Question

In the second step of glycolysis (the pathway that begins the oxidative breakdown of sugars), the enzyme phosphoglucose isomerase converts glucose 6-phosphate to fructose 6-phosphate. The equilibrium constant, K, for this reaction is 0.36. If ΔG° = –5.94 × log K, which conclusion can be made about this reaction?

Choose one:

A.   The ΔG° is negative, so the reaction must be coupled to the hydrolysis of ATP.

B.   The ΔG° is negative, so the reaction will always proceed in the forward direction.

C.   The ΔG° is positive, but in a cell that is burning sugars, the reaction can still proceed in the forward direction.

D.   The ΔG° is positive, so the reaction will proceed in the reverse direction.

Answer 1

O Opton c is correct.

Glucose - to phosphate & Phosphoghose 915 Fructose-6-phosphate isomerate keq = 0.36 – Agiren) ali 092 - 5.94 x dog given) This au= -594 x sop (0.36) short 2061 kilmset - 9 O Phosphoglucose isomerase do not favour any direction of reaction. O sa is positive this may favour reverse direction but it is not too positive or two large value. Thus, the reaction can still proved in forward direction Also it depends on cellulave tequirements