The first reaction in glycolysis is the phosphorylation of glucose: Pi + glucose → glucose-6-phosphate + H2O This is a thermodynamically unfavorable process, with ∆G°′ = +13.8 kJ/mol. (a) In a liver cell at 37 °C the concentrations of both phosphate and glucose are normally maintained at about 5 mM each. What would be the equilibrium concentration of glucose-6-phosphate, according to the above data? (b) This very low concentration of the desired product would be unfavorable for glycolysis. In fact, the reaction is coupled to ATP hydrolysis to give the overall reaction ATP + glucose → glucose-6-phosphate + ADP + H+ What is ∆G°′ for the coupled reaction? (c) If, in addition to the constraints on glucose concentration listed previously, we have in the liver cell ATP concentration = 3 mM and ADP concentration = 1 mM, what is the theoretical concentration of glucose6-phosphate at equilibrium at pH = 7.4 and 37 °C? The answer you will obtain is an absurdly high value for the cell and is never approached in reality. Explain why.
The first reaction in glycolysis is the phosphorylation of glucose: Pi + glucose → glucose-6-phosphate + H2O This is a
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