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MCAT Topical Tests Dear Future Doctor, The following Topical Test and explanations should be used to practice and to assess your mastery of specific topical information in test format This is an opportunity to practice the STOP, THINK, PREDICT methodology learned in the Kaplan classroom There are Discrete questions and Passage-based questions that test your ability to apply your foundation knowledge to MCAT-style questions, using 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 © 2004 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 Molecular Genetics Topical Test Time: 42 Minutes Number of Questions: 30 MCAT Molecular Genetics DIRECTIONS: Most of the questions in the following topical test are organized into groups, with a descriptive passage preceding each group of questions Study the passage, then select the single best answer to each question in the group Some of the questions are not based on a descriptive passage; you must also select the best answer to these questions If you are unsure of the best answer, eliminate the choices that you know are incorrect, then select an answer from the choices that remain Indicate your selection by blackening the corresponding circle on your answer sheet A periodic table is provided below for your use with the questions PERIODIC TABLE OF THE ELEMENTS H 1.0 He 4.0 10 Li 6.9 Be 9.0 B 10.8 C 12.0 N 14.0 O 16.0 F 19.0 Ne 20.2 11 12 13 14 15 16 17 18 Na 23.0 Mg 24.3 Al 27.0 Si 28.1 P 31.0 S 32.1 Cl 35.5 Ar 39.9 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 K 39.1 Ca 40.1 Sc 45.0 Ti 47.9 V 50.9 Cr 52.0 Mn 54.9 Fe 55.8 Co 58.9 Ni 58.7 Cu 63.5 Zn 65.4 Ga 69.7 Ge 72.6 As 74.9 Se 79.0 Br 79.9 Kr 83.8 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 Rb 85.5 Sr 87.6 Y 88.9 Zr 91.2 Nb 92.9 Mo 95.9 Tc (98) Ru 101.1 Rh 102.9 Pd 106.4 Ag 107.9 Cd 112.4 In 114.8 Sn 118.7 Sb 121.8 Te 127.6 I 126.9 Xe 131.3 55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 Cs 132.9 Ba 137.3 La * 138.9 Hf 178.5 Ta 180.9 W 183.9 Re 186.2 Os 190.2 Ir 192.2 Pt 195.1 Au 197.0 Hg 200.6 Tl 204.4 Pb 207.2 Bi 209.0 Po (209) At (210) Rn (222) 87 88 89 104 105 106 107 108 109 Fr (223) Ra 226.0 Ac † 227.0 Rf (261) Ha (262) Unh (263) Uns (262) Uno (265) Une (267) 58 59 60 61 62 63 64 65 66 67 68 69 70 71 * Ce 140.1 Pr 140.9 Nd 144.2 Pm (145) Sm 150.4 Eu 152.0 Gd 157.3 Tb 158.9 Dy 162.5 Ho 164.9 Er 167.3 Tm 168.9 Yb 173.0 Lu 175.0 90 91 92 93 94 95 96 97 98 99 100 101 102 103 † Th 232.0 Pa (231) U 238.0 Np (237) Pu (244) Am (243) Cm (247) Bk (247) Cf (251) Es (252) Fm (257) Md (258) No (259) Lr (260) GO ON TO THE NEXT PAGE as developed by Molecular Genetics Test Passage I (Questions 1–10) Acetylcholine (Ach) is one of the most commonly found neurotransmitters in almost all animals, from fruit flies to humans The two major receptors for Ach in humans are known as nicotinic and muscarinic When nicotinic receptors are stimulated at the neuromuscular junction muscle contraction is produced Muscarinic receptors can have various functions Acetylcholinesterase (AchE) is the enzyme that breaks down Ach into choline and an acetyl group, which can then be reabsorbed by neurons where they are used to synthesize new Ach molecules AchE deactivates Ach in a synapse, thus allowing a controlled length of stimulation and allowing the cells to reset themselves before the next impulse AchE follows the general formula for enzyme kinetics: k1 k3 ES E+S E+P k2 competitive Ach inhibitors, like Galantamine, less Ach is destroyed, and thus brain concentrations of the neurotransmitter increase Galantamine is a particularly useful drug because it not only inhibits AchE, but it also is an allosteric activator of nicotinic acetylcholine receptors, increasing the amount and effect of Ach in the synapse Other types of AchE inhibitors like Tacrine are also used to treat Alzheimer’s Table shows an experiment to determine the effects of Tacrine [Ach] (mM) Tube Tube Tube Tube Tube [E] (mM) -8 2.0 x 10 9.5 x 10-5 1.9 x 10-4 1.9 x 10-4 2.2 x 10-2 -5 2.5 x 10 2.5 x 10-5 2.5 x 10-5 2.5 x 10-5 2.5 x 10-5 [Tacrine] (mM) Vo (mM/min) 0 1.0 x 10-4 1.0 x 10-2 1.0 x 10-4 X 1.75 x 10-1 1.92 x 10-1 1.10 x 10-3 1.92 x 10-1 Table Formula Equation Where Km is the overall rate constant The velocity of the reaction is given by the Michaelis-Menten equation: vo = v max[ S ] KM + [ S ] and kcat = v max [E] AchE Activity Km = k2 + k3 k1 32˚ 37˚ 42˚ Figure Equation For acetylcholinesterase acting on Ach, Km = 9.5 x 10-5 and Kcat = 1.4 x 104 Kcat is a measure of the catalytic ability or the efficacy of an enzyme In addition to being an important tool in endogenous negative feedback loops, AchE is the target of a variety of drugs and synthetic compounds and its inhibition can have many very different impacts Organophosphate pesticides bind tightly to AchE and act as irreversible inhibitors, producing death in many organisms Nerve gases are also organophosphates and act in the same way These toxins can produce headaches, runny nose, narrowing of the pupils, and a tightened chest Lethal doses produce uncontrollable muscle contractions and death by asphyxiation GO ON TO THE NEXT PAGE On the other hand, AchE inhibitors are also used to treat a number of diseases, most notably Alzheimer’s Disease In the later stages of Alzheimer’s, patients have low levels of Ach in their brains By employing reversible KAPLAN MCAT Which of the following changes in acetylcholinesterase kinetics would occur after taking galantamine? A B C D Increased Km Decreased Vm Decreased k2 + k3 Increased Vm Some researchers hypothesize that manganese is a cofactor for acetylcholinesterase If this is true, which of the following might be effects of a deficiency of manganese in the diet? A B C D A Partial muscle paralysis, pupillary dilation, confusion B Difficulty breathing, muscle spasms, confusion C Vasodilatation, headaches, pupillary constriction D Bloody nose, difficulty breathing, headache 7.1 x 10-2 mM/min 3.4 x 10-21 mM/min 3.7 x 103 mM/min 7.4 x 10-5 mM/min Which of the following if true would strengthen the theory that some of the effects Alzheimer’s disease are associated with lower levels of acetylcholine in the brain? A Alzheimer’s patients frequently experience temporary partial paralysis B Alzheimer's plaques are found in areas where Ach concentration is high C Higher levels of nicotinic Ach receptors are found in Alzheimer’s affected brains D Levels of endogenous acetylcholinesterase inhibitors are higher in normal brains The acetylcholine in humans and insects are identical How would the graph in Figure change for insects? A B C D Which of the following is the velocity of the reaction in Tube 1? It shifts to the left It shifts to the right It would be narrower It would be wider Based on the information from the passage and table above, what type of effect does Tacrine demonstrate? A B C D Irreversible inhibition Reversible competitive inhibition Reversible noncompetitive inhibition Allosteric activation According to the passage, which of the following best explains the effect of Galantamine on nicotinic acetylcholine receptors? A Galantamine binds to the receptor protein and thus changes the conformation of the acetylcholine binding site B Galantamine binds to the receptor and activates a second messenger C When Galantamine binds to the receptor, it shifts its reaction velocity versus substrate curve to the right D Galantamine binds to the receptor and stabilizes its conformation before the receptor binds acetylcholine GO ON TO THE NEXT PAGE as developed by Molecular Genetics Test The competitive acetycholinesterase inhibitor Pyridostigmine bromide is used to effectively treat myasthenia gravis, a disease that causes debilitating muscle weakness Which of the following is a possible cause of the disease? 10 Which of the following graphs could represent an allosteric enzyme? (x-axis = Increasing [S], y-axis = Increasing reaction velocity.) Inhibited Uninhibited A Ineffective mutated acetylcholine secreted into the synapse at the neuromuscular junction B A loss of nicotinic receptors at the neuromuscular junction C Damaged dendrites of motor neurons innervating the muscles D Overproduction of the neurotransmitter and muscle spasms A Some molecules of acetylcholinesterase are predominantly hydrophobic and some are hydrophilic If a mutation occurs which prevents the production of hydrophilic acetylcholinesterase, which of the following might occur? A After acetylcholine is released into the neuromuscular junction, there will be a greater concentration of it in the middle region of the space than closer to either membrane B After acetylcholine is released into the neuromuscular junction, there will be a greater concentration of it on the nerve and muscle sides of the synapse than in the middle C More acetylcholine will be released into the neuromuscular junction, creating a stronger muscle contraction D More acetylcholine will be broken down, creating a weaker muscle contraction B C D GO ON TO THE NEXT PAGE KAPLAN MCAT Passage II (Questions 11–20) For many years it was thought that all biological catalysts were proteins However that changed when Thomas Cech discovered that the precursor of a ribosomal RNA in Tetrahymena could undergo self-splicing to produce an intron that was able to catalyze the transformation of other RNA molecules These RNA molecules with enzymatic activity were dubbed ribozymes This discovery provided a mechanism for enzymatic activity before the development of proteins early in the evolution of life Recently it has been hypothesized that ribozymes, in addition to being catalysts, might also serve a metabolic function by regulating the amount of enzymes produced Studies have suggested that the 3' end of mRNA of the gene glmS, which encodes for amidotransferase, can be cleaved into a ribozyme in a mechanism similar to what Cech observed in Tetrahymena Because Amidotransferase is an enzyme that catalyzes the formation of glucosamine-6phosphate from glutamine and fructose-6-phosphate, its splicing into a ribozyme would regulate the amount of enzyme present In order to enhance the understanding of the mechanism by which the glmS gene is converted into a ribozyme, students in a biology class performed an experiment using a strain of E Coli that overexpressed glmS gene as their control Three mutants were made by transforming the control with ampicillin resistant plasmids that overexpressed fructose-6-phosphate, glutamine, and both fructose-6phosphate and glutamine to respectively produce mutant 1, mutant 2, and mutant These cells were then plated in minimal media containing ampicillin and incubated for 24hrs at 37C A B C D 11 Which of the following would not have to be added to the RT-PCR mixture? A B C D Helicase Reverse Transcriptase Polymerase dNTPs 12 Which of the following represents dish B? A B C D Mutant Mutant Mutant Wild-type 13 How many of the four trials will develop a band in the southern blot? A B C D 14 If the formation of the ribozyme is triggered by the binding of glucosamine-6-phosphate to a binding site on glmS transcript: A B C D Southern for Muatant will have bands Southern for Muatant will have bands Southern for Muatant will have bands Southern for Wild-type will have bands Figure Following the 24-hour incubation period, the RNA from each dish was analyzed using reverse transcriptase polymerase chain reaction (RT-PCR), Southern Blotting and hybridization with labeled glmS gene 15 According to the passage, the formation of the ribozyme is an example of A B C D Positive regulation Negative regulation Inducible system Cannot be determined from the passage GO ON TO THE NEXT PAGE as developed by Molecular Genetics Test 16 Suppose that the mRNA for glmS is polycistronic (capable of coding for more than one peptide) How would this change the results obtained from the southern blot? A It would not change the results B Only half the number of ribozymes would be formed C It depends on the mechanism of ribozyme formation D It would change the results, the degree to which depends on the mechanism 19 Which of the following will not contribute to genetic variance in E.Coli? A B C D Translocation Transformation Transduction Conjugation 20 How is a ribozyme different from a protein catalyst? 17 Why is RT-PCR a more suitable method than RNA amplification? A B C D RT-PCR is RNA amplification RNA cannot be amplified RNA cannot easily be separated and detected DNA is more stable than RNA A A ribozyme is used up in the reaction whereas a protein is not consumed B A ribozyme can not form tertiary structures whereas proteins can for tertiary and quaternary structures C Proteins can have a greater variety of interactions with their substrates D Proteins are less stable than ribozymes 18 It is possible for the RNA of the gene glmS to lose its polarity when the 3’ end is cleaved off because A RNA is always single stranded and does not need to be polarized B RNA is normally single stranded and does not need to be polarized C Polarity in only necessary during replication D The RNA does not loose its polarity GO ON TO THE NEXT PAGE KAPLAN MCAT Passage III (Questions 21–30) Hormones have two main mechanisms of communicating with cells They may bind to a cell-surface receptor or they may interact with a cytosolic receptor Those that bind to cell-surface receptors are unable to diffuse into the cell; instead they activate receptors with cytosolic domains, which in turn activate a second messenger cascade that, depending on the hormone and receptor, will turn genes on or off Hormones that interact with cytosolic receptors diffuse through the cell membrane and bind to their receptor This hormone-receptor complex then translates into the nucleus to exert its effects directly One hormone that diffuses into the cell is estrogen Estrogen receptors are very similar to other intracellular receptors They have an N-terminal domain, which functions as the transcription-activation domain Near the center of the protein is the zinc-finger, which binds the DNA Estrogen binds at the C-terminal The estrogen receptor is capable of diffusing in and out of the nucleus In the cytosol, the receptor is attached to receptor-associated proteins Some of these receptorassociated proteins hide the DNA-binding domain of the estrogen receptor In the nucleus, in the absence of estrogen, the estrogen receptor suppresses estrogen-regulated genes by promoting histone deacetylation When estrogen binds to its receptor, the estrogen receptor dissociates from its chaperone proteins The estrogen-receptor complex dimerizes and translocates into the nucleus The exact mechanism of translocation is unclear but appears to occur through interaction of caveolin-1 (a cytosolic protein) with the estrogen receptor The dimer binds to the estrogen response element in DNA, and promotes hyper-acetylation of the histones The complex then recruits additional transcription factors, and the genes under control of estrogen are transcribed 21 The amino acid sequence of estrogen and testosterone receptors will be least similar in the: A B C D N-terminal domain Center of the protein C-terminal domain In the tertiary domain 22 Acetylation of histones most likely: A Relaxes DNA to allow transcription factor binding B Coils DNA to allow transcription factor binding C Relaxes DNA to prevent transcription factor binding D Coils DNA to prevent transcription factor binding 23 A scientist transforms a cell-line with estrogenregulated genes If the colony produced contains about 10,000 cells, and the scientist wishes to fully stimulate the production of the estrogen-regulated proteins without wasting estrogen, how many estrogen molecules must he add to the media? A B C D 10,000 20,000 30,000 60,000 24 Loss of the zinc-finger domain of the estrogen receptor will result in: A Constitutive transcription of estrogen-regulated genes B Inability to hyper-acetylate the histones of estrogenregulated genes C Deacetylation of estrogen-regulated genes D Inability of the estrogen-receptor dimer complex to translocate into the nucleus 25 Hormones that bind to intracellular receptors are typically: A B C D Lipophilic Anionic Amino acids Nucleic acids GO ON TO THE NEXT PAGE as developed by Molecular Genetics Test 26 Estrogen and testosterone are both produced from: A B C D Fatty acids Cholesterol Amino acids Glycogen 29 A mutation is made in which the estrogen receptor is unable to dissociate from its chaperone protein This mutation would most likely lead to: A Over-production of estrogen regulated genes B Inability of the estrogen-receptor complex to dimerize C Inability of the estrogen-receptor complex to bind to DNA D Inability of estrogen to bind to the receptor 27 The onset of action of steroids is typically: A Short, because the steroids diffuse directly into the nucleus B Short, because the steroids directly bind the DNA and stimulate transcription C Long, because the steroids must diffuse into the nucleus D Long, because the DNA must be transcribed and translated before the effect occurs 28 Loss of the C-terminal domain of the estrogen receptor leads to: A Constitutive activation of the estrogen response elements B Inability of the receptor to bind to DNA C Inability of the receptor to translocate into the nucleus D Constitutive deacetylation of the histones in estrogen response elements 30 A mouse model missing the caveolin-1 gene is created Females of this model will exhibit: I.Infertility II.An increase in FSH III.Enlarged ovaries secondary to increased estrogen production A B C D I only I and III only II and III only I, II, and III END OF TEST KAPLAN MCAT THE ANSWER KEY IS ON THE NEXT PAGE 10 as developed by Molecular Genetics Test ANSWER KEY: A B D C D 11 12 13 14 15 A D D C D 21 22 23 24 25 C A D B A 10 16 17 18 19 20 A D D A C 26 27 28 29 30 B D D C D KAPLAN C A B A D 11 MOLECULAR GENETICS TEST TRANSCRIPT Passage I (Questions 1–10) Choice A is the correct answer The last paragraph tells us that Galantamine is a reversible competitive inhibitor A competitive inhibitor is one that can be overcome by adding high concentrations of substrate, so with enough substrate the reaction will proceed just as fast as without the inhibitor Therefore the Vmax does not change Eliminate (B) and (D) However, the slope of the inhibited curve will be less steep In other words, to reach ½ Vmax the concentration of substrate must be greater with the inhibitor than without Km is the concentration at which ½ Vmax is reached Therefore in competitive inhibition Km will be greater than without (B) (C) (D) Distortion In competitive inhibition Vm does not change In non-competitive inhibition, however, Vm would decrease Opposite If k2 + k3 decreased then Km would decrease Km increases here, so k2 + k3 must also increase Distortion Vm does not change here Vm would never increase due to any type of inhibition Choice B is the correct answer If the cofactor for AchE were missing the enzyme would not function This effect would be similar to inhaling nerve gas, so look at paragraph Only (B) and (D) make sense here Though not mentioned in the paragraph, since Ach is common signal in the CNS, confusion is a likely side effect Answer (D) is very tempting since it almost matches the list in the paragraph, however, these effects were for nerve gas inhalation However (D) says bloody nose, whereas the list specifies runny nose Furthermore, the runny nose is unlikely to be a direct effect of AchE inhibition since acetylcholine does not directly effect the nose—the gas irritates the mucous membranes (B) is the best answer (A) (C) (D) Opposite The muscles will contract uncontrollably, but paralysis will not occur The pupils would contract, not dilate Distortion Though the other two occur, there’s no reason to believe that vasodilation would occur Vasoconstriction is more likely Distortion Though a very tempting choice, runny nose is an unlikely effect of manganese deficiency Choice D is the correct answer Don’t be fooled by the question stem; it’s asking about acetylcholinesterase, not acetylcholine All you need to know here is how cold blooded animals compare to warm blooded ones AchE in insects needs to work over a larger range of temperatures because the body temperatures of the insects change with the ambient temperature We cannot be sure which way the graph shifts—it would depend on the average temperature of the insect’s environment—but we know that it must be wider (D) is the correct answer (A) (B) (C) Distortion It may or may not shift, depending on environment; “cold blooded” does not mean their bodies are cold Distortion It may or may not shift, depending on environment Opposite The enzyme must function in various temperatures, so this graph would be wider than in humans The correct answer to this question is choice C Look carefully at the table In Tube 2, [S] = Km, so V = ½ Vmax Therefore when [Ach] is doubled in Tube 3, V would reach Vmax if not for the inhibitor (This is also another way to find Vmax = 0.35 mM/min) Now, if this were a competitive inhibitor, if sufficient substrate were added, Vmax would be reached However, if we look at Tube compared with Tube 3, [Ach] is increased by orders of magnitude, but the velocity does not change—Vmax for the enzyme with Tacrine has been reached and it’s smaller than the uninhibited enzyme’s Vmax This means that Tacrine must be noncompetitive (C) is the correct answer (A) Distortion The passage tells us that reversible inhibitors, not irreversible ones, are used to treat Alzheimer’s Tacrine is used to treat Alzheimer’s and thus it must not be irreversible Kaplan MCAT Biological Sciences Test Transcript (B) (D) Opposite In competitive inhibition, the inhibitor can be overwhelmed by adding excess substrate The inhibitor reaches a lower Vmax than the uninhibited enzyme FUD The passage says Tacrine is an inhibitor, so by definition it can’t be an activator Noncompetitive inhibitors are usually allosteric, however The correct answer is choice D Go back to the equations given in the passage To find Vo, we must combine the two equations Substitute kcat [E] for Vmax vo = = kcat[ E ][ S ] Km+[ S ] (1.4 × 104 )(2.5 × 10−5 )(2.0 × 10−8 ) 7.0 × 10−9 = 9.5 × 10−5 + 2.0 × 10−8 9.502 × 10−5 Now we need to approximate, = 7 ≅ = × 10−5 95, 020 100, 000 We increased the denominator, so the answer should be a bit bigger than x 10-5 Answer (D), 7.4 x 10-5 mM/min, is the correct answer (A) (B) (C) Miscalculation Order of magnitude error from miscalculating exponents Miscalculation Results from switching Km and kcat Miscalculation Results if the denominator is multiplied instead of added The correct answer is choice C If the levels of acetylcholine in the brain dropped, the body would attempt to mitigate the impact of such a change One way it might that is to produce more receptors for the Ach to bind to, thus increasing the efficacy of the remaining neurotransmitter (C) is the correct answer You can also answer this question by process of elimination-(A) (B) (D) Out of scope While Alzheimer’s does affect the brain severely, memory is its major victim You might arrive at this answer if you are misled by the fact that Ach is the primary neurotransmitter at the neuromuscular junction Partial paralysis could occur if the innervating neurons could not release a signal to the muscle, but we are specifically talking about the brain here Opposite If plaques are closely involved with either the cause or effect of the disease, then they should be associated with other causes/effects If the Ach levels are high in areas that the disease has effect, this would suggest that Alzheimer’s leads to increased levels of Ach Opposite This is a tricky option The third paragraph tells us that AchE is subject to negative feedback loops Less inhibition means more enzyme, which means less acetylcholine So this could strengthen the hypothesis However, if the body was producing less acetylcholine, it is likely that endogenous AchE inhibitors would increase in order to retain as much of the Ach as possible Therefore, lower levels of inhibitor could suggest that Ach levels in the brain are too high This discovery would not conclusively strengthen or weaken the theory Choice A is the right answer The last paragraph states that Galantamine is an allosteric activator of the nicotinic acetylcholine receptor This means that it binds to the protein at a secondary site and changes the receptor’s shape to give it a greater affinity for acetylcholine (A) is true and is the correct answer (B) Out of scope This is actually true When Galantamine binds to the nicotinic Ach receptor, it does activate a second messenger which causes the cell to express more receptors However, this is not discussed in the passage, and you are not expected to bring outside knowledge of Galantamine to the passage Kaplan MCAT Biological Sciences Test Transcript (C) (D) Opposite In allosteric activation, this graph would actually shift to the left, that is, the activator makes the protein more work with less substrate because it has a greater affinity for substrate Distortion Galantamine stabilizes the receptor-transmitter complex It makes the empty receptor less stable and thus increases its affinity for the transmitter—it needs the transmitter to become stable again The correct answer is choice B Myasthenia gravis is an autoimmune disease in which antibodies attack the nicotinic receptors on the postsynaptic membrane in the neuromuscular junction The antibodies prevent acetylcholine from binding to the receptors (and sometimes destroying them) and thus block the transmission and prevent muscle contraction Pyridostigmine bromide inhibits acetylcholinesterase and thus increases the concentration of the acetylcholine produced at the nerve terminal This helps to ease effect of lost (or blocked) receptors by making the most use of the available ones None of the other answers could both cause the symptoms and be eased by the treatment (B) is the correct answer (A) (C) (D) Distortion If the acetylcholine was not usable, preventing its breakdown and increasing its concentration would not ease the symptoms of the disease Distortion Again, this might cause the symptoms, but increasing Ach in the synapse will not make up for damage in transmission in the dendrites Opposite Too much Ach could result in spasms, which could result in weakness, but preventing breakdown of Ach would only increase its concentration and thus could not solve this problem Choice A is the correct answer This is a complicated question First we need to consider the differences between the two types of enzyme A protein that is largely hydrophobic is not going to be found in the cytoplasm or in the synapse—it will be imbedded in the cell membrane The hydrophilic enzyme will be free in fluid—likely in the synapse as paragraph discusses reabsorbing the metabolites from Ach So if the hydrophilic enzyme is knocked out, degradation of Ach would occur in the synapse close to the presynaptic or postsynaptic membranes More Ach would be present away from the enzymes that degrade it than near them Thus there would be a concentration gradient after neurotransmitter release, with more Ach in the middle of the gap than on either side (A) is the correct answer (B) (C) (D) Opposite The gradient would more likely occur in the opposite direction Distortion Less degradation will not mean that more neurotransmitter is released Opposite Less Ach will be broken down, most likely, because less enzyme will be present 10 The correct answer to this question is choice D Allosteric enzymes have multiple active sites and their reaction velocity depends on more than the concentration of one compound, hence the s-shaped curve (Think of Hb, an allosteric protein with cooperative binding.) Now to decide between (C) and (D) The choice is clear—an inhibited enzyme will reach Vmax at higher concentrations of substrate (A) (B) (C) Distortion This enzyme shows no coopertivity It is more likely to have only one active site Distortion This enzyme shows no coopertivity It is more likely to have only one active site Opposite Inhibition would result is less product produced per unit of concentration of substrate The inhibition curve must shift right Passage II (Questions 11–20) 11 The correct answer is choice A Since the RNA has to be reverse transcribed into DNA using RT-PCR; reverse transcriptase, polymerase, dNTPs and primers would be needed Helicase on the other hand is an ATP driven protein that unwinds the helix as the DNA is transcribed and is not necessary to carry out PCR (B) (C) (D) 180° RT-PCR requires reverse transcriptase to convert the RNA into DNA 180° Polymeraes is required to elongate the DNA chains 180° dNTPs are needed to form the DNA Kaplan MCAT Biological Sciences Test Transcript 12 The correct answer is choice D According to the passage, the mutants were made by transforming the wild-type bacteria with three different plasmids, each containing ampicillin resistant genes Since the wild-type bacteria was not transformed with such a plasmid, it would be unable to grow in a plate containing ampicillin (A) (B) (C) 180° Mutant was transformed with an ampicillin resistant plasmid 180° Mutant was transformed with an ampicillin resistant plasmid 180° Mutant was transformed with an ampicillin resistant plasmid 13 The correct answer is choice D In the southern blot, the RNA from the plates are reverse transcribed to DNA and then hybridized with the glmS gene Thus plates A, C and D which represent mutants 1,2, and (not necessarily in that order), all of which overexpress glmS DNA will develop bands in the southern blot In the event where mRNA were to be cleaved in order to produce a ribozyme, both the cleaved portions of the mRNA would hybridize with the labeled gene Wild-type however will not develop a band because it does not grow in the ampicillin-medium (dish B) 14 The correct answer is C Binding of glucosamine-6-phosphate to a binding site on glmS transcript will cause the glmS transcript to be spliced into two pieces, both of which will hybridize with the labeled glmS in the southern blot producing two bands Mutant is the only mutant that overexpresses both glutamine and fructose-6phosphate, both of which are necessary to produce glucosamine-6-phosphate and thus is the only mutant that will have bands in the southern (A) (B) (D) 180° Mutant only expresses fructose-6-phosphate, and so is not able to produce glucosamine6-phosphate Without glucosamine-6-phosphate, glmS is not split into 180° Mutant only expresses glutamine, and so is not able to produce glucosamine-6-phosphate Without glucosamine-6-phosphate, glmS is not split into 180° Wild-type does not express fructose-6-phosphate nor glutamine, and so is not able to produce glucosamine-6-phosphate Without glucosamine-6-phosphate, glmS is not split into 15 The correct answer is choice D According to the passage, the 3' end of mRNA of the gene glmS is spliced in a similar fashion to ribosomal RNA in Tetrahymena The passage does not discuss what induces this splicing and so we not know whether this is positive regulation or negative regulation Also, the passage does not discuss repressors or inducers, so answer choice C can be eliminated (A) (B) (C) Out of Scope Regulation of ribozyme production is not discussed in the passage Out of Scope Regulation of ribozyme production is not discussed in the passage Out of Scope Regulation of ribozyme production is not discussed in the passage 16 The correct answer is choice A mRNA that is polycistronic is RNA that can code for more than one polypeptide Since the sequence of nucleotides is the same it will be spliced in the same manner to produce a ribozyme Therefore, if mRNA for glmS is polycistronic, it would still produce the same strand of DNA when it undergoes RT-PCR to yield the same results The way in which mRNA is translated does not affect the results of the PCR (B) (C) (D) 180° The way in which mRNA is translated does not affect the results of the PCR 180° The way in which mRNA is translated does not affect the results of the PCR 180° The way in which mRNA is translated does not affect the results of the PCR 17 The correct answer is choice D Although there are methods to amplify RNA, RT-PCR is used to analyze that transcription because DNA is much more stable and easier to work with than RNA RT-PCR however does not technically amplify RNA, it transcribes the RNA to DNA then amplifies that DNA Thus answer choice D is the correct answer (A) Distortion RT-PCR however does not technically amplify RNA, it transcribes the RNA to DNA then amplifies that DNA Kaplan MCAT Biological Sciences Test Transcript (B) (C) 180° RNA can be amplified 180° RNA can be detected in a manner similar to Southern blot, called Northern blot 18 The correct answer is choice D All nucleic acids are polarized It is impossible for any nucleic strand to not have both a 3’ end and a 5’ end Thus answer choice D is the correct answer (A) (B) (C) 180° Polarity refers to the presence of a 3' and 5' end When the 3' end is cleaved, a new 3' end is created 180° Polarity refers to the presence of a 3' and 5' end When the 3' end is cleaved, a new 3' end is created 180° Polarity refers to the presence of a 3' and 5' end When the 3' end is cleaved, a new 3' end is created 19 The correct answer is choice A Translocation is the transfer of a chromosomal segment to a nonhomologous chromosome Since prokaryotes have only one circular chromosome, they cannot undergo translocation (B) (C) (D) 180° Transformation is the incorporation of a plasmid by a bacterial cell It leads to genetic variation by the incorporation of new genes into the chromosome 180° Transduction is the transfer of genetic material from one bacterial cell to another through a bacteriophage It leads to genetic variation by the spreading of new genes to new cells 180° Conjugation is the transfer of genetic material between two bacterial cells that are temporarily joined It leads to genetic variation by the spread of new genes to new cells 20 The correct answer is choice C A ribozyme is an RNA catalyst that catalyzes reactions without itself being consumed; therefore, it is a true enzyme Although a ribozyme can form tertiary structures it has only different types of nucleic acids whereas proteins have over 20 types of amino acids giving proteins a greater variety of possible interactions Finally, ribozymes comprised of RNA are much less stable than proteins (A) (B) (D) 180° Enzymes are not used up in the reaction Ribozymes are defined as enzymes, so they are not used in up in reactions 180° Ribozymes can form tertiary structures 180° Proteins are more stable than RNA Passage III (Questions 21–30) 21 The correct answer is choice C The passage states that the intracellular receptors for hormones are similar The N-terminal domains contain the transcription-activation domain, and the middle of the protein contains the DNA-binding domain The C-terminal domain is specific for the hormone-this is the site where the hormone binds Testosterone and estrogen receptors will be least similar in this domain (A) (B) (D) Opposite The N-terminal domains of the estrogen and testosterone receptors will be similar because they contain the transcription-activation domain Opposite The center of the estrogen and testosterone receptors will be similar because they contain the DNA-binding domain Out of Scope The tertiary structure of the proteins will probably be similar; however, the question asks about the amino acid sequence, which is the primary structure 22 The correct choice is A When DNA is relaxed it is called euchromatin, or true chromatin It is in this form that it can be transcribed The passage states that when the estrogen-receptor complex binds, the histones are acetylated and transcription is activated Therefore, acetylation promotes transcription factor binding (A) is the correct answer (B) Opposite Coiled DNA (heterochromatin) cannot bind transcription factors Kaplan MCAT Biological Sciences Test Transcript (C) (D) Opposite Relaxed DNA allows transcription factor binding Opposite Acetylation allows relaxation of the DNA into euchromatin to allow transcription factor binding 23 The correct choice is D Each estrogen-regulated gene requires molecules of estrogen to promote transcription because the estrogen-receptor complex must dimerize before it translocates into the nucleus If a cell contains estrogen-regulated genes, molecules of estrogen are required per cell For 10,000 cells, 60,000 molecules of estrogen are required (A) (B) (C) Miscalculation Only allows for molecule of estrogen per cell Miscalculation Allows for the required estrogen dimer, but does not account for the genes in each cell Miscalculation Allows for the genes in each cell, but does not account for the required estrogen dimer 24 Choice B is the correct answer The zinc-finger domain of the estrogen receptor is the DNA binding domain If the receptor does not have a zinc-finger domain then the receptor cannot bind to the DNA to acetylate (when bound to estrogen) or deacetylate (when not bound to estrogen) the histones So the answer is (B) (A) (C) (D) Opposite Constitutive transcription of estrogen-regulated genes would result if the estrogen receptor were always in the active conformation Loss of the zinc-finger domain would result in constitutive deactivation of the estrogen-regulated genes Opposite Estrogen receptor not bound to estrogen deacetylates the estrogen regulated genes However, it still requires the zinc finger domain to so Therefore, loss of the zinc-finger domain would result in inability to deacetylate the estrogen-regulated genes Out of Scope The passage does not discuss how the estrogen-receptor dimer translocates into the nucleus In fact, scientists not know how this translocation occurs 25 Choice A is the correct answer Hormones that bind to intracellular receptors must be lipophilic in order to cross the lipophilic cell membrane They cannot be charged, nor can they be very polar molecules (B) (C) (D) Opposite Anions (negatively charged ions) cannot cross the cell membrane Opposite Amino acids are polar molecules that cannot cross the cell membrane Opposite Nucleic acids are polar molecules that cannot cross the cell membrane 26 Choice B is the correct answer Steroid hormones, such as estrogen and testosterone, are produced from cholesterol (A) (C) (D) Distortion Fatty acids are used in energy production and storage Distortion Amino acids are used in the synthesis of proteins Distortion Glycogen is used for energy production and storage 27 Choice D is the correct answer Steroid hormones typically exert long-lasting, delayed effects because the proteins must be produced before the steroid takes effect It could therefore by several days before the release of the steroid into the bloodstream results in the production of proteins regulated by estrogen (A) (B) (C) Distortion Although the steroids diffuse directly into the nucleus, the onset of action is the time until the protein is produced, which is long Distortion The onset of action is the time it takes to produce the protein, which is long starting from transcription Distortion The length of time for the steroid to diffuse into the nucleus is not the ratedetermining step 28 Choice D is the right answer The C-terminal domain contains the estrogen-binding domain If the estrogen-binding domain is absent, the estrogen receptor will never be able to bind estrogen It will therefore always behave as it does when no estrogen is present and continuously deacetylate the histones in estrogen response elements Kaplan MCAT Biological Sciences Test Transcript (A) (B) (C) Opposite Constitutive activation of the estrogen response elements would result from a mutation that allowed estrogen to remain bound or a mutation that altered the conformation of the receptor to the estrogen bound state Distortion The zinc-finger domain is the DNA-binding domain Distortion The receptor is capable of diffusing into the nucleus in the absence of estrogen; loss of the C-terminal domain is unlikely to effect this 29 Choice C is the right answer The passage only gives one function for the estrogen receptor chaperone protein: hiding the DNA binding site Therefore, if the receptor is unable to dissociate from the chaperone, the most likely consequence will be inability to bind to DNA Although (B) and (D) may occur, the passage gives no evidence to that effect (A) (B) (D) Opposite A mutation in the chaperone would lead to underproduction of estrogen regulated genes Out of Scope Although this could occur, the passage gives no evidence or support of this mechanism Out of Scope Although this could occur, the passage gives no evidence or support of this mechanism 30 Choice D is the right answer If caveolin-1 is missing, the estrogen-receptor complex cannot reach the nucleus and exert estrogen's effects on protein production The body will perceive the defect as a lack of estrogen, and try to compensate by over-producing estrogen The pituitary will increase its production of FSH (follicle stimulating hormone) to augment estrogen production, and the ovaries may enlarge due to cell production in attempts to increase estrogen production However, oocytes and the endometrium will be unable to respond to the estrogen, resulting in infertility .. .Molecular Genetics Topical Test Time: 42 Minutes Number of Questions: 30 MCAT Molecular Genetics DIRECTIONS: Most of the questions in the following... developed by Molecular Genetics Test ANSWER KEY: A B D C D 11 12 13 14 15 A D D C D 21 22 23 24 25 C A D B A 10 16 17 18 19 20 A D D A C 26 27 28 29 30 B D D C D KAPLAN C A B A D 11 MOLECULAR GENETICS. .. conformation before the receptor binds acetylcholine GO ON TO THE NEXT PAGE as developed by Molecular Genetics Test The competitive acetycholinesterase inhibitor Pyridostigmine bromide is used
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