Succinate Equilibrium Concentration In the TCA cycle, one of the steps involves

Question

Succinate Equilibrium Concentration

In the TCA cycle, one of the steps involves the oxidation of succinate to fumarate, with the concomitant reduction of the cofactor FAD to FADH2. Given that G°´ for this reaction is 0 kcal/mol, what would be the equilibrium concentration of succinate (expressed in mM to the nearest tenth of a unit) if a biochemist started with a solution that contained 0.01 M each of succinate, FAD, and FADH2, along with 20.0 nM concentration of the enzyme that catalyzes the reaction?

Explanation / Answer

Given:

Conversion of succinate to fumarate

Delta G of the reaction is 0 Kcal /mol

[FAD]=[FADH2]

[Enzyme] =20.0 nM

Solution

calculation of equilibrium constant of the conversion.

Equilibrium constant relation with delta G

Delta G0 = - RTlnK

K = exp ( - Delta G / RT )

= exp (0 )   …..since delta G is zero

k =1

Equilibrium constant for this reaction is 1

ICE

          succinate          ------- > fumarate

I               0.01                                        0.01

C             -X                                            +X

E              (0.01-X)                                (0.01+X)

K = (0.01 +x) / ( 0.01-x)

0.01-x = 0.01+x

x = 0

Equilbrium concentrations of both will be = 0.01 M

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