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 1–4) The correct answer is choice C During transcription, DNA acts as a template for the synthesis of a strand of mRNA The segment of DNA that acts as the template for transcription is called the anti-sense strand, while the nontemplate strand of DNA, which is identical in sequence to the mRNA being transcribed (except for the presence of uracil in place of thymine) is called the sense strand The synthesis of mRNA is catalyzed by the enzyme RNA polymerase, which moves along the anti-sense strand in the 3' to 5' direction, and synthesizes the mRNA in the 5' to 3' direction Free nucleotides pair up with their complementary bases: guanine pairs with cytosine, thymine pairs with adenine, and adenine pairs with uracil So, to answer this question, you need to determine the base sequence of the DNA segment from which this mRNA was transcribed You should have immediately eliminated choices A and D because they both contain uracil, which, as you know, is only found in RNA To distinguish between choices B and C, you have to follow the basepairing rules and take the polarity of nucleic acid into account And, if you did so correctly, you should have obtained the following sequence of bases: CTCAGTCCTTTATCATA in the 5' to 3' direction, which is choice C If you chose choice B, it's because you forgot about the polarity So, choice B is wrong and choice C is the correct answer Choice C is the correct answer This question concerns the process of translation, in which an mRNA transcript is translated into an amino acid sequence Translation occurs at the ribosome Nucleotides are read in triplets known as codons, and each codon codes for a particular amino acid The chart of the genetic code matches the 64 mRNA codons with the amino acids to which they correspond The first letter, or base, of a codon is the left-most letter, in the 5' to 3' direction; the second base of a codon is in the middle; and the third base is the right-most letter As you can see, and as you should know, the genetic code is redundant; that is, an amino acid typically has more than one codon specifying it There are also four codons that don't strictly code for amino acids; the codon AUG codes for the amino acid methionine, but also signals initiation of translation Therefore, methionine, or Met, is always the first codon to be translated, and is therefore always the first amino acid in the peptide chain, prior to peptide modification and processing following translation During this process, the terminal Met is typically cleaved There are also three codons - UAA, UAG, and UGA - that signal termination of peptide synthesis and NOT code for any amino acid All of this is right there on the chart for you, so it's not like you needed to remember all of this stuff to answer the question A ribosome begins translation by reading the mRNA transcript from left to right, in the 5' to 3' direction, until the start codon - AUG is read You knew that this transcript must start with the codon AUG because you're told in the passage that this mRNA fragment contains the first coding nucleotide of v-src And the first coding nucleotide will be the first base of the first codon If you were told that this mRNA fragment did not contain the first codon, then you would start with the first three bases as the first codon, you wouldn't need to start with AUG In this particular sequence, A, U, and G are the second, third and fourth bases, respectively The second codon consists of the fifth, sixth, and seventh bases, and so on and so on Following AUG are the following triplets: AUA, which codes for isoleucine, AAG, which codes for lysine, GAC, which codes for aspartic acid, and UGA, which is a noncoding codon signaling the termination of translation Therefore, the sequence of amino acids in the peptide chain translated from this strand of mRNA will be: methionine, isoleucine, lysine, and aspartic acid So choice C is the correct answer If you chose choice A, then you probably forgot that the start codon AUG is necessary for initiation If you chose D, you probably misread the third codon, which is AAG, as AAC, which codes for asparagine Choice B is the sequence obtained if you translated the mRNA beginning with the first three bases, which are UAU, and so on, and then added on a methionine to the beginning of the sequence Again, the correct answer is choice C The correct answer to this question is choice B From the question stem you know that you need to find the experimental observation that most supports the model of cell transformation by v-src According to the passage, vsrc phosphorylates proteins with its kinase activity, but can only transform cells if it is bound to the plasma membrane Now that we've refamiliarized ourselves with the model, let's look at the answer choices Choice A is wrong because the substitution of the N-terminal amino acid of v-src, which rendered the cell incapable of kinase activity, should have prevented v-src from transforming cells Choice C is wrong for the exact same reason; it doesn't matter which of the terminal amino acids - the C or the N - is substituted as long as the end result is the same Choice B is correct because, assuming that v-src's ability to transform cells is dependent on its ability to bind to phospholipids, it makes sense that the loss of latter would result in the loss of the former If the protein reverts back to the wild type protein, as in choice D, then you would NOT expect v-src capable of cell transformation, since the wild-type protein is not a transforming protein How did we know the v-src even has a wild type form? Well, from the passage, you know that v-src is a transforming protein and that transforming proteins are mutants And since mutant proteins must have corresponding wild-type proteins, we know that it's perfectly normal for v-src to have a corresponding wild-type protein, c-src Anyway, the fact that the wild type version of v-src does not transform cells neither supports nor contradicts the model in that v-src must be bound to the plasma membrane and capable of phosphorylation for it to be capable of transforming cells Therefore, choice D is wrong Again, choice B is the correct answer Kaplan MCAT Biological Sciences Test Transcript The correct answer is choice D In a frameshift mutation the reading frame of a mRNA strand is altered because of the loss or addition of nucleotides, except in multiples of three Why? Because the remaining nucleotides are regrouped into a different set of triplets The protein that is translated as a result of a frameshift mutation is usually nonfunctional In this problem, deletion of the fifth base from the 5' end, which is an adenine, changes the second codon from AUA, which codes for the amino acid isoleucine, to UAA, which is one of the stop codons - a noncoding triplet that signals the termination of translation Remember, AUG is the codon that initiates translation If you are a little unclear about this point, listen to the explanation to question Thus, our frameshift mutation will result in the premature termination of translation With that in mind, let's take a look at the answer choices Choice A has the amino acids listed in the sequence as it would have been translated, including the stop codon, but this is pointless because synthesis halted immediately after the first codon was translated - which is why choice A is wrong and choice D is the correct answer If you chose any of the other answer choices, you were probably careless in your reading of the genetic code chart Again, the correct answer is choice D Passage II (Questions 5–12) The correct answer is choice B According to the passage, you know that in a hematocrit, the bottom layer Layer 1- consists of red blood cells; the middle layer - Layer - consists of white blood cells and platelets; and the top layer - Layer 3- consists of the liquid component of the blood, plasma If Layer of Patient A's hematocrit is found to have a greater-than-normal volume, then this must be due to an increase in the number of circulating platelets or an increase in the number of circulating white blood cells There are approximately 250,000 - 500,000 platelets per mm3 of blood, but this volume is fairly static There are approximately 5,000 - 10,000 white blood cells in circulation per mm3 of blood; however, this number increases substantially when the body is battling an infection, since white blood cells are one of the body's main defense mechanisms against foreign invaders such as bacteria and viruses During an infection, the body steps up its production of white blood cells and releases more white blood cells into circulation from its reservoirs Therefore, choice B is the correct answer, since Layer would be of greater-than-normal volume in a person with an infection Choice A, living at high altitudes, WOULD affect the volume of Layer - the red blood cells People living at high altitudes tend to have a greater vital capacity, a greater amount of red blood cells and hemoglobin, and an increased tissue vascularization than people living at lower altitudes or sea level All of these adaptations serve to increase the oxygen capacity of the blood, since the atmospheric partial pressure of oxygen is much lower at high altitudes Therefore, choice A is wrong Choice C, hemophilia, is a sex-linked genetic disease characterized by a deficiency or abnormality of a clotting factor, and would therefore affect Layer 3, since clotting factors are plasma proteins Choice D, sickle-cell anemia, would affect the volume of Layer 1, since this is the layer composed of red blood cells Again, the correct answer is choice B The correct answer is choice A You're told in the passage that normal hematocrit values range from 40 50% of total blood volume Looking at Patient B's hematocrit, we see that his red blood cell volume is approximately 25%, which is way below normal You're also told in the passage that a person suffering from sicklecell anemia has a low red blood cell count, because sickled red blood cells are very fragile and undergo lysis, thereby decreasing the volume of circulating red blood cells So choice A is the correct answer Looking at the other answers, we see that choices C and D can be eliminated since both infection and leukemia affect the volume of white blood cells in the body, and would be reflected by an above-normal volume in Layer of the hematocrit Choice B, overproduction of hemoglobin, is one of the consequences of living at a high altitude, as discussed in the explanation to question 5, and would most likely cause an INCREASE in the volume of Layer 1, not a decrease Again, the correct answer is choice A The correct answer is choice C Answering this question relied on your outside knowledge of the structure of the three different types of blood cells Red blood cells, which are the only cells found in Layer 1, NOT contain nuclei During the production of red blood cells, which is a process known as erythropoiesis, the nucleus of the primitive red blood cell shrivels and eventually disappears Likewise, platelets, which are found in Layer 2, NOT contain nuclei; a platelet is a fragment of bone marrow cell containing mitochondria, microtubules, vesicles, and granules, but no nucleus Plasma doesn't contain any cells; it consists mostly of solutes dissolved in water, but also contains respiratory gases, hormones, plasma proteins, waste products of metabolism, and dissolved nutrients Therefore, both Layer and Layer contain cells that lack nuclei; and so choice C is the correct answer The correct answer is choice D This is another question that relies on your outside knowledge - this time, on your knowledge of plasma components As stated in the passage, the top layer of a hematocrit, which is Layer 3, is composed of plasma Plasma consists of water, glucose, amino acids, inorganic ions and salts, fatty acids, respiratory gases, metabolic wastes, and plasma proteins - of which there are three main types: albumins, fibrinogen, and globulin Albumin plays a very important role in the maintenance of blood osmolarity, since albumin is too large to pass through capillary walls Therefore, choice D is the right answer Hemoglobin is the oxygen-carrying molecule of red blood cells, which are found in Layer 1, so choice A is wrong Choice B, macrophages, are large white blood cells filled with lysosomes that engulf and digest foreign material, and hence would be found in Layer 2, which contains both white blood cells and platelets Therefore, choice B is also wrong Likewise, leukocytes, which is just another way of saying white blood cells, are found in Layer 2, as I just said Again, choice D is the right answer Kaplan MCAT Biological Sciences Test Transcript The correct answer is choice C According to the passage, those individuals who are homozygous for the HbS allele are afflicted with sickle-cell anemia, and according to the question stem, the disease can be lethal Yet despite the high mortality of individuals with the genotype HbS HbS, the frequency of the HbS allele is as high as 40% in some areas of Africa This is an extremely high frequency for an allele that is essentially lethal in a double dose The explanation for this phenomenon lies in the fact that there is a heterozygote advantage; that is, individuals that have only one copy of the HbS gene have some advantage over normal individuals in areas where malaria is common The advantage is that heterozygotes rarely show any of the symptoms of sickle-cell anemia AND are less likely to die of malaria than normal homozygotes Why? Because the heterozygote's red blood cells not sufficiently support the growth of the malaria parasite Thus, those individuals possessing a single copy of the HbS allele have an advantage over those without - an indication that natural selection has favored the maintenance of the HbS allele frequency in that particular gene pool The number of heterozygotes that are resistant to malaria and survive to reproductive age more than outweighs the number of homozygotes that die young of sickle-cell anemia Thus, the HbS allele is maintained at a high frequency from generation to generation, and choice C is the correct answer Now let's take a look at the other answer choices Choice A, mutation, could not ALONE account for the high frequency of the HbS allele If mutation were the only evolutionary force at work in maintaining the high frequency of the HbS allele, this would imply that this mutation arose independently in 40% of certain African populations, which is highly, highly improbable Choice B, genetic drift, refers to changes in the gene pool of a small population due to chance, such as a freak natural disaster that kills everybody in the population that doesn't have a copy of the HbS allele It is not mere coincidence that the frequency of the HbS gene is highest in areas where malaria is common Choice D is wrong, because reproductive isolation refers to the incompatibility of genitalia, which prevents interbreeding, typically among different species Obviously, all human beings are capable of reproducing with one another, regardless of race, ethnicity, or creed Again, the correct answer is choice C, natural selection 10 The correct answer is choice D From the passage you know that when the gene coding for the beta-chain for both HbS and HbA are digested with the enzyme MstII, the HbA gene yields a 1.1-kb fragment because it contains an MstII site, while the HbS gene yields a 1.3-kb fragment because it lacks an MstII site Obviously, if you were to take the 1.1-kb fragment and hybridize it with a probe specific for the 1.1-kb fragment, the hybridization would be successful But what about trying to hybridize the 1.3-kb fragment with the same 1.1-kb fragment specific probe? Well since HbS results from a single amino acid substitution of HbA, this mutation most likely resulted from a point mutation A point mutation is the substitution of one nucleotide for another In other words, if am adenine is present instead of a thymine at the appropriate location within the gene, HbS will result instead of HbA This means that the two genes only differ by one nucleotide and this nucleotide must lie within the MstII site in the HbA gene So, yes, the 1.3-kb fragment will successfully hybridize with a probe specific for the 1.1-kb fragment Why? Because, every base in the probe that is complementary to the 1.1-kb fragment, except for the one that was substituted, will also be complementary to the 1.3-kb fragment, and therefore the probe will base pair to both fragments and the hybridization will be successful Now that we understand what is going on, let's look at the question From the question stem you know that after digestion of DNA sample with MstII, a 1.3-kb fragment is generated that does successfully hybridize with a probe specific for the 1.1-kb fragment This sounds exactly like what happens when the DNA coding for the betachain of HbS is digested by MstII But does this prove that this DNA sample contains the HbS gene? Well let's look at the answer choices Choices A and B are true - the HbS gene does NOT contain an MstII site, and the probe specific for the 1.1-kb fragment of HbA does bind to the 1.3-kb fragment of the DNA sample However, this does not prove that the DNA contains the HbS gene The only way to prove that the sample contains the HbS gene is to either match the sequence of the DNA with the known sequence of the HbS gene, or to prove that the DNA does indeed give rise to HbS protein Therefore, choices A and B are wrong O.K., how about C? Choice C is wrong because it's not true, the probe normally WOULD bind to both fragments Choice D, however, makes sense The only reason that the 1.3-kb fragment is generated instead of the 1.1-kb fragment is because the DNA sample lacks an MstII site But there is no reason why the only mutation that results in the elimination of this site has to be the mutation that produces the HbS gene Actually, any mutation that occurred within the MstII site of the HbA gene would produce a gene without the MstII site, resulting in the 1.3-kb fragment upon digestion by MstII And since the only change is in the MstII site, the probe specific for the 1.1-kb fragment of the HbA gene will still bind to the larger fragment, because almost all of the bases in the 1.3-kb fragment are still complementary to the probe Therefore, choice D is the correct answer 11 The correct answer is choice D, substitution HbS hemoglobin is structurally identical to HbA hemoglobin except for the substitution of a valine for a glutamate at a single point on the molecule In HbA the codon for glutamate is GAA; in HbS the codon for valine is GUA It is evident that in the RNA coding for HbS, adenine has been replaced by uracil, which means that, working backwards to the DNA from which the mRNA was transcribed, thymine has been replaced by adenine in the HbS gene This type of mutation is called a base-pair substitution, which is when one nucleotide is substituted for another Thus, choice D is correct Choice A, deletion, and choice B, insertion, involve the loss or addition of nucleotides, respectively Such mutations tend to have very serious effects on the protein synthesized, if one is synthesized at all, since nucleotides are read as a series of triplets known as Kaplan MCAT Biological Sciences Test Transcript codons Unless the addition or loss of nucleotides occurs in multiples of three, the reading frame of the RNA will be altered; this is known as a frame-shift mutation, which is choice C Again, choice D is the correct answer 12 The correct answer is choice B The probability that the firstborn child of two sickle-cell allele carriers will be a male carrier is 25% According to the passage, sickle-cell anemia is a recessively inherited autosomal disorder; it is NOT X-linked This means that the gene for hemoglobin is not on the X chromosome, and that the two characteristics in question - sickle-cell anemia and gender - independently assort during meiosis Therefore, by figuring out the probabilities of each independent event and multiplying their individual probabilities, you can determine the probability that the two events will occur simultaneously Okay, to figure out the percentage of offspring that are expected to be carriers in a cross between two carriers, let's a Punnett square The genotype is the same for both parents - HbA HbS times HbA HbS This is your basic 1:2:1 hybrid cross: 25% of the offspring will have the genotype HbA HbA and will express the normal phenotype; 50% of the offspring will have the genotype HbA HbS - and will be carriers of the trait; and 25% of the offspring will have the genotype HbS HbS and will express the sickle-cell phenotype So, one half of the offspring will be carriers Likewise, 50% of the offspring will be male, and 50% of the offspring will be female, because there is an equal chance of inheriting either the X or Y chromosome from the father, and a 100% chance of inheriting the X chromosome from the mother The product of these two probabilities, 1/2 times 1/2, is 1/4, or 25% Thus, the correct answer is choice B Passage III (Questions 13–17) 13 The correct answer is choice C This question asks you to figure out the most likely structure for compound A based on the information given in the passage The first classification test is the decolorization of a bromine solution, which is brownish red In case you forgot what this test is used for, we'll remind you: it detects double and triple bonds What happens is that if a double bond is present, the bromine attacks the bond and attaches itself to it, forming a dibrominated alkane from a double bond or a tetrabrominated alkane from a triple bond This uses up the bromine in the solution, making it turn clear Now, since all three compounds decolorize the bromine solution, they must all contain either double or triple bonds Further on in the passage, it says that compound A also produces a precipitate when treated with silver nitrate in ammonia, and it yields pentane when treated with excess hydrogen in the presence of platinum The silver nitrate test is quite specific and is used to detect triple bonds at the end of molecules Compounds with such terminal triple bonds produce precipitates, by a complicated reaction whose details you don't really need to know Anyway, since compound A did produce a precipitate, a terminal triple bond must be present here, and thus choice C is correct Choice A is incorrect because even though a triple bond is present, it's not at the terminal position, so the silver nitrate test would be negative Choice D is incorrect because it has two double bonds, which would also test negatively Choice B is incorrect for two reasons First of all, since it's a ring, this compound would not form pentane when it is reduced with hydrogen in the presence of platinum; the second is that it also wouldn't form a precipitate with silver nitrate Again, the correct answer is choice C 14 The correct answer 14 is choice C This question is similar to question 13 except that this one asks for the most likely structures for compound C, so you can tackle this question in the same way We know that Compound C decolorizes bromine in carbon tetrachloride, so it must be unsaturated, with either double or triple bonds We also know that compound C does NOT produce a precipitate when treated with silver nitrate, so it doesn't contain any terminal triple bonds Now, to figure out what the possibilities are for this compound's structure, calculate the hydrogen deficiency of the compound The formula for a fully saturated hydrocarbon is CnH2n+2, so since compound C has five carbons, it would have 12 hydrogens if it were fully saturated Compound C has hydrogens, so it is hydrogens UNDER a fully saturated hydrocarbon Now, each time a double bond or a ring is formed, two hydrogens are dropped out of the compound; when a triple bond is formed, four hydrogens are dropped out Since compound C has a deficiency of four hydrogens, the compound could contain either a ring and a double bond; two double bonds; or a triple bond The last bit of information that we get is that compound C does not form pentane when it is fully reduced with hydrogen in the presence of a platinum catalyst Instead, it forms a compound with the formula C5H10, which has a deficiency of two hydrogens, so we can conclude that compound C contains a ring and a double bond Of the choices that we're given, roman numerals I, III, and IV all contain a ring and a double bond, and therefore they are all possible structures for compound C Again then, choice C is the correct answer 15 The correct answer is choice D This question asks you to use the information in the passage to pick the structure which best represents compound B Let's backtrack a bit to find out what we already know about compound B We know compound B is unsaturated, and the fact that it is reduced by hydrogen in the presence of a platinum catalyst to give pentane shows that the original molecule was not cyclic Why? Because otherwise, a carbon-carbon bond would have to be disrupted in order to form a linear molecule like pentane, and hydrogenation won't accomplish that Therefore, choice B is incorrect We also know that compound B doesn't have a terminal triple bond, because it didn't give a positive result in the silver nitrate test So, by the same reasoning we used in the last question, the compound must have either two double bonds or a single, non-terminal triple bond The answer therefore lies between choices A, C, and D You should know that a strongly basic solution of potassium permanganate will replace a triple bond with a diketone If the diketone is then treated with an acidic solution, it will produce two carboxylic acids at the point Kaplan MCAT Biological Sciences Test Transcript between the two keto groups So, looking over the answer choices, you can see that choice D is the correct response, since it's the only one that contains a triple bond Be aware, that a mildly acidic solution of potassium permanganate oxidatively cleaves double bonds, replacing each of the double bonded carbons with a carboxyl group Anyway, knowing this bit of information you can easily eliminate choices A and C Again the correct choice is D 16 The correct answer is choice B In order to answer this question you need to remember the major uses of the different types of spectroscopy Let's briefly review those before we answer the question First we have infrared spectroscopy, or simply IR This is mainly used to determine what functional groups are present in a compound Next we have mass spectroscopy, which is generally used to determine a compound's molecular weight Then we have nuclear magnetic resonance spectroscopy, or NMR This is used to determine how the carbons are connected in the compound that is, what the carbon skeleton looks like Finally, there is ultraviolet spectroscopy, which is mainly used to determine a compound's level of saturation Okay, back to the question From the passage, we know that the molecular weights of all the compounds are the same, so we can't differentiate between them on the basis of their molecular weight; this rules out mass spectroscopy, and eliminates choices A and C right away We also know from the information in the passage that all three compounds are unsaturated, so theoretically we could use UV spectroscopy, choice D, to differentiate them However, even though this would tell us whether they had double or triple bonds, it wouldn't give us the exact structure For instance, it wouldn't reveal whether the multiple bonds were terminal or non-terminal Notice that the question asks which of the choices would be MOST useful; this means that even though you've found a possible answer, you should still check to see if there's another choice that's better Choice B, an IR in conjunction with an NMR, would be the most useful possible set of instruments, because it would reveal both the functional groups in the compounds AND the shape of the carbon skeleton Therefore, choice B is the best and correct answer 17 The correct answer is choice C This is a negative response question in that you have to pick the answer choice which would NOT be a possible structure for the compound with the formula C8H12O2 Well, if you draw out each answer choice it should be pretty evident that choice C will not have the formula C8H12O2 Choice C is a straight chain carbon molecule, so to begin with, you need to draw out carbons Then you have to draw in the aldehyde functionality at one end of the molecule You then need to draw in three carbon double bonds and a hydroxyl group Whether you draw the double bonds isolated, conjugated or cumulated, you should be able to see that the maximum number of hydrogens in the molecule is 10, and so the overall formula for this compound is C8H10O2, making it the correct answer If you the same sort of thing with answer choice A, but this time draw in a non-terminal triple bond along with a hydroxyl functionality and an aldehyde group, you can see that this molecule WILL have the formula C8H12O2 and so it is incorrect Choice B will also have the formula C8H12O2, even if you draw the carbonyl groups as two aldehydes, two ketones or one ketone and one aldehyde functionality Finally, choice D is also incorrect This one is a bit more tricky, as the straight chain will consist of carbons, due to the attachment of a cyclopropane group to one of the carbons Also interrupting this carbon stretch is an ester linkage You should be able to see that any combination of the functional groups whether they are drawn as terminal or non-terminal will lead to the formula C8H12O2 Again then, choice C is the correct answer Discrete questions 18 The correct answer is choice A Nondisjunction is defined as the failure of homologous chromosomes to properly separate during meiosis I, OR the failure of sister chromatids to properly separate during meiosis II The end result is the production of abnormal gametes; half of the gametes wind up with two copies of the same chromosome, the other half wind up with no copies of that chromosome If one of these gametes fertilizes a normal gamete, the resulting zygote will either have three copies of that chromosome, which is called trisomy, or one copy of that chromosome, which is called monosomy Since nondisjunction results in the movement of an entire, intact chromosome, the base sequence of the nondisjunctioned chromosome remains unchanged; so, choice A is the right answer Most monosomies and trisomies are lethal, causing spontaneous abortion during pregnancy One of the most common trisomies seen in humans is Down's syndrome, which results from the nondisjunction of chromosome 21 Choice B, base substitution, is exactly what its name implies, a substitution of one base for another, which obviously results in a change in the base sequence of DNA Therefore choice B is wrong Choice C, translocation, is when a chromosome breaks, and the fragment that breaks off joins with a nonhomologous chromosome This means that a new piece of DNA is added to a pre-existing chromosome, causing a change in the base sequence; therefore, choice C is also wrong Choice D, recombination, is by definition, the creation of new gene combinations by way of sexual reproduction and crossing over in eukaryotic organisms, and by way of transduction, transformation, and conjugation is prokaryotic organisms New gene arrangements necessarily implies that there is a change in base sequence, and so choice D is also wrong Again, choice A is the correct answer 19 The correct answer is choice D There are two reasons why the single bond length is shorter than expected in 1,3-butadiene First, the pi bonds that overlap and consequently form the two double bonds in 1,3-butadiene also ‘spill over’ into the single bond This results in partial double bond character and a shorter bond length Therefore, statement I is correct Kaplan MCAT Biological Sciences Test Transcript Orbital hybridization also goes some way to explaining why the bond length is shorter In the ethane molecule, and indeed any alkane, the carbon-carbon bonds are formed by the overlap of sp orbitals However, in 1,3butadiene, the carbon-carbon single bond is formed by the overlap of sp orbitals As a result, there is 33% s character in this bond, not 25% s character like in the sp bond This increase results in a shorter and stronger bond and so statement III is also correct Therefore, choice A is incorrect and again, D is the right answer Statement II is correct if you are describing an observable trend in the table However, it is incorrect if you are using it to reason why the 1,3-butadiene single bond length is shorter There are a greater number of carbons in 1,3-butadiene than in ethane and propene, but this does not account for the shorter bond length Therefore, choices B and C are wrong and choice D is the right answer 20 The correct answer is choice B This question is pure outside knowledge; there really isn't any reasoning involved at all The pancreas is both an endocrine and an exocrine organ As an endocrine organ, the pancreas secretes the hormones glucagon, insulin, and somatostatin As an exocrine organ, the pancreas secretes the enzymes trypsin, choice C, carboxypeptidase, choice D, lipase, choice A, and chymotrypsin, which is not one of our choices, into the small intestine Chymotrypsin, trypsin, and carboxypeptidase are actually secreted by the pancreas in their inactive forms, chymotrypsinogen, trypsinogen, and procarboxypeptidase, respectively, and are converted into their active forms in the small intestine Chymotrypsin and trypsin hydrolyze specific peptide bonds; carboxypeptidase hydrolyzes terminal peptide bonds at the carboxyl end; and lipase hydrolyzes lipids Therefore, choices A, C, and D are wrong On the other hand, aminopeptidase, choice B, is an enzyme secreted into the small intestine by the intestinal glands Aminopeptidase is also involved in protein digestion; it hydrolyzes terminal peptide bonds at the amino end Therefore, choice B is the correct answer 21 The correct answer is choice B This is another organic chemistry question, which asks about the number of acidic hydrogens in the four compounds listed Choice A is a diacid with two methoxy groups attached to the parent chain The two methoxy groups are on the two carbons adjacent to the carboxyl groups Since these methoxy groups are electron-donating, they destabilize the carboxylate anions, so these carboxyl groups have pretty low acidity Since choice A doesn't contain any very acidic hydrogens therefore, you can eliminate it Choice B is another dicarboxylic acid but it has an electron-withdrawing sulfonate substituent adjacent to both carboxyl groups This will stabilize both carboxyl anions making both carboxyl hydrogens highly acidic In addition, the hydrogen on the central carbon is also very acidic, because it's surrounded by the three electron-withdrawing groups Therefore, choice B contains three highly acidic hydrogens Choice C contains just one highly acidic hydrogen the one on the carboxyl group Finally, choice D also contains acidic hydrogen The hydroxyl proton in phenol is weakly acidic since the benzene ring can stabilize the negative charge of the phenoxide ion However, choice B still has the greatest number of acidic hydrogens, making it the correct answer 22 Choice A is the right answer This question tests your knowledge of eukaryotic, prokaryotic, and viral structure Viruses are composed of a nucleic acid encapsulated by a protein coat - period Bacteriophage are viruses that infect bacteria only Since the answer to the question is an organism characterized by the presence of a cell wall, you can rule out choice B, bacteriophage, and choice C, viruses, which leaves you with only two choices Choice D, fungi, are eukaryotic, heterotrophic organisms, and as such, have membrane-bound organelles Fungi have cell walls, which are composed of chitin, but they also have nuclei bound by nuclear membranes, so choice D is also wrong Which leaves us with choice A, bacteria Bacteria are members of the kingdom Monera - a group of unicellular, primitive, prokaryotic organisms Prokaryotes are characterized by a lack of membrane-bound organelles Bacteria typically have a cell wall, a cell membrane, ribosomes, and a circular chromosome located in a region of the cell known as the nucleoid Therefore, of the answer choices, it is the bacteria that is characterized by the presence of a cell wall but the lack of a nuclear membrane Again, choice A is the right answer Passage IV (Questions 23–27) 23 Choice D is the correct answer If you look at Figure you'll see that those athletes on a high protein and high fat diet did not recover to pre-exercise glycogen levels, EVEN after days of recovery The initial level of muscle glycogen was about 15 g/kg, and the level after days was about g/kg This seems to indicate that eating a high protein and fat diet incurs a great loss of glycogen Think about this for a minute The body needs ATP, and the majority of this ATP is generated from the catabolism of glucose during cellular respiration In the absence of glucose, the body draws on its other nutrient stores to compensate for this absence Proteins and fats are degraded into molecules that can enter the aerobic respiration pathway, and the stores of glycogen in the liver and muscle cells are converted back into glucose And this is exactly what happens to a person on the high protein weight loss diet described in the question; by depriving the body of carbohydrates, the stores of glycogen are converted into glucose, which causes a substantial weight loss early on in the diet Thus, choice D is the right answer Choice A is wrong because there is no reason to conclude that there is an increase in metabolic activity on a high-protein diet Choice B just doesn't make much sense; mammals are homeotherms, which means that a constant body temperature is Kaplan MCAT Biological Sciences Test Transcript maintained If there is an increase in body temperature not caused by fever, thermoregulatory mechanisms go to work to return internal temperature to normal Therefore, choice B is wrong Choice C is wrong because there is not a substantial loss of fat on an all protein diet; loss of fat is just wishful thinking on the part of the dieter Again, choice D is the correct answer 24 The correct answer is choice C You've got to determine which of the energy systems are used to supply the energy needed to sustain physical activity for 22.3 seconds, which is the amount of time it takes a world-class swimmer to swim 50 meters in the freestyle event You don't have to know anything about swimming to answer the question because the only thing that you need to know, which is the amount of time the activity takes, is given to you in the question stem And according to the passage, the phosphagen system can sustain up to 15 seconds of maximal activity But this energy system alone isn't enough to complete the 50-meter freestyle, so choice A is wrong Since the swim takes longer than 15 seconds, you know that the glycogen-lactic acid system must also be involved The glycogen-lactic acid system provides energy for 30 to 40 seconds in addition to the phosphagen system's 15 seconds Therefore the glycogen-lactic acid system must take over where the phosphagen system left off to enable the swimmer to complete the event Therefore, choice C is correct Since all physical activity begins with the phosphagen system, choice B is incorrect The aerobic system is not used for swimming a distance as short as 50 meters in 22.3 seconds, so choice D is wrong Again, choice C is the correct answer 25 Choice D is the correct answer After three hours of exercise, let alone two, it can be assumed that the phosphagen system of those athletes on the high protein and fat diet is completely used up, since this system can support activity for a maximum of 15 seconds, according to the passage So, choice A is wrong And, if you look at the graph, you'll see that after two hours of exercise, those athletes on the high protein and fat diet have completely depleted their muscle glycogen stores And if they exercise for yet a third hour, there's no recovery period during which the muscle glycogen supply can be replenished Well, this means that the glycogen-lactic acid system cannot provide any of the energy needed for the third hour of exercise, because the glycogen-lactic acid system supplies energy by breaking down muscle glycogen into lactic acid No more glycogen, no more glycogen-lactic acid system This means that you can eliminate choices B and C, because they both contain the glycogen-lactic acid system The only remaining energy system is the aerobic system, which is the only energy system capable of sustaining the body's energy needs during extended periods of strenuous exercise Therefore, choice D is the correct answer 26 The correct answer is choice D From the question stem you know that you need to find which of caffeine's effects would be deleterious to an athlete First of all, you should realize that you not need to know anything about caffeine to answer this question All you have to is evaluate the list of effects in the answer choices So let's look at the choices Is increasing the concentration of epinephrine, as in choice A, a good or bad thing? Epinephrine increases the conversion of glycogen to glucose in the liver and muscle tissue, causing a rise in blood glucose concentration and an increase in basal metabolic rate Well, if more glucose is made available to the muscle, the muscle will be able to generate more energy to perform the athletic activity Thus, this enhances the athlete's performance, and so choice A is incorrect What about systemic vasodilation, as in choice B? Vasodilation of the systemic system will allow more blood to flow into the organs and muscle tissue of the body, thereby providing an increased oxygen supply to the athlete’s' muscles Increasing the oxygen supply will increase the rate of cellular respiration, thereby generating more energy for the muscle So, this too is beneficial to the athlete, and thus choice B is wrong In humans, this vasodilation is short-lived, so this effect would only benefit athletes in the beginning of an event What about relaxing the muscles of the bronchi, as in choice C? The bronchi are the pair of respiratory tubes branching into either lung at the lower end of the trachea They subdivide into progressively finer passageways, the bronchioles, culminating in the alveoli The alveoli are the sites of gas exchange between the air in the lungs and the blood in the capillaries So by relaxing the bronchi muscles, more air is allowed to enter the lungs and thereby increase the partial pressure of oxygen in the blood As a result, more oxygen will be delivered by the blood to the muscles This increased oxygen supply will increase the rate of energy production Therefore, this effect of caffeine is also helpful to the athlete, so choice C is also incorrect So by the process of elimination, choice D is the correct answer ADH stimulates water reabsorption in the kidneys So if ADH secretion is inhibited by a diuretic such as caffeine, water reabsorption will decrease and dilute urine will be collected in the bladder If water reabsorption does not return to normal, dehydration can result Since dehydration would adversely affect the athlete's performance, choice D is the correct answer 27 The correct answer is choice A As you're told in the passage, the glycogen-lactic acid system converts the muscle's stores of glycogen into glucose, which is then converted into lactic acid in the absence of oxygen The lactic acid diffuses out of the muscle cells and into the bloodstream and interstitial fluid During the recovery period, this lactic acid must be removed This is accomplished in two ways First, some of the lactic acid is converted into pyruvate, which is metabolized by the Krebs cycle Second, most of the lactic acid is used to replenish the depleted glycogen stores in muscle cells This is done by first converting the lactic acid into glucose in the liver, and, in turn, this glucose is converted into glycogen And, if you look at the graph, you see that no matter what type of diet the athletes were on, it's obvious that the body tries to replenish its glycogen stores during the recovery period following exercise Therefore, choice A is the right answer You could also have gotten the right answer by the process of elimination, so let's take a look at the other choices for a moment I hope that you weren't fooled by choice B Kaplan MCAT Biological Sciences Test Transcript Although lactic acid might sound a little bit like lactose, it has nothing to with milk, or calcium, and is not stored in bone Choice C is wrong, because as already said, some of the lactic acid will be reconverted to pyruvate acid in the liver and fed into the Krebs cycle, which is a very long way from being stored as fat Choice D is wrong because in the glycolytic pathway, glucose is degraded into pyruvate, and in the absence of oxygen, pyruvate is converted into lactic acid The body's "goal" is to get rid of the lactic acid that it produced during the exercise, not create more of it Again, choice A is the correct answer ... given in the passage The first classification test is the decolorization of a bromine solution, which is brownish red In case you forgot what this test is used for, we'll remind you: it detects... at the terminal position, so the silver nitrate test would be negative Choice D is incorrect because it has two double bonds, which would also test negatively Choice B is incorrect for two reasons... wrong On the other hand, aminopeptidase, choice B, is an enzyme secreted into the small intestine by the intestinal glands Aminopeptidase is also involved in protein digestion; it hydrolyzes terminal