One of the important reactions in the biochemical pathway of glycolysis is the r

Question

One of the important reactions in the biochemical pathway of glycolysis is the reaction of glucose-6-phosphate (G6P) to for fructose-6-phosphate (F6P).

            G6P <----> F6P                        DG0= 1.7 kJ/mol

(a) Is the reaction spontaneous or non-spontaneous under standard thermodynamic conditions?

(b) Standard thermodynamic conditions imply the concentration of G6P and F6P to be 1 M, however, in a typical cell, they are not even close to these values. Calculate DG when the concentrations of G6P and F6P are 120 mM and 28 mM respectively, and determine the spontaneity of the forward reaction under these conditions. Assume the body temperature is 370C.

I know the answers are (a) nonspontaneous, (b) -2.1 kJ/mol, spontaneous

Explanation / Answer

a)

if dG0 is positive, this is NOT spontaneous! Since this favour products/reverse reaction

b)

In a cell, this actually occurs:

[F6P] = 28 mM = 28*10^-3 M

[G6P] = 120 mM = 120*10^-3 M

T = 37°C = 37+273 = 310 K

use equation:

dG = dG° + RT*ln(Q)

dG° = 1700 J/mol, R = 8.314 J/molK, T = 310 K, ln(Q) = ln(F6P / G6P)

dG = 1700  + 8.31*310*ln((28*10^-3)/(120*10^-3 ))

dG = -2048.96 J/mol = -2.048 kJ/mol --> -2.1 kJ/mol

this is negative, so this is spontaneous!

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