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

Question

Glycogen phosphorylase catalyzes the removal of glucose 1-phosphate from glycogen:

Glycogenn + Pi Glycogenn-1 + glucose 1-phosphate G°’ = +3.1 kJ/mol

a) Comment upon the standard free energy for this reaction.

b) Using the value of G°’, calculate the value of the equilibrium constant for this reaction (hint: removal of a glucose unit from glycogen does not change the concentration of glycogen).

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 G 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

a) Since the gibbs free energy change is +ve, the reaction is not spontaneous

b) delG0= -RTlnK ( for biochemical reaction T= 37 deg.c ( for biochemical reactions)

3.1*1000= -8.314*(37+273.15) lnK

lnK= -1.202

K=0.30053

Keq can be simplified to = glucose 1- phosphate/ PI =0.30053

PI/glocuse1 phosphate= 1/0.30053=3.32

When the observed ratios   of PI/ glycogen1 phosphate= 100:1

K= 1/100 =0.01

delG0= -8.314*310.15 ln (0.01)= 11874.8 J/mole.= 11.874 Kj/mole

T he high [PI ]/[glucose 1-phosphate] ratio in myocytes ( 100:1) means that [glucose 1-phosphate] is far below the equilibrium value.

This suggests that the rate at which phosphoglu-comutase removes glucose 1-phosphate (by conversion to glucose 6-phosphateis greater than the rate at which glycogen phosphorylase can produce it. Hence glycogen phosphorylase reaction is the rate-limiting step in glycogen breakdown

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