Question

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 ΔG°´ 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 0.011 M each of R-CH2-CH2-R', FAD, and FADH2 (but no R-CH=CH-R'), along with 20 nM concentration of the enzyme that catalyzes the reaction?

Answer 1

ΔG°´ for this reaction is 0 kcal/mol (given)

Here initial concentration of R-CH2-CH2-R', FAD, and FADH2 is equal to 0.011M
Concentration of enzyme = 20 nM

ΔG° = -RTlnK
0 = -RTlnK
0=lnK
ln1=lnK
Therefore,K=1
K is the equilibrium constant

                     R-CH₂-CH₂-R'    +    FAD    <---->   R-CH=CH-R'    +    FADH₂

initial concentration       0.011                 0.011                   0                        0.011
at equilibrium                      0.011-x              0.011-x                 x                        0.011+x

K= [R-CH=CH-R' ] [FADH2] / [R-CH2-CH2-R'] [FAD ]
1 =   (0.011+x) x/ (0.011-x) (0.011-x)
(0.011-x)² = 0.011x+x²
0.000121 + x² - 0.022x = 0.011x+x²
0.000121 = 0.033x
x = 0.00366 M

[ FADH₂]_{eq =} (0.011 + x) = 0.01466 M = 14.6* 10^-3 mM