1. Trang chủ
  2. » Giáo án - Bài giảng

AAMC MCAT test 7 a

34 413 0

Đang tải... (xem toàn văn)

Tài liệu hạn chế xem trước, để xem đầy đủ mời bạn chọn Tải xuống

THÔNG TIN TÀI LIỆU

Thông tin cơ bản

Định dạng
Số trang 34
Dung lượng 523,55 KB

Nội dung

Test 7R P Physic - 77 26 27 28 29 30 31 32 33 34 10 35 11 36 12 37 13 38 14 39 15 40 16 41 17 42 18 43 19 44 20 45 21 46 22 47 23 48 24 49 25 50 51 76 52 77 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 Test 7R Verbal 78 -137 78 103 79 104 80 105 81 106 128 82 107 129 83 108 130 84 109 131 85 110 132 86 111 133 87 112 134 88 113 135 89 114 136 90 115 137 91 116 92 117 93 118 94 119 95 120 96 121 97 122 98 123 99 124 100 125 101 126 102 127 Test 7R Biology 140 - 216 140 165 141 166 142 167 143 168 144 169 145 170 146 171 147 172 148 173 149 174 150 175 151 176 152 177 153 178 154 179 155 180 156 181 157 182 158 183 159 184 160 185 161 186 162 187 163 188 164 189 190 215 191 216 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 7R Key EXPLANATIONS for test 7R A nitrogen atom has single electrons and lone pair Hydrogen has one electron The structure in C correctly shows each nitrogen with three bonds and a one lone pair Each atom in this molecule has a full outer shell of electrons and all electrons are present Equation shows that the Raschig process requires the reaction of moles of ammonia to form mole of hydrazine The molar mass of ammonia is 17 g/mol (14 + + + 1) Therefore 34 g of ammonia is required to form mole of hydrazine The passage states that the formula for hydrazine hydrate is N2H4·H2O The formula weight for hydrazine is 32 g/mol (14 x + x 1) The formula weight for hydrazine hydrate is 50 g/mol (32 + x + 16) The percent by weight of hydrazine in the hydrate would be 32.0/50.0 x 100% The reaction shows the formation of hydrazine from its elements The value of ∆Hf° for a substance is the enthalpy change when one mole of that substance is formed at atm pressure and 25°C from the elements in their stable states at that pressure and temperature Table shows that ∆Hf° for hydrazine is 50.6 kJ/mol, therefore, the enthalpy change for the reaction shown would be 50.6 kJ/mol The basicity constant is a measure of the strength of a base It is the equilibrium constant for the reaction of the base with water The lower the value of Kb, the weaker the base In order for a reaction to be a spontaneous process, the value for ∆Go must be less than Because there are reactant moles, all liquids, and product moles, all gases, the entropy increases in this reaction Gas molecules have more freedom and randomness, and therefore higher entropy Since the ions are thousands of times more massive than the electron, answer B is justified (the hydrogen ion is the lightest ion and is nearly 2000 times more massive than the electron) An ion with so much mass compared to an electron will not be able to respond quickly because of its inertia The potential energy of an oscillatory motion is ½kx2 where x is the displacement Since the maximum displacement occurs at A and C, answer D is justified 10 The frequency is given by 9n0.5 With n = 1018, n0.5 = 109 Thus, frequency is *109, which is answer C 11 Wavelength is given by speed/frequency = *107/109 or 3*10-2 m This is the same as cm or answer A 12 This is basically a conservation of energy question As the electrons move from A to B they convert potential energy into kinetic energy (½mv2) and gain velocity and hence momentum (mv) as they so This momentum is just enough to allow the electrons to go from B to C; i.e the kinetic energy reconverts to potential energy—just equal to the amount of potential energy at point A 13 This is basically a conservation of energy question As the electrons move from A to B they convert potential energy into kinetic energy (½mv2) and gain velocity and hence momentum (mv) as they so This momentum is just enough to allow the electrons to go from B to C; i.e the kinetic energy reconverts to potential energy—just equal to the amount of potential energy at point A As the electrons move from A to B they convert potential energy (½kx2) into kinetic energy (½mv2), where x is the displacement from point B It is conservation of energy 14 The passage states that silicon cannot be purified by electrolytic techniques These techniques involve isolating silicon from silicate minerals by decomposing the minerals with an electrical current It would follow that the silicon could not be purified by electrolytic techniques if the minerals not decompose easily 15 Silicon is the fourth element in the third row of the periodic table and has 14 electrons The first 10 electrons would have the same electron configuration as Ne The next electrons would fill the 3s orbitals and the last electrons would begin to fill the 3p orbitals following Hund’s Rule (electrons will not pair up until each orbital in that sublevel has electron.) The unpaired electrons have parallel spins 16 According to VSEPR theory, electron pairs around an atom will result in a tetrahedral geometry 17 Elements in the same group share chemical properties Because potassium is a Group element, one would assume that sodium, another Group element would be the best substitute 18 Van der Waals forces are weak intermolecular attractive forces The relatively low boiling point of the SiCl3H indicates that the intermolecular forces are not strong and are most likely van der Waals forces 19 Fractional distillation can be used for purification when the components to be separated have different boiling points 20 A1 and A2 represent the nuclear masses of the fragments of fission This is given by the sum of their proton and neutrons Since the reaction started with 236 nucleons and three neutrons were = 233 mass units or−taken away the sum of the fragments must contain 236 nucleons 21 This reaction is exactly a reproduction of the statement of beta decay as given in the stem We see the new nucleus, a beta particle (an electron) and a neutrino The new nucleus is one atomic number higher because of the emission of the electron leaving an extra proton in the nucleus 22 The relative rate of decay for Ra compared to Pu is the inverse ratio of their halflives or 24000/1600 = 15 This is answer B 23 The total volume held by 106 kg of ash would be (106 kg/1000 kg/m3) = 1000 m3 This volume is held in a cube 10 m on a side since (10 m)3 = 1000 m3 This is answer B 24 The value for Eocell is equal to the sum of Eo of the half reactions -1.23 + 0.34 = 0.89 V 25 The ideal gas law makes the assumption that molecules have no volume This assumption is adequate when the gas is at atm, but when the pressure is increased to 500 atm the volume of the gas molecules is no longer negligible 26 velocity.×Power P is force Since speed v is fixed we must compare the forces F and Ft The force is the coefficient of friction between tiresµmg, where µon the level road F is by notingµand road We can find Now the force required to be overcome in going uphill Ft is The power in t his case, Pt, is Ft v The extra× power needed is then Pt minus the original power P or: Since cos 10o 1,≈ This is answer D 27 Light slows down because the index of refraction in the glass is greater than in the air The index is a measure of the ratio of the velocity in air to the velocity in the medium For sound the speed becomes greater because the speed of sound in a solid is much greater than in air (the glass has stiff rigid bonds which gives rise to a speed more than 10 times greater than in air) 28 The passage states that the reaction is zero order with respect to bromine indicating that bromine is not involved in the rate determining step If the reaction proceeds by the same mechanism with chlorine, then the reaction with chlorine will have the same rate as it does with bromine because it would be zero order with respect to chlorine 29 The passage states that the reaction is zero order with respect to bromine, therefore bromine can be removed from Equation A comparison of the results for Experiments 2, and (all with acetone concentrations of 0.80 M) shows that after dividing the rates of the reactions by the rate constants, the values obtained are directly proportional to the H+ concentration indicating the reaction is first order in H+.The passage states that the reaction is zero order with respect to bromine, therefore bromine can be removed from Equation A comparison of the results for Experiments 2, and (all with acetone concentrations of 0.80 M) shows that after dividing the rates of the reactions by the rate constants, the values obtained are directly proportional to the H+ concentration indicating the reaction is first order in H+ 30 The passage states that the reaction is zero order with respect to bromine, therefore bromine can be removed from the expression If the reaction is first order with respect to acetone and hydronium ion, then a = and c = 31 The passage states that the density of acetone is 0.791 g/mL and the molar mass is 58 g/mol The molarity of pure acetone can be calculated as follows: (0.791 g/mL)(1000 mL/L)(1 mol/58 g) = 13.6 mol/L 32 Light in the visible region of the spectrum ranges from approximately 400 nm to 700 nm The next region of light, below 400 nm is the ultraviolet region 33 This is found as the difference between 894 MHz and 824 MHz since these are given as the upper and lower frequencies This is 70 MHz and is answer C 34 Since power P is current I multiplied by voltage V, V = 12 Volts, and the maximum power is watts, then the current I must be 3/12 or 0.25 A This is answer C 35 This answer is implied by the passage The conversations must be on different frequencies to be unique and so no two frequencies can be used by different phones at the same time 36 Here we must calculate P/r2 The maximum value occurs for answer D The intensities are: A (0.6/4) W/mile2 = 0.15 W/mile2 B (0.6/9) W/mile2 = 0.067 W/mile2 C (3/25) W/mile2 = 0.12 W/mile2 D (3/16) W/mile2 = 0.19 W/mile2, making D the key 37 Light is a transverse wave with its electric and magnetic fields oscillating at right angles to the propagation vector where as sound is a longitudinal pressure wave This comes from prior knowledge brought to bear on the problem Answer A correctly states the result 38 The two plates of the capacitor collect charges of opposite sign As more charge arrives it is harder and harder to fill the plates until finally an equilibrium occurs, thus C is correct 39 As the capacitor discharges the voltage across it falls, thus to maintain a constant current, R must be proportionately reduced This is so from Ohm’s law, I = V/R To keep I fixed, R must fall with V This is answer B 40 Since energy is lost by heating the small resistor, r, the energy stored in the capacitor must be less than the work done by the battery during the charging process The battery supplies the energy for both processes and answer B is justified 41 The capacitor charges up and stores energy in the electric field between the places The energy stored is ½CVc2, where Vc is the voltage across the capacitor The battery is the source of energy for the circuit and thus is a store of energy The resistor is not a storage device for energy and answer C is the correct answer 42 With 12 Volts initially across the capacitor during its discharge (the capacitor will charge to the battery voltage of 12V) and a current of 0.002 A as found in Figure 2, then the initial This is answerΩresistance R must have been R = V/I = 12/0.002 = 6000 D 43 The rate law for Reaction is second order with respect to hydrogen ion At a pH of the hydrogen ion concentration is 1x10-1 When the pH is increased to 2, the hydrogen ion concentration is 1x10-2 Because all other conditions remain the same, the rate is decreased: (1 x 10-2)2/(1 x 10-1)2 or x 10-2 times The rate would therefore be (1.0 x 10-2 M/s) x (1 x 10-2) or x 10-4 M/s 44 No kinetic information is given for Reactions and 2, only equilibrium constants 45 Reaction requires that a reaction takes place between two anions which would experience electronic repulsion due to their negative charges Protonation of the oxygen would generate an electrically neutral species and the repulsion forces would be reduced 46 The passage states that the reaction of H218O and SO3 is fast Singly labeled SO4-2 would be prepared most quickly by reacting the unlabeled SO3 and the labeled H2O 47 In this reaction the oxygen is transferred from the chlorine to the nitrogen The transition state shown in Foil C is the only transition state that indicates the bond between the chlorine and oxygen breaking and a new bond between the oxygen and nitrogen forming 48 The passage states that the Sequences I and III in Reaction are fast and Sequence II is slow This indicates that Sequence II would have the highest energy barrier, as is shown in the energy diagram in Foil B 49 A substance boils when enough heat has been supplied to overcome the intermolecular forces The fact that ammonia has a higher boiling point that phosphine indicates that ammonia requires more heat to overcome the intermolecular forces than does phosphine and therefore the intermolecular forces in ammonia are stronger than those in phosphine 50 Density is mass/volume The densities of the four objects are: A.(1.5/0.50) = 3.0 g/cc B.(3.0/0.75) = 4.0 g/cc C.(4.5/1.00) = 4.5 g/cc D.(6.0/1.50) = 4.0 g/cc Thus, answer C with 4.5 g/cc has the highest density 51 Chlorine needs electron to fill its outer shell 52 The ammonium ion is an acid It is a proton donor 53 This is a Snell Law problem: where the n’s are the indices of refraction Taking the index of refraction of air to be nearly then: This is answer B 54 According to the information in Table 1, the solution would contain 37.7 g Pb(NO3)2/100 mL H2O The molality (mol solute/kg solvent) can be calculated as follows: (37.7 g Pb(NO3)2/100 mL H2O) x (1000 mL H2O/1 kg H2O) x (1 mol Pb(NO3)2/331 g Pb(NO3)2) = 1.14 mol Pb(NO3)2/kg H2 O = 1.14 m 55 The passage states that the freezing point depression constant for water is Kf = 1.86oC/m A solution that is 10.75 m ethylene chloride would lower the freezing point of water by 1.86oC/m x 10.75 m or 20.0oC The freezing point of water is 0oC, so the freezing point of the solution would be 0oC - 20.0oC or -20.0oC 56 The freezing point depression depends on the number of particles in the solution A 0.1 M solution of lead nitrate would have a particle concentration of 0.3 M (1 mole of lead ions and moles of nitrate ions for each mole of lead nitrate) while a 0.1 M solution of ethylene glycol would have a particle concentration of 0.1 M The lead nitrate would therefore have a 3-fold greater effect on the freezing point of water later notes that surveys of elderly showed that these promises largely proved to be true as the elderly found the friendship and leisure in these retirement communities that was advertised Therefore, the director’s statement supports the optimistic assumptions made by the author and the fact that many retirement communities have “evolved from mere developers’ tracts into communities with traditions of their own.” 120 The emphasis in the passage is that the current rate of longevity is unprecedented and that the comparison between now and seventeenth-century France documented this The author’s thesis is that the appearance of retirement communities is a kind of frontier: “Old age is hardly new, but for an entire generation to reach old age with its membership almost intact is new.” This comparison supports that thesis 121 The author, in discussing longevity, shows that people are not living appreciably longer in actual years but that more people are living into old age and that life expectancy depends on health in earlier years This statistic underscores this crucial difference between life expectancy, which has increased dramatically, and longevity, which has increased comparatively little 122 By referring to the school readiness theory of education as a “usual practice…of treating learning as an abstraction,” the author is implying that this theory is generally accepted The author points out that the outlook behind this theory was opposed by educational theorists Jean Piaget and A.L Gesell who advocated developmentally appropriate, personally meaningful education for children, a view supported by observers of modern education: “Most modern observers of children think that if a task is developmentally appropriate and has personal meaning for a child, it is approached as a pleasing challenge, not a struggle.” The author implies that any educational theory that goes counter to the kind advocated by Piaget and Gesell is “ineffective.” 123 The author states: “In the literature promoting their approach, the advocates of generalized readiness are clearly directing their appeal to school administrators.” 124 The author points out: “Developmentally appropriate instruction…appears to be a hard sell to decision makers concerned with uniformity.” This statement implies that uniformity is more convenient than an individualized approach, with the added benefit of “higher percentages on standardized tests.” 125 This analogy is most apropos because the author contends that, much like a form letter, school readiness is too impersonal and standardized whereas developmentally appropriate instruction, much like a personal note to a friend, is focused on the individual needs of children and is designed to be personally meaningful to them 126 The author discusses the importance of developing a test to validate curricular reforms and educational effectiveness 127 .The author would likely doubt the accuracy and quality of the report since that is how the author reacted to efforts to introduce school readiness The author also devotes the first two paragraphs to explaining why reformers should take care to prove their reforms are needed and credible 128 The objector does not say that politicians are better decision makers than anyone else, so the argument that politicians could “decide anything” is relevant The author reinforces that argument by saying: “A government is made up of individuals who are fundamentally similar to me, and to ‘err is human’ applies to us all.” The author concedes that a government leader or official may be more well informed or expert in a particular issue Therefore, an individual could accept the decision on the basis that the decision is likelier to achieve a better outcome, but even the decision to decide whether the government is right or wrong still is the responsibility of the individual 129 The author commits somewhat of a logical fallacy by asserting that it is impossible to surrender responsibility One could so freely—although the author is right in saying that once you make that choice you have “enslaved your will to some else’s will.” In the case of someone with the hypothetical psychological condition referred to in the question, a person surrenders responsibility involuntarily, because that person would be incapable of making the decision whether to be responsible or abdicate responsibility; nonetheless, that person surrenders it 130 This theme runs throughout the passage The author states: “You cannot hand over your autonomy willy-nilly to…the government or any one else.” The author also states: “No government, no body of people, no position, no individual can have moral authority over any individual.” The author also points out that “the final arbiter is the individual,” even if the government has good reason to require a particular behavior 131 This hypothetical finding theoretically represents a strong challenge to the assertion that the individual is the final moral arbiter, even if that assertion remains a valid argument, because it greatly increases the individual’s burden of proof and, hence, individual responsibility for decisions 132 Even if people drive on different sides of the road in different countries, the central thesis is unchanged by knowing this The author already gives the example of driving on the left side of the road; therefore, introducing the example of people driving on the right side of the road in another country neither weakens nor lends further support to the central thesis than is already in the passage At most this new information is simply a variation on the exact same argument The decision to drive on the right side of the road in a country where it is the law to so has the same obvious benefits in that country as it does in a country where it is the law to drive on the right side of the road A responsible individual would obey the law because of these obvious benefits, not because the government mandates driving on the left side, or right side, of the road 133 Even if people drive on different sides of the road in different countries, the central thesis is unchanged by knowing this The author already gives the example of driving on the left side of the road; therefore, introducing the example of people driving on the right side of the road in another country neither weakens nor lends further support to the central thesis than is already in the passage At most this new information is simply a variation on the exact same argument The decision to drive on the right side of the road in a country where it is the law to so has the same obvious benefits in that country as it does in a country where it is the law to drive on the right side of the road A responsible individual would obey the law because of these obvious benefits, not because the government mandates driving on the left side, or right side, of the road 134 The ethnic conflict in Bosnia reflects Gibbon’s view of history as well as view of human nature that extends from his historical analysis: “[Gibbon] instructs us that human nature never changes and that humanity’s predilection for factionalism, augmented by environmental and cultural differences, is the determinant of history.” 135 According to the author, Gibbon “was a conservative along the lines of his contemporary, Edmund Burke, who saw humankind’s best hope in moderate politics and flexible institutions that would not become overbearing.” The author also observes that a state or an empire can endure only if it generally limits itself to adjudicating disputes among its citizens.” The role of the U.S Constitution, with its checks and balances among the three branches of government, is to limit government in this way 136 Ethnic diversity, in Gibbon’s view, is one of the destabilizing factors in human history Therefore, a stable society that is ethnically diverse would most challenge Gibbon’s view—his view might hold up if this society were explained by a system of limited government that adjudicated these ethnic differences 137 The role of Milan and Nicomedia as functional capitals shows how historical changes occur gradually, because these roles came “decades before the formal division of the Empire into western and eastern halves and almost two centuries before Rome officially ceased to be the imperial capital….” 140 According to Theory II, the metabolic rates of mammals are directly proportional to the rate of generation of free radicals The information presented in the item stem, comparing humans, rats, and mice indicates that metabolic rate for humans is at 25cal/g/day, that of rats is at 150 cal/g/day, and that of mice is greatest at 180 cal/g/day Thus the generation of free radicals, and subsequent evidence of a free-radical-induced DNA damaged product in urinary output, is consistent with the graph shown in foil A Foil A shows a direct relationship between metabolic rate (x-axis) and the urinary output of a free-radical-induced DNA damaged product (y-axis) This is consistent with Theory II 141 According to Theory II, aging and cell death is caused by the accumulation of damaged DNA, RNA, and other molecules in the cell This damage is caused by free radicals This theory also indicates that addition of antioxidant vitamins, such as Vitamins C and E, can reduce the amount of damage caused by free radicals, and therefore slow the rate of aging and cell death Additionally, according to Theory I, connective tissue cells are known to divide a total of 50 times before dying, regardless of the conditions and treatments Assuming that connective tissues have already undergone 30 divisions prior to treatment with Vitamin E, Theory I would be supported if the cells undergo 20 more divisions before dying, but Theory II would be supported if the cells undergo more than 20 additional divisions If the cells undergo 40 divisions after treatment, this would clearly support Theory II 142 According to the item, the female octopus broods her eggs, but eats less than normal while caring for them, then dies after the eggs hatch If a certain endocrine gland is removed, the brooding behavior is eliminated, so the female resumes regular feeding and her life cycle is extended This provides the strongest support for Theory I because it shows that various characteristics of the organism’s life (brooding and feeding behavior) are hormonally controlled Thus, removing the endocrine gland, and source of hormones, can alter the behavior and lifespan of the organism 143 The item asks the examinee to identify the process that would not lead to free-radical-induced aging According to the passage, Vitamins C and E remove free radicals, thus rendering them harmless Thus, Vitamins C and E would not facilitate aging caused by free radicals 144 According to Theory II, lifespan is proportional to metabolic rate, and metabolic rate can be reduced somewhat by limiting dietary intake Thus, supporters of Theory II would predict that the rats in the “fasting” group would outlive the rats in the “fed” group by some finite amount The results presented under Theory II indicate that rats with a minimal diet live approximately 60 weeks longer than rats that are fed more food 145 According to the passage, Sarah was in excellent physical condition prior to her trip to the Caribbean Sea to go skin diving After her first diving experience, she noticed an elevated pulse rate and ventilation rate The most likely explanation for her body’s response was the activation of her sympathetic autonomic nervous system—the “fight or flight” response caused by adrenaline 146 According to the passage, Sarah went skiing in the mountains of Colorado At first, she noticed an elevated pulse rate and ventilation rate As the week progressed, these rates dropped, but were still higher than usual This prolonged increase in heart rate and breathing rate was most likely the cause of hypoxia (insufficient oxygen to the body cells) caused by insufficient blood hemoglobin to supply oxygen for exercise at the low oxygen pressure found at high altitudes 147 According to the passage, Sarah was in excellent physical condition prior to her trip to the Caribbean Sea to go skin diving After her first diving experience, she noticed an elevated pulse rate and ventilation rate According to the item, she also noticed that she produced more urine than usual The increased urine production can be explained by an increased blood pressure caused by adrenaline, released in response to excitement or anxiety—the fight or flight response 148 After Sarah’s accident, the physician detected myoglobin in Sarah’s urine Myoglobin is the substance that holds oxygen in the muscles and organs The physician’s observation is consistent with an injury to muscle or organs, but not bone 149 Control of heart rate, muscle coordination, and appetite is maintained by the brain stem, cerebellum, and hypothalamus, respectively 150 According to the item, Sarah noticed that her skin blood vessels were usually constricted to conserve body heat in the cold environment of the Colorado mountains where she went skiing Occasionally, however, her vessels would dilate for short periods of time to enable a sufficient supply of blood (and oxygen) to her cells Due to the physical exertion of skiing, her cells had an increased need for oxygen 151 The Cahn-Ingold-Prelog system, susbtituents are listed in order of decreasing atomic number The atoms directly bonded to the chiral carbon atom (x) are H, C, C, and N Nitrogen has the highest atomic number and therefore the NHCHO group has the highest priority 152 The analog of Compound would undergo a Wittig reaction with Ph3P=CH2 to form the double bond in Compound 153 The number of stereoisomers can be calculated using the formula 2n, where n is the number of stereocenters in the molecule Compound has stereocenters 154 The hydrogen on α-carbons c and d are most acidic It can be determined that the hydrogen at carbon c is removed because alkylation occurs at this carbon 155 The compound contains one ketone group, one ester group, and one carbon–carbon double bond 156 According to the passage, reverse transcriptase uses viral RNA as a template for making viral DNA inside the cell of the host This process is the reverse of RNA synthesis, a process that normally occurs in the host cell 157 According to the passage, one method of combating the AIDS virus is to interfere with the binding of the virus to the helper T cell This is done by producing antibodies to the gp120 proteins on the viral surface However, the variability of the antibody-evoking region of the gp120 protein makes it difficult for B cells to produce antibodies that neutralize the AIDS virus in the host because antibodies are very specific 158 AZT is effective for treating AIDS because it is missing a hydroxyl group on the 3’ carbon, a normal site for the bonding between a phosphate and sugar in the growing DNA polymer AZT becomes incorporated into the growing DNA chain in place of a regular nucleotide (which would have the hydroxyl group on the 3’ carbon), thereby interfering with the process of DNA polymerization 159 Antibiotics that are effective in interfering with bacterial (but not eukaryotic) ribosomes are ineffective at combating viruses because viruses typically lack ribosomes Thus, a drug that interferes with ribosome function would have no effect against a virus 160 According to the passage, during the infection of a helper T cell, gp120 proteins of the viral coat first bind to the CD4 antigens on the cell membrane Then the viral coat fuses with the membrane, dumping its RNA core into the cell Once dumped, this RNA core would be transformed into DNA by reverse transcriptase, and the viral DNA would then become incorporated into the host cell’s chromosomes until activated (replicated) at a later time Thus, when an AIDS virus has been incorporated into a CD4 cell, but has not yet been replicated, the viral genetic information is located in the CD4 cell’s nucleus—incorporated into the host’s DNA 161 The item asks which evolutionary mechanism would most likely explain the presence in humans of CD4 receptors on the helper T cells that bind to the gp120 proteins of the AIDS virus The most likely scenario is natural selection favoring chance mutation(s) of the virus This answer is supported by the information in the passage, indicating that there is a high rate of mutation in the gp120 protein of the virus In addition to having a high rate of mutation, the generation time for a virus is quite short, which enhances the ability of natural selection to make a difference on the viral population, even in a single generation time for humans 162 The item indicates that there is a virus similar to the AIDS virus, but that this virus infects only B lymphocytes (not the T cells invaded by AIDS virus) The function of a B lymphocyte is to produce and secrete antibodies when it encounters a foreign particle or substance Thus, a virus that infects only B lymphocytes would be expected to affect the production of antibodies 163 This reaction is an example of the formation of a hydrazone by the condensation of a hydrazine with an aldehyde 164 Tertiary alcohols react much more quickly with HCl than other types of alcohols 165 The graph shown in the item indicates that the concentration of cyclin rises and falls in a regular manner throughout the cell cycle, reaching a peak just at the beginning of mitosis, gradually declining during mitosis, reaching a minimum at the end of mitosis, and gradually increasing during interphase The mechanism that can best account for this oscillation in the concentration of cyclin is translation of cyclin mRNA (creating the protein from mRNA template) followed by proteolysis (destruction) of cyclin protein during mitosis 166 According to the item, embryonic mouse cells divide every 10 hours at o 37 C Under such circumstances, and starting with a single egg, x cells would be present after three days (or 24 x 3=72 hours) The number can be calculated by tracking the doubling time: → → → → 16 → 32 → 64 → 128 → 256 At the end of 70 hours (approximately days), there would be 128 cells, assuming that each cell underwent division 10 hours after its previous division 167 At C-5 the molecule is drawn so that the hydrogen, the group with lowest priority, is pointing away from the viewer The O has the highest priority, followed by C-4 and then C-6 The groups are oriented in clockwise order of decreasing priority indicating that C-5 has an R-configuration At C-7 again the molecule is drawn so that the hydrogen is pointing away from the viewer The O has the highest priority, followed by C-8 and then C-6 The groups are oriented in counterclockwise order of decreasing priority indicating that C-7 has an Sconfiguration 168 The Grignard reagent will add a methyl group to the ketone generating a tertiary alcohol 169 The Grignard reagent is CH3MgBr prepared by the reaction of CH3Br with Mg in diethyl ether 170 The passage states that Compound undergoes a regiospecific DielsAlder reaction indicating that Compound was added so that the Cl and CN groups are present in the position in the ring as shown in Compound 7, however both stereoisomers of this compound would be present 171 Step involves the addition of an acetyl group (CH3CO) to Compound which is an acetylation reaction 172 The passage presents information about inflammatory bowel disease, including Chron’s disease and ulcerative colitis, the latter of which is associated with inflammation of the colon The item asks what process would be most disrupted by an inflammation in the colon Since the primary process that takes place in the colon is absorption of water, then the absorption of water is the most likely process to be disrupted 173 The immune system is designed to attack foreign material in the body It avoids attacking tissues of its own body because it suppresses cells that are specific to its own body’s antigens (surface molecules that would otherwise initiate an immune response) 174 If an ulcer penetrated the walls of the intestine, this would allow the contents of the gastrointestinal tract to enter the peritoneal cavity Membranes surround this cavity, which would prevent further transport of the gastrointestinal contents through the rest of the body An ulcer in the small intestine would not allow the contents to enter the lumen because this is the normal place in which the contents are found 175 Assuming the genetic and autoimmune theories of inflammatory bowel disease are true, then the gastrointestinal antigen being targeted must be located on the surface of proteins encoded by the genes for the disease Antigens are carried on the surface of cells, not on the chromosomes, DNA segments, or RNA 176 Inflammatory bowel disease appears to have a genetic component, but it does not show clear evidence of Medelian inheritance This means that the trait cannot simply be “recessive” since, if it were, it would show Mendelian inheritance patterns 177 The passage describes a study in which the blood flow of participants was measured before and after 60 days of exposure to one of four treatments The treatments included placebo, placebo + vitamin E, fatty acid, and fatty acid + vitamin E The passage indicates that vitamin E is an antioxidant that reduces in vivo oxidation of ingested fatty acids As shown in Figure 1, the fatty acid group showed a reduction in blood flow by more than 40%, while the fatty acid + vitamin E showed an increase of about 10% The most likely explanation for the difference in blood flow between the fatty acid group and the fatty acid + vitamin E group is that the products of fatty acid oxidation (which would be formed when vitamin E is absent) reduce blood flow 178 One hypothesis suggests that the decrease in blood flow to the skin results from a change in the activity in the sympathetic nerves to the skin This hypothesis would be supported if researchers observed a change in the norepinephrine content of blood draining from the skin 179 An alternative method for examining the effects of fatty acids on blood flow would be to measure changes in blood pressure It blood pressure were measured, one would predict that blood pressure would be lowest in the capillaries, as compared to the heart, arteries, or arterioles Pressure would be even lower in the veins, but this option was not offered 180 The passage describes a study in which the blood flow of participants was measured before and after 60 days of exposure to one of four treatments—after collecting lifestyle information abut each subject The treatments included placebo, placebo + vitamin E, fatty acid, and fatty acid + vitamin E The passage indicates that vitamin E is an antioxidant that reduces in vivo oxidation of ingested fatty acids To interpret the results of the passage, researchers must assume that subjects not alter their lifestyle (eg eating habits) during the course of the experiment 181 According to the item, the fatty acid used was a 20-carbon polyunsaturated fatty acid The difference between saturated and unsaturated fatty acids is that saturated fatty acids lack carbon-carbon double bonds (because all carbons are saturated with hydrogen) 182 Researchers collected information about each subject’s age, dietary habits, etc They made sure that skin temperature was constant during the blood flow measurement The one variable that was not controlled or accounted for was skin blood flow This was dependent variable 183 The passage states that alkyl groups bonded to a carbocation center may stabilize the carbocation through hyperconjugation 184 The proton NMR shows equivalent H and an additional H downfield This is consistent with the structure of t-butyl alcohol The hydrogen on the methyl groups are all equivalent 185 If ethanol was heated with sulfuric acid the oxygen of the ethanol would be protonated in the first step with the alcohol acting as a base accepting a proton from the acid and forming water as a leaving group 186 The passage states that sec-butyl alcohol is the second most abundant product The peak in the chromatograph with area of would correspond to the sec-butyl alcohol The total area is (45 + + + 0.5 + 0.5) = 50 The percentage sec-butyl alcohol would therefore be 3/50 = 6% 187 All of the products are formed from the quenching of various carbocations, therefore all should have -OH groups In the infrared spectrum, bands characteristic of -OH groups are found near 3500 cm-1 188 An organism that causes human disease is isolated and studies Researchers would conclude that the organism is a bacterium rather than a virus if the organism reproduces in a culture medium lacking host tissue Bacteria not require host tissue to reproduce, whereas viruses 189 The smooth endoplasmic reticulum most resembles the Golgi apparatus in an intact eukaryotic cell when viewed under the microscope Both organelles appear to be membranes with many folds 190 If the ratio of R to R’ in the product mixture is 2:1, then the starting triglyceride must consist of two R-containing esters and one R’-containing ester 191 Lipases catalyze the hydrolysis of fats and other carboxylic esters— similar to fats, but not fats Lipase’s ability to catalyze the hydrolysis of fats and similar molecules reveals that some enzymes interact with several different substrate molecules that have similar chemical linkages 192 In horses, the genes for red coat color and white coat color are codominant Heterozygous horses have roan color If a roan-colored colt (CRCW) has a white mother (CWCW), the father’s coat must be either roan (CRCW) or red (CRCR) The mother could only give the white color gene, so the father must have given the red For the father to give the red color gene, he would have to carry one or two copies of it, meaning he must be either red or roan, respectively 193 The passage discusses the connection between H pylori infection and increased risk of gastric cancer If H pylori infection causes increased proliferation of mucosal cells in the stomach, this could lead to gastric cancer if genetic mutations occur in proliferating somatic cells that line the stomach 194 According to the passage, stomach ulcers and some forms of gastric cancer may be linked to H pylori infections in the stomach Without treatment by antibiotics, such infections can be persistent The most plausible explanation for why host antibodies are ineffective against H pylori is that antibody proteins may be denatured (destroyed) in the harsh (acidic) environment of the stomach If the antibody proteins are denatured, they will not function properly 195 According to the passage, there is more than one strain of H pylori One strain expresses the gene vacA that encodes a toxin The other strain encodes a gene cagA that leads to inflammation and might be related to the genesis of gastric cancer Thus, a significant difference between the strains is that the strains express different genes 196 According to the passage, expression of the cagA gene leads to inflammation of the mucosal lining of the stomach Thus, the cagA gene product triggers the movement of leukocytes into the mucosal tissue—because leucocytes gravitate toward an inflammation 197 According to the passage, infection by H pylori increases one’s risk of gastric cancer, but less than 25% of people affected ultimately develop such cancer Most people affected with H pylori not develop cancer because they tolerate the infection without developing tumors The DNA is not damaged and uncontrolled cell division does not occur 198 According to the passage, infection by H pylori increases one’s risk of gastric cancer, but less than 25% of people affected ultimately develop such cancer To be most effective, a gene therapy for gastric cancer should be directed against epithelial cells that give rise to tumors The infection originates in the epithelial tissue, which is the same location of the cancers—and the first place to focus preventative measures 199 Enzymatic activity in the stomach initiates the digestion of proteins Pepsin is the enzyme, secreted into the stomach, that catalyzes the hydrolysis of proteins into peptides and amino acids Pepsin works most efficiently at the low pH of the stomach—a pH that is obtained by secretion of HCl 200 According to the passage, the AB cell from the two-cell stage of the nematode, when kept in contact with the P1 cell, goes on to produce neurons, skin, and muscle In Experiment 1, researchers investigated the role of cell-to-cell communication by separating the cells of a two-cell embryo (AB and P1), and culturing 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 The observation that the fate of an isolated AB cell is different from that of an AB cell in an intact embryo supports the hypothesis that cell-to-cell communication is involved in the determination of cell fate 201 In Experiment 1, researchers investigated the role of cell-to-cell communication by separating the cells of a two-cell embryo (AB and P1), and culturing 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 The results of Experiment indicate that the direction of signaling between the blastomeres of a two-cell embryo is P1 → AB (since AB’s fate differed when isolated from P1, while P1’s fate was the same regardless of whether AB was present or not) 202 In Experiment 2, two-cell embryos were incubated in either cycloheximide (an inhibitor of translation) or actinomycin D (an inhibitor of transcription) The AB cells were then isolated and washed to remove inhibitors, and grown in culture AB cells of embryos treated with cycloheximide (the translation inhibitor, which would have prevented production of proteins at the ribosomes of both AB and P1 cells) produced only neurons and skin, while AB cells of embryos treated with actinomycin D (the transcription inhibitor, which would have prevented production of mRNA) produced neurons, skin and muscle—their normal fate These results indicate that the signaling interaction (between P1 and AB cells) at the two-cell stage probably involves protein, since proteins of the P1 cells could not have been produced to carry the necessary message(s) to the AB cells prior to isolation 203 The results of Experiment support the conclusion that gut specification during the four-cell stage requires cell-to-cell communication between P2 and EMS As indicated in Figure 2, the only case in which gut differentiation resulted, regardless of how long the cells were left in the four-cell stage, was the combination of EMS and P2 cells EMS and P2 cells that had zero minutes incubation in the four-cell stage still resulted in gut differentiation 204 If the zygote contains all unique cell contents that are necessary for gut differentiation, segregation of these substances during cell division would occur in the sequence of zygote to P1 to EMS to E This information comes directly from the flow chart in Figure 1, which shows that gut cells are derived from the following source: zygote → P1 → EMS → E 205 According to the information provided, the only somatic or visceral celltype tissue that derives from a single blastomere is gut This result is shown in Figure 2, which indicates that after 10 minutes in the four-cell stage, an isolated EMS cell can lead to gut differentiation In addition to this, the only cases in which gut differentiation occurred were cases in which EMS cells were present 206 The experiments indicate that nematode cells adopt different fates from those of their neighbors during development by both cell-to-cell signaling and segregation of cytoplasmic contents during cell division The importance of cellto-cell signaling is supported by the results of Experiment 1, in which the fate of an AB cell depended on whether it was cultured in the presence of P1 or not The importance of segregation of cytoplasmic contents during cell division is supported by the results of Experiment 3, in gut differentiation only occurred in cases where an EMS cell was present 207 The passage shows that Compound undergoes a rearrangement to form a ketone and the double bond in the side chain is retained If the compound shown in the stem reacts in the same way an aldehyde would form and the double bond in the chain would be retained 208 A negative entropy of activation indicates that the order of the transition state is increasing This is true of Chemist 1’s mechanism because the molecule would have to achieve the correct conformation for the reaction to take place 209 The rearrangement shown proceeding by Chemist 1’s mechanism can give only one product If Chemist 2’s reaction mechanism were correct there could possibly be cross products 210 The -OH stretching band appears in the IR at approximately 3500 cm-1 211 Based on the pedigree shown, the most likely pattern of inheritance for this rare disease (expressed in fewer than in 100,000 people) is sex-linked recessive The trait must be recessive since it does not appear in the parent’s generation (the parents are normal), but does appear in several of the children The trait is most likely sex-linked because only sons are affected, and not just sons from ONE cross, but sons from two crosses The mother had children with two men, and in both cases half of her sons had the disease The mother carries the trait XAXa, and passes the trait to half of her sons who inherit a Y from their father, thus leaving them with the disease (Xay) Daughters cannot inherit the disease because the father always gives a “normal” chromosome (XA) Half the daughters will be entirely unaffected, and half are expected to be carriers like their mother 212 The gall bladder stores bile produced by the liver and secretes it into the small intestine as needed Bile acts as an emulsifier, facilitating fat digestion When the gall bladder is removed, a patient will have reduced ability to digest fats 213 Unlike other organisms, the liver can partially regenerate after illness or damage This regeneration is accomplished by mitosis Mitosis is the process whereby human body cells (not gametes) reproduce 214 The N-H bond in Compound I would be capable of forming hydrogen bonds with water making it more soluble that the non-polar Compound II 215 In humans, cholesterol is a precursor to testosterone Cholesterol and testosterone are only two of several steroid-based natural molecules The body makes uses of the steroid backbone of cholesterol to produce the hormone testosterone 216 This item includes a description of an experiment in which activelydividing, synchronized cells were exposed to radioactively labeled 2deoxythymidine (the nitrogen base incorporated into DNA, but not RNA) After 30 minutes of exposure to the radioactively labeled substance, the cells were rinsed to remove unabsorbed label At various times thereafter, a group of cells from the culture were examined to determine the quantity of radioactive material in the nuclei The figure shows a peak in radioactivity between and 13 hours after treatment This peak represents DNA synthesis, since 2-deoxythymidine is a component of DNA, but not RNA or protein ... human land use Another implication in the passage and in final paragraph is that humans can know, appreciate, and become a part of the natural world and its symbiotic relationships—human awareness... unleashed a quake that was basically ready to go with a small provocation Answer B sounds good but the passage argues against elastic oscillations causing lasting deformations Answers C and D make... 20-carbon polyunsaturated fatty acid The difference between saturated and unsaturated fatty acids is that saturated fatty acids lack carbon-carbon double bonds (because all carbons are saturated with

Ngày đăng: 04/05/2017, 09:07

Xem thêm

TỪ KHÓA LIÊN QUAN

TÀI LIỆU CÙNG NGƯỜI DÙNG

TÀI LIỆU LIÊN QUAN

w