The isomerization of glucosc-6-phosphatc to fruclose-6-phosphatc is a key step i

Question

The isomerization of glucosc-6-phosphatc to fruclose-6-phosphatc is a key step in both glycolysis and gluconcogenesis. For this problem, assume a temperature of 37 DegreeC. A)Calculate G for conversion of glucose-6-phosphate to fructose-6-phosphate, if [G6P] = 0.083 mM and [F6P] = 0.014 mM and G^0 = +1.7 kj/mol. B)Do these cellular concentrations favor glycolysis or gluconeogcnesis ? C)The cell is a noncquilibrium environment. What would be the ratio of glucose-6-phosphate to fructosc-6-phosphate be if the reaction were allowed to reach equilibrium?

Explanation / Answer

A)

the reaction is

G6P ---> F6P

the reaction quotient is given by

Q = [F6P] / [G6P]

now

we know that

dG = dGo + RT lnQ

so

dG = dGo + RT ln [F6P] / [G6P]

using given values

we get

dG = ( 1.7 x 1000) + ( 8.314 x 310 x ln [ 0.014 / 0.083]

dG = -2.887 x 1000 J / mol

dG = -2.887 kJ/ mol

B)

we know that

for a reaction to be spontaneous

dG < 0

for this reaction

we obtained

dG = -2.887 kJ / mol < 0

so

these cellular concentrations favor glycolysis

C)

now

at equilirbium

dG = O

so

dGo = -RT lnKeq

1.7 x 1000 = -8.314 x 310 x ln Keq

Keq = 0.517

now

G6P ---> F6P

so the equilibrium constant is given by

Keq = [F6P] / [G6P]

so

[F6P] / [G6P] = 0.517

[G6P] /[F6P ] = 1.934

so

the ratio of G6P to F6P is 1.934

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