AAMC MCAT PRACTICE EXAM 2 | QUESTIONS & ANSWERS| ALREADY PASSED| VERIFIED & UPDATED

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) - 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) × (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. 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 - ANSWER 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. sp2 hybridized - ANSWER possess exactly one doubly bonded atom C/P: 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? Rate of reaction = 125 nM/s 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 - ANSWER A) 2.5 × 10-2 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. kcat, Vmax, [E] - ANSWER kcat = Vmax/[E] C/P: Absorption of ultraviolet light by organic mol