Biology secton test (12)

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 All rights are reserved pursuant to the copyright laws and the contract clause in your enrollment agreement and as printed below Misdemeanor and felony infractions can severely limit your ability to be accepted to a medical program and a conviction can result in the removal of a medical license We offer this material for your practice in your own home as a courtesy and privilege Practice today so that you can perform on test day; this material was designed to give you every advantage on the MCAT and we wish you the best of luck in your preparation Sincerely, Albert Chen Executive Director, Pre-Health Research and Development Kaplan Test Prep © 2003 Kaplan, Inc All rights reserved No part of this book may be reproduced in any form, by Photostat, microfilm, xerography or any other means, or incorporated into any information retrieval system, electronic or mechanical without the written permission of Kaplan, Inc This book may not be duplicated, distributed or resold, pursuant to the terms of your Kaplan Enrollment Agreement BIOLOGICAL SCIENCES TEST TRANSCRIPT Passage I (Questions 1–6) The correct answer is choice B This question requires you to determine which of the listed metabolic processes are involved in the production of uric acid, since hyperuricemia is defined as the excessive concentration of urate This information is found directly in the passage and its figure, and in essence makes this question a simple reading comprehension problem According to the passage, uric acid is formed from the breakdown of purines, and that implies that HYPERuricemia is caused by an INCREASE in the breakdown of purines In the event that you didn't know what purines are, you could have deduced that adenine and guanine are purines from Figure Adenine and guanine, along with cytosine, thymine, and uracil, are the nitrogen bases found in nucleic acid Adenine and guanine are the purine bases, and cytosine, uracil, and thymine are the pyrimidine bases An easy way to remember which are the pyrimidines bases is by using the pneumonic "CUT the PY." C, U, and T are the PYrimidine bases So, to answer the question, you need to determine which metabolic process would produce purines as its end product Let's look at our choices Fatty acids are comprised of triglycerides and long carbon chains So if fatty acids were broken down, or catabolized, they would not produce purines Since purines are not a component of fatty acids, choice A must be incorrect Choice C, amino acid synthesis, would obviously not yield purines; hence choice C is also incorrect Cholesterol synthesis, choice D, refers to the production of cholesterol, which is also not a purine, and therefore choice D is also wrong By the process of elimination, choice B, nucleotide degradation, is the correct answer Let's see why Nucleotides are components of both DNA and RNA, and contain three components: a nitrogen base, a sugar, and a phosphate group The base is either a pyrimidine or a purine So, when nucleotides are degraded, purines will be produced And as you're told in the passage, purines are further metabolized to uric acid Therefore, if you can limit or suppress the amount of purines produced by nucleotide degradation, you can limit the amount of uric acid produced and as a result alleviate hyperuricemia Therefore, choice B is the correct answer Choice C is the correct answer According to the passage, the drug allopurinol is an analog of hypoxanthine and an inhibitor of xanthine oxidase, and as such, is used to alleviate the elevated urate concentration associated with gout The question itself is concerned with allopurinol only in terms of its being a hypoxanthine analog First of all, what does being an analog mean? Well, since we're dealing with an enzyme - xanthine oxidase, and its substrate hypoxathanine, we're talking about a substrate analog By definition, a substrate analog is a substance with a structure similar to the natural substrate of an enzyme and which, because of this similarity, inhibits the action of the enzyme, as in competitive inhibition As you should know, a competitive inhibitor is a substance that competes with an enzyme's substrate for the enzyme's active site If the enzyme is "fooled" into binding the wrong substrate, it won't be able to it's job In terms of this question, the enzyme is xanthine oxidase and its natural substrate is hypoxanthine As an analog of hypoxanthine, allopurinol acts as a competitive inhibitor, binding to xanthine oxidase's active site, and inhibiting the conversion of adenine into urate Thus, choice C is the right answer Let's run through the other choices just to make sure Choice A is wrong because both allopurinol and hypoxanthine are the substrates - but substrates don't bind to each other, they bind to their enzyme's active site The first part of choice B and choice D is true - allopurinol does bind to xanthine oxidase However, this binding inhibits, not promotes, the conversion of hypoxanthine into xanthine; so choice B must be incorrect Furthermore, the conversion of adenine to hypoxanthine is mediated by a different enzyme not identified for you in the passage Therefore, the binding of allopurinol to xanthine oxidase does not block this reaction, and so choice D is also wrong Again, choice C is the correct answer Choice A is the correct answer The question requires you to calculate genetic probabilities for an X-linked recessive inheritance As you know, males have an X chromosome and a Y chromosome, while females have two X chromosomes This means that men produce X and Y gametes - in a 50 to 50 ratio, while women can produce only X-containing gametes Let's designate the recessive allele for HGPRT deficiency by X little h (Xh) and the normal allele by X Since the disease is recessive, a male with HGPRT deficiency would have to have the genotype X little h, Y, while a female with the deficiency would have to have the genotype X little h, X little h If the female was simply a carrier of the disease, she would have the genotype X little h, X The question tells you that the mother has an HGPRT deficiency So, as we just discussed, this means she must have the genotype X little h, X little h, and that all of her gametes will have the Xh chromosome You also know that the father is normal, meaning that his genotype is X, Y, which means that half of his gametes will have an X chromosome while the other half will have a Y If you couldn't figure out the results of this cross in your head, using a Punnett square will help you out The genotypic possibilities of this cross are: 50% X little h, X and 50% X little h, Y All daughters will be carriers and all sons will inherit the deficiency The question only asks about the female children, and since you were told that HGPRT deficiency is a recessive trait, there is a 0% chance that a daughter produced by this couple will inherit an HGPRT deficiency Thus, choice A is the correct answer The correct answer is choice D This is an organic chemistry question integrated into a biology passage Since organic chemistry appears in the biological sciences section of the MCAT, the MCAT writers try to demonstrate the connection between the two disciplines by placing a chemistry question in a biology passage or vice versa Anyway, back to the question This is really a very easy organic chemistry question, asking you to analyze a very common reaction Br2 can add to 2-hexene via a simple addition reaction In an addition reaction, the two molecules Kaplan MCAT Biological Sciences Test Transcript of bromine will add across the double bond, generating 2,3-dibromohexane What about the stereochemistry of this addition reaction? The addition of bromine to the pi bond of 2-hexene can only occur such that the bromine molecules may be added in an anti orientation, with one molecule of Br added to the pi bond from above and the other from below Upon addition of Br2 to trans-2-hexene, at each asymmetric carbon atom, one isomer has the R configuration while the other has the S configuration While upon addition of Br2 to cis-2-hexene, the R,S configuration is the same at both asymmetric carbon atoms And since the question tells you that Br2 is reacted with a mixture of cis and trans-2-hexene, R,R-, S,S-, R,S-, and S,R-2,3-dibromohexane can be produced Therefore, choices A, B, and C are accounted for by the addition reaction 2-bromohexane, choice D, cannot be synthesized based on the conditions of the question stem Br2 does not typically undergo a free radical reaction in the presence of an alkene The addition reaction between a halogen and an alkene is extremely favorable Since Br2 does not usually react as a free radical, the anti-Markovnikov product, 2-bromohexane, will not be formed The fact that urate is absent was only included to throw you off Since the question asks which one CANNOT be produced, choice D is the correct answer The correct answer is choice A This is one of those questions that can be answered without even reading the passage According to the question stem, the drug colchicine reduces inflammation caused by gout by inhibiting granulocyte migration Nevermind that you don't know off the top of your head what a granulocyte is - although you should know that it's a type of cell by the suffix -cyte the key word in that sentence is MIGRATION Migration implies movement, and cell movement is mediated by microfilaments and microtubules Microfilaments are microscopic filaments composed chiefly of actin and function in cell support and movement of the cell and the organelles within it Microtubules are microscopic tubular structures composed chiefly of tubulin and serve similar functions In addition to being found in the cytoskeleton, microtubules are found in cilia and flagella, and they form the spindle fibers of mitosis You might recall from your biology class that colchicine disrupts mitosis by inhibiting microtubule polymerization Well, since microtubules also mediate cell movement, colchicine inhibits that too In terms of alleviating inflammation due to gout, colchicine inhibits the migration, or movement, of granulocytes, which are the type of white blood cells responsible for the inflammatory response, among other things So, choice A is the correct answer And if you had quickly scanned the other answer choices, you'd have seen that none of them has anything to with cell movement Ribosomes, choice B, are the organelles responsible for translating mRNA transcripts into peptide chains; so choice B is wrong The nucleolus, choice C, is the suborganelle of the nucleus responsible for the synthesis of rRNA; so choice C is also incorrect Choice D is incorrect, because the Golgi apparatus is the organelle responsible for the modification and packaging of proteins Again, choice A is the correct answer Choice D is the correct answer This question basically asks you to infer what the effects of hypouricemia are, based on the information given to you in the passage You should have been able to figure out that HYPOuricemia means having an abnormally low concentration of urate if you knew that the root hypo means low, or little And if you didn't know that, you might have been able to deduce it, since the passage tells you that HYPERuricemia means too much urate, and therefore HYPOuricemia must be the opposite condition Well, you also know from the passage that too much urate results in gout, choice C, which is often characterized by arthritis, choice B So, choices B and C must be incorrect since they are a result of hypERuricemia In fact, if a person had hypOuricemia, they would not suffer from gout or any of its related symptoms And though there is a chance that the person might develop arthritis due to an unrelated factor, such as age, this question asks you choose your answer in light of the fact that the person has hypouricemia Thus, choice B is not the best answer According to the passage, urate lowers the rate of mutation by decreasing the rate of attack on DNA by highly reactive species such as free radicals and superoxide anions Since DNA mutations are the fundamental event in carcinogenesis, mutations dramatically increase the risk of cancer Therefore, if a patient has a low urate level, their rate of mutation will be increased, and hence the risk of cancer will also increase Therefore, choice D is the correct answer As for an increased risk for bacterial infection, choice A, the passage never discusses this topic nor does it mention any immune system deficiencies that might lead you to believe that this would be the correct answer Therefore, choice A is wrong, and choice D is the correct answer Passage II (Questions 7–12) The correct answer is choice C This question is really quite simple if you understood the last sentence of the passage and know a few basic definitions from introductory biology This is probably the most common type of MCAT question you need a little information from the passage and a little information from introductory biology to solve the problem The passage informs you that the net result of multicopy inhibition is that IN RNA is not allowed to attach to a ribosome Since we know that IN RNA is produced, this implies that transcription must have occurred Just to review for a minute, transcription is the production of RNA from a DNA template Therefore, choice B, transcription, is NOT repressed by multicopy inhibition, and is incorrect Replication, choice A, is the process by which new DNA is synthesized using the cell's original DNA as a template Since the question tells you that multicopy inhibition acts on RNA, not DNA, it must NOT block replication; hence choice A is also incorrect Let's look a little more closely at the attachment of an mRNA transcript to a ribosome Ribosomes are the site of protein synthesis, or translation, and consist of a small subunit and a large subunit Translation is initiated by the binding of an mRNA transcript to the small ribosomal subunit Therefore, if mRNA is not allowed to attach to a ribosome, Kaplan MCAT Biological Sciences Test Transcript translation cannot occur Hence, multicopy inhibition must repress translation, and choice C is the correct answer Choice D, post-transcriptional modification, refers to modifications made to an mRNA transcript AFTER transcription has occurred, as its name implies Post-transcriptional modifications include removing introns, adding a poly(A) tail or a 5' cap Since these occur PRIOR to the binding of the mRNA to a ribosomal subunit, multicopy inhibition does not block it, and thus choice D is wrong Again, choice C is the correct answer Choice A is the correct answer To answer this question all you need to is pick the answer choice that accounts for the increased rate of transposition observed in the new strain of cells The information needed to answer this question can be inferred from the passage The best approach to this type of question, which requires you to evaluate all of the answer choices, is to just go down the choices in order So let's start with choice A According to the passage, the dam methylation system regulates the rate of transposition and the activation of Tn10 with DNA replication You're told that the lack of methylation on the daughter strand of newly synthesized DNA activates the transposon by increasing the transcription of the transposase gene and enhancing the binding of this enzyme to the transposon Therefore, if a strain of cells lacked the dam methylation system, there wouldn't be any methylation of the strain's DNA, and the incidence of transposition would be expected to be much higher than in strains WITH the methylation system Therefore, choice A is the correct answer But let's take a look at the other answer choices just to make sure we've selected the best answer Choice B discusses the number of copies of OUT RNA and IN RNA, which means that choice B is concerned with multicopy inhibition OUT RNA is the mRNA transcript of the strong promoter, Pout, and IN RNA is the mRNA transcript of the weak promoter, Pin In a typical cell containing Tn10, there are 100 copies of OUT RNA per copy of IN RNA Thus, multicopy inhibition inhibits Tn10 transposition by ensuring that all IN RNA will be bound to OUT RNA and is therefore unavailable for translation If IN RNA was translated then transposition would occur Therefore, if the new strain of cells has an even higher ratio of OUT RNA to IN RNA, the chances of IN RNA being translated would be even lower Therefore, this would DECREASE the rate of transposition, not increase it and so choice B must be incorrect What about choice C? Choice C has to with the insertion of the transposon into host DNA and the absence of a repeated sequence on both sides of the transposon If you recall from the passage, following transposon insertion, the target sequence is duplicated and flanks the transposon However, the passage never relates the presence of this repeated sequence in the host DNA following transposon insertion to the rate of transposition or the functional ability of the transposon In fact, the passage never discusses ANY function for this repeated sequence Therefore, you would not be able to conclude that the lack of a repeated sequence in the host DNA would affect the rate of transposition Hence, choice C must be wrong Choice D refers to the overlap between the IN RNA and OUT RNA of Tn10 This overlap is what allows OUT RNA to complementary base pair to IN RNA, which in turn prevents the translation of the IN RNA If the overlap between the two transcripts were increased, then the interaction between the two molecules would be even stronger This would mean that it would take more energy to separate the IN RNA from the OUT RNA And this means that if anything, the IN RNA would remain paired to the OUT RNA and would be translated LESS frequently So by increasing the overlap between the transcripts from Pin and Pout, choice D, the rate of transposition would DECREASE and so choice D is also an incorrect choice Therefore, choice A is the correct answer Choice B is the correct answer The question stem specifically tells you that you will need to make an inference In other words, you will not find the exact information needed to answer this question right in the passage The passage discusses the issue of hemi-methylation in the context of the dam methylation system You're also told that following replication, the new strand, which is complementary based-paired to the parent strand, lacks methyl groups until the dam system has a chance to kick into action The reason why the parent strand has methyl groups attached to the DNA, while the new strand lacks them, is due to the semi-conservative mechanism by which DNA is replicated During DNA replication, the parent strands unwind and act as templates for complementary base-pairing with free nucleotides The end result is two daughter DNA helices - each helix consists of one of the original parent strands and one newly synthesized strand Thus, following replication of methylated DNA, each daughter helix will consist of one newly synthesized strand and one methylated parent strand - and there's how you get your hemimethylated DNA If you didn't know this, or couldn't infer it from the passage, you may have been able to derive the same information from the root of the word The prefix hemi means half, like in hemisphere So, hemi-methylated must mean half-methylated And since you know that DNA is double-stranded, hemi-methylated means that only half of this molecule is methylated; one strand has methyl groups and the other strand lacks methyl groups So now that we have reviewed methylation of DNA in the context of the passage, let's look at the answer choices Choice D states that hemi-methylated DNA will help transport proteins into and out of the nucleus Besides the fact that there is nothing in the passage that would lead you to conclude this, transportation across the nuclear membrane depends on carrier proteins, transport proteins, and the pores found in the nuclear membrane And although DNA is believed to lend a helping hand towards directing the transport process, this "help" is independent of its methylation status Therefore, choice D is incorrect The use of free methyl groups in the synthesis of amino acids and cofactors, which is choice C, is unrelated to the hemi-methylation of DNA The passage does not mention the source of the methyl groups used in the dam system, nor does it address other possible uses for methyl groups Basically, although methyl groups may be incorporated into amino acids and cofactors, there is no way to infer from the passage that the lack of complete methylation on a DNA molecule allows the cells to perform these other methylation reactions; hence choice C is also incorrect As for choice A: although the hemi-methylated DNA arises after replication, it does NOT affect the rate at which the cell replicates, or divides The replication of DNA, and hence the presence of hemi-methylated Kaplan MCAT Biological Sciences Test Transcript DNA, is only an indication of cell division; therefore, choice A is incorrect This leaves us with choice B as the correct answer Let's see why As previously mentioned, when DNA is replicated, it is done so in a semi-conservative fashion This means that one new strand and one parent strand complementary base pair to form a double-stranded molecule of DNA And from the passage you know that the new strand of DNA lacks methylation, while the parent strand is methylated Therefore, the cell is able to distinguish between the old strand - the strand with methyl groups, and the new strand - the one without methyl groups, which means that choice B is the correct answer Now you might be wondering why a cell would need to distinguish between the parent strand and new strand in a newly synthesized DNA helix Well, if there were any mutations or errors that occurred during replication, then the cell would know which of the strands - most likely the new strand - needed to be repaired Again, choice B is the correct 10 The correct answer is choice A This question requires you to extrapolate from the information provided in the passage Given the description of Tn10 activity, you're asked to determine if the transposon could have been responsible for the cell's death You know from the question stem that the cell died after replication occurred What is the significance of this statement? Well, from the discussion of dam methylation you should have realized that transposition is likely to occur following replication Therefore, the problem boils down to determining whether or not some event might occur during transposition that could be fatal to a cell So now let's look at the answer choices Choice A contends that, yes, the transposon could have been responsible for the cell death if it had inserted itself into a vital gene Since transposons insert at random sights in the host DNA, it IS quite possible that it inserted into a vital gene And if this were the case, the protein product of this gene would most likely be non-functional, and its inability to function properly could cause the cell to die Therefore, choice A seem plausible But let's look at the other choices Choice B also claims that Tn10 could cause cell death and relates it to multicopy inhibition It states that IN RNA inhibits the translation of OUT RNA But as you know from the passage, it is OUT RNA that inhibits the translation of IN RNA and thereby inhibits production of transposon-specific proteins Since choice B has the inhibition roles reversed, this is not a viable choice If the question stem had stated that the transposon was a mutant, then choice B could not have been so easily dismissed but this isn't the case, and so choice B is incorrect Choice C argues that since transposons are common, they cannot be lethal This, however, is faulty logic Commonness does not preclude lethalness Hence, choice C is also incorrect Choice D claims that Tn10 could not be responsible for cell death because every time the transposon inserts into the host DNA, it results in the duplication of the target sequence, which flanks the transposon However, there is no reason for you to conclude from the information in the passage that the presence of this repeated sequence has anything to with the viability of the cell Therefore, choice D is also incorrect Well, that leaves us with our original choice, A, as the correct answer 11 The correct answer is choice D This question is only marginally related to the passage The question stem discusses Tn10, but you don't need to know any information about transposons to answer this question The question is essentially asking you to choose the laboratory technique that would be MOST effective in determining whether or not a cell were resistant to an antibiotic Let's look at the answer choices Choice A, differential centrifugation, separates cells on the basis of weight Although if a cell did have a transposon it would be slightly heavier than an identical cell without Tn10, differential centrifugation is not sensitive enough to detect such a subtle difference in mass If this technique could separate cells that differed in molecular weight by only a few picograms, every cell would separate out at a different level because each cell would have a unique weight due to differing amounts of protein and RNA present in a cell at any given time Since differential centrifugation would not distinguish between cells with and without Tn10, and therefore would not distinguish between the ability to resist ampicillin, choice A is an incorrect answer If you were to hybridize the cells with radio-labeled antibody specific for E coli, as in choice B, you would only be able to determine which cells were E coli The antibody would bind only to a specific E coli protein This would not give you any information as to the cell's resistance to ampicillin Therefore, choice B is also wrong Choice C proposes the use of staining techniques and microscopy Although these techniques reveal a lot of information about cell morphology, there is no way to determine if a cell is resistant to certain drugs just by looking at it After all, can you tell if a person is immunized against polio simply by looking at them? So, choice C is incorrect This leaves us with choice D as the correct answer Incubating the cells on agar plates containing ampicillin is the only technique that directly exposes the cells to the antibiotic Any cells that contained Tn10 would be able to live in the presence of ampicillin; those that lacked Tn10 would die Therefore, choice D would allow you to distinguish between cells containing Tn10 and those lacking them, and so choice D is the correct answer 12 Choice C is the correct answer This is truly an easy question if you understand the rules of complementary base pairing and the differences between RNA and DNA RNA is a polynucleotide structurally similar to DNA, except that its sugar is ribose, it contains uracil, U, instead of thymine, T, and it is usually single-stranded For this question the important thing to remember is that in RNA, uracil is used instead of thymine So, in RNA, U pairs with A and C pairs with G As you know from the passage, in conjunction with the mechanism of multicopy inhibition, OUT RNA pairs with IN RNA But again, you don't need any information from the passage to actually answer the question; this is purely a knowledge-based question Since we know that we are dealing only with RNA molecules, and the base T never appears in RNA, you should have immediately eliminated choices B and D Also remember that every DNA and RNA molecule has a polarity; that is, each molecule has a 5' end and a 3' end And when two molecules complementary base pair, they line up in an anti-parallel orientation This means that the 5' end of one molecule pairs with the 3' end of the other molecule, and vice versa Therefore a strand of RNA with the sequence Kaplan MCAT Biological Sciences Test Transcript AUAUGCC in the 5' to 3' direction will complementary base pair with another strand of RNA of the sequence UAUACGG in the 3' to 5' direction But since all of the answer choices have the opposite polarity, you must read the transcript you just figured out from left to right, which gives you GGCAUAU in the 5' to 3' direction So choice C is correct Choice A is wrong, because although the sequence is correct, the polarity is wrong Again, choice C is the correct answer Passage III (Questions 13-17) 13 For question 13 the correct choice is D As I said before, even though this question refers to octane number, you can't get the answer directly from the passage You have to look to your outside knowledge of alkanes to answer this one, since the passage says nothing about isomerization, pyrolysis, or combustion So, how can you tell that this reaction is an example of isomerization? Well, if you count the carbons and hydrogens in both the starting material and the product, you can see that they total carbons and 14 hydrogens This should sound alarm bells right away, as you see that both compounds have the same number of carbons and hydrogens but differ in their molecular structure And isomers are defined as different compounds with the same molecular formula This is exactly what we have here: the two compounds have different names, the first being n-hexane and the product being 2,2-dimethylbutane; and they have the same formula Therefore, this reaction is an example of isomerization, and D is the correct choice Now for the wrong answers Let's start with choice A Combustion is a common process when dealing with alkanes In very simple terms, it involves heating a hydrocarbon in the presence of molecular oxygen to form water and carbon dioxide Usually, other products, such as carbon monoxide, are formed as well Anyway, the reaction shown in question is nothing like a combustion reaction, and so again, choice A is wrong Now for choice B Pyrolysis is also known as cracking and is characterized by the breakdown of a molecule when it is heated In the case of alkanes, pyrolysis results in the formation of smaller alkyl radicals, which can recombine to form a variety of other alkanes You cannot break n-hexane down into two alkyl radicals and then reform them as 2,2-dimethylbutane, so the reaction is not an example of pyrolysis So choice B is wrong Finally, choice C, polymerization, is also incorrect This process is defined as the joining together of small molecules to make a large molecule, just as in Reaction from the passage As you can see, however, for the reaction in question 13, there is no increase in size of n-hexane and so this is definitely not an example of polymerization Again, the correct answer is choice D 14 The correct answer here is choice A Remember that the passage defined the reference fuel as a mixture of nheptane and isooctane Also recall that the octane number of a given compound is equal to the percentage of isooctane in the reference fuel with that octane number In other words, if the octane number of a given compound is x, then there will also be x percent of isooctane in the reference fuel Therefore, a compound of octane number 50 will have an equivalent reference fuel containing 50% isooctane and 50% n-heptane For question 14, we know that the octane number of n-hexane is 25, so in the reference fuel we must have 25% isooctane And in our reference fuel, if we've got 25% isooctane, we'll have 75% n-heptane, and so choice A is the right answer The wrong answers are easy to discard B and D are wrong, since only one compound is named The text clearly states that the reference fuel is a mixture of isooctane and n-heptane not just one or the other Therefore, we can't really determine if choices B and D are reference fuels at all, because we not know their composition Choice C is wrong, because even though it is a reference fuel, its octane number is different from that of n-hexane Remember we said that the octane number of a reference fuel is equal to the percentage of isooctane? This means that the reference fuel given as choice C has an octane number of 75 And we're looking for one with an octane number of 25, which is that of n-hexane Again, choice A is the correct answer 15 The correct answer here is choice C This question requires that you have a lot more outside knowledge than the other questions The passage does provide a couple of helpful hints, though First, the reaction is drawn for you, making it easier to visualize what is happening Second, the paragraph preceding Reaction mentions that alkylation takes place, although it doesn't specifically say that this reaction is an example of alkylation Anyway, say you didn't spot the last clue; how can we determine whether Reaction is an example of Friedel-Crafts alkylation? Well, look at the reaction closely, and you can see that the benzene ring actually does become alkylated with a tertiary butyl group Based on your knowledge of aromatics, you should know that this is possible by using a catalyst Although it is not stated, a likely catalyst in this reaction would be the Lewis acid, AlCl3 This acts by forming AlCl4- and a butyl carbocation (CH3)3C+ This carbocation then acts as an electrophile and attacks the benzene ring to form a nonaromatic intermediate Aromaticity is regained by the loss of a proton, and the final result is substitution by an alkyl group Hence, this reaction is characteristic of Friedel-Crafts alkylation, and so choice C is the correct answer Let's now look at the wrong answers You could be fooled into thinking that choice D is the right answer, since the word substitution is mentioned However, so is the word nucleophilic, which, if you know your reaction mechanisms, should tell you that this is incorrect As we said earlier, the mode of attack is electrophilic, since a tertiary butyl carbocation is formed and is then attracted to the electron-rich benzene ring This carbocation substitutes an aromatic proton on the ring, so the overall reaction is electrophilic aromatic substitution Just as choice D is wrong due to the word nucleophilic, choice B is wrong because it includes the word addition The carbocation does add to the ring, but a proton has to be lost to regain aromaticity In other words, substitution takes place And finally there's choice A, which is also wrong An acyl group has the formula –COR, where R can be any alkyl group Since there are no oxygen groups in the final substituted product, there is no way that this can be Kaplan MCAT Biological Sciences Test Transcript acylation, even though the reaction does proceed by a Friedel-Crafts mechanism Once again, the correct answer is choice C 16 The correct answer is choice C To name this compound, you have to go through a series of steps First, you have to select the longest parent chain that contains the double bond If you look at the product in Reaction 2, you can see that the chain that satisfies this criterion is a pentene chain Next, the position of the double bond has to be numbered with the lowest possible value The double bond occurs between carbons and 2; notice that you don't call these carbons and So now you have 1-pentene The only answer choice that contains 1-pentene is choice C so this must be the correct response As a nomenclature review, let's continue with the explanation anyway You then have to name and number the alkyl substituents attached to the main parent chain If you look at the chain, you can see that all you have attached are three methyl groups; so this gives you a trimethyl alkene Where are the positions of these methyl groups? Well, you've got one at carbon in the chain and two at carbon 4, so the substituents can be termed 2,4,4-trimethyl If you then put everything together, the name of the compound is 2,4,4-trimethyl-1-pentene, which is choice C Choice D is wrong, because it does not follow IUPAC rules for the correct numbering of the double bond As we said before, the position of the double bond has to be numbered with the lowest possible value This is at carbon 1, and not carbon as this incorrect choice suggests By naming it 4-pentene, the positions of the methyl substituents are also incorrect, so this is not the right answer Choices A and B can be discarded since they break the first IUPAC rule for naming compounds Propene is not the longest carbon chain that contains the double bond; as we found earlier, pentene is the longest chain with the double bond For this reason, you shouldn't even have to look at the naming or numbering of the substituents, since the chain itself is incorrect 17 The correct answer here is choice B This is a passage-based question, and you can get by with hardly any chemistry knowledge for this one You need to look at Table closely to answer this one Look at the trend and see how 2-methylbutane fits into this pattern You should be able to see that the general trend is that octane number increases in the order "straight chain alkane is less than branched chain hydrocarbon." So, where would you classify 2-methylbutane in this table? Well, it's certainly not as branched as isooctane, so it would come above it in the table So for starters, you can narrow the octane number down to a value below 100 and you can therefore discard choice D Next, you need to look at the chain length of the hydrocarbons You can see that the chain length decreases from in the case of n-heptane down to in the case of 2-methylhexane It makes sense, then, that 2-methylbutane, a 4carbon chain, would follow 2-methylhexane in the table Now remember that 2-methylbutane would come before isooctane, since you have to take the degree of branching into consideration So in other words, 2-methylbutane would come between 2-methylhexane and isooctane in the table Then, remembering that octane number increases as hydrocarbon chain length decreases, we can conclude that 2-methylbutane has an octane number greater than 42 And since we already decided that its octane number is below 100, choice B is the correct answer By the way, this in fact the case, since the actual octane number of 2-methylbutane is 93 Now for the wrong answers As we said before, 2-methylbutane is not nearly as branched as isooctane and so according to the trend, it will not be higher than 100 Therefore, choice D is wrong 2-Methylbutane is just as branched as 2-methylhexane, but it contains fewer carbons Again, based on the trend that octane number increases with decreasing chain length, you can't really expect that the octane number of 2-methylbutane will be lower than that of 2-methylhexane In other words, the octane number of 2-methylbutane will not be less than 42 Therefore, you can discard choice C Choice A can be eliminated right away If it were correct, this would mean that the branched 2methylbutane would have a lower octane number than the straight-chained n-hexane Again, the trend in the table tells you that branched hydrocarbons have higher octane numbers than straight-chain hydrocarbons, so choice A is wrong Again, B is the correct choice Passage IV (Questions 18–22) 18 The answer to this question is choice C To answer this question you must apply some information from the passage and some outside information The three egg types used in the experiment were all mature and unfertilized: There was the whole egg, which consists of a nucleus, cytoplasm, and all the organelles, and which served as the control for the experiment; there was the half egg with the nucleus, which is identical to the whole egg except that it has a lot less cytoplasm; and finally, there was the half egg without a nucleus, which consists only of cytoplasm So, the major difference between the two half eggs is that the half without the nucleus doesn't have any DNA Well, you know from introductory biology that DNA is transcribed into mRNA, which is then translated into proteins So it seems kind of odd that the egg without the DNA was able to synthesize any proteins at all, let alone the SAME amount that the two nucleated eggs synthesized, on a gram-for-gram basis However, if mRNA transcription had occurred BEFORE the half eggs were created - that is, before a whole egg was split in two - that would account for the initial protein synthesis following artificial activation With this in mind, let's look at the answer choices Choice A is based on the transcription of mRNA AFTER activation But, as we just discussed, the egg without the nucleus does not contain DNA, and is thus incapable of transcription Therefore, choice A must be incorrect Choice B asserts that the levels of protein synthesis were equal in all three eggs types due to equal amounts of amino acid synthesis during the course of maturation Although an amino acid pool is necessary for protein synthesis, there is no reason why any of the eggs could not synthesize the amino acids after activation, since amino acid synthesis occurs Kaplan MCAT Biological Sciences Test Transcript OUTSIDE the nucleus Furthermore, it is the transcription of mRNA followed by its translation that directs protein synthesis A cell can have the largest pool of amino acids in the world, but these amino acids will not spontaneously form proteins in the absence of translation So even if all the eggs had synthesized the same amount of amino acid during maturation, on a gram-for-gram basis, this would NOT account for the observed protein synthesis; therefore, choice B is wrong Choice C is right on target If the eggs had transcribed the same amount of mRNA during maturation - BEFORE the split leading to the half eggs - then translation could begin in ALL of the eggs after activation, since translation occurs in the cytoplasm So, choice C would account for the observed protein synthesis and is therefore correct By the way, the actual number of mRNA transcripts in the whole egg was probably greater than the number of transcripts in the two half eggs; the half eggs probably had half as many as the whole egg Remember, however - we're talking about protein synthesis on a gram-for-gram basis Anyway, as for choice D, the half egg without the nucleus would NOT be able to synthesize any rRNA after activation, rRNA is synthesized in the nucleolus, which is found within the nucleus So choice D is wrong on that count Since rRNA is a structural component of ribosomes, and since ribosomes are found in the cytoplasm, this means that all three eggs have rRNA in their cytoplasm, which means that at the time of activation, all three eggs have the same protein synthesizing machinery present in their cytoplasm Thus, even if the two eggs with nuclei synthesized more rRNA after activation and made more ribosomes, this would not explain why protein synthesis was the same in all three eggs, on a gram-forgram basis So, choice D is wrong for this reason Again, choice C is the correct answer 19 Choice A is the correct answer To answer this question you must extrapolate from the data provided in Figure According to the passage and the figure, the levels of protein synthesis are approximately the same for the period of time shown in Figure The reason that the levels are the same is because the mRNA necessary for the early stages of development was transcribed in the eggs during maturation - BEFORE the eggs were split in two - and stored in the cytoplasm in an inactive form until the eggs were activated Once activation occurred, the mRNA was translated into the proteins necessary for the initial stages of embryonic development However, active mRNA transcripts have a very short lifespan - in order for protein synthesis to continue, more mRNA must be transcribed But in the egg without the nucleus, transcription is NOT a possibility, since nuclear DNA is a prerequisite So, if the experiment had continued for a longer period of time, you would expect the level of protein synthesis in the half egg without the nucleus to drop dramatically after the initial mRNA transcripts were translated Since the whole egg and the half egg with the nucleus are capable of transcription, you would expect protein synthesis to continue to increase in the pattern typical of the developmental chain of events This pattern is seen only in answer choice A, and so A is the correct choice 20 Choice B is the correct answer to this question Although this question is about fertilization, which is a topic related to the passage, there isn't any information in the passage to help you answer it According to the question stem, eggs have species-specific receptors on their outer surfaces to which sperm must attach for fertilization to occur The wording of this sentence implies that this binding is a necessary event of fertilization, and so if binding is blocked, the sperm would not be able to penetrate the egg cytoplasm and fuse with the egg nucleus So before an egg can be fertilized, the sperm must first bind to these receptors Based on this line of reasoning, a drug that blocks these receptors by binding to them irreversibly WOULD be an effective contraceptive Therefore, you can eliminate choices C and D because they support the notion that this would NOT be an effective contraceptive Besides, an effective contraceptive blocks fertilization in vivo - it doesn't prevent artificial activation in the lab - as in choice C Furthermore, the sperm nucleus can't fuse with the egg nucleus until the sperm has penetrated the outer layer; so choice D is wrong By eliminating choices C and D, you have increased your probability of getting the question correct to 50% Now all you have to is choose between A and B Choice A asserts that such a drug WOULD be an effective contraceptive because the binding of the receptors will cause the egg to atrophy There is NO information in the question stem that would lead you to this conclusion While it is true that if the egg is not fertilized it will eventually atrophy, this is not a direct consequence of blocked surface receptors An ovulated, unfertilized egg has a specific lifespan, independent of the accessibility of its surface receptors; therefore, choice A is incorrect Choice B is the correct answer, because as just discussed, sperm penetration is necessary for fertilization, and blocking these receptors will prevent penetration, which means that blocking these receptors will serve as an effective contraceptive Again, choice B is the right answer 21 The correct answer is choice B The purpose of this question is to test your understanding of some basic terms involving chromosomes If you knew what an autosomal cell was, then this question should have been real easy for you An autosomal cell is any cell that contains a full set of chromosomes - which consist of all cells except sex cells, or gametes In other words, an autosomal cell is a diploid cell So what's a diploid cell? A diploid cell is any cell in which there are two copies of every chromosome - one of maternal origin, one of paternal origin For example, in humans, a diploid cell, such as a liver cell or a pancreas cell, contains 22 pairs of autosomes and a pair of sex chromosomes So how the three egg types - or gametes - used in the experiment compare to an autosomal cell Well, most eukaryotic gametes, including those of sea urchins, are haploid cells This means that each gamete has only one of each pair of chromosomes in its nucleus - this is the result of meiosis; meiosis halves chromosome number, fertilization restores chromosome number, while mitosis maintains it So the whole egg is a haploid cell, and therefore contains half the number of chromosomes as an autosomal cell Likewise, the half-egg with a nucleus is also a haploid cell, since it also has a nucleus The half-egg without a nucleus has no chromosomes The question Kaplan MCAT Biological Sciences Test Transcript asks you to use the half egg with a nucleus as your basis for comparison Well, as we just discussed, the half-egg with a nucleus has the same number of chromosomes as the whole egg, and half the number of chromosomes as an autosomal cell Thus, choice A is wrong and choice B is the correct answer As for choices C and D - use of the terms "half" and "twice" have no meaning in a comparison of chromosome number between the half-egg with a nucleus and the half-egg without a nucleus, since the latter has no chromosomes So choices C and D are wrong Again, choice B is the right answer 22 Choice D is the correct answer to this question This is another example of a question that is peripherally related to the passage topic and cannot be answered on the basis of information in the passage itself Essentially, you're asked to choose the answer choice that best explains why the yolks of eggs that develop outside the body account for a larger PERCENTAGE of the egg's total volume than the yolks of mammalian eggs Remember, mammalian embryos develop internally, inside the mother's womb So let's take a look at all of the choices Choice A states that externally developing eggs are larger than internally developing eggs to protect the embryo from predators Although eggs that develop externally are susceptible to predation, it is typically factors such as camouflage and shell strength that protect the developing embryo, not egg size Furthermore, the question stem does not make an actual size comparison between externally developing and internally developing eggs; rather it makes a percent-of-volume comparison Thus, choice A can be eliminated As for choice B, animal body size is independent of the percent of total egg volume occupied by yolk whether the animal developed externally or internally Furthermore, given the total number of animal species that develop externally versus the total number of species that develop internally, there's a greater proportion of large species in the internal developers than in the external developers Think about fish, bird, insect, and reptile size, versus the size or your average mammal So, choice B is also incorrect Like choice B, even if choice C were true, it wouldn't account for the difference in yolk volume Furthermore, the term gestation applies to the duration of INTERNAL development, which is unique to mammals and marsupials; gestation is not used to refer to the amount of time it takes for an externally developing egg to hatch By the way, the period of time spent in the womb is generally much longer than the period of time it takes for an egg to hatch Human gestation is months, giraffe gestation is 14 months, and elephant gestation is 21 months, while it takes only 21 days for a chick to hatch! So, by the process of elimination, choice D must be the correct answer Yolk is rich in protein and serves as a store of nutritional reserves for the egg Since mammalian embryos are linked to their mothers via the placenta, they are able to obtain the majority of their nutritional requirements directly from the mother's blood throughout their development Therefore, yolk plays a small role in terms of nourishment for a mammalian embryo In an embryo that develops externally, however, the mother cannot provide additional nutrients to the developing embryo The only source of nourishment for an externally developing organism is whatever nutrients are stored within the egg itself Therefore, it makes sense that yolk should occupy a much greater percentage of total egg volume in an externally developing egg than in an internally developing egg Thus, choice D is the correct answer Discrete questions 23 The correct answer to question 23 is choice A Basically, all the information you need to answer this question is an understanding of resonance structures Notice that even though the question tells you that methoxymethyl chloride is very reactive under SN1 conditions, you don't need that information at all What you need to for this question is to identify valid resonance structures, not to explain anything about reaction mechanisms Anyway, a molecule is said to be a resonance hybrid when its structure can be represented by two or more Lewis structures The individual structures are called resonance contributors, and the molecule behaves like an average of these contributors Molecules that are resonance hybrids are more stable than would be expected You need to be aware that these resonance contributors not actually exist the molecule's nature cannot be described by any one of the structures taken alone Be sure to review the rules for drawing correct Lewis structures if you got this question wrong Let's look at the answer choices The first thing that you should notice is that each answer choice starts with the same structure This structure is the one that we would expect to see if the chlorine atom on methoxymethyl chloride dissociated from the molecule Notice that there is a positive charge on the carbon that had the chlorine atom attached to it and that there are two pairs of non-bonding electrons on the ether oxygen So if the first part of each answer choice is the same, which of the other structures shows a valid rearrangement of electrons? Choice A has a double bond between the oxygen and the CH2 carbon There is only one pair of non-bonding electrons on the oxygen and it has three bonds; so the oxygen must have a positive charge on it And, in fact, it does have a positive charge on it So, the oxygen has been drawn correctly let's take a look at the other atoms All the hydrogens have one bond for two electrons, the two carbons have four bonds for eight electrons each, and the oxygen has three bonds for six electrons and two non-bonding electrons for a total of eight This is a valid resonance structure; so this choice is the correct response Let's look at the wrong ones Choice B has a double bond to a hydrogen That puts four electrons on hydrogen exceeding its valence by two Choice C can be eliminated for similar reasons The far carbon has five bonds and ten electrons total Again, the valence of an atom is exceeded and the choice is wrong Choice D doesn't exceed the valence on any of the atoms, but one of the hydrogens from the CH2 has been moved to the oxygen The structure of the molecule has changed in this choice, something that we know makes this an invalid resonance structure Thus, the only set of structures that follows all of rules of drawing resonance structures is choice A Kaplan MCAT Biological Sciences Test Transcript 24 The correct answer is choice B This question requires you to know how to recognize the presence of functional groups on the basis of their characteristic IR absorption frequencies Just briefly, let's review infrared spectrometry Molecular bonds are constantly stretching, contracting, bending, and the like This activity can absorb infrared light, which gives rise to peaks on a spectrum For instance, peaks in the range of 4000 cm−1 to 2500 cm−1 usually represent the hydrogen stretching frequencies of a molecule The range from 2500 cm−1 to 2000 cm−1 shows triple-bonded species If you have a peak in this area, it's safe to say that you are dealing with either an alkyne or a nitrile From 2000 cm−1 to 1500 cm−1 are the double bonds Here is where you'll find evidence of carbonyls (from 1670 cm−1 to 1780 cm−1) and alkenes (1640 cm−1 to 1680 cm−1) The wavenumbers lower than 1500 cm−1 represents the "fingerprint region." An identical match of spectra in this region pretty much invariably denotes the same compound You can find a table of some of the specific functional group vibration frequencies in your Organic Chemistry Home Study Notes What about the question? Well, we know that 1710 cm−1 falls inside the region that shows double bonds The only answer choice that has a double bond in it is choice B Carbonyl groups in carboxylic acids absorb at about 1710 cm−1 if they are associated to other acid molecules through hydrogen bonds or 1760 cm−1 if they are not The carbon-oxygen bond of a hydroxyl group, choice A, would absorb somewhere between 1050 cm−1 and 1150cm−1 Carboxylic acid hydrogens, choice C, absorb between 2500 cm−1 and 3300 cm−1, depending on the extent of hydrogen bonding and aromatic hydrogens, choice D, absorb at about 3030 cm−1 Again, the correct answer is B 25 The correct answer to this question is choice A In enzymatic reactions, reaction velocity and temperature are directly proportional; that is, as temperature increases, so does the rate of the reaction This relationship exists because as temperature increases, the kinetic energy of the reactant molecules also increase, causing the molecules to move faster And this increased velocity will increase the rate at which the reactants collide, thereby increasing the rate of the reaction Reaction velocity will increase with increasing temperature, and therefore increasing kinetic energy, until a maximal rate is achieved This is one limiting factor in enzymatic reactions The other limiting factor occurs at very high temperatures High temperatures can cause enzymes to denature, thus halting the reaction and causing a sharp decrease in reaction velocity Based on this information, the graph best describing the relationship between reaction velocity and temperature should begin with an upward slope that eventually plateaus, followed by a sharp descending slope The upward slope corresponds to the increase in velocity with increasing temperature; the plateau corresponds to the maximal velocity of the reaction, and the descending slope corresponds to the denaturation of the enzyme due to high temperatures The only graph consistent with this description is choice A Therefore, choices B, C, and D are wrong, and choice A is the correct answer 26 The correct answer is choice B The nucleus of a zygote contains all the genetic information needed by all future cells of an adult organism in the form of genes As the zygote divides and differentiates, each cell maintains ALL of its genetic information, but it does not EXPRESS all of these genes; that is, certain genes are selectively expressed, while others are repressed Gene expression is selectively turned on or off, depending on the type of cell This is very important, because even though a myocardial cell may contain the same genetic information as an osteoblast, you wouldn't want to have your myocardial cells making bone in the middle of your heart When certain genes that are normally turned OFF are mistakenly turned ON, cancer can develop But back to the question Here we have an experiment where the nucleus of a tadpole intestinal cell is removed and transplanted into a denucleated frog zygote The zygote develops normally after the nucleus transplant The experimental results suggest that a differentiated nucleus contains the same genetic material as a zygote's nondifferentiated nucleus The fact that the nucleus was originally in a differentiated cell also suggests that selective repression of DNA is possible This conclusion is fairly straightforward since, if the salvaged nucleus lacked genetic material essential to a developing tadpole, the zygote would develop into, at best, a mass of intestinal cells If certain genes were missing, the zygote would not be able to develop normally These conclusions about the experimental results are best represented by choice B Let's take a look at the wrong choices for a second just to make sure choice B is the best answer Choice A says that the results of the experiment suggest that cell differentiation is controlled by irreversibly shutting off genes not needed by that cell In fact, the experiment proves that this is NOT true and therefore choice A is incorrect Shutting off genes must be reversible, otherwise this transplanted nucleus would NOT be able to direct the zygote to divide and develop into all of the different cell types found in a frog Choice C suggests that the cytoplasm of the zygote contains all of the information needed for normal adult development in the form of RNA This choice is wrong because the experiment did NOT specifically address this question The results of the experiment not support or contradict this claim Although DNA is the known carrier of genetic information, if you wanted to test the RNA theory, you would perform a different experiment, such as seeing if a denucleated frog zygote develops normally Choice D is also incorrect because, as with choice C, this answer choice is not addressed by the experiment Therefore, the experimental observations not support or contradict this choice And besides, all eukaryotic ribosomes are the same and are found only in the cytoplasm of a cell, not the nucleus This fact also disqualifies choice D Again, choice B is the correct answer 27 The correct answer is choice C This question tests your understanding of red blood cells, the antigens they express on their cell surfaces, and the types of antibody production these antigens elicit There are four major blood Kaplan MCAT Biological Sciences Test Transcript choices B and D, loss of functional groups from the ring is very unlikely Choice C is the only answer in which the functional groups on the ring actually react and are not substituted; so again, C is the correct answer 32 The correct answer here is choice A This is a pretty simple question to answer, if you know your basic carboxylic acid + alcohol reactions Essentially, you need to understand the same underlying principles behind the formation of aspirin; so yet again, you're being tested on your understanding of Reaction Just like in question 31, salicylic acid acts as a carboxylic acid in this esterification reaction Under acidic conditions, the oxygen in phenol will nucleophilically attack and add to the carboxyl carbon in salicylic acid Through a series of intermediates, water is eventually lost from the molecule and the final result is substitution of the carboxyl hydroxyl by the aromatic ring In other words, an ester linkage has formed between the carboxyl group of salicylic acid and the alcohol group in phenol, resulting in the formation of phenyl salicylate The correct answer is A Now for the wrong choices You should be able to see that choice D is benzoic acid, and so would act as the acid in the reaction Salicylic acid would therefore act as an alcohol by the nucleophilic attack of its hydroxyl oxygen on the carboxyl carbon of benzoic acid As a result, the final product would be o-benzoylbenzoic acid and not phenyl salicylate Choice D is therefore incorrect You could have easily fallen into the trap of choosing answer C in this question, since the difference between this and the correct choice is merely an extra CH2 group Even though this molecule would act as an alcohol in the reaction, phenyl salicylate would not be produced A similar compound-tolyl(o-hydroxy)benzoate would result instead, so choice C is also wrong Choice B is completely wrong Benzene is very unreactive, and under these conditions, it would not react with salicylic acid Even under more extreme conditions such as raised temperature, reaction between the two would be highly unlikely Again, the correct answer is choice A 33 The correct answer here is choice C Since aspirin is an ester it will react with water in the same way that all esters the ester linkage is cleaved forming an alcohol and a carboxylic acid In moist air, aspirin can be acidically cleaved to yield salicylic acid and acetic acid Acid is present in the form of H3O+; and through a series of intermediates, the ester linkage in aspirin is broken, resulting in the formation of salicylic acid (the alcohol) and acetic acid (the carboxylic acid) It is in fact the latter that causes the odor of vinegar, so you should be able to narrow the answer choices down to at least A or C Choice A is wrong because aspirin does not break down into citric acid and acetic acid You could have been tricked into choosing this one because citric acid also has a distinguishing odor; however, it's not of vinegar, but of lemon juice In addition, in order for citric acid to be formed, the ester would have to be aliphatic as opposed to aromatic Aspirin contains an aromatic ring and so upon hydrolysis, an aromatic functionality should be present in one of the products Neither citric nor acetic acid contains this functionality, so choice A can be discarded Choice B is incorrect since acetic anhydride will not be formed Although it is commonly used in the synthesis of aspirin, it will not be formed in the reaction with moist air Acetic anhydride is the dehydration product of two molecules of acetic acid yielding one molecule of water at the same time However, the question states that aspirin breaks down in a water saturated environment and so the hydrated product will exist, not the dehydrated product In other words, acetic acid, not acetic anhydride, will be formed with salicylic acid Choice D is also incorrect To begin with, there should not be any residual salicylic acid in aspirin Even if there were trace levels of salicylic acid present, it should be emphasized that salicylic acid would not react with moist air to form citric acid Passage VI (Questions 34–38) 34 The correct answer is A We already have a pretty good idea of the reagent's identity from reading the passage If you don't remember Figure 1, which describes the synthetic scheme for 1,3-cyclohexadiene, it's always there to look back at; but since we know that cyclohexane goes through two bromination and elimination steps to create two double bonds, and that Reagent X is used for the second bromination, we may not need to look back at the passage We already know that Reagent X is a source of bromine radicals since a bromine has added at the allylic position in the presence of hν That's enough information to go through and eliminate some choices Well, we can tell that choice A has bromine in it; so it passes our first requirement And although you may not know it off the top of your head, NBS is an allylic brominating reagent But if you didn't know that, you'd move along to choice B Lithium aluminum hydride (LAH) is a common, powerful reducing agent that can reduce electrophilic organic compounds However, it only works on electrophiles The electron density of alkenes makes them nucleophiles, so LAH won't even reduce the alkene to cyclohexane In addition, there is no bromine in LAH, so we can eliminate choice B Choice C is no better Tetramethylethylene diamine (TMEDA) also has no bromine in it Its main use in organic chemistry is the creation of nucleophilic allylic lithium reagents that can be used in chain lengthening In any case, we can eliminate this answer Choice D, phosphorous tribromide, has bromine in it, so we can't eliminate it right off, but PBr3 is a source of nucleophilic bromine It is used in SN2 reactions that brominate compounds Since we know that the reaction described in Figure is a radical reaction, not a nucleophilic substitution, we know that PBr3 cannot be Reagent X So the only reagent named here that fits all of our requirements for Reagent X is NBS, choice A Notice that you can eliminate half of the choices just from seeing that Reagent X has to have bromine in it The hard part, really, is doing exactly what the MCAT loves to make you They give you some of the information 12 Kaplan MCAT Biological Sciences Test Transcript you need, but ask you to come up with the rest of it That's often how your chemistry is tested on the MCAT Again, the correct answer is choice A 35 The correct answer to question 35 is A It isn't connected to the passage as directly as question 34 It requires you to rely a little more on your previous knowledge Since this question deals with the products of Reaction 1, that's probably a good place to start We know that raising the temperature will shift the composition of the product mixture toward the 3,6-dibromocyclohexene What's happening here is that the activation energy needed to get from the intermediate to the 3,4 isomer is lower than that needed to get to the 3,6 isomer, so it is formed more readily However, the 3,6 isomer has a lower overall energy, which makes it more stable That means that the increase in temperature makes it easier to go back to the carbocation and into the thermodynamically stable 3,6 form In other words, the 3,4 isomer is formed faster, that is, favored kinetically, but the 3,6 isomer is more stable, or favored thermodynamically That means that the 3,4 isomer is the kinetic product and the 3,6 isomer is the thermodynamically favored product Anyway, since the 3,6 isomer is the thermodynamically favored product, that means that statement I is true and III is not true the 3,4 isomer can't be the thermodynamically favored product, too Roman numeral II states that the 3,6 isomer is the kinetically favored product, you know this statement is incorrect Roman numeral IV states that the 3,4 isomer is the kinetic product This, in fact, is a true statement; however, it is not necessary to answer this question The reason the 3,4 isomer converts to the 3,6 isomer when it is heated, is that the 3,6 isomer is the thermodynamically favored product, period! So, the answer choice that states "I only" is the correct response, and that is choice A 36 The correct answer is B Question 36 takes us back to the last paragraph, Reaction 2, and the mechanism of electrophilic aromatic substitution Notice that the passage never called the mechanism in Reaction an electrophilic aromatic substitution; but we know that the aromatic compound, benzene, is having a hydrogen substituted by a bromine electrophile Well, let's recall what the passage did reveal about Reaction We know that Reaction starts when the catalyst polarizes the bromine molecule to make it a strong electrophile that can successfully bond with the electrons in benzene's pi system A carbocation is formed after the electrophile has bonded, which then releases a hydrogen ion to get rid of its positive charge That's the mechanism in a nutshell Which answer best describes the intermediate? Well, you can eliminate choices A and D right off because they aren't carbocations With those two eliminated, we look at choice B, which says that the intermediate is a doubly allylic carbocation Well, if something is allylic, it has a double bond in it, so if something is doubly allylic, it must have two double bonds in it Remember that the electrophile and the pi system bonded That means that a double bond has been broken, so there are only two double bonds left and the intermediate is not aromatic Choice C names an aromatic carbocation as the intermediate That might seem right; but remember, we just said that the electrophile breaks into the pi system and the bonds are no longer conjugated over the entire ring An aromatic ring must be fully conjugated In fact, the loss of the aromaticity and the desire to get it back is what drives the substitution reaction The benzene will eject a hydrogen and reform the aromatic ring The intermediate carbocation is not aromatic, or else it would not have that drive to substitute rather than add So choice B is the best answer 37 The correct answer is B This question, like so many others, requires you to understand the passage and bring in your outside chemistry knowledge You have to take what you can get from Reaction and combine it with your knowledge of benzoic acid Benzoic acid is simply a benzene ring with a carboxylic acid –COOH group substituted at one of its carbons You have to have a good understanding of the electrophilic aromatic substitution reaction from the passage, but you must also remember the effects that substituents have on further benzene substitution, so that you can tell which product is formed in quantity If you don't remember the substituent effects, they're reviewed in the Organic Chemistry Home Study Notes in the chapter on aromatic compounds Basically, the substituents on benzene affect the stabilization of the carbocations formed when an electrophile tries to substitute onto the aromatic ring Substituents can either activate or deactivate the benzene ring, making it easier or harder to substitute Ortho/para- directing groups either have pairs of non-bonding electrons that can stabilize the carbocation intermediate, like the halogens, or they have an electron donating nature that can stabilize the intermediate, like alkyl groups Meta- directors, on the other hand, have a positive or partial positive charge on the atom directly bonded to the aromatic ring That means that when an atom substitutes at the ortho- or parapositions, the resonance of the resulting carbocation will place a positive charge adjacent to the positively charged substituent The close proximity of these two positive charges destabilizes the carbocation Substitution at the metaposition does not set up this conflict between positive charges, so substitution here creates the most stable carbocation The carboxylic acid group has a partial positive charge on the carbon attached to the ring, so it is a metadirector In that respect, if we were to replace the benzene in Reaction with benzoic acid, when the electrophile did bond to the ring, it would most likely be at the meta- position Therefore, we'd expect to see m-bromobenzoic acid more than the o- or p- isomers, which makes B a better choice than A and C, but what about D? Well, we'd only get that if the benzene ring were substituted at both meta- positions However, the first substituted bromine is slightly deactivating because it withdraws electrons from the benzene ring, making it less appealing to electrophiles; therefore, we wouldn't expect to see too much benzoic acid that had substituted twice Thus, choice B is definitely the best answer Again, the correct answer is choice B 13 Kaplan MCAT Biological Sciences Test Transcript 38 The correct answer to question 38 is choice B This question asks you about the characteristics of benzene You have to rely on your outside knowledge for this one, because all the passage does is describe a substitution reaction of benzene Recall that benzene is aromatic Let's briefly talk about the structure of benzene As an aromatic compound, we know that it is a planar hexagon In fact, it is a perfect hexagon with 120 carbon-carbon bond angles The six hydrogen atoms project out from the six carbon atoms and are all in the same plane as the carbon ring As an aromatic compound, there must be p orbitals available for overlap That means that each of the six carbons must be sp hybridized so that the six pi electrons can delocalize over the entire ring in electron clouds above and below the plane of the ring This delocalization associates all six carbons into a coherent whole, right down to the bond lengths All of the carbon-carbon bond lengths are 1.39 Å, which falls between the lengths of single carboncarbon bonds and double carbon-carbon bonds With this in mind, let's look at statements one through four Roman numeral I says that all the carbons in the benzene ring are sp hybridized Well, remember that benzene must have p orbitals on each carbon atom in the ring to contribute to orbital overlap We just discussed that all the carbons must be sp hybridized So I is a true statement Looking at the answers, we see that they all have statement I in them So unfortunately, we can't eliminate any answers Roman numeral II says that benzene does not participate in heterolytic substitution reactions Here's where we need some outside chemistry A heterolytic reaction is one in which pairs of electrons are exchanged Reaction tells us that an electrophile attaches itself to the benzene ring The positively charged electrophile is electron deficient, so both electrons in the bond have to come from the benzene The reaction is finished when both electrons in a carbon-hydrogen bond stay with the benzene and an H+ is released So the reaction is heterolytic since all the electrons are traded in pairs Therefore, statement II is wrong That eliminates choice D Statement III says that benzene can be hydrogenated When a hydrocarbon is hydrogenated, multiple bonds are broken and replaced by the appropriate number of carbon-hydrogen bonds Are there any conditions severe enough to turn benzene into cyclohexane? There are a few catalytic methods of benzene hydrogenation, requiring platinum or palladium catalysts at high pressures around 2000 psi or powerful catalysts like rhodium on carbon But since it is possible, statement III does in fact characterize benzene That allows us to eliminate choice A, and now we're down to choices B and C Statement IV states that benzene contains alternating single and double bonds Well, that isn't true Spectroscopic data has shown that all carbon-carbon bond lengths are equal in benzene, just as I mentioned earlier Even though the molecule is often drawn with alternating single and double bonds, in reality it consists of one type of carbon-carbon bond somewhere in between a single and a double bond Anyway, that's just a fact about benzene that you have probably encountered a lot in your studies So statement IV is not true Now we can get rid of choice C And we are left with the correct answer, choice B Passage VII (Questions 39–44) 39 Choice B is the correct answer Before I jump right in and discuss this question, let me backtrack for a moment and talk a bit about antigens, antibodies, and the carrier effect When the immune system is exposed to an antigen, which is a substance that it recognizes as foreign, both helper T-cells and B-cells bind to the antigen The helper T-cells help the B-cells to produce antigen-specific antibodies that bind to the antigen and facilitate its destruction This is known as the primary response Meanwhile, there's another group of B-cells and helper T-cells that become primed to the antigen - they remember it - so if the immune system is re-exposed to the same antigen, its second response will be greater and quicker The helper T-cells that are primed to the antigen will stimulate the proliferation of the primed B-cells into antibody-secreting B-cells This is known as the secondary response However, sometimes the antigen is so small that the immune system doesn't even notice it, which means that this antigen will not elicit antibody production But if you take this antigen, complex it with a larger antigen, and inject an animal with it, THEN UPON REINJECTION, there will be antibody produced specific for the tiny antigen at a level comparable to a secondary response This is the essence of the carrier effect The tiny antigen is called a hapten, and the larger antigen, which is usually a foreign protein, is known as a carrier When an animal is primed with a hapten-carrier complex, you don't get any anti-hapten antibody production Rather, you get B-cells that are primed to the hapten and helper T-cells that are primed to the carrier If the helper T-cells were able to recognize the hapten, then you'd get anti-hapten antibody production immediately When the animal is reinjected, or boosted, with the same hapten-carrier complex, THEN you get anti-hapten antibody production Now that that's out of the way, let's answer the question That's right, Question 39 is one of those reading comprehension-type questions In fact, it would be giving this question too much credit by calling it reading comp rather, it's a "were-you-conscious-while-reading-this-passage" type of question If the answer is yes, then you should've gotten this one right According to the passage, in order to get a secondary response to a hapten, an animal must first be primed with a complex consisting of the hapten and a foreign protein known as a carrier There is no primary response to the hapten But upon reinjection of the same hapten-carrier complex, there will be anti-hapten antibody produced - this is the secondary response This effect is clearly illustrated by the results of Experiments and The results of Experiment illustrate which lymphocytes have the "memory" necessary for antibody production Even if most of this somehow eluded you, you still should've been able to eliminate ALL of the wrong choices because none of the experiments boosted the animals with a carrier only, a hapten only, or both free carrier and free hapten Therefore, choices A, C, and D are wrong, and choice B is the right answer to this question 14 Kaplan MCAT Biological Sciences Test Transcript 40 Choice D is the correct answer to this question This question deals with the results of Experiments and - either there WAS anti-DNP antibody production or there WASN'T Since anti-DNP antibody production is noted in Table by a simple yes or no, there is no way for you to determine the strength of this response when BSA is used as the carrier relative to the response when OA is used as the carrier Therefore, you should have immediately eliminated choices A and B As for choice C - while it is true that the secondary response DOES require a lower dose of hapten-carrier complex than the primary response, this is a moot point in regards to the experiments, since the results of the experiments not illustrate this in any way, shape, or form, since there is no data regarding quantities of hapten-carrier complexes used in either priming or boosting So choice C is also incorrect So we're left with choice D, which states that the results demonstrate that the secondary response requires priming and boosting with the same carrier Rabbit was primed and boosted with DNP-BSA and Rabbit was primed and boosted with DNP-OA, and both rabbits produced anti-DNP antibody Rabbit was primed with DNP-BSA but boosted with DNP-OA and didn't produce any anti-DNP antibody So far, this supports the claim of choice D But what about Rabbit 4? Well, Rabbit was primed with both DNP-BSA and free OA, boosted with DNP-OA and it produced anti-DNP antibody Why? Anti-DNP was produced because the DNP-BSA of the primer injection resulted in helper T-cells primed to BSA and B-cells primed to DNP, while the free OA of the primer resulted in helper T-cells primed to OA (The free OA also resulted in B-cells that synthesized anti-OA antibodies and B-cells primed to OA, but that's not relevant to the experimental design.) So when Rabbit was boosted with DNP-OA, the OA-primed helper T-cells recognized the OA in DNP-OA and stimulated the DNP-primed B-cells to proliferate into anti-DNP antibody secreting cells Remember, the passage tells you that it takes carrier-primed helper T-cells and hapten-primed B-cells to produce a secondary response to the hapten Therefore, choice D is the correct answer 41 Choice D is the correct answer This is yet another question testing your understanding of the carrier effect as detailed in the passage and illustrated by the results of the experiments Priming and boosting a rabbit with DNPHA is no different from priming and boosting a rabbit with DNP-BSA or DNP-OA Human albumin, HA, is a carrier just like BSA and OA; HA is a protein foreign to the rabbit's immune system Therefore, priming the rabbit with DNP-HA results in DNP-primed B-cells and HA-primed helper T-cells, and when boosted with the same complex, these T-cells will stimulate these B-cells to produce anti-DNP antibodies Hence, choice D is the right answer Choices A and B are wrong because the B-cells don't produce antibodies specific for the hapten-carrier complex as a unit Choice C is wrong because the B-cells are incapable of producing anti-hapten antibodies following the initial injection, because according to the passage, the hapten is too small to trigger antibody production Again, choice D is the correct answer 42 Choice B is the correct answer Let's analyze Experiment 3, using our newfound knowledge of the carrier effect Rabbit was primed with DNP-BSA, resulting in DNP-primed B-cells and BSA-primed helper T-cells Rabbit was primed with free OA, resulting in OA-primed B-cells and OA-primed helper T-cells Poor Rabbit was irradiated, which rendered it incapable of mounting an immune response of its own, since radiation kills the lymphocytes B-cells from Rabbit 5, which we know are DNP-primed, and helper T-cells from Rabbit 6, which we know are OA-primed, were transferred into Rabbit When Rabbit was boosted with DNP-BSA, anti-DNP antibody was NOT produced Why? Because Rabbit didn't have any BSA-primed helper T-cells, and we know that BSA-primed helper T-cells would be needed to stimulated the DNP-primed B-cells to proliferate into anti-DNP producing cells If Rabbit 5's BSA-primed helper T-cells had been transferred to Rabbit in conjunction with its DNP-primed B-cells, THEN a secondary response would have been elicited When Rabbit was boosted injected with DNP-OA, anti-DNP antibody was produced Why? Because we know that we need OA-primed helper T-cells to get anti-DNP producing B-cells when OA is the carrier molecule used in the booster injection Therefore, choice B is the correct answer Choice A is wrong not only because it's false, as just discussed, but because we're only concerned with the cell types that WERE transferred to Rabbit Choice C is wrong because, as we previously said, the B-cells transferred to Rabbit from Rabbit WERE DNP-primed Choice D is just plain silly; and besides, Rabbit was boosted with DNP-BSA PRIOR to being boosted with DNP-OA; so choice D is wrong Again, choice B is the correct answer 43 Choice A is the correct answer This question requires a bit of analysis and a complete understanding of the results of Experiment So if you are even the slightest bit confused about this experiment, please listen to the discussion of the previous question Okay, in this question, poor Rabbit is primed with DNP-BSA following irradiation; we're told that the rest of the experiment remains unchanged In other words, B-cells from DNP-BSAprimed Rabbit 5, along with helper T-cells from OA-primed Rabbit 6, were transferred to Rabbit Given this information, would you now expect Rabbit to produce anti-DNP antibody when boosted with DNP-BSA? Well, in the original version of Experiment 3, Rabbit was unable to produce a secondary response when boosted with DNPBSA because it lacked BSA-primed helper T-cells So we have to figure out if priming Rabbit with DNP-BSA following irradation would result in BSA-primed helper T-cells And the answer is NO! Irradiation prevents Rabbit from mounting an immune response, thus preventing the production carrier-primed helper T-cells and hapten-primed B-cells Rabbit still has to rely on the powers of Rabbit 5's B-cells and Rabbit 6's helper T-cells to produce an immune response Therefore, choice A is the correct answer Choice B is wrong because an irradiated rabbit can't produce antibodies to the transferred cells In fact, that's one of the reasons why Rabbit was irradiated in the initial version of Experiment - so that Rabbit wouldn't reject the donor cells from Rabbit and Rabbit Choices C and 15 Kaplan MCAT Biological Sciences Test Transcript D are wrong Rabbit doesn't have its own DNP-primed B-cells and BSA-primed helper T-cells to produce any antiDNP antibody, so that little "Yes" at the start of these choices eliminates them Again, choice A is the correct answer 44 Choice C is the correct answer According to the question stem, IgM is the predominant class of antibody produced during a typical primary response, while IgG is the predominant class of antibody produced during a typical secondary response The primary and secondary immune responses differ in other ways as well, such as the rate of antibody synthesis and the affinity of the antibody for the antigen, but you're not required to know any of this to answer the question I hope that you immediately ruled out choices A and B, since the question stem told you that a given immune response is predominated by either IgM or IgG - no response is ever 50-50 IgM to IgG Furthermore, choice B is wrong because Rabbit didn't produce any anti-DNP antibodies - IgM or IgG - after being boosted with DNP-BSA It's not really made clear in either the passage or the question stem whether the anti-hapten antibody production elicited in the carrier effect is parallel to the antibody production of the primary response or the secondary response However, it's not necessary to know this to know that choices A and B are wrong For choices C and D, we need to look at Rabbit 4, which was primed with both DNP-BSA and free OA and boosted with DNP-OA This means that while its unclear whether Rabbit produced anti-DNP IgM or IgG after being boosted with DNP-OA, it is clear that Rabbit did NOT produce anti-DNP IgM after being PRIMED Remember, we don't get anti-DNP produced BEFORE the booster injection - only after Therefore, choice D is wrong So choice C must be the right answer Priming Rabbit with free OA results in the production of anti-OA IgM, since free OA is recognized by the rabbit's immune system upon first exposure Remember, it's the hapten - not the carrier - that fails to trigger antibody production when injected alone So when Rabbit is boosted with DNP-OA, its OA-primed helper T-cells stimulate its OA-primed B-cells to proliferate into anti-OA IgG Even though OA is complexed with a hapten, its presence still elicits anti-OA production Therefore, you get more anti-OA antibodies produced following the booster injection, and this time around the predominant class of anti-OA antibodies synthesized is IgG Therefore, choice C is correct Passage VIII (Questions 45–49) 45 Choice C is the correct answer According to the passage, congestive heart failure leads to a decrease in cardiac output, which in turn can result in kidney, or renal, failure Renal failure, in turn, results in a decreased urine output Let's a little reasoning here If urine volume decreases, this means that the failing kidneys are reabsorbing a greater volume of water than healthy kidneys And if all of this water is reabsorbed instead of excreted, then intravascular volume will increase Since an increase in venous volume returning to the right side of the heart - an increased preload - is one of the causes of congestive heart failure, it's plausible that a person suffering from renal failure would experience the signs and symptoms of congestive heart failure, despite having a healthy heart Let's take a look at the answer choices Choices A and B should be immediately eliminated, because, as just discussed, renal failure results in decreased urine output, which means that intravascular volume will INCREASE, not decrease And since, according to the passage, increased preload, not decreased, is associated with congestive heart failure, choice D must be wrong and choice C is the correct answer 46 Choice B is the correct answer According to the question stem, CHF patients often experience shortness of breath, especially when lying down, and you're asked to determine why this phenomenon occurs According to the passage, in CHF, the veins become congested and there is an increase in venous pressure And from your outside knowledge of introductory biology, you should know that venous circulation is driven by the squeezing of adjacent skeletal muscle, that backflow of blood is prevented by valves within the veins, and that venous flow is affected by gravity Arterial circulation, on the other hand, is driven by the blood pressure generated by the heart Due to the force of gravity, venous blood tends to pool in the lower extremities of a person standing upright; when in a recumbent position, there is a decreased pooling because gravity is not forcing the blood toward the lower extremities, which means that there is GREATER venous return to the heart In a CHF patient already suffering from shortness of breath, increasing venous return by lying down will further stress a failing heart Therefore, choice B is the correct answer and choice A is wrong Choice C is true but irrelevant; it doesn't account for the increased shortness of breath in the recumbant position Compared to a person lying down, a person standing up DOES have a reduced right atrial pressure because of the blood pooling in the extremities, which leads to a greater cardiac output As for choice D, whether it's true or not - and you don't know because this issue is not discussed in the passage - it does not account for the increase in shortness of breath experienced by a CHF patient when lying down Again, choice B is the correct answer 47 The correct answer is choice B To answer this question you need to extract some information from the passage and then apply it to your knowledge of the circulatory system From the passage you know that in right-sided heart failure the veins become congested From this piece of information you know that the answer must be a vein and not an artery Therefore you can immediately eliminate choice A, the coronary arteries, and choice D, the aorta, which is the largest artery in the body Well this leaves us with two veins, the jugular veins, which are systemic veins, and the pulmonary veins So you need to determine which of these veins would be congested in right-sided heart failure You should know that the right side of the heart pumps deoxygenated blood into the lungs via the pulmonary arteries, where carbon dioxide is exchanged for oxygen Oxygenated blood returns to the left side of the heart via the pulmonary veins It is then pumped into the aorta, which sends it into circulation Deoxygenated blood is returned to 16 Kaplan MCAT Biological Sciences Test Transcript the right side of the heart via the systemic veins Since the pulmonary veins enter the left side of the heart, they would not be congested in right-sided heart failure; thus choice C is incorrect and choice B is correct The jugular veins are major veins of venous circulation and are located in the neck; like all peripheral veins, they carry deoxygenated blood Thus, if the right side of the heart fails, the jugular veins are just some of the many veins that would become congested Again, choice B is the correct answer 48 Choice A is the correct answer to this question To answer this question you need to compare the information from the passage about left-sided and right-sided heart failure Let's briefly review the similarities and differences Right-sided heart failure is associated with systemic venous congestion, peripheral edema, and low cardiac output Left-sided heart failure is associated with pulmonary venous congestion, pulmonary edema and also low cardiac output The low cardiac output characteristic of congestive heart failure caused by failure of either or both ventricles can lead to kidney failure and shortness of breath Now let's look at the answer choices Pulmonary venous congestion, choice A, is a characteristic of left-sided heart failure only, and is thus the correct answer Systemic venous congestion, choice B, is a characteristic of right-sided heart failure only; thus, choice B is wrong As we just discussed, kidney failure is due to low cardiac output And since low cardiac output is characteristic of BOTH right-sided and left-sided heart failure, choice C is incorrect Choice D is incorrect, since the movement of fluid out of blood vessels into interstitial spaces is the broad definition of edema, under which both peripheral edema and pulmonary edema fall Again, choice A is the correct answer 49 Choice D is the correct answer Answering this question was dependent on your knowledge of basic metabolism According to the question stem, a person suffering from that congenital defect with that weird name tetralogy of Fallot - prevents about 75% of the blood from becoming oxygenated This means there will be a 75% decrease in the amount of oxygen delivered by the blood to oxygen-needy cells If a cell's metabolic demands are not being met by the body, then the cell resorts to anaerobic metabolism to generate the ATP needed for daily activity This is parallel to the oxygen debt incurred by muscle cells during intense activity such as exercise; these cells revert to anaerobic metabolism During anaerobic metabolism, glucose is still converted into pyruvate via the reactions of glycolysis - since glycolysis doesn't require oxygen In addition to generating two molecules of ATP per molecule of glucose, molecules of NADH are also produced In the presence of oxygen, the pyruvate is converted into acetyl CoA, which then enters the Krebs cycle and the electron transport chain In the absence of oxygen, glycolysis is the only respiratory option available to the cell; but in order for glycolysis to continue NAD+ must be regenerated And NAD+ is regenerated by way of lactate fermentation: the pyruvate is reduced to lactate and NADH is oxidized back to NAD+ So in the absence of oxygen, lactate builds up The oxygen debt is the amount of oxygen needed by a cell to remove the lactate and restore the cell to normal functioning Therefore, choice D is the correct answer Glucose is the starting molecule of glycolysis, which occurs in both anaerobic or aerobic respiration Therefore, in a patient suffering from tetralogy of Fallot, you would expect the level of glucose to remain the same or even decrease Therefore, choice A is wrong Since the conversion of pyruvate to acetyl CoA requires oxygen, you would expect the concentration of acetyl CoA to decrease when oxygen delivery is compromised Thus, choice B is incorrect Ethanol, choice C, is also produced under anaerobic conditions But ethanol, which is the end product of alcohol fermentation, occurs only in yeast and some bacteria, not humans Therefore, choice C is incorrect If humans could produce ethanol during periods of oxygen debt, people would get drunk while exercising Anyway, choice D is the correct answer Discrete questions 50 The correct answer to this question is choice D This question deals with the exchange of material into and out of cells More specifically it deals with the concept of osmosis This is a very important question because osmosis is an MCAT favorite So make sure that you completely understand this concept Albumin is an osmotically active plasma protein and is one of the primary determinants of the blood's osmotic pressure Osmotic pressure is the tendency of water to flow from an area of lower solute concentration to an area of higher solute concentration If a membrane is impermeable to a particular solute, then water will flow across the membrane until the differences in the solute concentrations have been equilibrated This movement is defined as osmosis Remember, in osmosis, water always flows from a region of lower osmotic pressure to a region of higher osmotic pressure The question stem tells you that albumin is infused into arterial blood If you know that capillaries are pretty much impermeable to albumin, you should have had an easy time picking the right answer If you don't know that, the fact that albumin is defined as a plasma protein might have tipped you off Since increasing or decreasing a solute's plasma concentration via infusion does not normally affect a cell's permeability to that solute, you can eliminate choices A and B The albumin infusion would create a greater albumin solute concentration in the capillaries than in the interstitial fluid that bathes the capillaries Therefore, you expect an increase in the movement of water from the interstitial fluid into the capillaries Or, in other words, a decrease in the movement of water from the capillaries to the interstitial fluid This will reduce the osmotic pressure in the capillaries Therefore, choice C is incorrect and choice D is the correct answer 51 The correct answer is choice A This is another question that may seem very intimidating upon first glance But remember that the MCAT only requires you to know information from introductory biology Although the 17 Kaplan MCAT Biological Sciences Test Transcript question stem describes several symptoms, the bottom line is that the boy has elevated blood glucose To answer this question correctly, all you have to is decide which of the answer choices best accounts for this symptom Here is where your knowledge of biology comes into play All of the answer choices are hormones at low concentrations So, to answer the question correctly, you need to have memorized what functions these hormones control Insulin, which is released from the pancreas, lowers blood glucose and increases the storage of glycogen Basically, insulin promotes the conversion of glucose to glycogen So, if insulin levels were low, very little glucose would be converted into glycogen, leading to elevated blood glucose Therefore, choice A is the correct answer Let's look at the other choices quickly ACTH, adrenocorticotropic hormone, is released from the anterior pituitary and stimulates the adrenal cortex to synthesize and secrete glucocorticoids Glucocorticoids raise blood glucose levels and decrease protein synthesis Low levels of ACTH will retard the secretion of glucocorticoids, thereby LOWERING blood glucose Thus, choice B is incorrect Glucagon, another hormone secreted by the pancreas, stimulates the conversion of glycogen to glucose in the liver Essentially, this hormone has the opposite effect of insulin It promotes the conversion of glycogen into glucose, and therefore raises blood glucose So if glucagon levels were low, very little glycogen would be converted into glucose, and blood glucose would be low; thus, choice C is wrong Cortisol, a steroid hormone, has both mineralocorticoid and glucocorticoid activity Mineralocorticoid activity increases water reabsorption in the kidneys As for the glucocorticoid activity, as we just discussed with ACTH, low glucocorticoid levels will result in low blood glucose Therefore, choice D does not account for the increased blood glucose in the patient and is therefore incorrect Although you didn't really have to know anything about the symptoms of the 10-year-old boy in the question stem, let's discuss them briefly A young person complaining of thirst, polyuria, and weight loss is the classic presentation of Type I, or juvenile, diabetes mellitus Juvenile diabetes occurs when, for an unknown reason, the beta-cells of the pancreas stop producing enough insulin Again, choice A is the correct answer 52 The correct answer is choice C A single action potential, known as an impulse, is a change in electrical potential across an axon membrane conducted along a neuron When a stimulus acts upon the axon, the action potential exhibits an "all-or-nothing" phenomenon This means that once the neuron membrane is depolarized to its threshold potential, an action potential of a given magnitude and speed is generated Sodium ions rush into the axon through voltage-sensitive channels; these gates detect the change in membrane potential and respond by opening Following depolarization, the sodium ion gates shut, but potassium ion gates open, and potassium ions rush out of the neuron, thereby returning the neuron to its resting potential Nothing, not even the intensity of the stimulus initiating the action potential, can change the duration or magnitude of an in vivo action potential Therefore, choices A and B are incorrect The duration and magnitude of the action potential are not variable If an action potential is generated, then it will have a set duration and magnitude So how we differentiate between different intensities of pain, for example? The only way that a neuron can transmit an increase in stimulus intensity is by increasing the frequency at which action potentials are fired The more the neuron fires, the greater the perceived intensity of the stimulus Therefore, choice C must be the correct answer The threshold membrane potential is the strength of the stimulus needed to depolarize the membrane and generate an action potential As stimulus intensity increases, the threshold potential will be met, and an action potential will be generated But the threshold potential is a constant It does not change with the intensity of the stimulus Therefore, choice D is incorrect Again, choice C is the correct answer 53 The correct answer to this question is choice A This question is a little different from most genetics questions found on the MCAT in that instead of calculating a probability or percentage of offspring with a given trait, you are given the phenotypic ratios of the offspring and asked to determine the unknown parental genotype The principles needed to solve this type of problem are identical to those used in a standard genetics question The cross in this question is your basic testcross, in which a parent of an unknown genotype is crossed with a phenotypically recessive parent Since the genotype of the phenotypically recessive individual is necessarily homozygous, a testcross is used to analyze the phenotypic ratios of the offspring, and then, on the basis of these results, determine the unknown genotype of a parent When the homozygous recessive hamster, with the genotype little s little s, little w little w, is crossed with the hamster of unknown genotype, we're told that the F1 progeny are in a : : : phenotypic ratio of rough brown to smooth brown to rough white to smooth white The fact that both the dominant AND the recessive traits are expressed in the progeny indicates that the unknown parent must be heterozygous for both fur type and color, since two copies of a recessive gene are necessary for a recessive trait to be expressed in the offspring Hence the correct answer is choice A, big S little s, big W little w 54 The correct answer is choice A According to the law of independent assortment, the probability of a particular cross producing an individual of a particular genotype is equal to the PRODUCT of the individual probabilities of getting the particular genotype specified for each allelic pair In this question, we have two individuals who are heterozygous for each of the three genes, A, B, and C, and we want to know the probability of producing an individual that is homozygous dominant for each trait The way to this is to look at each gene individually Both parents are heterozygous for gene A; that is, their genotype is big A little a The ratio of their offspring with respect to gene A is 1/4 big A big A, 1/2 big A little a, and 1/4 little a little a This is the typical 1:2:1 ratio for all heterozygous crosses Since both parents are also heterozygous for genes B and C, the probability of their offspring being big B big B is 1/4, and the probability of their offspring being big C big C is also 1/4 Therefore, the 18 Kaplan MCAT Biological Sciences Test Transcript probability that a particular offspring will have the genotype big A big A, big B big B, big C big C, is equal to the product of the individual probabilities - 1/4 times 1/4 times 1/4 is equal to 1/64; so, choice A is the correct answer Passage IX (Questions 55–59) 55 The correct answer to this question is choice D This type of question is very common in persuasive argument passages - it requires you to understand both the similarities and the differences between the two theories presented in the passage Specifically, to answer this question, you need to determine which of the conformational states presented as answer choices are acceptable in the sequential model but NOT in the concerted model From the passage, you know that the sequential model does NOT assume equilibrium between the two conformational states In addition, you're told that the conformational state of one subunit in a protein can be different from the conformational state of another subunit In other words, intermediate forms of the protein can exist: one subunit can be in the R form while others remain in the T form In the concerted model, on the other hand, an equilibrium between the two conformational states does exist and all of the subunits of a particular protein must be in the same conformational state at any given time This means that if one subunit of a particular protein is in the R form, all of the subunits in that protein molecule must be in the R form The definitions of the T and R forms hold true for both models; the T form has low substrate-binding affinity and the R form has high substrate-binding affinity So the major difference between the two models is that the sequential model allows a mixture of R's and T's, while the concerted model does not Since we are dealing with a protein that consists of two identical subunits, the possible configurations are TT, RR, and RT; therefore, you should have immediately ruled out choice C, since this would be a configuration for a protein consisting of four subunits Of the three acceptable configurations, RT is the only one that could be found in the sequential model but not the concerted model Therefore, choice D is the correct answer Choice A, RR, and choice B, TT are wrong because these configurations are consistent with BOTH models Again, choice D is the correct answer 56 Choice C is the correct answer to this question This question may have seemed very difficult at first glance, since it looks like it involves complicated mathematical computations But remember, the MCAT asks few complicated computational questions, and most of these are in the physical sciences section Most questions are designed to make you think, not math problems This question deals only with the concerted model, so you can ignore the information in the passage about the sequential model You're dealing with an enzyme with two active sites that obeys the concerted model This means that at any given time, both binding sites on a given molecule must be in the same conformational state You're told that in the absence of substrate, the ratio of the T form to the R form is x 105 And from the passage, you know that in the absence of substrate nearly all of the molecules are in the T form Thus, this ratio simply means that in the absence of substrate, there are x 105 molecules in the T form for every molecule in the R form Therefore, the addition of saturating amounts of substrate will shift the T-R equilibrium, and almost all of the molecules will be now be in the R form So this means that there are now x 105 molecules in the R form for every in the T form Therefore, the ratio of R to T in the presence of saturating substrate is also x 105 Hence, the correct answer is choice C 57 The correct answer is choice B To answer this question you have to interpret the meaning of the two curves and then decide which of the two models best accounts for this meaning and why The curve labeled f sub R represents the fraction of the ATCase active sites in the R form, while the curve labeled Y represents the fraction of ATCase active sites with succinate bound So what would we expect these two curves to look like if ATCase conforms to the sequential model? We would expect that every active site occupied by succinate would be in the R form, while the other subunits would remain in the T form, until their active sites bound succinate In other words, you would expect the fraction of active sites in the R form to equal the fraction of active sites with substrate bound, which means that the two curves - f sub R and Y - would be identical Since this is clearly not the case, choices C and D must be wrong When a protein obeys the concerted model, this implies that when one of the subunits binds substrate - thereby converting it to the R form - all of the other subunits in the protein also convert to the R form In other words, you'll have subunits in the R form WITHOUT bound substrate Therefore, at non-saturating succinate concentrations, there will be a greater fraction of ATCase subunits in the R form - which includes R forms with and without succinate bound, than the fraction of ATCase with succinate bound How would this translate onto a graph? The f sub R curve should be to the left of the Y curve On the other hand, as you near saturating concentrations of succinate, all of the active sites are filling with substrate, and therefore the fR and Y curves should begin to converge And when all of the active sites are 100% filled, the two curves should meet Why? Because at this point, every subunit in the R conformation has substrate bound Therefore, choice B is the correct answer Choice A is wrong because you know from the passage that cooperative binding is consistent with both the concerted model and the sequential model Again, choice B is the correct answer 58 Choice C is the correct answer to this question To answer this question you must figure out which of the experimental observations would support only the sequential model So you need to understand the differences between the two different models Basically, the major differences between the models are: 1) all the subunits of a given molecule must be in the same conformation in the concerted model, while the subunits of a given molecule can 19 Kaplan MCAT Biological Sciences Test Transcript have different conformations in the sequential model; 2) in the concerted model, the T and R forms are in equilibrium, while no such conformational equilibrium exists in the sequential model; and 3) in the concerted model, the conformational change elicited by the binding of substrate to one subunit increases the substrate-binding affinity of the other subunits in a molecule, while in the sequential model, the conformational change elicited by the binding of substrate to one subunit can increase OR decrease the substrate-binding affinity of the other subunits in a molecule Now let's look at the answer choices Choice A supports the concerted model but not the sequential model, because the concerted model is based on a conformational equilibrium Therefore, choice A is incorrect Choice B supports BOTH models, since the addition of a saturating concentration of substrate would mean that all subunits would have substrate bound, which means that all subunits would be in the R form Therefore, choice B is also incorrect Choice C supports the sequential model only, since binding of substrate to one subunit can either increase OR decrease substrate-binding affinity of the other subunits in the molecule In the concerted model, the substratebinding affinity of other subunits can ONLY INCREASE after one subunit has bound substrate Therefore, choice C supports the sequential model, but not the concerted model; thus choice C is the correct answer Choice D doesn't really support either model In fact, denaturation of ANY enzyme would most likely prevent it from binding substrate Since protein denaturation really has nothing to with the allosteric models discussed in the passage, choice D is wrong Again, choice C is the correct answer 59 Choice B is the correct answer This is the type of question where you have to combine information provided in the passage and the question stem with your own outside knowledge In this case, you have to take what you know about the concerted model for allosteric proteins and apply it to the characteristics of hemoglobin behavior in various parts of the circulatory system Although you don't have to know that hemoglobin consists of four subunits, I'm going to use this fact in the following discussion, because it will make things easier to understand According to the question stem, the cooperative binding of oxygen to hemoglobin conforms to the concerted model This means that hemoglobin can exist in either the T form or the R form; there aren't any intermediate forms After the first subunit of a particular hemoglobin molecule binds oxygen and converts to the R form, the other three subunits also convert to the R form The T form has a lower affinity for oxygen than the R form, and in the presence of oxygen, the conformational equilibrium will shift in the direction of the R form Think about all of this for a moment This implies that the more oxygen present, the greater the ratio of R form to T form So the ratio of R to T hemoglobin molecules would be greatest when the partial pressure of oxygen is the greatest, and lowest when the partial pressure of oxygen is lowest Now let's run through a brief review, and I mean brief, of the circulatory system to remind ourselves where the partial pressure of oxygen is the greatest Deoxygenated blood is delivered to the right atrium by the superior and inferior venae cavae and the coronary veins The right atrium sends this blood to the right ventricle, which pumps it to the lungs by way of the pulmonary arteries So this is one of those rare instances where an artery is carrying deoxygenated blood The blood is oxygenated in the lungs and returned to the left atrium by the pulmonary veins The left atrium sends this blood to the left ventricle, which it pumps into systemic circulation by way of the aorta, and to cardiac muscle by way of the coronary arteries Therefore, of the four answer choices, it's clear that the left ventricle has the greatest partial pressure of oxygen and therefore the greatest ratio of R to T hemoglobin molecules So choices A, C, and D are all incorrect and choice B is the correct answer Passage X (Questions 60–64) 60 The correct answer to this question is choice A This question requires you to apply your knowledge of stereochemistry The diagram of the basic structure of a steroid is the most helpful piece of information in the passage As you can see in the diagram, the carbon skeleton consists of four cyclic systems attached to each other Three of these are cyclohexane rings, while the fourth is a cyclopentane ring Since the structure of the cyclopentane ring is the same in all four answer choices, we know we should focus on the conformation of the cyclohexane rings The most stable form that a 6-membered carbon ring can adopt is the chair conformation, which minimizes all types of strain Angle strain becomes negligible since the carbon-carbon bonds can adopt a tetrahedral angle of 109 the most stable angle possible for these sp hybridized carbons The puckered structure of the chair ensures that torsional strain is minimal, while steric strain is avoided because the hydrogens don't compete for the same positions in space With this in mind, you can narrow your choices to A or B, which each shows all three cyclohexane rings in the chair conformation Now we have to look at the substituents attached to the rings The key difference between choices A and B is that the methyl group and hydrogen in between the first and second ring are trans to each other in A, while they are cis to each other in B The trans relationship is lower in energy and so the formation of this is more favored If they were cis to each other, there would be more torsional strain in the molecule, which would make it less stable Remember that the question asks you which would be the most stable conformation Choice A is therefore the correct answer Now let's look at the wrong answers Choice B is wrong for the reason just described: the two substituents in between the first and second ring are cis to each other and so are subjected to more torsional strain As a result, choice B would be more unstable than A and is thus incorrect Choice C is wrong because one of the cyclohexane 20 Kaplan MCAT Biological Sciences Test Transcript rings is in the boat conformation as opposed to the chair conformation The boat conformation is much more strained than the chair conformation since it has eclipsed hydrogens Consequently, the conformation in choice C would have more torsional strain than the conformation in choice A Choice D is wrong for much the same reason as choice C The third cyclohexane ring is in the boat conformation and so is more highly strained than its chair counterpart in choice A Again, A is the right answer 61 The correct answer to this one is choice B This question deals with basic reaction mechanisms If you look at the reaction scheme and, more specifically, step 7, you can see that a proton is actually being lost from the molecule This is also mentioned in the notes following the reaction The question itself, however, states an important characteristic of E1 reactions namely, the dependency of the reaction on the substrate so make sure you pay special attention to this So, why is choice B correct? Well, as a related example, let's just briefly review the process of elimination in an alkyl halide This occurs by two steps loss of a halide to form a carbocation, followed by loss of a proton from the carbocation to form a double bond In cholesterol biosynthesis, the actual carbocation is formed way back in step 2, when the epoxide ring is opened and the first double bond in the molecule nucleophilically attacks a now electrondeficient carbon The key to answering the question, though, is being able to identify that a proton is lost in step 7, characteristic of the second stage of a typical elimination reaction Okay, we've now established elimination, but how can we say that it's unimolecular? The big clue here is in the wording of the question It states that the rate of step depends only on the concentration of the substrate This is in fact characteristic of unimolecular elimination If it were an E2 mechanism, the loss of a proton would be dependent on the concentration of the enzyme AND the substrate Consequently, you can rule out choice D in favor of choice B Choice A describes substitution and can therefore be discarded Unimolecular nucleophilic substitution and the E1 reaction are similar in that they share the same initial step namely, the formation of a carbocation SN1 differs, though, in that an incoming nucleophile adds to the carbocation to form a substituted product in the second step You can see from the nature of the product that there has been no substitution in the cyclohexane ring; so again, choice A is wrong Choice C is the most unrealistic answer for two reasons First, as we said before, the question suggests a unimolecular mechanism, and the SN2 reaction is bimolecular that is, dependent upon an incoming nucleophile and the substrate Second, this answer choice states that substitution takes place But, as we found, elimination takes place, and so again, choice B is the correct answer 62 The correct answer here is choice D Again, the most useful part of the passage is the reaction scheme If we look at step in greater detail, we can see that two processes actually occur First, the epoxide ring is opened and a hydroxyl functional group is formed; and second, cyclization of part of the alkene chain occurs In order to form the hydroxyl functional group, one of the carbon-oxygen bonds in the epoxide ring is broken Electrons from this bond are donated so a new oxygen-hydrogen bond can be formed In effect, the oxygen is protonated, and breakage of the carbon-oxygen bond means that the carbon becomes positively charged and susceptible to nucleophilic attack The double bond attacks the positively charged carbon and part of the molecule cyclizes to form a cyclohexane ring Therefore, the two processes occurring in step are protonation and nucleophilic attack, making choice D the correct answer Now let's look at the wrong choices Choice A states that protonation and cleavage take place This is technically correct, but it really only constitutes one of two processes so is not the best answer Protonation results as the epoxide ring is cleaved, and so one is the result of the other The answer choice fails to include the second important step, which is nucleophilic attack on the positively charged carbon, so choice A can be eliminated Choice B is also wrong Of course, ring opening does take place, but electrophilic attack is way off the mark You should know that donation of electrons from the carbon-oxygen bond will result in the formation of a positively charged carbon; so it can only be attacked nucleophilically, not electrophilically Finally, choice C is also wrong, since the molecule does not undergo hydration If hydration were to occur, this would involve the addition of molecule of water to the positively charged carbon Both hydration and nucleophilic attack cannot occur on the positively charged carbon, so choice C is incorrect Again, choice D is the right answer to question 62 63 The correct answer here is choice D This is pretty much an outside knowledge question in that you need to apply what you know about alcohols and their conversion to alkenes But are there any clues to help you? Well, the emphasis on high temperature and the presence of a strong acid hints to a dehydration reaction involving the hydroxyl group It is in fact the secondary alcohol group on the "A" ring that reacts with the acid This functional group will only react under severe conditions such as a temperature of 100 C in the presence of a strong acid The overall process involves loss of water to form a double bond Okay, what's happening here? Well, first the oxygen in the hydroxyl functionality is protonated to form an OH2 group This then comes off as a neutral water molecule, leaving a positive charge in its place The resulting carbocation can then eliminate a proton from either adjacent carbon to form the double bond The proton is actually eliminated from below the positive charge and so a conjugated diene In other words, there are now two double bonds separated by a single bond Elimination takes place at this position because conjugated dienes are more stable than non-conjugated dienes Therefore, you should be able to eliminate choice C for the reasons I have just mentioned a proton will be eliminated from below the positive charge to form the thermodynamically favored conjugated product Choices A and B are incorrect as well Vicinal diols are synthesized by the oxidation of an 21 Kaplan MCAT Biological Sciences Test Transcript alkene with cold potassium permanganate In case you had forgotten, a vicinal diol has hydroxy groups attached to adjacent carbons Geminal diols are usually formed by a base-catalyzed hydration of a carbonyl neither the reaction conditions nor the structure of cholesterol make this very likely A geminal diol has the two hydroxy groups attached to the same carbon Again, choice D is the correct response 64 Finally, let's look at question 64 The correct answer here is choice D As I said before, you have to draw on your knowledge of alkenes and stereochemistry to answer this one Aside from giving you the structure of cholesterol, the passage doesn't much to help you So, if you now look at the cholesterol molecule you can see that there are two major functionalities that can react a hydroxyl group and a double bond Your chemistry knowledge should tell you that Br2 will react with the double bond and not with the hydroxyl group You should also know that bromine will add across the double bond to form a dibromo product; so right away, you can discard answer choice B, which contains only one bromo substituent In the reaction, the bromine molecule approaches the cholesterol double bond and becomes polarized The double bond then attacks the positively polarized region of the molecule namely Br+ This adds to the double bond to form a bromonium ion intermediate Br− then approaches the opposite face of this cyclic cation and adds anti to the first bromine This means that in the final product, the bromines are trans to each other, not cis Now you can eliminate choice A, in which the bromines are cis to each other As I just said, the mechanism of addition is such that the bromines can only be trans to each other, so choice A is wrong Okay, we've now established that a dibromo product is formed; but what would the stereochemistry be like? Br2 approaches the molecule away from the methyl group because this is the less hindered side As a result, the bromonium ion intermediate is formed on the underside of the molecule Br− then adds from above the molecule, and the trans product pictured in choice D is formed Choice C is wrong because even though the bromines are trans to each other, a Newman projection along the carbon-carbon single bond attached to the bromine and methyl groups reveals that these two groups are eclipsed A staggered relationship, as in choice D, is more energetically favored; so choice C is incorrect Again, choice D is the correct answer Passage XI (Questions 65–72) 65 The answer to this question is choice C Before we begin looking at this question specifically, let's briefly review the passage and in particular Table If you don't understand the table, then the questions in this passage will be extremely difficult to answer The experiment can be summarized as follows: non-activated T-cells are grown both by themselves and with various compounds, either Tat, PHA or both, and the mRNA from these cells are isolated Remember that mRNA is transcribed from DNA So by examining the concentration of mRNA, you can extrapolate the amount of gene expression The mRNA is hybridized with a radio-labeled TNF probe, which will base pair with all mRNA transcripts from the TNF gene The more transcripts, the greater the amount of radioactivity Now that we understand the experiment a little bit better, let's look at the table There are really two sets of data - one set for the cultures involving Tat transcribed from Exon 1, and one set for the cultures involving Tat transcribed from Exon Although Culture a, cells alone, and Culture b, cells plus PHA, are part of each set of cultures, neither culture contains Tat But then why are there values for these cultures? Because those values represent the relative intensities of TNF gene expression in the T-cells In other words, T-cells express the TNF gene in the absence of Tat This makes sense, since TNF is a host gene and not a viral gene Let's return to Table For each set of cultures, there are values reported for two different quantities of mRNA extracted from the T-cells And if you look at the numbers in the table, you will see that the values reported for the 2.5 microgram samples of mRNA are approximately double the values reported for the 1.25 microgram samples This makes sense if you think about it If we use twice as much mRNA in the experiment, it stands to reason that there should be twice as many transcripts for any given gene, in this case, TNF, and therefore, twice as much radioactivity Instead of analyzing the entire table now, let's start doing the questions The interesting thing about Question 65 is that it's a Roman numeral question, which means that more than one item can be correct, and so you need to select all the correct items After you have identified one correct item, you can eliminate any answer choices that don't contain that item You're asked to identify the control cultures used in this experiment A control is a standard against which experimental observations may be evaluated, and as such, is identical in experimental procedure except for the ABSENCE OF THE FACTOR BEING STUDIED Based on this definition, let's look at Roman numeral I, Culture a, which is the cells alone This is obviously a control because unless you know the level of TNF present in nonactivated cells alone, you have no frame of reference to judge what effects the other compounds - PHA and Tat - are having on the cells Therefore, Culture a must be a control Since we have identified Roman numeral I as a correct choice, we can automatically eliminate choice B, Roman numeral III only Since III doesn't appear in any of the other answer choices, we can't eliminate any other answers Since the whole point of the experiment is to determine what effect Tat has on host immune response, any culture containing Tat would not be a control, but a rather an experimental variable This means that Cultures c1, c2, d1, and d2 are not controls So, in addition to eliminating Roman numeral III, we can eliminate IV as well; and so choice D must also be wrong Let's just look at Roman numeral IV for a second - the two d cultures PHA plus Tat is NOT a control because it shows the effect of the Tat peptide on activated cells We know this because we're told that PHA activates cells Well what about Roman numeral II, Culture b, which has the cells plus PHA First of all, since it's known that PHA induces T-cell activation, 22 Kaplan MCAT Biological Sciences Test Transcript and that TNF is involved in T-cell activation, you would expect HIGHER TNF levels when PHA is added to the cells than in Culture a And as you see from Table 1, this is indeed true In addition, Culture b serves as a control for the d cultures, which contain both PHA and Tat There is no way to determine what effect Tat has on activated cells, which is what the cells of Culture b are, if you don't know what the level of TNF is in activated cells in the absence of Tat So Roman numeral II is also a control Because I and II are correct, choice C is the correct answer 66 The correct answer is choice D All you have to to answer this question is read values off of Table After looking at the table, ask yourself which of the four samples listed in the answer choices has the fewest TNF transcripts? How can you determine this from the table? Well, "TNF transcripts" refers to mRNA transcripts of the TNF gene isolated from the cultures And a radio-labeled probe specific for TNF was used to figure out the number of TNF transcripts Although it's not possible to determine the absolute number of transcripts of TNF from the table, you can deduce the relative quantities In other words, you can tell if, for a given sample size, Culture a has twice as many transcripts of TNF as does Culture b, for example According to the passage, the intensities of the bands are directly proportional to TNF gene expression This means that the lower the value in the table, the less TNF gene expression, and therefore the fewer TNF transcripts So, to answer this question, all you have to is look up the values for each answer choice and select the answer with the lowest number According to the table, the values are: choice A, 33.70; choice B, 164.78; choice C, 69.98; and choice D, 20.54 Since choice D has the lowest value, this means that there are the fewest number of TNF transcripts present in this culture Therefore, choice D is the correct answer 67 Choice B is the correct answer In order to answer this question you must interpret the values in Table This makes the question a little harder than the previous one By looking over the answer choices, you see that one of the distinguishing features is this concept of activated versus non-activated cells So before we anything else, let's figure out what this means You're told that the experiment starts with non-activated T-cells from HIV-negative donors So all the cells are initially non-activated However, according to the passage, PHA, which is this substance known as a mitogen, activates T-cells This means that Cultures a, c1, and c2 contain non-activated cells, while Cultures b, d1, and d2 contain activated cells, since PHA is present in these cultures The next thing you have to is determine whether TNF gene expression is repressed, stimulated, or unaffected by Tat Repressed, or stimulated, or unaffected compared to what? Well, this is where the control cultures come into play: you need to compare the experimental cultures to the control cultures The appropriate comparisons for this experiment are: Culture a is the control for Cultures c1 and c2, and Culture b is the control for Cultures d1 and d2 If you are confused about how we determined which were the control cultures, please review the explanation to question 65 To determine the effect of Tat on non-activated cells, Cultures c1 and c2 are compared to the control, Culture a And to determine the effect of Tat on activated cells, Cultures d1 and d2 are compared to the control, Culture b Now that we've established this, let's see if we can discern any trends in the table Remember that the values in the table are directly proportional to TNF gene expression In the comparison of Culture c1 and c2 to Culture a, you see that Tat from Exon stimulates TNF gene expression in non-activated cells, while Tat from Exon does not In the comparison of Cultures d1 and d2 to Culture b, you see that Tat from both Exon and Exon represses TNF gene expression in activated cells Since the question only specifies Tat peptide from Exon 1, you can ignore any data about Tat from Exon With this information, let's see if we can eliminate any answer choices Well, since Tat DOES affect TNF gene expression, choice D must be incorrect In addition, we just determined that Tat stimulates TNF gene expression in non-activated cells Therefore, choice A is incorrect We also know that Tat represses TNF gene expression in activated cells Based on this, you can rule out choice C So, by the process of elimination, choice B is the correct answer Notice that we only had to look at the first part of the answer choices to find the correct answer The second half of the answer may have helped to identify the correct answer, but it wasn't needed Anyway, this is the very mechanism by which Tat is believed to operate - by activating non-activated cells, Tat paves the way for HIV infection Again, choice B is the correct answer 68 The correct answer is choice D To solve this question, you have to apply your interpretation of the experimental data to a new situation You're asked to determine which of the clinical observations best corresponds to, or most supports, the results of the experiment Right off the bat, you can eliminate choices A and C because they refer to HIV-negative patients Remember, Tat is an HIV protein and would therefore be found ONLY in HIVpositive patients Although the experiment uses the T-cells of HIV-negative donors, exposing these cells to Tat in vitro simulates the events that occur in an HIV-positive patient when HIV emerges from the latent state Therefore, there could NOT be any clinical observations in an HIV-negative patient that would relate to the results of this experiment Well, that leaves you with two choices, B and D - so you've already increased your chances of getting the question right to 50% Let's look at choice B, which has an HIV-positive patient with a reduced concentration of Tcells You're probably familiar with the fact that active HIV infection is characterized by a low T-cell count However, if knowing this led you to choose choice B, then you made the mistake of bringing your outside knowledge into this question Answer choices like choice B are known as distracters, or "seducers"; don't be fooled by them Remember, you're asked to choose the observation that supports the results of the experiments outlined in the passage Since the experiment does not address T-cell viability in HIV infection or the effect of Tat protein on T-cell viability, such an observation would neither support nor contradict the experiment Therefore, choice B is incorrect and choice D is the correct answer The T-cells of an HIV-positive patient suffering from the flu would be activated 23 Kaplan MCAT Biological Sciences Test Transcript You know this because the passage stated that antigens, such as the influenza virus responsible for causing the flu, activate T-cells So this scenario is similar to that of Cultures d1 and d2 - cells treated with Tat and PHA, which as a mitogen, activates T-cells Since the cells in Cultures d1 and d2 were activated by PHA, their TNF levels were low in comparison to Culture b, the control culture Therefore, this clinical observation would be consistent with the experimental data and would therefore support it Even if you didn't see the connection between having the flu and activated T-cells, you could have still found the right answer by the process of elimination Again, choice D is the correct answer 69 Choice C is the correct answer This is one of the easier questions in this passage, since the answer can be found directly in the passage According to the last two sentences of the passage, the values reported in Table are relative intensities directly proportional to the amount of TNF gene expression Since gene expression can be measured by the quantity of mRNA transcripts of the TNF gene, choice C is the correct answer Let's look at the other answer choices and see why they're wrong Choice A, the concentration of the Tat peptides, is part of the experimental design ng/mL and 100 ng/mL concentrations were used to determine the effect that the peptides have on host immune response as measured by TNF gene expression Remember, the experiment was measuring the effect of Tat, not measuring Tat itself Therefore, choice A is incorrect Choice B is also incorrect The experiment was not concerned with T-cell viability, nor were there any means employed in the experiment to measure the number of viable T-cells; so choice B is also incorrect Choice D, the number of HIV-infected cells, is also wrong To begin with, all of the cells used in this experiment were HIV-negative Furthermore, only a single protein from the HIV virus was used - the Tat protein is not solely responsible for HIV infection - viral nucleic acid and other viral proteins are also a requirement Hence, the non-infected cells used in the experiment could not become infected with HIV simply by exposure to Tat Again, choice C is the correct answer 70 Choice A is the correct answer To begin with, you have to discern the trend observed in the Cultures c1 and c2 treated with Tat from Exon - that is, the difference between the effect of 100 ng/mL of Tat from Exon on nonactivated T-cells vs the effect of ng/mL of Tat from Exon on non-activated T-cells If you look at Table 1, you see that at either 1.25 micrograms or 2.5 micrograms of mRNA, there is GREATER TNF gene expression with 100 ng/mL of Tat than with ng/mL of Tat Therefore, it can be inferred that the GREATER the concentration of Tat from Exon added to non-activated cells, the GREATER the concentration of TNF mRNA In other words, in nonactivated cells, the amount of TNF gene expression is DIRECTLY proportional to the concentration of Tat from Exon Well now that we know this, all we have to is find the most contrary, or opposite, trend Since in nonactivated cells, the greater the Tat concentration the greater the TNF expression, you might have expected Tat to have the same effect on the activated cells of Cultures d1 and d2 However, if you look at the experimental results for the d cultures, the opposite occurs: in activated cells, the GREATER the Tat concentration the LESS TNF expression In other words, there IS greater repression of TNF gene expression at the LOWER concentration of Tat from Exon This is contrary to the effect of the differing Tat concentrations on Cultures c1 and c2 Therefore, choice A is correct and choice B is wrong While choices C and D are true - there is greater TNF expression in the larger mRNA sample sizes and in Culture b than in Culture a - neither is unexpected in light of the effect of Tat concentration on nonactivated cells So choice C and choice D are both incorrect Again, choice A is the correct answer 71 The correct answer to this question is choice A This type of question requires you to extract information from the passage and combine it with your introductory biology knowledge This question deals with cell growth curves, which should be a familiar topic A growth curve plots number of cells versus time The rate of growth can be determined by looking at the slope of the curve The steeper the curve, the faster the growth, while the more gradual the curve, the slower the growth Next, you need to figure out how fast the cultures described in the question stem will grow relative to one another How can this be determined? Well, according to the passage, TNF acts as a growth factor for helper T-cells And since the data from Table indicates that Tat from Exon increases TNF expression in non-activated cells, while Tat from Exon does not seem to affect TNF expression, this information can be applied to the study of growth in cell cultures containing helper T-cells Why? Because if Tat affects TNF expression, and TNF expression affects helper T-cell growth, then it follows that there is a relationship between these three factors The more TNF, the greater the rate of helper T-cell growth So, if T-cells plus Tat from Exon results in greater TNF expression than in T-cells alone, we would expect helper T-cells plus Tat from Exon to result in greater helper T-cell growth than in helper T-cells alone Likewise, since T-cells plus Tat from Exon results in the same amount of TNF expression as in T-cells alone, we would expect helper T-cells plus Tat from Exon to result in the same amount of TNF expression as in helper T-cells alone So, this means that the curves for the cells alone and the cells plus Tat from Exon should be identical and to the right of the curve for the cells plus Tat from Exon 1, which should also be steeper than the other curve since its growth will be faster Therefore, choice A is the correct answer 72 Choice B is the correct answer In the question stem you're told that the Tat protein found circulating in the blood of an HIV-positive patient has a molecular weight less than half that of the peptide synthesized from the Tat gene inserted in a prokaryotic cell Why the difference in molecular weight? The key to this is found in the first paragraph of the passage - the Tat protein is encoded by two exons spliced together Well the only way that you get two exons spliced together is if the RNA that linked them together was excised Now you don't have to know the 24 Kaplan MCAT Biological Sciences Test Transcript specifics of HIV genetics to answer this question, but I'm going to give you a little background information to make things easier to understand The two exons that code for Tat are separated by a sequence of DNA that codes for other viral protein products When Tat needs to be synthesized, ALL of this DNA is transcribed - the Tat exons plus the DNA that links them However, since the desired protein product is Tat, the RNA coding for the other proteins is excised and the two Tat exons are spliced together All of this occurs within the nucleus of the eukaryotic host cell remember, you're dealing with a retrovirus, which must integrate into the host genome before the viral genes can be expressed Therefore, the initial mRNA transcript is longer than the transcript from which Tat is translated Prokaryotes not posttranscriptional modification The initial mRNA transcript is directly translated there isn't any excision and splicing going on here Prokaryotes are INCAPABLE of such processing Remember, in prokaryotes, there is no nucleus to separate transcription from translation Therefore, if the viral DNA that contains the Tat exons was inserted into a prokaryote, the mRNA transcribed and translated by the cell would consist of the codons coding for several viral proteins - not just Tat And this means that the protein synthesized from this mRNA transcript will be much longer and much heavier than the protein synthesized in the HIV-positive patient Therefore, this prokaryotic protein couldn't rightfully be referred to as Tat So now let's take a look at the answer choices The molecular weight of the protein would not be influenced by interaction with TNF - there's no information in the passage that would lead you to believe this; so choice A is wrong Choice B basically sums up in a nutshell what we were just discussing about the differences between eukaryotic and prokaryotic protein synthesis, and is obviously the correct answer Random mutations during synthesis could have caused the protein to have the same, a greater, or a smaller molecular weight depending on the nature of the mutation However, because mutations are a RANDOM event, it is highly unlikely that all of the prokaryotic DNA would undergo mutations that resulted in a peptide twice the size of its eukaryotic counterpart This is clearly NOT the most likely explanation, and so choice C can be dismissed As for choice D - nowhere in the question stem does it say that the prokaryotic cells were exposed to PHA Furthermore, exposure to PHA does not affect Tat weight or the weight of any proteins, for that matter; PHA activates lymphocytes Therefore, choice D is incorrect Again, choice B is the correct answer Discrete questions 73 The correct answer is choice C Answering this question correctly was dependent on your knowledge of cyanide Cyanide is a poison that interferes with the electron transport chain of the inner mitochondrial membrane by binding to one of the electron transfer complexes In doing so, cyanide completely inhibits the flow of electrons and effectively paralyzes the transport chain As a consequence, the proton pump comes to a stop, and ATP cannot be generated aerobically In addition, electron carriers such as NADH and FADH2 can't deliver their high energy electrons to the electron transport chain because it has become backed up This means that NAD+ and FAD are not regenerated, and aerobic respiration can't continue Getting back to the question; energy is required for a bacteriophage, or any organism for that matter, to replicate DNA replication, RNA transcription, and protein synthesis all require ATP and are processes necessary for viral replication, as well as for host cell function If aerobic ATP formation is inhibited, there will not be enough ATP for normal cell functions OR for viral replication Therefore the correct answer is choice C Choices A, B, and D are incorrect because they not correctly describe cyanide's mechanism of action Again, choice C is the right answer 74 The correct answer is choice D At first glance, this might have seemed like a complicated question - it has a long question stem and some unfamiliar terms But in actuality, this is a pretty simple question This is a common trick used on the MCAT - don't be fooled by the perceived level of difficulty Anyway, back to the question If you remove the duodenum, which is the first segment of the small intestine, and tie off the ducts that drain into it, what would you be missing? Let's look at each answer choice in turn How would pepsin be affected by duodenum removal? The answer - there would be NO effect at all since pepsin is synthesized and secreted in the stomach Therefore, choice A is incorrect Choice B, water absorption, would not be affected very much, since water absorption is mainly a function of the latter two segments of the small intestine - the jejunum and the ileum - as well as a function of the large intestine So, choice B is also incorrect Mechanical digestion, choice C, would not be affected by duodenum removal, since this is a function of the chewing action of your teeth and the churning of your stomach Therefore, choice C must be wrong Choice D, fat digestion, WOULD be affected by duodenum removal, because with the ducts to the duodenum tied off, the intestines would not receive bile from the gallbladder or lipase from the pancreas Both bile and lipase are major players in fat digestion Bile is produced in the liver and stored and concentrated in the gallbladder before being released into the duodenum by means of the common bile duct Bile emulsifies fats, thereby making them more susceptible to the action of lipase, the pancreatic enzyme that digests fats Lipase enters the duodenum by means of the pancreatic duct So with the duodenum missing and its ducts tied off, fat digestion would be dysfunctional Therefore the correct answer is choice D 75 The correct answer for question 75 is choice B For this question you need to find the answer choice that is false, not true Make sure that you read questions stem carefully if you're not careful, you could easily pick the first "true" answer choice that you come to Before we look at the answer choices, let's just clarify what a bromonium ion is It's actually a cyclic cation where two adjacent carbons are attached to a bromide ion as well as each other The 25 Kaplan MCAT Biological Sciences Test Transcript bromonium ion is formed when the nucleophilic double bond induces a positive charge on a bromine atom of a bromine molecule This electrophilic bromine atom adds to one of the carbons of the double bond forming a positive charge on the other carbon The bromine then bonds to this carbon and assumes the positive charge A nucleophile then adds to one of the carbon atoms of the bromonium ion ANTI to the bromine atom In other words, a trans compound is ALWAYS formed So how we know that choice B will not proceed by this intermediate? Well if you know your alkenes you should know that HBr will add to propene by a Markovnikov mechanism Firstly, hydrogen adds to the carbon with the fewest alkyl substituents which would obviously be the double bonded CH2 As a result, the intermediate formed is CH3CH+CH3 Br− then adds to the central carbon to form the alkyl halide, 2bromopropane So, the intermediate formed here is in fact a carbocation not a bromonium ion making choice B the correct answer The reactions in answer choices A, C, and D all proceed by bromonium ion intermediates making them all incorrect If you're not sure of the chemistry, another way to look at the answer choices is to see that all of the choices except B have Br2 as one of the reactants Since you are looking for the one reaction that is incorrect, B is a good one to choose 76 Let's now look at question 76 The correct answer here is choice D This question tests your knowledge of stereochemistry If you look at the two compounds, you can see that the difference between them is the orientation of the methyl group on the ring In the first compound, the methyl group is in an axial position whereas in the second one, it is in an equatorial position The tertiary butyl group that is attached to the ring is very bulky and tends to remain in the equatorial position in order to avoid steric strain Anyway, if you look at the relationship between the two ring substituents you can see that in the first case they are cis to each other, whereas in the second case they are trans to each other In other words, these two compounds are cis-trans isomers and so choice D is the correct answer Now lets look at the wrong answers Choice A is wrong as the definition of an enantiomer is a nonsuperimposable mirror image If you take a closer look at the molecules you can see that yes, they are nonsuperimposable but they are not mirror images of each other If they were mirror images then the methyl groups would either be in equatorial or axial positions in both molecules Therefore, choice A is wrong Choice B is wrong as well Even though geometric isomers are fundamentally cis and trans isomers, they are cis and trans according to the position of the substituents attached to a double bond There are no double bonds present in these molecules, and so they can only be described as cis-trans isomers and not geometric isomers Finally, lets look at choice C Meso compounds are molecules that contain chiral centers and are superimposable on their mirror images Since the compounds in question are not mirror images of each other, they are not meso Again the correct response to this question is D 77 The correct answer to this question is choice D To answer this question correctly you've got to be familiar with the rules of base pairing in DNA Adenine always pairs with thymine, and guanine always pairs with cytosine With this information in mind, let's look at the question Treatment of DNA with 2-aminopurine causes an adenine to be substituted for a guanine This means that after treatment with 2-aminopurine, the guanine-cytosine base pair is replaced with an adenine-cytosine base pair However, the cytosine in the complementary strand does not correctly base pair with the adenine So after replication, the cytosine will be replaced with a thymine This substitution of a thymine for a cytosine will allow the adenine to base pair with thymine This situation is an example of how a single point mutation, the adenine for a guanine, is incorporated into future generations If the mutation had been corrected before replication, there would have been no change in the DNA base sequence Again, choice D is the correct answer 26 ... experiment in the question stem sets up four test tubes, numbered I through IV, with each test tube containing a different blood group To each of these test tubes, type A red blood cells were added... definitions from introductory biology This is probably the most common type of MCAT question you need a little information from the passage and a little information from introductory biology to solve the... of these test tubes will form precipitate upon the addition of type A red blood cells Precipitation occurs when the blood antigens react with their specific antibodies Therefore, the test tubes