AAMC FL2 C-P Questions Review 2023

AAMC FL2 C-P Questions Review 2023 What 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 - This question is A because AMC is attached to the peptide on the carboxyl side. This suggests that an amide linkage involving the N atom in AMC is used to covalently attach the fluorophore to the peptide. The passage says AMC is removed from a peptide via peptide bond hydrolysis. Thus, AMC must form a peptide bond with the peptide. Peptide bonds are between amine and carboxylic acid groups to form an amide. There is only one amine in AMC. It also helps to know that chymotrypsin cleaves on the carboxyl side of an aromatic amino acid, meaning the functional group on AMC that binds is the amine. 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) - 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) × (3 × 10 8) × [1 / (360 × 10 −9) − 1 / (440 × 10 −9)] J per photon is converted to other forms between the excitation and fluorescence events. Compared to the concentration of the proteasome, the concentration of the substrate is larger by what factor? A. 5 × 101 B. 5 × 102 C. 5 × 103 D. 5 × 104 - The answer to this question is D. The proteasome was present at a concentration of 2 × 10-9 M, while the substrate was present at 100 × 10-6 M. The ratio of these two numbers is 5 × 104. Put the given concentrations into a ratio. The question asks by how much the substrate concentration is bigger, so put the substrate concentration on top. This is [S]/[E] = (100 microM)/(2 nanoM).Simplify the ratio by dividing. Since math and keeping track of units isn't my forte, I personally like to rewrite all units in SI units with scientific notation, like this: (100 x 10-6 M)/(2 x 10-9 M). Then I divide the numbers (100/2 = 50), and then I divide the powers of ten (10-6 / 10-9 = 103). So the answer is 50 x 103 M, which equals one of the answer choices, 5 x 104 M. 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 - The answer to this question is A. The fact that the rate of product formation did not vary over time for the first 5 minutes implies that the enzyme was saturated with substrate. Under these conditions, kcat = Vmax/[E] = (125 nM/s)/5.0 μM = 2.5 × 10-2 s-1. 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 - The answer to this question is B. 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. C. Different effects dominate at different frequencies. With IR, there isn't enough energy in the photon to excite electrons, so the energy goes into vibrations. With UV, there is always enough energy in the photon to cause an excitation of the electron, and that's where the energy goes, rather than vibration. D. This is an example of the photoelectric effect. You can absorb energy without causing emission. Continues...