1) glycogen phosphorylase catalyzes the removal of glucose from glycogen. the ∆g°’ for this reaction is 3.1 kj/mole 1. calculate the ratio of [pi]/[glucose-1-phosphate] when the reaction is at equilibrium. assume a temperature of 37 °c. (hint: the removal of glucose units from glycogen does not change the glycogen concentration. and yes, i do mean pi/glucose-1-phosphate, not the other way around2) the measured ratio of [pi]/[glucose-1-phosphate] in myocytes (muscle cells) under physiological conditions is more than 100: 1. this system is not at equilibrium under physiological conditions. 2. why would a reaction in a cell (and especially a reaction that is part of a larger pathway) not exist at equilibrium? 3)what does the ratio of [pi]/[glucose-1-phosphate] tell you about the relative rates of glycogen phosphorylase and phosphoglucomutase in myocytes?

1. G° = -RT ln (G1P/P)

3.1 = 8.314 × 310 × ln (G1P/P)

3.1 / 2577.34 = ln (G1P/P)

0.0012 = ln (G1P/P)

0.0012 = (log G1P/P)/log 2.71828

0.4342 × 0.0012 = log G1P/P

0.00052 = log G1P/P

G1P/P = 10^0.00052 = 1.0012

P/G1P = 1/1.0012 = 0.9988

2. The cleavage of glycogen phosphorolytically is beneficial for the cell to conduct the process as the discharged glucose is phosphorylated. A general hydrolytic cleavage would give rise to only a glucose, which has to be phosphorylated again with the help of ATP.  

Another merit of phosphorylated glucose is that it comprises the negative charge and cannot diffuse out of the muscle cell. Thus, the reaction will not be at equilibrium under the physiological conditions and always encourages the generation of the products. The formation of products will amend the change in free energy in such a manner that the reaction will always carry in the forward direction.  

3. Greater the ratio of [Pi]/[glucose-1-phosphate], higher will be the relative rate of glycogen phosphorylase in comparison to the phosphoglucomutase as the transformation of Glu-1-P becomes slow because of lesser accessibility of substrate.  

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