Detailed answers to practice test 3R by AAMC

Physical Science Part 1-77

1. Reaction 4 is shown in the following equation, which is answer choice A.

Answers B and D do not show a reaction involving PbCO3 (s), as required by Reaction 4. Answer C shows an implausible and unbalanced equation. Thus, answer choice A is the best answer.

2. Reaction 1 is shown in the following equation.

Compound A, the white solid, is PbSO4 (s ). Neither the reactant Pb(NO3)2 nor the product NaNO3 can precipitate because all nitrates and sodium salts are water soluble. PbI2 cannot precipitate because iodide is not present. Thus, answer choice D is the best answer.

3. The dissolution of Pb(OH)2(s ) is represented by the following equation.

At pH 9, the concentration of OH-(aq) is greater than the concentration of OH-(aq) at pH 7. According to Le Chtelier's principle, the additional common ion, OH-(aq ), will shift the position of equilibrium to the left, and less Pb(OH)2 will dissolve. Thus, answer choice A is the best answer.

4. The reactions described in the passage show that lead(II) is successively precipitated as PbSO4, PbI2 , and PbCO3 . This sequence shows (assuming equal anion concentrations, as must be done here) that PbCO3 is

less soluble than PbI2 , and PbI2 is less soluble than PbSO4 . The order in which the anions precipitate Pb2+

is: CO32- then I- then SO4

2-. When this sequence is applied to the question, answer choice B is in the correct order, and answers A, C, and D are all in the opposite order. Thus, answer choice B is the best answer.

5.

6. The liquid and vapor phases coalesce at point D of Figure 2, where the densities of liquid and gaseous CO2 are equal. Thus, answer choice D is the best answer.

7. The question does not compare CO2 to a specific solvent, so we are looking for an inherent property of CO2 that makes it a good solvent for an organic oil. Supercritical CO2 is similar to a liquid and can be easily removed by evaporation because it changes into a gas when the pressure is lowered. Answers A,B, and C are not true of CO2, and answers A and C are not desirable properties of an extraction solvent. Thus, answer choice D is the best answer.

8. Polar water molecules are held together by relatively strong hydrogen bonds; whereas, the linear, nonpolar molecules of CO2 are held together at room temperature by weak London dispersion forces. Thus, answer choice D is the best answer.

9. The critical point, shown as a dot () in Figure 1, is near 30oC and 80 atm. Answer choice C is the best

answer.

10. According to the principle of like dissolves like, the covalent compound CO2 is a better solvent for a covalent compound than it is for an ionic compound. Diethyl ether, C2H5OC2H5, is a covalent compound and NaCl, NH4NO3, and KOH are ionic compounds. Thus, answer choice B is the best answer.

11. The emission peaks P1 and P2 are described in the passage as due to an electron from an outer energy

level filling a vacant inner energy level, resulting in emission of an X-ray photon. These photons have discreteenergies, and therefore discrete wavelengths, so they appear in the spectrum as peaks. Thus, answer choice C is the best answer.

12. The power P , supplied by the battery to accelerate the electron beam is given by the formula P = I

V , where I is the beam current and V is the potential difference between the cathode and anode. Therefore P = (5 x 10-3 A) x (105 V) = 5 x 102 W. Therefore, answer choice A is the best answer.

13. The emitted X- ray has the positive energy difference between the atomic energy levels as measured bytheir ionization potentials. For Pb n = 2 to 1, that is (1,400 x 10-17 J) - (240 x10-17 J) = 1160 x 10-17 J = 1.16 x 10-14 J. Thus, answer choice C is the best answer.

14. According to the passage, bremsstrahlung is produced when electrons are accelerated during collisions

with ions. All the choices of answers are ions except He, a neutral atom. Therefore, answer choice A is the correct answer.

15. To increase the kinetic energy of the electrons, they must be accelerated by a higher voltage between

the cathode and anode, thus the voltage of HV was increased. Thus, answer choice A is the best answer.

16. The probability of an X-ray emission event at a given wavelength is measured by its intensity in the spectrum. In Figure 2, P2 has a higher intensity than does P1. Thus, answer choice D is the best answer.

17. The dipole moment of a molecule is the vector sum of all of the bond moments. According to the data

in Table 1, the dipole moment of SnBr4 is zero; therefore, its bond moments add to zero or cancel. Thus, answer choice D is the best answer.

18. The 2s electron cloud in NO is in a bonding molecular orbital (MO) that forms by the overlap of the

2s orbital of an oxygen atom with the 2s orbital of a nitrogen atom. Because O is more electronegative than N, the electron cloud in the resulting 2s MO is larger around the oxygen atom than it is around the nitrogen atom. Thus, answer choice D is the best answer.

19. Table 1 gives the dipole moment of HF as 1.82 D. Chlorine is just below fluorine in the periodic table;

therefore, the electronegativity of chlorine, though significant, is less than that of fluorine. Chlorine is less effective than fluorine in creating a separation of charge when bonded to hydrogen, and the dipole moment of HCl is slightly less than that of HF. Thus, answer choice B is the best answer.

20. HCl is polar covalent because H and Cl share a pair of bonded electrons that are more strongly attracted

to the chlorine atom. The higher effective nuclear charge (i.e., the charge of the nucleus minus the shielding caused by extranuclear electrons) of chlorine accounts for its greater electronegativity. Thus, answer choice D is the best answer.

21. Carbon dioxide, O=C=O, is linear. Therefore, the two CO dipoles cancel because they are in opposite

directions. If one of the oxygen atoms is removed, the resulting CO will have a dipole because the species is linear and comprised of two different atoms. Thus, the dipole moment will change from zero in CO2 to a positive value in CO. Thus, answer choice D is the best answer.

22. An analysis of the two structures shows that the bond moments in PCl5 add to zero; whereas, those in

PCl3 do not. As shown in the figure, PCl3 is pyramidal not planar. Thus, answer choice B is the best answer.

23. Work is the product of the force on an object and the distance the object moves in the direction of the applied force. In this case, work = 20 N x 10 m = 200 J. Thus, answer choice C is the best answer.

24. Evaporation occurs when a molecule attains sufficient speed or kinetic energy to overcome the

attractive forces of a liquid. Resonance, surface tension and potential energy all relate to molecules that are not in motion. Thus, answer choice B is the best answer.

25. The relation between distance, acceleration, and time is: d=(1/2)at2 . To solve for the time it takes the runner to use t=(2 d /a )1/2=(2 * 3/1.5)1/2=2s. Therefore, answer choice C is the best answer.

26. A body is in transitional equilibrium when the components of all external forces cancel. For the sheet:

Fcos=4N, Fsin =3N. The magnitude of F is found by adding the squares of the components: F2cos2 + F2

sin2 =F2=4 2 + 3 2=25N2.

Therefore F=5N. The F vector points in the proper direction since tan =0.75= 3/4. Thus, answer choice C is the best answer.

27. The only experimental difference in Trial 1 vs. Trial 2 is that, in Trial 2, the test tube is placed in water (20oC) to cool rather than in air (also 20oC). In other words, only the surroundings were different. Thus, answer choice B is the best answer.

28. The melting point of acetamide is 80oC; therefore, acetamide will melt when it is in a test tube that is

placed in a water bath at 90oC. The temperature of the water in the bath, not the amount of water in the bath, determines whether or not the acetamide will melt. The period of time for acetamide to melt, starting at 90oC, is more than the corresponding period, starting at 100oC (i.e., the temperature of boiling water). Thus, answer choice D is the best answer.

29. Without controlling the temperature (i.e., raising the temperature of the water bath above 80oC), the

experimenter could not have observed melting or freezing. Without monitoring the time, the experimenter could not have determined the period of time for the samples to melt or freeze. The temperature of melting (freezing) of a pure substance such as acetamide is independent of the amount melted, and Experiment 2 shows that the surroundings control the period of time for freezing to occur. Thus, answer choice A is the bestanswer.

30. The time period of melting is independent of the time intervals used by the experimenter to record

temperatures. The sample would freeze completely after 23 min regardless of the time interval used by the experimenter to record temperatures. Thus, answer choice B is the best answer.

31. After Experiment 1, the sample was removed from a hot water bath as a liquid. Subsequently, the

sample froze during Trial 1. Therefore, the sample had to be reheated in a water bath above its melting point to start Trial 2 as a liquid. Thus, answer D is the best answer.

32. If the data for Trial 1 were plotted, the temperature would drop to 80oC and remain at this melting

temperature for 23 min (or 23 min x 60 sec/min = 1380 sec). The line at 80oC would not slope downward at all in the figure, and it would extend well past 270 sec, the maximum time shown in the figure. Thus, answer choice A is the best answer.

33. The tone with the shortest period has the shortest wavelength. In Figure 1a, the period of the third

harmonic (the curve with the smaller dashes) is seen to be shorter than the other two harmonics. Thus, answer choice C is the best answer.

34. The three curves in Figure 1a intersect at three points in time. The second intersection occurs in the

middle of the time axis. At that point all three curves have zero displacement. Therefore, answer choice C is the best answer.

35. The period T and frequency f of a tone are related by T=1/ f. If the first harmonic has a frequency of

100 Hz, then the second harmonic has a frequency of 200 Hz. The period corresponding to 200 Hz is 1/200 s-1=0.005 s. Thus, answer choice A is the best answer.

36. The amplitudes of the three harmonics can be compared in Figure 1a. The first harmonic is seen to be

largest while the other two have equal amplitudes. Answer choice A best represents these observations.

37. A fourth harmonic would have a shorter period than the other three. Since T=1/ f, the fourth harmonic would have a higher frequency than the third harmonic. Therefore, answer choice D is the best answer.

38. The waveform in Figure 1c begins to repeat at the zero displacement point near the end of the time

axis. This is the same time period as the first harmonic as seen in Figure 1a. Thus, answer choice A is the best answer.

39. The relation of wavelength, frequency, and wave velocity is f=v. For light, v=3 x 108m/s. The

wavelength for Material B at an efficiency of 0.42 is read from Table 1 as =1.06 x 10-6m. The frequency of this light is 3 x 108 m/s divided by 1.06 x 10-6 m, giving 2.8 x 1014Hz. Thus, answer choice C is the best answer.

40. The passage states "a coating that maximized the absorption of light." Therefore, the coating also

maximizes the conversion efficiency. Thus answer choice C is the best answer.

41. Identical voltage sources connected in parallel produce the same output voltage as a single source. (Whereas if they were connected in series, the source voltages would be added.) Therefore, answer choice D is the best answer.

42. The equation K=hf given in the passage implies that the photon energy must be greater than the

work function of the material in order to liberate an electron, i.e. for K to be positive. Thus, answer choice A is the best choice.

43. Applying a coating that makes independent of the wavelength means is the same for all . Thus,

is constant when plotted versus , a horizontal line. Therefore, answer choice C is the best answer.

44. According to the data in Table 1, both structure and molecular weight (i.e., molar mass) affect the melting point of a compound. In order to assess the effect of molecular weight or mass alone, any other effectssuch as obvious structural differences must be minimized. This is best done by comparing two compounds that are structurally similar. Because the structures of propionic acid and butyric acid (Answer C) differ by only a CH2 group, they best show that melting point increases with molar mass. All of the other answer choices compare two compounds that differ significantly in structure. Therefore, the melting points of these compounds include both molar mass and structural effects. Thus, answer choice C is the best answer.

45. Figure 1 shows the pH of the solution to be about 3 before any NaOH(aq) is added.

pH=- log[H3O

+]

3 =-log[H3O+]

[H3O+]=10-3 M=0.001 M= Answer A

46. The freezing point depression of an aqueous solution is a colligative property (i.e., it depends on the number of solute particles in a given volume of water.) Given two solutions, the one with the greater number of solute particles per liter of solution freezes at the lower temperature. Answer C is the only answer that relates a larger number of solute particles directly to a lower freezing point. Oxalic acid is diprotic and ionizesin accord with the pKa values in Table 1 to a greater extent than does crotonic acid. Subsequently, oxalic acid requires more NaOH than does crotonic acid to reach a pH of 4.7, and oxalic acid produces a larger number ofparticles in solution.Thus, answer choice C is the best answer.

47. In a titration of RCOOH, the concentrations of RCOOH and RCOO- are equal at the mid-point of

the titration. This is often called the half-equivalence point. From the expression for the equilibrium constant of a weak acid HA, when [HA]=[A-], then [H3O

+]=Ka and pH=pKa. Table 1 shows the pKa value for a

monoprotic acid to be 4.69 4.88. Answer choice A (4.8) lies in this range, the other choices do not. Alternatively, Figure 1 shows the pH at the half equivalence point of a weak acid to be about 4.8. Thus, answer choice A is the best answer.

48. The first sentence of the passage states that the unknown was a liquid at room temperature (20oC).

Table 1 shows that the melting point of crotonic acid is 71.6oC, which means it is a solid at room temperature (i.e., it melts 51.6oC above room temperature). Thus, answer choice C is the best answer.

49. In general, catalysts lower the activation energy of the slowest step in a reaction. Thus, they increase

the rate of the reaction without increasing the number of collisions, the kinetic energy of the reactants, or the Keq of a reversible reaction. Thus, answer choice C is the best answer.

50. Conservation of linear momentum requires: mradon vradon=mhelium vhelium with helium identified

as the alpha particle. The nuclear masses can be approximated by their mass numbers (222 and 4). Thus, the recoil speed of the radon is (4/222) * 1.5 x 107 m/s= 2.7 x 105 m/s. Therefore, answer choice B is the best answer.

51. The overall order of a reaction is the sum of the exponents of the concentrations in the rate law. The

exponent of [NO2] is 1 and that of [F2] is 1, and their sum is 1+1=2. Thus, the overall order is two or second order. Thus, answer choice C is the best answer.

52. The problem gives Hf

o for HCl as -92.5 kJ/mol. This means that the formation of one mole of HCl from its elements liberates 92.5 kJ of heat, as shown in the following equation.

Therefore, the formation of two moles of HCl liberates twice this amount or 185.0 kJ.

The question asks for the enthalpy change (H) for the reverse reaction. When the reaction is reversed, the sign of H is changed from to +. Thus, the reverse reaction requires +185.0 kJ = Answer D.

53. The ratio of object to image distance equals the ratio of object to image height. The ratio of image to object height is found by rearranging the ratios to give 4f/(4/3)f=1/3. The image is demagnified by a factor of 3. Thus, answer choice A is the best answer.

54. In the artery, the product of blood speed and artery cross-sectional area is everywhere constant because

the volume flow rates are equal. At the point with half the normal area, the speed must double so that the samevolume of blood passes through the constriction as does through the normal part of the artery. Thus, answer choice C is the best answer.

55. A stronger B field increases the magnetic force, Fm = q v B. The electric force must also increase to

achieve equilibrium. This implies a larger electric field in the artery and a larger voltage across the artery. Thus, answer choice C is the best answer.

56. Volume flow rate is the product of blood speed and artery cross-sectional area: (0.20 m/s)( /4)

(1.0 x 10-2 m)2 = 5 x 10-6 m3/s. Therefore, answer choice B is the best answer.

57. Since volume flow rate is proportional to blood speed, it doubles when v doubles. Therefore, answer choice B is the best answer.

58. A magnetic force acts on a moving charge in a direction that is perpendicular to both the velocity of the

charge and the direction of the magnetic field. This is a basic law of the interaction of electric currents and magnetic fields. Thus, answer choice D is the best answer.

59. The passage states that sulfuric acid reacts with Cu(s) to produce Cu+ and SO2. Thus, sulfuric acid is

converted into sulfur dioxide, or H2SO4SO2. The oxidation number of sulfur in H2SO4 can be found by assigning oxidation numbers of +1 for hydrogen and -2 for oxygen. For the formula H2SO4 to be neutral, the sum of the oxidation numbers must be zero. If x is the oxidation number of sulfur in H2SO4, then: 2(1) + 4(-2) + x=0, and x=+6. Likewise, for SO2: 2( - 2) + x=0, and x=+4. The change in oxidation number is from +6 to +4, which is answer choice B.

60. The boiling point of HNO3 is given in the question as 86oC. Because HNO3 must boil out of the flask

and be trapped in the tube, the temperature of the flask must be above the boiling point of HNO3 (i.e.,

65. The first ionization of sulfuric acid, H2SO4, is normally 100% in water. However, under conditions of low water content, all of the H2SO4 cannot ionize. Qualitatively, the only source of SO4

2- is HSO4-. The Ka for the second ionization step of a parent acid is a few orders of magnitude smaller than that of the first step; therefore, SO4-2 must be the least abundant species, because it is only formed in the second ionization step. Thus, answer choice A is the best answer.

Quantitatively, the mass relationship in 100g of 98% H2SO4 is 98g H2SO4 and 2g H2O. The molar mass of H2SO4 is 98 g/mol and of water is 18 g/mol. Thus, in 100 g of 98% H2SO4 there is one mole of H2SO4 and 2/18 or 1/9 mole of H2O. In excess water, H2SO4 would ionize completely. However, in this case (i.e., very low water content), only about

1/9 of a mole can react stoichiometrically with water to form H3O+ and HSO4

-. Of this 1/9 mol, only a small fraction of

the HSO4- further ionizes to H3O

+ and SO4-2, because HSO4

-2 is a weak acid. Therefore, SO4-2 is the chemical with

the lowest concentration. Thus, answer choice A is the best answer.

66. The reflected radar signal has a frequency shift due to the Doppler effect. Mercury must have been moving toward the Earth for the shift to be to a higher frequency. Thus, answer choice A is the best answer.

67. An equatorial bulge of the Sun would be caused by some process that is not the same in all directions.

Of the four answers, only the Sun's rotation about an axis fulfills this condition. The bulge is related to the balance between the centripetal force and the gravitational force. Therefore, answer choice A is the best answer.

68. The perihelion will have moved through an angle =t after a time t has elapsed. The conversion of

angles is given by the following. An arcsecond is 1/60 of an arcminute. An arcminute is 1/60 of a degree. The time for 500 arcseconds per century to accumulate to =360 degrees is 360/(500/(60 x 60)) = 360 x 60 x (60/500) centuries. Thus, answer choice C is the correct answer.

69. The orbit of Mercury around the Sun is, by Kepler's and Newton's laws, an ellipse with the Sun at one

fixed focus. Therefore, answer choice C is the best answer.

70. The time for the radar signal to travel from the Earth to Venus and return is the total distance divided bythe velocity of light: 2 x 5 x 1010 m / 3 x 108m/s = 3.3 x 102s = 300 s. Thus, answer choice B is the correct answer.

71. The ellipse precesses around the Sun at one focus. Any fixed point on the ellipse is a constant distance

from the focus. Therefore it traces out a circle around the focus. Answer choice A is the best answer.

72. As written, the equation shows the reduction of Zn2+ to Zn. The negative reduction potential for Zn2+ means that Zn has a positive oxidation potential (i.e., Zn is easily oxidized.) Zinc will displace hydrogen, which has a zero standard reduction or oxidation potential (i.e., hydrogen is the standard against which other substances are measured.). As given in the problem, ionic zinc is Zn2+. Therefore, zinc metal liberates hydrogen gas and produces ZnCl2(aq). Answer choice B is the best answer.

73. Power is defined as the rate of doing work. For the automobile, the power output is the amount of work

done (overcoming friction) divided by the length of time in which the work was done. Therefore, answer choice D is the best answer.

74. The bonds labeled C and D in the figure are of equal length but shorter than bond B. This is because

two resonance structures can be drawn: in one resonance structure, bond C is a double bond, and in the secondresonance structure, bond D is double bond. Double bonds (bond order = 2) are shorter than single bonds (bond order = 1), so the bond order for bonds C and D is about 1.5 and for bond B about 1. Bond B is longer than bond A because of the small atomic radius of hydrogen compared to nitrogen. Thus, answer choice B is

the best answer.

Alternatively, the figure shows that C = D, so answer choices C and D can be eliminated. Also, A is clearly shorter than B; therefore B is the longest. Thus, answer choice B is the best answer.

75. Conservation of energy requires that the 15.0 eV photon energy first provides the ionization energy to

unbind the electron, and then allows any excess energy to become the electron's kinetic energy. The kinetic energy is this case is 15.0eV - 13.6 eV = 1.4 eV. Thus the correct answer is A.

76. In radioactive decay, the sum of the mass numbers A and atomic numbers Z, before and after decay,

must balance. The numbers for beryllium undergoing positron decay are: mass (7= 7 + 0) and atomic (4 = 3 + 1). The resulting nucleus is 7

3Li. Thus, answer choice B is the best answer.

77. Work is the product of force and distance. The easiest way to calculate the work in this pulley problem

is to multiply the net force on the weight mg by the distance it is raised: 4 kg x 10 m/s2 x 5m = 200J. Therefore, answer choice D is the best answer.

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Writing Sample

Time: 2 items separately timed at 30 minutes each

138. Consider this statement: An understanding of the past is necessary for solving the problems of the present. Write a unified essay in which you perform the following tasks. Explain what you think the above statement means. Describe a specific situation in which solving a current problem might not require an understanding of the past. Discuss what you think determines whether or not the past should be considered in solving the problems of the present. Sample Essay: Understanding the Past History is an integral part of the learning process. By studying events of the past, we can analyze the repercussions of certain behavior and action patterns. It is a fundamental way to lay the groundwork and predict the outcomes of future events. History is governed by human behavior. Although times have changed, and technology and knowledge has advanced, people are still driven by the same needs, desires, and insecurities of ages past.

One area in which the study of history is essential is in the conflict between disputing nations. During the Gulf War in 1991, America was at first unsure of its potential role. This country did not want to repeat the tragic losses of the Vietnam War, but at the same time could not let injustices occur before its very eyes. By studying previous military strategy, impetus, and conditions, the United States was able to enter the war without suffering a humiliating defeat. Civil rights issues have also used historical experiences to determine proper conduct. The civil rights movements go back to the 1960's, when black leaders were just beginning to assert and articulate their arguments, as well as achieve their goals. The recent racial riots in Los Angeles, while violent, showed how people can learn from the past. There were definite and inspiring examples of concern crossing racial borders while before, the conflict was African-Americans against whites, we saw examples of multi-racial groups banding together to protect stores, homes, and families. Many of those people did not want to repeat the horrifying events of the past. On the other hand, some problems exist today that are totally independent of any historical event. The current issue of AIDS prevention, treatment, and search for its cure has generated a whole new set of rules and etiquette. Our world has never before had to deal with the devastating effects of the AIDS virus, not with the quickly increasing numbers of infected people. Looking at the past could give us no knowledge on the workings of this disease, nor on its cure. It seems to have bypassed every known strategy used before in defeating a virus. In fact, looking to the past could even cause problems. It was the past, and even ongoing, sexual practices that allowed AIDS to spread so quickly. Instead of looking to the past for new information, we must reform our histories to stop this disease. When, then, is the past crucial to our understanding of current events? It is important only, and especially, when it relates to the present situation. History can lay the groundwork for a course of action. But, of course, this is only true when the courses of action are similar. There must be some common threads tying the past and present together. With racial tension in mind, the commonalities stem from common catalysts for anger and feeling in injustice and equality. Moreover, these events are mediated by human behavior. Also, conflicts between nations arise because people disagree. In fact, people, and the involvement of people, may also be the common thread tying the past and the present together. But, with something like the AIDS virus, this crisis is not governed by any set of rules or behavior. No previously established fundamental law of virus behavior exists to dictate its actions, for it proceeds with a total disregard and lack of emotion. It just keeps changing and slipping through our fingers, with no historical example to give us a guideline as to its future actions. History is crucial to understand. It can provide clues to our future, and help us solve certain problems. But, this can only be true if these problems, or similar ones, existed before and were governed by similar mechanisms. SCORE POINT: 6 ANNOTATIONS: This paper, which demonstrates clarity, depth of analysis, and a recognition of the complexity of the issue, is thoroughly developed with relevant and specific historical examples. The writing is noteworthy as well, showing a superior command of language, particularly in terms of syntactic variety. Considering the time limitations of the test, which do not allow for much proofreading or editing, the fluency is remarkable. Few errors in sentence structure, grammar, or mechanics occur in the paper. The paper begins with a simple restatement: "History is an integral part of the learning process." This sentence

announces the writer's central idea but, at the same time, allows for the possibility that history cannot teach us everything we need to know. This subtle but clearly stated proposition prepares the reader for a series of related examples that serve as illustrations of the writer's themes and ideas. The first examples are of the Gulf War and the Civil Rights movement. The writer uses the Gulf War to demonstrate how the United States has applied the lessons of Vietnam (establishing that we can learn from our past mistakes) to a contemporary problem of international relations. Then the writer supplies a parallel example of a domestic problem, suggesting that individual behavior during the Los Angeles riots was the result of lessons learned from the Civil Rights movement. The writer notes that "people did not want to repeat the horrifying events of the past." These two well-developed examples explain the meaning of the topic and expand on the idea offered in the introduction ("By studying events of the past, we can analyze the repercussions of certain behavior and action patterns"). In the next paragraph, the writer observes that "some problems exist today that are totally independent of any historical event." For an example, the writer chooses to discuss the issue of AIDS, in particular the related issues of "prevention, treatment, and the search for its cure." The writer grants that researchers have attempted to use their understanding of the past to find successful cures and treatment, but the disease "seems to have bypassed every known strategy used before in defeating a virus." This leads the writer, quite logically, to consider the question suggested by the third writing task: "When, then, is the past crucial to our understanding of current events?" The writer's conclusion, which is consistent with the ideas expressed in the opening paragraph of the paper, is that "history is crucial to understand" because it can supply "clues to our future, and help us solve certain problems." The writer stresses the importance of learning from the past while, at the same time, recognizing that an understanding of the past cannot guide us to the solution of all current problems. The specific criterion the writer presents is that we can solve problems that have issues with "some common thread." especially those involving human behavior (and the writer returns to the issue of tension between the races, commenting that "commonalities stem from common catalysts for anger and feelings of injustice and equality"). Throughout this final paragraph, the writer returns to the paper's main idea and previous examples. This unifies the essay structurally and thematically. This comprehensive and well-documented paper is impressive in its scope and ambition. The command of language, combined with a successful organization plan, results in a sustained, clear, and effective paper.

139. Consider this statement: Politicians too often base their decisions on what will please the voters, not on what is best for the country. Write a unified essay in which you perform the following tasks. Explain what you think the above statement means. Describe a specific situation in which a politician might make an unpopular decision for the good of the country. Discuss the principles you think should determine whether political decisions should be made to please the voters or to serve the nation. Sample Essay: Politicians In a representative democracy, representatives are selected by the voters to convey their ideas and values in the government. These representatives are voted for by citizens according to their degree to which they will uphold these ides and values. Citizens would obviously not vote someone into office

who believes in the opposite of the citizens on several issue. The representatives will be re-elected in the same manner; the degree to which the citizens ideas and values were upheld. It is not surprising that politicians will base their decisions on what will please the voters and not on what is best for the country. The politicians must maintain the popularity of the voters and the best method to achieve that is to please them with the actions made in governmental circles. The politicians however are not merely carbon copies of the citizen's consensus opinions. The politicians will have opinions of their own and occasionally this may conflict with those of the voters. At this time the politicians may make an unpopular decision for what they feel is for the good of the country. One example is often seen with the petition of Nazi groups to march. While an exceptionally high majority of citizens would never want to see this march occur, many politicians would have no choice but to let the march preceede for the greater good, in this instance it is the right to free speech guaranteed by the 1st Amendment to the constitution. From this ideal, much of this country was founded and it would by hypocritical to deny it to another group regardless of how unpopular this group was to the voters. While this is an extreme case of politicians displeasing the voters for the good of the country, there exists a great range of "grey" area where politicians and voters do not meet eye to eye. So what should be considered when making a decision to please the voters or serve the nation? Fortunately, forthe most part, the voters will also have the best interest of the nation at heart but trouble can still arise. One major problem is the building of new prisons or landfills. For most voters, there is no question that they are needed, but none of the voters wants to see the prison or landfill wind up in their backyards. To deal with such problems and still remain in good standing with the voters, the politicians must learn to make concessions. For instance, the same district where a new prison is built, a new High School and Industrial Park is set up to better education and increase jobs and the local economy. A politicians must weigh the potential degrees of disfavor that they may incur when determining whether to serve the nation at the risk of the voters. Politicians are unable to please all of the voters all of the times, but by ensuring that unfavorable decision are accompanied by many favorable ones, the politicians can balance on the treacherous tightrope between serving their country and serving their voters. SCORE POINT: 5 ANNOTATIONS: This paper clearly addresses the three elements of the rhetorical assignment, examining the issue of political decision making in a democracy as it pertains to voter input and perceptions about politically sensitive issues. The presentation is coherent, focused, well developed and unified thematically as the writer presents specific examples to illustrate the complexity of the issue. The explanation and commentary on the examples, especially the discussion of the prison/landfill dilemma in the next-to-last paragraph, engage the reader and provides ample evidence to support the writer's position. There are weaknesses in the presentation of the paper, especially minor lapses in diction and mechanics, but there is good syntactic variety and little repetition or redundancy. Sentences are well formed and purposeful. Effective work choice and imagery are used (politicians not being "carbon copies" of citizens' views, politicians keeping their footing on the treacherous tightrope"). Overall, there is a strong command of language. The first two paragraphs address one of the inherent problems in representative democracy: what should politicians do

when their opinions, conscience, and sense of what is awful happen to differ from the views of the people who elect and reelect the politicians to promote their viewpoints? The writer explains the elements of representative democracy in the first paragraph. In the second paragraph, the writer illustrates the dilemma by offering an example: Nazi groups desiring to demonstrate in a community where the vast majority of citizens oppose Nazi marches. The writer admits, in the next paragraph, that the example is an extreme one. This declaration works as a transitional device, allowing the writer to proceed to a more common and ambiguous example. By noting that the construction of landfills is favored by voters but that they seldom want them in their own communities, the writer explores the predicament facing elected offices. The writer says politicians "must weigh the potential degrees of disfavor" and acknowledges that politicians may not be able "to please all the voters all of the time." Thus, compromise is required, and the writer offers a creative solution to the hypothetical situation" build a prison and "a new High School and Industrial Park." The strength of the paper is demonstrated by the complex level at which the issues are discussed. Ideas are developed by comparing examples and reflecting on the situations and solutions presented. The lapses in language control are minor, especially considering the time constraints of the test. The writer's success in language control are minor, especially considering the time constrains of the test. The writer's success in expressing ideas and exploring issues is apparent in the attention to an execution of the rhetorical assignment.

Biological Science 140-216 in 100 min

140. The Regulative Hypothesis proposes that each cell contains complete information for construction of the multicellular organism. Choice D is just a restating of this hypothesis. It can not be a correct answer, because the Regulative Hypothesis contradicts the Mosaic Hypothesis. If a single egg produces identical twinsor triplets, identical plans for each individual must have been transferred from the egg in accordance with the Regulative Hypothesis, so A is also a wrong answer. For an embryo to have two normal - sized heads, identical construction information must have been introduced into both parts as stated in the egulative Hypothesis, so B is also wrong. The Mosaic Hypothesis, states that the parts of the plan are handed out to the cells that need them at the time of division. The production of partial embryos is consistent with the mosaic theory because it implies that each embryo received just part of the total construction plan. Answer choice C is the only one of the four statements that supports the Mosaic Hypothesis.

141. Construction information for an organism must be coded in the form of the DNA in the nucleus of the egg. If the nucleus of an egg cell can be replaced by the nucleus of a gut cell and still direct the production of a complete individual, then the gut cell must contain all the construction information an egg nucleus contains. This is an example of the Regulative Hypothesis. This narrows the choice of answers to either A or B.

Answer choice A suggests that we should decide that the experiment supports the Regulative Hypothesis because an environmental factor induced the development of the egg, but, in fact, the way the egg was activated to begin development is irrelevant to the question of the source of the information guiding development. Only answer choice B points out a relevant fact: that the egg with a gut nucleus developed into a complete organism.

142. The development of incomplete embryos, described in choice A, falsifies the Regulative Hypothesis. Answer choice B states that the fate of a cell depends on its own internal factors, but this would be true whether the Mosaic Hypothesis or the Regulative Hypothesis was true. Cells must have genes to guide their fate. The question is whether they have a full complement of the organism's genes or just a part. Answer choice D restates the Mosaic Hypothesis, that genes are distributed unevenly to daughter cells during development. Answer choice C, the correct answer, however, supports the Regulative Hypothesis. If all cells contain identical information for construction of the organism, then some external factor must tell different parts what they should become. If specific courses of development were shown to be dependent on the position of the cell in the embryo, it would explain this aspect of the Regulative Hypothesis. By supplying an answer to the question of how the parts of an animal develop differently when they have the same genes, that information buttresses the Regulative Hypothesis. Thus, answer choice C is the best answer.

143. When the position of the cells was switched, the cells took on a fate consistent with their new positions. Answer choice A is the correct answer. It would be hard to deny that there must have been "position - dependent cellular interactions" involved. Answer choice B can not be true because the switched A cells did not develop the way A cells normally do, but instead developed the way B cells developed. They took on B cell characteristics, which would be impossible if they had a different set of genes. Answer choice C is not relevant to the experiment and is true of development in general. It is a statement of the obvious: that embryogenesis is a process dependent on the organism's individual cells. Choice D is contradicted by the experiment. Development did vary when the position of the cells was switched. There fore, answer choice A is the best answer.

144. After meiosis, human germ cells contain half the genetic material of other cells in the body, but they still retain one copy of each gene. The normal complement of genes is re-established at fertilization, before development of the embryo begins. The facts stated in answer choice A are therefore irrelevant. Answer choice B is not good evidence for the Mosaic Hypothesis. It just suggests that all genes are not active at all times. The Regulative Hypothesis fits our understanding of human development best and because cell positiondoes have an effect on development in humans. The correct answer must be answer choice D.

145. Answer choice D is the only choice that shows that all cells of the body have the same genetic makeup, because it is the only one in which the chromosome content at the 64 cell stage matches the chromosome complement at the two cell stage. Answer choices A and B show a reduction in the number of chromosomes and answer choice D shows an increase. Thus, answer choice D is the best answer.

146. Data that showed that all embryonic cells have the same developmental potential (answer A), that

interactions occur among cells (answer B) or that the fate of cells depends on external conditions (answer C), would constitute evidence for the Regulative Hypothesis not the Mosaic Hypothesis. The only choice that is consistent with the Mosaic Hypothesis is answer choice D. Evidence that the fate of cells depends on their position disproves the theory that their fate depends on restrictions in the number of genes the cell contains. Cells can not acquire features based on their position unless they already have genes for those features.

147. The passage states that an insoluble form of casein forms as the pH decreases from 6.6 to 4.6. Sour milk

is acidic. Thus, answer choice A is the best answer.

148. The passage mentions three specific classes of compounds: (1) Calcium caseinate, which precipitates at low pH as casein; (2) Proteins (lactalbumins and lactoglobulins), which precipitate when denatured by heat; and (3) Lactose, which hydrolyzes to galactose and glucose. In the procedure mentioned in the passage, the powdered calcium carbonate is probably added to neutralize any remaining acetic acid. Solid 1 (casein) forms

on the addition of acetic acid and is removed by filtration, so it is not present when the CaCO3 is added. Solid 2 (denatured and precipitated proteinaceous material) is insoluble in the CaCO3. Thus, answer choice A is the best answer.

149. As seen in the equation, the hydrolysis of lactose results in the net addition of H2O during the cleavage

of the disaccharide. Thus, answer choice B is the best answer.

150. Solid 2 forms after the solution is heated to boiling, a condition that denatures and precipitates the lactalbumins and lactoglobulins. Thus, answer choice C is the best answer.

151. Patient A in group 3 received a placebo, and so was untreated. Nevertheless, a response was observed, a

reduction in symptoms (lower rating on the 5 - point scale). The correct answer, therefore, is choice A, that the response was one not associated with treatment. The other answer choices, B, C and D, would only be plausible if the patient in question had received treatment with a histamine blocker, an acetylcholine blocker or drug with both properties. Thus, answer choice A is the best answer.

152. Drug A blocks histamine receptors and acetylcholine receptors. The group that received Drug A

reported the lowest level of symptoms. This suggests that decreases in the effect of both histamine and acetylcholine should be the optimum treatment. Answer choice D represents this combination. The histamine component of the symptom reduction is achieved by blocking its histamine receptors with an antihistamine. Introducing acetylcholinesterase, an enzyme that catalyzes the degradation of acetylcholine, reduces the acetylcholine component. The question assumes that the examinee knows that acetylcholinesterase breaks down acetylcholine or at least recognizes that the term refers to an enzyme for which acetylcholine is a substrate. Thus, answer choice D is the best answer.

153. The passage states that drug B blocks histamine receptors only. Group 2 received Drug B. Of all the

subjects in group 2, subject C reported the lowest level of symptoms. Answer choice D is therefore the correctresponse: that the person who received the most benefit from blocking of histamine receptors alone was subject C in group 2.

154. Epithelial cells cover all the free surfaces of the body forming an interface between body tissues and the

external environment. When fluid is released from the body it must cross an epithelial surface. Answer choice A is therefore the correct choice. The cells that secrete mucous are primarily found in mucosal glands made up of ectodermal epithelium continuous with the epithelial lining of the nasal cavities. None of the other typesof tissue named, connective (answer B), contractile (answer C) or neurosecretory (answer D) impinge on the external environment.

155. Acetylcholinesterase is an enzyme that catalyzes the degradation of the neurotransmitter acetylcholine

into choline and acetic acid. An acetylcholinesterase inhibitor prevents the breakdown of acetylcholine. This causes acetylcholine to remain in the synaptic cleft for a longer period of time, stimulating the nasal epithelial cells to secrete. The correct response to the question is therefore answer choice C, that acetylcholinesterase inhibitors must increase parasympathetic activity at the acetylcholine receptors. Answer choice D states the opposite, that it decreases activity. Answer choices A and B are incorrect because they describe alternatives that would decrease rather than increase synaptic conduction.

156. The passage states that cholesterol is a precursor of steroid hormones. Of the compounds listed,

estrogen is a steroid hormone. Therefore, the key is D.

157. Evidence that familial hypocholesterolemia (HC) is caused by a genetic disorder comes from two facts: (1) it is relatively common in some families and absent in others, and (2) it develops independent of diet. This combination is described in choice A. Answer choices B and C imply that it is related to diet. Answer choice D suggests that this consistent familial protein defect might be independent of a DNA defect. Because DNA codes protein structure and DNA is the molecular basis of heredity, this cannot be a correct choice. Therefore,answer choice A is the best answer.

158. The cause of HC presented in the passage is the inability of membranes of affected individuals to bind

the blood-protein that transports cholesterol, the low-density lipid binding protein, LDL. The drug described in answer choice B would be the best model of HC. Such a drug could, by preventing membrane binding of LDL, disrupt normal functioning in the same way as proposed for HC. Increasing absorption of lipids by the small intestine (answer A) would not mimic HC as well; it would just increase all types of lipids in the blood. Lowering the bile in the small intestine (answer C) would prevent lipid emulsification, hence absorption, but would not be a good model. Decreasing plasma membrane protein production (answer D) would not be a good model because it would produce widespread effects on the body that would be difficult to untangle from the effect of loss of membrane binding proteins alone. Thus, answer choice B is the best answer.

159. Both the father and mother have cholesterol levels of 3.0 mg/mL, which is in the moderately affected range. This suggests they have one copy of the HC gene and one copy of the normal (wild-type) gene and thatthe two versions of the gene (alleles) are co-dominant. When a gene is recessive, its effect (or lack of effect) ismasked by a corresponding dominant gene. The child has a cholesterol level of 7.0 mg/mL. This is in the severely affected range according to the passage. This child appears to have acquired one HC gene from each parent for a total complement of two HC genes and no normal genes. This is consistent with the idea that the normal and HC genes are co-dominant. Therefore, the best answer is answer choice B.

160. If the lacteals of the intestinal villi were to contract, it would reduce fat absorption because ingested fats

collect in the lacteals for transport to the venous (portal) circulation. The correct answer therefore is answer choice A. If the capillaries of the intestinal smooth muscle constricted, the rhythmic contractions of the intestines would be reduced, but there would be no specific effect on fat absorption. Any constriction of bloodvessels associated with the peritoneal covering of the intestine would have no effect on fat absorption, becausethis covering is on the outside of the organ, opposite the luminal surface where fat absorption takes place.

161. The strongest support for the hypothesis that HC has a genetic rather than an environmental origin

comes from the fact that it develops independent of diet. Answer choice B is the correct statement: that HC develops when diet is held steady. Answer choice A does not address the question of whether affected and unaffected families have different diets. Answer choices C and D address issues that do not speak to the question of whether the disorder is genetic or environmental.

162. According to the stem, in pea plants "short" (t) is recessive to "tall" (T). In a cross of heterozygotes (Tt

x Tt) one would expect that 75% would be "tall" (Tt or TT) and 25% would be "short" (tt). Given that 787 plants in the next generation were "tall," the closest prediction of the number of small plants is B, 277.

163. Breakdown of glucose proceeds first by glycolysis, then by oxidation in the citric acid (Krebs or tricarboxylic acid) cycle. The enzymes for the former process are located in the cytoplasm and those for the latter are in the matrix of mitochondria. The 14C label, therefore, would first appear in the cytoplasm, then in the mitochondria, as stated in answer choice D.

164. The "central dogma" of molecular biology is that DNA encodes RNA, which encodes proteins. An early(pre-Watson and Crick) version of the dogma was "one gene, one enzyme." While this question appears to be asking about blood types, the answer is best discovered by relying on the entral dogma. Answer choices A andB can be eliminated because neither RNA nor genes code directly for carbohydrates. Answer choice D can be eliminated because whether genes contain carbohydrates or not is irrelevant to the outcome of protein manufacture. The only answer choice that is consistent with the central dogma of molecular biology is answerchoice C: that the gene codes for an enzyme that is responsible for attaching the appropriate carbohydrate. People with blood type B and AB have an enzyme that people with other blood types do not have. The enzyme adds an extra galactose monosaccharide to the O glycolipid expressed on the red blood cell surface.

165. In starvation, the body uses up its stores of carbohydrate and lipids, and then begins to break down bodyproteins for metabolic energy. A byproduct of the metabolism of the amino acids from protein is nitrogen. Thecorrect answer is therefore answer choice B: the nitrogen in the urine comes from breakdown of the body's proteins. Neither lipids, nor carbohydrates such as glycogen, contain nitrogen, so the excess nitrogen could not come from those sources. The reason animals have kidneys is to provide a way of eliminating nitrogenous waste products. Incomplete reabsorption of nitrogenous products is the goal of the kidney, not necessarily a pathological process.

166. The primitive eukaryotic cells are described in the passage as anaerobes, so it is likely that they could breakdown sugars by glycolysis. That means answer choice A is incorrect. The bacteria are described as aerobic, so the passage suggests that the ability to use oxygen was acquired with the bacteria. Electron transport in mitochondria and bacteria is described and the electron transport system is needed for eukaryotes to use oxygen. The acquisition of aerobic bacteria as partners would have provided them with the ability to carry out aerobic metabolism through use of the Krebs cycle and electron transport. Answer choice B is therefore the correct choice. The anaerobic, eukaryote-precursor cells must have engaged in cell division and the processes of transcription and translation, so answer choices C and D are incorrect.

167. The endosymbiotic theory described in the passage is widely accepted today, so there must be some way of using the theory to reconcile the fact that mitochondrial proteins are made in the cytoplasm. We know,for example, that lateral transfer of genes from one genome to another is widespread among living organisms. It would have been an easy matter for mitochondrial genes to be acquired by the host cell nucleus and to disappear from the mitochondrial genome. It is likely that this exchange would have resulted in a net gain in efficiency for the composite organism. Answer choice A is therefore the most plausible answer. Despite manysimilarities between mitochondria and bacteria, the question of cytoplasmic synthesis of mitochondrial proteins is relevant to the question of the origin of mitochondria and needs to be answered. The fact that mitochondria do have all the machinery (ribosomes, t- RNA, etc.) for protein synthesis eliminates answer choice C from consideration.

168. Hydrogen ions (H+) are protons. The provision of a channel for proton flow across the membrane wouldallow hydrogen ions to flow across the membrane until equilibrium had been achieved between the concentrations on each side of the membrane. Because ATP synthesis is driven by a flow of hydrogen ions down a concentration gradient, ATP production will decrease and eventually stop as equilibrium is established (not increase as suggested in answer choice A or remain unchanged as suggested in answer choice D). The decrease has nothing to do with the rate of hydrogen ion donation by NADH, answer choice B. Answer choice C is the correct answer.

169. When a eukaryotic cell ingests a particle in the process known as endocytosis, the particle ends up in thecytoplasm inside a vesicle, a hollow membrane bounded intracellular organelle. The membrane of the vesicle is formed from a closed-up patch of membrane detached from the cell's own plasma membrane. Prokaryotes also have their own plasma membranes. It seems likely that the two membranes of the mitochondrion consist of an inner prokaryote-derived membrane and an outer eukaryote-derived one, this arrangement having been established when some early cell engulfed a bacterium through the process of endocytosis. Answer choice C describes this arrangement best. Because the primitive eukaryotic-precursor cell did not use electron transport,their membranes lacked the complicated protein machinery for this process. The inner membrane of mitochondria contains this machinery while the outer one lacks it. Answer choices B and D both suggest eukaryotic membranes make up the inner membrane, so they must be wrong. If answer choice A was true, the electron transport proteins would be found in both membranes of mitochondria.

170. Answer choice D, is the correct answer. It presents an example of bacteria inhabiting eukaryotic cells that is analogous to the hypothesized relationship between mitochondria and their eukaryote hosts. Because it

shows that such a relationship is possible, it provides stronger evidence than the other answer choices in support of the theory presented in the passage. That the number of genes in a mitochondrion is less than that of the typical bacterial cell (answer A) does not support the hypothesis in the passage, it refutes it. That mitochondria have hundreds of different enzymes (answer B) neither supports the theory or refutes it since thenumber in bacteria and mitochondria are not compared. That mitochondria have diameters that are the same asbacteria is dubious (answer C) since both mitochondria and bacteria are very diverse in their sizes and shapes.

171. Answer choice C is the correct answer because it states that the fact that mitochondria and bacteria both have circular DNA is a reason to think they are related. Such a characteristic is likely to have been highly conserved over time since it involves the basic material of life. Changes in the topology of the DNA molecule would involve substantial changes in the way the molecule replicated and the way it was transcribed. The fact that both have circular DNA supports the symbiotic hypothesis presented in the passage. Alternative choices A and B presented are not actually true characteristics of bacteria. Bacteria do have 70S ribosomes, not the 80S ribosomes of eukaryotes as stated in answer choice A and they do reproduce by binary fission, which choice B denies. The loss of ability to carry out anaerobic respiration (answer D) might be a product of the long history of association with the host cell. It is not unusual for symbiotic organisms to lose abilities that arecompensated for by host functions.

172. Any disruption of mitochondria is likely to decrease ATP production since they are a major cellular source of that molecule. Answer choices C and D can not be right because they propose an increase in ATP production. There is no information in the question to suggest that valinomycin will cause K+ to compete withH+ for an active site on ATP synthetase. Furthermore, one would suspect that ATP or precursor molecules such as phosphate and ADP would occupy the active site on the ATP synthase molecule. Answer choice A, therefore, does not seem plausible. The question does provide the information that valinomycin increases the flow of K+ across the membrane. An influx of another positively charged ion into the compartment would disrupt the electrochemical gradient responsible for the necessary flow of protons. Answer B, therefore, seemsmore plausible than any other choice.

173. The carbonyl group (C=O) in carvone makes carvone more polar and has a higher boiling point than thehydrocarbon limonene. Thus, answer choice A is the best answer.

174. The passage mentions that the ebulliator was added to introduce small air bubbles into the system. This is the same function provided by a boiling chip at atmospheric pressure. The air bubbles break the surface tension of the liquid being heated and prevent superheating and bumping. Thus, answer choice C is the best answer.

175. We presume that the separation of limonene and (+)-carvone can be improved by modifying either the apparatus shown in Figure 1 or the procedure (i.e., the separation as described in the passage did not achieve atotally satisfactory separation.) The separation of the two liquids takes place in the fractionating column as thetwo liquids vaporize and condense, with the lower-boiling liquid distilling first. If the fractionating column is shortened (Answer D), the liquids will vaporize and condense fewer times (i.e., there will be fewer theoretical plates) and the degree of separation will worsen. Cooling the condenser with ice water (Answer C) will have no effect on the degree of separation because condensation takes place after the separation has occurred. Creating a lower pressure inside the distilling apparatus (Answer B) will lower the boiling points of both liquids and narrow their difference in boiling point, making it necessary to also increase the length of the fractionating column in order to achieve the same degree of separation as obtained originally. Lengthening thefractionating column was not part of answer choice B. Heating the distillation flask (i.e., increasing the temperature) at a slower rate (Answer A) will allow both liquids more time in the fractionating column (increase the number of theoretical plates, allowing liquid and vapor to equilibrate); therefore, of the four

options, this one is most likely to improve the degree of separation of the two compounds. Thus, answer choice Ais the best answer.

176. Carbon 5 is the only stereogenic carbon atom (chirality center); carbons 2 and 7 are not stereocenters. Thus, answer choice B is the best answer.

177. The boiling point of a liquid is the temperature at which the vapor pressure of the liquid equals the surface pressure. The normal boiling point is measured at 1 atm pressure. The vapor pressure of a liquid increases with increasing temperature. Hence, the boiling point of a liquid decreases as the pressure on the surface of the liquid is decreased. If a leak develops in the apparatus, the surface pressure will increase, as willthe boiling points of both liquids. Thus, answer choice A is the best answer.

178. To solve this genetics problem, we will take a systematic five-step approach. First, we need to assign symbols to the genes involved. The question asks us to assume that the defective hormone response is X-linked. So, symbolize the gene that causes a lack of response to the X hormone as XX and the normal (wild type) gene as just X. The second step is to list the possible types of haploid gametes the father can produce and the possible types the mother can produce. Since the gene is on the X chromosome, we will leave out all chromosomes except the sex chromosomes X and Y. The father is affected in this case. He must therefore have the XX gene. The father's genotype is therefore XXY. No information is given about the mother, but since the trait is very rare, chances are the mother does not carry the defective gene. We will assume that the mother's genotype is XX. The third step is to enter the relevant maternal and paternal haploid gamete genotypes into a Punnett square, putting XX and Y across the top and X and X down the side. These are the all the possible haploid genotypes of the father's spermatozoa and the mother's ova. The fourth step is to fill inthat square with all possible combinations of those gamete genotypes that would be produced by fertilization. The fifth step is to examine the genotypes in the squares and determine which would express the genetic disorder, that is, which would be unable to respond to hormone X. Since the passage tells us that no females are affected, the XX X genotype does not express the trait. From the diagram therefore it appears that none of the offspring, male or female, will be affected. This is the common result in many X-linked disorders such as hemophilia. Males predominantly have the disorder, but do not pass it on to their offspring. Females rarely have the disorder, but can be carriers.

179. The passage tells us in statement 4 that protein R is capable of phosphorylating Protein P. Phosphorylation is the addition of high-energy phosphate groups by a kinase to another protein. A molecule such as ATP (adenosine triphosphate) donates the phosphate group. The only protein that utilizes ATP is protein R (answer B). Protein R must split ATP, releasing ADP (adenosine diphosphate) and attaching a phosphate molecule to Protein P. Thus, answer choice B is the best answer.

180. Because phosphorylated Protein P was necessary and sufficient to increase the rate of Ca 2+ into the cell(statement 6 in the passage), Protein P is probably a membrane channel, a pore in the membrane that allows

Ca 2+ to pass freely. Thus, answer choice C is the best answer.

181. The phosphorylation of Protein P must alter it in some way that allows it to increase Ca2+ influx. The most immediate way to stop the influx of Ca2+ therefore should be to dephosphorylate Protein P (answer D). Ca2+ would continue for a time in all the other answer choices A-C.

182. According to statement 5, Protein P from affected males from Family 2 could not be phosphorylated. This means that the genetic defect in Family 2 must be in the gene that codes for Protein P. Therefore, the correct answer choice is B, that the alteration of one amino acid in Protein P is the likely result of the mutationin the DNA of affected males.

183. If a carbene was the intermediate, then Compound 4 and Compound 5 would form identical intermediates that would go on to yield identical product distributions. The passage states that the chemists ruled out Mechanism A because Compounds 4 and 5 formed varying ratios of Compounds 6 and 7. Thus, answer choice C is the best answer.

184. As seen in Reaction 1, OCH2CH3 replaces Br. Thus, answer choice B is the best answer.

185. By either mechanism, ethoxide abstracts a proton to form ethanol, EtOH, and a bromide ion is liberated.The spectator ion, Na+, joins with the Br- to form NaBr. Thus, answer choice A is the best answer.

186. Neither Compound 2 nor Compound 6 is optically active; therefore, the specific rotation of both compounds is zero. Thus, answer choice D is the best answer.

187. Compound 1 contains only sp2 hybridized carbon atoms and does not contain any sp or sp3 hybridized carbon atoms. Compound 2 does not contain any sp3 hybridized carbon atoms but does contain sp and sp2 hybridized carbon atoms. The carbon-carbon double bond that is not part of the two benzene rings in Compound 1 involves sp2 hybrid orbitals. This double bond undergoes transformation in Reaction 1 resulting in a carbon-carbon triple bond in Compound 2. The carbon-carbon triple bond involves sp hybrid orbitals. Thus, Choice A is the best answer.

188. The question asks about the effect of an increase in the level of albumin, one of the major plasma proteins. Because albumin has nothing to do with the immune response, answer choice A is incorrect. The plasma proteins can not cross the walls of blood vessels, but water molecules can. The wall of the artery acts as a semipermeable membrane setting up the conditions needed for osmosis to occur. An increase in plasma albumin will upset the osmotic balance because the blood will become hypertonic with respect to the tissue. Water will have to flow into the bloodstream to reestablish equilibrium. One of the causes of edema, increasedfluid in body tissues, is a decrease in the plasma protein level. This occurs, for instance, in starvation when thebody is forced to use its albumin as an energy source. An increase in the plasma protein level would have the opposite effect: fluid would enter the bloodstream (answer D).

189. The taxonomic levels of living organisms, in order, from the groups consisting of the least closely

related to most closely related are: kingdom, phylum, class, order, family, genus, and species. Distantly related organisms may be in the same kingdom, but as one descends the hierarchy the taxonomic groups consist of smaller and smaller numbers of more and more closely related organisms. The question asks which taxonomiclevel groups organisms that are the most closely related. The answer is genus, choice B. No other answer choice is a taxonomic level below genus. There are many mnemonics that students use to remember the order of the taxonomic levels, such as, "King Phillip came over from Great Spain." Another way to identify the lowest level of grouping is to recall that the Linnean binominal nomenclature for a species consists of its genus name and its species name (Homo sapiens, for instance). Individual species can be uniquely named using this combination of names because these are the two lowest levels of the taxonomic series.

190. In imprinting, the perception of an object is enhanced in some way during a critical period early in development. The term was first used to describe the following behavior of ducklings, so the correct answer isanswer choice A. Habituation, answer choice B, is a form of non-associative learning in which the response toa stimulus decreases over time, perhaps due to some kind of neural fatigue. Conditioning, answer choice C, is a form of associative learning. In classical conditioning, for instance, an animal may come to associate the sound of a bell with the presentation of food. In operant conditioning, the performance of an act with the anticipation of a reward, discrimination (answer D) is a perceptual process, not a learning process.

191. The potential energy of a cycloalkane is a function of its torsional, steric, and ring strains. In order to make comparisons between cycloalkanes with different numbers of methylene units, divide the heats of combustion by the number of methylene units in the cylcoalkane to get the potential energy per methylene. The potential energy of cyclohexane per methylene is at a minimum when compared to the other cycloalkanes, because it can adopt a conformation (i.e., a chair) that minimizes the torsional, steric, and ring strains. When it undergoes combustion, cyclohexane yields less heat per methylene than the other cycloalkanes, because it is initially at a lower potential energy per methylene. Thus, answer choice C is the best answer.

192. The compound shown is aromatic and contains tertiary amino groups. Amines can act as bases or nucleophiles because of the lone pair of electrons on the nitrogen of the amino group. However, the nitrogens of the tertiary amino groups in the compound shown are sterically hindered. As a result this compound is not agood nucleophile. Thus, Choice D is the best answer.

193. The glomerulus is a tuft of capillaries that bulges into the capsular space (also referred to in some texts by its eponym, Bowman's space), which is a potential space lined with simple squamous epithelium. Fluid is expressed from the blood across the capillary endothelium. It enters first the capsular space, then the kidney tubule. The part of the tubule closest to the glomerulus is the proximal tubule, the next part is the U-shaped loop of Henle and the last part of the tubule is the distal tubule. Suppose one gradually increased the rate at which fluid was expressed from the bloodstream (the glomerular filtrate rate) and measured the concentration of some substance reabsorbed by the kidney as it left the distal tubule as urine. The concentration might not rise at first, because the cells lining the tubule might be completely reabsorbing the substance and putting it back into the blood stream. Eventually, however, the rate of flow would reach a point at which it exceeds the rate at which the tubule cells could reabsorb the substance. The fluid would be flowing too rapidly through thetubule for the cells to reabsorb all the substance. The rate of flow through the tubule at which the substance begins to be observed in the urine is Tm. At that point the rate of flow of fluid through the tubule begins to exceed the capacity of the kidney tubule cells to reabsorb the substance. This description of Tm is given only in answer choice C, the correct answer.

194. The amount of glucose in the fluid in the capsular space as it enters the tubule system would be the same as the amount of glucose in the plasma, but by the time the fluid reaches the distal tubule much of the

glucose will have been reabsorbed. If the amount being filtered per minute is less than Tm, all of it will have been reabsorbed. If it is more than Tm, then the amount will be linearly related to the amount in the plasma. In the case proposed in this question, the amount being filtered per minute (125mg/min) is less than the Tm for glucose given in the passage (320 mg/min), so all of the glucose should be reabsorbed. The correct answer, therefore, is 0 mg/min, answer choice A.

195. Low blood pressure could have an effect on ADH levels (as suggested in answer choices B and D), which could, in turn, affect the amount of substance A in the urine, but there is not enough information in the passage to decide if such an effect exists and, if it does, whether its effect would overcome the certain affect of lowering the blood pressure. The best answer is answer choice C: that low blood pressure decreases the glomerular filtration rate, allowing more time for reabsorption and decreasing the amount of substance A in the urine. Blood pressure is the source of the energy that forces fluid into the capsular space. If the heart stopped and the blood in the glomerular capillaries had no hydrostatic pressure, fluid in the space around the glomerulus would flow back into the capillary bloodstream. This would occur because the protein-rich blood would be hypertonic with respect to the protein-poor fluid in the capsular space so that the fluid would flow down the osmotic gradient into the blood.

196. The Tm is a characteristic of the individual substances in the tubule system and a measure of how efficiently each substance can be reabsorbed. A high Tm indicates a high capacity for reabsorption of substances in the kidney tubules. In figure 1, the Tm for each substance can be read as the concentration in plasma when the concentration in the urine is zero. In this case it looks like the Tm for Substance A is a little over 6 mg/mL and that of glucose is 10 mg/mL. So substance A has a lower Tm. This means the tubules will not reabsorb it very efficiently. Much of it will be spilling into the urine, thus being eliminated from the body.The question asks which will clear from the blood more rapidly at a concentration of 8 mg/mL and why. The answer is that all glucose will be reabsorbed at that concentration, none will appear in the urine and none will be cleared from the plasma. Glucose has such a high Tm, that all of the glucose will be reabsorbed into the bloodstream, perhaps to reenter the kidney tubule again. A rate of 8mg/mL is above the Tm of substance A, sothere will be some substance A in the urine at this plasma concentration. The answer to the question depends on the value of Tm, not the slope of the clearance line, so answer choices A and D can be eliminated. Substance A will clear more rapidly than glucose, therefore, answer choice B is correct.

197. Above 10 mg/mL, glucose begins to be found in the urine. The Tm for glucose is therefore 10 mg/mL, answer choice B. This can be read from the graph by looking at the concentration in the plasma where the concentration in the urine is zero. In other words, where the clearance line for glucose crosses the axis.

198. The best answer is that increased blood pressure will affect the glomerular filtration rate, answer choice A. Tm is a characteristic that depends on the characteristics of the cells lining the renal tubules and independent of blood pressure, so answer choice B is not correct. Water reabsorption and concentrating abilityare the same, so answer choices C and D are essentially the same. Increasing blood pressure should increase flow of fluid through the kidney system and decrease, rather than increase, water reabsorption, so these answer choices are incorrect.

199. The glomerulus functions as a filter, removing fluid from the capillary blood and leaving behind molecules and formed elements too large to pass through. This fluid is called glomerular filtrate. The capsular space is lined with simple squamous cells unsuited for reabsorption, so answer choice A is incorrect. The first reabsorptive segment of the tubule is the proximal tubule. Most of the glucose should be reabsorbed in the

proximal tubule since it is the first segment to process the filtrate. In fact, normally, all glucose reabsorption takesplace in the proximal tubule (which has been calculated to remove half a pound of glucose a day from the glomerular filtrate). The proximal tubule is lined with simple columnar cells with a brush border on the luminal side consisting of microvilli. The microvilli give the cell an extensive surface area in contact with the filtrate to facilitate reabsorption. Therefore, the correct answer is answer choice B.

200. The correct answer is answer choice A: that women have lower bone density than men. Men and women have the same number of vertebrae, and there is no evidence presented in the passage to suggest that there are sex differences in calcium uptake or vitamin D production. Lower initial bone density would increase the impact of any subsequent bone loss.

201. Estrogen and progesterone are actively secreted by the ovaries of pre-menopausal women and act to maintain the uterine cycle. With advancing age the ovary becomes less responsive to pituitary gonadotropins and cyclical changes in the endometrium of the uterus disappear. The menstrual cycle can be re-established byadministration of estrogen and progesterone in a regimen that approximates the rise and fall of hormone levelsin pre-menopausal women. So the correct answer is answer choice C, periodic menstruation will resume. To the extent that the therapeutic dosages are typical of pre-menopausal women, side effects such as breast tissue atrophy (answer A), vaginal tissue drying (answer B) and lactation (answer D) would not be expected to occur.

202. Calcium levels in the blood need to be kept constant. Parathyroid hormone and calcitonin regulate bloodlevels of calcium. The passage gave the information that calcitonin analogs inhibit osteoporosis. If calcitonin inhibits osteoporosis, it must function to take calcium out of the blood and into the bone while preventing the loss of calcium from bone into the blood. High levels of blood calcium available for deposit into bone should stimulate this process rather than inhibit it. Thus answer choice B, calcitonin, must be incorrect. Because growth hormone (answer A) and thyroid hormone (answer C) are not involved in the regulation of blood calcium levels, they are also incorrect. The correct answer must be parathyroid hormone (answer D) which is inhibited by high levels of calcium. A mnemonic students use to remember which hormone puts calcium into bone and which gets rid of it is: Calcitonin - in, parathro id-rid.

203. The passage only states that post-menopausal women show the accelerated bone loss and the acceleration slows after eight to ten years. From this it seems likely that the acceleration has something to do with withdrawal of estrogen. Administration of estrogen to men would then have no therapeutic effect on bone loss, since they should not be suffering from withdrawal of the hormone, having never had high levels. Men given estrogen should have the same chance of developing osteoporosis as a control population, answer choice C.

204. Osteoporosis involves two factors, decreased formation and increased breakdown of bone. Osteoblasts mediate the former process and osteoclasts the latter. Decreased osteoblast activity and increased osteoclast activity result in bone loss, answer choice D.

205. The question states that the product of the aldol self-condensation is a -hydroxy ketone, meaning that the condensation is not complete, in that the product has not lost water. (Note: An aldol addition reaction forms a -hydroxycarbonyl compound; whereas, an aldol condensation reaction implies that the addition product has lost water.) Because the addition product is a dimer of the starting compound, the product has a molar mass that is twice that of the starting compound. In this case, the molecular weight of the starting compound is 144/2 = 72 g/mol = Answer B.

206. This question involves a reaction at equilibrium. Thus, Le Chtelier's principle is in effect, and the

position of equilibrium can be shifted toward Product A by removing it as it forms. Catalysts (Answers A and B) do not affect the position of equilibrium, and heating acetone to the boiling point (Answer D) would remove reactant and shift the equilibrium to the left. Thus, answer choice C is the best answer.

207. The compound at answer choice D is a constitutional isomer of Product A; the other compounds are not.Thus, answer choice D is the best answer.

208. Product B contains a double bond, because it results from the dehydration of Product A. Thus, the addition of a drop of Br2/CCl4 to Product B will result in the red color of bromine disappearing as bromine adds to the double bond. Thus, answer choice B is the best answer.

209. The six hydrogen atoms in acetone are magnetically equivalent and their 1H NMR signal appears as a singlet near =2 ppm. Thus, answer choice A is the best answer.

210. All three compounds are methyl ketones. The passage states that Product A gives a positive iodoform test, so we can conclude that all three compounds give a positive iodoform test. Thus, answer choice D is the best answer.

211. The inner membrane of a mitochondrion is analogous to the plasma membrane of a prokaryote. The enzymes for oxidative phosphorylation are embedded in the inner membrane. The endosymbiotic theory suggests that mitochondria are descendents of prokaryotes that were engulfed by endocytosis into a vesicle lined with a membrane derived from the cell membrane of the eukaryote host. This is the outer membrane. The inner membrane is the plasma membrane of the endosymbiotic prokaryote, so answer choice D is correct.Answer choice A is incorrect because the cell wall of bacteria does not resemble a plasma membrane. Ribosomes have nothing to do with oxidative phosphorylation, so answer choice B is incorrect. Bacteria lack a nuclear membrane, thus answer choice C is also incorrect.

212. Nucleosomes consist of histones and DNA and are a basic structure of eukaryotic chromosomes. Therefore, the key is B.