the first reaction in glycolysis is the phosphorylation of glucose. one possible reaction might be:

pi + glucose ↔ glucose− 6−phosphate

this is a thermodynamically unfavorable process with ∆g' o = 3.3 kcal mol−1

a. in a liver cell at 370 c, the concentration of both phosphate and glucose are normally maintained at about 5 mm each. what would be the equilibrium concentration of glucose− 6−phosphate, if the 3 reaction proceeds as outlined above? does this reaction represent a reasonable metabolic step for the catabolism of glucose? explain why?

b. in principle, at least one way to increase the concentration of g6p is to drive the equilibrium to the right by increasing the intracellular concentrations of pi and/or glucose. assuming a fixed concentration of pi at 5.0 mm, how high would be the intracellular concentration of glucose have to be to give an equilibrium concentration of glucose−6−phosphate of 0.25 mm (the normal physiological concentration)? would this route be reasonable, given the maximum solubility of glucose is less than 1 m?

c. in fact, in vivo the reaction is coupled to atp hydrolysis (∆g' o = -30.5 kj mol−1 ) to give the overall reaction: atp + glucose ↔ glucose− 6−phosphate + adp write the two reactions which when coupled yield the above reaction. calculate ∆g' o and keq' for the coupled reaction.

d. if, in addition to coupling of phosphorylation reaction to atp hydrolysis, we also have in the liver cell 3 mm atp and 1 mm adp, what is the in vivo equilibrium concentration of glucose−6−phosphate?