AAMC MCAT Practice Exam 2 2024 with 100% correct answers, Exams of Physics

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AAMC MCAT Practice Exam 2

2024 with 100% correct

answers

C/P: What expression gives the amount of light energy (in J per photon) that is converted to other forms between the fluorescence excitation and emission events? "intensity of fluorescence emission at 440 nm excitation at 360 nm) was monitored for 20 minutes" A) (6.62 × 10-34) × (3.0 × 108) B) (6.62 × 10-34) × (3.0 × 108) × (360 × 10-9) C) (6.62 × 10-34) × (3.0 × 108) × [1 / (360 × 10-9) - 1 / (440 × 10-9)] D) (6.62 × 10-34) × (3.0 × 108) / (440 × 10-9) - correct answer C) (6.62 × 10-34) × (3.0 × 108) × [1 / (360 × 10-9) - 1 / (440 × 10-9)] The answer to this question is C because the equation of interest is E = hf = hc/λ, where h = 6.62 × 10 − 34 J ∙ s and c = 3 × 10 8 m/s. Excitation occurs at λe = 360 nm, but fluorescence is observed at λf = 440 nm. This implies that an energy of E = (6.62 × 10 −34) × ( × 10 8) × [1 / (360 × 10 −9) − 1 / (440 × 10 −9)] J per photon is converted to other forms between the excitation and fluorescence events. C/P: Compared to the concentration of the proteasome, the concentration of the substrate is larger by what factor? "purified rabbit proteasome (2 nM) was incubated in the presence of porphyrin...the reaction was initiated by addition of the peptide (100 uM)" A) 5 × 101 B) 5 × 102 C) 5 × 103

• D) 5 × 104 - correct answer D) 5 ×

C/P: Four organic compounds: 2-butanone, n-pentane, propanoic acid, and n- butanol, present as a mixture, are separated by column chromatography using silica gel with benzene as the eluent. What is the expected order of elution of these four organic compounds from first to last? A) n-Pentane → 2-butanone → n-butanol → propanoic acid B) n-Pentane → n-butanol → 2-butanone → propanoic acid C) Propanoic acid → n-butanol → 2-butanone → n-pentane D) Propanoic acid → 2-butanone → n-butanol → n-pentane - correct answer A) n-Pentane → 2- butanone → n-butanol → propanoic acid The answer to this question is A. The four compounds have comparable molecular weights, so the order of elution will depend on the polarity of the molecule. Since silica gel serves as the stationary phase for the experiment, increasing the polarity of the eluting molecule will increase its affinity for the stationary phase and increase the elution time (decreased Rf). C/P: The half-life of a radioactive material is: A) half the time it takes for all of the radioactive nuclei to decay into radioactive nuclei. B) half the time it takes for all of the radioactive nuclei to decay into their daughter nuclei. C) the time it takes for half of all the radioactive nuclei to decay into radioactive nuclei. D) the time it takes for half of all the radioactive nuclei to decay into their daughter nuclei. - correct answer D) the time it takes for half of all the radioactive nuclei to decay into their daughter nuclei. The answer to this question is D because the half-life of a radioactive material is defined as the time it takes for half of all the radioactive nuclei to decay into their daughter nuclei, which may or may not also be radioactive. C/P: A person is sitting in a chair. Why must the person either lean forward or slide their feet under the chair in order to stand up? A) to increase the force required to stand up B) to use the friction with the ground

C) to reduce the energy required to stand up D) to keep the body in equilibrium while rising - correct answer D) to keep the body in equilibrium while rising The answer to this question is D because as the person is attempting to stand, the only support comes from the feet on the ground. The person is in equilibrium only when the center of mass is directly above their feet. Otherwise, if the person did not lean forward or slide the feet under the chair, the person would fall backward due to the large torque created by the combination of the weight of the body (applied at the person's center of mass) and the distance along the horizontal between the center of mass and the support point. C/P: The side chain of tryptophan will give rise to the largest CD signal in the near UV region when: A) present as a free amino acid B) part of an a-helix C) part of a B-sheet D) part of a fully folded protein - correct answer D) part of a fully folded protein The answer to this question is D because tryptophan has an aromatic side chain that will give rise to a significant CD signal in the near UV region if it is found in a fully folded protein. C/P: Which amino acid will contribute to the CD signal in the far UV region, but NOT the near UV region, when part of a fully folded protein? "Asymmetry resulting from tertiary structural features causes the largest increase in CD signal intensity in the near UV region of peptides. The side chains of amino acid residues absorb in this region. The peptide bond absorbs in the far UV region (190-250 nm). The CD signals of these bonds are dramatically impacted by their proximity to secondary structural elements." A) Trp B) Phe C) Ala

D) +1 - correct answer C) 0 The answer to this question is C because at pH 7.2, the N-terminus will be positively charged and the C- terminus will be negatively charged. In addition, the lysine side chain will carry one positive charge and the glutamic acid side chain will carry one negative charge. C/P: In designing the experiment, the researchers used which type of P-32 labeled ATP? A) aP32-ATP B) BP32-ATP C) γP32-ATP D) δP-32 ATP - correct answer D) δP-32 ATP The answer to this question is C because the phosphoryl transfer from kinases comes from the γ- phosphate of ATP. Therefore, the experiment should require γ32P-ATP. C/P: When used in place of spHM, which peptide would be most likely to achieve the same experimental results? "This experiment was repeated in the presence of a synthetic peptide that mimics the HM domain (sHM) of Ser/Thr kinases with the amino acid sequence FLGFTY. Phosphorylated sHM (spHM) was also used in place of sHM." A) FLGFAY B) FLGFQY C) FLGFGY D) FLGFEY - correct answer D) FLGFEY The answer to this question is D because the phosphorylated threonine would most likely be mimicked by glutamic acid in terms of size and charge. C/P: Based on the information in the passage, PDK1 catalyzes the addition of phosphate to what functional group?

"This experiment was repeated in the presence of a synthetic peptide that mimics the HM domain (sHM) of Ser/Thr kinases with the amino acid sequence FLGFTY. Phosphorylated sHM (spHM) was also used in place of sHM." A) Hydroxyl B) Amine C) Carboxyl D) Phenyl - correct answer A) Hydroxyl The answer to this question is A because reactions involving either Ser or Thr would involve the hydroxyl group in the side chain of these amino acids. C/P: Which statement about the cooperativity of RIα/C activation and RIα protein folding is supported by the data in figures 2 and 3? A) Both activation and folding are cooperative. B) Activation is cooperative, but folding is not. C) Folding is cooperative, but activation is not. D) Neither activation nor folding is cooperative - correct answer A) Both activation and folding are cooperative. The answer to this question is A because both curves have a sigmoidal shape, which is indicative of cooperative processes. C/P: A patient puts on a mask with lateral openings and inhales oxygen from a tank. Which phenomenon causes static air to be drawn into the mask when oxygen flows? A) Doppler effect B) Venturi effect C) Diffusion

C/P: Which property of a substance is best used to estimate its relative vapor pressure? A) Melting point B) Boiling point C) Molecular weight D) Dipole moment - correct answer B) Boiling point The answer to this question is B because of the properties listed, the boiling point of a substance will give the best estimate of its relative vapor pressure. C/P: What are the structural features possessed by storage lipids? A) Two fatty acids ester-linked to a single glycerol plus a charged head group B) Three fatty acids ester-linked to a single glycerol C) Two fatty acids ester-linked to a single sphingosine plus a charged head group D) Three fatty acids ester-linked to a single sphingosine - correct answer B) Three fatty acids ester- linked to a single glycerol The answer to this question is B because triacylglycerols are neutral storage lipids. They consist of three fatty acids ester-linked to a single glycerol. C/P: In the overall electrochemical reaction: N2(g) + H2(g) --> NH3(g) Half reactions: H2(g) → 2H+ + 2e- N2(g) + 6H+ + 6e- → 2NH3(g) A) nitrogen is oxidized at the anode, and hydrogen is reduced at the cathode. B) nitrogen is reduced at the cathode, and hydrogen is oxidized at the anode.

C) nitrogen is reduced at the anode, and hydrogen is oxidized at the cathode. D) nitrogen is oxidized at the cathode, and hydrogen is reduced at the anode. - correct answer B) nitrogen is reduced at the cathode, and hydrogen is oxidized at the anode. The answer to this question is B because oxidation always occurs at the anode and reduction at the cathode of an electrochemical cell. Since nitrogen decreases in oxidation state during the reaction, it is reduced. Hydrogen, on the other hand, increases in oxidation state and is, therefore, oxidized. ANOX REDCAT - correct answer In an electrochemical cell: oxidation occurs at the anode reduction occurs at the cathode C/P: In industrial use, ammonia is continuously removed from the reaction mixture. This serves to drive Reaction 1 because of: N2(g) + H2(g) --> NH3(g) A) Boyle's law B) Charles's law C) Heisenberg's principle D) Le Châtelier's principle - correct answer D) Le Châtelier's principle The answer to this question is D because removing a product as it forms causes a displacement from the equilibrium condition. The system will respond by shifting more reactants to the product side. This is an example of Le Châtelier's principle. Boyle's law - correct answer a law stating that the pressure of a given mass of an ideal gas is inversely proportional to its volume at a constant temperature

"SCY conductors are favored for this use because their proton conductivities increase substantially with temperature." A) SCY temperature higher than electrode temperature B) SCY temperature lower than electrode temperature C) SCY temperature the same as electrode temperature D) The temperature of the components does not make a difference. - correct answer A) SCY temperature higher than electrode temperature The answer to this question is A because the proton conductivity of SCY increases with increasing temperature, while the favorability of reaction decreases with overall temperature. It is therefore beneficial to maintain the SCY conductor at a higher temperature. C/P: At 25°C, the formation of [Cu(NH3)4]2+ according to Equation 1 is most likely a: "The value of the formation constant of [Cu(NH3)4]2+ is 5.6 x 1011 at 25°C." A) spontaneous process with positive ΔG°. B) spontaneous process with negative ΔG°. C) nonspontaneous process with positive ΔG°. D) nonspontaneous process with negative ΔG°. - correct answer B) spontaneous process with negative ΔG°. The answer to this question is B because the equilibrium constant for the reaction is very large (much greater than 1). This necessarily means that ΔG° is negative and the reaction is spontaneous. C/P: In the stepwise formation of [Cu(NH3)4]2+ from [Cu(H2O)4]2+, which of the following ions would form in the second step? "The formation of [Cu(NH3)4]2+ takes place in a stepwise manner with one water molecule being replaced by an ammonia molecule in each step."

A) [Cu(H2O)2(NH3)2]2+ B) [Cu(H2O)2(NH3)3]2+ C) [Cu(H2O)3(NH3)]2+ D) [Cu(H2O)3(NH3)2]2+ - correct answer A) [Cu(H2O)2(NH3)2]2+ The answer to this question is A. The reaction was stated to proceed in four sequential steps, with each step involving the replacement of one water molecule by a molecule of ammonia. As a result, the product of the second step is [Cu(H2O)2(NH3)2]2+ C/P: Which of the following best describes the bonds between Cu2+ and the nitrogen atoms of the ammonia molecules in [Cu(NH3)4]2+? A) Ionic B) Covalent C) Coordinate ionic D) Coordinate covalent - correct answer D) Coordinate covalent The answer to this question is D because the Lewis acid-base interaction between a metal cation and an electron pair donor is known as a coordinate covalent bond. ionic bond - correct answer complete transfer of valence electron(s) between atoms covalent bond - correct answer chemical bond that involves the sharing of electron pairs between atoms coordinate covalent bond - correct answer one atom donates its electrons to form the covalent bond without the other atom contributing C/P: Consider the reaction shown in Equation 1 at equilibrium. Would the concentration of [Cu(NH3)4]2+ increase if the equilibrium were disturbed by adding hydrochloric acid?

B) I and III only C) II and III only D) I, II, and III - correct answer C) II and III only The answer to this question is C because Reaction 1 is a Lewis acid-base reaction. The fact that the reaction proceeds in the forward direction indicates that NH3 is a better Lewis base toward Cu2+ than H2O. This also means that NH3 donates a lone pair of electrons more readily than does H2O. C/P: Which of the following atoms will be expected to have the smallest second ionization energy? A) Na B) C C) O D) Ca - correct answer D) Ca The answer to this question is D. Metals have lower ionization energies than non- metals as long as the ionization event involves a valence electron. Since Na is an alkali metal, it has only one valence electron and has a large second ionization energy. Ca is an alkaline earth metal and has two valence electrons. It will therefore have the smallest second ionization energy of the four atoms listed, which include Na and two non-metals. C/P: Which of the following species has the largest mass percent of oxygen? A) H2O B) CaCO C) CO D) HCO3- - correct answer A) H2O The answer to this question is A because mass percent of oxygen is calculated by multiplying the coefficient for oxygen in the formula of the substance by 16 and then dividing by the molar mass of the substance. Water has the highest percentage of oxygen by mass of the compounds listed (16/18) × 100 = 89%.

C/P: What is the pH of a buffer solution that is 0.2 M in HCO3- and 2 M in H2CO3? (Note: The first pKa of carbonic acid is 6.37.) A) 4. B) 5. C) 6. D) 7.37 - correct answer B) 5. The answer to this question is B. The pH of the solution can be calculated using the Henderson- Hasselbach equation: pH = pKa + log([base]/[acid]). Plugging in the values provided in the question gives pH = 6.37 + log(0.2/2) = 5.37. Henderson-Hasselbalch equation - correct answer pH = pka + log [base]/[acid] pOH = pkb+ log [acid]/[base] C/P: What is the concentration of Ca2+(aq) in a saturated solution of CaCO3? (Note: The solubility product constant Ksp for CaCO3 is 4.9 × 10-9.) Ca2+(aq) + 2HCO3-(aq) --> CaCO3(s) + CO2(g) + H2O(l) A) 2.4 × 10-4 M B) 4.9 × 10-5 M C) 7.0 × 10-5 M D) 4.9 × 10-9 M - correct answer C) 7.0 × 10-5 M The answer to this question is C. The solubility product constant expression for CaCO3 is Ksp = [Ca2+] [CO32-]. Since equal quantities of Ca2+(aq) and CO32-(aq) are produced when CaCO3 dissolves, this expression reduces to 4.9 × 10-9 = x2, or 49 × 10-10 = x2. This can be solved directly by taking the square root of each side. C/P: An inflatable cuff was used to temporarily stop blood flow in an upper arm artery. While releasing the pressure to deflate the cuff, a stethoscope was used to listen to blood flow in the forearm. The

A.5.1 × 1011 protons added to it. B.5.1 × 1011 electrons removed from it. C.2.0 × 1010 protons added to it. D.2.0 × 1010 electrons removed from it. - correct answer D.2.0 × 1010 electrons removed from it. The answer to this question is D because the number of charges in excess can be computed as +3.2 × 10 -9 C/1.6 × 10 -19 C = +2.0 × 10 10. This means that the rod has an excess of positive charge, created by removing a number of +2.0 × 10 10 electrons from the material, as it is not possible to add protons in a manner described in this question. C/P: Which single bond present in nitroglycerin is most likely the shortest? "The average bond energies in kJ/mol of the C-H, C-O, C-C, and O-N single bonds present in nitroglycerin are 413, 358, 347, and 201, respectively." A) C-H B) C-O C) C-C D) O-N - correct answer A) C-H The answer to this question is A. All of the bonds listed are single bonds. Since hydrogen has a much smaller atomic radius than second period elements, the covalent bond between C and H is shorter than any of the other bonds listed. C/P: Based on the passage, the magnitude of ΔH° (in kJ) for the decomposition of 2 moles of nitroglycerin at 25°C is closest to which of the following? A) 500 B) 1000 C) 2000 D) 3000 - correct answer D) 3000

The answer to this question is D. The value of ΔH° can be calculated using the data provided in Table 1 and applying Hess's Law. Two moles of nitroglycerin produce 6 moles of CO2(g) and 5 moles of H2O(g). The value of ΔH° for this amount of nitroglycerin combusted is 2(364.0) - 6(393.5) - 5(241.8) = -2842 kJ/mol. C/P: At STP, the volume of N2(g) produced by the complete decomposition of 1 mole of nitroglycerin would be closest to which of the following? 4C3H5N3O9(l) → 12CO2(g) + 10H2O(g) + 6N2(g) + O2(g) A) 5 L B) 10 L C) 20 L D) 30 L - correct answer D) 30 L The answer to this question is D. Based on the balanced equation provided, 4 moles of nitroglycerin produce 6 moles of N2(g). Therefore, 1 mole of nitroglycerin will produce 1.5 moles of N2(g). At STP 1.5 moles of N2(g) will occupy 33.6 L since the molar volume of an ideal gas at STP is 22.4 L/mol. C/P Which single bond present in nitroglycerin is the LEAST polar? A) C-H B) C-O C) C-C D) O-N - correct answer C) C-C The answer to this question is C because bonds between identical atoms are likely to be the least polar. This is true in this case since each C atom is bonded to one nitro group only. C/P: What is the average power consumed by a 64-year-old woman during the ascent of the 15-cm-high steps, if her mass is 54 kg? Time to climb up 30 steps (s): 27