# Te glucose / glucose-6-phosphate substrate cycle involves distinct reactions of glycolysis and gluconeogenesis that interconvert these two metabolites. Assume that under physiological conditions, [ATP] = [ADP] and [Pi] = 1 mM.Consider the following glycolytic reaction catalyzed by hexokinase:ATP + glucose ADP + glucose-6-phosphate Go = - 16.7 kJ/mol(a) Calculate the equilibrium constant (K) for this reaction at 298

Te glucose / glucose-6-phosphate substrate cycle involves distinct reactions of glycolysis and gluconeogenesis that interconvert these two metabolites. Assume that under physiological conditions, [ATP] = [ADP] and [Pi] = 1 mM.
Consider the following glycolytic reaction catalyzed by hexokinase:
ATP + glucose ⇆ ADP + glucose-6-phosphate ∆ Go′ = - 16.7 kJ/mol
(a) Calculate the equilibrium constant (K) for this reaction at 298 K, and from that, calculate the maximum [glucose-6-phosphate] / [glucose] ratio that would exist under conditions where the reaction is still thermodynamically favorable.
(b) Te reverse of this inter conversion in gluconeogenesis is catalyzed by glucose-6-phosphatase:
Glucose-6-phosphate + H2O ⇆ glucose + Pi ∆Go′ = - 13.8 kJ/mol
K = 262 for this reaction. Calculate the maximum ratio of [glucose]/ [glucose-6-phosphate] that would exist under conditions where the reaction is still thermodynamically favorable.
(c) Under what cellular conditions would both directions in the substrate cycle be strongly favored?
(d) What ultimately controls the direction of net conversion of a substrate cycle such as this in the cell?

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