Item I: Yes. The author draws this distinction in lines 31-34 in order to emphasize the claim that formalist
D: Yes. While this evidence supports the claim that there was a trigger external to both the Iceland and Laurentide ice sheets (see the explanation for choice C), it gives no reason to believe that this trigger was a third ice sheet
137. B
A: While the author indicates that researchers believe that Heinrich events occur about every 7000 years (lines 2-5), none of their findings or interpretations of evidence are based on this time frame.
B: Yes. If the debris layer came from asteroids rather than from ice sheets, it would seriously call into question both the Laurentide explanation and the criticism of that explanation. All of the research described in the passage is predicated on the assumption that the debris was left behind by icebergs discharged from ice sheets. If that assumption is false, the conclusions drawn at first about the action and effects of the Laurentide sheet, and later about the synchronized behavior of the Icelandic sheet, would be without foundation.
C: Evidence about how tightly glaciers adhere, and presumably then about how fast they might move, has no direct relevance to any of the research reported in the passage. Glacial movement and friction were factors in Bond’s original explanation (lines 31-38), but as far as we know, he did not claim any particular rate of speed that would be
contradicted by this evidence.
D: The research in the passage, as described, is not based on the assumption that all cores must show debris. This does not call into question the relevance of samples with debris that have been found, or the conclusions based on those
findings.
BIOLOGICAL SCIENCES
Passage I
138. C. Difficulty in exhaling would lead to an accumulation of CO2, since that is the gas that is exhaled. CO2 is converted by carbonic anhydrase to carbonic acid, which dissociates into a hydrogen ion and a bicarbonate ion; the resulting
accumulation of H+ would lead to acidosis (specifically, respiratory acidosis, since it is due to a failure in the respiratory system). O2 would not accumulate, and even if it did, this would not be converted to an acid and would have no effect on pH (choice B is false). The text of the question states that inhalation is not a problem, so there would not be a decrease in O2 intake (choice D is false). Note that choice A, while true, is irrelevant since it does not answer the question.
139. C. Neutrophils are phagocytic white blood cells that typically move to sites of inflammation via chemotaxis.
Erythrocytes are red blood cells and do not participate in mediating inflammation (choice A is false), thrombocytes are platelets and play a role in blood clotting, not inflammation (choice B is false), and myocytes are not even a white blood cell—they are muscle cells (choice D is false).
140. B. This sequence is described in the first paragraph of the passage. Choice C also describes events in the proper sequence but lists events that occur several steps after T-lymphocyte activation, making B a better choice. Choices A and D are false; histamine release leads to vasodilation, not vasoconstriction or bronchodilation.
141. A. The second paragraph states, “One therapeutic strategy would be to target a particular subset of T lymphocytes known as T-helper (TH) cells,” and goes on to describe how TH1 cells can inactivate the TH2 cells that are correlated with asthma. Administering IgE may sensitize mast cells, leading to an asthma attack (choice B is false), and neither Zileuton nor colchicine is mentioned in the passage (so choices C and D are false).
Passage II
142. D. Figure 2 shows that Drug C inhibits the binding of cells to both Fg and FN. Choice D is the only choice that shows this.
143. A. Referring again to Figure 2, Drug A inhibits the binding of cells to Fg, but does not affect the binding of cells to FN. This is the same effect as Drug D, described in the question text.
144. C. Figure 1 shows that A5 cells can bind to both Fg and FN (statements I and II are true), but Drug C inhibits binding of A5 cells to FN (statement III is false).
145. B. Since one would not expect water to affect the binding of cells to Fg or FN, and since Drug B does not affect the binding of cells to Fg or FN (choices A and C are false), it seems logical to assume that Drug B might have been water (choice B is true, and D is false).
146. C. Since Drug C affects the binding of cells to both Fg and FN, it seems that this drug would be the one most likely to cause problems. Drug A only affects the binding of cells to Fg (choice A is false), and Drug B has no effect whatsoever (choices B and D are false).
147. D. Based on Figure 2, neither Drug A nor Drug B has an effect on the binding of cells to FN (choices A and B are false), and Drug C decreases the percentage of bound cells (choice D is true, and C is false).
Passage III
148. A. From Table 1, we note that the reaction rate is dependent on both the concentration of Compound 1 (Student group 1 vs. 3) and the concentration of sodium ethoxide (Student group 1 vs. 2). The rate-determining step is therefore
bimolecular (E2, since this is an elimination reaction). Because E2 reactions involve a transition state in which two s bonds break to form one p bond in a concerted fashion, the students came to the correct conclusion.
Br CH3
H H
OCH2CH3
Br
H
OCH2CH3
CH3
‡
149. D. The students described the cleavage of two s bonds in a concerted, one-step transition state to form a p bond. This is characteristic of an E2 reaction.
150. C. Compound 1 produces two unique alkenes due to the removal of a b-proton from the methyl group (to produce Compound 3) or from the ring (to produce Compound 2). The only alkyl halide that has two unique b-hydrogen atoms and can therefore produce two alkenes is choice C. Note that choice A would yield a single terminal alkene, while choices B and D do not generate alkenes under these conditions since either the bromine (choice B) or b-hydrogen (choice D) is bonded to an sp2-hybridized carbon.
151. B. Since this is an elimination reaction, the carbon atom bearing the leaving group (bromine) rehybridizes from sp3 to sp2.
152. B. Without a base (NaOCH2CH3) to promote an elimination reaction, Compound 1 will undergo a substitution reaction with ethanol (a nucleophile) to produce choice B. Since the alkyl halide is tertiary, it can ionize to form a relatively stable carbocation, making this substitution reaction unimolecular (SN1).
Passage IV
153. B. These bacteria are described in the passage as producers of toxins that are “superantigens”; i.e., antigens that produce an exaggerated immune response. The passage does not discuss their reproduction (choice A is false) or the inhibition of metabolic enzymes (choice D is false). Autoimmune reactions, by definition, are abnormal immune reactions to normal cellular antigens, not to bacterial antigens (choice C is false).
154. B. The primary structure of proteins is their amino acid sequence. The secondary structure of proteins involves interactions between the backbone of the peptide chain to form structures such as a-helices and b-sheets (choice C is false), and the tertiary structure of proteins involves interactions between amino acid side chains to form the overall three
dimensional structure of the protein (choice D is false). The isoelectric point of amino acids has nothing to do with their sequence in a particular protein (choice A is false).
155. C. The passage states that TSS is characterized by a high fever, hypotension (circulatory system), and a rash (skin).
Fever is a non-specific disease defense mechanism, and of the four systems listed as answer choices, the lymphatic system is most directly involved with disease defense.
156. C. The passage states the conventional antigenic stimulation activates 1 in 100,000 T lymphocytes, or 0.001%.
Superantigens activate 20% of T lymphocytes, or 20,000 of 100,000 cells.
Independent Questions
158. A. Adding a repressor protein that binds to the superantigen gene would prevent its transcription (and its subsequent translation; choice B could prevent synthesis and can be eliminated). Adding a complementary nucleic acid sequence that can bind to superantigen mRNA would prevent its translation (choice C could prevent synthesis and can be eliminated).
Adding a stop codon within the superantigen gene would result in a stop codon in the superantigen mRNA upon
transcription. This would terminate protein synthesis prematurely when the mRNA was translated (choice D could prevent synthesis of the complete protein and can be eliminated). However, tRNA nucleotides that bind to both mRNA and ribosomes are normal—this is how mRNA is translated. Adding more of these nucleotides would not prevent (and may enhance!) translation of the superantigen protein (choice A would not prevent synthesis and is the correct answer choice).
159. B. The O at position a is the same as the corresponding O in the hemiacetal, so choice A is eliminated. The hydroxyl group in the hemiacetal is released and replaced by the OCH3 (methoxy) group of methanol. So, the oxygen at position b comes from the 18O labeled methanol, while the oxygen at position c comes from the released hydroxyl group.
160. D. The terminal amine of phenylhydrazine (recall that hydrazine is H2NNH2) will react with the carbonyl shown in the question (similar to imine formation). The first step of this addition–elimination reaction is addition.
CH3 O
H2N NHPh phenylhydrazine OH
CH3 HO
H+
HN NHPh
N CH3
addition elimination NHPh
161. D. Parthenogenesis (choice D) is the activation and subsequent development of an egg without the use of a sperm, for example by pricking it with a fine needle. Isogamy (choice A) is a condition in which male and female gametes are visibly indistinguishable, but this does not preclude their ability to join in sexual reproduction. Hermaphroditism (choice B) and pseudohermaphroditism (choice C) are states in which a single animal produces both sperm and eggs and is able to self- fertilize; although unusual, they are examples of sexual reproduction.
162. C. If there are 12 nucleotide base pairs for every complete 360∞ turn of the helix, there must be one base pair for every 30∞ of the turn. This most closely corresponds with the data for the Z conformation of DNA given in the question text.
Passage V
163. C. As stated in the passage, isoprene is a 5-carbon unit; two isoprene groups form a terpene. Cembrene (Compound 7) has 20 carbon atoms, making it a diterpene (20 carbon atoms = 4 isoprene units = 2 terpene units). b-Elemene (Compound 9) has 15 carbon atoms (3 isoprene units = 1.5 terpene units), making it a sesquiterpene.
164. C. Conjugated p bonds produce strong UV absorptions. While Compound 5 has three double bonds, they are not conjugated, and will therefore not produce a strong UV absorption (eliminating choice A). While hydroxyl groups (choice B) and carbonyl groups (choice D) are prominent in IR spectroscopy, they do not produce strong UV absorptions.
165. A. (–)-b-Elemene (Compound 9) is shown in Figure 1 in the passage. Its enantiomer, (+)-b-elemene, will have the opposite stereochemistry at all three stereocenters. Note that choices B and D have only two of the three stereocenters inverted, and choice C has only one.
166. D. Since the procedure must convert an ester functional group (–OAc) into an alcohol functional group (–OH), choices A and C are eliminated, and condensation (choice B) does not apply here. The answer is D: Saponification is the hydrolysis of an ester using a base.
167. A. In thin-layer chromatography (TLC), the stationary phase is highly polar, while the mobile phase is less polar.
Compounds that are more polar therefore have greater affinity for the stationary phase and a slower migration. Compound 5, with its two hydroxyl groups, is the most polar of the four compounds listed in the answer choices and will therefore migrate the slowest.
Passage VI
168. A. Referring to Figure 2, it is clear that since the gene is being expressed in some tissues and not in others, it is not
“spreading widely throughout the fetus” (choices C and D are false). Furthermore, since some of the organs in which the gene is expressed are large (such as the liver) and some are smaller (such as the pancreas), it is not “only expressed in the largest organs” (choice B is false). However, all of the organs in which the gene is expressed are abdominal organs, making choice A the best choice.
169. D. The passage describes the cationic lipids as being positively charged, so DNA must be negatively charged (choices A and B are false). Furthermore, the whole reason the DNA must be coated with lipids in the first place is because it is too hydrophilic to easily cross the cell membrane on its own (choice D is true, and C is false).
170. B. The cell membrane does not contain DNA. In eukaryotes, DNA is found in the nucleus, and in prokaryotes it is found in the cytoplasm (choices C and D are false). Furthermore, the cell membrane is not a pure lipid bilayer; it contains proteins in addition to phospholipids, cholesterol, and carbohydrates (choice B is correct, and A is false).
171. C. The only cells that are passed from parent to offspring are the germ line cells (eggs and sperm). All other cells are somatic cells and are not passed on to the next generation; thus, a gene introduced to a somatic cell could not be passed on to offspring (choice C is true, and choices A and B are false). Furthermore, any DNA contained within a germ line cell, be it original cellular DNA or introduced DNA, will be passed on (choice D is false).
172. A. All cells in a multicellular organism contain the same genome, inherited from the original egg and sperm that fused during fertilization (choices C and D are false). What makes one type of cell different from another type of cell is not the DNA it contains, but the specific genes within that DNA that are or are not expressed. Since all cells do not express the same genes, cells can vary widely in their appearance and function (choice A is true, and B is false).
Passage VII
173. A. The passage states that fighting big fires is a frightening experience. Frightening and/or stressful experiences activate the sympathetic division of the autonomic nervous system, which, among other effects, acts to increase the heart rate (choices C and D are false). Since the passage goes on to state that Steve’s experience was especially frightening, one can assume that his heart rate would be more affected than Joe’s (choice A is true, and B is false).
174. C. Since Joe has suffered extreme skin scarring (as described in the passage), he would have more difficulty cooling by sweating than Steve. Older males have lower metabolic rates than younger males and would therefore tend to have lower body temperatures (choice A is false), and if Steve has a greater percentage of body fat than Joe, Steve would tend to overheat more easily than Joe (choice B is false). Furthermore, if Steve weighs half as much as Joe (as stated in the passage), then he has a greater surface-to-volume ratio than Joe (choice D is false).
175. C. The oxygenation of blood can be affected by many things, including the rate and depth of breathing (increases in these values can lead to greater oxygenation, choice A is false), hemoglobin concentration (increases in [Hb] can increase
176. D. Since glomerular filtration rate (GFR) is directly proportional to blood pressure, increases in blood pressure would lead to increases in GFR. Therefore, if Steve’s blood pressure increased more than Joe’s, one would expect Steve’s GFR to increase more than Joe’s (choice D is true, and A is false). Reabsorption rate is not affected by blood pressure (choices B and C are false).
177. C. Since Joe weighs twice as much as Steve, he would require more energy to perform the same work tasks (choice C is true and A is false). Choice B is contradictory; an individual with a greater basal metabolic rate would not consume less energy. Choice D is irrelevant, since the question asks about the energy involved in performing work, not the energy involved in maintaining the body at rest (basal metabolic rate).
178. D. Since the liver is not involved in the production of digestive enzymes (choice A, the job of the pancreas), antidiuretic hormone (choice B, the job of the hypothalamus), or new blood cells (choice C, the job of the bone marrow), damage to the liver should not affect their production. However, since the liver is directly involved in the production of bile salts, liver damage should reduce their production.
179. B. The blood pressure of the firefighters is given in the passage as 125/70 mm Hg; this represents the systolic blood pressure over the diastolic blood pressure. The systolic blood pressure (125 mm Hg) is the pressure measured when the heart is contracted, and the diastolic blood pressure (70 mm Hg) is the pressure measured when the heart is relaxed (choice A is false, and B is true). Note that both choices C and D refer to the pressure when the blood is entering an artery (in other words, when the heart is contracting), thus these measurements must be systolic measurements (choices C and D are false).
Passage VIII
180. A. The aldol/decarboxylation reaction will begin (in the presence of a base) with the deprotonation of the diacid to produce a nucleophilic enolate. The primary function of pyridine is to facilitate this deprotonation (choice A). [Note that pyridine will also deprotonate the carboxylic acid functional groups of the diacid (choice B), but this does not result in their neutralization, as the answer choice says.]
181. B. SOCl2 converts carboxylic acids into acid chlorides (and alcohols into alkyl chlorides).
182. C. Since the methyl groups replace all four carbonyl a-protons, enolization is not possible (choice C).
183. D. By analogy to the passage, methyl propenoate would be converted to 1-propanol:
OCH3
O
OH
methyl propenoate 1-propanol
OCH3
O methyl propanoate
LiAlH4 condition e condition d
H2/Pd/C
Condition d results in the reduction of the double bond, while condition e (followed by an acidic workup) reduces the ester to an alcohol.
184. B. The passage states that Compound 6 was reacted with 2-methylpropanoic acid in the presence of the strong base lithium diisopropylamide (LDA). LDA deprotonates 2-methylpropanoic acid (first at the carboxylic acid, then at the a- proton) to generate an enolate. This enolate then reacts with Compound 6 to give choice B:
OH
O
O–Li+
O–Li+
CO2–Li+ CO2–Li+
2-methylpropanoic acid
LDA Compound 6 H+ choice
B
2 equivalents
1 equivalent
Independent Questions
185. D. The function of antidiuretic hormone (ADH) is to allow the kidney tubules to reabsorb more water, thus retaining water in the body. After being dehydrated for two days, the hiker’s ADH levels would be expected to be high.
Glucocorticoids (choice A) have various jobs, including reducing inflammation and raising blood glucose during stress, aldosterone (choice B) increases Na+ retention, and insulin (choice C) lowers blood glucose.
186. A. Endocytosis is the uptake of extracellular material. Phagocytic cells of the immune system, such as macrophages (choice A) and neutrophils, would be expected to have the highest rates of endocytosis. Erythrocytes (choice B) are red blood cells and function in oxygen transport, osteoblasts (choice C) are bone-building cells, and neurons (choice D) send nerve impulses between the brain, spinal cord, and body organs.
187. C. Antibody proteins are synthesized by B-cells and secreted into the blood. Secreted proteins are translated by ribosomes attached to the endoplasmic reticulum (specifically, the rough ER). The nucleus and mitochondria are not associated with protein translation (choices A and B are wrong), and while the Golgi apparatus is involved in the secretory pathway (the proteins would interact with the Golgi), it is not specifically involved in the translation (C is a better answer choice than D).
188. C. Transmission at the neuromuscular junction (NMJ) is chemical in nature. Neurons release ACh onto muscle cells;
the ACh binds to receptors on the muscle cells and triggers their depolarization (and thus their contraction). The ACh is then degraded by a cholinesterase found at the NMJ. Addition of a cholinesterase blocker would prevent the degradation of ACh, thereby prolonging the stimulation of the muscle and preventing it from repolarizing in order to receive another impulse (choice A is true and eliminated). Addition of a toxin that prevents the release of ACh would certainly interfere with impulse transmission; if the neurotransmitter can’t be released, the muscle cells cannot be stimulated (choice B is true and eliminated). Addition of a substance that binds to ACh receptors would prevent ACh from binding and prevent impulse transmission to the muscle cell (choice D is true and eliminated). The only choice that would not interfere with
transmission is choice C; an increase in ACh receptor sites would enhance impulse transmission.
189. C. The fact that the lens fails to form in the absence of the optic cup indicates that the optic cup is necessary for lens development, and may in fact induce it. This says nothing about the timing of neurulation relative to gastrulation (choice A is false), or about the specific timing of eye development (choice B is false). Choice D is a false statement in addition to being irrelevant; cell differentiation is a gradual process, not an all-or-none event, and in any case, this says nothing about