Question

Go'=30.5 kJ/mol

a) Calculate the equilibrium constant for this reaction at 25C.

b) Because deltaGo' assumes standard pH of 7, the equilibrium constant calculated in (a) corresponds to Keq'= [oxaloacetate] [NADH]/ [L-malate] [NAD+]

The measured concentration of L-malate in rat liver mitochondria is about 0.20 mM when [NAD+]/[NADH] is 10. Calculate the concentration of oxaloaceteate at pH 7 in mitochondria.

c) To appreciate the magnitude of the mitochondrial oxaloacetate concentration, calculate the number of oxaloacetate molecules in a single rat mitochodrion. Assume the mitochondrion is a sphere of diameter 2.0 micro meters.

Answer 1

a. At standard state, dG^o = -RT ln K_eq

or, 30.5 kJ/mole = - 8.314 J.K^-1.mole^-1 * 298 K * ln K_eq

or, K_eq = 4.5 * 10^-6

b. Keq'= [oxaloacetate] [NADH]/ [L-malate] [NAD+]

or, 4.5 * 10^-6 = [oxaloacetate]/(0.20 mM)*10

or, [oxaloacetate] = 9*10^-6 mM = 9*10^-9 M

c. Radius of mitochondrion = 1 micrometer = 10^-6 m = 10^-5 dm

volume of mitochondrion = (4/3) π r³ = (4/3)π (10^-5 dm)³ = (4/3)π(10^-15) dm³ = (4/3)π(10^-15) liter

According to the definition of molarity we can say:

1 liter of mitochondrion plasma contains 9*10^-9 M oxaloacetate

So, (4/3)π(10^-15) liter of mitochondrion plasma contains [(4/3)π(10^-15) * 9*10^-9 M] oxaloacetate

or, (4/3)π(10^-15) liter of mitochondrion plasma contains [(4/3)π(10^-15) * 9*10^-9 * 6.023*10²³ ] oxaloacetate molecules or 23 oxaloacetate molecules.

Ans: 23 molecules.