## AAMC Practice Test 4 - Kaplan Test Prep

MCAT*. *MCAT is a registered trademark of the Association of American Medical Colleges. KAPLAN EXPLANATIONS FOR. AAMC Practice Test 4 ...
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kaplan explanations for

AAMC Practice Test 4

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Physical Sciences

1. D

14. A

27. B

40. B

2. A

15. D

28. C

41. A

3. B

16. A

29. D

42. B

4. D

17. C

30. C

43. A

5. D

18. C

31. D

44. C

6. A

19. A

32. C

45. B

7. C

20. C

33. A

46. C

8. A

21. C

34. D

47. D

9. B

22. A

35. B

48. C

10. D

23. D

36. A

49. A

11. A

24. D

37. C

50. B

12. C

25. B

38. A

51. B

13. A

26. A

39. D

52. D

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AAMC Test 4 Explanations

AAMC Test 4 Physical Sciences—Kaplan Explanations Passage I (Items 1–5) Passage Map Passage Type: Experiment Topic: Static friction Scope: Experiment to determine static friction ¶1: Static friction was studied using blocks with same mass and base area but made up of different building materials. ¶2: Pulley system was used to determine static friction of blocks on wooden board. Fig. 1: Pulley system used to determine static friction of block Table 1: Threshold mass for each type of block at different block base areas 1. (D) Glancing at the answer choices, we confirm that the item requires us to apply our knowledge of conservation of energy. The answer choices consist of two parts, such that we must determine the type and source of the energy responsible for the kinetic energy of the sliding block. To answer the question, we must understand the experimental setup, so let’s start by reviewing the passage and focus on the details that might help us determine the energy of the block. According to our passage map, paragraph 2 discusses the methods used to measure the threshold mass. We are told that increasing the mass hung on the hook eventually causes the block to slide. When the mass is hanging from the hook, it has potential energy due to gravity. When the mass and the block system begin to move, this hanging block loses potential energy as it gets closer to the ground. Where does this energy go? Since the sliding block is moving, its increased kinetic energy originates from the hanging block’s initial potential energy. Although some of the energy will be lost through friction, the flow from gravitational to kinetic energy will still occur. Thus, we can conclude that the kinetic energy of the block comes from the gravitational potential energy of the mass on the hook, which matches with choice (D). (A) Distortion. The kinetic energy of the string also comes from the gravitational potential energy of the mass. The string is not the origin of this energy, but rather, it is the gravitational potential energy of the mass on the hook that makes the string and the block move along. (B) Distortion. The board remains stationary; therefore, it doesn’t have any kinetic energy. (C) Distortion. Although selecting a “zero” for the gravitational potential energy is a relative process, since the block does not change its height while it moves, we can correctly assume that there is no change in potential energy. 2. (A) The item asks us to use the table provided in the passage so as to determine the relationship between static friction and base area. Indeed, all of the answer choices attempt to provide a mathematical proportionality between the two values—it is our immediate task to use the data in Table 1 to establish this relationship. According to Table 1, base area is provided in the first column. However, which column is indicative of the static friction measurement? Since the experiment relies on a mass hanging from a hook to determine the weight at

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AAMC Test 4 Explanations

the system accelerates. In other words, we would expect the block and the mass to experience a net acceleration. Since the net force is constant, the net acceleration will also be constant. A constant acceleration will increase the velocity at a linear rate—only choice (D) mirrors our prediction and is thus the correct response. (A) Distortion. Because the force is continually applied, the block accelerates. The speed would be constant only if there was no acceleration, that is, if the net force was equal to zero. However, the weight of the mass provides the system with a net force downwards, indicating a non-constant velocity. (B) Distortion. Constant, non-changing acceleration causes a linear increase in velocity. (C) Distortion. Because the force is continually applied, the block accelerates, suggesting that the velocity is never constant. 5. (D) This item presents us with a novel situation in which a sliding board is used to measure the force of static friction. By changing the angle of inclination of the board, the researchers are able to determine when the block begins to slide. Why? In this scenario, rather than having a hanging mass supplying the force required to move the block, it is the force of gravity pulling the object down along the surface of the board that supplies the driving force. The influence of gravity on the object depends entirely on the angle of inclination of the plane with respect to the horizontal (if we were to draw a force diagram for an object on an inclined plane at an angle θ, the component of gravity parallel to the plane would be equal to mg sin θ). Basically, changing the angle in this new experiment is equivalent to changing the mass used in the original experiment presented in the passage. From the given answer choices, only choice (D) matches our prediction. (A) Distortion. The time it takes for the block to slide down the board depends on the kinetic friction, not the static friction. (B) Distortion. The distance the block slides depends on where the block starts, and is irrelevant when considering the static friction. (C) Distortion. The mass of the board does not affect how the block moves; it simply provides a surface for the block to slide on.

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Passage II (Items 6–9) Passage Map Passage Type: Information Topic: Gases Scope: Chemicals produced that are either gaseous or produced by gaseous reactions ¶1: Sulfuric acid (H2SO4) is produced by gaseous reactions. Rxn. 1: Sulfur and oxygen combine to form sulfur dioxide. Rxn. 2: Sulfur dioxide and oxygen react to form sulfur trioxide. Rxn. 3: Sulfur trioxide and water react to form sulfuric acid. ¶2: NH3 forms coordination compounds in which it forms covalent bonds with a transition metal. Rxn. 4: Nitrogen and sulfur combine to form ammonia. ¶3: Nitrogen and oxygen are prepared by fractional distillation of air; nitrogen separates first. 6. (A) The item asks us to determine why NH3 is able to form coordination compounds with transition metal ions. The answer choices are short explanations, but they do not lead us in any particular direction, so let’s use our map of the passage to determine where coordination compounds were discussed. Paragraph 2 tells us that a coordination compound forms when NH3 forms covalent bonds with a transition metal. In order for a covalent bond to form, the two atoms involved in bonding must share a pair of electrons. Transition metals are relatively electron-deficient. Because of their partially-filled d orbitals, they have a large valence shell to fill and relatively few electrons to fill them. On the other hand, NH3 has a lone pair of electrons offered by the nitrogen atom. Therefore, we expect NH3 to donate an electron pair to form a covalent bond with a transition metal. This bonding reaction is essentially a Lewis acid-base reaction: NH3 (an electron pair donor, i.e., a Lewis base) donates an electron pair to a transition metal (an electron pair acceptor, i.e., a Lewis acid). The answer choice that mirrors our prediction is choice (A). (B) Faulty use of detail. Although NH3 is capable of hydrogen bonding, such bonding is due to intermolecular forces, whereas the question specifically asks us about coordination complexes resulting from covalent bonds, which are intramolecular forces. (C) Faulty use of detail. Although NH3 is a weak base, its ability to donate a pair of electrons is what allows it to form coordination compounds. (D) Faulty use of detail. Although it is true that the oxidation number of nitrogen is –3, this does not aid its ability to form coordination compounds. 7. (C) This item tests an all time favorite MCAT concept—boiling point. Based on the answer choices, our task is to determine the main difference that explains why nitrogen has a lower boiling point. Let’s start by quickly reviewing what the boiling point assumes about the nature of a compound. In order for a liquid to boil, or enter the gas phase, the intermolecular forces between molecules must be broken. In our stem, for oxygen and nitrogen, the only intermolecular forces that exist are van der Waals forces. These forces are caused by temporary shifts in the

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electron density of the molecule—a temporary dipole moment of the molecule is created by more electrons being present on one side of the molecule than the other. These temporary dipole moments create attractive forces between individual molecules. Van der Waals forces are stronger when they occur between larger molecules because such molecules will have a greater dipole moment and a greater polarizability (the separation of charge will increase due to the size of the molecule). For the purposes of our question, we are told that nitrogen has a lower boiling point, which implies that it has weaker van der Waals forces, and thus weaker intermolecular interactions to overcome than oxygen. This, in turn, implies that the nitrogen molecule is smaller than oxygen— this is true since the molar mass of nitrogen is 28, while that of oxygen is 32. As a result, nitrogen has a lower boiling point because less energy is required to overcome the van der Waals forces between these relatively smaller molecules—choice (C) is the correct answer. (A) Distortion. The reactivity of a molecule does not influence its boiling point. (B) Distortion. Electronegativity does not affect boiling point. (D) Distortion. Intermolecular, not intramolecular, forces determine boiling point. 8. (A) This item asks us to examine Reaction IV as described in the passage and apply our knowledge of reaction equilibria so as to increase the formation of the products—this question tests us on an MCAT favorite, Le Châtelier’s principle. Le Châtelier’s principle tells us that when a reaction is subjected to a stressor (a change in concentration of the reactants or products, or a change in temperature or pressure), the equilibrium of the reaction will shift so as to eliminate the stressor. For example, the addition of reactants and the removal of products will shift a reaction towards the product side of the reaction. Returning to our question, we can infer that in order to make more NH3, we must either remove the product, NH3, or increase the concentration of one of the reactants, N2 or H2. Since the reaction is exothermic (ΔH is negative), we could also decrease the temperature (heat is on the right side of the reaction, since it is a product) to increase the production of NH3. Lastly, increasing the pressure will favor the side with the fewer moles of gas, so our reaction equilibrium will shift towards the product side. Any one of these stressors would work so as to increase the formation of NH3—only choice (A) agrees with our prediction. (B) Opposite. Decreasing pressure will shift the reaction towards the side with a greater number of gas moles, which in this case would cause the concentration of reactants to increase. (C) Opposite. Because heat is a product, increasing the temperature will shift the reaction towards the side of reactants, so as to “use up” the excess heat. (D) Distortion. Catalysts only alter the rate of a reaction, not the state of equilibrium. 9. (B) This item prompts us to analyze Reaction I, and determine which of the three given choices is the limiting reagent—let’s start by reviewing this concept. The limiting reagent of a reaction is the reagent that is completely consumed first. This implies that the limiting reagent is one of the reactants, not the products, which allows us to eliminate choice (C). In order to determine the limiting reagent, we must compare the number of moles present for each reactant to the molar ratio given by the balanced reaction. In examining Reaction I, we see that one mole of S reacts with one mole of O2 to form SO2. Given the masses of the two reactants, let’s calculate the number of moles present for S and O2. The molecular weight of S is 32 g, and the molecular weight of O2 is 32 g. Therefore, if 36 g of S and 32 g of O2 are used, this means that just over 1 mole of S and exactly 1 mole of O2 are supplied. Since the reactants react in a 1:1 ratio, O2 will be consumed before S. As a result, we will be left

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with an excess of S. We predict, therefore, that O2 is the limiting reagent since it will be completely consumed first—choice (B) is the correct answer. (A) Distortion. More moles of S are available than O2, making O2 the limiting reagent. (C) Distortion. SO2 is a product and therefore cannot be a limiting reagent. (D) Distortion. More moles of S are available than O2, making O2 the limiting reagent.

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AAMC Test 4 Explanations

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Physical Sciences

as dense as the first liquid, we expect that the pressure in the liquid is twice as big. The pressure of 50 N/m2 at zero depth must be accounted for. At a depth of 10 cm, therefore, the pressure exerted by the second liquid will be 2 × 400 + 50 = 850 N/m2. Again, choice (C) is the correct answer. (A) Opposite. This halves instead of doubles the pressure. (B) Distortion. The pressure would not remain constant. (D) Distortion. This quadruples instead of doubles the pressure. 18. (C) Since we are asked to determine the specific gravity of an object, we know that we are being tested on our understanding of hydrostatics. The question presents us with a short experimental design. Our first task is to understand what we are asked, so let’s start by defining specific gravity. Specific gravity is the ratio of the density of a substance to the density of water. In order to determine the specific gravity of the object, therefore, we need to calculate its density. First, let’s analyze what happens when the object is submerged in the liquid. When immersed in benzene, the object seems to lose 5 grams of mass, so that its mass is now 10 grams. This is not to say that it dissolves, but rather, its mass, if it were to be measured, would be 5 grams less than we would expect. Since the mass (and weight) is related to the forces that the object experiences, we can infer that something must be pushing the object upwards. Indeed, according to Archimedes’ principle, when an object is submerged in a liquid, it will experience a buoyant force that pushes up, opposing the force of gravity. The buoyant force on an object immersed in a fluid is equal to the weight of the fluid displaced by the object. With this information in mind, we can deduce that the apparent loss of mass that the object experiences is due to the buoyant force from the benzene that pushes the object upwards. Relating this to the values given in the question stem, we can infer that the true mass of the object is 15 grams, while the mass of the benzene displaced is 5 grams. Knowing the masses of each of the two variables, we can now calculate the density of the object. The question stem provides us with the specific gravity of benzene. Now, we must calculate the density of the object in relation to that of benzene. The specific gravity (SG) of benzene is:

ρbenzene SGbenzene = ________ ​  ρ  ​    water

and the specific gravity of the object is:

ρobject ​ ρ  ​  SGobject = ______ water

Since we do not know the density of water, we can combine the two equations so as to eliminate ρwater. From the first equation, the density of water can be calculated as:

ρ ρwater = _________ ​  benzene   ​  SGbenzene

Replacing this value in the second equation, we obtain the specific gravity of the object:

ρobject SGobject = ________ ​ _______     ρ ​ SGbenzene     ​ benzene

​ or

SG SGobject = ρobject × _________ ​  ρ benzene ​    benzene

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AAMC Test 4 Explanations

We have already been given the specific gravity of benzene in the question stem, so we now have to calculate the densities of the object and benzene using the definition of density.

mobject m ρobject = ______ ​   ​  and ρbenzene = ________ ​  benzene ​  Vobject Vbenzene

Plugging in the density values back into the original equation, we obtain the following:

mobject SG SGobject = ______ ​   ​  × __________ ​  mbenzene      ​ Vobject ______ ​  benzene   ​ Vbenzene

In this question, since the object is completely immersed under the liquid, the volume of the benzene displaced is equal to the volume of the object submerged, so we can rewrite the last equation so as to cancel out the two:

mobject SG SGobject = ______ ​   ​  × __________ ​  mbenzene      ​ Vobject ______ ​  benzene    ​  Vobject

SGobject = mobject

SG × _________ ​ m benzene ​  benzene

We already established that the mass of the object is 15 grams, the mass of benzene is 5 grams, and the specific gravity of benzene is 0.7, so let’s plug these values in:

SGobject = 15 g × ____ ​ 0.7 ​ = 3 × 0.7 = 2.1 5g

Our result matches with choice (C). A quicker way to calculate this answer, although perhaps not as intuitive, is to realize that since the object suffers an apparent loss of a third of its true mass, its density has to be three times bigger than that of benzene. Since the density of benzene is 0.7, the density (and therefore specific gravity) of the object must be 2.1—again, choice (C) is the correct answer. (A) Miscalculation. This is twice the specific gravity of benzene. (B) Miscalculation. As always, be careful with your calculations. (D) Miscalculation. As always, be careful with your calculations. 19. (A) The item asks us to determine what happens when a metal that is at its melting temperature is heated up for a short time. Glancing at the answer choices, we notice that we are specifically looking at phase changes and the temperature associated with them. Let’s start by reviewing what happens when a solid is heated up. If the temperature applied is higher than the melting point of the substance, the solid will eventually melt. Melting is a process during which the intermolecular forces are broken. While the temperatures of the solid and the melted liquid increase as the temperature increases, during the actual melting process, the input heat is used to break the intermolecular bonds in the solid so as to form a liquid. As such, during that brief period in which the solid melts, the temperature of the substance remains constant until all of it is melted. More specifically, this occurs because the latent heat of fusion is needed to convert a substance from solid to liquid. Returning to our question, we are told that the metal is at the exact melting temperature. By holding it for a fraction of a second under a Bunsen burner, we are applying a small amount of heat to the metal. However, the small amount of energy supplied by the Bunsen burner during less than a second is too small to provide the heat of fusion for 100 g of

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metal; it may be enough to provide the heat of fusion for a portion of the metal. We would expect, therefore, that a portion of the metal will melt; however, the temperature of the metal will not change until all of the metal is melted. As a result, a small amount of the metal will turn to liquid but the temperature of the entire system will remain the same—our prediction matches with choice (A), the correct answer. (B) Distortion. Only a small amount of the metal will turn to liquid; the energy from the Bunsen burner is not enough to melt all of the metal. (C) Distortion. The temperature will remain constant until all of the metal melts. (D) Distortion. The temperature will remain constant until all of the metal melts.

Passage IV (Items 20–26) Passage Map Passage Type: Experiment Topic: Volumetric Analysis Scope: To determine the molecular weight of weak acids using volumetric analysis ¶1: Volumetric analysis is used to determine molar mass of an unknown weak acid. Table 1: Names and formulae of weak acids ¶2: 8 g of NaOH was dissolved in 2 L of water. Eq. 1: Solvation of NaOH in water ¶3: NaOH solution was standardized against KHP. Eq. 2: Reaction of NaOH with KHP – standardization of NaOH ¶4: Unknowns were dissolved in water/water-ethanol and titrated with NaOH. Eq. 3: Reaction of benzoic acid with NaOH ¶5: Table 2 contains experiment data collected by students. Table 2: Partial data of experiments 20. (C) This is a stoichiometry item, since we are asked to calculate the number of moles of a substance using the data from the passage. The item stem prompts us to analyze the results of student A. According to our passage map, the best place to search for such information is Table 2. In Table 2, we see that student A uses 0.5500 g of KHP. In order to calculate the number of moles, however, we need not only the mass, but also the molecular weight of the compound, so let’s also search for this in the passage. According to our passage map, KHP is mentioned in paragraph 3. Here, we are told that KHP has a molar mass of 204.2 g/mol. To determine the number of moles present, we have to divide the mass by the molar mass of KHP as follows: m nKHP = _______ ​  KHP   ​  MWKHP 0.5500 g nKHP = ​ ____________       ​ 204.2 g/mol

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AAMC Test 4 Explanations

We can speed up our calculations by using scientific notations: –2 55 × 10 nKHP = ​ ____________      2 ​ mol 2.042 × 10

Also, the results in the answer choices differ by an order of magnitude, so we can safely use approximations:

× 10–2 ​  __________ nKHP ≈ ​ 55   mol 2 × 102

nKHP ≈ 25 × 10–4 mol ≈ 2.5 × 10–3 mol

The answer choice that most closely matches our result is choice (C). (A) Miscalculation. As always, be careful when using scientific notations in your calculations. (B) Miscalculation. This is an error of magnitude. (D) Miscalculation. As always, be careful when using scientific notation in your calculations. 21. (C) This item tests our understanding of state functions, so let’s start by defining this term. A state function is a property whose magnitude depends only on the initial and final states of the system, and not on the path of the change. Examples of state functions are enthalpy, entropy, free energy, pressure, temperature, and volume. Specifically, we have to determine how the enthalpy, entropy, and Gibbs free energy changed during Equation 1. This is a Roman numeral question, and while generally we would evaluate the statement that is most common in the answer choices first, in this situation, we can analyze them in order, since each statement appears in exactly two answer choices. Glancing back at the passage, we notice that Equation 1 shows the dissolution of NaOH in water. What other information does the passage give us about this reaction? According to our map, paragraph 2 describes this reaction—we are told that the temperature of the solution rose during the dissolution of NaOH in water. This means that heat is released, meaning that the reaction is exothermic. The enthalpy of an exothermic reaction is less than zero (ΔH < 0); therefore, statement I is true—we can eliminate choices (B) and (D). Since neither choice (A) nor choice (C) includes statement II, we can skip it, moving instead on to statement III. Whenever a solid is dissolved in water, the disorder (and therefore the entropy) increases. Therefore, the entropy change for Equation 1 is greater than zero (ΔS > 0), so statement III must be true. Answer choice (C) is thus the correct response. Roman numeral I – True: Since heat is a product of the reaction, the system is exothermic (ΔH < 0). Roman numeral II – False: According to the Gibbs free energy equation, ΔG = ΔH – TΔS, because the enthalpy is less than 0 and the entropy is greater than zero, the free energy must be less than zero. Roman numeral III – True: Whenever a solid is dissolved in water, the disorder (and therefore the entropy) increases (ΔS > 0). (A) Distortion. Statement III is also true. (B) Distortion. Statement II is false. (D) Distortion. Statement II is false.

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22. (A) This is a straightforward item that asks us to determine the molarity of a solution. Where does the passage discuss the solution prepared by the instructor? According to our passage map, paragraph 2 describes the solution as being prepared by dissolving 8 g of NaOH in 2 L of water. To determine the molarity (in moles/liter), we must first calculate the number of moles of NaOH, and take into consideration the volume of the solution. m nNaOH = ________ ​  NaOH   ​  MWNaOH 8g nNaOH = ​ _________     ​ 40 g/mol nNaOH = 0.2 mol We have 0.2 mol of NaOH in 2 L of water. In order to calculate the molarity, we have to divide the number of moles of NaOH over the total volume of water as such: mol _______  ​  = 0.1 mol/L = 0.1 M Molarity = ​ 0.2   2L The answer choice that matches our response is choice (A). (B) Miscalculation. This is the number of moles of NaOH, but since 2 L of water were used, the molarity is half of this value. (C) Miscalculation. This would be the molarity of the solution if 0.5 L of water were used. (D) Miscalculation. This answer would be incorrectly obtained if you divided the number of grams of NaOH by the volume of water. 23. (D) This item tests our understanding of titrations and pH. Glancing at the answer choices, we notice that we do not have to calculate an exact value, but rather, we need to make a rough estimate of whether the solution will be acidic, basic, or neutral. As a quick check, we can remind ourselves that a solution with a pH of 7 is neutral, a pH lower than 7 is acidic, whereas a pH higher than 7 is basic. We are prompted to look back at Equation 3 to determine the pH of the solution at the equivalence point. The equivalence point is the point where the number of acid equivalents is equal to the number of base equivalents. At the equivalence point of a titration of benzoic acid with sodium hydroxide, the solution is a mixture of sodium benzoate and water. Knowing this, at equivalence, the solution contains water (neutral by nature) and a base (pH > 7), which leads us to predict that the solution has a pH above 7—choice (D) is the correct answer. (A) Opposite. This answer is representative of an acidic solution; the solution in Equation 3 is basic. (B) Opposite. This answer is representative of an acidic solution; the solution in Equation 3 is basic. (C) Distortion. This answer is representative of a neutral solution; the solution in Equation 3 is basic. 24. (D) This is a straightforward item testing our knowledge of acids and their conjugate bases. An acid and its conjugate base differ by a single proton. The molecular formula for chlorobenzoic acid is shown in Table 1. The difference between chlorobenzoic acid (HC7H4ClO2) and chlorobenzoate is a single proton. The conjugate base of chlorobenzoic acid is therefore C7H4ClO2–—the correct answer is choice (D).

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AAMC Test 4 Explanations

(A) Distortion. OH– is the conjugate base of water. Hydroxide ions are produced by Arrhenius bases (such as NaOH, KOH, or other compounds that contain a hydroxyl group). (B) Distortion. Water is the conjugate base of H3O+. (C) Distortion. This formula is lacking an oxygen atom. 25. (B) This item requires us to interpret the results of one of the experiments performed in the passage. Specifically, we have to determine what student E concluded about the titration of succinic acid with NaOH. We can categorize the answer choices into two groups, which claim that succinic acid is either diprotic or triprotic. According to Table 1, the molecular formula of succinic acid is H2C4H4O4. Since this acid has only two protons, it is diprotic—we can now eliminate choices (C) and (D). How many moles of base will a diprotic acid require to be neutralized? Considering that NaOH provides only one mole of hydroxide ions per mole of solution and that succinic acid has two protons, we would expect that each molecule of succinic acid will require two hydroxide ions in order to be neutralized. Thus, succinic acid requires twice as much NaOH than a monoprotic acid. The only answer choice that mirrors our prediction is choice (B). (A) Opposite. Succinic acid requires twice the number of moles of NaOH expected for a monoprotic acid. (C) Distortion. Succinic acid has two protons; it is diprotic. Also, it requires twice the number of moles of NaOH expected for a monoprotic acid. (D) Distortion. Succinic acid has two protons; it is diprotic. Also, it requires twice the number of moles of NaOH expected for a monoprotic acid. 26. (A) This item tests our understanding of the passage and stoichiometry. In the item stem, we are given a hypothetical situation in which the standardization of NaOH with KHP was calculated incorrectly by including water in the total mass of KHP. Glancing at the answer choices, we notice that our specific task is to determine if the molarity of NaOH calculated was too high, too low, or the same as compared to the actual value. Let’s quickly review the passage to find where the standardization of NaOH is discussed. According to paragraph 3 and Equation 2, in the titration of KHP with NaOH, the number of moles of KHP titrated is equal to the number of moles of NaOH needed to reach the equivalence point. If the KHP contains moisture, then the mass of the KHP is greater than the mass of dehydrated KHP. Therefore, because water was included when measuring the mass of KHP, the number of moles of KHP actually present is less than the theoretical number of moles of KHP used in the calculation. For example, if 204.2 g of KHP is used, the calculated number of moles will be 1.0. However, if the KHP contains any moisture, less than 1.0 mole of KHP will be present in solution, even though the student might think that she has 1.0 mole. The molarity of NaOH is calculated by determining the volume of NaOH needed to reach the equivalence point. Because the actual number of moles of KHP titrated is less than the number used in calculation, the amount of NaOH actually needed to reach the equivalence point will also be lower. This results in a falsely elevated calculated molarity of NaOH. The only answer choice that describes this situation is choice (A). (B) Opposite. The calculated molarity of NaOH would be too high since the student thought she had more moles of KHP than she actually did. (C) Opposite. The calculated molarity of NaOH would be too high since the student thought she had more moles of KHP than she actually did. (D) Distortion. The student took into consideration the number of moles of KHP as calculated from the mass of KHP. This answer choice would be true if we were looking strictly at molarity.

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Physical Sciences

Passage V (Items 27–31) Passage Map Passage Type: Experiment Topic: Potassium nitrate Scope: Properties of potassium nitrate and its solution ¶1: Properties of potassium nitrate ¶2: A 20% by mass solution of KNO3 in water was prepared. Table 1: Properties of a 20% by mass solution of KNO3 in water ¶3: Definition of condosity 27. (B) Before we analyze the item, let’s characterize the answer choices. They are all expressions, so there is no need in carrying out any calculations. We are required to set up the mole fraction of KNO3 in the solution prepared in paragraph 2. The mole fraction of a substance is the number of moles of that substance divided by the total number of moles of all species in the solution. Knowing this, the correct answer choice should have the number of moles of KNO3 in the numerator and the number of moles of KNO3 plus the number of moles of water in the denominator. In order to calculate the mole fraction of KNO3, we must first determine the number of moles of potassium nitrate and the number of moles of water in the solution. Where in the passage can we find this information? According to our passage map, Table 1 gives us the solute concentration (or the potassium nitrate concentration) as 226.5 g/L. So, in 1 L of solution, there are 226.5 g of potassium nitrate present in the solution. According to the first sentence in the passage, the molecular weight of potassium nitrate is 101.1 g/mol. From this information, we can calculate the number of moles of potassium nitrate as being the mass of solute divided 226.5 by its molecular weight: ​ ____ ​. This number should appear in both the numerator and the denominator of the correct 101.1   answer—we can eliminate choices (A) and (C). To determine the rest of the denominator, we must determine the number of moles of water present. According to Table 1, there are 906.1 g/L of water present. In 1 L, then, there are 906.1 g of water. To determine the number of moles, we must once again divide the mass by the molecular weight of water, which is 18 g/mol. As such, the second component of the denominator is 906.1/18. We can 226.5 now predict that the numerator of the correct answer is ​ ____ ​, whereas the denominator is ____ ​ 226.5 ​ + ____ ​ 906.1 ​. The only   101.1   101.1   18    answer choice that matches our prediction is choice (B). (A) Miscalculation. This is the mole fraction of water in the solution. (C) Miscalculation. This gives the number of moles of water divided by the number of moles of KNO3. (D) Miscalculation. The number of moles of potassium nitrate must be included in the denominator. 28. (C) This item tests our understanding of precipitate formation and saturated solutions. The answer choices offer a range of compounds—only one will not cause a precipitate to form. A saturated solution of KNO3 contains the maximum number of potassium ions and nitrate ions that can be solvated in solution. In a saturated solution, no more potassium or nitrate ions can be dissolved. Again, our task is to determine the compound that will not influence the solubility of KNO3. What factors will influence the solubility of a compound? According to Le Châtelier’s principle and the common ion effect, addition of a compound containing either one of the original ions in solution

17

AAMC Test 4 Explanations

(in this case, potassium or nitrate) will cause a precipitation of crystals. If we add a compound that contains one of the ions already present in solution, we are directly shifting the equilibrium towards the formation of solid reactant, so that KNO3 will precipitate. To answer this question, however, any compound that does not contain either potassium or nitrate could be added to the saturated solution without causing precipitation. From the given choices, only NH4Cl does not contain either potassium or nitrate, making choice (C) the correct answer. (A) Opposite. The nitrate ions will cause KNO3 to precipitate. (B) Opposite. The nitrate ions will cause KNO3 to precipitate. (D) Opposite. The potassium ions will cause KNO3 to precipitate. 29. (D) This item tests our understanding of the passage. Where did we read about condosity? According to our passage map, the last paragraph of the passage discusses this term. The condosity of a solution is defined as “the molar concentration of sodium chloride that has the same specific conductance as the solution.” The answer choices refer to the four different points on the graph of conductance versus molarity. Since the question stem asks us to determine the point on the graph that represents the solution of KNO3, we must determine the conductance and molarity of this solution and compare these values to those of NaCl. Let’s search the passage for the relevant information. According to Table 1, the molarity of the potassium nitrate solution is 2.241 mol/L and the condosity is 2.49 mol/L. Since the condosity of the solution is defined as the molar concentration of sodium chloride that has the same specific conductance as the solution, the potassium nitrate solution has the same conductance as a 2.49 M solution of NaCl. With this information in mind, let’s analyze the graph. On the plot, only points B and D represent solutions that have the same conductance as a 2.49 M solution of NaCl—we can eliminate choices (A) and (C). The molarity of the potassium nitrate solution is 2.241 mol/L. Since this is lower than the 2.49 M solution of NaCl, the appropriate point has to be to the left of the NaCl point, or point D. The correct answer for this question, therefore, is choice (D). (A) Distortion. This point represents a solution with the same molarity as NaCl but a higher conductance. (B) Distortion. This point represents a solution with the same conductance as NaCl but a higher molarity. (C) Distortion. This point represents a solution with the same molarity as NaCl but a lower conductance. 30. (C) This is a straightforward question that requires us to calculate the mass of KNO3 in 100 mL of solution. According to our passage map, we can obtain all the necessary information from Table 1. In this table, we are told that the concentration of potassium nitrate (solute) is 226.5 g/L. While this value is reported in liters, we need to calculate the mass of KNO3 in a tenth of this volume, that is, in 100 mL. The amount present in 100 mL (0.1 L) is: mKNO3 = 226.5 g/L × 100 mL = 226.5 g/L × 0.1 L

mKNO3 = 22.65 g

The answer choice that matches our prediction is choice (C), the correct response. (A) Miscalculation. This is half the mass of KNO3 in 100 mL of solution, or the mass in 200 mL. (B) Miscalculation. Be careful with your calculations as this answer is close to the actual mass. (D) Miscalculation. This is the solubility of KNO3 in 100 g of water.

18

Physical Sciences

31. (D) This item requires us to use the information presented in the passage so as to calculate the number of potassium ions in the KNO3 solution. The answer choices are expressed in powers of ten, so we are more interested in the order of magnitude rather than an exact value. Let’s examine the KNO3 molecule to determine how many moles of K+ are in solution. Each mole of KNO3 gives rise to one mole of K+. According to Table 1, the molarity of the student’s solution is 2.241 mol/L, whereas the volume of the solution is 1 L. As such, there are 2.241 moles of KNO3 and therefore of K+ in the solution. To determine the number of potassium ions, we have to use Avogadro’s number, which expresses the number of molecules in one mole. We are only interested in the order of magnitude, so we can make safe approximations. # of K+ ions = 2.241 mol × 6 × 1023 ions/mol

# of K+ ions = 12 × 1023 ions

# of K+ ions = 1.2 × 1024 ions

The correct answer is therefore choice (D). (A) Miscalculation. As always, be careful when using scientific notation in your calculations. (B) Miscalculation. As always, be careful when using scientific notation in your calculations. (C) Miscalculation. As always, be careful when using scientific notation in your calculations.

Discrete Set II (Items 32–35) 32. (C) The item presents us with a hypothetical reaction and asks us to determine the partial pressure of one of the products. Our first step is to write down and balance the given reaction: CH4 + O2 → CO2 + H2O Balancing the equation, we obtain the following mole ratios: CH4 + 2O2 → CO2 + 2H2O We are told that the reaction reaches completion and has a final pressure of 1.2 torr. The only compounds present at the end of the reaction are carbon dioxide and water—our task is to calculate the partial pressure of water. In order to do this, we must use the mole ratios of the products and the total pressure by applying Dalton’s Law, which states that the partial pressure of a gas is determined by the mole fraction, X (moles of water divided by total moles of products), of that gas multiplied by the total pressure of the system: PH2O = Ptotal × XH2O Since we have one mole of CO2 and two moles of water, the total number of moles at the end of the reaction is 3. Out of this, two thirds are water, so the mole fraction of water is ​ _23 ​. Plugging this value back in Dalton’s equation, we obtain that the partial pressure of water is: PH2O = 1.2 torr × __ ​ 2 ​  3 2.4 torr _______ PH2O = ​   ​    3 PH2O = 0.8 torr The correct answer, therefore, is choice (C).

19

AAMC Test 4 Explanations

(A) Miscalculation. This is the partial pressure of carbon dioxide. (B) Miscalculation. This answer may be obtained if you incorrectly balanced the equation. (D) Miscalculation. This is the total pressure of the system. 33. (A) This item tests our understanding of chemical formulas. We are provided values for the masses of different atoms in a particular compound and asked to determine how many atoms of each kind are present in the molecule. Our first step is to define the “empirical formula” as the lowest whole number ratio of moles of each element in a compound. In order to do this, we must calculate the number of moles of each element by dividing the mass by the molecular weight. Based on the answer choices, we then have to determine the molar ratio of C:H:O. The molecular weight of carbon is 12 g/mol, so if the molecule has 12.0 g carbon, it contains one mole of carbon. In addition, two moles of hydrogen are present (2.0 g of hydrogen at 1.0 g/mol) and one mole of oxygen (16.0 g of oxygen at 16.0 g/mol). Establishing the molar ratio of C:H:O, we obtain 1:2:1, so the empirical formula is CH2O—only choice (A) mirrors our prediction. (B) Miscalculation. This would be the empirical formula for a compound that contains 72.0 g carbon, 1.0 g hydrogen, and 128.0 g oxygen. (C) Miscalculation. The empirical formula is the lowest whole number ratio of moles of each element in a compound. (D) Miscalculation. The empirical formula is the lowest whole number ratio of moles of each element in a compound. 34. (D) From the information provided in the item stem, we know that we are being tested on our knowledge of standing waves, that is, waves found in pipes. Specifically, given a pipe open at both ends with a length of 1 m and a diameter of 0.1 m, we have to determine the fundamental wavelength of the wave. The answer choices are all numerical, so we have to determine the mathematical result. Our first task is to understand what the resonant wavelength is. The fundamental wavelength is the highest wavelength that the pipe can support. Generally, we talk about fundamental frequency as the lowest frequency that can be observed in a pipe. The wavelength of a pipe with both ends open can be determined using the formula: λ = ___ ​ 2L n ​ , where L is the length of the pipe and n the harmonic. Since the item specifically asks us to determine the fundamental resonant wavelength, we can correctly assume that we are analyzing the first harmonic, so n is equal to 1. Plugging in the values for the length of the pipe and the harmonic number, we obtain: × 1 ​ m  ________ λ = ​ 2   =2m 1 This implies that the fundamental resonant wavelength is twice as long as the length of the pipe. Our result matches with choice (D). Notice that the diameter of a pipe is not needed when working with standing waves, since the wavelength is independent of the thickness of the pipe. (A) Miscalculation. This is the diameter of the pipe. (B) Miscalculation. This answer would be incorrectly obtained if you used the diameter of the pipe in your calculations.

20

Physical Sciences

(C) Miscalculation. This is the length of the pipe. 35. (B) This item tests our knowledge of mechanical waves (e.g., water, sound, etc.). Glancing at the answer choices, we notice that we have to determine if mechanical waves transport matter, energy, both, or neither, so let’s start by reviewing the properties of mechanical waves. Mechanical waves cause a local oscillation of material. In the case of a transverse wave on water, the material that makes up the wave (the water particles) oscillates up and down. No material is transported as a result of mechanical waves, so we can eliminate choices (A) and (C). Additionally, the waves cause a propagation of energy. This energy propagates in the same direction as the wave. Any kind of wave (mechanical or electromagnetic) has a certain energy. The correct answer must be choice (B). We can also think of a specific example of a mechanical wave, such as sound, where the wave is made up of the vibrations of atoms in air. While energy is propagated by a sound wave, the air particles themselves are not. On the other hand, in an electromagnetic wave, such as light, photons are transported across a medium along with the energy of the wave. Since mechanical waves only transport energy, not matter, the correct answer is, again, choice (B). (A) Distortion. Mechanical waves do not transport matter. (C) Distortion. Mechanical waves do not transport matter. (D) Distortion. Mechanical waves transport only energy, not matter.

Passage VI (Items 36–40) Passage Map Passage Type: Information Topic: Geological Events Scope: Timing of collision between North America and Africa ¶1: Timing of geologic events is determined by magnetic measurements and radioactive dating ¶2: Radioactive dating is done by determining amount of

40K

and its decay product

40Ar.

¶3: Simplified model used to determine timing of collision 36. (A) This is a straightforward item that tests our understanding of collisions. Specifically, we have to determine which of the four measurements given in the answer choices would be conserved during a collision. We can infer from the first paragraph of the passage that the collision of the continents was an inelastic event, since the continents changed their shapes. In inelastic collisions, only momentum is conserved—the correct answer is therefore choice (A). (B) Distortion. During inelastic collisions, the kinetic energy decreases. (C) Distortion. In inelastic collisions, the potential energy is not conserved. (D) Distortion. The impulse is the change in momentum, and since the momentum is conserved, the change in momentum is zero.

21

AAMC Test 4 Explanations

22

Physical Sciences

but maintaining the atomic mass. We predict, therefore, that during the decay of emitted—choice (D) is the correct response.

40K

into

40Ar,

a positron is

(A) Distortion. Gamma decay does not change the mass number or atomic number of the atom. It simply lowers the energy of the atom and results in the emission of a photon. (B) Distortion. An alpha particle consists of two protons and two neutrons (a helium atom). This would decrease the atomic mass—in the radioactive decay of potassium, the atomic mass is conserved. (C) Distortion. Beta decay, or emission of an electron, occurs when a neutron splits into a proton and an electron. The mass number is unchanged but the atomic number increases. 40. (B) In spite of the length of the item stem, this is a rather straightforward item. Although it initially appears that we are tested on our knowledge of collisions, we are in fact asked to determine the forces acting on the continents. We are told that one continent exerts a force F on the other continent, and we have to determine what the reaction force will be equal to. What do we know about forces opposing each other? According to Newton’s third law, the force exerted by the smaller continent on the larger continent is equal in magnitude to the force exerted by the larger continent on the smaller continent. The mass is not a factor when determining action and reaction forces. Regardless of the mass, size, volume, or any other variable of two objects, when they come in contact, the forces that the two objects exert on one another are equal in magnitude and opposite in direction. Therefore, the force of the smaller continent is of magnitude F, making answer choice (B) the correct response. (A) Distortion. According to Newton’s third law, the forces of action and reaction are equal in magnitude and opposite in direction. (C) Distortion. According to Newton’s third law, the forces of action and reaction are equal in magnitude and opposite in direction. (D) Distortion. According to Newton’s third law, the forces of action and reaction are equal in magnitude and opposite in direction.

Passage VII (Items 41–47) Passage Map Passage Type: Information Topic: Vacuum photodiode Scope: Structure and function of vacuum photodiode ¶1: Describes photoelectric effect/ work function ¶2: Construction of vacuum photodiode: E = ∆V/L Fig. 1: Sketch of vacuum photodiode detector 41. (A) This item tests our understanding of the experimental setup. In glancing at the answer choices, we are asked to determine how we can change either L or R so as to decrease the electric field between the electrodes. According to our passage map, paragraph 2 states that the electric field between the electrodes is equal to the electric

23

AAMC Test 4 Explanations

voltage difference divided by L. In turn, we should know that the electric voltage difference that occurs at the vacuum photodiode is equal to the voltage of the battery minus the voltage drop across the resistor such that ΔV = V – IR. Therefore, the electric field is equal to: –  ​ IR  ______ E = ​ V   L In order to decrease E, we can either decrease the numerator or increase the denominator—in other words, we can either increase R or L. Knowing this, we can eliminate choices (B) and (D). Doubling L will halve the electric field. However, since we are subtracting the resistance from the voltage, if we increase R by 2, we decrease the electric field, but not by a factor of 2. Thus, increasing L by a factor of 2 will decrease the electric field between the electrodes by the greatest amount—choice (A) is the correct answer. (B) Opposite. This will increase the electric field. (C) Distortion. The resistance in the circuit affects the voltage of the battery reaching the anode for a given current. Increasing resistance does decrease the electric field, but not by as much as changing the length. (D) Distortion. This will increase the electric field. 42. (B) We have to determine the change in potential and kinetic energy that occurs when an electron is freed from the cathode and travels to the anode—a glance at the answer choices suggests that we should understand whether the energy increases, remains relatively the same, decreases, or is zero. Where is this discussed in the passage? According to our map, the first paragraph describes the photoelectric effect—the ejected electron will have a kinetic energy equal to the photon’s energy minus the work function. The item stem states the energy of the photon is only slightly greater than the work function. What effect does this have on the electron’s energy? This means that the kinetic energy of the electron as it is ejected is very small and that most of the electron’s energy is still stored as potential energy. As the electron is propelled from the cathode to the anode, its potential energy is converted to kinetic energy. Since the electron had very small kinetic energy to begin with, we would expect that the final kinetic energy should be roughly equal to its starting potential energy—energy should be conserved, since there is no mention of any nonconservative forces. Our prediction matches with the correct answer, choice (B). (A) Distortion. According to the conservation of energy, the final kinetic energy cannot be more than the initial potential energy. (C) Distortion. Since there are no nonconservative forces acting on the system, the final kinetic energy cannot be less than the initial potential energy. (D) Distortion. The final kinetic energy cannot be zero. 43. (A) This item once again tests our understanding of the photoelectric effect. What happens when more photons are incident on the metal? We are told in the first paragraph of the passage that when a photon with an energy greater than the work function of the metal strikes the cathode, an electron is released. Indeed, each photon with an energy greater than the work function that strikes the cathode releases one electron. We predict, therefore, that increasing the number of photons contacting the metal will also increase the number of electrons ejected—the answer choice that mirrors our line of reasoning is choice (A). (B) Distortion. The potential energy of the ejected electron depends on the electric field (U = QV).

24

Physical Sciences

(C) Distortion. The magnitude of the electric field between the electrodes depends on the battery and the resistor—not the number of incident photons. (D) Distortion. The speed of the electrons at the anode is affected by the kinetic energy of the electron when it is ejected, which is in turn dependent on the energy of the photon, not the number of photons. 44. (C) This item requires us to have a good understanding of the photodiode setup. We are asked to describe the movement of the electron from the cathode to the anode. In paragraph 2, we find that the cathode is connected to the negative end of the battery, while the anode is connected to the positive end of the battery. Because of this, we can expect these two electrodes to act like a parallel plate capacitor. This being true, the anode (positively charged) exerts an attractive force on the negatively charged electrons. Since the electron experiences a net force towards the anode, according to Newton’s second law (F = ma), we can infer that the electron accelerates. Thus, the ejected electron leaves the cathode with a given velocity and, under the influence of the acceleration generated by the attractive force of the anode, its velocity increases as it approaches the anode. The answer choice that mirrors our prediction is choice (C). (A) Out of scope. There is no mention of collisions in the passage. (B) Distortion. The electron experiences a net force towards the anode, and therefore accelerates—its velocity will increase with time. (D) Out of scope. We are asked to describe the movement of the electron as it travels from the cathode to the anode. The passage does not mention anything about holes in the vacuum photodiode. 45. (B) This is a straightforward item that tests our knowledge of the definition of power in a circuit—a quick glance at the answer choices shows that calculations will be necessary. We are given the current through the circuit and asked to calculate the power. Power is determined by the formula: 2 P = IV = ___ ​ V  ​ = I2R R

From the given formulas, P = I2R is the easiest to work with in our situation, since paragraph 2 informs us that the resistance is 100 Ω. Using scientific notation to speed up our calculations:

P = (1 × 10–32) × 102

P = (1 × 10–6) × 102

P = 1 × 10–4 W

The correct answer is choice (B). (A) Miscalculation. This is an error of magnitude. (C) Miscalculation. As always, be careful when using scientific notation in your calculations. (D) Miscalculation. As always, be careful when using scientific notation in your calculations. 46. (C) The item tests our understanding of circuits. We are asked to determine what happens when electrons flow from the cathode to the anode—the answer choices make reference to different variables in the circuit, such as the

25

AAMC Test 4 Explanations

current, the resistance, or the voltage. What does the movement of negative charge result in? Electrons that are ejected from the cathode are replaced by electrons from the battery and anode. Thus, the movement of electrons from the battery to the cathode creates a current. The only effect on the apparatus of electrons being ejected from the cathode is the formation of a current in the circuit—our rationale best matches with choice (C). (A) Distortion. There is no reason to believe that the voltage will reverse polarity. (B) Distortion. The voltage difference is determined by the current and the resistance in the circuit. In fact, paragraph 2 states that the potential difference between the two electrodes is equal to the potential difference across the battery. (D) Distortion. The total resistance is determined by the resistors—the movement of electrons in the circuit does not affect the resistance. 47. (D) This item requires us to have a deeper understanding of the photoelectric effect, so let’s start by quickly reviewing paragraph 1. The answer choices provide direction to our focus by referring to either an increase or decrease in the number of photons or the speed of the electrons ejected. After glancing at paragraph 1, we confirm that when photons are incident on the cathode, electrons are ejected (not photons); we can thus eliminate choices (A) and (B). Furthermore, we are told that the energy of the photon that is directed at the cathode is, in part, used to overcome the work function of the material, while the rest of the photon’s energy is converted into the kinetic energy of the ejected electron. Increasing the frequency of the photon will increase the energy of the photon (E = hf). Therefore, increasing the energy of the photon implies that the kinetic energy that the electron will acquire will also increase. Since the answer choices discuss speed, and not energy, we must take one last step to relate these two variables. Because KE = _​ 12 ​mv2, as kinetic energy increases, velocity also increases—the answer choice that matches our line of reasoning is choice (D). (A) Distortion. Electrons, not photons, are ejected from the cathode. (B) Distortion. Electrons, not photons, are ejected from the cathode. (C) Opposite. An increase in the photon’s energy means that more energy will be transferred to the electron, making the kinetic energy, and thus speed, of the electron higher.

Discrete Set III (Items 48–52) 48. (C) This item tests our understanding of the Bohr model of the atom—specifically, what causes an emission of radiation, or energy. Let’s start by quickly reviewing the Bohr model. According to this theory of an atom, protons and neutrons exist in the nucleus while electrons orbit around the nucleus. Furthermore, Bohr postulated that there are several orbitals that electrons may fill, with each orbital having a characteristic energy. Electrons can only assume positions in these specific orbitals, and not in between. In other words, electrons will have specific, quantified energy values. To move between these orbitals, an electron must gain or lose a discrete amount of energy. When an electron moves from a lower to a higher energy orbital, it must absorb some energy—enough to account for the increased energy of the orbital. As such, when it moves from a higher to a lower energy orbital, the electron will release energy. In general, energy released can be observed as radiation (most often, a photon or a light ray). Radiation is emitted whenever an electron moves from a higher to a lower energy orbital—choice (C) is the correct answer.

26

Physical Sciences

(A) Distortion. Energy is emitted when an electron moves from a higher to a lower energy orbital; it may not occur each time an electron changes orbitals. (B) Out of scope. Electrons move with a constant velocity around the nucleus, much like a planet orbits around the sun. (D) Opposite. An orbital of larger radius will actually have a higher energy since the orbital is farther away from the nucleus. When an electron moves to a higher energy orbital, it requires the absorption of energy. 49. (A) This is a straightforward item that tests our understanding of the heat of fusion. In the answer choices, we are given four phase changes and the corresponding substances undergoing them; our task is to determine in which one of these four instances we can measure the heat of fusion. The heat of fusion of a substance is the energy required to convert a solid into a liquid or to melt a solid—the correct answer must be choice (A). (B) Distortion. This would be the heat of sublimation. (C) Distortion. This would be the heat of vaporization. (D) Distortion. This would be the heat of vaporization. 50. (B) This question tests our understanding of circuits—the answer choices suggest some numerical analysis. Although the schematic looks quite complicated, let’s break it down into smaller pieces. There are four resistors, three of them in parallel with the rest of the circuit, and one of them (the 3-ohm resistor) in series. What do we know of resistors in series and in parallel? Resistors in series have the same current running through them, but different voltage drops depending on the magnitude of the resistance. So the current going through the 3-ohm resistor should be the same as the one going through the 2- and 4-ohm resistors added together. The question stem tells us that the current in the 2-ohm resistor is 2 A. What does this tell us about the 4-ohm resistor? Since the current entering this parallel split must be conserved, and since the current preferentially takes the path of least resistance, we know that the current traveling through the 2-ohm resistor is double that traveling through the 4‑ohm resistor—the current through the 4-ohm resistor must be 1 A. Summing up the currents, we see that, right before entering the parallel split, the current was 3 A. Since current is conserved throughout the circuit, 3 A of current also runs through the 3-ohm resistor—choice (B) is the correct answer. (A) Miscalculation. This is the current in the 2-ohm resistor. (C) Miscalculation. This is double the current in the 2-ohm resistor. (D) Miscalculation. This is double the current in the 3-ohm resistor. 51. (B) This is a straightforward item that tests our understanding of plane mirrors. Let’s start by quickly reviewing how an image is formed in this type of mirror. A plane mirror produces an image behind it (virtual image) at a distance equal to the object distance in front of the mirror. Our task is to calculate a numerical value for the minimum distance away from the mirror at which a far-sighted person could see his own reflection. The image must be at least 300 cm away from the person—this can be accomplished by having the person stand 150 cm from the plane mirror, since the reflection will place the virtual image another 150 cm behind the mirror. Therefore, the minimum distance that this person must stand from the mirror is at least 150 cm—the correct answer is choice (B).

27

AAMC Test 4 Explanations

(A) Miscalculation. At this distance from the mirror, the image will only be 150 cm away and the person will not be able to see clearly. (C) Miscalculation. The image will be 600 cm away—the person could stand closer. (D) Miscalculation. The image will be 1,200 cm away—the person could stand closer. 52. (D) This item requires us to use our knowledge of pH and molar concentrations—the nature of the answer choices suggests some calculations are required. Given the pH of a strong acid, we are asked to calculate the concentration of protons in solution. The pH of a solution is related to the proton concentration by the following equation: pH = –log[H3O+]

In order to calculate the proton concentration, we have to raise the left side of the equation to the power of 10 so as to eliminate the log function:

[H3O+] = 10–pH

If the pH of the given solution is 6.0, the proton concentration is:

[H3O+] = 10–6

Consequently, choice (D) is the correct answer. (A) Opposite. The negative sign in the exponent is missing. (B) Miscalculation. This is half the concentration of protons in solution. (C) Miscalculation. This is double the concentration of protons in solution.

28

Verbal Reasoning

Verbal reasoning ANSWER KEY 53. D

64. A

75. A

86. D

54. B

65. C

76. B

87. D

55. D

66. D

77. C

88. A

56. A

67. D

78. B

89. B

57. D

68. B

79. C

90. C

58. B

69. A

80. C

91. D

59. A

70. B

81. B

92. C

60. B

71. A

82. D

61. A

72. A

83. A

62. D

73. B

84. D

63. D

74. D

85. A

29

AAMC Test 4 Explanations

30

Verbal Reasoning

31

AAMC Test 4 Explanations

57. (D) Where does the author allude to forbidden acts? Although this detail may not appear directly in our passage map, a quick paragraph scan reveals that the author makes some relatable assertions in paragraph 5. Here, the author adamantly argues that “there is no convincing argument in favor of torture, of arbitrary execution, of keeping sections of a population or a whole in a state of slavery.” The correct answer choice should agree with outlawing any one of these mentioned acts. Of those mentioned, only torture appears in the answer choices, making choice (D) the correct answer. (A) Out of scope. Private property is not discussed in the passage. (B) Out of scope. The right to vote is not discussed in the passage. (C) Out of scope. Suspension of certain freedoms in times of national crisis is not discussed in the passage. 58. (B) As a “which of the following” item, it is difficult to make a substantial prediction. Instead, let’s first review the purpose of the passage and then move onto the answer choices. According to the passage map, the author discusses efforts to extend universal human rights. The correct answer, therefore, won’t agree with this aspect of the passage, or any resulting argument that stems from this claim. In working through the answer choices, choices (A), (C), and (D) are out of scope, leaving choice (D) as the correct answer. Indeed, an assertion that argues that countries should be able to run internal affairs without outside influence most strongly challenges the idea of universal human rights. Choice (B) is thus the correct answer. (A) Out of scope. Unjust laws are beyond the scope of this passage, which focuses on universal human rights. (C) Out of scope. Speed of implementation of human rights laws is not discussed in the passage. (D) Out of scope. The timeliness of ending oppression is not discussed in the passage.

Passage II (Items 59–64) Passage Map Passage Type: Humanities Topic: Women in literature Scope: Address the usage of the term “woman writer” Purpose: To argue that the use of such a term is disparaging to the efforts made by women to reach equality in the literature world Mapping the Passage: ¶1: Definition of the term “woman writer”—a politics of sex term ¶2: Political terms, such as “woman writer,” are not part of writers’ language. ¶3: The author disagrees with the use of this term, since the experience of writer is universal (not sex-specific). ¶4: Feminism opposes segregation, but this world of “women writers” actually furthers the segregation cause. ¶5: The sociological/political message of the “woman writer” ¶6: The use of such division will solidify the idea of segregation in writing.

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33

AAMC Test 4 Explanations

be considered writers. From the given choices, only answer choice (D) matches with our prediction and is thus the correct response. (A) Out of scope. We don’t know if these women will be called classical feminists, as this conclusion is not discussed in the passage. (B) Faulty use of detail. Although political involvement is mentioned in paragraph 5, the passage does not imply that these women will become political—only that they will gain political power. (C) Out of scope. The passage does not say that the human component of literature will disappear. 63. (D) How does the author feel about women writers? Throughout the passage, the author criticizes this designation, as it furthers the idea of segregation. How would the new information presented in the item stem affect the author’s argument? If the author admired the work of a woman writer, then this would be discrepant with the author’s main argument. In paragraph 2, the author explains this negative viewpoint on the term “woman writer,” arguing that “the language of politics is not the writers’ language.” By embracing the literature of a woman writer, the author would contradict his own assertion. Choice (D) nicely mirrors our prediction and is thus the correct answer. (A) Out of scope. This new information would have no bearing on the idea that classical feminism was created to bring an end to intellectual segregation. (B) Out of scope. In paragraph 3, the author argues that women writers can be nourished by other writers, regardless of sex. (C) Out of scope. The admiration of the author for women writers would not be discrepant with value judgments about male writers. 64. (A) Where does the author define literature? According to paragraph 3, the author asserts that “[t]here is a human component to literature that does not separate writers by sex but that…engenders sympathies from sex to sex….” The correct answer choice, therefore, will be the one example of literature in which both sexes are treated equally. With this in mind, only choice (A) matches our prediction, and is thus the correct answer. (B) Out of scope. An essay about the women’s liberation movement would not be considered literature by the author, as it is gender-specific. (C) Out of scope. A novel that attacks men for their ignorance would not be considered literature by author, as it is gender-specific. (D) Out of scope. A political tract arguing for a woman president would not be considered literature by the author, as it is gender-specific.

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Verbal Reasoning

Passage III (Items 65–70) Passage Type: Social Sciences Topic: Privacy Scope: Patient-physician privacy for prisoners Purpose: To argue that if there is no danger; patient privacy supersedes Mapping the Passage: ¶1: Hypothetical situation presented on physician-patient confidentiality for prisoners ¶2: Prisoners do not lose their right to privacy in the doctor-patient relationship. ¶3: Physicians need to have clinical autonomy to preserve this privacy. ¶4: Situation of finding contraband: if dangerous, doctor has a duty to report it; if not, privacy supersedes ¶5: When no danger, patient comes first ¶6: Reinforces the need for informed consent, provides examples emphasizing author’s argument 65. (C) Where does the author discuss prisoners and X-rays? A quick passage scan shows that this is addressed in paragraph 6. Here, the author argues the importance of providing informed consent to prisoners. In paragraph 2, the author conveys the idea that “patients feel distrust because physicians are often employed by the incarcerating institution.” Without a document providing a means for patient-physician trust, it is logical to conclude that prisoner patients will have greater distrust for the attending physicians. The only choice that mirrors our prediction is choice (C), the correct answer. (A) Opposite. If the inmates refuse the X-rays because they do not have any informed consent, this would show that they actually believe they do NOT have a Constitutional right to privacy. (B) Out of scope. The comparison of concern for health between prisoners and nonprisoners is not discussed in the passage. (D) Out of scope. The inmates might be carrying weapons for self-protection, but the refusal to have an X-ray is not proof that they think this way. 66. (D) What does the passage say about this sort of situation? A quick review of the passage map reminds us that in the case of inmates, safety comes first, whereas privacy is secondary. In the case of a hidden weapon, the doctor must report a violation of safety and will report the incident no matter what. The only choice that matches our prediction is choice (D), the correct answer. (A) Opposite. The physician must report the incident regardless of whether or not a threat was made. (B) Opposite. The physician must report the incident even if consent was not obtained. (C) Opposite. The physician must report the incident regardless of whether or not the patient denied possessing the weapon.

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AAMC Test 4 Explanations

67. (D) What is the author’s argument regarding the physician-patient relationship in a prison setting? According to our passage map, paragraphs 2 and 3 argue that prisoners do not lose their right to privacy in the doctor-patient relationship, and that physicians need to have clinical autonomy to preserve this privacy. If confidentiality is essential to good medical practice, then any breach of such confidence would also lead to a breach in the medical practice. Knowing this, choice (D) must be the correct answer. (A) Out of scope. The passage does not make any claims regarding the connection of quality of care to confidentiality. (B) Out of scope. The clinical autonomy addressed in the passage is not accomplished by the refusal of physicians to reveal information. (C) Out of scope. There is nothing in the passage that relates the quality of care to the respect of patients’ confidence. 68. (B) Where does the passage use this phrase? A quick passage scan reveals its location in paragraph 2. In this paragraph, the author uses the term “necessary information” in discussing health-related information that is personal but related to and helpful for medical investigations. Indeed, the term “necessary information” refers to that information which is medically relevant. With this is mind, we can safely match our prediction with choice (B), the correct answer. (A) Out of scope. Past criminal activity is not regarded as medically relevant information. (C) Out of scope. The intent to harm others is not considered to be health-related information. (D) Out of scope. Psychiatric history is not medically relevant information. 69. (A) As a “which of the following” item, a substantial prediction is difficult to make. Instead, let’s start by reviewing the purpose of the passage and then move onto the answer choices. According to our passage map, the author’s intent is to argue that in a prison setting, if there is no danger, patient privacy supersedes. The correct answer must be a necessary truth which agrees with this overall purpose. Since the nature of a prison-employed physician conflicts with the idea of unbiased confidentiality, it is logical to conclude that in a private setting, physician confidentiality is less impaired, making choice (A) the correct answer. (B) Out of scope. The status of patients, whether incompetent or competent, is beyond the scope of this passage. (C) Faulty use of detail. According to the passage, physician confidentiality is only threatened when there is no informed consent for procedures such as X-rays. (D) Faulty use of detail. Physician confidentiality is not a Constitutional right that is lost by prisoners—it is still retained. 70. (B) What is the primary argument for the importance of this covenant? Contingent on the idea of confidentiality of treatment is the premise that prisoners should have privacy with their doctors because this ensures good medical care. Why is this idea of confidentiality questioned by prisoner patients? According to paragraph 2, “physicians are

36

Verbal Reasoning

often employed by the incarcerating institution.” Indeed, this fact casts a shadow of doubt over physician-patient confidentiality. However, if the prisoners assume that physicians are independent of the institution, then they would have greater reason to trust their relationship with their physician. If this were true, the author’s argument regarding the physician-prisoner relationship loses its crucial assumption, and is thus weakened. Based on this, choice (B) is the correct answer. (A) Opposite. This knowledge would strengthen the author’s argument. (C) Opposite. In this case, the action by prison officials would strengthen the case made by the author. (D) Out of scope. Whether or not the prisoners understand their Constitutional rights is irrelevant to the author’s argument.

Passage IV (Items 71–77) Passage Type: Social Sciences Topic: Evolution Scope: The inclusion of altruism and kin selection into theories of evolution Purpose: The inclusion of kin selection into theories of evolution extends genetic potential into kindness Mapping the Passage: ¶1: Freud: paradox of human social life—we are by nature selfish and aggressive, but need to suppress that for the good of man. ¶2: How can altruism fit into evolution? ¶3: Resolution: altruism works to increase gene potential—kin selection ¶4: Kin selection and humans—effect of free will and other factors ¶5: Kin selection is favorable because it considers the capacity for kindness as a biological potential. 71. (A) Where does the author mention this phrase? A brief skim of the passage reveals its location in paragraph 1. Here, the author asserts that “what we prize and strive for (with pitifully limited success), we consider as a unique overlay.” What do we consider as uniquely human? From this same paragraph, we can infer that our ability “to act altruistically for the common good and harmony” is what we consider to be uniquely human. Knowing this, choice (A) is the correct answer. (B) Opposite. We try to suppress these biological inclinations (animalistic behavior), instead attempting to embrace a uniquely human characteristic—altruism. (C) Opposite. We try to suppress these biological inclinations (selfishness), instead attempting to embrace a uniquely human characteristic—altruism. (D) Out of scope. The ability to be self-critical is beyond the scope of this passage. 72. (A) As a “which of the following” item, it is difficult to make a specific prediction. Instead, we’ll have to keep referring back to the passage to determine which answer choice is not supported by an example or reference to an authority. In general, assertions made in the passage that aren’t further supported are located either at the end

37

AAMC Test 4 Explanations

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Verbal Reasoning

76. (B) How does this Benjamin Franklin quote relate to the passage? Based on its principles, the advice given by Franklin echoes the fundamental concept that is the basis for kin selection. Indeed, paragraph 3 teaches us that “animals evolve behaviors that endanger themselves only if such altruistic acts increase their own genetic potential by benefiting kin.” Therefore, the passage centers on the idea of altruism and this quote serves to emphasize this. (A) Out of scope. Biological determinism is not a concept mentioned in the passage. (C) Opposite. The concept presented in this quote mirrors the idea of kin selection—an example of altruistic behavior, not animal aggression. (D) Out of scope. Genetic mutation is not a concept mentioned in the passage. 77. (C) Where are Freud’s ideas discussed? According to our passage map, in paragraph 1, Freud argues that altruism is a result of the suppression of our biological inclinations. How does this relate to the remaining ideas presented in the passage? The rest of the passage describes kin selection as an example of altruistic behavior displayed in animals that is a result of genetic factors. Indeed, if altruism can be attributed to genetic factors, then it is not necessary to overcome biological instincts to exhibit altruistic behavior, since altruism itself is a biological instinct. The only answer choice that fits this rationale is choice (C), the correct answer. (A) Opposite. This idea is actually attributed to Freud in the passage and would strengthen his argument. (B) Opposite. This statement strengthens Freud’s argument. (D) Out of scope. Agonizing dilemmas are beyond the scope of this passage.

Passage V (Items 78–82) Passage Map Passage Type: Natural Sciences Topic: Pollination Scope: The development of flower features to encourage pollination Purpose: To discuss the development of pollination via wind, beetles, and finally bees Mapping the Passage: ¶1: In the Mesozoic time, most pollination was accomplished by the wind. ¶2: Beetles discovered sap and resins and began feeding on them, acting as pollinators. ¶3: Adaptations of the plants to beetle pollination ¶4: Result was an increase in plant fertility. ¶5: Plants begin to develop into flowers to assist in pollination. ¶6: Flowers developed specific features to allow pollination to occur only by certain species.

39

AAMC Test 4 Explanations

40

Verbal Reasoning

in helping the flowers evolve their tubular corolla structures. Since the author considers the development of the tubular corolla a “great step forward in floral design,” this new evidence would destroy an integral part of the author’s hypothesis—the role of the long-tongued insects. Knowing this, we recognize that only choice (B) can be the correct answer. (A) Out of scope. Although this may be true, this finding does not compromise the hypothesis about the development of the flowers. (C) Out of scope. The appearance (or nonappearance) of a fungus does not compromise the author’s conclusions. (D) Out of scope. Even if other flowers without corollas can be chemically stimulated to form this structure, this finding does not weaken the author’s ideas for this particular development. 82. (D) Where does the passage mention tubular corollas? A brief skim of the passage reveals this discussion is located in paragraph 6. Here, we read that “in flowers pollinated by the long-tongued insects, the petals became fused into a tubular corolla…[which] tended to screen out the beetles…and to restrict visitors to those insects…that fly regularly from flower to flower of the same species.” The author’s main idea in this part of the passage is that the corolla developed to screen out less effective pollinators, selecting instead for the better insects that could fly selectively between flowers of the same species. We are told that the tubular corolla was originally developed so as to select the insect pollinators which travel among flowers of the same species. In order to hypothesize that these tubular corollas further evolved to selectively accommodate only a certain type of insect species, we need to understand what evolutionary advantage this provides to the flowers. Choice (D) matches our line of reasoning and is thus the correct answer. (A) Out of scope. The lack of development of long tongues in beetles is not relevant to the hypothesis of how one single species became the pollinator. (B) Out of scope. Because this hypothesis deals with one particular species, the evolution of other species is out of the scope of this part of the discussion. (C) Out of scope. We are only concerned with the present state of the flies at this point in the flowers’ development.

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AAMC Test 4 Explanations

Passage VI (Items 83–87) Passage Map Passage Type: Social Sciences Topic: Evolution Scope: Civilization evolution versus genetic evolution Purpose: To show how Homo sapiens possesses qualities that allow it to undergo cultural evolution faster than genetic evolution Mapping the Passage: ¶1: There are two qualities unique to human evolution that separate it from animal evolution. ¶2: The two qualities: consciousness and culture; language is a vital part of culture ¶3: These two qualities combine to create civilization and allow people to be concerned with issues beyond subsistence. ¶4: There are changes in both the animal kingdom and human civilization. ¶5: Cultural evolution is faster than genetic evolution. ¶6: Cultural evolution allows ideas to transform entire societies very quickly, as opposed to genetic evolution, in which change occurs very slowly. 83. (A) Where does the author discuss animals without consciousness? Our passage map directs us to paragraph 2, where the discussion of consciousness can be found. We read that “an animal without consciousness may ‘know’ the world it inhabits, but only the human animal knows it knows.” In other words, humans, who have consciousness, are self-aware, while animals are not. Indeed, animals without this quality are not aware of their individuality, making choice (A) the correct answer. (B) Faulty use of detail. Learning experience is introduced in the discussion of culture, which comes later in the passage. (C) Out of scope. The discussion of consciousness does not deal with expressions of knowledge. (D) Opposite. In fact, according to the passage, an animal without consciousness may still be aware of its surroundings. 84. (D) What is unique about culture that makes human evolution distinctive? According to the purpose of the passage as summarized in the passage map, the author shows how Homo sapiens possess qualities (consciousness and culture), which allow them to undergo cultural evolution faster than genetic evolution. The correct answer must provide a premise which further supports this argument. If genetic mutations do not cause rapid social change, then the author’s conclusion comparing cultural evolution to genetic evolution is correct. Choice (D) is therefore the correct answer. (A) Faulty use of detail. Although Darwin’s theory, according to the passage, did revolutionize the conception of human nature, the author’s argument regarding culture and human evolution is not dependent upon this premise.

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Verbal Reasoning

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AAMC Test 4 Explanations

(B) Out of scope. Nonhuman language is outside of the scope of this discussion. (C) Out of scope. The resolution of complex problems through visual imagery is not discussed in this passage.

Passage VII (Items 88–92) Passage Map Passage Type: Humanities Topic: Zoroastrianism Scope: Zurvanite heresy Purpose: To show how the Zurvanite heresy is inconsistent with basic Zoroastrian thought Mapping the Passage: ¶1: Story of Zoroaster (founder of Zoroastrian religion); Ahura Mazda is the one uncreated god. ¶2: Zoroaster said that the uncreated god was actually two beings—one good and one evil. ¶3: Zoroaster: history has an end, which contrasts with contemporary Babylonian thought of cyclical time ¶4: The development of Zurvanite heresy: Zurvan (time) was the father of A. Mazda and A. Mainya (the two beings). ¶5: Zurvanism contradicted the major tenet of Zoroastrianism and eventually disappeared. 88. (A) Where does the author address the idea of two primary spirits? According to our passage map, the author discusses these beings in paragraph 2. Here, the author asserts that “the two primal beings each made a deliberate choice between good and evil, an act that prefigures the identical choice that all persons must make for themselves in this life.” Since both humans and these primary spirits had to choose between good and evil, choice (A) is the correct answer. (B) Opposite. Zoroaster believed that at the end of history, the good spirit would prevail. (C) Distortion. According to the passage, Ahura Mazda is one of these two beings, not the creator. (D) Distortion. Zoroastrians do not believe in the superiority of Zurvan (time) over the other two beings. 89. (B) Where does the author mention Indian influences on the Zurvanite concept of time? Since it isn’t mentioned in the passage, we don’t know how this evidence will influence the author’s argument, making choice (B) the correct answer. (A) Opposite. The passage does not discuss Indian conceptions of time, so we cannot infer the effect on the author’s argument. (C) Opposite. The passage does not discuss Indian conceptions of time, so we cannot assume the author’s argument was weakened. (D) Opposite. The passage does not discuss Indian conceptions of time, so we cannot assume the author’s argument was disproved.

44

Verbal Reasoning

45

Biological Sciences

Biological sciences ANSWER KEY 95. A

108. D

121. A

134. A

96. A

109. C

122. C

135. A

97. D

110. B

123. C

136. C

98. C

111. B

124. C

137. D

99. A

112. D

125. B

138. C

100. D

113. C

126. B

139. C

101. C

114. D

127. D

140. A

102. A

115. A

128. C

141. D

103. A

116. C

129. A

142. A

104. D

117. A

130. C

143. A

105. D

118. B

131. B

144. D

106. A

119. A

132. A

145. B

107. B

120. C

133. D

146. D

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AAMC Test 4 Explanations

AAMC Test 4 biological Sciences—Kaplan Explanations Passage I (Items 95–100) Passage Map Passage Type: Information Topic: Autonomic nervous system Scope: ANS and regulation of pupil size ¶1: Divisions of the autonomic nervous system and their characteristics Fig. 1: Structure of the autonomic nervous system, shows innervation of iris muscles ¶2: Circular muscle innervated by PS and radial muscles innervated by S ¶3: Pupillary light reflex; PS – causes pupil constriction, S – causes dilation 95. (A) The item asks us to determine the effect of morphine on the pupils. Glancing at the answer choices, we notice that our task is twofold: first, we have to decide whether the pupils dilate or constrict; secondly, we need to determine if the neurotransmitter responsible for such signaling is acetylcholine or norepinephrine. Let’s tackle the first step. The item stem tells us that morphine increases impulse traffic on the PS nerves to the iris; in other words, morphine stimulates the PS division of the ANS. According to our map, paragraph 2 tells us that the PS division innervates the circular muscle of the iris, while paragraph 3 indicates that such stimulation is responsible for constriction. Thus, stimulating the PS pathway will cause the pupil to constrict—we can eliminate choices (C) and (D). Furthermore, the first paragraph describes the neurotransmitter of the PS division as being acetylcholine, not norepinephrine (which is the neurotransmitter for the sympathetic division of the ANS). Since morphine increases the amount of the neurotransmitter acetylcholine to cause the pupils to constrict, choice (A) is the correct answer. (B) Distortion. According to the first paragraph, norepinephrine is the neurotransmitter of the sympathetic division, not the parasympathetic. (C) Opposite. Based on the information in paragraph 3, the pupils constrict as a result of PS activity. (D) Opposite. Based on the information in paragraph 3, the pupils constrict as a result of PS activity. Additionally, norepinephrine is the neurotransmitter of the sympathetic division, not the parasympathetic. 96. (A) This a rather dense item stem, so let’s break it down into pieces. According to the answer choices, we are asked to decide if the mechanism through which atropine works is active or passive. Additionally, we have to decide which division of the ANS is affected by this drug. The item stem tells us that atropine prevents acetylcholine from attaching to its receptors. From paragraph 1, we know that acetylcholine is the neurotransmitter of the PS division—impeding the attachment of acetylcholine to its receptors effectively blocks the effect of the PS system, so we can eliminate choices (C) and (D). Furthermore, the item stem tells us that blocking the effects of one of the divisions of the ANS is considered a passive mechanism—choice (A) is the only answer choice that matches our prediction.

48

Biological Sciences

49

AAMC Test 4 Explanations

nervous system connect with each other to form a sympathetic trunk. This interconnection of the ganglia allows multiple pathways to be stimulated with one stimulus. In contrast, the parasympathetic division has ganglia which do not interconnect; hence, any stimulation will be isolated to that particular area. We would thus expect the interconnection of the multiple pathways in the S division to produce a more rapid response. In addition, it can be inferred that the sympathetic nervous system would produce a whole-body response, since it is involved in the “fight or flight” mechanism which, by nature, needs to elicit a fast, response to danger. The correct answer choice must be choice (A). (B) Faulty use of detail. It is true that the S division secretes norepinephrine, but there is no indication that norepinephrine produces a faster response than acetylcholine. (C) Opposite. The separation of the ganglia slows the response. (D) Faulty use of detail. It is true that the PS division secretes acetylcholine, but there is no indication that acetylcholine produces a faster response than norepinephrine. 100. (D) At first sight, this item may appear complicated, but breaking it down into smaller pieces will make it more manageable. According to the answer choices, we must decide if this drug will cause pupil dilation or constriction, due to increased or decreased acetylcholine levels. In the item stem, we are told that physostigmine inhibits the breakdown of acetylcholine by inhibiting acetylcholinesterase. Without acetylcholinesterase activity, the breakdown of acetylcholine is prevented—the effect of physostigmine is really an increase in the levels of acetylcholine in the body, so we can eliminate choices (A) and (C). Our next task is to determine what happens with the pupil in response to acetylcholine. Based on the first paragraph, we know that acetylcholine is the neurotransmitter of the PS division. According to our passage map, the third paragraph informs us that the parasympathetic nervous system causes pupillary constriction. With the increased levels of acetylcholine, we would expect physostigmine to cause the pupils to constrict—our prediction matches with choice (D). (A) Opposite. Physostigmine increases acetylcholine levels, causing pupillary constriction. (B) Opposite. Increased acetylcholine levels cause pupillary constriction. (C) Opposite. Physostigmine increases acetylcholine levels by inhibiting its breakdown.

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Biological Sciences

Passage II (Items 101–107) Passage Map Passage Type: Experiment Topic: Freeze tolerant organisms Scope: Glucose as freeze-protection ¶1: Some animals have a mechanism in which only extracellular water freezes, making them freeze tolerant ¶2: Accelerated glucose release may protect tissues by depressing the freezing point, thus facilitating cell dehydration Fig. 1: Response of body temperature, heart rate, ice content, tissue glucose, and tissue water in a freeze-thaw cycle ¶3: Experiment – injection of frogs with saline or glucose; blood analyzed for hemoglobin content after freeze-thaw cycle Fig. 2: Survival and plasma hemoglobin after freeze-thaw cycle in frogs injected with saline, glucose 101. (C) How would high glucose levels promote cellular dehydration, or loss of water? All of the answer choices make reference to glucose, so let’s first determine where this is discussed. According to our passage map, paragraph 2 of the passage states that high glucose concentration (hyperglycemia) promotes dehydration of the cells. We know that the cellular plasma membrane is selectively permeable such that glucose cannot cross the cell membrane, but water can. Thus, increasing the concentration of glucose outside the cell increases the osmotic pressure of the extracellular fluid. The cell will respond by trying to reestablish equilibrium with the extracellular environment so as to maintain homeostasis. As a result, water leaves the cell to decrease the osmotic pressure of the extracellular fluid. We predict, therefore, that hyperglycemia causes cell dehydration because glucose increases the extracellular plasma osmolarity, which, in turn, causes water to rush out of the cells. The answer choice that matches our response is choice (C). (A) Distortion. The plasma membrane does not perform osmotic work. (B) Distortion. Water can freely cross the plasma membrane, so the dehydration of cells does not require ion exchange pumps. (D) Distortion. Water can diffuse freely across the membrane, so there is no need for water to be exchanged for glucose molecules. 102. (A) How can we support the hypothesis that these animals undergo deliberate hyperglycemia? Since hyperglycemia refers to high glucose, the correct answer choice has to reflect a change in the organism that occurs in the regulation of blood sugar so as to allow an increase in glucose levels. Typically, insulin is secreted in response to high blood sugars. However, if abnormally high blood sugar levels are desired, then the insulin response must be suppressed. The only answer choice that matches our prediction is choice (A), the correct response. (B) Opposite. Glucagon is secreted in response to low blood sugars. This response does not need to be suppressed; rather, it needs to be exaggerated, so as to encourage further glucose secretion.

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AAMC Test 4 Explanations

(C) Opposite. Glycogen catabolism is the source of the hyperglycemia, so slowing it down would not cause increased levels of glucose. (D) Opposite. The pancreatic endocrine response to hyperglycemia (insulin secretion) must be suppressed. 103. (A) This item requires us to have a good understanding of the cryoprotective mechanism in these frogs. How would a continued heart beat be important to the glucose-mediated protection? A pulse, and thus a beating heart causes circulation of blood throughout the body. Since glucose has a cryoprotective role in the frogs’ body, we would expect a need for it to be present throughout the body of the animal. Since the glucose is being released in the liver (the site of glycogen catabolism), it must be distributed throughout the remainder of the body so as to protect all organs and cells. As such, we would predict that circulating the blood aids in the distribution of glucose throughout the body tissues—the answer choice that best mirrors our prediction is choice (A). (B) Distortion. This does not relate to glucose. (C) Distortion. This does not relate to glucose. (D) Faulty use of detail. Although it is true that a beating heart requires a constant supply of glucose, this would not explain why a continued heart beat is crucial in the cryoprotective role of glucose. 104. (D) The item tests our understanding of the passage. We are asked to determine where ice will be found in a freeze tolerant frog. Since this is a Roman numeral question, the most efficient strategy is to analyze the statement that appears the most often in the answer choices. Thus, the order in which we will analyze the statements is II, III, and then I. Where in the passage did we read about the freezing mechanism? According to our map, the first paragraph explains that intracellular ice is lethal but that freeze tolerant animals allow the formation of extracellular ice. Statement II is true, since blood plasma is an extracellular fluid; we can eliminate choice (B). The lymph is also found outside of the cells, so statement III must be true. Based on this, the only answer choice that can be correct is choice (D). Indeed, statement I, the cytoplasm, is not true based on the information in the passage, since intracellular ice results in cell death. Again, a freeze tolerant frog will contain ice in extracellular fluids such as lymph and blood plasma, making choice (D) the correct answer. Roman numeral I – False: Cytoplasm is located on the inside of a cell—according to paragraph 1, intracellular ice results in cell death. Roman numeral II – True: Ice can be found in any extracellular fluids—including blood plasma. Roman numeral III – True: Ice can be found in any extracellular fluids—including lymph. (A) Distortion. Statement III is also true. (B) Distortion. Statement II is also true. (C) Distortion. Statement I is false. 105. (D) Glancing at the answer choices, we notice that the question is asking us to determine a specific time interval during which the frogs performed only anaerobic respiration. Anaerobic respiration occurs in the absence of oxygen. The heart and lungs are responsible for supplying oxygen to the cells of the body. When the heart is no longer beating, the cells of the body are no longer receiving oxygen—under such circumstances, the cells must

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rely solely on the energy provided through anaerobic respiration for survival. According to our map, Figure 1 provides us with important clues about the heart rate of the frogs. The second diagram shows that after 12 hours of the freeze-thaw episode, the heart slows down and eventually stops beating. Furthermore, between hours 12 and 26 of the freeze-thaw episode, the heart does not beat at all. During these hours, the cells are relying entirely on anaerobic respiration, since there is no source providing oxygen to the tissues (as would be needed for aerobic respiration). The answer choice that matches our prediction is choice (D). (A) Opposite. According to the second graph in Figure 1, the heart is beating and supplying oxygen to the cells at this point—aerobic respiration is occurring. (B) Opposite. According to the second graph in Figure 1, the heart is beating and supplying oxygen to the cells at this point—aerobic respiration is occurring. (C) Opposite. According to the second graph in Figure 1, the heart is beating and supplying oxygen to the cells at this point—aerobic respiration is occurring. 106. (A) The item asks us to analyze the results of Figure 2 to determine if the data supports the idea that glucose plays a cryoprotective role. We can categorize the answer choices into two groups: those that either support or deny the glucose hypothesis. Let’s turn our attention to Figure 2. First, we notice that survival improves in frogs receiving higher doses of glucose, suggesting that glucose is protective against freezing. Secondly, the plasma hemoglobin level is lower in frogs receiving glucose. Normally, hemoglobin is found on the inside of erythrocytes in the blood and carry oxygen/carbon dioxide. When erythrocytes lyse, or break, they would be expected to release hemoglobin into the blood. As such, plasma hemoglobin levels can be used as a marker of hemolysis, or cell death. Therefore, we can conclude from Figure 2 that exogenous glucose promotes survival by protecting against hemolysis. This indeed supports the hypothesis that glucose plays a cryoprotective role—the only answer choice that matches our line of reasoning is choice (A). (B) Faulty use of detail. Although it is true that death caused by freezing is directly proportional to the extent of hemolysis (more hemolysis, lower survival rate), this statement does not explain the correlation of survival with levels of exogenous glucose. (C) Opposite. Injected glucose lowered plasma hemoglobin levels, indicating that survival rates (avoiding cell death via lysis) are related to treatment. (D) Opposite. Injected saline did not promote hemolysis, but rather failed to protect against it. Saline was used in this experiment as a placebo treatment. 107. (B) This item requires a thorough understanding of the passage, since we are asked to determine one factor that would increase a frog’s chances of survival. According to the passage, glucose has a cryoprotective role—Figure 2 shows that the presence of exogenous glucose increases a frog’s survival rate. What remains is determining what mechanism would most drastically facilitate an increase in endogenous glucose levels in frogs. According to paragraph 2, hyperglycemia is attained by glycogenolysis of hepatic glycogen stores. In order to generate higher levels of plasma glucose, the frogs would need to have a high level of glycogen stored in their liver. Although a high tissue glucose concentration is the end result of glycolysis, tissue glucose is consumed eventually. If there is no glycogen store available to provide a steady supply of glucose, the high tissue glucose concentration will eventually fall, leading to hemolysis (death) of the organism’s cells. The only answer choice that matches our line of thought is choice (B), the correct response.

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(A) Distortion. Hyperglycemia causes dehydration of cells, not dehydration of the organism. (C) Faulty use of detail. The plasma hemoglobin was measured in the experiment as a marker of hemolysis, not as a variable that changes survival. (D) Distortion. Although a high tissue glucose concentration is the end result of glycolysis, tissue glucose is consumed eventually. If there is no glycogen store available to provide a steady supply of glucose, the high tissue glucose concentration will eventually fall, leading to hemolysis (death) of the organism’s cells.

Passage III (Items 108–112) Passage Map Passage Type: Experiment Topic: Characteristics of unknown compounds Scope: Solubility and spectroscopic characteristics of unknown compounds ¶1: Characteristics of unknown Compound A, base hydrolysis into Compounds B and C ¶2: Characteristics of Compound B ¶3: Characteristics of Compound C Table 1: Possible structures and melting points of Compound C 108. (D) The item asks us to determine the functional groups present in Compound C based on its solubility data. The answer choices offer us different combinations, each containing two functional groups. Where in the passage can we find more information about the solubility of Compound C? According to our passage map, paragraph 3 tells us that Compound C is soluble both in dilute acid and dilute base. Compounds that are soluble in bases tend to be acids—only two of our answer choices contain an acid (carboxylic acid), so we can eliminate choices (B) and (C). Conversely, compounds that are soluble in acids tend to be bases. Amines are slightly basic and are thus soluble in acid solutions, whereas amides are not—the only answer choice that fits is choice (D). (A) Distortion. Amides are not soluble in either slightly acidic or basic solutions. (B) Distortion. Amides are not soluble in either slightly acidic or basic solutions. (C) Distortion. Nitro groups are not soluble in either slightly acidic or basic solutions. 109. (C) This is a straightforward item that tests our knowledge of IR spectroscopy. What do all of the compounds in Table 1 have in common? The only similarity that exists is that each of the compounds in Table 1 appears to contain a carbonyl (C=O) functionality. The carbonyl group produces a strong, sharp band between 1,650 cm–1 and 1,800 cm–1 on an IR spectrum—the correct answer is choice (C). (A) Distortion. The N–H bond of amines produces sharp bands between 3,100 cm–1 and 3,500 cm–1, but not all compounds in Table 1 contain this functionality. Alcohols produce broad bands between 3,100 cm–1 and 3,500 cm–1.

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(B) Distortion. The C–C triple bond of alkynes produces a band at 2,200 cm–1—none of the compounds contains a triple bond. (D) Distortion. Peaks in the 1,000–1,500 cm–1 range tend to be indistinct; this area is known as the fingerprint region. 110. (B) The item presents us with a hypothetical situation in which two of the compounds in Table 1 get mixed together. Glancing at the answer choices, we notice that our task is to determine the new melting point of this mixture, and whether or not it will be a broad or a narrow range. Generally, impurities have two effects on melting point. First, the melting point is no longer found at a specific value, and thus appears as a broad range of temperatures, rather than a sharp point. This allows us to eliminate choices (C) and (D). Secondly, based on our knowledge of colligative properties, we know that the presence of another compound will result in a melting point depression (we generally refer to it as freezing point depression, but the temperature at which a compound freezes is the same as that at which it melts, so the two terms have the same value). In other words, the solid melts at a lower temperature than the pure compound. According to Table 1, 4-aminobenzoic acid generally melts at around 188–189°C. When contaminated with camphoric acid, we would expect the melting point to have a broad range below 188°C. The answer choice that matches our prediction is choice (B). (A) Opposite. Impurities depress the melting point. (C) Opposite. Impurities change the melting point into a broad range of temperatures. (D) Opposite. Impurities change the melting point into a broad range of temperatures. Also, impurities depress the melting point. 111. (B) This item requires us to have an understanding of the passage description of Compound A. The answer choices offer us four different possible functional groups, but only one of them will fit with the findings regarding Compound A. Let’s go back to the first paragraph and review the information that we are given. We know that base-catalyzed hydrolysis (refluxing in NaOH) of Compound A results in the formation of Compound B and Compound C. Compound B is an alcohol, whereas Compound C is a carboxylic acid. As such, base-catalyzed hydrolysis of Compound A results in an alcohol and a carboxylic acid. The only hydrolysis that results in an alcohol and a carboxylic acid is that of an ester. The reaction mechanism is illustrated below: O–

O

O +H+

+

O O

OH

OH

OH

OH– ester

carboxylic acid

alcohol

Therefore, we predict that Compound A is an ester, which matches with choice (B). (A) Distortion. Cleavage of an ether usually takes place in the presence of either HI or HBr to produce a haloalkane.

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(C) Distortion. A hydration reaction of a ketone will form a ketal. (D) Distortion. A hydration reaction of an aldehyde will form an acetal. 112. (D) After characterizing the answer choices, we understand that the item asks us to determine which physical property can best be used to make a distinction between the compounds in Table 1. We do not have enough information about the solubility behaviors or the compound reactivities towards alcohols other than that which we might be able to infer based on the chemical structures. Yet, since all of the compounds in Table 1 are acids, we expect them to have similar solubility and reactivity behaviors. Knowing this, we can eliminate choices (A) and (B). In addition, the melting points are very close to each other (providing some overlap), thus making it hard to distinguish among the different compounds. Thus, choice (C) cannot be the correct answer either. Lastly, we notice that each compound has a vastly different structure and molecular formula, giving them very different molecular weights. The molecular weight of camphoric acid is 200 g/mol; hippuric acid is 179 g/mol; diethylbarbituric acid is 184 g/ mol, and 4-aminobenzoic acid is 137 g/mol. We predict that the characteristic that will best allow us to distinguish the four compounds is their molecular weight—choice (D) is the correct answer. (A) Distortion. Since all of the compounds are acids, they have similar solubility behaviors. (B) Distortion. Since all of the compounds are acids, they have similar reaction profiles. (C) Distortion. The table tells us that the melting points exist between 183°C and 191°C, with many of the compounds containing overlapping temperatures.

Passage IV (Items 113–119) Passage Map Passage Type: Information Topic: Bacterial reproduction Scope: Bacterial reproduction and conjugation; importance in antibiotic resistance ¶1: E. coli – lives in colon and reproduces asexually; Plasmids confer resistance ¶2: Conjugation ¶3: Man with ruptured appendix received ampicillin and kanamycin; later contracted M. tuberculosis that did not respond to either drug 113. (C) How can a bacterium that has certain genes give rise to a daughter cell that lacks these genes? The answer must be in the passage. Since we are discussing bacterial reproduction, the words “sex pilus” should prompt us to look back at the second paragraph. Paragraph 2 tells us that the genes for sex pilus construction are contained on a plasmid, not the bacterial chromosome. As such, choices (A) and (B) can be safely eliminated. Furthermore, paragraph 1 tells us that plasmids may not be equally distributed to daughter cells. The reason why the daughter cell lacks the genes for the sex pilus construction is that it never received the plasmid that contains these genes. Our prediction matches with the correct answer, choice (C).

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119. (A) This item asks us to compare bacterial conjugation to sexual reproduction in eukaryotic organisms. First, let’s make sure that we understand the process of conjugation—our map indicates that we should review paragraph 2. During conjugation, genes from one bacterium are transferred to another bacterium. These genes are typically located on plasmids. However, since plasmids can also recombine with the bacterial genome, occasionally the sex pilus gene can be incorporated into the bacterial chromosome. As a result, during conjugation pieces of the bacterial genome can be transferred to new bacteria. This sequence of events ensures genetic variation and recombination in the bacteria—a desirable goal, since it increases the survival chances of the species. Let’s compare this with eukaryotes. In eukaryotes, the goal of sexual reproduction is to produce genetically unique offspring so as to promote genetic variation, which, once more, aids in the adaptation of a species to the environment. Thus, both bacterial conjugation and sexual reproduction culminate in producing genetic variability—the only answer choice that matches our prediction is choice (A), the correct response. (B) Distortion. This situation is called aneuploidy. Since bacteria do not have nuclei, this choice cannot be true. (C) Distortion. Conjugation does not result in daughter cells, just a recombination of the genomes of two bacteria. (D) Opposite. According to paragraph 2, conjugation can occur between members of different species.

Discrete Set I (Items 120–122) 120. (C) The question asks us to calculate the formal charge on the atoms present in the figure. However, glancing at the answer choices, we notice that we are only supposed to look at the first and seventh nitrogen atoms, and that the possible charges are ±1. The formal charge of an atom represents the number of valence electrons of that atom minus the number of electrons present in bonds or as free electron pairs. An easy way to calculate formal charge is to determine the number of valence electrons of the element based on the period it is in, and then subtract the number of sticks and dots that surround the atom in the Lewis structure. A stick represents a bond (so for every bond, we only look at the electron donated by this particular atom), and two dots represent a free electron pair. Let’s apply this strategy to our question. Nitrogen is in the fifth group, so it has 5 valence electrons. If we analyze N-1 and N-7 from the given diagram, we obtain the following:

FC (formal charge) = valence electrons – sticks – dots

FCN = 5 – sticks – dots

The formal charge on N-1 is:

FCN-1 = 5 – 2 – 4 = –1

The formal charge on N-7 is:

FCN-7 = 5 – 4 – 0 = +1

The formal charges are –1 and +1 for N-1 and N-7, respectively. This result matches with the correct answer, choice (C). (A) Distortion. N-7 has a +1 formal charge. (B) Distortion. N-1 has a –1 formal charge. (D) Opposite. N-7 has a +1 formal charge, and N-1 has a –1 formal charge.

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AAMC Test 4 Explanations

121. (A) This diagram represents a neuronal pathway between a stimulus and the muscle, and, more specifically, a picture of a reflex pathway. A pain stimulus is transmitted through a sensory neuron to the spinal cord, which interprets the message and then sends a motor neuron to the muscle (which makes the muscle contract so as to withdraw from the painful stimulus). In our diagram, Roman numeral I represents the sensory neuron which is activated when the painful stimulus is in contact with the skin, Roman numeral II is the synapse between this neuron and the spinal cord, Roman numeral III is the synapse between the spinal cord and the motor neuron, and Roman numeral IV is the neuromuscular junction between the motor neuron and the muscle. If we want to add more neurons so as to illustrate the pathway through which pain is detected, we have to involve the brain since feeling pain is not a reflex, but rather a perception that needs cortical input. Thus, we would have to add neurons at the level of the spinal cord, so that the signal can ascend to the brain. The sites where these neurons should be added are the level of the spinal cord input and output, that is, at positions II and III. The correct answer, therefore, is choice (A). (B) Distortion. IV is the site of the neuromuscular junction where the motor neuron acts on the muscle to make it contract. (C) Distortion. IV is the site of the neuromuscular junction where the motor neuron acts on the muscle to make it contract. (D) Distortion. I is the site of detection of the painful stimulus so the sensory neurons are there already. Additionally, IV is the site of the neuromuscular junction where the motor neuron acts on the muscle to make it contract. 122. (C) The ratio of affected individuals, recessive genes, and heterozygous individuals are all key ideas that help us identify this as an item that tests our understanding of the Hardy-Weinberg equation. This concept allows us to express the distribution of dominant and recessive phenotypes within a population such that:

p2 + 2pq + q2 = 1

p+q=1

where p is the frequency of the dominant gene, q is the frequency of the recessive gene, p2 is the frequency of the dominant homozygotes, q2 is the frequency of recessive homozygotes, and 2pq is the frequency of the heterozygotes. The stem gives us the number of individuals who are recessive homozygotes, and asks us to calculate the number of individuals who are heterozygous. In other words, we are given q2 and asked to calculate 2pq. The question tells 160 16 us that the number of recessive homozygotes (people exhibiting the recessive trait) is ​ ____    ​, or ​ ___ ​. Therefore, q is 1000  100    equal to:

q2 = 0.16

q = 0.4

This is the frequency of the recessive gene in a population. In order to calculate the number of heterozygous individuals, we need to determine the frequency of the dominant gene:

p+q=1 p = 0.6

Our task now is to calculate the number of heterozygous individuals, that is, 2pq, as follows:

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2pq = 2 × 0.4 × 0.6 = 0.48

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This means that about 48% of individuals within this population are heterozygous. The question stem mentions that the population is composed of 1,000 people, so the number of expected heterozygous individuals is 480. Choice (C) is the correct answer. (A) Miscalculation. This is q2. (B) Miscalculation. This is the value of q, multiplied by 1,000. (D) Miscalculation. This is the value of p, multiplied by 1,000.

Passage V (Items 123–128) Passage Map Passage Type: Experiment Topic: Endothelium Scope: Endothelium role in vasodilation ¶1: Vasodilation of arteries in response to acetylcholine was inhibited if endothelial layer was damaged. ¶2: Relaxation of intact and damaged endothelium in response to different concentrations of acetylcholine were compared Fig. 1: Effect of acetylcholine on dilation in aorta with and without endothelium, in presence and absence of norepinephrine ¶3: L-NMMA inhibits NO synthesis, which is a vasodilator. 123. (C) The item asks us to interpret the results illustrated in Figure 1 in order to explain the necessity of the endothelium for smooth muscle relaxation. Our first step, therefore, is to analyze the graph. Looking at Figure 1, the top chart describes the change in tension in the aorta ring containing a functional endothelium. Initially, norepinephrine is added and the tension in the ring increases. After addition of acetylcholine up to a concentration of 10–7 M, the tension decreases. We conclude from here that acetylcholine does indeed contribute to smooth muscle relaxation. The second chart shows the tension in the aorta ring without endothelium. Just as in the first graph, norepinephrine is added and the tension in the ring increases. Addition of acetylcholine, however, does not result in smooth muscle relaxation, since the tension does not decrease until the washout begins. This allows us to conclude that in the absence of the endothelium, acetylcholine does not have any effect on smooth muscle relaxation. We predict that the main difference between the two charts in Figure 1 is that acetylcholine decreases tension at a concentration of 10–7 M in the presence of endothelium, but does not decrease tension in the absence of endothelium. This observation leads to the conclusion that intact endothelium is necessary for smooth muscle relaxation in the presence of acetylcholine. The only answer choice that matches our prediction is choice (C). (A) Faulty use of detail. Although it is true that tension increases when norepinephrine is added, this has nothing to do with relaxation in the presence of ACH. (B) Opposite. The tension decrease does not occur in the ring without endothelium until washout. (D) Distortion. The tension decreases during washout in both rings.

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or constriction of blood vessels, often referred to as the resistance to blood flow. We predict, thus, that blood pressure is determined by cardiac output and resistance to blood flow—the correct answer is choice (B). (A) Distortion. L-NMMA and norepinephrine will both alter the constriction and dilation of blood vessels, altering the resistance component of blood pressure, but they are not the two primary factors which determine blood pressure. (C) Distortion. L-arginine is used to synthesize NO, but cannot be said to be a primary factor that determines blood pressure. (D) Distortion. Heart rate and stroke volume determine cardiac output, which is only one of the components needed to determine blood flow—we also need to take into consideration the resistance of the blood vessels. 127. (D) This item is a pseudo-discrete because we do not need any information from the passage to answer it. The stem is essentially asking us to determine when it would be beneficial to reduce blood flow to certain areas. Why is vasoconstriction utilized in the cardiovascular system? Vasoconstriction is important in directing blood flow away from different parts of the body—most important when that area of the body is losing large amounts of blood. By constricting, the area of the blood vessel decreases, thereby maintaining a more normal blood pressure, which is essential for the body’s systems. Thus, vasoconstriction is the body’s mechanism of maintaining a normal blood pressure when there is a decrease in blood volume—choice (D) is the correct response. (A) Opposite. Vasodilation causes a decrease in blood pressure. (B) Opposite. Vasodilation of the arteries in the muscles increases the blood flow to muscles. (C) Opposite. Vasodilation of the skin vessels causes blushing. 128. (C) This item tests our understanding of a detail in the passage, namely the effect of L-NMMA on aortic ring relaxation. Where in the passage can we find this information? According to our map, paragraph 3 tells us that L-NMMA inhibits NO synthase. What does NO synthase do? NO synthase synthesizes NO, which promotes relaxation of smooth muscle. Thus, when L-NMMA inhibits the synthase, it directly inhibits the production of NO, thus decreasing the amount of NO available. If there is less NO, there will be less smooth muscle relaxation, and, consequently, more muscle contraction—or increased tension in the aorta ring. This is our prediction—only choice (C) fits with our rationale, and is therefore the correct answer. (A) Distortion. NO and acetylcholine use two unrelated mechanisms for smooth muscle relaxation. (B) Opposite. L-NMMA inhibits the production of NO. Since NO relaxes smooth muscle, the absence of NO will increase muscle contraction. (D) Distortion. NO and acetylcholine use two unrelated mechanisms for smooth muscle relaxation.

Discrete Set II (Items 129–132) 129. (A) The item asks us to select the least polar compound from the four given choices. The polarity of a molecule depends on the polarity of the intramolecular bonds and the symmetry of the molecule. Glancing at the answer choices, we notice that the only two types of bonds that we have to analyze are C–H and C–Br, both polar bonds.

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AAMC Test 4 Explanations

In order for the molecule to be least polar, it has to have bonds whose polarities cancel each other out; that is, it has to be symmetrical. A clear example of this is a methane molecule that has a tetrahedral shape and four hydrogens surrounding a central carbon atom; because of the symmetry of the molecule, methane is not polar, despite its polar bonds. Taking this concept and applying it to our question, we realize that when four bromine atoms are equally distributed around the carbon atom, the vector sum of the polarity of each individual C-Br bond cancels out for a total molecule polarity of zero. Therefore, the molecule that has the smallest polarity is tetrabromomethane, CBr4, choice (A). (B) Distortion. The vector sum of bond polarities of this molecule is greater than zero because the molecule is asymmetrical with respect to the polar bonds. (C) Distortion. The vector sum of bond polarities of this molecule is greater than zero because the molecule is asymmetrical with respect to the polar bonds. (D) Distortion. The vector sum of bond polarities of this molecule is greater than zero because the molecule is asymmetrical with respect to the polar bonds. 130. (C) This is a rather straightforward question that tests our knowledge of Organic Chemistry vocabulary. We are given an equation that shows the migration of a hydrogen atom from an alpha-carbon to an oxygen atom, causing a change in the hybridization of the carbon and oxygen atoms. This is the definition of tautomerism. The tautomers shown are keto-enol tautomers, because they change their functional groups from a ketone to an alkene and alcohol functional group (enol). The correct answer, therefore, is choice (C). (A) Distortion. Reduction is a change in the formal charge of the molecule—no such change occurs in this equation. (B) Distortion. Resonance is the movement of electrons within a molecule, but the bonds formed between atoms remain unchanged. (D) Distortion. This term refers to sugars and their ability to change conformations between the α and β anomers. 131. (B) This item tests us on a content detail pertaining to fungal spores. Fungi form spores in harsh environmental conditions, allowing the species to survive unfavorable environments. Therefore, spores are metabolically inactive forms that reactivate only when environmental conditions improve. Spores are also haploid (n). The correct answer is choice (B). (A) Opposite. Fungal spores are metabolically inactive and haploid. (C) Opposite. Fungal spores are resistant to environmental changes. (D) Opposite. The spores are encased in a material resistant to environmental conditions. 132. (A) As a typical genetics item, we will have to appeal to our knowledge of Punnett squares. The stem tells us that hemophilia is the result of a sex-linked gene. We also know that this gene must be recessive because the mother is a carrier but does not show signs of the disease. We are asked to determine the probability that all three sons will have hemophilia. The first step in this type of question is to determine the genotypes of the parents. Let’s denote the hemophilia gene as Xh. The man has normal blood clotting. Since he has only one X gene, this

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gene must be normal; otherwise, he would show signs of the disease. The father’s genotype is therefore XY. The woman has normal blood clotting, but her father had hemophilia. Since the father had only one X gene, it must have been an Xh and his daughter had to inherit it. Since the woman does not express symptoms of hemophilia, we assume that she has to have one normal X gene. The mother’s genotype must be XXh. So the cross is XXh x XY. Let’s plug these values in a Punnett square to determine the probability of a son having hemophilia:

X Y

X XX XY

Xh XhX XhY

There are two possibilities for a male child from this cross: hemophilia and normal. So the chance of having one son with hemophilia is ​ _12 ​. Each son has a ​ _12 ​chance of having hemophilia. To get the chance of having three sons with hemophilia, we must multiply the individual probabilities of all three sons having hemophilia. ​ 1 ​  Probability of three sons with hemophilia = __ ​ 1  ​× __ ​ 1  ​× __ ​ 1  ​ = __ 2 2 2 8 The answer choice that matches our calculation is choice (A). (B) Miscalculation. This is the chance of either having one child (either sex) with hemophilia, or having two sons with hemophilia. (C) Miscalculation. This answer would be incorrectly obtained if you added the probabilities of all three sons having hemophilia. (D) Miscalculation. This is the chance of having one son with hemophilia.

Passage VI (Items 133–136) Passage Map Passage Type: Information Topic: Septic shock Scope: Two possible treatments of septic shock ¶1: Lipopolysaccharide antigens (endotoxins) on gram-negative bacteria cause septic shock syndrome. Fig. 1: Cascade of events leading to septic shock, begins with endotoxins ¶2: 2 treatment plans to reduce mortality rate of septic shock ¶3: Treatment 1 – general anti-inflammatory drug to block activation of lymphocytes ¶4: Treatment 2 – antibody to block the endotoxin-binding protein complex on macrophages 133. (D) Glancing at the answer choices, we notice that the question asks us to analyze Treatment 2 to determine the role of the antibody on T-cell or macrophage activation and production. Let’s begin by reviewing the description of the treatment. The passage states that Treatment 2 blocks the attachment site for the endotoxin-binding protein complex on macrophages. What is the purpose of doing this? If the protein complex does not attach to macrophages, then the macrophages will not be activated. According to Figure 1, macrophage activation leads

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136. (C) This is a straightforward question that tests our knowledge of the cell cycle—it is a pseudo-discrete, since we do not need any information from the passage to answer it. Basically, the stem asks us to identify the phase of the cell cycle during which DNA is replicated. If a drug interferes with DNA replication, the cell cycle would halt in the stage where DNA replication takes place. DNA replication occurs in the S, or synthesis, phase of the cell cycle. Therefore, the correct answer is choice (C). (A) Distortion. The G0 phase is a non-dividing phase; it is a stage of biochemical activity. (B) Distortion. The G1 phase is a phase of intense biochemical activity and growth. When a cell proceeds through this stage and passes a restriction point, it is committed to dividing. (D) Distortion. The G2 phase occurs after the S phase; the cell continues to grow until mitosis commences.

Passage VII (Items 137–140) Passage Type: Information Topic: Esters Scope: Esters are synthetic precursors for methyl ketones and carboxylic acids. ¶1: Acetoacetic ester and malonic ester are used to prepare methyl ketones and carboxylic acids. Scheme 1: Acetoacetic ester/malonic ester; conversion of malonic ester into carboxylic acid ¶2: Alkylation of acetoacetic ester ¶3: Acetoacetic ester may be deprotonated into a dianion and then alkylated. ¶4: Compounds that may be synthesized by acetoacetic ester or malonic ester synthesis Table 1: Compounds that may be synthesized by acetoacetic ester or malonic ester synthesis 137. (D) This item asks us to determine which of the four given halides can be a reactant in the synthesis of Product 5. According to Table 1, Product 5 is 4-methyl-2-pentanone. CH3 H3C

O CH3

Next, we need to figure out how acetoacetic ester is converted into a ketone. The acetoacetic ester synthesis is similar to the malonic ester synthesis shown in Scheme 1. However, the final product in Scheme 1 is a carboxylic acid, while the acetoacetic ester reaction results in a methyl ketone. This is due to the absence of a second ester group. Let’s review the reaction mechanism:

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AAMC Test 4 Explanations

O O

O

O 1. NaOEt 2. RX

OEt

O

1. NaOH 2. H+

R

OEt R To make this generic product mirror the 4-methyl-2-pentanone, the R group must be an isopropyl halide. The only answer choice that refers to an isopropyl halide is choice (D), the correct answer. (A) Distortion. This will produce 2-butanone. O

O X

2-Butanone

CH3

R

Methyl Ketone

(B) Distortion. This will produce 2-pentanone. O

O XH2C

2-Pentanone

CH3

R

Methyl Ketone

(C) Distortion. This will produce 2-hexanone. O

O

X 2-Hexanone

Methyl Ketone

R

138. (C) The item asks us to determine why the product of the saponification is acidified. A glance at the answer choices suggests that our task is to determine the reactant and product of the acidification reaction. Since the acidification reaction follows saponification, we need to understand what the result of the saponification reaction will be—since it will be the acidification reactant. Where is saponification? Saponification is the hydrolysis of a triacylglycerol, resulting in glycerol and carboxylate salts as products. Therefore, the acidification reaction converts the salt (the reactant) into some product. Returning to the answer choices, only one mentions a salt as the starting product for the acidification—choice (C) must be the correct answer. To complete our analysis of this reaction, we read

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in paragraph 1 that the acidification of the carboxylate salt results in a carboxylic acid (the end product). Again, choice (C) is the correct answer. (A) Distortion. Although the saponification followed by acidification results in the conversion of an ester into an acid, the saponification product—the reactant used in the acidification reaction—is a salt. (B) Opposite. This is the opposite, since a salt is acidified so as to convert it into an acid. (D) Distortion. Saponification results in the conversion of an ester into a salt. 139. (C) Glancing at the answer choices, we notice that the question asks us to determine the synthetic pathway that will lead to the formation of 3-ethyl-2-pentanone. Let’s start by quickly reviewing where the acetoacetic ester 1 reactions are discussed. According to our map, paragraph 2 tells us that a second alkylation step of acetoacetic ester 1 can be incorporated into the synthesis, so let’s try rewriting the product as a methyl ketone with two R groups. This is important, since the formation of 3-ethyl-2-pentanone from acetoacetic ester 1 requires two alkylation steps: O

O R

+

H3C

CH3

R 3-ethyl-2-pentanone

Methyl Ketone

R

Rewriting this as a successive alkylation, we notice that the synthesis requires an ethyl group to be added twice—this can be accomplished by alkylating 2-acetoacetic ester with an ethyl halide. The answer choice that matches our prediction is choice (C), the correct response. (A) Distortion. Using isobutyl iodide will produce 3-methyl-2-pentanone. O

(B) Distortion. 3-methyl-2-butanone would be produced by successive alkylations with methyl iodide. O

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AAMC Test 4 Explanations

(D) Distortion. Alylkation of the dianion with isopropyl iodide: O

O

O–

O–

OEt

OEt

O O–

O–

O

I +

OEt

OEt

140. (A) This item tests our understanding of the second alkylation of acetoacetic ester. According to our passage map, this is discussed in paragraph 2, so let’s quickly review the mechanism. The reaction occurs because the hydrogen atoms on the carbon between the two carbonyl carbons are slightly acidic (alpha hydrogens adjacent to carbonyl carbons are quite acidic). When the acetoacetic ester has been alkylated once, the remaining carbon is less acidic. Because of this, a stronger base is necessary to successfully facilitate the second alkylation. The only answer choice that mirrors our prediction is choice (A), the correct response. (B) Opposite. Addition of an alkyl group makes the remaining carbon less acidic. (C) Distortion. Since the product is sterically hindered, a smaller base will be more effective at abstracting the proton. (D) Distortion. The monoalkylated product is less soluble in ethanol.

Discrete Set III (Items 141–146) 141. (D) The item tests us on an important detail in the mammalian respiratory system. What accompanies the inflation of the lungs? First of all, the inflation of the lungs corresponds to inhalation. More specifically, the inflation of the lungs is accomplished by the contraction of the diaphragm, which enlarges the chest wall cavity. As the volume of the chest wall cavity is increased, the lungs (attached to the chest wall by the pleura) expand as well. Expansion of the lungs creates a negative pressure, sucking air into the lungs by a negative pressure pumping action. Our prediction matches with the correct answer, choice (D). (A) Distortion. Gas exchange is accomplished by diffusion of gases. (B) Distortion. Gas exchange at the level of the lungs is accomplished by passive transport. (C) Opposite. This characterizes the exhalation stage, when the lungs are deflated and air is forcefully pushed out.

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Biological Sciences

142. (A) This item asks us to identify a primary difference between bacteria and viruses—the answer choices confirm this by offering cellular criteria. Let’s start by quickly reviewing each of these two organisms. Bacteria are prokaryotic cells that may contain cell walls, lack membrane-bound organelles, and have all the basic cellular compounds (DNA, RNA, protein). Bacteria reproduce via fission or budding. Viruses, on the other hand, are nonliving particles that may be encased in a protein coat similar to a cell wall. All viruses contain some type of genetic material (RNA or DNA) and the proteins and enzymes necessary to incorporate their genetic material into the host. Viruses reproduce by using the host cell’s machinery, followed by budding of the viral progeny. Based on this information, we can now look at the answer choices. The correct answer must be choice (A), since bacteria and viruses can be distinguished by their mode of reproduction. (B) Distortion. Both bacteria and viruses can have a cell wall. (C) Distortion. Both bacteria and viruses lack a nuclear membrane; they do not contain membrane-bound organelles such as the nucleus. (D) Distortion. Both bacteria and viruses contain RNA and protein. 143. (A) The item asks us to determine if aldosterone is secreted in response to increased sodium chloride. Categorizing the answer choices, we notice that our task is to predict a yes/no response followed by an explanation. To answer this question, we must first review the function of aldosterone. Aldosterone increases sodium reabsorption at the level of the kidney. Based on this, we can immediately eliminate choices (B) and (D), since aldosterone does not cause sodium secretion. When is aldosterone used? Aldosterone is typically secreted in response to low blood volume—the consequent reabsorption of salt into the circulatory system causes water to be also pulled into the blood, effectively relieving the low blood volume/pressure. Since aldosterone increases sodium reabsorption, we would expect its secretion to be decreased in the presence of excess salt—the body will try to eliminate the excess salt from the body via excretion. Any secretion of aldosterone would interfere with this homeostatic response, as the consequent salt reabsorption would prevent the body from removing the excess salt. Knowing this, we do not expect aldosterone levels to be increased in response to excess salt—choice (A) is the correct answer. (B) Distortion. Aldosterone increases sodium reabsorption at the level of the kidney. (C) Opposite. Following the ingestion of excessive quantities of NaCl, the plasma aldosterone concentration should be low, so as to facilitate the excretion of the excess salt. (D) Opposite. Aldosterone increases sodium reabsorption at the level of the kidney. 144. (D) Why can HIV reproduce in the host cells? We are given that HIV is an RNA virus that can insert itself into the human genome, which is composed of DNA. In order for the viral RNA to be incorporated into the human DNA genome, the RNA must first be converted to DNA. The enzyme necessary to do this is reverse transcriptase, so-called because normally the transcriptase enzyme transcribes DNA to RNA. A successful RNA virus would require reverse transcriptase in order to transcribe its RNA to DNA, effectively allowing insertion into the host DNA—choice (D) is the correct answer. (A) Distortion. HIV destroys T-cells but this does not allow it to insert itself into the host genome. (B) Distortion. We are told that HIV only contains RNA. (C) Distortion. HIV has core proteins, but this does not allow it to insert itself into the host genome.

71

AAMC Test 4 Explanations

145. (B) This is a straightforward question that requires knowledge of the menstrual cycle to determine the relationship between several organs. To begin with, the hypothalamus stimulates the anterior pituitary to secrete and release GnRH in a cyclical manner, allowing the release of LH and FSH. Based on this information alone, we can immediately eliminate choices (A), (C), and (D). LH and FSH stimulate the development of ovarian follicles, leading to the secretion of progesterone and estrogen by the ovary. Estrogen causes feedback inhibition on LH and FSH. The rise and fall of these hormones coordinates ovulation and the menses. The answer choice that correctly illustrates this interplay of organs is choice (B). (A) Distortion. The thyroid is not involved in the menstrual cycle. (C) Distortion. The thyroid is not involved in the menstrual cycle. (D) Distortion. The adrenal glands are not involved in the menstrual cycle. 146. (D) This is quite a complicated question so let’s break it down into steps. The first task is to realize that this is a genetics problem requiring a Punnett square in order to determine the number of different phenotypes resulting from two successive crosses. The F1 generation is produced by crossing homozygous long, white plants with homozygous short, red plants. We’ll represent the gene coding for long stems as L, and that for short stems as l; also, the gene for white color can be written as r, and that for red color can be written as R. Knowing this, the F1 generation will be the result of an LLrr × llRR cross. As such, all of the offspring in the F1 generation have the genotype LlRr (long/pink). Crossing the offspring from the F1 generation amongst themselves results in LlRr × LlRr. To make things more manageable, we’ll rely on the use of a Punnett square:

LR

LR LLRR

Lr LLRr

lR LlRR

lr LLRr

Lr lR lr

LLRr LlRR LlRr

LLrr LlRr Llrr

LlRr llRR llRr

Llrr llRr llrr

We notice that the phenotypes created in the F2 generation are long/red (LLRR, LlRR), long/pink (LLRr, LlRr), long/white (LLrr, Llrr), short/red (llRR), short/pink (llRr), and short/white (llrr)—there are six different phenotypes, making choice (D) the correct answer. (A) Miscalculation. This would not be the number of phenotypes occurring in the F2 generation. (B) Miscalculation. This would be the number of phenotypes resulting from a monohybrid cross of the F1 generation. (C) Miscalculation. This would not be the number of phenotypes occurring in the F2 generation.

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