Question

The hydrolysis of glucose-6- phosphate to give glucose plus phosphate has a ∆G o = - 3.3 kcal/mol. In the liver glucose concentrations are extremely high during the fed state because of the proximity of the liver to the portal vein, and thus hydrolysis should be unfavorable. What is ∆G for this hydrolysis reaction at 0.5 M glucose, 0.01 M phosphate and 10 -6 M for glucose-6- phosphate? (Assume 2.303RT = 1360 cal/mol.)

A. -3460 cal/mol

B. -1730 cal/mol

C. 0 cal/mol


D. +1730 cal/mol

E. +3460 cal/mol

Answer 1

glucose-6- phosphate+ H2O(l) -------> glucose + phosphate

    kc   = [glucose][phosphate]/[glucose-6-phosphate]

          = 0.5*0.01/10^-6   = 5000

G     =G⁰ + RTlnKc

G⁰    = -3.3Kcal/mole = -3300cal/mole

G     =G⁰ + RTlnKc   

           = G⁰ + 2.303RTlogKc  

           = G⁰ + 1360logKc       (Assume 2.303RT = 1360 cal/mol.)

G      = -3300+1360log5000

              = -3300+ 1360*3.6989

              = 1730.5cal

D. +1730 cal/mol >>>>>answer