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FL 5 AAMC MCAT QUEST ION
WITH COMPLETE SOLUTI ON
Question : What quantity of Compound 1 must be
provided to prepare 100.00 mL of solution with a concentration equal to Ki?
- 48.4 mg
- 24.2 mg
- 5.64 mg
- 2.92 mg
"Compound 1 (molar mass: 483.5 g/mol ) has been shown
to inhibit HIV-1 protease with Ki = 60.3 μM (Table 1). Ki is the dissociation constant for the enzyme-bound inhibitor, which is either EI or ESI, depending on the type of inhibitor."
Correct answer: D
In 100.00 mL solution, 60.3 μM Compound 1 contains 6.03 μmol, which when converted to mol and multiplied by the molar mass, yields 0.00292 g or 2.92 mg.
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Question : What functional group transformation occurs
in the product of the reaction catalyzed by Na+-NQR?
A. RC(=O)R → RCH(OH)R
- ROPO32- → ROH + Pi
C. RC(=O)NHR'→ RCOOH + R'NH2
D. RC(=O)OR'→ RCOOH + R'OH
"The electron transport pathway in Na+-NQR is composed of four flavins (FAD, FMNc, FMNb, and riboflavin) and a [2Fe-2S] center, with electrons flowing
in the direction: NADH → FAD → [2Fe-2S] → FMNc
→ FMNb → riboflavin → ubiquinone. Two electrons are transferred from NADH to FAD in the first step of the cycle, but all subsequent steps are one-electron transfers."
Correct answer: A
This is two-electron reduction of a ketone to an alcohol, which is the reaction catalyzed by Na+-NQR.
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Question : flavin
Correct answer:
Question : ubiquinone
Correct answer: coenzymeQ - Biologically active quinone
(electron acceptor in photosynthesis and aerobic respiration) - Reduced to ubiquinol upon the acceptance of electrons. - Long alkyl chain = lipid soluble = act as an electron carrier within the phospholipid bilayer.
Question : What is the ratio of cation to enzyme in the
spectroelectrochemical experiments described in the passage?
A. 1:2
B. 2:1
C. 20:1
D. 200:1
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"...the researchers used spectroelectrochemistry to investigate the chemical changes that take place during electron transfer and how these changes are impacted by the presence of various cations. Na+-NQR was diluted to a final concentration of 0.75 mM in 0.150 M LiCl, NaCl, KCl, RbCl, or NH4Cl (each solution also contained redox active mediators) and placed in a glass instrument cell with CaF2 windows."
*ratio of various cations (0.150 M) to Na+-NQR (0.75mM)
Correct answer: D
The ratio can be found by noting that the enzyme concentration was 0.75 mM, while the concentration of cations was 0.150 M = 150 mM. The ratio is therefore
200:1.
*units!