Glycogen phosphorylase catalyzes the removal of glucose 1 -phosphate from glycog

Question

Glycogen phosphorylase catalyzes the removal of glucose 1 -phosphate from glycogen: Glycogen_n-1 + Pi Glycogen_n-1 + glucose 1-phosphate Delta G degree' = +3.1 kJ/mol a) Comment upon the standard free energy for this reaction. b) Using the value of delta G degree', calculate the value of the equilibrium constant for this reaction c) Using the value of K, calculate the equilibrium ratio of [Pi] to [glucose 1 -phosphate]. d) The observed physiological ratio of [Pi] to [glucose 1-phosphate] is 100:1. Under these conditions, calculate AG under physiological conditions. e) Comment on the difference between the equilibrium and physiological ratios, assuming that all of the glucose 1 -phosphate formed eventually enters the glycolysis pathway.

Explanation / Answer

For the given reaction,

a) The standard free energy change is +ve. that means the reaction is non-spontaneous.

b) dGo = -RTlnK

With T = 298 K we get,

3100 = -8.314 x 298 lnK

K = 0.286

c) K = 3.494 = [glycogen][glucose1-phosphate]/[glycogen][Pi]

[Pi]/[glucose-1-phosphate] = 1/0.286 = 3.494

d) [Pi]/[glucose-1-phosphate] = 100/1

dGo = -8.314 x 298 ln(1/100) = 11.41 kJ/mol

e) The oberved standard free energy change at physiological ratio is much higher than for the reaction shown above.

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