Biology secton test (14)

Since growth was NOT induced in vivo ciliary neurons treated with CNTF, then choice A CANNOT be concluded, and must therefore be wrong.. Choice D, pupil dilation, is the only sympathetic

MCAT Section Tests

Dear Future Doctor,

The following Section Test and explanations should be used to practice and to assess your mastery of critical thinking in each of the section areas Topics are confluent and are not necessarily in any specific order or fixed proportion This is the level of integration in your preparation that collects what you have learned in the Kaplan classroom and synthesizes your knowledge with your critical thinking Simply completing the tests is inadequate; a solid understanding of your performance through your Score Reports and the explanations is necessary to diagnose your specific weaknesses and address them before Test Day

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Passage I (Questions 1–6)

1 The correct answer is choice C The main concept in this passage is that the formation of synaptic connections in the nervous system is a complex procedure, somewhat comparable to shooting in the dark Since neurons have no way of sensing each other, the probability that a single neuron growing towards a target neuron will form a synaptic connection is quite low To compensate for this, the developing nervous system has evolved the following mechanism: it initially produces an abundance of neurons that grow towards a single target neuron Once the proper synapse has formed, the superfluous neurons degenerate Compare this to someone trying to shoot an arrow at a bullseye The chances that one arrow will hit the center are much greater if many arrows are shot The arrows that don't hit the target merely fall by the wayside This analogy is appropriate because, as you're told in the passage, the growth of the neurons toward the target is not completely random; it is aimed, to some degree, by the neurotrophic agents that control neuron growth This implies that without neurotrophic agents, synapse formation would occur much less efficiently; thus, choice C is right Let's take a look at the other choices Choice A is incorrect because, as we just explained, the process of synapse formation certainly does NOT have predictable results Choice B

is incorrect because we also know that the process of synapse formation isn't completely random; it's under the influence of neurotrophic agents Finally, choice D is incorrect because the passage states that proper nerve pathways are established trough synaptic connections This means that synapse formation IS absolutely essential for the formation of complex nervous pathways Again, the correct answer is choice C

2 Choice D is the correct answer The assays were performed in order to determine whether CNTF increases the growth of ciliary neurons and spinal motor neurons both in vivo and in vitro As a control, both types of neurons were also treated with the fluid used to dilute CNTF in the experiment again, under both in vivo and in vitro conditions The dilution fluid was NOT expected to increase neuron growth; that's why it was used as the control Anytime CNTF promoted neuron growth, it provided support for the hypothesis behind the assays Okay, so now let's look at the answer choices to determine which one is supported by the experimental data Since growth was NOT induced in vivo ciliary neurons treated with CNTF, then choice A CANNOT be concluded, and must therefore be wrong And although it might be concluded that in VITRO growth of ciliary neurons is dependent on CNTF, since 25% of the neurons in that assay experienced growth, as just discussed this is NOT true for in VIVO ciliary neurons Therefore, choice B is also incorrect Since 30% of spinal motor neurons treated in vivo with CNTF experienced growth, one would conclude that this growth WAS dependent on CNTF; thus choice C is also wrong Choice D, however, is supported by the data because NONE of the control assays were expected to cause neuron growth, and none of them did Therefore, it can be concluded that neither in vivo nor in vitro growth of spinal motor neurons, or for that matter of ciliary neurons, is dependent on the dilution fluid used in the preparation of CNTF Thus, choice D

is the correct answer

3 The correct answer is choice D This question doesn't actually have a whole lot to do with the passage; it takes a tiny little piece of information from the second paragraph and uses it as a springboard To answer it, you have

to know the differences between the parasympathetic and sympathetic divisions of the autonomic nervous system These two divisions often elicit antagonistic responses when they innervate the same organ In general, the parasympathetic division elicits physiological responses that conserve and gain energy, such as slowing down heart rate, vasoconstriction blood vessels in skeletal muscle, constricting pupils, and decreasing metabolic rate; the sympathetic division, on the other hand, stimulates responses that expend energy and prepare an organism for action, such as inhibiting digestive processes, increasing heart rate, vasodilating blood vessels in skeletal muscle, dilating pupils, and increasing metabolic rate These responses are commonly known as the "flight-or-fight" responses Based

on this, it can be inferred that if a developing chick embryo were given an NGF inhibitor, then the axons of sympathetic neurons would not grow in abundance, and the proper synaptic connections would not be formed And this means that some of the nervous pathways leading to the physiological responses associated with sympathetic innervation might very well be compromised in an adult chicken Choice D, pupil dilation, is the only sympathetic response among the four choices; therefore, it's the only one that might be compromised, and so choice D is the correct answer

4 The correct answer is choice A The passage tells you that in this experiment, it was expected that anywhere that ciliary neurotrophic growth factor was introduced, neuron growth would be induced Thus it's logical that anywhere CNTF was omitted, as in the four control experiments with dilution fluid, growth should not have been induced However, in the in vivo assay of ciliary neurons plus CNTF, you notice that even though growth factor was introduced, no neuron growth occurred This is an unexpected result, especially since growth was induced both in vivo and in vitro for the spinal motor neurons, and in vitro for the ciliary neurons Therefore, choice A is the correct answer There are many possible explanations for this result, one is that the experimental conditions may simply have gone awry Another is that perhaps CNTF simply does not work this way within the body that is, CNTF's in vitro neurotrophic effects may not be due to the same mechanism that normally causes ciliary neurons to grow in vivo, so that even though it may serve as a neurotrophic agent for embryonic ciliary neurons in vitro, it doesn't in vivo In any

case, none of the remaining answer choices are correct, because they're all expected results Again, the correct answer

is choice A

5 The correct answer is choice D As was explained in the explanation to question 1, the process by which a target neuron forms a synapse with a developing neuron is largely determined by the probability that a neuron growing toward a target neuron actually synapses with it The chances of this occurring would be quite low were it not for neurotrophic agents, which increase the number of neurons growing toward a single target So during development, a large number of neurons are overproduced so as to ensure that at least one of them forms the proper synapse Once the synapse has formed, the remaining neurons, which are no longer needed, will degenerate Of the four answer choices, this process is most analogous to the fertilization of an ovum by a spermatozoan

Sperm swimming through the female reproductive tract towards the egg are somewhat directed by chemical signals, just as neurotrophic agents direct neurons in their growth Yet there is still the possibility that the ovum won't

be fertilized, just as there's a chance that a single developing neuron won't synapse with its target neuron To increase this probability, nature has dictated that millions of spermatozoa simultaneously attempt to fertilize the same egg, much like the initial overproduction of neurons increases the probability that the proper synapse will form Thus, choice D is correct Choice A is incorrect because there is no question of probability in the migration of chromosomes during cell division This process is precisely controlled; the chromosomes are guided towards their objective - the cell poles - by spindle fibers that are attached to their centrioles Choice B is incorrect because the packaging and exocytosis of a secretory protein is not in any way similar to the development of a neural pathway It's not as if a thousand molecules of the same protein are synthesized in the hopes that one of them makes it into a secretory vesicle and out of the cell Finally, choice C is also incorrect because hormones are released into the circulatory system in minute quantities, and then travel to their target cells, where they bind with surface receptors specific for them Hormones are NOT produced in excess, nor is there only a single cell with a single receptor that is receptive to the hormones' effects Again, the correct answer is choice D

6 Choice C is the correct answer According to information in the question stem, when the same exact procedures that had been performed on ciliary neurons and spinal neurons of embryonic chicks were performed on adult chickens instead, CNTF failed to induce any growth Well, the only difference between this set of experiments and the ones described in the passage, is the age of the chickens A quick glance at the answer choices reveals that all four of them deal with this age discrepancy, so it makes sense to evaluate each of them to determine which best explains the experimental results Choice A says that ciliary and spinal motor neurons degenerate during maturation Not only is this false, but there's absolutely no evidence for it in the passage What the passage DOES tell you is that the superfluous neurons produced by the developing embryonic nervous system degenerate AFTER a synapse has properly formed Thus, choice A is incorrect If choice B, which says that adult chickens can form new synapse, were actually true, then you would expect that CNTF WOULD have induced neuron growth in the chickens Since this did not occur, and since it's not clear from the passage whether adult chickens can or can't form new synapses, choice B must also be wrong Choice C, however, does provide a plausible explanation for the lack of CNTF-induced neuron growth in the adult chickens: neuron growth can only be induced during a critical period of embryonic development

In other words, there is a limited period of time during which CNTF can induce the growth of ciliary and spinal motor neurons, and this period occurs during embryonic growth After this critical period has passed, ciliary and spinal motor neurons are no longer sensitive to the effects of CNTF So, choice C is correct As for choice D: While the claim that CNTF can't induce growth of embryonic ciliary neurons in vivo IS supported by the results of the experiments described in the passage, this does NOT account for the lack of ciliary neuron growth and spinal motor neuron growth in ADULT chickens treated with CNTF; therefore, choice D is wrong Again, choice C is the correct answer

Passage II (Questions 7–14)

7 The correct answer to question 7 is choice D The method of protecting carbonyl groups given in the passage is to convert them to ketals by reacting them with alcohols; the structure of a ketal can be seen in Reaction 1 All four choices are alcohols, so you need to figure out which one will form the best ketal The passage states, in the first paragraph, that carbonyls are protected more effectively by conversion to cyclic ketals As you can see from Reaction 1, a non cyclic ketal is formed when a carbonyl group reacts with the hydroxyl groups of two alcohol molecules Therefore, it is reasonable to suppose that a cyclic ketal can be formed by the reaction between a carbonyl group and a diol So, choice C, which contains just one hydroxyl group, would form a non-cyclic ketal and can be rejected right away All the other choices are diols choice B is a trans unsaturated 1,4-diol, and choices A and D are 1,2-diols In choice B the hydroxyl groups are not adjacent, and the mobility of the carbon skeleton is quite limited due to the presence of the trans double bond, therefore it would be impossible to form a cyclic ketal Both remaining choices are 1,2-diols, but choice A has four isopropyl substituents so it is more sterically hindered, and therefore less reactive Hence, choice D is the correct answer

8 The correct choice answer is choice B The passage states that lithium aluminum hydride will reduce the ester group to an alcohol and would also react with the ketone if left unprotected Therefore, choice A, which shows the ketone group still intact, is wrong Next you should be able to eliminate choice C The passage and the diagram

of Compound Y both indicate that the ester group is reduced to a CH2OH alcohol group So choice C, which has it converted to a carboxyl group, is wrong Finally, to choose between choices B and D, you have to think about the effect of the reducing agent on the ketone group You should know that lithium aluminum hydride is NOT a strong enough reagent to completely remove the oxygen atom and reduce the compound to the alkane form Therefore, choice D is wrong and choice B is the correct answer

9 The correct answer is choice C The phrase "based on the information in the passage" in the question-stem indicates that you need to look for analogies between the reaction shown in this question and the passage material You should be aware that thiols react with carbonyl groups in much the same way as alcohols In addition, given that oxygen and sulfur occupy adjacent positions in Group 6 of the Periodic Table, it is reasonable to assume that a sulfide group will react similarly to a ketone group You can see that compound II is a cyclic thioketal, so just as a diol was

used to form a cyclic ketal in the passage and in question 7, a dithiol would be needed to produce a cyclic thioketal

Thus, choices A and B can be rejected right away Choice D is also wrong, because the passage says that ketals are stable under basic conditions, but are hydrolyzed under acidic conditions Therefore, assuming that thioketals behave similarly, compound II wouldn't be affected by base but would be hydrolyzed by acid to produce compound III Therefore, choice C is the correct answer

10 The correct answer to question 10 is choice A The amino group nitrogen, which has a lone electron pair, is nucleophilic As for phenylacetyl chloride, since oxygen is strongly electronegative, the C-O double bond is polarized, giving the oxygen a partial negative charge, and the carbon a partial positive charge As a result of this charge distribution, the carbon is attacked by the nitrogen atom, resulting in displacement of the chloride and the formation of an amide link So choice A is correct Choice B is wrong because, as we've just seen, the amino group

is electron-rich and therefore nucleophilic, so it can't be an electrophile Likewise, choice C is wrong because the carbonyl oxygen is nucleophilic, not electrophilic Finally, the replacement of the chloride by the amino group is a simple nucleophilic substitution; there's no oxidation, or for that matter reduction, involved in the process at all So choice D is wrong and again, the correct answer is choice A

11 For question 11, the correct answer choice is A You're looking for a way to make the carboxyl group more reactive, which will facilitate peptide bond formation Peptide bonds form by the nucleophilic attack of the amino group of an amino acid on the carboxyl group of another The reaction of thionyl chloride with a carboxylic acid substitutes a chlorine atom for the carboxyl hydroxyl group, producing an acyl chloride An acyl chloride is the most reactive of the various kinds of carboxylic acid derivatives, and certainly more so than the corresponding carboxylic acid This is because the chloride ion is a much better leaving group than the hydroxide anion and is therefore more easily displaced by the nucleophilic attack of the amino group Thus, choice A is correct Choice B is wrong because heating the solution will have no effect on the structure of the carboxyl group Increasing the temperature may or may not promote the reaction, depending on the thermodynamics of the reaction, but NOT by activating the carboxyl, which requires some kind of chemical modification Choice C is also wrong, because adding aqueous base to compound I would result in the formation of a carboxylate anion This would be even less reactive towards an amino nucleophile than the acid Finally, esterification of the acid choice D would replace the hydroxyl group by an alkoxide group (RO) This alkoxide group would be a poor leaving group, and therefore the ester would not be reactive enough to form the peptide bond, so choice D is also wrong Once again, the correct response is choice A

12 The correct answer to question 12 is choice B The passage states that ketals are stable in basic solution and are hydrolyzed by an acidic solution Because of this pH dependence, other functional groups in the molecule can be transformed in basic solution, leaving the ketal untouched; the ketal can then be removed, using a dilute acid solution,

to give the original ketone Anyway, getting back to the question, why are ketals the preferred protecting groups? They are preferred over ethers because ethers are particularly unreactive compounds they're pretty stable in both basic and acidic solution Consequently, while ketals can be cleaved by adding dilute acid, cleavage of ethers would require higher temperatures and a more concentrated acid choice B the correct response

13 The correct answer is choice C As with any nonpolar amino acid, the acidic, neutral, and basic forms of alanine roughly correspond to acidic, neutral, and basic values of pH At their isoelectric point, amino acids exist predominantly in the zwitterion form: positively charged ammonium groups and negatively charged carboxylate groups As the pH falls, the carboxylate groups gain hydrogen ions and become neutral, while the ammonium groups retain their positive charges, so the molecules exist predominantly as cations In contrast, as the pH rises above the isoelectric point, the ammonium groups lose their extra hydrogens to form neutral amino groups, whereas the carboxylate groups remain negatively charged, so that the molecules become anions The form of alanine shown in the passage is an anion, and as the highest pH value is represented by choice C, this is the correct answer

14 The correct answer to question 14 is choice D Compound X is a complicated polyfunctional molecule which you have to name according to IUPAC rules The first thing you should do is identify the functional groups present in the molecule Compound X consists of a cyclopentane ring with three functional groups: a methyl group, a ketone group, and an ester group Choice A has all the right functional groups mentioned in the name: "methyl",

"oxo", which is used in the middle of names to refer to ketones, "cyclopentyl", and "ethanoate" However, on closer

inspection you should be able to see that choice A is wrong the ester functional group has been given the highest priority and so numbering the carbon ring starts with the carbon attached to the ester group In choice A the ketone functionality is numbered 2, and the methyl group is numbered 3; either way you count, these numbers can't be right if the ester substituted carbon is labeled 1 Secondly, the convention for esters is to name them after the alcohol and the acid which can be thought of as having reacted to form the ester bond The alcohol is named first, as the subsidiary group, and the acid is named second and has either the "oate" suffix or the word carboxylate added to it Here, the ethyl group in Compound X is the alcohol part of the molecule and should be named first, but it is in fact named as the acid part of the molecule, so choice A is also wrong for this reason Choice B also names the compound after the ester group However, the methyl and oxo substituents are again labeled 3 and 2 Also, the name does not follow the basic convention for naming esters that I just outlined Normally, "ethoxy" would denote an ether, not an ester so choice B is wrong Choice C is also wrong as the ketone is given the highest priority and Compound X is named as a substituted cyclopentanone So, by elimination, we are left with choice D Let's just take a look and see why D is the correct name for Compound X As I said, the highest priority group in compound X is the ester The first part of the name of an ester is the alcohol part of the molecule In this case, the alcohol from which the ester is formed is ethanol, so the first word in the name should be "Ethyl" Next we have to decide the name of the acid part of the molecule Well, the main carbon chain of the acid is a cyclopentane ring The ring has two substituents, the ketone oxygen and the methyl group Of these two, the oxygen takes priority, and so the carbons of the ring are numbered counter-clockwise, with the one attached to the ester group being number 1 This makes the ketone carbon number 3, and the methyl carbon number 4 Therefore, the acid from which the ester is formed is "4-methyl-3-oxocyclopentane carboxylic acid" So, putting these two parts of the molecule together, the full name of the ester is ethyl (4-methyl-3-oxocyclopentane) carboxylate, which is choice D

Passage III (Questions 15–19)

15 The correct answer is choice B The key to this question is the fact that whereas simple RNA viruses are poor vectors for gene therapy, retroviruses are good ones So the factor that makes simple RNA viruses bad vectors must

be something that is NOT true of retroviruses The major difference between the two is that retroviral genomes get reverse transcribed into DNA Remember, the passage tells you that the genome of a simple RNA virus is directly transcribed into messenger RNA If a simple RNA virus were used as the vector for a therapeutic gene, then the gene would not be replicated and passed on to daughter cells, because an RNA gene cannot become integrated into the cell's DNA genome Integration is a necessary step of gene therapy and can only occur if the gene is in the form of DNA Since integration cannot occur, the viral RNA is rapidly degraded by the cell Choice A is wrong because there

is no evidence in the passage that RNA viruses have smaller genomes than either DNA viruses of retroviruses, or that they might therefore not be able to accommodate a therapeutic gene Choice C is also incorrect, because although it's true that certain viruses can infect only specific cell types, this is true for all viruses, not just simple RNA viruses, so

it would not be a factor that would distinguish simple RNA viruses in particular And, in any case, the problem of cell type specificity on the part of the virus is not even discussed in the passage, so there's no evidence to support this choice Choice D is incorrect since there is also no evidence in the passage to support the statement that incorporation

of a therapeutic gene into an RNA genome would render it too unstable to be used as a vector Instead, the passage implies that it is possible to produce vectors carrying therapeutic genes from all kinds of viruses, but that retroviruses make the best vectors for some other reason - which as we've just discussed, is the reason described by choice B, the correct answer

16 Choice C is the correct answer The passage described microinjection as being a time-consuming technique requiring a high level of expertise, an electroporation as a technique that is traumatic for the cells Neither of these criticisms is applied to retroviral delivery, which is described instead as occurring via a "normal" infection mechanism Normal retroviral infection does not require any technique more complicated than adding the virus to the cells to be infected, and it does not harm the cells during the actual infection process, since the virus enters a lyso genic cycle Thus, choice C is correct Choice A, which states that the site of gene integration can be more precisely controlled using a retroviral delivery system than by using physical methods, is wrong because the passage states that the retroviral DNA integrates into the cellular DNA at a random location - that is, not precisely at all! Choice B is also incorrect, since there is no evidence in the passage to suggest that a retroviral delivery system allows a greater proportion of its target cells to integrate the therapeutic gene than does electroporation or microinjection And although you are not expected to know this, microinjection is probably the most efficient method of gene therapy in terms of the proportion of treated cells that are infected with the therapeutic gene Finally, choice D is a wrong answer since the passage states that provirus integration occurs during RNA replication thus, only cells that can divide are amenable to retroviral gene therapy Again, the correct answer is choice C

17 The correct answer is choice C As you're told in the passage, a retrovirus integrates its DNA into cellular DNA during DNA replication, mainly during cell division, so it can only integrate into the chromosomes of cells that are capable of dividing Of the four cell types listed as answer choices, only neuronal cells cannot divide Let's look at the other choices Choices A and B are both wrong, and for similar reasons Liver cells would actually be GOOD targets for retroviral gene therapy because they divide continuously Continuous division means that the therapeutic

gene would be replicated along with the cellular DNA and inherited by the daughter cells, thus ensuring a continuous

supply of the therapeutic gene product; thus, choice A is wrong Likewise, though dead skin cells are continually sloughed from the surface of the skin, they are replaced by living ones through cell division The retrovirus could only successfully infect live cells, and once it had infected these cells, the therapeutic gene would NOT be lost, but rather inherited by the daughter skin cells Choice D is incorrect because bone marrow cells are good targets for gene therapy involving a retroviral vector, for the very reason that they CAN be successfully removed from the body, infected, and then replaced Moreover, bone marrow cells DO divide and eventually produce important blood cell types, which makes them particularly fruitful subjects for gene replacement therapy As you may be aware, bone marrow replacement has been used in recent years to treat lymphomas So, choice C is the correct answer

18 Choice A is correct For a retrovirus carrying a therapeutic gene to successfully infect and integrate into its target cell's genome, several events must occur The first stage of infection is the binding of the retrovirus' protein envelope to receptors on the surface of the target cell This facilitates the entry of the retrovirus into the cell, which consists of RNA, is first transcribed into DNA by the enzyme reverse transcriptase You might therefore have been tricked by choice C, which actually says that the retroviral genome must be TRANSLATED by reverse transcriptase; however, translation is a different process The stage of protein synthesis during which a strand of messenger RNA TRANSCRIBED from the genome is used to produce a strand of amino acids is translation Thus, choice C is also wrong Choice D would not occur during successful gene therapy; the retroviral proteins encoded by the genes gag, pol, and env are not synthesized after integration has already occurred Gag and env, which code for the retroviral core proteins and protein envelope, respectively, would be synthesized after infection only if the retrovirus entered a LYTIC cycle And although the pol region codes for reverse transcriptase, which is necessary for integration, reverse transcriptase is synthesized BEFORE integration occurs; so choice D is incorrect Again, choice A is the right answer

19 The correct answer is choice D This one can easily be answered by the process of elimination There is no evidence whatsoever in the passage for choices A, B, or C, but let's go through them anyway Choice A is just plain ridiculous Non-disjunction is either the failure of sister chromatids to properly separate during mitosis, or the failure

of homologous chromosomes to properly separate during meiosis - the net result being that some daughter cells inherit multiple copies of one chromosome, while others lack the chromosome entirely In any case, nondisjunction is not at all desirable, and is in fact typically lethal Therefore it is improbable that a therapeutic gene integrated into its target cell's DNA would normally cause a nondisjunction even; this would NOT be expected to correct a genetic defect, and if a nondisjunction event did somehow occur, it would not be therapeutic; thus, choice A is wrong Choice

B is wrong because the host's immune system would have no opportunity to come into contact with and recognize the retroviral DNA as foreign since the viral DNA integrated along with the therapeutic gene into the host cell's DNA Choice C is wrong because in order for replication of the retroviral DNA and formation of infectious virions to occur, the virus would have had to enter a Iytic cycle, which does not occur in successful gene therapy Remember, if the retrovirus was Iytic, then the host cell would be killed, and this is clearly not the goal of retroviral gene therapy

The goal is to introduce a healthy gene into a genetically defective cell and produce its protein product WITHOUT

causing an infection, which necessarily implies that the retroviral DNA persists in the host DNA in an noninfectious form Thus, choice C is wrong and choice D is correct

Passage IV (Questions 20–24)

20 The correct answer is choice C This question requires you to examine the results of Test 2, in which stool samples were taken from all of the patients both before and after the administration of a drug that enhances bile production and secretion The table gives values for the levels of fat and certain vitamins in the patients’' stools The most noticeable thing about these values is that compared to Patients G, H and K, and especially to Patient L who is healthy and can be assumed to have the normal levels of fat and vitamins in this stool sample, Patient F has abnormally high levels of both fat and vitamins in his stool sample, Patient F has abnormally high levels of both fat and vitamins A, D E, and K in the stool samples taken prior to Test 2 From this it can be inferred that Patient F probably suffers from fat malabsorption and malabsorption of these vitamins So basically, this question is concerned with what would most likely happen when someone who has the same malabsorption symptoms as Patient F is administered bile The results of administering bile would logically be similar to the results of administering a drug that enhances bile production and secretion; therefore, you can base your answer on Patient Fs Test 2 results You may have remembered that bile aids in fat digestion; otherwise, you could have deduced it from the date, which show that Patient F's stool fat level declined sharply after administration of the drug It might also have helped if you recalled that vitamins A, D E, and K are all fat-soluble vitamins, and that these are absorbed along with fats in the small intestine Consistent with this, the data shown that after the drug treatment, the fat-soluble vitamin levels in Patient F's stool decreased markedly, in step with the fat levels, and approached the levels for the other patients From this, it can be inferred that administering bile to the patient described in the question, whose symptoms are similar to those of Patient F, would most likely enhance the intestinal absorption of both fat and vitamin A, as well as the other fat-soluble vitamins Thus, choice C is the correct answer

21 The correct answer is choice C As discussed in the previous explanation, the results of Test 2 indicate that bile enhances the absorption of both fat and the fat-soluble vitamins, since the concentrations of these substances in Patient F's stool sample decreased after the administration of a drug that enhances bile production and secretion In

addition, you should remember from your knowledge of human digestion that fat and fat-soluble vitamins are absorbed in the small intestine, NOT in the colon, which is also known as the large intestine Therefore, the fact that there are bacteria in the colon that actively metabolize bile salts neither supports nor contradicts the results of Test 2

in any way, because bile is produced in the liver and functions in the small intestine Since digested food only reaches the colon AFTER it passes through the small intestine, the bacteria in the colon that metabolize bile salts do not affect the production, secretion, or functioning of bile These bacteria are, in fact, beneficial to humans, since the

metabolism of bile salts allows 95% of the salts to be reabsorbed and recycled Choice D is incorrect because, first of

all, Test 2 is not concerned with calcium absorption, nor is calcium absorption discussed anywhere in the passage; and secondly, bile is not involved in calcium absorption And finally, calcium ions are actually small enough to be absorbed in the small intestine without the aid of any enzyme or other substance Again, choice C is correct

22 The correct answer is choice A Given the results of Test 1, it can be deduced that Patient H most likely suffers from lactase deficiency Since lactase is the enzyme that hydrolyzes lactose into galactose and glucose, a lactase deficiency would lead to an abnormally high concentration of lactose in the intestinal lumen and consequently

in the stool sample, especially following a high lactose meal The reason for this is that lactose is a disaccharide, and disaccharides cannot be absorbed by the intestinal villi; sugars are only absorbed once they've been broken down into monosaccharides Thus, lactose as such is NOT found in the blood; so the expression "low blood lactose" is meaningless, and therefore choice C, while tempting, is wrong The same goes for other disaccharides such as maltose and sucrose Any oligosaccharides that are not degraded and absorbed in the small intestine become osmotically active substances - that is, by remaining in the intestinal lumen, they increase the concentration of solutes in the intestinal lumen and, as a result, less water is reabsorbed Decreased intestinal water absorption results in watery stools - that is, diarrhea Thus, choice A is correct and choice B, constipation as you're probably aware is the opposite of diarrhea, is wrong Finally, fatty stools, choice D, is a symptom that you would expect in a patient suffering from fat malabsorption, and according to the results of Test 2, Patient H clearly does NOT suffer from this problem, Thus, choice D is also wrong Again, choice A is the correct answer

23 The correct choice is D As stated in the question stem this question is NOT concerned with the results of Test 3, but rather with the blood calcium levels PRIOR to the administration of a high-calcium meal And if you looked at the answer choices before actually trying to answer the question, you might have realized that the question requires you to consider the feedback mechanism regulating the concentration of calcium in the blood, and use that to interpret the pretest data in the column for Test 3 The piece of knowledge you need to answer this question is that an organism responds to a DECREASE in blood calcium by increasing the parathyroid gland's secretion of the parathyroid hormone Parathyroid hormone increases blood calcium by increasing bone resorption, and by converting vitamin D into its active form, which increases calcium absorption in the small intestine On the other hand, the thyroid gland responds to an INCREASE in blood calcium by increasing its secretion of CALCITONIN, calcitonin is the hormone that decreases blood calcium by increasing its excretion and deposition in bone Getting back to the question: since Patient K had the lowest blood calcium prior to Test 3, it can be inferred that, compared to the other patients, Patient K most likely has the HIGHEST blood level of parathyroid hormone and the LOWEST blood level of calcitonin By the same line of reasoning, it is likely that compared to the other patients, Patient H has the lowest blood level of parathyroid hormone and the highest blood level of calcitonin So, choice D is the correct answer

24 Choice D is the correct answer Although pancreatic insufficiency as a cause of malabsorption syndrome is only mentioned in passing in the first paragraph of the passage, there is enough information in the question stem itself for you to be able to answer the question Okay, from the results of Test 2, which is the only one of the three tests related to fat absorption, you can generalize the following: anything that compromises fat digestion in the small intestine will cause an increase in the amount of fat excreted in a person's stool The fact that patients suffering from pancreatic insufficiency have fatty stools suggests that there is some substance excreted by the pancreas that it is essential to fat digestion So, let's take a look at the answer choices A deficiency in amino acid absorption is definitely a possible symptom of pancreatic insufficiency, since the pancreas secretes proteases such as trypsin, chymotrypsin, and carboxypeptidase, and without these enzymes, amino acid absorption in the small intestine would

be compromised, however, a deficiency in amino acid absorption is NOT a possible cause of fatty stools, so choice A

is wrong As for choice B: insulin is a secretion of the endocrine pancreas; insulin is the hormone that decreases blood glucose by stimulating the conversion of glucose to glycogen when blood glucose is too high Therefore, choice B is also incorrect, since insulin deficiency does not cause fat malabsorption And pancreatic amylase is an enzyme secreted by the pancreas into the small intestine; it hydrolyzes carbohydrates into disaccharides Therefore, choice C is also incorrect Well, by the process of elimination, choice D must be the right answer Pancreatic lipase is the enzyme MOST integral to fat digestion in the small intestine; pancreatic lipase works in conjunction with bile to hydrolyze fats into fatty acids and monoglucerides If you didn't remember this offhand, you might have been able to guess that this enzyme breaks down fat from its name: "lip" for lipid, another name for fat, and "ase" as the suffix denoting an enzyme Thus, a deficiency in pancreatic lipase secretion would most definitely cause a person to produce fatty stools, and so choice D is the correct answer

Discrete Questions

25 The correct choice is B Vaccines consist of attenuated - that is, weakened or inactive bacterial or viral forms Vaccines are specifically designed to "fool" the body into synthesizing antigens against a particular pathogen, WITHOUT actually causing the disease typically associated with that pathogen Therefore, choice A is wrong, because

a smallpox vaccine containing vaccinia virus does NOT cause smallpox The reason vaccinia is used in the smallpox vaccine is that its protein coat contains antigens similar enough to those found on the smallpox virus that they stimulate the proliferation of B lymphocytes that will then produce antibodies specific for both vaccinia virus and smallpox virus The vaccine recipient is thus protected against a future smallpox infection This type of acquired immunity is known as active immunity Active immunity has two phases: first, the B lymphocytes differentiate into either plasma cells or memory cells The plasma cells immediately start to synthesize antibodies; the memory cells remain inactive, but retain surface receptors specific for the vaccine's antigens This is known as the primary response Upon subsequent exposure to the same antigen, such as a second vaccination or an exposure to the infectious agent, these same memory cells elicit a greater and more immediate proliferation of B lymphocytes; and this is known as the secondary response If you take a look at the graph, you'll see that the curve following the first smallpox vaccination corresponds to the primary response, and the curve following the second smallpox vaccination corresponds to the secondary response Thus, choice B is correct As for choice C, although the increase in the serum level of smallpox antibody CAN be attributed to the synthesis of smallpox antibodies, they are synthesized by the recipient's B Iymphocytes, NOT by vaccinia virus Being a virus, vaccinia can't synthesize anything but the proteins and nucleic acid it needs to replicate itself, and it can only do that with the use of a host cell's genetic machinery; it certainly can't synthesize antibodies, which are only produced by multicellular organisms So, choice C is wrong And there's no evidence in the question stem to support the claim that the smallpox vaccine or any other vaccine requires a minimum

of 45 days to confer active immunity, so choice D is also wrong Again, the correct answer is choice B

26 The correct choice is C This question requires you to understand the sequence of events that lead to muscle contraction A neuromuscular junction is composed of the presynaptic membrane of a neuron and the postsynaptic membrane of a muscle fiber (which is called the sarcolemma) The two are separated by a synapse In response to an incoming action potential, the presynaptic membrane of a neuromuscular junction releases the neurotransmitter acetylcholine into the synapse Acetylcholine diffuses across the synapse and binds to acetylcholine receptors on the sarcolemma, causing the membrane to depolarize and generating an action potential The action potential causes the sarcoplasmic reticulum to release large amounts of calcium ions into the sarcoplasm - which is the cytoplasm of a muscle fiber This in turn leads to the shortening of the muscle fiber's sarcomeres, which contracts the muscle According to the question stem, myasthenia gravis causes the body to produce antibodies that trigger the removal of the acetylcholine receptors found on muscle fibers Without these receptors, acetylcholine can't bind to the membrane and trigger its depolarization And if this doesn't occur, then neither can any of the events that normally follow it So,

of the four choices, the one that would be directly impaired by myasthenia gravis would be the conduction of an action potential across the sarcolemma: therefore, choice C is correct Choices A and B are wrong, because although they too are impaired by myasthenia gravis, they're dependent on the conduction of an action potential, so it can't really be said that they are DIRECTLY impaired Choice D, acetylcholine synthesis, is wrong because it's not at all affected by myasthenia gravis; acetylcholine is synthesized by the neuron itself, and this process is independent of whether or not acetylcholine receptors are present on the postsynaptic membrane Again, choice C is the right answer

27 The correct choice is A Muscular hypertrophy is an increase in muscle mass, which is due to an enlargement

of its constituent cells The most likely cause of hypertrophy in any muscle is an increased workload Therefore, the most likely cause of hypertrophy of the left ventricle would be an increase in workload for the left ventricle The left ventricle is responsible for pumping blood into the aorta and supplying the force necessary to propel it through the rest of the systemic circulation therefore, an increase in systemic blood pressure would increase the resistance of systemic circulation and force the left ventricle to work harder to overcome it; and as a result, the left ventricle would gradually hypertrophy Thus, choice A is right and choice B is wrong Choices C and D are wrong because pulmonary circulation is propelled by the RIGHT ventricle, and so any changes in pulmonary blood pressure would primarily affect the right ventricle Again, choice A is correct

28 The correct answer is choice B Water reabsorption in the kidneys is directly proportional to the osmolarity

of the interstitial tissue, relative to the osmolarity of the filtrate When the osmolarity of the filtrate is lower than the osmolarity of the kidney tissue, the tendency is for water to diffuse OUT OF the nephron; when the osmolarity of the filtrate is higher, the tendency is for water to diffuse INTO the nephron Infusing the nephron of a healthy person with

a concentrated sodium chloride solution increases the filtrate osmolarity; thus, water will diffuse into the nephron to try and compensate for this change So, you can now rule out choices C and D And since the volume of urine excreted is inversely proportional to the amount of water reabsorption, there will be a concomitant increase in urine volume Thus, choice A is wrong and choice B is correct

29 The correct choice is A To answer this question correctly, you need to understand the genetics of the ABC blood system; that is, you need to know that the A and B alleles are codominant to the O allele Thus, a person with the genotype AA or AO has a type A blood; a person with the genotype BB or BO has type B blood; a person with the genotype 00 has type O blood Now, getting back to our question: A man with type AB blood has the genotype AB and can therefore produce with either the A allele or with the B allele A woman with type O blood has two O alleles

and therefore only produce gametes of the O allele So if this couple has children, the only two genotypes their children can possible inherit with respect to the blood groups are AO and BO, which corresponds to the phenotypes type A blood and type B blood, respectively Thus, choice A is the correct answer

Passage V (Questions 30–33)

30 The correct answer to question 30 is choice C The reaction we're talking about is the conversion of the

epoxide group into a trans-1,2-diol that is, a ring opening The chemical method that's commonly used to open an

epoxide ring is treatment with acid and water This causes the epoxide oxygen to become protonated making one of the adjacent carbons susceptible to nucleophilic attack If that nucleophile is water, the result is the formation of a

trans-1,2-diol just as with the enzymatic reaction, so choice C is correct None of the other choices will open the

ring Choices A and B, potassium permanganate and osmium tetroxide, are both used to create diols from alkene

functional groups by adding two hydroxyl groups simultaneously The products of these reactions are actually cis

diols, because both hydroxyl groups are added from the same side of the double bond But, more importantly, neither one would convert an epoxide into a diol, so A and B are both wrong Choice D, a mixture of nitric and sulfuric acids, forms the highly electrophilic nitronium ion, and is used for the nitration of benzene or other aromatic rings It would probably add NO2 groups in various places on the conjugated ring system, producing a mixture of products, none of which would be what we're looking for Again, only the acid-catalyzed ring opening will create the desired trans 7,8-diol, and thus the correct choice is C

31 The correct answer to question 31 is choice A This is a biology question stuck into the middle of this organic chemistry passage and since the question asks for the most LIKELY explanation, you have to think through each answer choice carefully and figure out the right answer by elimination First, let's see what we know about the pathway shown in Figure 2 According to the passage, the net result of this series of reactions is to produce a carcinogen from benzo[a]pyrene It's not clear from the passage whether benzo[a]pyrene is carcinogenic before the transformation; however, the molecule clearly doesn't have all the functional groups that are present in the carcinogen, and based on Figure 3, this means that it can't bind to DNA and cause mutations by the same mechanism

So, it seems likely that benzo[a]pyrene is actually LESS carcinogenic, meaning that the biotransformation process actually makes the compound MORE deleterious Now let's have a look at the answer choices Choice B is wrong

on a technicality: the mutation mechanism described at the end of the passage substitution of an incorrect base for the correct base during the formation of a new DNA strand is NOT frameshift mutation, but point substitution A frameshift mutation is where one or more bases are inserted into or deleted from a strand, changing the codon reading frame and thereby totally altering the amino acid sequence that is encoded; on the other hand, a simple substitution like the one described here will at most change ONE codon Due to the redundancy of the genetic code, it is possible for a point substitution to yield the same amino acid; at worst, one amino acid will be substituted for another, leaving the rest of the amino acids intact Moving right along, choice A suggests that the pathway toxifies some compounds and detoxifies others, with the net effect being beneficial, presumably meaning that detoxification predominates This theory is consistent with the fact that the pathway apparently makes benzo[a]pyrene more carcinogenic; and it also provides a justification for the pathway's existence Moreover, it's consistent with the general properties of enzymes and with the function of the liver: many enzymes can act on a variety of related compounds, and this seems particularly likely for those in the liver, since the liver processes a wide variety of substances So, choice A is a likely answer However, we still have to consider choices C and D A compound, such as Structure 3, that is mutagenic and carcinogenic definitely qualifies as a toxin, so choice C, which says that the pathway DETOXIFIES benzo[a]pyrene, must be incorrect However, choice D is not completely implausible: the biotransformation of benzo[a]pyrene COULD once have served some useful function that no longer exists, and the negative effects might not yet have selected against it to the point of inactivating the pathway However, since DNA structure has been the same since very early in evolutionary time, the NEGATIVE effects of benzo[a]pyrene biotransformation, which are based on its attacking guanine residues, must always have been considerable So, since the passage gives no evidence that there were ever any POSITIVE effects, choice D is an outside chance at best Choice A is the most likely choice, and therefore the correct answer

32 The correct answer to question 32 is choice C Here you simply need to remember the general rules for solubility If you look at the two compounds you're comparing, you will see that they have almost the same ring structure, but the carcinogen has additional functional groups that benzo[a]pyrene lacks two hydroxyl groups and one epoxide group, both located at the same end of the molecule Since oxygen atoms are extremely electronegative, the electron density will be pulled towards them making the molecule polar On the other hand, benzo[a]pyrene has

no electronegative groups; and since it is conjugated, its pi electrons are evenly distributed throughout the molecule, making it nonpolar You should remember that as a general rule, polarized or ionizable molecules are more soluble than nonpolar molecules in a polar solvent like water in other words, "like dissolves in like" Therefore, choice C correctly explains why structure 3 is more soluble in water than benzo[a]pyrene Choices A, B, and D are all wrong Each of these choices defines a relationship between solubility and some other property of a compound While it is true that the molecular weight, the number of substituents, and the amount of saturation can all influence solubility, it

is not a direct correlation By far the most important factor determining solubility is the STRUCTURE of the molecule and whether or not it is similar to the structure of the solvent So again, the correct response is choice C

33 For question 33, the correct answer is choice D To answer this question you have to look at the stereochemistry of the ring opening and think about the mechanism involved After the reaction, one of the hydroxyl groups is oriented above the ring, and one of them is below This is due to the fact that the epoxide is protonated and then one of its carbon atoms is attacked by water When the water molecule attacks, it displaces the epoxide oxygen from the carbon it attacks and leaves that oxygen bonded to the other carbon In other words, there is an SN2 reaction

and so the configuration of that carbon is inverted, forming a trans-diol Getting back to the question; in Figure 2 the

epoxide ring is shown as being above the plane of the main, flat, conjugated part of the molecule, so you should be able to see from Figure 3 that the OH group on carbon 7, which is still above the six-membered ring, is the one that's derived from the old epoxide oxygen Thus, the OH group at carbon 8 must be the one derived from the attacking water molecule that is, the one containing the new oxygen atom, which must have attacked from BELOW the six-membered ring Therefore, choice D is correct

Passage VI (Questions 34–40)

34 The correct answer is choice C To answer this question you need a basic understanding of membrane construction Membranes are composed of a phospholipid bilayer, with the hydrophilic phosphatases of the phospholipid molecules arranged so that they face outwards toward the intracellular matrix and the extracellular matrix, both of which are aqueous The hydrophobic lipid moieties are "sandwiched' between the phosphatases so that they don't interact with these aqueous environments If you have trouble remembering the meanings of hydrophilic and hydrophobic, you can use the roots of the words to help you: "hydro" means water, "philic" means loving, and

"phobic" means fearing; thus literally, hydrophilic means water-loving and hydrophobic means water-fearing Anyway,

as the name implies, a transmembrane protein is one that spans the width of the entire membrane and extends through both sides Therefore, all transmembrane proteins, including the transmembrane glycoproteins of erythrocytes, have a region exposed to the aqueous intracellular matrix, a region in contact with the inner lipids of the membrane, and a region exposed to the extracellular matrix As a result, transmembrane proteins must contain both hydrophilic and hydrophobic domains; thus, choice C is correct Even if you didn't know this, you should have been able to choose the correct answer by the process of elimination, since the other three choices can easily be ruled out Choice A is wrong simply because it's not true and there is no evidence for it in the passage While the sodium-potassium pump is in fact

a transmembrane glycoprotein, it does NOT have negatively charged sialic acid residues bound to its extracellular surface, and is a completely different protein, with a completely different function, from the transmembrane glycoprotein that is discussed in the second paragraph of the passage Choice B is wrong because not only is there no evidence in the passage to support it, but also, as you should remember, erythrocytes do not, and cannot, divide As you're told, an erythrocyte is an nucleated cell, which means that it lacks a nucleus and thus DNA; and without DNA,

a cell can't go through the cell division process As you're told in the passage, red blood cells have an average life span

of 120 days, after which they are removed from circulation by the spleen, without leaving any descendant cells Choice C is wrong because, even if you didn't know that HEMOGLOBIN is the oxygen carrying component of erythrocytes, there's no evidence in the passage that would lead you to think that the transmembrane glycoproteins functioned as such Again, choice C is correct

35 Choice B is the correct answer To answer this question, you needed to extrapolate from a piece of information in the first sentence of the second paragraph: "Erythrocytes are nucleated cells that lack all membranous organelles." This means that in addition to lacking a nucleus, erythrocytes also lack Iysosomes, endoplasmic reticulum, Golgi bodies, and mitochondria, all of which are membranous organelles So what does all of this mean in terms of citrate? Well, citrate is one of the compounds produced by the citric acid cycle, also known as the Krebs cycle or the TCA cycle, which is one of the three stages of aerobic respiration; the other two stages are glycolysis and

electron transport Since mitochondria are the site of the citric acid cycle, and since erythrocytes lack mitochondria, it

follows that it is NOT possible for erythrocytes to produce citrate Because of this, erythrocytes are anaerobic and can produce ATP only from glycolysis Thus, choice B is correct and choice D is wrong The claims that erythrocytes are nucleated and that they lack endoplasmic reticulum are both true, but these facts do not affect the erythrocytes' ability

to produce citrate one way or another, so choices A and C are incorrect Again, choice B is the correct answer

36 The correct answer is choice A To answer this question you needed to know that anemia is a condition in which there is a deficiency of red blood cells If you knew that, then this one was pretty easy for you Blood is composed of liquid plasma and solid components Isolated plasma does not contain any cells Since anemia affects the number of red blood cells and not the volume of plasma, you would expect that plasma volume would be fairly similar in a normal person and in an anemic Therefore, it can be inferred that if 40% of a normal person's blood volume is occupied by red blood cells, then LESS THAN 40% of an anemic person's blood volume must be occupied

by red blood cells Choice A 25%, is the only hematocrit value less than 40%, and therefore it must be the right answer

37 Choice A is the correct answer This is a basic Mendelian genetics question Since you're told that hereditary spherocytosis is an autosomal dominant disease and that the man described in the question is heterozygous for it, that means that he has one copy of the hereditary spherocytosis allele The woman that he marries is normal, which means