Question

Many metabolic reactions are coupled reaction. Such as the first step in glycolysis drawn below: Glucose + Pi ↔ glucose-6-phosphate + H2O

a. The ΔG°’ is 14 kJ•mol-1. What is the ratio of [G6P]/[glucose][Pi] at equilibrium at 25°C?

b. The above reaction is coupled to the hydrolysis of ATP. In muscle cells at 37°C, the steady-state ratio of [ATP]/[ADP] is 12 & the ΔG°’ for ATP hydrolysis is -30.5 kJ•mol-1. Assuming that glucose and G6P achieve equilibrium values in muscle cells, what is the ratio of [G6P] to [glucose]?

Answer 1

a) dGo = -RT ln K     where K = [G6P] /[Glucos] [Pi]

14 x 1000 = - 8.314 x 298 ln [ G6p]/[Glucose] [Pi]

[G6P] /[Glucose] [Pi] = 2.235 x 10^ -6

b)   ATP + H2O ---> ADP + Pi    is added to GLucose + Pi ---> G6P + H2O

we get   ATP + GLucose <----> ADP + G6P    , dH = -30.5+14 = - 16.5 KJ /mol

dG = -RT ln K    where K = [G6P] [ADP]/ [ATP] [glucise]

- 16.5 x1000 = -8.314 x 310 ln ( [G6P]/[Glucosse] 12)

[G6P]/[Glucose] = 7236