The conversion of 1 mol of fructose 1, 6-bisphosphate to 2 mol of pyruvate by th

Question

The conversion of 1 mol of fructose 1, 6-bisphosphate to 2 mol of pyruvate by the glycolytic pathway results in a net formation of a. 1 mol of NAD+ and 2 mol of ATP. b. 1 mol of NADH and 1 mol of ATP. c. 2 mol of NADH and 2 mol of ATP. d. 2 mol of NAD+ and 4 mol of ATP. e. 2 mol of NADH and 4 mol of ATP. When a mixture of glucose 6-phosphate and fructose 6-phosphate is incubated with the enzyme isomerase, the final mixture contains twice as much glucose 6- phosphate as fructose 6-phosphate. Which one of the following statements is most nearly correct, when applied to the reaction below (R = 8.315 J/mol-K and T = 298 K)? Glucose 6-phosphate doubleheadarrow fructose 6-phosphate a. delta G^is -1.7 kJ/mol. b. delta G^is incalculably large and positive. c. delta G^is incalculably large and negative. d. delta G^is + 1.7 kJ/mol. e. delta G^is zero.

Explanation / Answer

Ans. #1. F-1,6-bisP + 2NAD+ -------glycolysis-----> 2 Pyruvate + 2 NADH + 4 ATP

[see glycolysis figure]: Formation of 1 mol pyruvate from F-1,6-bisP produces 1 mol NADH and 2 mol ATP.

Correct option. E.

#2. Given, in the final reaction mixture, [G-6-P] is twice of [F-6-P].

            Or, [G-6-P] = 2 x [F-6-P].

So, if [F-6-P] = X,      then [G-6-P] = 2X

#3. G-6-P --------> F-6-P  

Equilibrium constant for forward reaction, K = [F-6-P] /[G-6-P] = X/ 2X = 0.5

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

            Where,

                        dG0 = Standard Gibb's free energy

                        R = Universal gas constant = 0.008315 kJ mol-1 K-1

T = Temperature in kelvin

K = Equilibrium constant

Putting the values in equation 1-

            dG0 = - (0.008315 kJ mol-1 K-1) x 298 K x ln (0.5)

            or, dG0 = - 2.47787 kJ mol-1 x (- 0.693)

            or, dG0 = + 1.7 kJ mol-1

Thus, for the given reaction, dG0 = +1.7 kJ mol-1

Thus, correct option = D.

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