Given this scenario, what would be the equilibrium concentration of FADH2 (expre

Question

Given this scenario, what would be the equilibrium concentration of FADH2 (expressed in mM to the nearest tenth of a unit) if a biochemist started with a solution that contained 0.008 M each of R-CH2-CH2-R', FAD, and FADH2 (but no R-CH=CH-R'), along with 23 nM concentration of the enzyme that catalyzes the reaction? Full question details are provided below:

Edit: The equilibrium constant = [products]/[reactants]

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Question 9 1 pts Imagine that in a metabolic pathway, one of the steps involves the oxidation of saturated compound to an unsaturated compound, with the concomitant reduction of the cofactor FAD to FADH2 as shown below: R-CH2-CH2-R' FAD R-CH CH-R' FADH2 Imagine further that AGO for this reaction is 0 kcal/mol. Given this scenario, what would be the equilibrium concentration of FADH2 (expressed in mM to the nearest tenth of a unit) if a biochemist started with a solution that contained 0.008 Meach of R-CH2-CH2-R, FAD, and FADH2 (but no R-CH CH-R), along with 23 nM concentration of the enzyme that catalyzes the reaction?

Explanation / Answer

gibbs free energy=-2.303 x nRT log Keq.

0= -2.303 x 2x 8.314 x 298 x log [products]/[reactants].

0=-2.303 x 2 x 8.314 x 298 x log[R-CH=CH-R'][FADH2]/[RCH2CH2R'][FAD]

0=-2.303 x 2 x 8.314 x 298 x log (1-0.008)(FADH2)/(0.008)(0.008)

0=2.303 x 2 x 8.314 x 298x log(0.992)(y)/0.000064

0=-2.303 x 2 x 8.314 x298 x 10-5 x log(15500)(y)

0=-2.303 x 2 x 8 x 298 x log(15500)(y).

log(15500)(y)=2 x 2.3 x 8 x 298x 10-5.

log(15500)(y)=(0.114)

(15500)(y)=antilog(0.114)

y=1.30/15500

y=1.30/15500=0.000083=8.3 x 10-5M.

where y is the equilibrium concentration of FADH2.

y=8.3 x 10-5 x 103 mM=0.083mM---answer--equilibrium concentration of FADH2

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