AAMC FL 2 Chem/Phys correctly answered 2023

AAMC FL 2 Chem/Phys correctly answered 2023What atom is the site of covalent attachment of AMC to the model tetrapeptide used in the studies? A. I B. II C. III D. IV - correct answer A. I basically asking where does the bond between amc and Y break. since you know they are attached by a peptide bond, which means the peptide chain attaches to the amino group of the next amino acid( that's why an amino acid runs from the N-terminal to the C-terminal), the covalent attachment is at site 1 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? 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)] E = hf = hc/λ Excitation occurs at λe = 360 nm, but fluorescence is observed at λf = 440 nm so I guess it's excitation - fluorescence ? One double bonded atom is... - correct answer sp2 The concentration of enzyme for each experiment was 5.0 μM. What is kcat for the reaction at pH 4.5 with NO chloride added when Compound 3 is the substrate? A. 2.5 × 10-2 s-1 B. 1.3 × 102 s-1 C. 5.3 × 103 s-1 D. 7.0 × 105 s-1 - correct answer A. 2.5 × 10-2 s-1 -rate of product formation did not vary over time for the first 5 minutes implies that the enzyme was saturated with substrate kcat = Vmax/[E] = (125 nM/s)/5.0 μM = 2.5 × 10-2 s-1 Saturating refers to.... - correct answer when adding more substrate does not increase the rate. That would refer to the top of the Michaelis-Menten curve or where the y axis is Vmax. The hyperbolic curve levels off remember that. So at a saturating amount of substrate the rate is kcat or the turnover number which is the maximum rate substrate can be converted to product Absorption of ultraviolet light by organic molecules always results in what process? A. Bond breaking B. Excitation of bound electrons C. Vibration of atoms in polar bonds D. Ejection of bound electrons - correct answer B. Excitation of bound electrons -The absorption of ultraviolet light by organic molecules always results in electronic excitation. -Bond breaking can subsequently result, as can ionization or bond vibration, but none of these processes are guaranteed to result from the absorption of ultraviolet light 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 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) 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 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 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 -In order to stand up, the person has to maintain a rotational equilibrium. -If they don't there will be a torque because their center of gravity will be at a distance from fulcrum of rotation (in this case their knees/legs), and their weight will produce a torque making it hard to stand up. -The answer is poorly worded because it just says equilibrium, there are two types of equilibrium, rotational and translational. -In this case, if the person doesn't lean forward or slide their feet under the chair, they won't have rotational equilibrium. -So leaning forward puts their center of mass above the fulcrum, making the torque zero (because Torque=distance from fulcrum x force sin (theta), and in this case that distance would be zero). -The other answers are wrong because A: obviously the goal isn't to increase the force required to stand up B: There will always be friction with the grown, even if you don't adjust your center of mass to stand up, and C: energy is a state function, if we are going from point A to point B (in this case sitting to standing) the amount of energy required to do so doesn't change. 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 α-helix. C. part of a β-sheet. D. part of a fully folded protein. - correct answer D. part of a fully folded protein the passage says that asymmetry due to tertiary structure gives rise to signal in the near UV region and that aromatic rings absorb in this region as well. A fully folded protein is expressing tertiary structure and tryptophan has an aromatic ring hence answer choice D. 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? A. Trp B. Phe C. Ala D. Tyr - correct answer C. Ala Says side chains with amino acids absorb in near region In designing the experiment, the researchers used which type of 32P labeled ATP? A. α32P-ATP B. β32P-ATP C. γ32P-ATP D. δ32P-ATP - correct answer C. γ32P-ATP -because the phosphoryl transfer from kinases comes from the γ-phosphate of ATP. -Therefore, the experiment should require γ32P-ATP. When used in place of spHM, which peptide would be most likely to achieve the same experimental results? A. FLGFAY B. FLGFQY C. FLGFGY D. FLGFEY - correct answer D. FLGFEY -The question is basically asking "which sequence is most similar to FLGFTY". All the answers are replacing T (threonine with R = OH) with some other amino acid. The goal is to find which substitution is most similar to Threonine -We are NOT looking for a sequence similar to FLGFTY, we are looking for a sequence similar to the PHOSPORYLATED spHM. Glutamate is the right choice because it is charged, like the phosphate would be Based on the information in the passage, PDK1 catalyzes the addition of phosphate to what functional group? A. Hydroxyl B. Amine C. Carboxyl D. Phenyl - correct answer A. Hydroxyl -because reactions involving either Ser or Thr would involve the hydroxyl group in the side chain of these amino acids 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 -A because both curves have a sigmoidal shape, which is indicative of cooperative processes From the data presented in Figure 3, which RIα variant is the most stable? A. L203A B. I204A C. Y229A D. R241A - correct answer A. L203A -because a higher melting temperature is indicative of a more stable protein, as more energy is needed to unfold the protein. L203A has an approximate Tm of 50°C, therefore, it is the most stable and even more stable than the WT protein -Remember Tm is the halfway point Based on the data presented in figures 2 and 3, what is the most likely role of Y229 in protein stability and cAMP activation? A. Y229 is important for protein stability but not critical for cAMP activation. B. Y229 is important for cAMP activation but not critical for protein stability. C. Y229 is important for protein stability and critical for cAMP activation. D. Y229 is not important for protein stability and not critical for cAMP activation. - correct answer A. Y229 is important for protein stability but not critical for cAMP activation. -because the thermal melt shows that removal of Y229 decreases stability, therefore Y229 is important for stability. -Removal of Y229 has little effect on protein activation, as the activation curve is similar to WT activation Venturi Effect - correct answer the reduction in fluid pressure that results when a fluid flows through a constricted section of a pipe. What phenomenon causes static air to be drawn into the mask when oxygen flows? A.Doppler effect B. Venturi effect C. Diffusion D. Dispersion - correct answer B. Venturi Effect because oxygen pressure is the sum of the oxygen static pressure P and the oxygen flow pressure ρv2/2. In the area of the mask openings, Pair = P + ρv2/2, thus Pair > P. Air enters the mask because the static pressure of the air is larger than the static pressure of the oxygen in flow What causes duplex DNA with a certain (A + T):(G + C) ratio to melt at a higher temperature than comparable length duplex DNA with a greater (A + T):(G + C) ratio? A. Stronger van der Waals forces of pyrimidines B. Stronger van der Waals forces of purines C. Increased π- stacking strength D. Reduced electrostatic repulsion of phosphates - correct answer C. Increased π- stacking strength -G-C base pairs form stronger π-stacking interactions than A-T base pairs, thereby creating the most thermal stability. -This disparity has often been used to explain the increased melting temperature of DNA rich in GC