Question

Consider the malate dehydrogenase reaction from the citric acid cycle. Given the following concentrations, calculate the free energy change for this reaction at 37.0 �C (310 K). ?G�\' for the reaction is +29.7 kJ/mol. Assume that the reaction occurs at pH 7.

[malate] = 1.35 mM

[oxaloacetate] = 0.150 mM

[NAD ] = 310 mM

[NADH] = 120 mM

Answer 1

Ans. Balanced Reaction:

            Malate + NAD⁺------MDH-----> OAA + NADH        , dG⁰ = +29.7 kJ/mol

Rate constant, K = ([NADH] [OAA]) / ([Malate] [NAD⁺])

                                “All concentrations in terms of molarity, 1 M = 1000 mM”

                Or, K = (0.120 x 0.00015) / (0.00135 x 0.310)

                Hence, K = 0.043

Now, using: dG = dG⁰ + RT lnK         - equation 1

            Where,

                        dG = Gibb's free energy under experimental conditions =?

                        dG⁰ = Standard Gibb's free energy = 29.7 kJ/mol

                        R = Universal gas constant = 0.008314 kJ mol^-1 K^-1

T = Temperature in kelvin = 310 K    ; [at 37⁰C]

K = Equilibrium constant = 0.043

Putting the values in equation 1-

            dG = 29.7 kJ/mol + (0.008314 kJ mol^-1 K^-1) (310 K) ln 0.043

            or, dG = 29.7 kJ/mol + 2.57734 kJ/mol x (-3.147)      ; [ ln 0.043 = -3.147]

            or, dG = 29.7 kJ/mol + (- 8.11 kJ/mol)

            Hence, dG = 21.59 kJ/mol

Therefore, free energy change for the reaction, dG = 21.59 kJ/mol