Question

After the glucose-phosphate species forms in the metabolic process, the isomerization of glucosephosphate to fructose-phosphate occurs. If the equilibrium constant for the isomerization reaction at 37°C (body temperature) is 0.65, what is the value of ΔG when the ratio of glucose-phosphate to fructosephosphate is 10 to 1?

glucose-phosphate → fructose-phosphate

Keq = 0.65

Answer 1

glucose-phosphate → fructose-phosphate

the ratio of glucose-phosphate to fructosephosphate is 10 to 1.

Keq = 0.65 at 37°C or 310K

ln K eq = - dG⁰ / RT

First calculate the dG⁰ as follows:

ln 0.65 = - dG⁰ / 0.008314 kJ mol^-1 K^-1* 310 K

-0.430= - dG⁰ / 2.577 kJ mol^-1

dG^{0 =}    1.11   kJ mol^-1

To calculate the dG at 310 K use the following expression

dG = dG^o + R T ln Q.

dG = 1.11 kJ/mol + 0.008314 kJ mol^-1 K^-1* 310 K ln 1/ 10

dG =1.11 kJ/mol + 0.008314 kJ mol^-1 K^-1* 310 (-2.30)

dG = -4.82 kJ/mol