Question

NEED HELP WITH PART C ONLY (NO SCAM PLEASE, I JUST NEED HELP WITH PART C) THANK YOU

The first reaction in glycolysis is the phosphorylation of glucose:

Pi+glucose⟶glucose−6−phosphate+H2O

This is a thermodynamically unfavorable process, with ΔG∘′=+13.8kJ/mol.

Part A

In a liver cell at 37 ∘C the concentrations of both phosphate and glucose are normally maintained at about 5 mM each. =>the equilibrium concentration of glucose-6-phosphate = 1.2*10^-7 M

Part 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+

ΔG∘′ for the coupled reaction is -18.4 kJ/mol

NEED HELP WITH PART C ONLY (NO SCAM PLEASE, I JUST NEED HELP WITH PART C) THANK YOU

PART 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 glucose-6-phosphate at equilibrium at pH=7.4 and 37 ∘C?