MCAT AAMC practice exam Chem Phys exam questions fully solved & verified for accuracy 2024

What quantity of Compound 1 (483.5g) must be provided to prepare 100.00 mL of solution with a concentration equal to Ki (60.3uM)? M=mol/v 1 umol=0.000001 moles (1x10^-6) 2.92mg If only [I] is increased, then [ESI] or [EI] increases. This is an example of: A. the Bose-Einstein Principle. B. the Heisenberg Uncertainty Principle. C. the Le Châtelier's Principle. D. the Pauli Exclusion Principle. Solution: The correct answer is C. A. The Bose-Einstein Principle states that a collection of atoms cooled close to absolute zero will coalesce into a single quantum state. B. The Heisenberg Uncertainty Principle states that one cannot know both the momentum and position of an object with absolute certainty. C. Le Châtelier's Principle states that in a reversible process, the application of stress to the system will cause the system to respond in a way that will relieve this stress. In this case, the reversible process is E + I forming EI or ES + I forming ESI. In either case, increasing [I] induces stress of the system, and the system relieves that stress by converting I to either more EI or ESI. D. The Pauli Exclusion Principle states that two or more identical fermions, e.g. electrons, cannot occupy the same quantum state. Which structural change to Compound 1 would make it more water soluble? A. Replacing benzene CH with N in the ring B. Replacing C=O with C=CH2 C. Replacing N-N=N with CH-CH=CH D. Replacing NH with NCH3 Solution: The correct answer is A. A. Nitrogen in the benzene ring would have a lone pair that could accept a hydrogen bond from water, thus increasing the solubility of the compound. B. Replacing C=O with C=CH2 would decrease the polarity of the compound and make it less water soluble. C. Replacing N−N=N with CH−CH=CH would decrease the polarity of the compound and make it less water soluble. D. Replacing NH with NCH3 removes a hydrogen bond donor, thus decreasing water solubility. In μM•s-1 and μM, what should the approximate values of kcat/KM and Ki be, respectively, when [I] = 180 μM? Neither is affected by a change in I, therefore they will stay the same (150 and 60.3). Ki is the equilibrium constant and is never affected by increasing I. What functional group transformation occurs in the product of the reaction catalyzed by Na+-NQR (NADH and ubiquinone)? A. RC(=O)R → RCH(OH)R B. ROPO32- → ROH + Pi C. RC(=O)NHR'→ RCOOH + R'NH2 D. RC(=O)OR'→ RCOOH + R'OH Solution: The correct answer is A. A. This is two-electron reduction of a ketone to an alcohol, which is the reaction catalyzed by Na+-NQR. A. This is the reaction catalyzed by a phosphatase. A. This is the reaction catalyzed by a protease or amidase. A. This is the reaction catalyzed by an esterase. What is the chemical structure of a component found in four of the five cofactors (flavins) used by Na+-NQR? Solution: The correct answer is B. A. This is the structure of adenine. It is only found in FAD. B. This is the structure of flavin, found in four of the five cofactors used by Na+-NQR. C. This is the structure of ubiquinone. It is a substrate, but not a cofactor. D. This is the structure of histidine. It is an amino acid, not a cofactor. What is the ratio of cation (0.150M) to enzyme (0.75mM) in the spectroelectrochemical experiments described in the passage? A. 1:2 B. 2:1 C. 20:1 D. 200:1 0.150M=150mM 150/0.75= 200: 1 (D) The reaction between NADH and ubiquinone is exergonic, but the reaction, when catalyzed by Na+-NQR, does not generate much heat in vivo. What factor accounts for this difference? The reaction catalyzed by Na+-NQR in vivo: A. is more exothermic as a result of the lower activation energy. B. occurs sequentially in several small steps. C. maintains a large separation between the reacting centers. D. is coupled to the movement of a charged particle against a concentration gradient. Solution: The correct answer is D. A. This is impossible. Even if it were true, this would make the heat generation larger, not smaller, for the catalyzed reaction. Catalysis does not change thermodynamics. B. By Hess's Law, the heat of reaction will sum and be the same. The fact that the reaction can be broken down into steps will not change the overall thermodynamics. C. This is also impossible. The reactants ultimately must be close together to react. D. The movement of a charged particle against its concentration gradient is energetically costly. Coupling the two processes: the redox reaction between NADH and ubiquinone and the movement of Na+ up a concentration gradient makes the overall process less exothermic. Two open flasks I and II contain different volumes of the same liquid. Suppose that the pressure is measured at a point 10 cm below the surface of the liquid in each container. How will the pressures compare? A. The pressures will be equal. B. Pressure in flask I will be less. C. Pressure in flask II will be less. D. The pressures cannot be compared from the information given. Solution: The correct answer is A. A. The pressure at a point 10 cm below the surface of the liquid is the same in both flasks because the pressure is equal to the liquid density multiplied by the gravitational acceleration multiplied by 10 cm. B. The pressure at a point 10 cm below the surface of the liquid in flask I is the same as the pressure in flask II at 10 cm below the surface because the pressure is equal to the liquid density multiplied by the gravitational acceleration multiplied by 10 cm. C. The pressure at a point 10 cm below the surface of the liquid in flask II is the same as the pressure in flask I at 10 cm below the surface because the pressure is equal to the liquid density multiplied by the gravitational acceleration multiplied by 10 cm. D. The pressures can be compared because both pressures are calculated according to the hydrostatic pressure formula p = ρgd, where ρ is the liquid density, g is the gravitational acceleration, and d is the depth where pressure is measured. What is the identity of an atom that contains six protons and eight neutrons? A. Nitrogen B. Carbon C. Oxygen D. Silicon Solution: The correct answer is B. A. The atom cannot be nitrogen because nitrogen contains seven protons. B. Carbon contains six protons because it also contains six electrons as a neutral atom. C. Oxygen contains eight protons because as a neutral atom it contains eight electrons. D. The silicon atom contains 14 protons. Which of the following substances is polar? A. NF3 B. CCl4 C. CO2 D. Li2 Solution: The correct answer is A. A. The geometry of trifluoroamine is impacted by the lone pair on nitrogen, making it trigonal pyramidal. No bond dipoles cancel; this results in a polar molecule. B. While each C-Cl bond in carbon tetrachloride is polar, the sum of the dipole moments cancel as a result of its tetrahedral molecular geometry. C. Carbon dioxide is a linear molecule. The bond dipole moments of each C=O bond cancel as they are in opposite directions. D. This molecule is necessarily non-polar as it is comprised of two identical atoms. If all else is held constant, which of the following changes would NOT double the volume of a gas? A. Doubling the pressure B. Doubling the absolute temperature C. Halving the pressure D. Doubling the number of gas molecules Solution: The correct answer is A. A. Based on Boyle's Law, P is inversely proportional to volume, thus, doubling the pressure of a gas sample will decrease, not increase, the volume. B. Doubling the absolute temperature of a gas sample will double the volume because T is directly proportional to V for an ideal gas according to Charles' Law. C. Halving the pressure of a gas sample will double its volume because P is inversely proportional to V for an ideal gas. D. Doubling the number of gas molecules present will double the volume according to Avogadro's Law. What is the number of neutrons in the nucleus of the atom used to produce laser radiations? A. 48 B. 49 C. 50 D. 51 86Kr+ 36 protons, so 86-36=50 (C) What is the molecular formula of the heterocyclic aromatic compound pyrrole? A. C2H3N B. C4H5N C. C6H7N D. C8H9N Solution: The correct answer is B. A. This is the molecular formula of the non-aromatic heterocycle 2H-azirine. B. This is the molecular formula of pyrrole, a five-membered aromatic heterocycle containing one nitrogen atom. C. This is the molecular formula of various forms of azepine, a seven-membered heterocycle, none of which are aromatic. D. There is a cyclohexene-fused pyrrole with this molecular formula, but not pyrrole itself. Approximately how many moles of Kr+ are contained in the laser tube at 0°C and 1 atm? V= 11cm^3 A. 3 x 10-7 B. 2 x 10-6 C. 4 x 10-5 D. 5 x 10-4 D 11cm^3= 11x10^-3L 1 mole of gas at STP is 22.4 L 11/22.4= 0.5 (x10^-3), therefore 5e-4 is the answer The radiation of wavelength 605 nm CANNOT be used to produce the fluorescence radiations depicted in Figure 3 because: A. the energy of the absorbed radiation must be larger than the energy of the fluorescence radiation. B. the energy of the absorbed radiation must be smaller than the energy of the fluorescence radiation. C. the 605-nm radiation has more energy than the 407-nm radiation. D. the 605-nm radiation is not visible. Solution: The correct answer is A. A. Fluorescence can occur when the absorbed radiation has a photon energy larger than the photon energy of the radiation emitted through fluorescence. The photon energy is inversely proportional to the radiation wavelength, thus the 605-nm wavelength radiation cannot produce the entire fluorescence radiation spectrum shown in Figure 3 as its photon energy is below that of the fluorescence radiation of wavelength 604 nm. B. The photon energy of the absorbed radiation must exceed the photon energy of the fluorescence radiation because the energy difference cannot be created from nothing, per the energy conservation principles. C. The photon energy is inversely proportional to the radiation wavelength according to the formula Energy = Planck's constant multiplied by the speed of light divided by the wavelength, thus the 605-nm wavelength radiation has less photon energy than the 407 nm wavelength radiation. D. The 605 nm wavelength radiation is visible, but this feature is unrelated to fluorescence. The CD spectroscopy signal that was used to generate the data in Figure 1 arises from the chirality of the: A. α carbon. B. amide nitrogen. C. carbonyl carbon. D. β carbon. Solution: The correct answer is A. A. In a protein, each amino acid residue, except Gly, has a chiral α carbon. B. Amide nitrogen is achiral. C. Carbonyl carbon is achiral. D. With the exception of Ile and Thr,the β carbon in amino acid residues is achiral. CD spectra arise from the massively greater presence of α carbon chirality. Which physical property does NOT change with the amino acid substitution made in TPMT*5? (L49S) A. Molecular weight B. Hydrophobicity C. Hydrogen bonding capability D. Net charge Solution: The correct answer is D. A. The side chain of the L49S variant (TPMT*5) at position 49 has a different molar mass than the side chain of Leu. B. The L49S variant (TPMT*5) has a hydrophilic residue at position 49, but wild-type TPMT has a hydrophobic residue at that position. C. The L49S variant (TPMT*5) can hydrogen bond, but wild-type TPMT cannot. D. The L49S variant (TPMT*5) has the same charge as wild-type TPMT because the amino acid residues do not have charge. Samples from various time points of the proteolysis of TPMTwt were subjected to SDS-PAGE under reducing conditions. Which figure best depicts the expected appearance of the gel? (Note: The arrow indicates the movement of the protein through the gel.) As befits proteolysis, the number of lower molecular weight bands with time increases and the original protein band at the highest molecular weight diminishes with time. If the combined mass of the TPMT (transferase) substrate and cofactor was determined before the enzymatically catalyzed reaction and then compared to the combined mass of the product and the cofactor after the reaction, the net change in molecular weight will be: A. +15 g/mol. B. 0 g/mol. C. -15 g/mol. D. -16 g/mol. 0 g/mol bc TPTP is a transferase that transfers methyl group from cofactor to substrate The ligand of hMPRα (progesterone) is derived from which compound? A. Glucose B. Phenylalanine C. Glycerol D. Cholesterol Solution: The correct answer is D. A. Glucose is a simple sugar, which are not precursors to steroids. B. Phenylalanine is an amino acid, which are the components of proteins. C. Glycerol is a the trihydroxyl backbone for triacyl glycerols, which are fats. D. The ligand for hMPRα is the steroid progesterone. Steroids are a class of lipids that are derived from cholesterol. The second purification step (His tags) is which type of chromatographic separation? A. Affinity B. Size exclusion C. Cation exchange D. Anion exchange Solution: The correct answer is A. A. Displacement of the protein from the column in this step involved disrupting the binding of the (His)6 tag to the column. This is a classic example of affinity chromatography. B. Size-exclusion chromatography separates proteins by molecular weight, not selective column binding. C. Cation-exchange chromatography separates proteins with different positive charges (or positive versus negative/neutral charge). Its separation method is not based on selective binding to the column based on a functional group-specific chemical reaction or ligand-receptor interaction. D. Anion-exchange chromatography separates proteins with different negative charges (or negative versus positive/neutral charge). Its separation method is not based on selective binding to the column based on a functional group-specific chemical reaction or ligand-receptor interaction. What structural feature(s) is(are) most important to the functioning of this compound as described in the passage? A. Specific configuration of numerous chirality centers B. Multiple hydrolysable linkages C. Combination of large hydrophobic and hydrophilic regions D. Presence of a reducing sugar Solution: The correct answer is C. A. Chirality is important for specific binding to other compounds. This is not a requirement of a detergent. B. Multiple hydrolysable linkages would facilitate metabolism. Compound 2 was a detergent used to help isolate membrane proteins. C. Compound 2 was used as a detergent. It liberated a protein from a membran