has come into electrical contact with alive wire would minimize the damage inflicted if:

I. The person is wearing insulating rubbershoes.

II. The voltage of the circuit is 120 V.

III. The current of the wire is below 10 mA.

A. I only

B. II only

C. II and III only

D. I and III only

21. An individual with dry skin picks up a live wire andexperiences a current of 1 A. If the wire has a resis-tance of 15 �, how much total energy is dissipatedafter 45 seconds?

A. 15 J

B. 675 J

C. 10 kJ

D. 451 kJ

22. Suppose an individual with dry skin has come in con-tact with a 1.5 A DC circuit with wires of negligibleresistance. In an attempt to rescue the individual,another person, with wet skin, makes contact with thefirst. Consequently, the amount of time required for10 C of charge to pass through the circuit will:

A. increase.

B. decrease.

C. remains the same.

D. There is not enough information to determine achange in current.

23. One would expect a person with high cellular ionconcentration to be:

A. less susceptible to damage from an electricshock.

B. more susceptible to damage from an electricshock.

C. just as susceptible to damage from an electricshock.

D. The passage does not provide enough informa-tion to determine who would be more suscepti-ble to damage from an electric shock.

9

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24. Which of the following is implied in the passageabout a current of 70 mA passing through the heart?

A. It will cause less severe damage than a currentof 65 mA, since a larger current reduces the riskof death.

B. It will cause less severe damage than a currentof 1.5 A because the effect on the heart wouldnot be as great.

C. It will cause greater damage than a current of1.5 A because it will cause the heart to stop,resulting in death.

D. It causes the heart to acquire an irregularrhythm, resulting in improper blood flow.

Questions 25 through 28 are NOT based ona descriptive passage.

25. Which of the following species does not have a full octet?

A. F-

B. Rb+

C. Mg2+

D. P2-

26.

The front end of a 5-m car is even with the back endof a 10-m truck. Both are traveling in adjacent lanesof a highway, the car at 40 m/s and the truck at 30m/s. How much time elapses until the back end of thecar is even with the front end of the truck?

A. 0.20 s

B. 0.25 s

C. 1.00 s

D. 1.50 s

27. In a circuit with a 12-V battery and two resistors (5-� and 7-�) connected in parallel, what is the currentthrough the 5-� resistor?

A. 1 A

B. �1

5

2� A

C. �1

7

2� A

D. �1

3

4

5

4� A

28. Place these four species in order of increasing atomicradius: Cl-, Ar, K+, Br-.

A. K+< Ar< Cl–< Br–

B. Ar< Cl–< Br–< K+

C. Cl–< Ar< Br–< K+

D. Cl–< Ar< K+< Br–

car

truck

40 m/s

30 m/s

10

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Passage V (Questions 29-33)

Chlorofluorocarbons (CFCs) have been used as refrig-erants and propellants since their development in the1930s. However, due to catalytic destruction of ozone bychlorine, the use of CFCs has become less desirable. Vari-ous alternatives have been considered; of these, the mostsignificant is dimethyl ether.

Dimethyl ether is easily formed from two methanolunits via the following reaction:

Reaction 1

Dimethyl ether is attractive as an alternative propellantfor aerosol spray cans and as a refrigerant because it isenvironmentally unreactive and is easily degraded in thetroposphere.

In addition to its use as a propellant and a refrigerant,dimethyl ether has also proved useful as an ultracleandiesel. As a fuel, dimethyl ether is ideal because of its highoxygen content (35%) and the ease with which it can becompressed to a liquid. The skeletal combustion reactionfor dimethyl ether is shown below.

a H3COCH3 (g) + b O2 (g) → c CO2 (g) + d H2O(l)

Reaction 2

Dimethyl ether can also undergo oxidation with the aidof ammonia monooxygenase to form formaldehyde:

Reaction 3

This reaction is used industrially in the production offormaldehyde. The methanol byproduct is isolated for useas a solvent in chemical laboratories, making the overallreaction very efficient.

Various compounds and their boiling points are foundin Table 1.

Compound Boiling point (ºC)Carbon dioxide -78.6Dimethyl ether -25Diethyl ether 35

Methanol 64.5Formaldehyde 96

Water 1004-chlorobiphenyl 2914-bromobiphenyl 310

Table 1

29. What is the rate expression for Reaction 1?

A. k1 [CH3OH]

B. k1 [CH3OCH3][H2O]

C. k1 [CH3OH]2

D. k1

30. When Reaction 2 is balanced, what are the values forthe coefficients?

A. a = 1; b = 2; c = 3; d = 3

B. a = 1; b = 3; c = 2; d = 3

C. a = 2; b = 2; c = 3; d = 2

D. a = 2; b = 2; c = 2; d = 3

31. Which one of the following conditions will favor theproducts in Reaction 1?

A. High pressure

B. Reduced pressure

C. A moist environment

D. The presence of a desiccant

[CH3OCH3][H2O]��

[CH3OH]2

CH3OCH3(g) CH3OH(l) + CH2O(l)

1

ammoniamonooxygenase

O2(g)2

2 CH3OH(g) CH3OCH3(g) + H2O(g)k1

k-1

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32. Which of the following best explains why the boilingpoint of formaldehyde is anomalous with respect tothe other compounds in Table 1?

A. The boiling point of formaldehyde is lower thanthat of 4-chlorobiphenyl or 4-bromobiphenylbecause of its lower molecular weight.

B. The boiling point of formaldehyde of formalde-hyde is higher than that of dimethyl etherbecause of its lower molecular weight.

C. The boiling point of formaldehyde is higherthan that of methanol because of its ability tohydrogen bond.

D. Formaldehyde has a carbon-oxygen doublebond, making its boiling point higher than thatof compounds with similar molecular weights.

33. Which of the following would increase the percentyield of Reaction 3 while suppressing Reaction 2?

A. Limiting the amount of oxygen in the reactionmixture

B. Performing the reaction at low pressure

C. Removing water from the reaction mixture

D. Performing the reaction in the presence ofexcess methanol

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Passage VI (Questions 34-39)

A chemical engineer devises two experiments to deter-mine the density and viscosity of a large sample of anunknown liquid compound:

Experiment 1

In order to determine the density of the compound, theengineer conducts an experiment to measure the accelera-tion of various materials as they sink in the compound.These test pieces have equal volume, and include cubesmade of wood, concrete, and iron. The engineer places 750mL of the compound in a 1.0-L beaker. Using a high-speedcamera, the engineer is able to record the speed with whicheach test piece sinks as a function of time. The measure-ments of the cubes’ positions are always performed usingthe bottom edge of the cube. The results are summarized inTable 1:

Table 1

Speed Wood Concrete Ironafter time: �=0.4�103 �=2.3�103 �=7.8�103

kg/m3 kg/m3 kg/m3

0.0 sec 0.0 m/s 0.0 m/s 0.0 m/s0.5 sec 0.0 m/s 3.2 m/s 4.4 m/s1.0 sec 0.0 m/s 6.4 m/s 8.8 m/s1.5 sec 0.0 m/s 9.6 m/s 13.2 m/s

Experiment 2

The engineer drills a hole of radius 4.5-cm into the sideof a 3.0-L beaker. The engineer then fills the beaker withthe unknown liquid compound, removes the stopper andmeasures the velocity of the exiting liquid. Using the fluidat the top of the beaker as a reference, the engineer uses theBernoulli equation to predict the speed of the fluid as itleaves the hole:

P1 + �12� ��2 + �gh = constant,

where, P is the pressure of the atmosphere, � is the densityof the liquid, � is the velocity of the liquid, h is the heightof the liquid, and g is the acceleration due to gravity. Bycomparing the predicted results to the actual results, theengineer concludes that the unknown liquid compound hasnegligible viscosity.

34. As the engineer switches from the wood test piece tothe iron test piece the percent increase in mass is:

A. 18.5%

B. 29.5%

C. 1.85 � 103%

D. 2.95 � 103%

35. According to the experimental data, which of the fol-lowing statements is true?

A. Some volume of unknown compound weighsmore than the same volume of concrete.

B. The ratio of the density of the unknown com-pound to the density of wood is greater thanone.

C. The specific heat of the unknown substance islarger than that of iron.

D. The work done by the unknown liquid as a testpiece sinks equals the change in the kineticenergy of that test piece.

36. Which best explains why the engineer usedBernoulli’s equation to determine relative viscosityof the unknown liquid compound?

A. Only a viscous fluid exerts a constant pressurethroughout the fluid.

B. The real behavior of a non-viscous fluid devi-ates greatly from that predicted by Bernoulli’sequation.

C. The average kinetic energy of the fluid is givenby 1/2��2, and is directly proportional to theviscosity of the fluid.

D. Bernoulli’s equation does not hold when fric-tional forces within the fluid are high.

37. Suppose the engineer’s second experiment isrepeated at a higher altitude. Compared to the exper-imental velocity obtained by the engineer, the veloc-ity of the exiting liquid at a higher altitude isexpected to:

A. decrease.

B. increase.

C. remain the same.

D. increase at first, then decrease when the pres-sure of the liquid decreases.

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38. In a third experiment, the engineer pours the liquidthrough a funnel into a test tube, as shown below.

At which point is the velocity of the liquid the greatest?

A. A

B. B

C. C

D. D

39. Suppose that in Experiment 1, the iron test piece washeated to 100ºC and then submerged into the liquidcompound. Assuming negligible heat transfer to theliquid, which of the following will be true?

A. The volume of the liquid compound willdecrease.

B. The weight of the iron test piece will increase.

C. The weight of the iron test piece will decrease.

D. The buoyancy force on the test piece willincrease.

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Passage VII (Questions 40–45)

Maple syrup is prized for its unique flavor and consis-tency. First harvested and used by the natives of Canadabefore the arrival of European explorers, it is still exportedon a large scale. In order to produce maple syrup, a num-ber of steps must be carried out. First, the maple tree mustbe “tapped”. This is done by boring a hole from the out-side of the tree trunk to the sap-carrying core of the tree,about 5 centimeters deep. The sap can then flow into abucket for collection. Once the sap has been collected, itmust be boiled to remove excess water; sap is only 3.0%sucrose by mass, whereas syrup is 64% sucrose. In addi-tion to water and sucrose, syrup contains malic acid,potassium and calcium salts as well as trace amounts ofphenolic compounds, amino acids and vitamins. The struc-ture of malic acid is shown in Figure 1.

Figure 1

To optimize the syrup production process, severalexperiments were conducted. In the first experiment, thesap distillation is performed at different temperatures, withthe following results shown in Table 1. Purity values arebased on desirable trace element levels.

Pressure (atm) Purity of resultant syrup0.78 0.960.82 0.910.90 0.831.0 0.751.17 0.64

Table 1

In an effort to further optimize the syrup mixture, anadditional experiment is conducted which is designed tostandardize the sucrose level. Sucrose is oxidized by dini-trogen tetroxide in the presence of vanadium (V) oxideaccording to Reaction 1:

Reaction 1

By measuring the volume of NO gas evolved, the levelof sucrose in the syrup is determined. When 100 millilitersof syrup are allowed to react with an excess of dinitrogentetroxide at STP, 3.36 moles of NO are collected.

To determine malic acid levels, an oxidizing agent isused to convert malic acid to oxaloacetic acid. (The struc-ture for oxaloacetic acid is shown in Figure 2.)

Figure 2

The reaction progress is monitored using IR spec-troscopy. Various group frequencies are shown in Table 2.

Bond Type of Compound Frequency Range,cm-1

C=C Alkene 1610-1680C-H Alkane 2850-2970

1340-1470O-H Monomeric alcohols, 3590-3650

phenolsC-O Alcohols, ethers, 1050-1300

carboxylic acids, estersC=O Aldehydes, ketones, 1690-1760

carboxylic acids, esters

Table 2

40. During the conversion from sap to syrup at 1 atm, thetemperature of the solution is:

A. less than 100ºC, because sucrose boils at alower temperature than water.

B. 100ºC, because that is the boiling point of water.

C. greater than 100ºC because the concentration ofsucrose is higher than that of other solutes in thesap solution.

D. greater than 100ºC because a solute elevates theboiling point of a solution.

C12H22O11(aq) + 9 N2O4(s)V2O5

6 H2C2O4(aq) + 18 NO(g) + 5H2O(l)

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41. Which of the following is the orbital hybridization ofthe starred atom in malic acid?

A. sp2

B. sp3

C. sp3d2

D. s2p2

42. Based on the values given in Table 1, the purityshows:

A. an inverse relationship with pressure because aspressure decreases, the boiling point of the sapsolution increases, causing other impurities toboil off with the excess water.

B. an inverse relationship with pressure because aspressure decreases, the boiling point of the sapsolution decreases, protecting the delicate traceelements from denaturation.

C. a direct relationship with pressure because aspressure increases, trace elements are forced tocondense and stay in solution.

D. a direct relationship with pressure because aspressure increases, the boiling point of the sapsolution increases, causing the sucrose to breakdown into its components, glucose and fructose.

43. Which of the following is the limiting reactant forReaction 1?

A. Vanadium (V) oxide

B. Dinitrogen tetroxide

C. Sucrose

D. Water

44. How much NO was collected from the oxidation ofsucrose in the syrup?

A. 22.4 L

B. 33.6 L

C. 75.4 L

D. 224 L

45. When all of the malic acid from the syrup has beenconverted to oxaloacetic acid, what will the change inthe IR spectrum of the solution?

A. A peak will disappear from 1050-1300 cm-1

because malic acid loses alcohol functionality tobecome oxaloacetic acid.

B. A peak will form at 1690-1760 cm-1 because aC=O double bond is formed during the conver-sion from malic acid to oxaloacetic acid.

C. A peak will disappear from 2850-2970 cm-1

because the C-H bond in malic acid is convertedto a C=O bond in oxaloacetic acid.

D. A peak will disappear from 3590-3650 cm-1

because malic acid loses alcohol functionality tobecome oxaloacetic acid.

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Passage VIII (Questions 46-50)

Engineers are interested in studying the various waysin which heat transfers from one place to another. In gen-eral, heat energy flows from one place to another in one ofthree ways:

Conduction

On an atomic scale, heat conduction is the transfer ofkinetic energy between molecules. Energy transfers frommore energetic molecules to less energetic molecules bycollision. Just as some materials are better at conductingelectricity than others, some materials conduct heat betterthan others. Good heat conductors have a relatively highthermal conductivity, k. The rate of transfer of heat energydepends on the thermal conductivity of the material, thecross-sectional area A, the temperature difference �Tacross its length, and the length L.

Convection

Under conditions of constant pressure, raising the tem-perature of a gas means increasing its volume as well.Since this causes the density to drop, the gas will risehigher, carrying away the increased energy. Heat energymoved in this manner is said to have been transferred byconvection.

Radiation

Most objects continuously radiate energy in the formof electromagnetic waves. Radiation from the sun providesover one hundred times our daily energy needs. Objectswith emissivity e = 0 are perfect reflectors: they neitherabsorb nor emit energy. On the other hand, objects with e = 1 are called black bodies: they absorb all the energyincident upon it, and radiates energy away at the same rate.Objects with emissivities between zero and one are imper-fect absorbers and radiators. An object with emissivity e attemperature T radiates energy with power

P = � AeT 4, where � = 5.67 × 10–8 and

A is the surface area of the object in square meters.

46. Under what condition does heat energy flow fromone position to another?

A. There is an entropy difference between the twopositions.

B. The space between the two positions is filledwith material that conducts heat.

C. There is a temperature difference between thetwo positions.

D. It is physically meaningless to discuss heatenergy as traveling from one place to another.

47. A physical system is surrounded by a double wall ofinsulation, and the space between the walls is com-pletely evacuated. By using a vacuum as insulation,which types of heat flow are prevented?

I. Conduction

II. Convection

III. Radiation

A. I only

B. II only

C. III only

D. I, II and III

48. If heat energy radiates away from an object with awavelength of 200 nm, what is the frequency of thewave?

A. 1.5 � 10-1 Hz

B. 1.5 � 106 Hz

C. 5.0 � 1014 Hz

D. 1.5 � 1015 Hz

W�m2 K4

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49. Which of the following is true of an isolated gas con-tained in a fixed volume that is heated by convection?

A. The entropy of the gas increases, and the inter-nal energy of the gas increases.

B. The entropy of the gas increases, and the inter-nal energy of the gas decreases.

C. The entropy of the gas decreases, and the inter-nal energy of the gas increases.

D. The entropy of the gas decreases, and the inter-nal energy of the gas decreases.

50. How does the rate of conduction of heat energythrough a material depend the length L throughwhich the heat travels, the cross-sectional area A, andthe temperature difference �T between the two ends?

A. The rate of conduction of heat energy is directlyproportional to A and �T, and inversely propor-tional to L.

B. The rate of conduction of heat energy is directlyproportional to A, �T, and L.

C. The rate of conduction of heat energy isinversely proportional to A, �T, and L.

D. The rate of conduction of heat energy is directlyproportional to L and �T, and inversely propor-tional to A.

Questions 51 through 56 are NOT based ona descriptive passage.

51. B2H6(g) + H2O(l) → B(OH)3 (aq) + H2 (g)

Consider the above unbalanced equation. For thisreaction, how many grams of diborane are required toproduce 13.4 L of H2 at STP?

A. 27.6 g

B. 13.8 g

C. 4.6 g

D. 2.76 g

52. If an electron, starting from rest, accelerates from endto end across a 30-meter region whose endpointshave a potential difference of 5.0 kV, what is thestrength of the electric field in the region?

A. There is no electric field in the region.

B. 6.0 × 10–4 N/C

C. 1.7 N/C

D. 1.7 × 102 N/C

53. Which of the following compounds has the highestionic character?

A. LiH

B. NaI

C. RbCl

D. CBr4

54. When a parent nucleus undergoes alpha-decay, thedaughter nucleus:

A. has 2 fewer protons and 2 fewer neutrons thanthe parent nucleus.

B. has 2 fewer protons and 4 fewer neutrons thanthe parent nucleus.

C. has the same number of protons and neutronsthan the parent nucleus.

D. has 1 fewer proton and 2 fewer neutrons than theparent nucleus.

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55. Which of the following is true of the Gibbs freeenergy, �Gº?

A. �Gº = �Hº+T�Sº

B. �Gº = �Sº T�Hº

C. �Gº = RT ln KeqD. �Gº = RT ln Keq

56. If a 3-kg object moving at 5 m/s strikes a wall, andthen rebounds away from the wall at 3 m/s, howmuch kinetic energy did the object lose?

A. 0 J

B. 13.5 J

C. 24 J

D. 37.5 J

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Passage IX (Questions 57-62)

An assortment of portable electronics relies on theLeclanché cell, or Zn-C dry cell, for voltage. The cell usesan outer layer of zinc to provide electrons for the innerMnO2 and graphite core. Under the zinc layer is a mixtureof ZnCl2 and NH4Cl, allowing ion movement to neutralizethe overall charge moving toward the inner core. The over-all reaction of the cell is:

Zn(s) + 2 MnO2(s) + 2 NH4+(aq) → Zn2+(aq) + Mn2O3(s) +

2 NH3(aq) + H2O(l); ∆E° = 1.71 V

Equation 1

While the Leclanché cell remains common, two otherbattery types are widely used, as well. The alkaline dry cellfunctions in an identical manner to the Leclanché cell,except that in the alkaline dry cell NH4Cl is replaced byKOH. An entirely different battery type, the rechargeable,or Ni-Cd, cell is popular since it has the ability to act as anelectrolytic cell, after which it can be reused. Therechargeable battery gives an electromotive force of 1.4 Vand as a galvanic cell, the anode reaction for a Ni-Cd bat-tery is

Cd (s) + 2 OH- (aq) → Cd(OH)2 (s) + 2 e-

Equation 2

57. Which best describes the reaction of a Ni-Cd batterywhile it is being recharged?

A. Cd(OH)2(s) + 2 Ni(OH)2(s) → Cd(s) + 2 NiO(OH)(s) + 2 H2O(l)

B. Cd(OH)2(s) + 2 Ni(OH)2(s) → Cd(s) + 2 H2O(l)

C. Cd(s) + 2 Ni(OH)2(s) → Cd(OH)2(s) + 2 NiO(OH)(s) + 2 H2O(l)

D. Cd(s) + 2 NiO(OH)(s) + 2 H2O(l) → Cd(OH)2(s) + 2 Ni(OH)2(s)

58. An engineer attempts to recharge a Ni-Cd batteryusing a Zn-HgO cell. If the reduction potential at theanode of the Zn-HgO cell is -0.762 V, in order to suc-cessfully recharge the battery the reduction potentialat the cathode must be:

A. less than 0.638 V.

B. greater than 0.638 V.

C. less than 2.162 V.

D. greater than 2.162 V.

59. One disadvantage of a Leclanché cell is that the con-centration of aqueous species changes with use. Overthe life of a working Leclanché cell:

A. �E for the cell will increase, �G for the cell willdecrease.

B. �E for the cell will decrease, �G for the cellwill decrease.

C. �E for the cell will increase, �G for the cell willincrease.

D. �E for the cell will decrease, �G for the cellwill increase.

60. Which of the following statements is true of theLeclanché cell and alkaline cell while in operation,but NOT true of the Ni-Cd cell while it is beingrecharged?

A. The anode is the site of reduction.

B. The cathode is the site of reduction.

C. The anode is positive.

D. The cathode is positive.

61. Suppose a Leclanché cell were missing the mixtureof ZnCl2 and NH4Cl. Which of the following is mostlikely to occur?

A. The redox reaction will not occur since there isno reducing agent.

B. The redox reaction will not occur since there isno oxidizing agent.

C. The cell will not function since there is no saltbridge.

D. The cell will function normally as long as Znand MnO2 are present.

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62. Suppose another type of battery uses a metal X(s) inthe anode reaction. If the cell has a voltage of 1.2 V,what is the charge of the oxidized X ion if the batteryhas a �Gº = -347 kJ?

A. +1

B. +2

C. +3

D. +4

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Passage X (Questions 63-67)

The longbow as a weapon of war is one of the greatesttechnological advances in military history, allowing Eng-land to become the foremost power in Europe during the14th and 15th centuries. Its power is stored in the elastic

bowstring, with a potential energy PE � �12�kx2,where x is

the amount the string is retracted and k is a measure of thestrength of the bow.

To test the strength of different bows, an archer shootsarrows at a target 2 m above the ground and 100 m awayfrom where he stands. The archer holds each bow at aheight of 2 m from the floor and at an angle of 30 degreeswith respect to the horizontal. The bowstring is pulled backa distance x along the bow’s line of sight (i.e. 30 degreesfrom the horizontal) and then released. The arrow has amass of 100 g. By evaluating the amount the string must beretracted in order to strike the target, one can make a com-parison of the strengths of the different bows. The bowstring retractions are listed in the following table for thebows used in the test. Air resistance may be neglected.

Table 1

Bow Distance of string retraction1 50 cm2 60 cm3 80 cm4 100 cm

63. If the bow string is drawn back the same distance forall four bows, which bow would store the greatestamount of potential energy in the string?

A. Bow #1

B. Bow #2

C. Bow #3

D. Bow #4

64. In the experiment, the ratio of the velocity of thearrow fired from bow #4 to the velocity of the arrowfired from bow #1 is:

A. 4:1

B. 2:1

C. 1:2

D. 1:1

65. If one bow needs to have its string pulled back twiceas far in order for the arrow to reach the same targetas another bow, it is approximately:

A. twice strong as the other bow

B. four times as strong as the other bow

C. twice as weak as the other bow

D. four times as weak as the other bow

66. When the archer draws back the string of a bow, thearrow:

A. gains elastic potential energy and gains gravita-tional potential energy.

B. gains elastic potential energy and loses gravita-tional potential energy.

C. loses elastic potential energy and gains gravita-tional potential energy.

D. loses elastic potential energy and loses gravita-tional potential energy.

67. What is the effect of wind resistance on the trajectoryof an arrow released from a bow?

I. The maximum height of the trajectory isreduced.

II. The maximum distance of the trajectory isreduced.

III. The time required for the arrow to fall tothe ground is reduced.

A. I and II only

B. II and III only

C. I and III only

D. I, II, and III

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Passage XI (Questions 68-72)

Zero-emissions vehicles often use electric power bat-teries because these can generate power without burningany fuel. Until recently, lead storage batteries were oftenused in automobiles, but these are very heavy and canrelease poisons into the environment if not cared for prop-erly. However, a promising battery for zero-emissionsvehicles has become available: the sodium sulfide battery.It makes use of the following half-reactions to generatepower:

2 Na → 2Na+ 2e-

Equation 1

3S + 2e– → S32-

Equation 2

The development of the sodium sulfide battery pre-sented a number of difficulties. Metal sulfides do not dis-sociate appreciably in aqueous solution, due to the stronglybasic nature of the sulfide ion. Therefore, instead of focus-ing on the Ksp of a metal sulfide, chemists have chosen todetermine a “Kspa”; a Ksp in acidic solution. The genericdissolution equilibrium for a metal sulfide (MS) is shownin Equation 3.

MS(s) + 2H3O+(aq) M2+(aq) + H2S(aq) + 2H2O(l)

Equation 3

Ksp and Kspa values are shown for a variety of metalsulfides in Table 1.

Metal Sulfide Ksp KspaHgS 2.0�10-53 2�10-32

CuS 7.9�10-37 6�10-16

PbS 3.2�10-28 3�10-7

ZnS 2.0�10-25 2�10-4

FeS 7.9�10-19 6�102

MnS 3.2�10-14 3�107

Sulfide is a relatively large, polarizable anion. Thisquality is referred to as “soft” whereas a “hard” anion hasa small, non-polarizable electron cloud. Soft anions preferto bond with soft cations, while hard anions bond stronglyto hard cations.

68. What is the net reaction when Equations 1 and 2 arecombined in a sodium sulfide battery?

A. 2 Na + S → Na2S

B. 2 Na + 3 S → Na2S3C. Na + S → NaS

D. 2 Na + 3 S → 2 NaS3

69. Where do the oxidation and reduction reactionsoccur in the Na-S battery?

A. The sodium half-reaction is the oxidation reac-tion and occurs at the anode; the sulfur half-reaction is the reduction reaction and occurs atthe cathode.

B. The sodium half-reaction is the reduction reac-tion and occurs at the anode; the sulfur half-reaction is the oxidation reaction and occurs atthe cathode.

C. The sulfur half-reaction is the reduction reactionand occurs at the anode; the sodium half-reac-tion is the oxidation reaction and occurs at thecathode.

D. The sulfur half-reaction is the oxidation reactionand occurs at the anode; the sodium half-reac-tion is the reduction reaction and occurs at thecathode.

70. Why is the Ksp for HgS smaller than that of ZnS?

A. The electron cloud for a mercury ion is smallerthan that of zinc.

B. Mercury is a liquid at room temperature, whilezinc is a solid.

C. Zinc has filled s orbitals, while mercury doesnot.

D. Mercury appears two rows below zinc in theperiodic table.

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71. Which of the following is most relevant in explainingwhy the Kspa values for metal sulfides are so muchlarger than their Ksp values?

A. All ionic compounds dissolve better in acidicsolution than in aqueous solution.

B. Kspa values are always larger than Ksp values.

C. OH- and HS- are strongly basic.

D. The charges on the ions are +2 and –2.

72. Why does the Kspa decrease from MnS to FeS toCuS?

A. As you move from left to right within a period,the cations become harder.

B. As you move from left to right within a period,the cations become softer.

C. Smaller cations are softer than large ones.

D. As you move down a column, the cationsbecome harder.

Questions 73 through 77 are NOT based ona descriptive passage.

73. A and B are converted to AB by means of a three-stepmechanism.

B+B → B2 (fast)A+B2 → AB2 (slow)AB2 +A → 2AB (fast)

The rate expression for this process is:

A. rate = k[A][B]

B. rate = k[A]2[B]2

C. rate = k[A][B]2

D. rate = k[A][B2]

74. A block of ice is reduced in temperature from -5º Cto -10º C. As a consequence, which of the followingis true?

A. The entropy of the block of ice has increased.

B. The total entropy of the universe has increased.

C. The temperature of the air surrounding theblock of ice has decreased.

D. The average kinetic energy per molecule of thewater has increased.

75. An object with mass 2m, moving at speed v, collideswith a stationary object of mass 3m. If the collision istotally inelastic, what is the final speed of the objectwith mass 3m?

A. 0

B. �25��

C. �

D. �32��

76. A student is instructed to prepare a solution with thehighest pH possible by mixing two substances. Thestudent should mix a:

A. strong base with a strong acid.

B. strong base with a weak acid.

C. weak base with a weak acid.

D. weak base with a strong acid.

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77. A monochromatic beam of light with a wavelength of400 nm strikes a detector. If 3100 eV of energy is col-lected, how many photons struck the detector?

(h = 4.15�10-15 eVs, c = 3�108 m/s)

A. 100

B. 500

C. 1000

D. 10000

STOP. IF YOU FINISH BEFORE TIME IS CALLED,CHECK YOUR WORK.YOU MAY GO BACK TO ANYQUESTION IN THIS SECTION ONLY.

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Verbal ReasoningTime: 85 MinutesQuestions 78–137

DO NOT BEGIN THIS SECTION UNTIL YOU ARE TOLD TO DO SO.

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Passage I (Questions 78-83)

From the outset of his dramatic poem Samson Ago-nistes (1671), John Milton establishes and expands upon ahero/antihero dichotomy that has its roots in the Book ofJudges. Samson is the “epic hero,” a tragic figure whofalls, despite tremendous personal strength. In prison,Samson’s thoughts and words are melancholy and self-effacing. He compares his body to a vessel, tragicallysteered off course and consequently wrecked by his lust forDalila.

The chorus of friends visiting him in prison does notallow their hero to take full blame, but placates him withthe androcentric consolation: “wisest Men / Have err’d,and by bad women been deceiv’d.” In this, the chorus’trope of woman-as-deceiver (and thus logical repositoryfor blame) is much more simplified and essentialist than isSamson’s view. While he clearly despises Dalila for lead-ing him into such a trap (“That specious Monster, myaccomplisht snare”), he also implicates himself in his cap-ture.

Samson’s confession of his culpability is, in a way,analogous to the scene in Milton’s more widely read Par-adise Lost in which Adam takes partial responsibility forthe Fall. However, despite the similarities of the “falls” inSamson Agonistes and Paradise Lost, the overall genderrelations in the two texts are not so simply analogous.Adam and Eve are co-creators of the Fall; Eve is not thedeceiver but rather the deceived. She beseeches Adam totaste the fruit not out of malice or hope for worldly gain,but instead in an attempt to share the “wisdom” that shefalsely believes she has acquired. Dalila, on the other hand,is much more cognizant of her deception; in fact, she rev-els in it as a means of gaining wealth and renown. Shereceives the forgiveness of neither her husband, Samson,nor her author, Milton. Samson may be aware of his cul-pability, but his anger toward and hatred of Dalila is notquelled by this self-awareness.

When Dalila first approaches Samson in the prison, shefeigns contrition, telling Samson that she did not realizeher deed would cause him so much agony—that shewishes to make amends for her “rash but more unfortunatemisdeed.” It is, however, difficult to believe that someoneheretofore positioned as the deceiver would do anythingbut deceive. Samson rebukes her, regretting the lust thatdrew him to her side; he no longer wishes to be “entangl’dwith a posynous bosom snake.” This imagery calls to mindnot only the deadly asp of another femme fatale, Cleopa-tra, but also Satan as the serpent in the Garden. This iden-tification links Dalila more explicitly with the serpent thanwith Eve. Eve and Dalila, though both responsible to somedegree for a “fall,” are, in Milton’s eyes, two very differentwomen.

It is interesting to note that, while Milton avoids thefallacy of overt stereotyping, by describing Dalila as a cer-tain “type” of woman, Dalila herself employs essentialismto relieve herself of some culpability and ingratiate herselfwith Samson. She attempts to pass off her treachery ascommon to all women, saying, “it was a weakness/ In me,but incident to all our sex.” She uses antifeminist rhetoricto exonerate herself in the same way Samson used it toincriminate her. In both instances, the fallacy of sexualessentialism is clear.

When Samson rejects her advances, Dalila quicklyreverts to the cold persona that the reader has come toexpect. She gloats over the downfall of the great Samsonand her “public marks of honor and reward”—self-aggran-dizing glee that certainly does not endear Dalila to Sam-son, to Milton, or to the reader, but does reemphasize hercleverness and power.

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VERBAL REASONING

DIRECTIONS: There are nine passages in the VerbalReasoning test. Each passage is followed by severalquestions. After reading a passage, select the bestanswer to each question. If you are not certain of ananswer, eliminate the alternatives that you know to beincorrect and then select an answer from the remain-ing alternatives. Indicate your selection by blackeningthe corresponding oval on your answer document.

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78. The primary purpose of this passage is to:

A. provide a detailed comparative study of genderroles in two of Milton’s dramatic poems, Sam-son Agonistes and Paradise Lost.

B. examine the motivations and actions of a some-times oversimplified character.

C. advocate a feminist reading of the character ofDalila in Samson Agonistes.

D. compare and contrast Milton’s version of theSamson and Dalila story with that found in thebiblical book of Judges.

79. The passage suggests that which of the following isNOT a tactic employed by Dalila in order to manip-ulate Samson?

A. appropriating patriarchal stereotypes in order tofurther her argument

B. deceiving under the guise of romantic and sex-ual interest

C. arguing the superiority of the female intellect tothe male

D. giving false apologies in order to win back Sam-son’s trust

80. The author most likely mentions the serpent (line 45)in order to:

A. make a comparison between Milton’s portrayalof Dalila and Shakespeare’s portrayal ofCleopatra.

B. remind the reader of the similar dichotomybetween Adam/Eve and Samson/Dalila.

C. emphasize Milton’s view of Dalila by compar-ing her to a creature traditionally associatedwith deception.

D. differentiate between the asp in the story ofCleopatra and the serpent of biblical tradition.

81. According to the passage, Samson Agonistes and Par-adise Lost share all of the following characteristicsEXCEPT:

A. the “fall” of a male character as the result of adecision made by a female character.

B. a similar gender hierarchy in which the womanwields the power of deception.

C. an acceptance of culpability by the male charac-ter.

D. the literary expansion of a biblical story.

82. Which of the following, if true, would be LEASTconsistent with the author’s claim that Milton viewsEve and Dalila as two clearly different “types” ofwomen?

A. The introduction of textual evidence that provesSamson’s punishment was as devastating asAdam’s

B. Evidence from Paradise Lost that proves Evemade Adam taste the fruit so that she might havesome amount of power over him

C. The introduction of evidence that, in the Judgesversion of the story, Samson was neither asintelligent nor as self-reflective as he is in Mil-ton’s retelling

D. The introduction of evidence that, at the begin-ning of both the Judges story and Samson Ago-nistes, Dalila was in love with Samson

83. It can be reasonably inferred that the author of thispassage is:

A. a biblical scholar doing comparative research onthe similarities and differences between an OldTestament story and a literary retelling.

B. a feminist scholar researching examples of fem-inist characters in English literature.

C. a historian researching societal attitudes towardstrong female figures.

D. a literary scholar exploring the personality andmotivations of a significant female character.

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Passage II (Questions 84-90)

Remote karst islands off the southern coast of Chilecontain some of the world’s most inaccessible caves. Ascattered and relatively rare geologic feature, karst formswhen acidic groundwater and rainwater create tunnels,caverns, underground streams, and caves in limestone.Paleogeologists studied the region’s geologic history byusing plate tectonics to identify how these islands settled intheir present positions.

Scientists dated the islands by examining the limestonefound on Tarleton, an island in the group. The Tarletonlimestone contains a fossil fauna rich in foraminifera,chiefly marine protozoans. Using these fossils, scientistsdetermined that the limestone formed near the Equator ina tropical or subtropical environment during the late Car-boniferous and early Permian geologic time periods, 315to 240 millions years ago. The foraminifera represent thecarbonate cover deposited on submarine mountains orseamounts, most likely as a coral reef. Since the subpolarenvironment of these Chilean islands is inhospitable tomarine animals of warmer seas, it would have been impos-sible for the limestone to have formed while the islandswere at their present location.

Scientists know that during the Carboniferous period,Chile was not located in its present position but at a lati-tude where coral reefs would not have formed. Continentaldrift and plate tectonics explain the movement of theEarth’s crustal plates. Imagine the upper regions of theEarth as a conveyor belt. The lithosphere, about the top 60miles or so of the mantle, moves over the athenosphere.The lithosphere contains all the continental and oceanicplates. As the plates collide against each other, one platesubducts beneath its neighbor. Mantle rocks rise to the sur-face while others are pushed down into the athenosphere.The Earth’s internal heat keeps the rocks in the atheno-sphere in a partial liquid or plastic state, and the continen-tal and oceanic plates ride on top of this layer. Earthquakesand volcanic eruptions take place on plate margins.

Paleogeologists reviewed the Earth’s geography whenthe foraminifera were living. The Earth looked drasticallydifferent than it does today. Four hundred million yearsago the supercontinent Pangaea formed, becoming com-pletely assembled in the early Permian (280 to 240 millionyears ago), the same period as the Tarleton limestone. Thelimestone, exotic to the continental plate, was not alwayspart of South AmericA. The rocks were transported pas-sively on top of one of the tectonic plates beneath what isnow the Pacific Ocean. As the plate subducted under theedge of the South American plate in the Triassic or earlyJurassic period (230 to 160 millions years ago), the lime-

stone accreted to the mainland. The faulted limestone islocated along with sediments derived from the erosion ofthe approaching continent and small basaltic slices of theold Pacific ocean floor. The compressive tectonic pro-cesses recrystallized the limestone. Regional metamor-phism and glacial periods also shaped the limestone.Strong winds and corrosive rains worked to form the karstlandscape seen today. Limestone of similar origin has beenfound along the margins of the modern Pacific Ocean inCanada, Japan, and New Zealand, providing further evi-dence that as the ocean floor moved, it carried pieces ofthis near-equatorial reef all over the globe.

84. The author’s description of the supercontinent Pan-gaea supports the scientific hypothesis that:

A. foraminifera mainly consist of marine proto-zoans.

B. rocks from the athenosphere travel down to thelithosphere.

C. plate tectonics explains the formation of karst.

D. plate tectonics explains how rocks formed inone environment can be found in the oppositeenvironment.

85. It can be inferred from the passage that by the Trias-sic period:

A. Pangaea was just assembling as a super-conti-nent.

B. Pangaea had broken up.

C. the Tarleton limestone was in the Tropics.

D. the Tarleton limestone remained above theEquator.

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86. According to the passage, plate tectonics explainshow protozoans that formed near the Equator came tobe found in rock deposits now located at latitudes notconducive to their growth. Which of the followingwould most weaken the author’s conclusions abouthow the limestone came to be off Chile’s coast?

A. Dating of the Chilean mainland to a differentperiod than the Tarleton limestone

B. Evidence that the assimilation of the super-con-tinent of Pangaea was a gradual process

C. Evidence that the Chilean coast never experi-enced volcanic or earthquake activity

D. The association of abundant sunshine with coralreef formation

87. The author discusses the Tarleton limestone in order to:

A. introduce terminology for understandingforaminifera.

B. determine the age of a group of islands off thesouthern coast of Chile.

C. explain the terrain of southern Chile.

D. explain plate tectonics.

88. According to the passage, Tarleton island is:

I. exotic to the South American Continent.

II. mainly composed of a coral reef.

III. the product of early Carboniferous platecollision.

A. I only

B. II only

C. I and II only

D. I and III only

89. It can be construed from the passage that one of thecontributions of the theory of plate tectonics to theunderstanding of paleogeology is:

A. an explanation for how marine protozoans formlimestone.

B. the ability for scientists to predict earthquakesand other seismic occurrences.

C. an explanation for continental drift.

D. the confirmation for the depths of differentstrata in the Earth’s mantle.

90. Which of the following discoveries would best sup-port the author’s argument that an equatorial coralreef was widely dispersed by plate tectonics?

A. Madagascar and Chile are the same age.

B. Madagascar was once part of India.

C. Fossilized protozoans the same age as the fossilforaminfera on Tarleton Island were found onMadagascar.

D. Fossils of the same chemical make-up as theTarleton limestone can only be found off Chile’scoast.

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Passage III (Questions 91-97)

The relationship between the fiction and life of a nov-elist is a tangled one, complicated in many cases by ourlack of precise details regarding the novelist’s life. In theinfamous case of Jane Austen, for example, her devotedfamily worked overtime to destroy her correspondence andpersonal writings, leaving an enigmatic figure that speaksof her life to us only through fragments. Other novelistspersonally extirpated their most intimate writings. Nor,even when such writings survive, enhanced by other bio-graphical material, is the relationship between fiction andbiography clear. The recently published journals of Cana-dian novelist Lucy Maud Montgomery may serve to illu-minate certain aspects of this relationship.

Prior to the publication of the journals, Montgomerywas renowned mostly for her novel Anne of Green Gables,a bright and cheerful novel with dark undercurrents. Shewrote 20 other similar, ostensibly sunny novels. Yet someare serious studies of the changing gender roles of womenin the early twentieth century, others are thinly veiled criesof rebellion and anger, while two are deeply neuroticworks that suggest extreme psychological torment, a tor-ment emphasized in the journals covering this period.

Montgomery began writing her journals at the age of10 or so; the journals published to date begin at the age of14 and appear to give a day-to-day account of her life untilher mid-sixties. (A final volume, covering the last troubledyears of her life, is scheduled for publication in the nearfuture.) Seemingly, events in the novels mirrored those inthe journals, which purport to be an accurate day to dayrecord of her life, detailing her impressions of those times,and thus, allowing us to “see” the inspiration behind cer-tain scenes in the novels. Yet we lack the original journals.The journals now available are Montgomery’s copies,which she began making at the age of 40, and that sheclaims contain exact transcripts of her original words.Close study of the journals, however, casts doubt on thisclaim. First, physical evidence shows us that Montgomerycut pages from her copies and inserted others, suggestinglater authorial censorship. (One of these cuts appears in thepages where she first discusses her husband in depth; laterknowledge of him may well have caused her to rewritesome of her initial impressions.) In volumes one and two,Montgomery claims that she burnt the journals she wrotebetween the ages of 10 and 14. Later entries in the journal,however, suggest that this comment may be disingenuous,since, in volume four, she seems to quote from journals shewrote at the age of ten. It is possible, of course, that sheremembered the precise words that she had written at theage of ten; both the journals and other sources testify to herprodigious memory. But it is equally probable, given the

physical evidence of the journals themselves, that Mont-gomery instead decided only late in life to destroy her ear-liest writings, and inserted comments throughout thejournal to make it seem as if all of them had beendestroyed years before. Thus the seamless nature of Mont-gomery’s journal narrative appears to be less a result of acarefully ordered life, and more the result of her innateurge to shape her life into a narrative format.

Hence, the journals do not, on deeper examination,provide us with a wholly unvarnished factual narrative.Indeed, rather than casting light upon the influence of facton fiction, the journals instead appear to delineate theinfluence of fiction on fact, suggesting that that the writer’surge to shape events in a narrative format may even extendinto autobiography.

91. It can be inferred from the passage that the authorwould consider which of the following importantwhen determining the accuracy of the journals of anovelist:

I. Physical changes made to the original workof the journals.

II. The resemblance of the journals to theauthor's novels.

III. The author’s innate urge to shape his or herlife story to a narrative format.

A. I only

B. II only

C. I and III only

D. I, II and III

92. The information in the passage suggests that whenauthors prepare their journals for publication, theymay:

A. reject material that they may consider to be tooboring to hold a reader’s attention.

B. deliberately alter facts in order to present a bet-ter image of themselves to readers.

C. carefully sculpt their life stories using novelistictechniques.

D. decide to write an autobiography using events,incidents, or sentences culled from their novels.

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93. In discussing the fact that Montgomery removed andinserted a page where she first discusses her husbandin depth, the author of the passage assumes that:

A. the original page recorded Montgomery’s then-impressions of her future husband.

B. Montgomery found her original impressions ofher husband to be misleading and inaccurate.

C. Montgomery’s experiences with her husbandcaused her to reevaluate his personality traits.

D. Montgomery wished to show a different imageof her future husband instead of the one sheoriginally created.

94. It can be inferred from the passage that the authorbelieves that the final volume of Montgomery’s jour-nals will:

A. fully explicate the relationship between awriter’s life and the fiction she produces.

B. contain more information about the editing pro-cess Montgomery used in her journals.

C. contain stories similar to events in Mont-gomery’s novels written in the same period.

D. shed more light on Montgomery’s psychologi-cal problems.

95. The author most likely mentions Jane Austen in orderto:

A. form a contrast between the great volume ofinformation known about Jane Austen and theincomplete journals left by Lucy Maud Mont-gomery.

B. critique Jane Austen’s family for destroyingmany of Austen’s letters and other autobio-graphical material.

C. highlight one difficulty in tracing the exact rela-tionship between fiction and the daily events ofa novelist’s life.

D. note that scholars have paid little attention to therelationship between Austen’s fiction and herremaining letters.

96. Which of the following would the author be mostlikely to use to determine the relationship between anauthor’s published journals and his or her bestsellingnovel?

A. Interviews with the friends of the author todetermine the veracity of the published journals

B. A textual comparison of the novel to contempo-raneous journal entries

C. Interviews where the author explains the genesisof the novel

D. A comparison between any rough drafts of thejournals to the published text

97. Assume that a literary critic were to find the roughdrafts of journals that Montgomery wrote at the ageof ten, only to realize that Montgomery gave differ-ent versions of the events recorded in these journalswhen writing about them at the age of 60. The authorof the passage would most likely regard this as:

A. further proof that Montgomery made laterchanges to her journals.

B. evidence of Montgomery’s writing habits.

C. indication that Montgomery’s memory was notas excellent as she liked to claim.

D. irrelevant to the argument made in the passage.

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Passage IV (Questions 98-103)

The advent of the Internet inadvertently caused theemergence of a new “literary genre”, fan-fiction. In fan-fiction, amateur writers compose stories, usually sciencefiction or fantasy, based on the work of well-knownauthors or television shows. In some cases, the writers usealready created characters; in other cases, they devise newcharacters, but use the same setting – say a group of inter-stellar explorers attacked by the Romulans from Star Trek.The stories range from astounding to execrable. Surpris-ingly, even the most abysmal have attracted a wave offierce defenders– along with equally fervent detractors.

The genre could not be described as entirely new.Medieval writers routinely penned new stories aroundestablished characters, such as Arthur or Roland, while thenineteenth century saw several pastiches and homages topopular writers, as well as unauthorized, but profitable,sequels to popular works composed by anonymousauthors. In the uncertain copyright era of the time, manywriters, especially British ones publishing in America,found themselves helpless to halt such pirated works andprinting. Twentieth century international copyright laws,intended to halt this, generally prevented many piratedworks, novel continuations, and otherwise unfair andunauthorized use of a author’s work or characters.Undoubtedly some continued to write such works, but thedifficulty of publishing these proved enough to keep theseworks deeply underground, until the advent of the Net,which soon found itself hosting several fan-fiction web-sites.

Fan-fiction has proven extraordinarily difficult to regu-late, much less end. First, creators have the difficulty oflocating the sites in the first place, since sites proliferatewildly, making them difficult to track. Second, many coun-tries either lack or do not enforce copyright laws. Closingdown foreign sites has proven beyond the means of manyauthors. Third, as writer Anne McCaffrey found, closingdown these sites may cause a fan backlash. McCaffrey,exercising her legal rights, threatened several sites withlawsuits; the sites closed, but their owners bitterlyprotested McCaffrey’s actions, sending copies of her let-ters throughout the Net, and calling for a boycott of herwork.

This reaction may seem incredible: the fans are, afterall, breaking copyright law, and more, violating the rightsof authors and artists they profess to admire. Yet many fan-fiction authors do not view matters in this right. Somepoint to medieval and Renaissance authors, such as Shake-speare, who did much the same thing, noting that copyrightis in itself a relatively new legal concept. Others claim that

by writing fan-fiction, they are increasing interest in – andthus sales – for the author. A few professional authors haveadded merit to this claim, by stating that they have noobjections to fan-fiction, as long as the amateur authorssign documents agreeing to release all claim to ownershipof the characters and setting in the fan-fiction. In somecases, the professionals have even edited collections of theamateur fiction. Still others believe that some charactersand created universes have become so well known thatthey have entered the public domain, and are therefore fairgain for amateur authors. While this argument lacks meritin most cases, certain universes – Star Trek among them –may indeed have entered the public domain. Finally, ama-teur authors bring up their most reiterated argument:money, or the lack thereof. If fan-fiction is not a profitableventure, they say, what is the harm?

Few courts have been called upon to answer thesequestions, since amateur authors generally lack the neces-sary funds to engage in litigation. Thus, the professionalauthors have so far lacked the opportunity to make a clearlegal statement that would explain the harm.

98. When attempting to defend their practices by refer-ring to medieval and Renaissance writers, the fan-fic-tion writers assume:

A. medieval and Renaissance writers provide alegal precedent for their position.

B. the medieval and Renaissance writers had noobjections to other writers using their materials.

C. the writing of tales about established legendaryor historical figures is analogous to the writingof tales set in fantastic or science fictional uni-verses created by a living author.

D. current international copyright laws go againsthistorical precedent and thus must be changed.

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99. In discussing the lack of opportunity for professionalwriters to make a clear legal statement regarding theharm caused by fan-fiction, the author assumes:

A. a clear legal statement explaining the harmcaused by fan-fiction will cause amateur authorsto stop writing fan-fiction.

B. clear legal statements may be made only in acourtroom setting.

C. the high cost of court cases deters all amateurwriters from pursuing their arguments in court.

D. the threat or actuality of court cases are the onlylegal avenue available to professional writersconcerned about fan-fiction writing.

100. Based on the information on the passage, if an ama-teur writer wrote a story using the characters from aprofessional work of fiction, and disseminated thestory over the Net:

A. the amateur author would be breaking interna-tional copyright laws.

B. the amateur author would face expensive litiga-tion procedures.

C. the amateur author would not be breaking inter-national copyright laws.

D. the story may be difficult to remove from theInternet.

101. It can be inferred from the passage that writers in themedieval and Renaissance eras:

A. lacked the protection of international copyrightlaw.

B. preferred to work with previously created char-acters and settings.

C. were interested in making quick money out oftheir writings.

D. never created new characters and settings,instead choosing to rewrite previously writtenstories.

102. According to the passage, when determining whethera work of amateur fiction is breaking internationalcopyright law or not, which of the following mustalso be determined?

A. The country of publication of the amateur work

B. Whether the work of amateur fiction hasreceived monetary compensation

C. Whether the work of amateur fiction hasincreased the sales of the professional writing itis based on

D. Whether the work of amateur fiction is based ona professionally printed work

103. The argument of the fan-fiction writers, as expressedin the passage, would be most strengthened by:

A. evidence that professional writers are notharmed by amateur stories based on their pro-fessional fiction.

B. proof that the fan-fiction writers are not receiv-ing monetary compensation for their work.

C. an independent determination of the harmcaused by fan-fiction.

D. a clear legal statement listing which charactersand settings may be considered to be in the pub-lic domain.

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Passage V (Questions 104-110)

Although the majority of African literature is publishedin European languages, since the 1960s, the decade ofmost African nations’ independence, African writers andscholars have debated the appropriateness of Europeanlanguages for African authors. Some argue that Europeanlanguages, legacies of the colonial encounter, cannot ade-quately represent nor speak to African cultures or commu-nities. Others claim that European languages are now also“African,” and that their use allows an author to reach awider audience.

Written literature in Africa certainly predates colonial-ism; however, under European rule, perceptions of litera-ture in the continent changed. For the first time, worksproduced by indigenous authors were characterized as“African” – previously, authors had neither attempted toembody nor address the entire continent. Colonialism alsospurred an increase in written literature, both in Europeanand African languages. This increase has accelerated, par-ticularly since World War II. As the number of African lan-guages in print grew, so did the discussion of theappropriate language for African literature.

Nigerian novelist Chinua Achebe typifies one side ofthe debate, arguing that English is an African language. Heasserts, further, that as the national language, English uni-fies Nigeria, allowing writers to reach a much wider audi-ence than they would otherwise, since there are dozens ofindigenous language groups in Nigeria. Achebe argues thatif English disappeared, so would the nation of Nigeria.However, Achebe also writes in his home language, Igbo,and does not claim that European languages are in anyother way preferable to indigenous African ones.

The alternate stance is argued by Kenyan writer Ngugiwa Thiong’o, who has written in his native language,Gikuyu, since the 1960s. Ngugi (as he is known) arguesthat a writer has a responsibility to his people, the vastmajority of Africans, who don’t read a European language.The use of European languages, he writes, is not neutral:European languages are inextricably linked to the legacyof colonialism. Even in the late 1980s, he notes, Englishwas imposed as a “civilizing” influence in Africa. Accord-ing to Ngugi, scholars can take only one of two positions:on the side of oppression or on the side of resistance. Thereis no possible intermediate stance.

Nevertheless, many African authors have taken posi-tions in the debate, and some of them avoid the two poles.Those who use English appear to agree with Achebe thatthey use it to make their works accessible to larger num-bers. Those who write in African languages, however, do

so for a variety of reasons: some for their languages’expressive features, others to support their local communi-ties. Ngugi’s own contributions to the debate certainlyfocused the issue and increased attention to it, but itappears that not many authors followed Ngugi’s lead inswitching to African languages. Following the debate, onewonders what the African readers believe: whether theylean toward one side or another, or whether it doesn’taffect their choice of literature at all.

104. According to the passage, what effect has the debateover the appropriate language for African literaturehad on writers and scholars in Africa?

A. It has led to a significant increase in the numberof authors using European languages.

B. It has led to a significant increase in the numberof authors using African languages.

C. It has attracted greater attention to the questionof language use in literature.

D. It has increased the level of literacy of Africancommunities.

105. Ngugi wa Thiong’o would be most likely to disagreewith which of the following statements?

A. An author writing in both Gikuyu and Englishcan bridge the gap between colonizer and colo-nized.

B. The use of a European language may allow anauthor to reach a larger number of African read-ers than the use of any single African language.

C. Written literature is an appropriate arena forexpressing social and political conflict.

D. English is more expressive than Gikuyu forsome realms of life.

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106. Which of the following statements is best supportedby the passage?

A. Ngugi and Achebe have addressed each other’swriting directly.

B. Ngugi used to write in a European language, butnow writes in an African language.

C. Most African people prefer to read literaturewritten in a European language.

D. Most African authors have taken a stance on oneside of the debate or the other.

107. The author of the passage would be most likely toagree with which of the following statements?

A. The English language has had a civilizing effectin Africa.

B. In the debate about language use in African lit-erature, no intermediate stance is possible.

C. The colonial era stifled the production ofAfrican literature.

D. The opinions of African readers have not beenconsidered in this debate.

108. Achebe and Ngugi would be most likely to agree onwhich of the following points?

A. An author’s primary responsibility is to the peo-ple of his home community.

B. Anything that characterizes a contemporaryAfrican nation can be considered truly African.

C. Languages cannot be considered independentlyof social and political context.

D. The choice of a language for literature can bemade for purely expressive reasons.

109. According to the passage, European colonialism hadall of the following effects on literature in AfricaEXCEPT:

A. an increase in literature written in African lan-guages.

B. an increase in literature written in European lan-guages.

C. the introduction of written literature to Africa.

D. the new self-identification of writers as“African.”

110. According to the passage, the linguistic situation ofNigeria is best characterized by:

A. several important local languages all of whichalso serve as official languages.

B. many local languages, with English as the offi-cial state language.

C. universal reliance on English for everydayaffairs as well as political life.

D. contention between those who use Europeanlanguages and those who use African languages.

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Passage VI (Questions 111-118)

It is almost a truism that humans are the only animalswith genuine language. But in recent years, observations ofthe great apes in captivity have shown remarkable capaci-ties for language in non-human primates. These studiessuggest that humans are not so different from other ani-mals as we believe.

Though many animals communicate with each other,their communication systems are much more limited thanhuman language. Wild chimpanzees use a kind of commu-nication known as a call system. Call systems are com-posed of discrete units or calls, each of which has aparticular meaning, such as “food” or “danger.” The callsare produced as automatic responses to certain stimuli; achimpanzee cannot discuss food after eating, or talk aboutwhat he will eat tomorrow. Nor can he “trick” his troop-mates by using the call for food if there is none present.The calls cannot be modified or combined. If a chim-panzee encountered danger and food together, he wouldnot be able to convey both meanings through calls.

All human languages display several features not pre-sent in call systems. The most essential aspect of languageis symbolism: sounds and characters used in words have anarbitrary (but conventional) connection to the things theyrepresent. Additionally, languages display productivity,meaning that speakers can combine words or other sym-bols to form new meanings or “coin” new words. Lan-guages also involve displacement. We can talk about thingsor people that are not present, or that do not even exist.Finally, languages are shared by a community and trans-mitted through learning. The ability to use these features oflanguage suggests a qualitative difference between humanlanguages and primate call systems.

Several chimpanzees and gorillas have been taught lan-guages or symbolic systems in captivity. Although non-human primates lack the anatomical structures that enablespoken language, several researchers have taught apesvisual languages including American Sign Language(ASL). Primatologist Roger Fouts taught ASL to a chim-panzee named Lucy. Lucy not only was able to discussitems she couldn’t see, but she could also lie. When Foutsasked her one day who was responsible for dirtying herroom, Lucy blamed another human. Psychologist PennyPatterson worked with a gorilla named Koko, who learnedASL (or an adapted version called “Gorilla Sign Lan-guage”). Koko could combine previously learned signs toproduce new meanings, such as “finger bracelet” to refer toa ring. Lucy, Koko, and other apes who have learned ASLhave attempted to teach the language to other apes, such astheir babies or other captive apes.

A later researcher named Herbert Terrace attempted todiscredit these studies, claiming that the apes had merelybeen mimicking their trainers. Terrace taught anotherchimpanzee isolated ASL signs, to see if the ape couldlearn to combine them on his own. Terrace and otherhumans who interacted with this chimpanzee did not con-verse in ASL in his presence because this would suppos-edly enable him to copy them. Unlike Lucy and Koko, thischimpanzee did not learn language in any meaningfulsense. The results of this experiment actually support thenotion that apes can learn language as humans do, how-ever, since we learn increasingly complex linguistic struc-tures through observing and interacting with other humans.A human child exposed only to isolated words would prob-ably not learn language either.

The differences between humans and apes in the use oflanguage are outweighed by similarities in linguistic abil-ity. Although apes have used language only in captivity,their use of language draws on naturally present cognitivecapacities. All true students of humanity must admit thatthe great apes differ from humans only in the extent towhich they have used language; but the ability is present.

111. Imagine that a researcher discovered a group of wildchimpanzees who could combine calls, so that, forexample, they could tell their troop-mates that dangerand food were both present if a predator was near afood source. The author of the passage would proba-bly explain this new discovery as evidence that:

A. those chimpanzees had been in contact withhumans at some point.

B. chimpanzees’ linguistic abilities are part of theirbiological inheritance.

C. that group of chimpanzees had evolved morethan others.

D. call systems are true languages.

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112. According to the passage, the ability of Lucy to lie toRoger Fouts provides an example of linguistic:

A. productivity.

B. displacement.

C. insolence.

D. transmission.

113. Which of the following provides an example of sym-bolism, as defined in the passage?

A. Stop signs in North America are octagonal inshape, but in Europe stop signs are round.

B. The word “meow” is a transliteration of theactual sound that a cat makes.

C. On a mercury thermometer, a longer stripe ofmercury indicates a higher temperature.

D. Young children’s scribbles are often unrecog-nizable as the thing or person that the child isdrawing.

114. Which of the following accords best with the author’sview about call systems and languages?

A. Call systems are essentially the same as humanlanguages, only simpler.

B. True languages evolved out of call systems.

C. The differences between call systems and lan-guages are greater than the similarities.

D. The similarities between call systems and lan-guages are greater than the differences.

115. Why does the author include information about Ter-race’s research?

A. To disprove the theory that all chimpanzeeshave the ability to learn language

B. To criticize Terrace’s research methodology

C. To draw an analogy with children who fail tolearn language

D. To show that a potential weakness of theauthor’s argument is in fact a strength

116. The author would be LEAST likely to agree withwhich of the following statements?

A. Humans are not the only animals that are capa-ble of learning and using true language.

B. Humans and chimpanzees differ in the degree towhich they can learn and use language.

C. Humans have the ability to learn language, butthey require input from other humans in order tolearn it.

D. Human language is only one of many kinds oflanguage existing on earth.

117. The passage implies that which of the following con-ditions is necessary for an organism to learn lan-guage?

I. Childhood involvement with users of a lan-guage

II. A biologically-based cognitive propen-sity for language

III. A vocal apparatus capable of formingspeech sounds

A. II only.

B. I and II only.

C. I and III only.

D. I, II and III.

118. According to the information provided in the pas-sage, call systems and language are similar in that:

A. both are used by non-human primates in thewild.

B. both involve an arbitrary association betweensymbols and the things they represent.

C. both are made up of discrete units that cannot becombined.

D. both are natural systems used by primates forthe purpose of communication.

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Passage VII (Questions 119-124)

Imagine this scenario: a business owner is ordered topay several million in damages in a federal court. Theowner of the business moves to another state and immedi-ately purchases a house worth 2 million (or more) in cash;she then declares bankruptcy and the inability to pay thecourt ordered settlement. In the meantime, she takes a job,or lives off her pension and/or her IRA in comfort or evenluxury, and ignores the demands for payment from her dis-tressed creditors. She can take out a home-equity loan forany short-term emergencies. Alternatively, if she lives in ahouse with a guest house, she may rent out the guest housefor additional income – income that again remains safefrom her creditors. Myth? No, reality – if the said housewas built in Florida.

Florida’s Homestead law, passed in the early twentiethcentury, and seemingly invulnerable to change, protectsall in-state primary residences. As defined by the state leg-islature, a homestead consists of 160 contiguous acres ofland outside municipal areas, or one-half of an acre of landinside municipal areas and any buildings or improvementsupon that land. Homeowners can claim homestead exemp-tions on only one home; second homes within the state,whether used as rental properties or weekend residences,are treated as normal real property assets. The homesteadcarries several advantages. First, Florida tax law provideshomestead owners with a property tax exemption on thehomestead ($25,000) reducing property tax to nearinsignificant levels for many households. Second, Floridalaw ensures that no homeowner can lose title to the home-stead, even when declaring bankruptcy. While certainliens, including those for state and municipal taxes, mort-gages, and homestead improvements, remain valid, noother creditors may seize or force the sale of the homesteadproperty to collect moneys. Nor may household items,including furniture and utilities, be seized. Thus, all moneyinvested in the homestead remains safe from creditors.This remains true even if creditors can prove that moneywas shifted into the homestead in an attempt to shieldassets from creditors, either through direct purchase, orthrough transferring assets to pay off a mortgage.

In addition, Florida has no income tax. (The statereceives most of its tax revenue from taxes on other kindsof property, from sales tax, and from hotel taxes drawnfrom the tourist trade.) Nor may creditors attach debtors’wages; debtors may continue earning salaries and pensionswith no fear of court-ordered wage deductions. While thecombined laws serve to shelter numerous poor and work-ing class people from loss of home and utter financial dis-aster, unsurprisingly, others have come to use Florida’sliberal laws as a means to avoid paying debts or court-

ordered settlements. In general, creditors remain powerlessagainst this.

Individual state laws, as protected by the TenthAmendment, have remained sacrosanct in most U.S.Supreme Court rulings. But the growing rush of high-pro-file cases of people fleeing to Florida and sheltering assetsin homesteads has triggered a strong reaction from credi-tors, who are currently pressuring the Senate to changefederal law and prevent debtors from making these flights.The proposed federal law, however, raises numerous issuesabout the relationship of state and federal law. If, as manyargue, state law retains precedence in issues of real prop-erty, then the Senate proposal, if passed, may well prove tobe unconstitutional. Yet the Senate remains the best hopeof creditors. Florida receives an excess of tax revenue,making changes in the property tax laws unlikely—even ifFlorida voters, the vast majority of whom directly benefitfrom the Homestead law, could be persuaded to entertainsuch a change.

119. If a debtor seeking relief from creditors (other thanmunicipal or mortgage creditors) were to purchase atwo-acre estate in a municipal locality in Florida,based on passage information it would be most rea-sonable to expect that:

A. the creditors could force the sale of the one anda half acres of land not included in the half-acreportion of the homestead.

B. the estate would not be covered under the newlegislation being considered by the Senate.

C. the debtor would not be able to find relief fromthe creditors on the two acre estate.

D. only one-half acre of the estate would beimmune from the creditors.

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120. Which of the following assertions, would mostweaken the author’s argument that Florida voters areunlikely to change the current Homestead law?

A. Many Florida voters are actually creditors them-selves, and need a way to recover damages.

B. The Florida legislature would compensatehomestead owners for any added taxes incurredbecause of changes in the tax law.

C. The Florida legislature would, in exchange forchanges made to the homestead law, createother shelters for Florida residents seekingrefuge from creditors.

D. The state faces an unexpected shortage of taxrevenues thanks to a declining tourist trade, andneeds to raise funds quickly to provide neededstate services to voters.

121. Passage information indicates that, if a homesteadowner in Florida fails to pay for home improvementsupon the homestead property, the person(s) responsi-ble for the improvement may:

A. not force the sale of the homestead or a portionof the homestead in order to pay the debt on thehome improvement.

B. attach the wages of the homestead owner to col-lect the debt.

C. place a lien upon the homestead or a portion ofthe homestead.

D. demand payment from the homestead owner’sother assets.

122. It can be inferred from the passage that IRAs andcompany pensions:

A. provide a stable source of income for all debtorsseeking refuge from creditors.

B. may be used to make home improvements uponthe homestead.

C. are exempt from income tax.

D. cannot be attached by creditors seeking restitu-tion for debt.

123. The author would be LEAST likely to agree withwhich of the following opinions?

A. Some debtors may need the protection ofFlorida’s Homestead and tax laws.

B. Voters are unlikely to vote against what theyperceive as their best financial interests.

C. In cases where federal and state laws come intoconflict, the state law should always take prece-dence.

D. The high-profile cases of debtors fleeing toFlorida may be encouraging others to followtheir examples.

124. The author probably mentions the poor and workingclass residents of Florida in order to:

A. defend a portion of the Florida Homestead Law.

B. contrast them to poor and working class resi-dents of other states who lack the protection ofthe Florida Homestead Law.

C. explain why the Florida Homestead Law isunlikely to be overturned by the Florida legisla-ture.

D. give further examples of debtors unfairly hidingfrom creditors.

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Passage VIII (Questions 125-130)

Historians have long debated whether the Roman his-torian Tacitus wrote The Histories prior to composing hismasterpiece, The Annals of Imperial Rome, a work cover-ing a time period that ends shortly before The Historiestakes up the tale. Certainly, structural similarities suggestthat Tacitus wrote the books nearly contemporaneously.The books each contain a vivid and exciting narrative line,striking character portraits created with pithy statements, adecided bent for racy anecdote, and a strong antipathytowards the imperial system. Both books do not hesitate topass moral judgements upon historical figures, nor, for thatmatter, to paint highly unflattering portraits of imperialrulers such as Nero and Domitian. The books also make anattempt at historical objectivity, unusual in ancient texts. InThe Histories, for example, Tacitus manages some faintpraise for the rather worthless Vitellus, and in The Annals,he admits that the hated Tiberius did accomplish some use-ful deeds. Both books also glitter with the irony for whichTacitus became renowned. But with these, the similaritiesend. A close study of the differences between the worksstrongly suggests that The Annals is the later, more bitter,and more mature work – if not necessarily the better writ-ten.

Perhaps the most immediate and notable difference liesin the narrative format of both works. In The Histories,Tacitus occasionally interrupts his narrative line forlengthy discourses on the nature of men and the universe,or for multiple moral denunciations or approbations of thevarious emperors of the work. The more mature Tacituseschewed these digressions, instead allowing his tale toestablish these same points. In part, this simplicity maystem from the nature of his subjects in The Annals: theactions of such figures as Agrippina and Nero certainlydemonstrate the nature and corruptive influence of unlim-ited power, making further comment nearly futile. Thecharacters in The Histories never sank to this level ofmoral turpitude, and thus may have invited commentary,since their own characters could not make as strong a pointabout corruption. The earlier Tacitus would spend twoparagraphs calling Vitellus useless; the later Tacitusavoided this tactic, and instead simply pointed out howNero concentrated on writing terrible poetry whenevercritical decisions had to be made, sparing readers any fur-ther moral commentary.

It should not be thought that The Annals lack all narra-tive digression. It does not, but such digressions take a verydifferent format. The first type of digression in The Annalsconsists of lengthy (and doubtless fictionalized) deathbedscenes; the second type is caused by Tacitus’ insistenceupon retaining a rigid year by year narrative format in the

Annals, and by his discussion of world affairs, not simplymatters pertaining to the Roman Empire. The Annals dis-cusses the affairs of Britain and Gaul at length; The Histo-ries, in contrast, gives the lengthy saga of the Jewish revoltonly in brief, and only as part of the biography of Titus.

Nonetheless, the differences between the books shouldnot be overstated. The Tacitus of The Annals had gained,perhaps, a better understanding of the importance of grainshipments, and lost, perhaps, what little remained of hisgood opinion of human nature. But the highly intelligent,ironic Tacitus deploring the condition of the RomanEmpire during the Year of the Four Emperors is little dif-ferent than the later Tacitus still more forcibly deploringthe condition of the Roman Empire under Nero andTiberius. For Tacitus, in both works, the issue remained thesame: the inevitable corruption of humanity.

125. The author’s chief argument would be most weak-ened by evidence that:

A. a rigid year by year narrative format is actuallya more mature form of historical writing.

B. moral digressions in historical works form anessential part of their innate narrative structure.

C. historical works that focus on an overall world-view are less intriguing than works with a nar-row focus.

D. the ability to compose lengthy moral commen-taries stems from maturity, and rarely appears inearly writings.

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126. Which of the following would the author of the pas-sage be most likely to consider a mature work of his-tory?

A. A study of the French Revolution that narratedevents in a year by year format

B. A biography of Joseph Stalin that presentedanecdotes from the dictator’s life without fur-ther comment

C. A history of the American Revolution thatplaced its events in the context of other eventson the world stage

D. An analysis of the role of trade restrictions inthe decline of Imperial China

127. It can be inferred from the passage that the authorbelieves that, compared to Nero, the emperor Vitel-lus:

A. did not present as compelling a figure of moraldepravity.

B. had a less pernicious effect upon the RomanEmpire.

C. did little to assist the growth and power of theRoman Empire.

D. did not waste his time writing bad poetry.

128. The passage suggests that the Roman historian Taci-tus would be LEAST likely to agree with which ofthe following viewpoints?

A. Historical works should be filled with colorfulanecdotes illustrating the character of the histor-ical personas under discussion.

B. As the history of the Roman Empire progressed,its rulers became less corrupted by power.

C. The writing of poetry by Roman emperors mayindicate a less than devoted passion to theiractual jobs.

D. The emperors Nero and Tiberius may serve asemblems of human corruption.

129. Passage information would indicate that ancient his-torians, when writing about the Roman Empire:

A. chose to focus on figures renowned for theirmoral depravity.

B. largely focused on the affairs of the city ofRome, and not on events taking place in thegreater Empire.

C. rarely strove for historical objectivity, insteaddelivering their opinions upon the historical fig-ures they discussed.

D. used either a rigid year by year narrative format,or else a narrative format that followed histori-cal figures from birth until death.

130. Suppose that further writings by Tacitus were to bediscovered that were more bitter in tone than those ofThe Annals, contained multiple moral digressions,and lacked a rigid year by year format. The authorwould most likely:

A. claim that the newly discovered writings werewritten after The Annals.

B. claim that the newly discovered writings werewritten before The Histories.

C. use the newly discovered writings to help dateThe Histories and The Annals more precisely.

D. be unable to give a precise date to the composi-tion of the newly discovered writings.

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Passage IX (Questions 131-137)

Traditional science and philosophy have been built onan argument spiritual and scientific in equal measure.Philosophers have argued that there is a complete andunbridgeable separation between the nature of divinity andthe nature of the creations springing from that divinity.There is, they argue, division between soul and matter, andbetween the thought of an object and the actual objectitself. This dualistic argument has been held in roughlyequal balance by opposing sides that are in turn philo-sophical and empirical. In one era it leans to the side ofexperience and solidity, and in another to the less tangibleand immeasurable essences of the mind and the spiritualessence of the being that seeks to comprehend the object.On one side stands Immanuel Kant with his claim that theintangible soul is the cornerstone of existence, elevatingthe ego to a position of privilege over the material. On theother side are the Socratics with their claim that the onlyacceptable intangible is that the mind can know nothing onits own but must rely upon what it sees. Both extremesassume, however, the same idea of two essences, the mate-rial and the spiritual.

A great number of philosophers past and present con-tend that the understanding of an object is separate fromthe object itself. They argue that it would be wrong toclaim that the image of a fountain one has in mind is thesame thing as the fountain itself, or that the idea of “foun-tain” that the mind builds up after seeing thousands offountains can be linked to any fountain existing in actualreality. This Platonic argument would suggest that theremust be a gap between what is real and what is perceived.

Some would object, however, that this difference infundamental essences is not an actual difference at all, butis rather a flaw in the method of perception itself. Thephilosopher William James in his Essays in RadicalEmpiricism champions this materialist view and arguesthat the very idea of consciousness, of the ability to per-ceive a certain thing, is a useful but immaterial concept.James regards it as an arbitrary creation that must be rec-ognized as a model rather than an entity in and of itselfbefore the broader problem of dualistic philosophies canbe addressed.

James argues from the supposition that rather than twotypes of matter, one from which springs the spiritual, andone the material, there is only one type of substance thatexplains the same actions dualism explains but with moreefficiency. If there is only one substance from which every-thing springs, the philosopher argues, that realm of experi-ence called consciousness by dualistic philosophers can beexplained simply as an interaction between the objects per-

ceived and the one perceiving. When someone pushes anobject from one place to another, for example, one does notregard the act of pushing as separate from either the persondoing the pushing or the object being pushed. Rather, it isa junction between the two objects. It is a description ofthe two materials’ behaviors in relation to one another,with no need to call in a separate type of existence toexplain the action. James argues that the same is true ofconsciousness: It is the interaction of the individual andwhat he perceives rather than a separate and distinct entityunto itself.

Critics of James, however, raise at least two funda-mental objections to his argument. First, James seems toengage in circular reasoning when he argues from the start-ing point that only one type of matter exists. Secondly,those challenging James’ approach will point to the possi-bility that the interaction of two difference objects may inand of themselves create something new and entirely dif-ferent.

131. Which of the following is a claim made in the pas-sage but NOT supported by evidence, explanation, orexample?

A. Interaction between objects of the same materialmay result in the creation of something with adifferent nature

B. The understanding of a physical object can beseparated from the object that is being under-stood.

C. James presents arguments that may be open tovalid challenge on logical grounds.

D. What is known as “consciousness” can beexplained by interactions between individualsand physical objects.

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132. The author of the passage would be most likely toagree with which of the following statements?

A. The universe can be clearly delineated into twocategories of existence.

B. The idea of consciousness is an outdated con-cept that should be removed from modern phi-losophy.

C. Abstract ideas can provide insight into actualhuman practice.

D. Philosophies of existence can only be chal-lenged with practical example.

133. Which of the following statements, if true, wouldmost challenge the argument set forth by WilliamJames as described in the passage?

A. Children’s concepts of right and wrong can bealtered by exposure to some computer games.

B. A theory of consciousness that does not involvea spiritual element is more complicated than onethat separates consciousness and objects.

C. Individuals can lose self-awareness if certainareas of the brain are surgically removed.

D. Actions cannot be separated from the objectsthat they involve.

134. Suppose that a certain surgery is discovered to elim-inate the sense of right and wrong while retainingself-consciousness. Based on the information in thepassage, this discovery would:

A. support James’ theory.

B. support the dualistic theory.

C. support both theories.

D. support neither theory.

135. According to the author, implicit in the dualistic viewof consciousness is the assumption that:

A. The awareness of self is critical to humannature.

B. Self-consciousness is present only in animalswith spiritual capacity.

C. The existence of intelligence is essential to theexistence of objects.

D. Matter can exist in only two forms.

136. The passage implies that a philosophy built aroundthe idea of a strictly material universe:

A. precludes the idea of human consciousness.

B. assumes the existence of two types of matter.

C. may assume what it sets out to prove.

D. strengthens the arguments set forth by Kant.

137. The author most probably cites the Socratics in line17 in order to:

A. support the materialist view that there is no spir-itual element.

B. demonstrate the disparity between modern andancient philosophies.

C. provide an example of a dualistic philosophy.

D. counter James’ theory of a single essence.

STOP. IF YOU FINISH BEFORE TIME IS CALLED,CHECK YOUR WORK.YOU MAY GO BACK TO ANYQUESTION IN THIS SECTION ONLY.

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Writing SampleTime: 60 Minutes

2 Prompts, Separately Timed:30 Minutes Each

DO NOT BEGIN THIS SECTION UNTIL YOU ARE TOLD TO DO SO.

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WRITING SAMPLE

DIRECTIONS: This section is a test of your writing skills. Thesection contains two parts. You will have 30 minutes to completeeach part.

Your responses to the prompts given in the Writing Sample willbe written in the ANSWER DOCUMENT.Your response to Part 1must be written only on the answer sheets marked “1,” and yourresponse to Part 2 must be written only on the answer sheetsmarked “2.” You may work only on Part 1 during the first 30 min-utes of the test and only on Part 2 during the second 30 minutes.If you finish writing on Part 1 before the time is up, you mayreview your work on that part, but do not begin writing on Part 2.If you finish writing on Part 2 before the time is up, you mayreview your work only on Part 2.

Use your time efficiently. Before you begin writing a response,read the assignment carefully and make sure you understandexactly what you are being asked to do. You may use the spacebelow each writing assignment to make notes in planning yourresponses.

Because this is a test of your writing skills, your response to eachpart should be an essay composed of complete sentences andparagraphs, as well organized and clearly written as you canmake it in the allotted time. You may make corrections or addi-tions neatly between the lines of your responses, but do not writein the margins of the answer booklet.

There are six pages in your answer booklet to write yourresponses, three pages for each part of the test. You are notrequired to use all of the pages, but to be sure that you haveenough space for each essay, do not skip lines.

Essays that are illegible cannot be scored. In addition, essaysthat are not written in English will not be scored.

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Part 1Consider the following statement:

An item may be in the news only because nothing else happened that day.

Write a unified essay in which you perform the following tasks. Explain what you think the above statement means. Describea specific situation in which an item may be in the news although other events happened that day. Discuss what you thinkdetermines whether an item may be in the news.

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Part 2Consider the following statement:

Employees have a right to privacy in their workplace.

Write a unified essay in which you perform the following tasks. Explain what you think the above statement means. Describea specific situation in which employees do not have a right to privacy in their workplace. Discuss what you think determineswhether employees have a right to privacy in their workplace.

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Biological SciencesTime: 100 MinutesQuestions 138-214

DO NOT BEGIN THIS SECTION UNTIL YOU ARE TOLD TO DO SO.

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52

BIOLOGICAL SCIENCES

DIRECTIONS: Most of the questions in the BiologicalSciences test are organized into groups, with adescriptive passage preceding each group of ques-tions. Study the passage, then select the single bestanswer to each question in the group. Some of thequestions are not based on a descriptive passage; youmust also select the best answer to these questions. Ifyou are unsure of the best answer, eliminate thechoices that you know are incorrect, then select ananswer from the choices that remain. Indicate yourselection by blackening the corresponding circle onyour answer sheet. A periodic table is provided belowfor your use with the questions.

1

H

1.0

2

He

4.0

3

Li

6.9

4

Be

9.0

5

B

10.8

6

C

12.0

7

N

14.0

8

O

16.0

9

F

19.0

10

Ne

20.2

11

Na

23.0

12

Mg

24.3

13

Al

27.0

14

Si

28.1

15

P

31.0

16

S

32.1

17

Cl

35.5

18

Ar

39.9

19

K

39.1

20

Ca

40.1

21

Sc

45.0

22

Ti

47.9

23

V

50.9

24

Cr

52.0

25

Mn

54.9

26

Fe

55.8

27

Co

58.9

28

Ni

58.7

29

Cu

63.5

30

Zn

65.4

31

Ga

69.7

32

Ge

72.6

33

As

74.9

34

Se

79.0

35

Br

79.9

36

Kr

83.8

37

Rb

85.5

38

Sr

87.6

39

Y

88.9

40

Zr

91.2

41

Nb

92.9

42

Mo

95.9

43

Tc

(98)

44

Ru

101.1

45

Rh

102.9

46

Pd

106.4

47

Ag

107.9

48

Cd

112.4

49

In

114.8

50

Sn

118.7

51

Sb

121.8

52

Te

127.6

53

I

126.9

54

Xe

131.3

55

Cs

132.9

56

Ba

137.3

57

La *

138.9

72

Hf

178.5

73

Ta

180.9

74

W

183.9

75

Re

186.2

76

Os

190.2

77

Ir

192.2

78

Pt

195.1

79

Au

197.0

80

Hg

200.6

81

Tl

204.4

82

Pb

207.2

83

Bi

209.0

84

Po

(209)

85

At

(210)

86

Rn

(222)

87

Fr

(223)

88

Ra

226.0

89

Ac †

227.0

104

Rf

(261)

105

Ha

(262)

106

Unh

(263)

107

Uns

(262)

108

Uno

(265)

109

Une

(267)

*

58

Ce

140.1

59

Pr

140.9

60

Nd

144.2

61

Pm

(145)

62

Sm

150.4

63

Eu

152.0

64

Gd

157.3

65

Tb

158.9

66

Dy

162.5

67

Ho

164.9

68

Er

167.3

69

Tm

168.9

70

Yb

173.0

71

Lu

175.0

90

Th

232.0

91

Pa

(231)

92

U

238.0

93

Np

(237)

94

Pu

(244)

95

Am

(243)

96

Cm

(247)

97

Bk

(247)

98

Cf

(251)

99

Es

(252)

100

Fm

(257)

101

Md

(258)

102

No

(259)

103

Lr

(260)

PERIODIC TABLE OF THE ELEMENTS

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Passage I (Questions 138–143)

A chemist performs several experiments to determinethe identity of an unknown alcohol, A, which has a molec-ular formula of C6H14O.

Experiment 1Reaction of A with sulfuric acid (H2SO4) gives a mix-

ture of products which are analyzed by gas chromatogra-phy. The area for the three peaks obtained from thechromatogram indicates a mixture of products containing63% trans-4-methyl-2-pentene, X, 30% cis-4-methyl-2-pentene, Y, and 7% 4-methyl-1-pentene, Z.

Experiment 2The chemist reacts compound A with hydrochloric acid

which gives compound B having a molecular formula ofC6H13Cl. The same mixture of alkenes (X, Y, Z) isobtained upon reaction of compound B with potassiumhydroxide. However, the areas obtained by gas chromatog-raphy are different: 46% trans-4-methyl-2-pentene, 30%cis-4-methyl-2-pentene and 24% 4-methyl-1-pentene.

Experiments 1 and 2 are summarized in Table 1

Table 1Product Distribution

Experiment Reagents % X % Y % Z1 A + H2SO4 63 30 72 B + KOH 46 30 24

Experiment 3To determine the structure of A, the chemist oxidizes

the unknown alcohol to the ketone, C, having a molecularformula of C6H12O, by reacting it with bleach (NaOCl).

The boiling point of the unknown alcohol, A, is deter-mined to be 117°C and the refractive index is 1.3962. Fig-ure 1 summarizes the reactions of unknown alcohol, A,performed by the chemist.

Figure 1

C6H12O

C

NaOCl CH3COOH

OHHCl

C6H14Cl

A

X

Y

Z

+

+

B

C6H13Cl

Cl

O

HClKOH

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138. The difference in composition of the 3 alkenesbetween the two elimination reactions (reaction of Awith H2SO4 versus reaction of B with KOH) isbecause:

A. reaction with H2SO4 proceeds via the E1 mech-anism while reaction with KOH proceeds viaE2.

B. reaction with H2SO4 proceeds via the E2 mech-anism while reaction with KOH proceeds viaE1.

C. reaction with KOH proceeds via formation of acarbocation so the product ratio is favored ther-modynamically.

D. reaction with H2SO4 involves a concertedmechanism which leads to more of the less sub-stituted product.

139. Which of the following alkenes is obtained from thereaction of this dibromo compound with Zn inethanol?

A. trans-4-methyl-2-pentene

B. cis-4-methyl-2-pentene

C. 4-methyl-1-pentene

D. mixture of trans-4-methyl-2-pentene and cis-4-methyl-2-pentene

140. To determine the structure of the unknown alcohol,the student oxidizes it to the corresponding ketone:

In the NMR spectrum of the pure ketone, which sig-nal corresponds to c?

A. A singlet at 2.3 ppm

B. A doublet at 2.1 ppm

C. A multiplet at 0.9 ppm overlapped with a singlet

D. A multiplet at 2.3 ppm overlapped with a singlet

141. The best synthetic route to (CH3)3COCH3 is:

A. react (CH3)3COH with K, then add CH3I.

B. react (CH3)3COH with NaOH, then add CH3I.

C. react CH3OH with K, then add (CH3)3CI.

D. react CH3OH with NaOH, then add (CH3)3CI.

142. Which of the following is a significant differencebetween alcohols and ethers?

A. Hydroxide ions are better leaving groups thanalkoxide ions, so alcohols are prone to nucle-ophilic attack in the absence of strong acidwhile ethers are not.

B. Primary and secondary alcohols can be oxidizedto aldehydes or ketones while ethers cannot.

C. Alcohols can function as weak bases or nucle-ophiles while ethers cannot.

D. Alcohols have a hydrogen atom that may beremoved by acid while ethers do not.

143. Which of the following alkoxides is the most stable?

I. Cl3CCH2O-

II. F3CCH2O-

III. (CH3)3CCH2O-

IV. CH3CH2O-

A. I

B. II

C. III

D. IV

O

c

H

H

C2H5OH

Br

Br

Zn

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Passage II (Questions 144–149)

Epinephrine, more commonly known as adrenaline, isa neurotransmitter produced by the adrenal gland in theadrenal medulla. In humans, epinephrine is a stimulant ofthe sympathetic nervous system thereby acting to increaseheart rate, raise blood pressure, and suppress involuntarymuscle contractions of the digestive system.

The biosynthetic pathway of epinephrine can be seenbelow in Figure 1. The biosynthesis of epinephrine firstinvolves the hydroxylation of the corresponding aminoacid precursor, phenylalanine, by phenylalanine-hydroxy-lase to form tyrosine. Tyrosine is further hydroxylated to3,4-dihydroxyphenylalanine (L-DOPA) by tyrosinehydroxylase. L-DOPA is decarboxylated to dopamine. Athird hydroxylation yields norepinephrine. Methylation ofnorepinephrine’s amino group by S-adenosylmethionine(adoMet) produces epinephrine.

Figure 1 Biosynthetic pathway of epinephrine

The chemical synthesis of adrenaline, shown in Figure2, was first performed in 1904. When catechol is reactedwith chloroacetylchloride in the presence of phosphorusoxychloride and hydrochloric acid, 3,4-dihydroxy-chloroacetophenone is produced. Subsequent reaction withmethylamine produces adrenalone which is then reducedto yield a racemic mixture of adrenaline. Only the R enan-tiomer is biologically active.

Figure 2 Chemical synthesis of epinephrine

144. S-adenosylmethionine (adoMet) is a large, complexmolecule that acts as an electrophilic, methyl groupdonor in Step 5 of Figure 1. The structure of adoMetcan be seen below. What type of reaction mechanismtakes place during this particular step of epinephrinebiosynthesis?

A. Unimolecular nucleophilic substitution

B. Bimolecular nucleophilic substitution

C. Electrophilic addition

D. Electrophilic substitution

O

H

CH2

H

N

N

N

N

NH2

H

HO OH

H

S

CH3

CH2CH2CH

NH2

HOOC

S-Adenosylmethionine (adoMet)

HO

HOCl

Cl

O

POCl3 / HCl

HO

HO

Cl

O

H3CNH2 / HCl

HO

HO

HN

O

CH3

HO

HO

HN

OH

CH3

catechol

+

chloroacetylchloride

step 1

step 2

adrenalone

+ 2H+

adrenalin

step 3

3,4-dihydroxychloroacetophenone

COOH

NH2

COOH

NH2HO

COOH

NH2HO

HO NH2

HO

HO

NH2

HO

HO

OH

HN

HO

HO

OH

phenylalanine

step 1

step 2

L-DOPA

tyrosine

step 3dopamine

step 4

norepinephrine

step 5adoMet

CH3

epinephrine

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145. Upon hydroxylation of phenylalanine in Step 1 ofFigure 1, why does the hydroxyl group prefer to bepara to the alanine substituent on the benzene ring?

A. The alkyl component of alanine donates elec-tron density to the ring through inductanceresulting in the stabilization of the intermediateupon addition at the para position.

B. The amine functional group acts as a para direc-tor by donating a lone pair of electrons to thering through resonance stabilization of the inter-mediate.

C. The hydroxyl group acts as a para director bydonating a lone pair of electrons to the ringthrough resonance stabilization of the interme-diate.

D. The carboxylic acid functionality withdrawselectron density from the ring through both res-onance and inductive effects resulting inincreased stabilization of the intermediate uponaddition at the para position.

146. Which of the following reducing agents could beused to perform Step 3 in Figure 2?

I. Lithium aluminum hydride

II. Sodium borohydride

III. PCC (pyridinium chlorochromate)

A. I only

B. I, II and III

C. III only

D. I and II only

147. The IUPAC name for phenylalanine is:

A. 2-amino-3-phenylpropanoic acid.

B. 2-amino-3-propanoic benzene.

C. 1-phenylpropanamide.

D. 1-phenyl-2-aminopropanoic acid.

148. Which of the following structures best represents theintermediate formed upon reacting catechol withchloroacetylchloride in Step 1 of Figure 2?

149. What is the correct Fischer projection for R-epinephrine?

Ph

HO H

H

H N HCH 3

N HCH 3

H OH

Ph

H H

Ph

H N HCH 3

H

HO H

H OH

H

H NHCH3

PhA.

B.

C.

D.

HO

HO

H

O

Cl

HO

HO

Cl

Cl

OH

HO

HO

H

O

Cl

HO

HO

O

Cl

A.

B.

C.

D.

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Passage III (Questions 150-154)

Sensor-effector protein pairs are common in botheukaryotic and prokaryotic cellular signaling. These two-component regulatory systems allow cells to sense theirenvironment and internal metabolic state. One of the bestunderstood two-component systems is the signal transduc-tion pathway of bacterial chemotaxis, which is mediatedby a two-component sensory system that detects chemicalstimuli and transmits signals to the bacterium’s flagellarmotors. The direction of rotation of the motors determineswhether the cell swims smoothly or takes a tumble in itscourse. When a bacterium swims through an increasingconcentration gradient of an attractant (sugars, aminoacids, or small peptides), its flagella will have a smooth-swimming bias. The bacterium’s transmembrane chemore-ceptor proteins bind chemical attractants or repellentsoutside the cell and relay information about the bac-terium’s environment to a series of proteins inside the cell.

In the bacterium Escherichia coli, the CheA/receptorcomplex and CheY are the two major protein componentsof the signaling pathway for chemotaxis. CheA, a cyto-plasmic protein, forms complexes with the cytoplasmicportion of chemoreceptor molecules. The conformation ofthe transmembrane receptors when they are binding a lig-and affects CheA’s activity. CheA is a histidine autokinase,meaning that it can bind adenosine triphosphate (ATP) andthen phosphorylate itself at a histidine residue. An aminegroup on the histidine residue, the 48th residue from theamino terminus of CheA, serves as the substrate for the a-phosphate of ATP. ADP is released after the histidineresidue has been phosphorylated. CheA communicateswith another Che protein, CheY, by transferring phosphategroups to it. Phospho-CheY directly interacts with the bac-terium’s flagellar motors and induces a clockwise changein the direction of flagellar rotation, causing the cell totumble (Figure 1). The CheZ protein is a phosphatase forphospho-CheY, meaning that it hydrolyzes the phosphategroup from phospho-CheY.

Figure 1 Signaling pathway for chemotaxis

150. Amino acid side chains on the surface of CheA aremost likely:

A. charged or polar.

B. uncharged or non-polar.

C. histidine residues.

D. cysteine residues which readily interact withother proteins.

151. In order to inhibit bacterial chemotaxis in an infectedmammalian host, a drug would most likely affectwhich of the following cellular components?

A. The phospholipid bilayer

B. tRNA

C. Peptidoglycan cross-links

D. The mitochondria

152. A wild type strain of E. coli is streaked in an agarplate containing a concentration gradient of maltose.As the bacteria swim toward areas of high maltoseconcentration, which of the following occurs?

A. The relative activity of CheZ increases

B. CheA autophosphorylation decreases

C. The ratio of tumbling to smooth-swimmingincreases

D. The proportion of counter-clockwise flagellarrotation decreases

extracellular space

cytosolreceptor

CheW

CheA

CheA~P CheY~P

CheYCheZ

Pi

ATP clockwise = tumble

counter-clockwise = forward

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153. A mutant CheA with a greater affinity for ATP might:

A. create a stronger complex with the transmem-brane receptors.

B. create a weaker complex with the transmem-brane receptors.

C. increase the response of bacteria to externalstimuli.

D. decrease the response of bacteria to externalstimuli.

154. Which of the following cellular processes is mostsimilar to the signaling system in bacterial chemo-taxis?

A. Regulation of the glycolytic enzyme, phospho-fructokinase

B. Activation of a helper T-cell in response to aninfectious antigen

C. Stimulation of a G-protein by the hormoneepinephrine

D. The troponin/tropomyosin response to an influxof calcium ions

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Passage IV (Questions 155–160)

In the kidney, the force that drives filtration at theglomerulus is the hydrostatic pressure generated by theheart. As in all capillary beds, the pressure of the bloodentering the permeable capillary causes the filtration ofwater and small molecules until the osmotic pressure cre-ated by the large molecules remaining in the blood is suf-ficient to resist the outward flow of fluid. The filtrate thatenters the tubule contains glucose, amino acids, ions, andnitrogenous wastes in the same concentrations as in theblood plasma, but it lacks plasma proteins. As this fluidpasses down the tubule, its composition changes as thecells of the tubule actively reabsorb certain molecules fromthe tubular fluid and secrete other molecules into it. Theperitubular capillaries supply the cells of the tubules withO2 and nutrients, bring to them the molecules to besecreted into the tubules and carry away the molecules thatare reabsorbed from the tubules back into the blood.

In order to assess the overall renal function, renal clear-ance of various substances is often measured. Clearance isdefined as the amount of plasma that must be cleared of asubstance to account for appearance of that substance infinal urine. A clearance of 100 ml/min. for substance x, forexample, indicates that 100 ml of plasma will be com-pletely cleared from substance x each minute. Clearance ofa substance x (Cx) can be expressed by the following equa-tion:

Cx =

where [Ux] is the concentration of substance x in urine, Vis the urine flow rate and [Px] is the concentration of sub-stance x in the plasma. If a substance is freely filtered bythe kidney, and is neither reabsorbed, secreted nor metab-olized, then clearance of that substance is equal to theglomerular filtration rate (GFR). Secretion of a substanceis indicated when the clearance of that substance is greaterthan GFR. Similarly, absorption of a substance is indicatedwhen clearance of that substance is lower than GFR.

155. Which of the following will reduce the amount of fil-trate that is entering the Bowman’s capsule from theglomerular capillaries?

A. Increased plasma protein concentration

B. Increased plasma glucose concentration

C. Elevated arterial pressure

D. Decreased plasma ion concentration

156. What percent of urea is reabsorbed from the filtrate ifurea concentration in the urine is 100 times greaterthan urea concentration in the blood? (GFR= 200L/day, urine flow rate = 1L/day)

A. 10%

B. 25%

C. 50%

D. 75%

157. Penicillin is a drug that can be filtered and secretedinto the tubular fluid. However, when the drugprobenecid is administered together with penicillin,the secretion of penicillin is markedly reduced.Which of the following is a true statement concern-ing clearance of penicillin?

A. The clearance of penicillin is increased in thepresence of probenecid.

B. The clearance of penicillin is decreased in thepresence of probenecid.

C. The clearance of penicillin is unaffected in thepresence of probenecid.

D. The clearance of penicillin cannot be deter-mined in the presence of probenecid.

[Ux]V�[Px]

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158. Creatinine is a muscle metabolite that is easily fil-tered across the glomerular capillaries. However, theamount of creatinine excreted by each nephron isgreater then the amount of creatinine that enters theBowman’s capsule from the glomerular capillaries.The process that accounts for the difference in crea-tinine amounts most likely involves:

A. reabsorption of creatinine at the proximaltubules.

B. secretion of creatinine obtained from the per-itubular capillaries.

C. synthesis of creatinine in the cells of the tubules.

D. breakdown of creatinine in the tubular fluid.

159. Mannose is a monosaccharide that is freely filteredacross the glomerulus into Bowman’s capsule and isneither reabsorbed, nor metabolized by the cells ofthe nephron. What will be the most likely effect ofintravenous mannose administration on the volumeof urine produced?

A. Urine volume will increase

B. Urine volume will decrease

C. Urine volume will remain the unchanged

D. The effect of mannose on urine volume cannotbe predicted.

160. Which of the following conditions will favor themovement of water, ions, and glucose from the inter-cellular space of the proximal tubule into the per-itubular capillaries?

A. High capillary hydrostatic and low capillaryosmotic pressures

B. High intercellular hydrostatic and low capillaryosmotic pressures

C. Low capillary hydrostatic and high capillaryosmotic pressures

D. Low intercellular hydrostatic and high intercel-lular osmotic pressures

Questions 161 through 164 are NOT basedon a descriptive passage.

161. What is the mechanism of action of tropomyosin?

A. Calcium binds to tropomyosin molecules tocause the tropomyosin molecules to shift andexpose the myosin-binding sites.

B. Calcium binds to tropomyosin molecules tocause the troponin molecules to shift and exposethe actin-binding sites.

C. Calcium binds to troponin molecules to causethe tropomyosin to shift and expose the myosin-binding sites.

D. Calcium binds to troponin molecules to causethe troponin molecules to shift and expose theactin-binding sites.

162. In a woman who is not pregnant, which of the fol-lowing is true with regard to estrogen and proges-terone levels immediately preceding the beginning ofmenses?

A. Estrogen and progesterone levels both decrease.

B. Estrogen levels decrease while progesteronelevels increase.

C. Estrogen levels increase while progesterone lev-els decrease.

D. Estrogen and progesterone levels both increase.

163. The compound shown below is an example of thebroad class of organic compounds known as steroids.

The functional groups present in the compound are:

A. alcohol, alkyne, ester

B. alkyne, carboxylic acid, ether

C. alcohol, alkene, ester

D. alcohol, alkyne, ether

CH3CO

H3C

H3COH

C CH

O

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164. Iron(III) bromide (FeBr3) is used as a Lewis acid cat-alyst in a particular reaction of benzene. This meansthat the iron atom:

A. is electron rich.

B. is able to act as an electron-pair donor.

C. can act as a Brønsted acid.

D. is electron deficient.

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Passage V (Questions 165-170)

An electrophilic addition reaction results in the con-version of one �-bond and one �-bond into two �-bonds.The result of this change is usually energetically favorablebecause the heat involved in making two �-bonds exceedsthat needed to break one �-bond and one �-bond. Additionreactions are therefore usually exothermic.

Figure 1 General Electrophilic Addition Reaction

Hydrogen halides, for example, react with alkenes bydonating a proton to the �-bond. The overall result is atwo-step process involving the formation of a carbocationand a halide ion from the alkene and the hydrogen halide.Being highly reactive, the carbocation then combines withthe halide ion by accepting one of its electron pairs. Theo-retically, the addition of H-X to an unsymmetrical alkenecould occur in two ways. In practice, however, one productusually predominates.

Markovnikov’s Rule: In the ionic addition of anunsymmetrical reagent to a double bond, the positive por-tion of the adding reagent attaches itself to a carbon atomof the double bond so as to yield the more stable carboca-tion as an intermediate.

A student performed experiments in order to determinethe structure of three unknown compounds, X, Y and Z.Each unknown is a derivative of C6. All are cyclic hydro-carbons with varying degrees of unsaturation.

Experiment 1:

Each unknown compound was treated with a solutionof Br2 in CH2Cl2 as follows:

Beaker 1 (Unknown X + Br2): Addition of 5 drops ofreagent resulted in immediate color change to red.

Beaker 2 (Unknown Y+ Br2): Color change resulted onlyafter addition of 20 drops of reagent.

Beaker 3 (Unknown Z + Br2): Addition of 5 drops ofreagent result in immediate color change to red.

At the conclusion of the experiment, the beakers wereplaced in a darkened fumehood. The next day, 2 of thebeakers were empty while the other contained a colorlessliquid.

Experiment 2:

Each beaker was treated with 5 drops of a solution ofHBr and the pH was tested with litmus indicator.

Beaker 1 (Unknown X + HBr): red litmus

Beaker 2 (Unknown Y + HBr): no color change

Beaker 3 (Unknown Z + HBr2): Addition of 5 drops ofreagent resulted in immediate color change to red.

A small amount of FeBr3 was added to Beakers 1 and3 and another litmus test was performed.

Beaker 1 (Unknown X + HBr + FeBr3): red litmus

Beaker 2 (Unknown X + HBr + FeBr3): no color change

Beaker 3 (Unknown X + HBr + FeBr3): no color change

165. If each unknown is represented by one of the follow-ing structures, what are the corresponding identitiesof compounds X, Y and Z?

A. X = I Y = II Z = III

B. X = II Y = III Z = I

C. X = I Y = III Z = II

D. X = III Y = I Z = II

I. II. III.

C C + X Y

C C

YX

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166. Following Experiment 1, the beakers were placed ina fumehood and left overnight. The next morning,two of the beakers were empty while the other con-tained a colorless liquid. What was the liquid?

167. In Experiment 2, what information does the litmustest give us?

A. Red color indicates acidic solution therefore noreaction occurred.

B. No color change indicates neutral solutiontherefore no reaction occurred.

C. Red color indicates acidic solution thereforereaction occurred.

D. No color change indicates neutral solutiontherefore reaction occurred.

168. According to Markovnikov’s Rule, addition of waterto 1-butene should give a:

A. primary alcohol.

B. secondary alcohol.

C. tertiary alcohol.

D. mixture of primary and secondary alcohols.

169. Addition of Br2 to cyclopentene would give a prod-uct which is:

A. achiral.

B. racemic.

C. meso.

D. optically active.

170. Which of the following addition reactions occurs thefastest?

A. I, because a primary cation is formed in the rate-determing step.

B. II, because the cation formed in the rate-deter-mining step is stabilized through an inductiveeffect.

C. III, because a secondary cation which is reso-nance-stabilized is formed in the rate-determin-ing step.

D. II, because a secondary cation which is reso-nance-stabilized is formed in the rate-determin-ing step.

HClCl

+ Cl

OCH3 ClHCl OCH3 +Cl

OCH3

HClCl

Cl+

I.

II.

III.

Br

Br

BrH

H

Br

Br

HH

Br

Br

A.

B.

C.

D.

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Passage VI (Questions 171–175)

Pure double-stranded DNA (dsDNA) can be unwoundand the strands separated using either heat or variouschemicals. The most common technique for measuring therelative amounts of double-stranded DNA as compared tosingle-stranded, DNA (ssDNA) is to use spectroscopy tomeasure how much ultraviolet light at a particular wave-length is absorbed by the DNA sample in question. Ultra-violet (UV) light is much more easily absorbed by ssDNAthan by dsDNA, because of the tight packing of the helixin the double-stranded form.

The melting point, or Tm, of dsDNA is the point atwhich the two strands separate. It is determined by a vari-ety of factors, including the relative percentages of purinesand pyrimidines, the length of the double helix beingunwound, and the ionic concentration of the solution inwhich the DNA sits. The length of the DNA strands usedis a particularly important factor, as longer strands add tocomplexity and allow more stability even at high tempera-tures. A common ion interfering with dsDNA denaturationor separation is sodium (Na+), which increases Tm as itsconcentration in the solution increases, similar to the boil-ing point elevation caused when sodium is added to water.

Renaturation, the opposite of strand separation, is alsoaffected by temperature, the concentration of ions in thesolution, and strand length; however, the most importantfactor seems to be the overall concentration of the nucleicacid itself. When the concentration of the two complemen-tary strands to be renatured is high, the formation ofdsDNA from ssDNA is much faster.

Scientists studied the denaturation of dsDNA froma certain organism using UV light to determine thedegree of strand separation at various temperatures. Theresults are shown in Table 1 and Figure 1.

Temperature Relative Absorbance(°C) (� = 260 nm)70 1.0075 1.0380 1.0885 1.2090 1.4095 1.41

Table 1

Figure 1

The scientists noted on the graph that the meltingpoint, Tm, of the organism’s dsDNA was approximately85°C, where about half of the DNA had denatured andbecome single-stranded. Thus, below Tm, the organism’sDNA was primarily dsDNA, while above Tm it was pri-marily ssDNA.

171. The organism’s DNA being studied by the scientistshas its most single-stranded character at which of thefollowing temperatures?

A. 70°C

B. 75°C

C. 80°C

D. 85°C

172. DNA with which of the following characteristicswould be most easily denatured?

A. Complementary strands with an equal composi-tion of A, C, T, and G base pairs

B. Complementary strands which are identical toone another

C. Complementary strands which have a higher A-T content than G-C content

D. Any DNA found in a sodium-rich solution

0.9

1.1

1.3

1.5

65 75 85 95

Temperature (°C)

Rel

ativ

e ab

sorb

ance

(�

=26

0nm

)

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173. In a second experiment with the same organism’sDNA, scientists use very long strands of DNA thatthey attempt to melt apart from each other. Which ofthe following values would be the best estimation forthe Tm of these new, longer strands?

A. 75°C

B. 80°C

C. 85°C

D. 95°C

174. The main difference between ssDNA and dsDNA isthat:

A. dsDNA has bases that include thymine (T),while ssDNA has bases including uracil (U).

B. dsDNA has strands oriented in an antiparallelfashion, while ssDNA does not.

C. dsDNA cannot undergo translation whilessDNA can leave the nucleus to undergo trans-lation.

D. dsDNA has 5’ to 3’ polarity while ssDNA doesnot.

175. Given the data in the experiment described in the pas-sage, increasing the temperature of a solution ofunwound (ss) DNA a few degrees would most likely:

A. separate the DNA even further into its single A,C, G, and T nucleotides.

B. decrease the Tm of the DNA after it renatures.

C. slow the renaturing process so that dsDNA takeslonger to form.

D. have no effect on the solution at all.

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Passage VII (Questions 176–180)

When studying the human respiratory system, scien-tists often measure how much oxygen is consumed in aparticular amount of time under a variety of resting orexercising situations. Oxygen consumption is known asVO2 and is measured in milliliters (mL) of oxygen con-sumed per minute breathing. Normal, healthy humans havea resting VO2 that of approximately 240 mL O2/min. Thereis a different maximal oxygen intake for every individual.

Scientists studied the VO2 of several groups of patientsunder a variety of conditions. A group of normal patientswas compared to a group of patients with a chronicobstructive disease such as emphysema, an obese group,and a group with congestive heart failure. All three of theseabnormal conditions are characterized by a lack of normalair movement through the lungs and, consequently, lessoxygenation for red blood cells throughout the body.

Emphysema patients have damaged their alveoli, oftendue to smoking, so that the alveolar tissue is destroyed andno longer attached to the capillaries that are meant to sur-round each of these small air sacs. Obese patients find ithard to expand their lungs due to interference from largeamounts of fat surrounding the lung tissues. Congestiveheart failure patients suffer from weakly contracting ven-tricles in their heart, causing fluid to back up and build upwithin the alveoli. All three groups experience vastlyincreased resistance within their airways, with air flowseverely restricted and blood oxygenation much poorerthan normal.

The scientists collected data comparing VO2 againstVE, a measure of how much air leaves the lung everyminute (Figure 1). They noticed that for the patients withabnormal lungs, the maximum rate of air leaving the lungswas reached very quickly as the patients breathed fasterand faster. As they expected, these patients were all trap-ping air in their lungs and having difficulty expelling it.Compared to the normal patients, those with increasedresistance in their airways were not refreshing the airwithin their lungs as quickly, as shown by the fact that asthey consumed more and more oxygen (VO2) they rapidlyreached their maximum volume of air expelled per minute(VE).

Figure 1

176. According to the data, heavy exercise in congestiveheart failure patients results in approximately howmuch air exhaled per minute?

A. 10 L O2 / min

B. 20 L O2 / min

C. 30 L O2 / min

D. 40 L O2 / min

177. Compared to normal patients, one would expect thatobese patients at rest would:

A. breathe faster, since it is easier for them to movelarge amounts of oxygen in and out of theirlungs.

B. breathe faster, since they quickly attain theirmaximum VO2.

C. breathe more slowly, due to increased airwayresistance and the large amount of work it takesto breathe.

D. breathe more slowly, because they consume alot of oxygen with minimal effort.

178. Air entering the lungs must pass through which of thefollowing structures before reaching the alveoli?

A. Parietal pleura

B. Larynx

C. Visceral pleura

D. Alveolar capillaries

200

300

400

0 20 40 60 80

VE (L/min)

VO

2 (m

L/m

in)

Normal

Congestiveheart failure

Obesity

Emphysema

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179. The residual volume (RV) is the amount of air left inthe lungs even after forcible exhaling, since some airis always left behind. Total lung capacity (TLC)includes both the residual volume as well as theamount of air that can be forcibly inhaled or exhaled.In emphysema patients, one would expect:

A. RV to increase and TLC to decrease.

B. RV to decrease and TLC to increase.

C. both RV and TLC to increase.

D. both RV and TLC to decrease.

180. Nervous control of respiration originates at the:

A. cerebellum.

B. brain stem.

C. pleura.

D. sinoatrial node.

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Passage VIII (Questions 181-185)

Cyclic ethers behave much like their straight-chaincounterparts simply because they both possess the sameether functional group. However, the three-membered ringcompounds, also known as epoxides, possess a uniquechemical reactivity. Epoxides are normally prepared in thelaboratory by treatment of an alkene with a peroxyacid:

Figure 1

An intramolecular Williamson Ether Synthesis canalso result in the formation of epoxides. This occurs whenthe nucleophilic alkoxide ion and the electrophilic alkylhalide are both located on the same molecule. The reactionproceeds by way of a typical SN2 mechanism and the bestresults are obtained using a primary alkyl halide, sulfonateor sulfate. The general reaction mechanism for theWilliamson Ether Synthesis can be seen below in Figure 2.Although the rate for this reaction is slightly faster than therate for the comparable intermolecular reaction, both reac-tions rely on an SN2 mechanism.

Figure 2

When comparing the rates of formation of variouscyclic ethers, it has been determined that the formation ofan epoxide occurs very quickly despite the high amount ofring strain. In fact, the preparation of oxacyclopropane by2-bromoalcohol occurs five times more quickly than theformation of the corresponding oxacyclopentane. This isprimarily due to entropic effects and the fact that the rela-tive proximity of the nucleophile and leaving group resultsin a relatively small transition state energy thus allowing arapid ring construction.

As shown in Figure 3, epoxides may be cleaved bytreatment with acid. This treatment occurs under muchmilder conditions than with typical ethers, however. Forexample, dilute aqueous acid at room temperature is suf-ficient to cause the hydrolysis of epoxides to 1,2-diols.This acid-catalyzed cleavage begins with the protona-tion of the ether oxygen by the acid catalyst. The epox-ide then undergoes backside attack by water at the mosthindered electrophilic carbon resulting in a trans-1,2-diol. This particular reaction exhibits characteristics ofboth SN1 and SN2 pathways. The transition state for theacid- catalyzed mechanism relies on SN2-like geometrywhile also possessing SN1-like carbocationic character.

Figure 3

Unlike typical ethers, epoxides may be cleaved by anucleophilic base such as an alkoxide or hydroxide ion (Fig-ure 4). This occurs through a typical SN2 mechanism wherethe nucleophilic attack occurs at the most accessible epox-ide carbon. Although an ether oxygen is a poor leavinggroup in an SN2 reaction, the elevated reactivity of the epox-ide is sufficient to allow cleavage by basic nucleophiles.

Figure 4

ROROH

C C OH +RO

C C

O

+ C C O-RORO

C C

O

+ H+

- H+C C

O

H

H O H

C CHO- H+

C C OHHOO H

R3

R2

H

Br

R1 OH

NaOHH2O

R3

R2

H

Br

R1 O-

O

C C

R3

R2

R1

H

+ H2O + HBr

C

C

HH

H H

OO R

OH

C C

H

H

H

H

O

+R OH

O

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181. Which of the following factors contributes to theincreased reactivity of epoxides?

I. Epoxides are unable to adopt a nonplanarconformation resulting in considerable tor-sional strain between neighboring C-Hbonds.

II. The orbitals involved in the bond formationof epoxides overlap in an angular fashionresulting in a weaker and a more reactivemolecule.

III. The bond angles of the epoxide ring are-compressed and therefore deviate greatlyfrom the normal tetrahedral value.

A. I and II only

B. I and III only

C. II and III only

D. I, II, and III

182. Which of the following reaction conditions wouldnot result in the formation of an epoxide?

A. I and II only

B. II and III only

C. II and IV only

D. III and IV only

183. Acid-catalyzed epoxide cleavage takes place by SN2-like attack of a nucleophile on the protonated epox-ide resulting in a trans-substituted product. Thismechanism is similar to the mechanism of which ofthe following electrophilic addition reactions?

184. Suppose one reacted butylmagnesium bromide withethylene oxide. The reaction mechanism might becharacterized by which of the following mechanisticpathways?

A. SN1 only

B. SN2 only

C. both SN1 and SN2

D. There would be no reaction.

185. When comparing the formation of oxacyclobutane tooxacyclopropane, one can predict that:

A. the rate of formation will increase due to anincrease in entropic effects.

B. the rate of formation will decrease due to anincrease in ring strain.

C. the rate of formation will increase due to adecrease in ring strain.

D. the rate of formation will decrease due to adecrease in entropic effects.

A.

B.

C.

D.

Br2

CCl4

H2

Pd /C

H

H H

H3C 1) Hg(OAc)2/ H2O

HCl

2) NaBH4

I.

II.

III.

IV.

Cl OH

O

C H2Cl2, 25oC

1)

Cl2

H2O

2)

NaOH

H2O

OH

Cl

NaOH

HOBr

NaOH

H2O

H2O

O

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Questions 186 through 190 are NOT basedon a descriptive passage.

186. Which of the following regarding somatostatin istrue?

I. Is increased by high blood glucose levels

II. Decreases insulin secretion

III. Decreases glucagon secretion

A. I only

B. II only

C. I and II only

D. I, II, and III

187. In a particular population, for a trait with two alleles,the frequency of individuals expressing the recessivephenotype is 0.36. What is the frequency of individ-uals expressing the homozygous dominant pheno-type?

A. 0.16

B. 0.6

C. 0.36

D. 0.64

188. Patent ductus arteriosus is a condition that resultsfrom a failure of the ductus arteriosus to close com-pletely after birth. Patent ductus arteriosus wouldresult in blood flow from the:

A. right atrium to the left atrium.

B. right atrium to the right ventricle.

C. liver to the inferior vena cava.

D. aorta to the pulmonary artery.

189. In a particular species of plant, the alleles for flowercolor are codominant. When a pure-breeding plantwith white flowers is crossed with a pure-breedingplant with red flowers, the resultant plants bear pinkflowers. If two plants with pink flowers are crossed:

A. 100% of the resultant plants will produce redflowers.

B. 100 % of the resultant plants will produce pinkflowers.

C. 50% of the resultant plants will produce redflowers, while the other 50% will produce whiteflowers.

D. 25% of the resultant plants will produce redflowers, 50% will produce pink flowers, and25% will produce white flowers.

190. The product of the following reaction will be:

A. optically pure and have the R configuration.

B. optically pure and have the S configuration.

C. a racemic mixture of enantiomers.

D. an achiral compound.

Cl

H OHNaI

?acetone

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Passage IX (Questions 191–195)

Some population ecologists maintain that almost allspecies fall neatly into one of two categories based on theirreproductive strategies and population growth: an r-selected category and a K-selected one. Scientists refer tor-selected species as those that reproduce very quickly,have many of offspring at once, and show little or noparental care for the offspring. Offspring of r-selectedspecies mature quickly and are able to reproduce relativelysoon after birth.

In contrast, K-selected species produce a few, large,well-developed young, who exhibit a lengthy and delayedmaturation and a delayed reproductive maturity. To com-pensate, parents of K-selected species usually remain withtheir young as they develop. The trade-offs in these twodifferent reproductive strategies, r-selection and K-selec-tion, are clear: lots of offspring, little parental care, andlow survival rate of offspring versus fewer offspring, moreparental care, and a greater chance of offspring survival.

Scientists studying a species of water python, Liasisfuscus, have found that the python exhibits distinct differ-ences in its reproductive timing simply as a result of smallchanges in its nest-site location. Female pythons use eithernest-sites with cooler temperatures and frequent tempera-ture variations or nest-sites with higher and more stabletemperatures. Pythons using the cool nests take longer toreproduce and, as a consequence, must wait up to twoyears before being able to reproduce again. Those in thewarmer nests reproduce more quickly, have offspringwhich survive better, and can reproduce again one yearlater. Yet, pythons laying eggs in the cooler nests remainwith their eggs till hatching, whereas pythons in thewarmer nest-sites generally desert their eggs within a fewdays of laying them. Scientists disagree about how to cat-egorize Liasis fuscus.

Hypothesis I

The water python, Liasis fuscus, should be classified astwo separate subspecies. The cool nest pythons, whichremain with their young and have fewer offspring becausethey can reproduce only every second year, should be clas-sified as a K-selected subspecies. The warm nest pythons,which reproduce more often and generally abandon theiryoung, should be classified as an r-selected species. Evi-dence suggests that the cool nest pythons reproduce slowlyover time and gradually plateau at a population numberappropriate for their habitat, while warm nest pythons areable to reproduce more quickly. This may be because thewarmer temperatures allow warm nest pythons to investless energy in reproduction than the cool nest pythons. As

a result, warm nest pythons rarely maintain a steady popu-lation number but instead demonstrate boom and bustcycles of birth and death. These are characteristics of K-selected and r-selected species respectively.

Hypothesis 2

The water python, Liasis fuscus, cannot be categorizedinto the conventional groups of r-selected or K-selectedspecies, for it does not fall neatly into either of these cate-gories. Both the cool nest and the warm nest pythonsexhibit qualities that show a mix between characteristicsnormally associated with only one or the other of thesepopulation categories. Evidence suggests that the pythonsvary their nest site placement according to the prevailingweather patterns and changing physical conditions in theirenvironment. In addition, the availability of food sources,such as rats, seems to help determine where the femaleslay their eggs during any one breeding season.

191. Some scientists have found that the type of nest sitechosen by the pythons is, in fact, genetically encodedfrom birth such that warm nest pythons producewarm nest progeny, and the cool nest pythons pro-duce cool nest progeny. This finding:

A. strengthens Hypothesis 1, because it suggeststhat the cool nest and warm nest pythons may begenetically distinct from each other.

B. weakens Hypothesis 1, because it suggests thatall pythons share similar genes for nest siteselection and should not be classified as sepa-rate subspecies.

C. strengthens Hypothesis 2, because it reinforcesthe idea that the pythons do not fit into the con-ventional r-selected or K-selected groups.

D. weakens Hypothesis 2, because it invalidates ther-selected and K-selected classificationschemes.

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192. Before the recent python studies, scientists believedthat maternal nest attendance, and the amount of timea mother spends guarding the nest and raising theyoung, did not vary across python species. This beliefsupports:

A. Hypothesis 1 only.

B. Hypothesis 2 only.

C. both Hypotheses 1 and 2.

D. neither Hypothesis 1 nor Hypothesis 2.

193. The main difference between the two hypothesesstated above is that:

A. Hypothesis 1 suggests that cool nest and warmnest pythons cannot be considered the samespecies, while Hypothesis 2 suggests thatbecause both types of python have the samefood sources, they should be considered as asingle species.

B. Hypothesis 1, if true, would explain how tem-perature differences in embryos could affectmaturation, while Hypothesis 2 would not.

C. Hypothesis 1 suggests that cooler temperaturesallow the formation of K-selected species ingeneral, while warmer temperatures allow theformation of r-selected ones; Hypothesis 2 con-tradicts this idea.

D. Hypothesis 1 implies that the two types ofpythons would fit neatly into r- and K-selectioncategories if viewed as separate subspecies, whileHypothesis 2 argues against the ease of groupingthese two types of reproductive behaviors.

194. If a group of females pythons were found who laidlarge numbers of eggs during successive reproductiveseasons one year apart and remained with these eggsuntil hatching, this finding would support:

A. Hypothesis 1 only.

B. Hypothesis 2 only.

C. both Hypothesis 1 and Hypothesis 2.

D. neither Hypothesis 1 nor Hypothesis 2.

195. Hypothesis 1 suggests that:

A. temperature plays an important role in deter-mining the ability of female pythons to leavetheir nests alone.

B. cool nests are more likely to result in out-of-control population growth than are warm nests.

C. food sources, such as rats, do not play a role indetermining nesting success.

D. boom and bust cycles of birth and death areextremely common in this species of python.

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Passage X (Questions 196-202)

Many of the body’s metabolic functions are very sensi-tive to pH, and normal function can occur only within avery narrow pH range. To maintain normal pH range, addi-tion of acid or base to the body must first be buffered in theplasma by the buffer system. The bicarbonate buffer sys-tem is one of the major buffers of the mammalian bloodthat uses carbonic anhydrase to catalyze the followingreaction:

H+ HCO3- CO2 + H2O

Buffers, however, only temporarily solve the problemand the body must ultimately excrete excess acid or base.The kidneys and lungs are important organs involved inmaintaining blood’s pH and buffering capacity. The kid-neys reabsorb filtered HCO3

- as well as excrete excessacid, produced during metabolism, through the process ofH+ secretion by the cells of the tubule.

The cellular process involved in HCO3- reabsorption

primarily uses a membrane protein called the Na+/H+

antiporter that couples absorption of Na+ to secretion ofH+ (Figure 1). The antiporter uses the energy of the Na+

gradient to secrete H+ into the lumen of the nephron. In thetubular fluid, H+ combines with HCO3

- in a reaction cat-alyzed by carbonic anhydrase to form CO2 and water. BothCO2 and H2O diffuse into the cells where carbonic anhy-drase catalyzes the reverse reaction, forming HCO3

- andH+. HCO3

- can then exit the tubular cells from the side ofthe membrane facing interstitial tissues and return to theblood.

Figure 1 The Na+/H+ antiporter

The lungs are involved in regulating plasma pH byregulating the level of CO2 in the blood; this is accom-plished by regulating the rate of ventilation (Figure 2).This response is mediated by chemoreceptors in thebrainstem that sense changes in blood pH and modifythe rate and depth of breathing.

Figure 2

196. If people on a meat containing diet generate moreacid than base, which of the following responses islikely to occur in people with increased meat con-sumption?

A. The activity of the tubular Na+/H+ antiporterwill decrease

B. The rate of HCO3- reabsorption will increase

C. The rate of ventilation will decrease

D. The rate of filtration will increase

197. Tubular cells maintain a low intracellular concentra-tion of Na+ largely through the action of Na+/K+

ATPase located in the plasma membrane facing theinterstitial tissues. What would occur if the action ofNa+/K+ ATPase is blocked?

A. The rate of H+ secretion will increase

B. The rate of HCO3- reabsorption will decrease

C. Tubular cells will shrink in volume

D. The pH of the blood will rise

-10

10

30

50

7.15 7.25 7.35 7.45

pH

Ven

tila

tion

rat

e (l

iter

s/m

in)

blood

lumen

H+ + HCO3- carbonicanhydrase

CO2 + H2O

H+ + HCO3-

carbonicanhydrase

CO2 + H2O

K+ Na+

HCO3-Na+

proximal tubule

Na+/H+ antiporter

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198. A condition known as metabolic acidosis can resultfrom increased acid production or acid administra-tion into the blood. All of the following adaptationscan be useful in restoring plasma pH duringmetabolic acidosis EXCEPT:

A. expression of proton pumps on the luminal sur-face of the tubular cells.

B. rapid stimulation of brainstem chemoreceptors.

C. decrease in the activity of carbonic anhydrase inthe tubular cells.

D. increase in the rate and the depth of breathing.

199. Which of the following terms best describes the pro-cess of H+ secretion through the action of the Na+/H+

antiporter?

A. Facilitated diffusion

B. Simple diffusion

C. Active transport

D. Secondary active transport

200. If an individual voluntarily elevates ventilation rateby increasing the rate and the depth of breathing,then which of the following renal compensatorymechanisms would be helpful for maintainingplasma pH?

A. Decreased filtration of HCO3-

B. Increased urine pH

C. Decreased absorption of HCO3-

D. Increased urine volume

201. Which of the following will increase blood’s buffer-ing capacity?

A. Increased blood volume

B. Doubled concentration of HCO3-

C. Administration of carbonic anhydrase inhibitors

D. Increased concentration of H+

202. What would be the effect of a drug that selectivelyinhibits carbonic anhydrase in the lumen of thetubule without affecting carbonic anhydrase else-where?

A. Increased amounts of H+ will be secreted butlower amounts of HCO3

- will be reabsorbed.

B. Increased amounts of H+ will be reabsorbed butlower amounts of HCO3

- will be secreted.

C. Both the amount of H+ secreted and the amountof HCO3

- reabsorbed will be decreased.

D. Both the amount of H+ secreted and the amountof HCO3

- reabsorbed will be increased.

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Passage XI (Questions 203–209)

Proteins are important biological macromoleculesmade up of amino acids. They function to add cellularstructure and support, signal transduction pathways, facil-itate oxygen transport, and provide defense frompathogens. Because of the extreme length of many pro-teins, the chemical synthesis of polypeptides in good yieldis difficult. Furthermore, amino acids are highly-function-alized molecules. All amino acids contain carboxylic acidand amide groups and many contain additional functionalgroups on their side chains. Because of their complexstructure, prevention of undesirable side reactions is oftendifficult.

Protein synthesis became much easier and more effi-cient during the 1960’s when R. Bruce Merrifield pio-neered the development of solid-phase peptide synthesisand protecting group technology. In solid-phase synthesis,the protein is synthesized unidirectionally from the C-ter-minal to the N-terminal (the opposite of ribosomal proteinsynthesis in cells). The growing peptide chain is attachedto a bead of Merrifield Resin (Figure 1) through an esterlinkage.

Figure 1 Merrifield resin

The chain is lengthened by the addition of Boc (tert-butyloxycarbonyl)-protected amino acids and N,N-dicy-clohexylcarbodiimide (DCC) as shown in Figure 2.

Figure 2 Solid phase protein synthesis

The growing polypeptide is then deprotected, and thenext protected amino acid is added to the chain. Finally,

the completed polypeptide is removed from the resin withhydrofluoric acid, which also removes the remaining Bocgroup. Because each step occurs with nearly 100 percent-efficiency, the polypeptide can be extended for about 100residues before the yield of the desired polypeptidebecomes too low. Longer chains can be synthesized byjoining two shorter chains (the N-terminal one of whichmust have a thioester on its C-terminal) through the Kentreaction, shown in Figure 3.

Figure 3 Kent reaction

203. The Kent reaction in Figure 3 is an example of whichof the following reactions?

A. SN2

B. Double nucleophilic acyl substitution

C. Transesterification

D. Michael addition and Robinson annulation fol-lowed by ring-opening

HN

SR

R

O

NH2

O

HS

+

HN

S

R

O

NH2

O

HN

NH

R

O

HS

O

O NH

OH H2NO

O

R2

O R1

O

+

+ DCCBoc group

O NH

HN

O

O

R2

O R1

O

+ DCU

CH CH2n

Cl

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204. Why might a chemist who has synthesized lysozyme,an enzyme that catalyzes the hydrolysis of bondsbetween sugars, have difficulty obtaining theenzyme’s active form in the presence of the reducingagent 2-mercaptoethanol?

A. 2-mercaptoethanol reversibly inhibits thelysozyme by binding to its active site

B. 2-mercaptoethanol reduces the carbonyls in thepeptide backbone to amines, thus inactivatingthe enzyme.

C. The Boc protecting group is cleaved by 2-mer-captoethanol, and protein synthesis will conse-quently be less efficient.

D. The lysozyme cannot fold correctly because thedisulfide bonds between cysteine residues in theenzymes are reduced in the presence of 2-mer-captoethanol to produce free thiols.

205. Which of the following is the most likely mechanismfor removal of a Boc protecting group?

A.

B.

C.

D.

O NH

O R1

OH+

HO N

H2

O R1

O

NH2

R1

O

+

O NH

O R1

O

H+

O NH2

O R1

O

NH2

R1

O

O CF3

O

O

CF3

O

O O CF3

O O

+

O NH

O R 1

OH+

OHNH2

R 1

O

O NH2

O R1

OOH2

- H+

OH

OH

O - H+

+

+

O NH

O R1

OH+

CF3 O

O

CF3 O

O

O NH2

O R1

O

NH2

R1

O

+

76

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206. Why does the newly added amino acid need to beprotected?

A. Side chains of the amino acids in the polypep-tide chain may react with the amine of the newlyadded amino acid.

B. The unprotected amino acid could react withitself rather than with the resin-bound chain.

C. The Boc group renders the amino acid morereactive.

D. Unprotected amino acids are insoluble in thesolvents used in solid-phase peptide synthesis.

207. What is the pI of aspartic acid, given the pKas for thefollowing functional groups: peptide carboxylic acid1.88, side-chain carboxylic acid 3.65, amide 9.60?

A. 2.76

B. 3.65

C. 5.04

D. 6.63

208. The structure of lysine is shown below. Why is theBoc group an inappropriate protecting group for theside chains of lysine residues?

A. The basic amine side chain of lysine would reactwith the Boc group to inactivate it.

B. The Boc group cannot be removed from a sidechain amine by the same mechanism as from amain chain amine.

C. After deprotection, the lysine side chain canreact with incoming amino acids, leading tounwanted side products.

D. During deprotection, the amine side chainwould undergo SN1 reaction with the Boc groupto form a secondary amine.

209. Which of the following statements are true of solid-phase polypeptide synthesis?

I. The linkage between th Merrifield resinand the C-terminal amino acid is weakerthan a peptide bond.

II. A polypeptide synthesized on solid-phasewill always be fully active.

III. The Kent reaction is equally useful with allN-terminal amino acids.

A. I only

B. I and II only

C. III only

D. II and III only

NH3+

NH2

O

OHLysine

77

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Questions 210 through 214 are NOT basedon a descriptive passage.

210. Color blindness is an X-linked recessive disorder inhumans. A woman who is color-blind marries a manwith normal color vision. If the couple has three chil-dren, what is the probability that all three childrenwill have normal color vision?

A. 12.5%

B. 25%

C. 50%

D. 100%

211. Which of the following amines is the most basic?

212. Digestion of carbohydrates takes place in the:

I. mouth.

II. stomach.

III. small intestine.

A. I and II only

B. I and III only

C. II and III only

D. I, II, and III

213. The antigen binding site of an antibody:

A. is nonpolar.

B. cannot contain hydrogen bonds.

C. is determined by the antibody's three-dimen-sional structure.

D. is the same from antibody to antibody.

214. Bacteriophages are viruses that infect bacteia. A bac-teriophage becomes integrated into a host bacteriumas part of:

A. translation.

B. transformation.

C. the lytic cycle.

D. the lysogenic cycle.

STOP. IF YOU FINISH BEFORE TIME IS CALLED,CHECK YOUR WORK.YOU MAY GO BACK TO ANYQUESTION IN THIS SECTION ONLY.

A.

CH2NH

B.

N

CH3

C.

CH

NH2

D.

N

H

78