Passage I
143. A. The question states that morphine increases parasympathetic impulse traffic to the iris, and this, as indicated in the last paragraph of the passage, would cause constriction of the pupil (choices C and D can be eliminated). Furthermore, the passage states that acetylcholine is the neurotransmitter released by the parasympathetic system at the organ level (choice B can be eliminated).
144. A. Since atropine is preventing acetylcholine (the normal parasympathetic neurotransmitter) from binding to its receptor, it is blocking the effects of the parasympathetic system, and this is described in the question as a passive mechanism (choices B, C, and D can be eliminated).
145. D. Acetylcholine is the parasympathetic neurotransmitter, thus if it is removed from circulation faster than
norepinephrine (the sympathetic neurotransmitter), parasympathetic effects must turn off faster than sympathetic effects.
Choices A, B, and C all describe sympathetic effects; only choice D, stimulation of digestive secretion, is a parasympathetic effect, and would be inactivated the fastest.
146. C. Any nerve fiber that causes an effect on an organ is a motor fiber; sensory fibers only pick up information from an organ and send it to the brain. Thus nerve fibers that cause cardiac slowing must be motor fibers (choices B and D can be eliminated). Specifically, the fibers that reduce the heart rate are parasympathetic fibers (choice A can be eliminated).
147. A. The type of neurotransmitter released is irrelevant (choices B and D, while true, do not answer the question and can be eliminated). The interconnected ganglia of the sympathetic system allow for rapid systemic response, because if one ganglion gets stimulated, it can rapidly stimulate all ganglia in the system. If the ganglia are not interconnected, as in the parasympathetic system, this is not possible (choice D can be eliminated).
148. D. Physostigmine in this case is acting as an “inhibitor of an inhibitor.” Acetylcholinesterase breaks down acetylcholine, thus decreasing its levels. If physostigmine inhibits acetylcholinesterase, then acetylcholine would not be degraded, and its levels would rise (choices A and C can be eliminated). The passage states that the effect of acetylcholine on the pupil is to cause constriction (choice B can be eliminated).
Passage II
149. C. First, glucose is never used directly as an energy source; it is oxidized and the energy released is stored in the form of ATP. In any case, osmosis is a passive process and does not require an energy source (choice A can be eliminated), nor does water cross the plasma membrane through “ion exchange pumps” (choice B can be eliminated). Choice C is true.
Additional glucose molecules would raise the osmotic pressure (think “concentration”) of the extracellular space; this in turn would promote the movement of water out of the cell by osmosis (cellular dehydration). Choice D may or may not be true. The passage implies that tissue glucose increases, but whether this occurs by exchange with water molecules is not clear. This makes C a better choice than D.
150. A. Hyperglycemia is high extracellular or plasma glucose levels. Insulin’s job is to reduce plasma glucose levels, so suppressing insulin secretion could lead to higher plasma glucose, and would support the hypothesis. Glucagon has the opposite effect of insulin. Its job is to raise plasma glucose; suppression of this hormone could not lead to hyperglycemia (choice B can be eliminated). Glucose released from the liver into the plasma comes from the catabolism (breakdown) of glycogen. If this process were slowed, it could not cause hyperglycemia (choice C can be eliminated). Lastly, increased sensitivity of all pancreatic endocrine responses would lead to increased levels of both insulin and glucagon in the blood.
While excess glucagon could certainly lead to hyperglycemia, excess insulin could not — the hormones have opposite effects (choice D is eliminated).
151. A. The question specifically asks for the relationship of a beating heart to the cryoprotective role of glucose; since temperature equilibration and a reduction in the rate of ice formation are not glucose-related, choices B and C can be eliminated on the basis of irrelevance. Choice D is just false; a beating heart requires a constant supply of ATP, not glucose, as an energy source. Choice A is true and has the most relevance with respect to the question being asked.
152. D. The passage states that only extracellular water freezes. Cytoplasm is intracellular fluid, so Item I is false and choice C can be eliminated. Blood plasma is extracellular; Item II is true, and choice B can be eliminated. Lymph is also extracellular; Item III is true, and choice A can be eliminated.
153. D. The frog’s tissues would be forced to rely solely on anaerobic respiration once the heart stopped beating and no longer circulated blood and oxygen throughout the body. Based on the graph of Heart Rate in Figure 1, this occurs between 12 and 24 hours after the onset of freezing.
154. A. Figure 2 shows that frogs injected with glucose had a much higher rate of survival than frogs injected with saline, and that the rate of survival increased with increased glucose. Thus it is clear that glucose has some cryoprotective role in frogs (choices C and D can be eliminated).
155. B. Choices A, B, and D to some extent are true. However, the main “theme” of the passage is that wood frogs can survive freezing episodes by increasing their extracellular glucose concentrations, and that the primary mechanism by which this occurs is accelerated glucose release from glycogen stores in the liver. Therefore, it seems that the most effective physiological condition for surviving freezing episodes would be to have ample liver glycogen stores (choice D is true, but B is better). Choice A might be tempting, since the role of the excess glucose is to cause cellular dehydration, but simple dehydration in the absence of glucose is not described in the passage as a survival mechanism. Choice C is not discussed anywhere in the passage as being a method of surviving freezing.
Passage III
156. D. The passage states that Compound C is soluble in both dilute acid and base. These data suggest that Compound C has both a basic and an acidic functional group. Of the choices given, only D contains both a basic (amine) and an acidic (carboxylic acid) functional group.
157. C. Since all four compounds in Table 1 contain a C=O double bond, they will all have a strong, sharp band at 1700 cm–1 in their IR spectra.
158. B. While pure compounds typically have sharp melting points, impure mixtures tend to melt over a broad temperature range. This eliminates choices C and D. Choice A can be eliminated since the melting point of a mixture is never higher than that of the components of the mixture.
159. C. The passage states that Compound B is a water-soluble alcohol. It is most likely water soluble due to the hydrogen-bonding ability of its hydroxyl group.
160. B. The fact that Compound A slowly dissolves in refluxing aqueous NaOH to result in the formation of two new compounds is consistent only with choice B. An ester is hydrolyzed under such conditions to form a carboxylate and an alkoxide.
161. C. Only choice C, 4-aminobenzoic acid, is an amine. Hippuric acid and diethylbarbituric acid are amides, not amines.
162. D. Since three of the four compounds in Table 1 contain carboxylic acids, they cannot be distinguished based on solubilities or reactivities toward alcohols. Thus, we can eliminate choices A and B. Since all four compounds melt in the range of 183–191°C, they cannot be unambiguously distinguished based on their melting points. They can only be distinguished by their molecular weights.
Passage IV
163. C. The passage states that the genes for conjugation are carried on a plasmid (choice A can be eliminated), and that during cell division, plasmids may not be equally distributed among daughter cells. It may be that one of the daughter cells failed to receive the plasmid that carried the genes for conjugation. The cell membrane has nothing to do with moving the replicated chromosomes apart (choice B can be eliminated), and bacteria are prokaryotic and do not contain membrane-bound organelles like lysosomes (choice D can be eliminated).
164. D. Antibiotics do not lead to mutations in bacteria (choice A can be eliminated), nor can bacteria develop an “immune reaction” to antibiotics. An immune system is required for that, which bacteria do not have! (Choice B can be eliminated.) There is nothing in the passage to support choice C, so it can be eliminated. It is most likely that, due to random mutation, some of the patient’s E. coli were resistant before treatment. Since they were not killed by the antibiotics, they continued to reproduce.
165. A. Of the choices given, the best answer is choice A, although this is still a relatively unlikely possibility. However, antibiotics do not induce mutation (choice B can be eliminated), the rate of reproduction has nothing to do with antibiotic sensitivity (choice C can be eliminated), and modification of metabolism would not alter antibiotic sensitivity (choice D can be eliminated).
166. C. The appendix is found at the beginning of the colon, so if it were to rupture, the E. coli that normally inhabit the colon could enter the abdominal cavity and cause serious problems (choice C is true, and choices B and D are false).
M. tuberculosis causes tuberculosis, not appendicitis (choice A can be eliminated).
167. A. The persistence of an organism in any environment is determined primarily by the number of surviving offspring it can produce (A is true). The ability to produce more vitamins (choice B) or to metabolize glucose faster (choice C) do not affect E. coli’s ability to reproduce, nor does their lack of pathogenicity (choice D).
168. B. No digestive enzymes are produced by the bacteria, and no nutrient absorption occurs in the colon (choices A and D can be eliminated). Choices B and C actually are equally likely to occur; the “jobs” of E. coli in the large intestine are to produce vitamins and to reduce the growth of other, pathogenic bacteria. However, since the passage specifically mentions vitamin production and does not mention reduced growth of other bacteria, choice B is a better answer than C.
Independent Questions
169. B. Of the choices given, actin will bind only to myosin molecules. (Don’t be tempted by choice A; during muscle contraction, ATP binds to myosin, not to actin.)
170. C. N-1 has two covalent bonds and two lone pairs, so it is similar to a deprotonated amine and has a (–) charge.
N-7 has four covalent bonds and is thus an ammonium ion and has a (+) charge.
171. A. Beginning with the neuron that senses the painful stimulus, the neurons, in order, are a sensory neuron, an interneuron, and a motor neuron. In order to perceive the pain, the sensory information would have to be relayed to the brain and processed. Additional neurons could be placed anywhere around the interneuron (which is found in the central nervous system at the spinal cord level); in other words, at either synapse II or synapse III.
172. C. First, always remember that the phrase “randomly mating population” is a tip off for using Hardy–Weinberg in some way. The question states that 160 members of a 1000-member population exhibit a specific recessive trait. A
homozygous recessive genotype is required to exhibit a recessive trait, thus 16% of the population is homozygous recessive.
In the equation for genotype frequency, p2 + 2pq + q2 = 1, the value of q2 is 0.16; thus, q = 0.4. Plugging this value for q into the equation for allele frequency, p + q = 1, we find that p = 0.6. The question asks for the number of individuals in the population that are carriers of this trait. Carriers are those individuals heterozygous for the trait; they “carry” the allele, but don’t express it because of their dominant allele. In the equation for genotype frequency, these are the “2pq” individuals.
Plugging in the values we obtained previously, we get 2pq = 2 × 0.6 × 0.4 = 0.48, or 48% of the population is made up of
Passage V
173. A. Blood cells generally have short life spans; if they were not replaced continuously by stem-cell division and differentiation, we would quickly run out of blood cells, and the number of cells remaining would be incompatible with life.
Blood cells are stored in the spleen (choice B is false), they do not exit the body through the urinary and digestive systems (choice C is false), and they do not differentiate into other cell types (choice D is false).
174. A. The development of hormone receptors on the cultured stem cells implies that their differentiation will be controlled by the hormones that bind there. More rapid division (choice B) does not address the differentiation issue; in other words, the cells are dividing more rapidly, but is their differentiation random or controlled? Random differentiation supports the stochastic view (choice C can be eliminated), and choice D does not seem to support either the deterministic view or the stochastic view.
175. B. Stem cells that express hormone receptors support the deterministic view, but if the expression of the receptors is random, this also supports the stochastic view. If hormone X causes the formation of erythrocytes, this supports only the deterministic view (choice A can be eliminated). If the receptors are being expressed in response to specific signals, this also supports the deterministic view (choice C can be eliminated). Choice D does not seem to support either view.
176. D. If all blood-cell types need to be replaced, then a totipotent stem cell (one that can differentiate into all cell types) needs to be introduced (choices A and B can be eliminated). Furthermore, the greatest advantage would come from a totipotent stem cell that differentiated deterministically, as this would allow it to produce cell types according to the specific needs of the new host, as directed by the host’s hormones and other external signals (choice C can be eliminated).
Passage VI
177. C. While hydration and oxymercuration–demercuration make the more-substituted alcohol, hydroboration–oxidation makes the less-substituted alcohol. Therefore, choice C is correct.
178. A. After Step 1, the organomercurial alcohols formed have the –OH and the Hg(OAc) groups trans on the ring. This relative stereochemistry is most likely achieved through the intermediate in choice A wherein the 3-membered ring is opened by H2O, as shown below:
HgOAc
OH2
HgOAc HgOAc
OH
H2O – H
179. C. This reaction sequence places the –OH group on the less-substituted carbon atom of the C=C double bond, and as stated in the passage, does not involve skeletal rearrangements.
180. D. The O–H stretch of alcohols occurs at 3500 cm–1 in IR spectra.
181. A. Since Items II and III are two-step synthetic reactions, they are not likely to be reversible. Therefore, only choice A can be correct.
182. C. Since the two isomeric organomercurial alcohols in Equation 2 are non-superimposable mirror images of each other, they are enantiomers.
Passage VII
183. C. The fact that relaxation due to ACh only occurred in the ring with intact endothelium supports the conclusion that intact endothelium is necessary for this process. The tension increases in both rings upon addition of NE (intact
endothelium and damaged endothelium), so this gives us no information about the role of the endothelium specifically, and in any case NE is not ACh! (Choice A can be eliminated.) Tension decrease in both rings is approximately equal, and if anything, is slower in the ring without endothelium (choice B can be eliminated), and of course tension decreases during washout in the ring without endothelium. The NE that caused the increase in tension in the first place is being removed.
Still, this gives us no information about the role that the endothelium plays in response to ACh (choice D can be eliminated).
184. C. Since we have no data on the effects of ACh in the absence of NE, we cannot make any conclusions about NE’s effect on the smooth muscle’s sensitivity to ACh (choices A and B can be eliminated). Clearly ACh, which would cause a reduction in tension, has no effect at 10–8 M, so it is untrue to say that its effect is greatest at this concentration (choice D can be eliminated). Since concentrations of ACh above 10–6 M were not tested, and since the ring with endothelium responded to an ACh concentration of 10–7 M, it is fair to say that the presence of endothelium caused at least a 10-fold increase in sensitivity of the aortic muscle.
185. B. Muscle tension should decrease in the presence of ACh. This only occurred in the ring with endothelium; the ring without endothelium was not sensitive at all to ACh at any concentration tested (choice D can be eliminated). Since muscle relaxation occurred at 10–7 M ACh, choice B is the best response.
186. B. The two major factors that determine blood pressure are the cardiac output and the peripheral resistance. L-NMMA is not a naturally-occurring substance; the passage describes it as one of the enzyme inhibitors that “were developed”
(choice A can be eliminated). While blood volume does play a role in blood pressure, not all the L-arginine consumed will be converted to NO, so the amount of L-arginine in the diet cannot tell us anything about the status of the blood pressure (choice C can be eliminated). Choice D is tempting because it mentions the two factors that determine cardiac output (heart rate and stroke volume) and we know that cardiac output is a major determinant of blood pressure. However, choice D fails to take into account the other major determinant of blood pressure, the peripheral resistance.
187. A. The somatic nervous system deals only with the stimulation and contraction of skeletal muscle, not blood vessels (choices C and D can be eliminated). Furthermore, since ACh causes vasodilation (a parasympathetic effect) and NE causes vasoconstriction (a sympathetic effect), choice B can be eliminated.
188. D. Vasoconstriction would lead to an increase in blood pressure (choice A can be eliminated) and a reduction in blood flow (choices B and C can be eliminated). It would be most helpful in a situation where blood pressure was rapidly dropping and needed to be increased, such as during a hemorrhage.
Passage VIII
189. A. The bond angle in an equilateral triangle is 60°.
190. A. According to the passage, hydrolysis of an epoxide under acidic conditions results in inversion of stereochemistry.
This means that a trans diol must result, eliminating choices C and D. Choice B can be eliminated because it makes no sense; one cannot have a diol that is both axial and equatorial.
191. A. Since both carbon atoms of the epoxide are equally substituted, the regiochemistry is dictated by sterics. Since R is much larger than R′, attack of Nu: will be primarily at the carbon bearing R′. This eliminates choice C. Choice B can be eliminated because the stereochemistry of the left carbon in the product is drawn incorrectly (it is inverted). Choice D can be eliminated because the passage states that the configuration of the carbon that is attacked by the nucleophile undergoes inversion. This leaves choice A as the answer.
192. C. A 5-membered ring results from intramolecular hydrogen bonding in a 1,2-diol:
O H O
Ph
H
. . . . 2 1 3 4
5
193. D. Since styrene oxide acts as an electrophile in these ring-opening reactions, the greater nucleophilicity of diethylamine over ethanol explains its higher reactivity.
Independent Questions
194. A. CBr4 is the least polar of the bromomethanes because the four C–Br bond dipoles cancel each other out due to the tetrahedral symmetry of the molecule. This results in no dipole whatsoever for CBr4.
195. C. Tautomerism is a structural equilibrium and usually involves movement of a proton, as in this keto–enol equilibrium.
196. B. Remember the basic themes about fungi: they have a haploid life cycle (choice A can be eliminated), they possess a cell wall (choice D can be eliminated), and they can reproduce asexually by spore formation. Spores are, in a sense, like seeds. They are surrounded by a tough coat that helps them survive environmental extremes. When they have “weathered the tough times” and are again in a favorable environment, they will germinate (choice C can be eliminated).
197. B. The heart is derived from mesoderm, and so is the skeletal system. The eye and spinal cord are derived from ectoderm (choices A and C can be eliminated), and the liver is derived from endoderm.
198. A. A man with normal blood clotting has the genotype XY. Because women receive an X chromosome from their fathers, a woman with a hemophiliac father must carry at least one X chromosome with the recessive allele for hemophilia on it. However, because she displays normal blood clotting, her other X chromosome must be normal; in other words, she is heterozygous and has the genotype XhX. Here’s the Punnett square:
X X
X X X XX
Y X Y XY
h
h
h
From the Punnett square we can see that the probability of having a son with hemophilia (genotype XhY) is 1/2. For the probability of all three of their sons having hemophilia, use the Rule of Multiplication: 1/2 × 1/2 × 1/2 = 1/8.
Passage IX
199. D. Since T cells are not discussed in the passage anywhere, and since the antibody binds to a macrophage protein, choices B and C can be eliminated. From the diagram, it is clear that the events leading to septic shock involve activation of macrophages. If this is to be prevented, macrophage activation must be inhibited (choice A can be eliminated).
200. A. Drugs that decrease inflammation would not affect platelet or red blood cell counts (choices B and C can be eliminated). The inflammatory response depicted in the diagram shows hypotension, a decrease in blood pressure, as part
200. A. Drugs that decrease inflammation would not affect platelet or red blood cell counts (choices B and C can be eliminated). The inflammatory response depicted in the diagram shows hypotension, a decrease in blood pressure, as part