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AAMC FL2 C/P EXAM 2024 QUESTIONS WITH COMPLETE SOLUTIONS 100% CORRECT!!
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AAMC FL2 C/P EXAM 2024 QUESTIONS WITH COMPLETE SOLUTIONS 100% CORRECT!!
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AAMC FL2 C/P EXAM 2024 QUESTIONS WITH COMPLETE SOLUTIONS Evatee 6/27/24 AAMC FL2 C/P AAMC FL2 C/P EXAM 2024 QUESTIONS WITH COMPLETE SOLUTIONS 100% CORRECT!!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 Answer - 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) Answer - 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 Answer - 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.
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