Biology secton test (12)

Choice C, amino acid synthesis, would obviously not yield purines; hence choice C is also incorrect.. The best approach to this type of question, which requires you to evaluate all of th

MCAT Section Tests

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Passage I (Questions 1–6)

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 1 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

2 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 do 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

3 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

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

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

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

5 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 do 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

6 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)

7 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,

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

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 do 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

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 hemi-methylated 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

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 do 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

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 6 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

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 2 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 heptane 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

n-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 do 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 1 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 1 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 non-aromatic 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

acylation, even though the reaction does proceed by a Friedel-Crafts mechanism Once again, the correct answer is choice C

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 1 and 2; notice that you don't call these carbons 4 and 5 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 2 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 4 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 1 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 7

in the case of n-heptane down to 6 in the case of 2-methylhexane It makes sense, then, that 2-methylbutane, a carbon 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

4-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 2-methylbutane 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

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-for-gram basis So, choice D is wrong for this reason Again, choice C is the correct answer

Figure 1 According to the passage and the figure, the levels of protein synthesis are approximately the same for the period of time shown in Figure 1 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 yo u have to do 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

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 do 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

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 9 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

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 do 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 do 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

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 non-differentiated 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 do 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 do 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

groups: A, B, AB, and O There are three alleles determining these four groups: alleles A and B are codominant, while allele O is recessive The A allele codes for the A antigen on the red blood cell surface, the B allele codes for the B antigen, and the O allele does not code for any antigen So, a person with type A blood has the genotype AA or

AO, a person with type B blood is either BB or BO, a person with type AB blood has the genotype AB and a person with type O blood is OO A person's blood serum does not contain antibodies to any of its own antigens, but does contain antibodies to the other blood antigens This means that a person with type A blood will have antibodies to the

B antigen in their blood serum and a person with type B blood will have antibodies to the A antigen in their blood serum What about a person with type AB blood? Well, they would have neither anti-A nor anti-B antibodies since their red blood cells carry both antigens People with type O blood, on the other hand, will have both anti-A and anti-

B antibodies since their red blood cells have neither the A nor B antigens The presence or absence of certain antibodies allows for a quick and easy determination of blood groups

The 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 To answer this question, you must determine which 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 containing blood serum with anti-A antibodies will form precipitate The two blood types that will produce anti-A antibodies are type B and type O blood Therefore, precipitate will form in test tube II, which contains type B blood, and test tube IV, which contains type O blood Precipitate will not form in the other blood types because they do not contain the anti-A antibody So the correct answer is choice C

Passage V (Questions 28–33)

28 Choice A is the correct answer To answer this, you need to know your practical stuff and, more specifically, why sodium bicarbonate is frequently used in both syntheses and extractions You obviously need to search for the point in the passage where sodium bicarbonate is added, although the passage doesn't explain the purpose of this step

In the experiment, aspirin is formed when the solution of acetic anhydride, salicylic acid, and sulfuric acid is heated When cold water is added, it becomes pretty clear that the crude white precipitate contains aspirin The fact that it is a precipitate indicates that aspirin is relatively insoluble in water

If we take a look at the structure of aspirin in greater detail, we can see that it has two functional groups those of a carboxylic acid and an ester Therefore, upon dissolving product X in sodium bicarbonate, what would you expect to happen? You should be able to see that aspirin will dissolve in this solution, as the carboxylic acid functionality will dissociate from –COOH into –COO−Na+ This is possible since the hydrogen on the carboxyl group is slightly acidic In other words, aspirin has been converted into its corresponding sodium salt You should be aware that salicylic acid is present in product X as an impurity and will also be converted to its corresponding sodium salt it is removed later in the recrystallization step By filtering, the aspirin salt is then isolated in the filtrate and can be converted back to its solid form by acidification with hydrochloric acid So again, the correct choice is A

Now for the wrong choices Choice B is incorrect, since the contaminant cannot form a water soluble sodium salt Even if you don't know what the nature of the contaminant is, you should already know that aspirin will form a water soluble sodium salt due to reasons already discussed If the contaminant could be dissolved in water as its corresponding salt, it would mix with aspirin Consequently, aspirin would not be isolated, and this step would be useless Choice C is also incorrect Since Reaction 1 is an equilibrium, there will be some salicylic acid present in product X any remaining acetic anhydride is removed during the water washings Consequently, any residual salicylic acid will go on to form a sodium salt and so go into the filtrate along with soluble aspirin As I said earlier, salicylic acid is actually removed during the recrystallization step and not in the sodium bicarbonate step, so choice C

is wrong Finally, let's look at choice D This answer choice states that the purpose of using sodium bicarbonate is to prevent the hydrolysis of aspirin back to salicylic acid We already know that all sodium bicarbonate will do is convert aspirin into its corresponding sodium salt Hydrogen ions are present at most stages of the reaction, even after sodium bicarbonate has been added As long as these hydrogen ions are present, there will always be trace levels

of salicylic acid due to this backward reaction, which we discussed earlier Consequently, sodium bicarbonate will not prevent hydrolysis, and so choice D is wrong Again, choice A is the correct answer

29 The correct answer here is choice B For this question, you really need to look closely at the structure of salicylic acid and think of other possible reactions it could undergo in acidic conditions, since the contaminant must

be a derivative of it The biggest clue in the passage is the nature of the functional groups in salicylic acid

A key problem in the synthesis of aspirin is that salicylic acid possesses two functional groups that not only can react with other molecules, but can also react with each other This is exactly what happens; the contaminant is the product of the hydroxyl group of one salicylic acid molecule forming an ester linkage with the carboxyl group of another How does this mechanism work? Well, you should be able to see that the oxygen belonging to the hydroxyl group in salicylic acid can nucleophilically attack and add to the carboxyl carbon of another salicylic acid molecule, thus forming a protonated intermediate Water can then be lost from this intermediate, and so an ester linkage forms between the two molecules This then leaves a hydroxyl and carboxyl group at each end free to react further Theoretically, an infinite number of molecules can react to form this repeating unit, which consists of a benzene ring attached to an ester moiety In practical terms however, the length of the polymer is influenced by a number of factors such as depletion of the substrate

If you draw out part of the polymer, you should be able to mark off the repeating units Instead of drawing them all out, you can simply draw out one that can then be boxed off and designated the letter n This letter stands for how many repeating units are in the molecule, and as I said earlier, can be anything from 1 to infinity You don't have

to worry about this, though; basically, if you can identify that a polyesterification reaction has taken place between adjacent salicylic acid molecules, you have nearly cracked it All you need to do now is take a look at what lies outside the repeating unit It should be obvious that those functionalities that lie outside the repeating unit are a hydroxyl group attached to the benzene ring and a hydrogen attached to the repeating ester group Added together, these give water, which is not an integral part of the polymer If you remember that the reaction acid plus alcohol forms an ester plus water, this all adds up to choice B

Choice A is wrong because the repeating unit does not contain an ester linkage but a carbonyl linkage instead The reaction between both functionalities, as explained earlier, results in the formation of an ester linkage and not this carbonyl linkage; so there is a subtle difference between choice A and choice B In this answer choice, the repeating unit omits a single oxygen, which incorrectly lies outside the brackets Consequently, choice A is wrong Choices C and D are also wrong To form either of these polymers, there would have to be 3 functional groups attached to the benzene ring Additionally, if there were 3 functional groups, they would all be likely to react with adjacent groups Therefore, you wouldn't have these straight chain polymers but rather branched polymers This observation is incidental, though The point here is that for these polymers to exist, the salicylic acid molecule would have to gain an additional hydroxyl group in the case of choice C, or an additional carboxyl group as in choice D However, you would not get these additional groups attached to the ring, since the reaction conditions aren't right for aromatic substitution Again, choice B is the correct answer

30 The right answer here is choice C In answering this question, it would probably be helpful if you tried to draw out the reaction, since it is easier to visualize on paper The only part of the passage that is really useful in answering this question is Reaction 1 You have to look at the structure of the reactants and products and independently decide what type of reaction has occurred Let's look at the reaction in more detail In the synthesis of aspirin, 1 molecule of salicylic acid reacts with 1 molecule of acetic anhydride As in the process of polyesterification described in question 29, the nucleophile in this reaction is the oxygen in the hydroxyl functionality

of salicylic acid This oxygen attacks and adds to one of the carbonyl carbons in acetic anhydride Since acetic anhydride is a symmetrical molecule, it doesn't matter which carbon is attacked In order to facilitate this attack, the carbonyl oxygen of acetic anhydride is protonated by sulfuric acid, which makes it even more prone to nucleophilic attack A series of intermediates are formed, eventually leading to the expulsion of acetic acid and the formation of aspirin So, overall, we can see that the proton on the phenol group of salicylic acid has been replaced by an acyl group obtained from acetic anhydride The oxygen of the phenol group is the nucleophile This constitutes nucleophilic acyl substitution, and again, choice C is correct Choices A and B can be discarded simply because the word aromatic is included You should be able to see that a reaction occurs not on the benzene ring, but rather with the functional groups attached to it Under these relatively mild reaction conditions, the benzene ring is unlikely to be attacked electrophilically or nucleophilically Therefore, choices A and B are wrong Choice D is also wrong since

no addition reaction takes place From the nature of the product, it is again obvious that the hydrogen on the phenol group has been replaced by an acyl group Thus, this is classed as nucleophilic substitution and not nucleophilic addition

31 The correct answer here is choice C Just as in question 30, this tests your knowledge of nucleophilic acyl substitution, and so you cannot really find the answer in the passage The most helpful part of the passage is probably Reaction 1 again, which you can relate to the situation described in the question

By now, you are fully aware that salicylic acid has two different functional groups an alcohol and a carboxylic acid, both of which can undergo esterification reactions In the synthesis of aspirin discussed in the passage, it is the alcohol functional group of salicylic acid that reacts with acetic anhydride to form an ester The carboxylic acid functionality participates in esterification when it is reacted with an excess of alcohol in an acidic environment In this case, the oxygen in methanol nucleophilically attacks the carboxyl carbon in salicylic acid In other words, salicylic acid is now acting as a carboxylic acid in the reaction This reaction mechanism is quite similar

to that of salicylic acid and acetic anhydride, in that a series of protonation and deprotonation steps eventually leads to the formation of the ester, methyl salicylate Again, this process can be classed as nucleophilic acyl substitution, but this time it is the hydroxyl on the carboxylic acid functionality that has been replaced, not the hydrogen on the hydroxyl group

For this particular question, the alternatives are very easy to discard Let's take a look at choice D first This states that benzoic acid will be formed upon reaction of salicylic acid with methanol In order for this to occur, the hydroxyl group would have to be removed from the aromatic ring This is highly unlikely, because the functional groups would react in preference to the actual aromatic ring; so the hydroxyl functionality will remain attached to the ring and benzoic acid will not be produced Likewise, if choice B were correct, not only would the hydroxyl group have to be removed from the ring, but the carboxyl group would also have to be reduced to an aldehyde The reaction conditions are far from ideal for either of these processes to occur, so choices B and D are wrong Choice A is also incorrect If phenol were formed, salicylic acid would have to undergo decarboxylation Again, under these reaction conditions, the process of decarboxylation is highly unlikely At high temperatures, only 1,3-dicarboxylic acids and β-keto acids spontaneously decarboxylate Salicylic acid is neither, and so would not lose its carboxyl group Just as in

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

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 1 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 do 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)

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 1 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 do They give you some of the information