AAMC MCAT test 7

Printing Guide Use this printing guide as a reference to print selected sections of this practice test To print, click the PRINTER icon located along the top of the window and enter one of the following options in the PRINT RANGE section of the print dialog window: To Print Enter Print Range Options Complete Practice Test Click ALL radio button Physical Sciences Section Click PAGES FROM radio button and enter pages to 24 Verbal Reasoning Section Click PAGES FROM radio button and enter pages 25 to 45 Writing Sample Section Click PAGES FROM radio button and enter pages 46 to 48 Biological Sciences Section Click PAGES FROM radio button and enter pages 49 to 75 Periodic Table Click PAGES FROM radio button and enter page 50 to 50 Answer Sheet Click PAGES FROM radio button and enter page 76 to 76 This document has been encoded to link this download to your member account The AAMC and its Section for the MCAT hold the copyrights to the content of this Practice Test Therefore, there can be no sharing or reproduction of materials from the Practice Test in any form (electronic, voice, or other means) If there are any questions about the use of the material in the Practice Test, please contact the MCAT Information Line (202-828-0690) MCAT Practice Test Physical Sciences Time: 100 minutes Questions: 1-77 Most questions in the Physical Sciences test are organized into groups, each containing a descriptive passage After studying the passage, select the one best answer to each question in the group Some questions are not based on a descriptive passage and are also independent of each other If you are not certain of an answer, eliminate the alternatives that you know to be incorrect and then select an answer from the remaining alternatives Indicate your selected answer by marking the corresponding answer on your answer sheet A periodic table is provided for your use You may consult it whenever you wish This document has been encoded to link this download to your member account The AAMC and its Section for the MCAT hold the copyrights to the content of this Practice Test Therefore, there can be no sharing or reproduction of materials from the Practice Test in any form (electronic, voice, or other means) If there are any questions about the use of the material in the Practice Test, please contact the MCAT Information Line (202-828-0690) Periodic Table of the Elements H He 4.0 10 1.0 Li Be B C N O F Ne 6.9 9.0 10.8 12.0 14.0 16.0 19.0 20.2 11 12 13 14 15 16 17 18 Na Mg Al Si P S Cl Ar 23.0 24.3 27.0 28.1 31.0 32.1 35.5 39.9 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 39.1 37 40.1 38 45.0 39 47.9 40 50.9 41 52.0 42 54.9 43 55.8 44 58.9 45 58.7 46 63.5 47 65.4 48 69.7 49 72.6 50 74.9 51 79.0 52 79.9 53 83.8 54 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe 85.5 55 87.6 56 88.9 57 91.2 72 92.9 73 95.9 74 (98) 75 101.1 76 102.9 77 106.4 78 107.9 79 112.4 80 114.8 81 118.7 82 121.8 83 127.6 84 126.9 85 131.3 86 Cs Ba La* Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn 132.9 87 137.3 88 138.9 89 178.5 104 180.9 105 183.9 106 186.2 107 190.2 108 192.2 109 195.1 197.0 200.6 204.4 207.2 209.0 (209) (210) (222) Fr Ra Ac† Unq† Unp Unh Uns Uno Une (223) (226) (227) (261) (262) 58 (263) 59 (262) 60 (265) 61 (267) 62 63 64 65 66 67 68 69 70 71 Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu 140.1 90 140.9 91 144.2 92 (145) 93 150.4 94 152.0 95 157.3 96 158.9 97 162.5 98 164.9 99 167.3 100 168.9 101 173.0 102 175.0 103 Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr 232.0 (231) 238.0 (237) (244) (243) (247) (247) (251) (252) (257) (258) (259) (260) * † Th Passage I Thousands of tons of hydrazine (N2H4) are produced each year for commercial uses, including the production of agricultural chemicals At room temperature, hydrazine is a volatile liquid that exists in hydrogen-bonded networks similar to those found in liquid water Hydrazine may be prepared by the Raschig process, the reaction of ammonia with sodium hypochlorite, as shown in Equation 2NH3(g) + NaOCl(aq) → N2H4(aq) + NaCl(aq) + H2O(ℓ) Like ammonia, hydrazine is a base in aqueous solution Figure shows the equilibria reactions of ammonia and hydrazine in aqueous solution NH3(aq) + H2O(ℓ) NH4+(aq) + OH-(aq) Keq = 1.8 × 10-5 N2H4(aq) + H2O(ℓ) N2H5+ + OH-(aq) Keq = 8.5 × 10-7 N2H5+ + H2O N2H62+ + OH-(aq) Keq = 8.9 × 10-16 Figure Equilibria (Keq = equilibrium constant) Equation Hydrazine usually is shipped as the hydrate (N2H4 · H2O) because it is easier to handle and can be easily dehydrated to form the anhydrous compound Hydrazine and its chemical derivatives are good rocket propellants For example, hydrazine reacts with dinitrogen tetroxide (N2O4) to produce gaseous nitrogen and water Equation shows the reaction and the enthalpy change N2H4(ℓ) + N2O4(ℓ) → N2(g) + H2O(g) ∆H° = -1040 kJ mol-1 Which of the following Lewis structures best represents hydrazine? A) B) C) Equation Some thermochemical data for hydrazine and dinitrogen tetroxide are given in Table Table Properties of Hydrazine and Dinitrogen Tetroxide at 298 K Property N2H4(ℓ) N2O4(g) ∆Hf° (kJ mol-1) 50.6 9.2 -1 ∆Gf° (kJ mol ) 149.2 97.9 -1 -1 S° (J K mol ) 121.2 304.3 D) How many grams of ammonia are required to make one mole of hydrazine by the Raschig process? A) 8.5 g B) 17.0 g C) 32.0 g D) 34.0 g Sharing or reproducing this material in any form is a violation of the AAMC copyright What is the percent by weight of hydrazine in hydrazine hydrate? A) 6.0/50.0 x 100% B) 18.0/32.0 x 100% C) 18.0/50.0 x 100% D) 32.0/50.0 x 100% What is the enthalpy change (∆Ho) for the reaction shown above? A) 50.6 kJ mol-1 B) 149.2 kJ mol-1 C) (149.2 + 298 x 121.2) kJ mol-1 D) (149.2 - 50.6) kJ mol-1 As a result of being a weaker base than ammonia, hydrazine: The formation of hydrazine from its elements is NOT a spontaneous process at 25oC and atm because: A) ∆So for the reaction is > B) ∆Ho for the reaction is < C) ∆Go for the reaction is > D) So for hydrazine is > The entropy change (∆So) for the reaction shown in Equation is: A) < because the moles of gaseous products > the moles of gaseous reactants B) < because water is a product of the reaction C) > because the moles of gaseous products > the moles of gaseous reactants D) > because water is a product of the reaction A) has a smaller acidity constant (Ka) than does ammonia B) has a smaller basicity constant (Kb) than does ammonia C) can be protonated twice to form N2H62+ D) forms hydrogen bonds in aqueous solution Sharing or reproducing this material in any form is a violation of the AAMC copyright Passage II A gas of electrically charged and neutral particles is called a plasma Plasma physics is a broad term applicable to such diverse areas as space physics, gas lasers, gaseous electronics, and controlled thermonuclear fusion A plasma has the ability to oscillate and propagate waves These waves can be excited by applying an oscillating electric field to the plasma The simplest oscillation is a high-frequency oscillation of the plasma electrons Consider a plasma that is electrically neutral, consisting of positive ions immersed in a “sea” of electrons If the electron sea is slightly displaced from the ionic background, electric fields act to restore the electrons to their original equilibrium positions The electron sea subsequently moves toward the equilibrium position, overshoots, and oscillates back and forth These oscillations are so rapid that the positive ions seem to be fixed in the background (see Figure 1) The frequency f at which these oscillations occur for a given number density, n (electrons per cubic meter), is 1/2 f = [kne /(πm)] 1/2 ≈ 9.0n where e and m are the elementary charge and electron mass, 1.6 x 10-19 C and x 10-31 kg, respectively The constant k = x 109 Nm2/C2 occurs in Coulomb’s law The approximation on the right side of the equation gives the frequency in Hz, when n is expressed in m-3 Figure Positive ions surrounded by a sea of mobile electrons (gray denotes the electron sea) A and C denote the oscillation extremes Why can the positive ions be considered to be fixed during the electrons’ oscillations? A) The ions are bound together with strong nuclear forces B) An ion is much more massive than an electron C) The ions experience no force when the electron sea is displaced D) Coulomb’s law prohibits the motion of the ions In Figure 1, the maximum electrical potential energy occurs at: A) A only B) B only C) C only D) A and C only Sharing or reproducing this material in any form is a violation of the AAMC copyright 10 The density of a typical laboratory plasma is 1018 m-3 This value leads to plasma oscillations at: 12 As the Figure electrons oscillate through equilibrium point B, they move on to C because of: A) x 1018 Hz B) x 1012 Hz C) x 109 Hz D) x 106 Hz A) the momentum gathered as they moved from point A B) Coulomb forces pulling on the electron sea C) magnetic forces of attraction between the positive ions and the electron sea D) the large potential energy they have at point B 11 A plasma wave moving through a plasma has a frequency of 109 Hz and a speed of 3.0 x 107 m/s What is the wavelength of this wave? A) 3.0 cm B) 3.0 m C) 3.3 cm D) 3.3 m Sharing or reproducing this material in any form is a violation of the AAMC copyright 13 What best describes changes that occur as the electron sea moves from position A to position B in Figure 1? A) Kinetic energy is transformed into potential energy B) Potential energy is transformed into kinetic energy C) Power is dissipated as heat D) Turbulence brings the electron sea to rest Passage III Silicon, the second most abundant element in the earth’s crust, is found combined with oxygen in a variety of silicate minerals The most common is silica (SiO2), which is a network solid Silicon cannot be purified by electrolytic techniques When elemental potassium became available in the nineteenth century, it was used in a silicon purification procedure Today, silicon is produced commercially by the reaction of silica with carbon or calcium carbide in an electric furnace at 2000°C (Equation 1) The product is about 98% pure, with impurities of iron, oxygen, aluminum, and other elements Further purification is achieved by halogenating the silicon, purifying the resulting gas by fractional distillation, and then reducing the halogenated silicon compound (Equations 2-3) SiO2(s) + C(s) → Si(ℓ) + CO(g) Equation Si(s) + HCl(g) → SiCl3H(g) + H2(g) Equation SiCl3H(g) + H2(g) → Si(s) + HCl(g) 15 What is the electron configuration for a groundstate silicon atom? A) [Ne] 3s B) [Ne] 3s C) [Ne] 3s D) [Ne] 3s ↑↓ ↑↓ ↑↓ ↑↓ 3p ↑↑ 3p ↑ ↓ _ 3p ↑↓ _ 3p↓ ↓ ↓ 16 According to valence shell electron pair repulsion (VSEPR) theory, what is the geometry around silicon in SiCl3H? A) Linear B) Tetrahedral C) Trigonal bipyramidal D) Octahedral 17 Which of the following elements could best substitute for potassium in the purification of silicon? A) H2 B) Na C) Mg D) Ca 18 SiCl3H has a normal boiling point of 33oC What are the predominant forces between SiCl3H molecules? Pure silicon is a hard, brittle, nonreactive substance with a metallic luster A) Ionic forces B) Covalent bonds C) Hydrogen bonds D) van der Waals forces 14 The purification of elemental silicon was difficult to achieve because it: 19 SiCl3H is purified by fractional distillation Why does this procedure effect a purification? A) is a rare element B) is too reactive to isolate easily C) exists in minerals that not decompose easily D) does not crystallize A) SiCl3H is not water soluble B) SiCl3H is decomposed by water C) SiCl3H has a lower boiling point than the solid impurities D) SiCl3H has a lower melting point than the impurities Equation Sharing or reproducing this material in any form is a violation of the AAMC copyright Passage IV The production of electrical power via nuclear fission reactions often provokes heated discussions about nuclear waste disposal In a typical uranium fission, a uranium nucleus absorbs a neutron and undergoes fission, as illustrated in the reaction The superscript denotes the atomic mass and the subscript the atomic number The nucleus U-236 (i.e., 236U) decays immediately into two fission fragments X and Y, along with the release of two or three neutrons Energy is produced in the fission process by the conversion of nuclear mass into energy This conversion is described by Einstein’s famous relation E = mc2, where c is the speed of light x108 m/s, m is the mass that is converted, and E the resulting energy released An analysis of the reaction shown reveals that about 1/1000 of the original starting mass of U-235 is missing after the reaction This missing mass accounts for the energy produced in the reaction The fission fragments X and Y constitute the radioactive waste from uranium fission These fragments then undergo beta and/or gamma decay The resulting fragments themselves may be radioactive, resulting in further decays until a stable isotope is reached Hundreds of years must pass before these radioactive fragments decay to nonradioactive nuclei 20 If three neutrons are produced in the U-235 fission reaction discussed in the passage, what relation must the atomic masses A1 and A2 obey? 21 If fission fragment X undergoes beta decay, then one neutron in the nucleus is converted into a proton, an electron and a neutrino (the electron and neutrino v′exit the atom) If the new fission fragment is called X ′, the betadecay reaction would be written as: A) B) C) D) 22 Half-lives are useful indicators of how dangerous a radioactive substance is The halflives of Pu-239 and Ra-226 are 24,000 yrs and 1600 yrs, respectively In comparison to atoms of Pu-239, atoms of Ra-226 will decay at a rate: A) times faster B) 15 times faster C) times slower D) 15 times slower 23 A standard coal-burning power plant produces about 106 kg of fly-ash every week Assuming that the density of fly-ash is 1000 kg/m3, what would be the length of the side of a fly-ash cube made from this waste? A) m B) 10 m C) 100 m D) 1000 m A) A1 + A2 = 92 B) A1 + A2 = 232 C) A1 + A2 = 233 D) A1 + A2 = 236 Sharing or reproducing this material in any form is a violation of the AAMC copyright 10 Passage VI Crohn’s disease and ulcerative colitis are two forms of inflammatory bowel disease, which differ in several respects Crohn’s disease may occur in any part of the gastrointestinal tract while ulcerative colitis is confined to the colon (large intestine) Crohn’s disease may involve all layers of the tract while ulcerative colitis affects only the mucosa, the inner lining of the colon Abdominal pain and diarrhea following a meal are signs of Crohn’s disease, but a progressive loosening of a bloody stool is the first symptom of ulcerative colitis Management of inflammatory bowel disease is achieved by drug therapy to suppress the inflammation which leads to diarrhea, but there is no known cure The cause of inflammatory bowel disease is controversial Genetic, pathogenic, and immunogenic theories have all been advanced Inflammatory bowel disease tends to run in families, with 20% of patients having a relative with the disorder But if inflammatory bowel disease is genetic, it is not inherited in a simple Mendelian way Some research suggests that inflammatory bowel disease is an autoimmune disease An antigen in the body, perhaps in the digestive tract, is recognized as foreign by the immune system This antigen may then stimulate the body’s defenses to produce an inflammatory response that continues without control 173 Normally the immune system avoids attacking the tissues of its own body because: A) a special intracellular process recognizes only foreign antigens B) the body does not make any antigens that the immune system could recognize C) it changes its antibodies to be specific only to foreign antigens D) it suppresses cells specific to the body's own antigens 174 An ulcer that penetrated the wall of the intestine would allow the contents of the gastrointestinal tract to enter: A) the perineum B) the peritoneal cavity C) the pleural cavity D) the lumen of the intestine 175 If the genetic and autoimmune theories of inflammatory bowel disease are true, then the gastrointestinal antigen being targeted by the immune system is probably on: A) the chromosomes carrying the genes for the disease B) part of the DNA segments constituting the genes for the disease C) stretches of the mRNA's coded for by the genes for the disease D) the surface of the proteins encoded by the genes for the disease There is also a lingering suspicion that the inflammation is triggered by some bacterium or other organism that takes up residence in the gastrointestinal system This theory has had a resurgence since the discovery that the bacterium Helicobacter pylori may play a causal role in gastric ulcers 176 The fact that there appears to be a genetic component to inflammatory bowel disease, but that it does not show clear Mendelian inheritance ratios suggests any of the following, EXCEPT: 172 What process would be most disrupted by an inflammation of the colon? A) the gene for the disease has incomplete penetrance B) the gene for the disease has limited expressivity C) the disease is polygenic D) the gene for the disease is recessive A) Digestion B) Absorption of nutrients C) Absorption of water D) Secretion of digestive enzymes Sharing or reproducing this material in any form is a violation of the AAMC copyright 62 Passage VII Fats are known to affect blood flow Research was conducted to examine the effects of a 20-carbon polyunsaturated fatty acid on blood flow through the skin Skin was chosen because blood flow could be measured easily and without discomfort to the subjects Male college students ranging from 18-28 years of age were randomly divided into treatment groups of 10 subjects each Prior to participation in the study, subjects were screened for health conditions, medications, and/or dietary practices that would bias the collected data Subjects were instructed not to alter their habits or lifestyles during the experiment Each treatment group was given of dietary supplements (Table 1) Table Dietary Supplements placebo placebo + vitamin E fatty acid fatty acid + vitamin E Vitamin E, an effective anti-oxidant, was given to groups to reduce in vivo oxidation of the ingested fatty acids Both the fatty acid and the placebo were given at g/10 kg body weight/day; vitamin E was given at 100 mg/10 kg body weight/day The supplements were packaged in identical gelatin capsules and taken before meals Subjects were unaware of which supplement they were ingesting Blood flow through the skin of each subject’s arm was measured twice; immediately before and immediately after 60 days of dietary supplementation Each blood-flow measurement was taken at the same time of day and at a skin temperature of 32oC Subjects were required to fast for 12 hours and rest quietly in the laboratory for 30 minutes before skin blood flow was measured The results are shown in Figure Sharing or reproducing this material in any form is a violation of the AAMC copyright Figure Effect of dietary supplements on skin blood flow 177 The most likely explanation for the difference in skin blood flow between the fatty acid group and the fatty acid + vitamin E group in Figure is that: A) vitamin E alone reduces skin blood flow more than fatty acids alone B) vitamin E alone increases skin blood flow more than fatty acids alone C) the products of fatty acid oxidation reduce skin blood flow D) unoxidized fatty acids reduce skin blood flow 178 It was hypothesized that the decrease in blood flow to the skin resulted from a change in the activity of the sympathetic nerves to the skin Which of the following observations would support this hypothesis? A) A change in the norepinephrine content of blood draining from the skin B) In vitro contraction of the smooth muscle in skin blood vessels in response to acetylcholine C) A lack of epinephrine receptors in skin blood vessels D) In vivo dilation of the skin blood vessels 63 179 An alternative method for examining the effects of fatty acids on blood flow would be to measure changes in blood pressure If blood pressure were measured, one would predict that it would be lowest in which of the following? A) Heart B) Arteries C) Arterioles D) Capillaries 180 To interpret the results, the researchers must assume that: A) fatty acids have no effect on skin blood flow B) vitamin E reduces skin blood flow C) the subjects did not alter their habits during the study D) blood pressure differed between the measurements Sharing or reproducing this material in any form is a violation of the AAMC copyright 181 A 20-carbon polyunsaturated fatty acid was used in this study Unsaturated fatty acids differ from saturated fatty acids in that saturated fatty acids contain: A) more carbon atoms B) fewer hydrogen atoms C) no carbon-hydrogen bonds D) no carbon-carbon double bonds 182 In the design of the experiment, all of the following factors were controlled EXCEPT: A) skin temperature B) age of the subjects C) diurnal rhythms in physiological responses D) skin blood flow 64 Passage VIII Organic carbocations may be generated from alcohols in the presence of strong acids or from alkenes by the addition of a proton Carbocations are stabilized through the inductive effect and through resonance Alkyl groups bonded to the cation center also may add stability through the partial overlap of filled orbitals with an empty orbital, a concept often referred to as hyperconjugation Such alkyl groups have one filled orbital aligned with the empty orbital of the cation The overlap of these orbitals allows the pair of electrons in the filled orbital to reduce the electron deficiency of the cation center In an early research study designed to examine the properties of carbocations, an ethyl carbocation was produced from the reaction of ethyl fluoride with SbF5, as shown in Reaction I (unbalanced) The 4-carbon molecules were hypothesized to arise from the following steps: • The ethylcarbocation loses a proton to form ethene • Another CH3CH2+ carbocation reacts with the ethene to form a 4-carbon carbocation • Rearrangements contribute to the formation of the final products 183 Which of the following concepts may be used to explain why the stability of carbocations increases as the number of alkyl groups attached to the carbocation increases? A) Resonance B) Steric interference C) Hyperconjugation D) Ion pairing 184 Which of the following compounds is the major compound isolated from Reaction 1? A) Reaction I The products of Reaction I were quenched with water, then separated by gas chromatography, and identified by 1H NMR The resulting mixture of products consisted primarily of molecules with four carbon atoms The proton NMR spectrum of the major product is shown below in Figure B) C) D) 185 The ethyl carbocation may be formed when ethanol is heated with sulfuric acid The first step of this reaction is: Figure The next most abundant product was an optically inactive sample, later determined to be sec-butyl alcohol Sharing or reproducing this material in any form is a violation of the AAMC copyright A) the elimination of water from ethyl alcohol B) the protonation of the hydroxyl oxygen of ethyl alcohol C) the loss of a proton by ethyl alcohol D) the loss of the hydroxyl group of ethyl alcohol 65 186 The gas chromatograph trace from the workup of Reaction is shown below with the integrated areas indicated 187 All the products that were isolated from the quenching of Reaction should contain an absorption in the infrared spectrum near: A) 1600 cm-1 B) 1700 cm-1 C) 2250 cm-1 D) 3500 cm-1 What percentage of sec-butyl alcohol was isolated? A) 1.5% B) 3% C) 5% D) 6% Sharing or reproducing this material in any form is a violation of the AAMC copyright 66 These questions are not based on a descriptive passage and are independent of each other 188 An organism that causes a human disease is isolated and studied Researchers conclude that the organism is a bacterium rather than a virus because the organism: A) undergoes mutation B) lacks a nuclear membrane C) contains protein in its outermost covering D) reproduces in a culture medium lacking host tissue 189 Which of the following organelles most resembles the Golgi apparatus when an intact eukaryotic cell is viewed under the electron microscope? A) Nucleolus B) Mitochondrion C) Plasma membrane D) Smooth endoplasmic reticulum 190 If mol of a pure triglyceride is hydrolyzed to give mol of RCOOH, mol of R'COOH, and mol of glycerol, which of the following compounds might be the triglyceride? 191 The lipases catalyze the hydrolysis of fats and other carboxylic acid esters The lipases illustrate the fact that: A) some enzymes are molecules other than proteins B) most enzymes interact with only one specific substrate molecule C) some enzymes interact with several different substrate molecules that have similar chemical linkages D) some enzymes interact with many biologically active substrate molecules of dissimilar structures and linkages 192 In horses, the genes for red coat color and for white coat color are codominant Heterozygous horses have roan-colored coats Consider a roan-colored colt that has a white mother What could be said about the coat color of the colt’s father? A) It must be red B) It must be roan C) It could be either red or roan D) It could be either red or white A ) CH2OC(O)R | CHOC(O)R | CH2OC(O)R B) CH2OC(O)R′ | CHOC(O)R′ | CH2OC(O)R′ C) CH2OC(O)R | CHOC(O)R′ | CH2OC(O)R D) CH2OC(O)R | CHOC(O)R′ | CH2OC(O)R′ Sharing or reproducing this material in any form is a violation of the AAMC copyright 67 Passage IX Until recently, conventional medical wisdom attributed stomach ulcers to an excess of acid, and the treatment for ulcers consisted primarily of antacids and dietary modification However, a pathogenic bacterium, Helicobacter pylori, is now implicated in most cases of stomach ulcers Current treatment employs antibiotics directed against these bacteria and often is successful in eradicating persistent infections Several important questions about H pylori remain unanswered It is unclear why this bacterium causes chronic infections in some individuals but not in others; many infected persons not develop ulcers The mode of transmission is also unknown, although people in developing countries are more frequently infected with H pylori than are people in developed countries with good sanitation There is a relationship between H pylori infection and cancer Infected individuals have a two-fold increased risk of gastric cancer, although >75% of patients with active infections not develop cancer Genetic studies of H pylori have identified genes that are expressed in different strains of this bacterium One gene, vacA, encodes a toxin Expression of another gene, cagA, leads to inflammation and may be related to the genesis of gastric cancer Although many individuals develop antibodies against H pylori antigens, these antibodies rarely eradicate the infection; evidently, this pathogen has developed effective ways to elude host defenses 193 H pylori infection may cause increased proliferation of mucosal cells in the stomach This may lead to gastric cancer if: A) genetic mutations occur in proliferating germ cells B) genetic mutations occur in proliferating somatic cells C) the immune system fails to recognize bacterial antigens D) crowded mucosal cells are likely to remain in interphase Sharing or reproducing this material in any form is a violation of the AAMC copyright 194 Which of the following statements explains most plausibly why host antibodies are ineffective against H pylori? A) Antibody proteins may be denatured in the harsh environment of the stomach B) Antibodies are not generally effective against bacteria C) H pylori infection may suppress the activity of the immune system D) Antibodies are not secreted from host tissues into extracellular spaces 195 One difference between different strains of H pylori is that they: A) attack different hosts B) express different genes C) exhibit different degrees of resistance to antibiotics D) exist in either developed or developing countries 196 According to the passage, the cagA gene product will cause: A) the disruption of host cell enzymatic activity B) the disruption of host cell protein synthesis C) the movement of leukocytes into mucosal tissue D) the vasoconstriction of arterioles in the mucosal layer 197 Most people infected with H pylori not develop gastric cancer because they: A) not incorporate bacterial genes in their chromosomes B) have robust immune systems that defeat early cancers C) eradicate the infection before any tumors develop D) tolerate the infection without developing tumors 68 198 To be most effective, a gene therapy for gastric cancer should be directed against: A) the stomach epithelial cells that give rise to tumors B) the antibody producing cells of the immune system C) all the cells in the host’s body D) all known genes in the H pylori genome Sharing or reproducing this material in any form is a violation of the AAMC copyright 199 Enzymatic activity in the stomach initiates the digestion of: A) lipids B) ethanol C) polysaccharides D) proteins 69 Passage X Experiment Nematode development has been used as a model system to investigate the differentiation of cells from their neighbors (see Figure 1) Three experiments are described here Two-cell embryos were incubated in the presence of either cycloheximide, an inhibitor of translation, or actinomycin D, an inhibitor of transcription The AB cells were then isolated, washed to remove the inhibitors, and grown in culture AB cells from embryos treated with cycloheximide produced only neurons and skin, whereas those from embryos treated with actinomycin D produced neurons, skin, and muscle Experiment To investigate the role of cell-to-cell communication, researchers separated the cells of a two-cell embryo and cultured them independently The cultured AB cells produced neurons and skin, but no muscle, whereas the cultured P1 cells gave rise to all of the tissues produced by P1 cells of an intact embryo Sharing or reproducing this material in any form is a violation of the AAMC copyright Experiment To investigate specification of the gut, which was thought to result from the segregation of cytoplasmic contents during early cell divisions, cells were isolated at various intervals (as indicated on the time lines of Figure 2) during the 15-minute four-cell stage, cultured individually or recombined in pairs, and allowed to develop If gut differentiation occurred, proteases were released into the culture medium The results are shown in Figure 70 Figure Early development, showing one-, two-, and four-cell stages (left), and segregation of fate (right) in the nematode The blastomeres and tissues formed by the founder cells are shown Figure Results of Experiment At right center are shown the cell recombinations, and the time lines at far right indicate when the cells were separated and recombined, as well as the results Sharing or reproducing this material in any form is a violation of the AAMC copyright 71 200 Which result of Experiment supports the hypothesis that cell-to-cell communication is involved in the determination of cell fate? 203 The results of Experiment indicate that gut specification during the four-cell stage requires cell-to-cell communication between: A) The fate of an isolated AB cell differs from that of an AB cell in an intact embryo B) The fate of an isolated P1 cell is indistinguishable from that of a P1 cell in an intact embryo C) At the two-cell stage, isolated blastomeres can divide and differentiate D) Several different blastomeres can produce both neurons and muscle tissue A) P2 and EMS B) AB1 and EMS C) P1 and AB1 D) AB1 and AB2 201 The results of Experiment indicate that the direction of signaling between the blastomeres of a two-cell embryo is: A) AB → P1 B) P1 → AB C) P1 → P2 D) zygote → AB 202 The results of Experiment indicate that the signaling interaction at the two-cell stage probably most involves which class of macromolecules? A) DNA B) Messenger RNA C) Ribosomal RNA D) Protein Sharing or reproducing this material in any form is a violation of the AAMC copyright 204 If the zygote contains unique cell contents that are necessary for gut differentiation, segregation of these substances during cell divisions would occur in the sequence of zygote to P1 to: A) AB B) EMS to E C) P2 to EMS to E D) both P2 and EMS 205 The only somatic or visceral cell-type tissue that derives from a single blastomere is: A) neuronal B) muscle C) gut D) germ cell 206 These experiments indicate that nematode cells adopt fates different from those of their neighbors during development by: A) mechanisms that not involve transcription B) cell separation followed by independent development of the blastomeres C) both cell-to-cell signaling and segregation of the cytoplasmic contents during division D) the separation of their three primary tissue layers 72 Passage XI An example of the Claisen rearrangement is shown in Equation Heating Compound yields Compound Two chemists propose alternative mechanisms for this reaction Equation Chemist The rearrangement proceeds through a concerted, cyclic transition state in which a new carbon-carbon bond is formed between carbon-3 of the side chain and the ortho-carbon of the aromatic ring, while the oxygen—C-1 bond is broken The intermediate ketone enolizes to the observed product, as shown in Equation Equation Chemist The oxygen—C-1 bond undergoes a heterolytic cleavage to form a resonance-stabilized intermediate carbocation and a phenolate anion The intermediate carbocation attacks the orthocarbon of the aromatic ring, to give a ketone which enolizes to the observed product as shown in Equation Sharing or reproducing this material in any form is a violation of the AAMC copyright Equation Independent experiments reveal that no cross products are obtained when a mixture of Compound and an analog with a different aryl group and a substituted side chain undergo rearrangement in the same solution Chemist performs the Claisen rearrangement on Compound in which C-1 is labeled with radioactive carbon-14 The rearranged product containing the label is degraded by treatment with osmium tetroxide (OsO4), followed by oxidation with periodic (HIO4) acid The degradation products are monitored for radioactivity 207 Which of the following compounds will be produced from heating the compound shown below if it reacts in the same way as Compound 1? CH2=CH-O-CH2CH=CH2 A) CH3CH2CH2CH2CHO B) CH2 = CHCH2CH2CHO C) CH2 = CHCH2OCH2CH3 O D) || CH2 = CHCCH2CH3 73 208 The entropy change between a reactant and its activated complex is called the entropy of activation (∆S≠) Which chemist’s mechanism is supported if ∆S≠ < for the reaction shown in Equation 1? A) Chemist 1’s because a negative ∆S≠ indicates a transition state that is more ordered than the reactant B) Chemist 1’s because a negative ∆S≠ indicates a transition state that is less ordered than the reactant C) Chemist 2’s because a negative ∆S≠ indicates a transition state that is more ordered than the reactant D) Chemist 2’s because a negative ∆S≠ indicates a transition state that is less ordered than the reactant Sharing or reproducing this material in any form is a violation of the AAMC copyright 209 The absence of cross products supports the mechanism of: A) both chemists B) Chemist only C) Chemist only D) neither chemist 210 The IR spectrum of Compound displays a band for the -OH group near: A) 1000 cm-1 B) 1500 cm-1 C) 2500 cm-1 D) 3500 cm-1 74 These questions are not based on a descriptive passage and are independent of each other 211 The following pedigree shows the occurrence of a very rare disease that is expressed in fewer than in 100,000 individuals 214 Which of the compounds shown below is more soluble in water? (CH3CH2)2NH Compound I (CH3CH2)2CH2 Compound II A) Compound I B) Compound II C) They are about equally soluble D) Neither has appreciable water solubility 215 In humans, cholesterol is a precursor to: A) insulin B) glycogen C) testosterone D) DNA Which of the following is the most likely pattern of inheritance for this disease in these individuals? A) The mother carried a sex-linked dominant allele for the disease B) The mother carried a sex-linked recessive allele for the disease C) Each father carried an autosomal recessive allele for the disease D) Each father carried a sex-linked recessive allele for the disease 216 In an experiment on the phases of the cell cycle, cultures of actively dividing, synchronized cells were exposed to radioactively labeled 2-deoxythymidine for 30 minutes, then rinsed to remove the unabsorbed label At various times thereafter, groups of cells were removed from the cultures and the nuclei examined to determine their content of radioactive material Results are shown in the figure below 212 After the gall bladder is removed from a patient, the patient will most likely have reduced ability to digest: A) protein B) starch C) sugar D) fat 213 The liver is different from many other organs in that it can at least partially regenerate following illness or damage This regeneration is accomplished primarily through: A) fission B) meiosis C) mitosis D) cell growth Sharing or reproducing this material in any form is a violation of the AAMC copyright Based on the figure, what process was occurring during hours 3-13 after treatment with radioactive 2-deoxythymidine? A) Mitosis B) Meiosis C) DNA synthesis D) RNA synthesis 75 MCAT Practice Test Answer Sheet Physical Sciences (A) (B) (C) (A) (B) (C) (A) (B) (C) (A) (B) (C) (A) (B) (C) (A) (B) (C) (A) (B) (C) (A) (B) (C) (A) (B) (C) 10 (A) (B) (C) 11 (A) (B) (C) 12 (A) (B) (C) 13 (A) (B) (C) 14 (A) (B) (C) 15 (A) (B) (C) 16 (A) (B) (C) 17 (A) (B) (C) 18 (A) (B) (C) 19 (A) (B) (C) 20 (A) (B) (C) 21 (A) (B) (C) 22 (A) (B) (C) 23 (A) (B) (C) 24 (A) (B) (C) 25 (A) (B) (C) 26 (A) (B) (C) 27 (A) (B) (C) 28 (A) (B) (C) 29 (A) (B) (C) 30 (A) (B) (C) 31 (A) (B) (C) 32 (A) (B) (C) 33 (A) (B) (C) 34 (A) (B) (C) 35 (A) (B) (C) 36 (A) (B) (C) 37 (A) (B) (C) 38 (A) (B) (C) 39 (A) (B) (C) 40 (A) (B) (C) 41 (A) (B) (C) 42 (A) (B) (C) 43 (A) (B) (C) 44 (A) (B) (C) 45 (A) (B) (C) 46 (A) (B) (C) 47 (A) (B) (C) 48 (A) (B) (C) 49 (A) (B) (C) 50 (A) (B) (C) 51 (A) (B) (C) 52 (A) (B) (C) 53 (A) (B) (C) 54 (A) (B) (C) 55 (A) (B) (C) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) Verbal Reasoning 78 (A) (B) (C) (D) 79 (A) (B) (C) (D) 80 (A) (B) (C) (D) 81 (A) (B) (C) (D) 82 (A) (B) (C) (D) 83 (A) (B) (C) (D) 84 (A) (B) (C) (D) 85 (A) (B) (C) (D) 86 (A) (B) (C) (D) 87 (A) (B) (C) (D) 88 (A) (B) (C) (D) 89 (A) (B) (C) (D) 90 (A) (B) (C) (D) 91 (A) (B) (C) (D) 92 (A) (B) (C) (D) 93 (A) (B) (C) (D) 94 (A) (B) (C) (D) 95 (A) (B) (C) (D) 96 (A) (B) (C) (D) 97 (A) (B) (C) (D) 98 (A) (B) (C) (D) 99 (A) (B) (C) (D) 100 (A) (B) (C) (D) 101 (A) (B) (C) (D) 102 (A) (B) (C) (D) 103 (A) (B) (C) (D) 104 (A) (B) (C) (D) 105 (A) (B) (C) (D) 106 (A) (B) (C) (D) 107 (A) (B) (C) (D) 108 (A) (B) (C) (D) 109 (A) (B) (C) (D) Sharing or reproducing this material in any form is a violation of the AAMC copyright 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) Writing Sample 138 139 Biological Sciences 140 (A) (B) (C) (D) 141 (A) (B) (C) (D) 142 (A) (B) (C) (D) 143 (A) (B) (C) (D) 144 (A) (B) (C) (D) 145 (A) (B) (C) (D) 146 (A) (B) (C) (D) 147 (A) (B) (C) (D) 148 (A) (B) (C) (D) 149 (A) (B) (C) (D) 150 (A) (B) (C) (D) 151 (A) (B) (C) (D) 152 (A) (B) (C) (D) 153 (A) (B) (C) (D) 154 (A) (B) (C) (D) 155 (A) (B) (C) (D) 156 (A) (B) (C) (D) 157 (A) (B) (C) (D) 158 (A) (B) (C) (D) 159 (A) (B) (C) (D) 160 (A) (B) (C) (D) 161 (A) (B) (C) (D) 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (A) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (B) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (C) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) (D) 76 ... 45 58 .7 46 63.5 47 65.4 48 69 .7 49 72 .6 50 74 .9 51 79 .0 52 79 .9 53 83.8 54 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe 85.5 55 87. 6 56 88.9 57 91.2 72 92.9 73 95.9 74 (98) 75 101.1 76 102.9... 101.1 76 102.9 77 106.4 78 1 07. 9 79 112.4 80 114.8 81 118 .7 82 121.8 83 1 27. 6 84 126.9 85 131.3 86 Cs Ba La* Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn 132.9 87 1 37. 3 88 138.9 89 178 .5 104 180.9... 65 66 67 68 69 70 71 Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu 140.1 90 140.9 91 144.2 92 (145) 93 150.4 94 152.0 95 1 57. 3 96 158.9 97 162.5 98 164.9 99 1 67. 3 100 168.9 101 173 .0 102 175 .0 103