Question

A.) Let’s assume that the hexokinase reaction proceeds in the following manner as opposed to what we see in A) and B). Given this what is the new Keq?

fructose-6-phosphate + ATP ↔ F16BP + ADP deltaG° = -14.2 kJ/mol

B.) If we assume that the [ATP] is 10x greater than the [ADP] what is the new molar ratio of [F16BP]:[fructose-6-phosphate] at equilibrium?

C.) If the cellular concentrations of fructose-6-phosphate and F16BP are 0.028mM and 0.014 mM respectively what does this and the answer in B.) tells us about the equilibrium status of the reaction PFK-1 catalyzes under cellular conditions? Given this would you expect PFK-1, and the reaction it catalyzes, to have a small or large flux control coefficient?

Answer 1

Answer-

A. The new equilibrium constant will be 1

B.

If ATP is 10x , then then, the new molar ration of [F16BP]: [fructose-6-phosphate] at equilibrium will be 10 x -14.2 = 142

C. The concentration of F5P and F16BP indicates that fructose-6-phosphate is twice more concentated in cellular level.

Flux control coefficient is the degree of control that each enzyme exerts on a metabolic pathway flux.

Phosphofructokinase-1 (PFK-1) is one of the most important regulatory enzymes of glycolysis. PFK-1 catalyzes the important "committed" step of glycolysis, the conversion of fructose 6-phosphate and ATP to fructose. It is an allosteric enzyme made of 4 subunits and controlled by many activators and inhibitors.

The cellular level of PFK1 are correlated with glycolytic flux. The first reaction that commits glucose to the glycolytic pathway is catalyzed by the enzymephosphofructokinase-1 (PFK-1) and is tightly regulated. One of the most potent activators of PFK-1 is fructose 2,6 bisphosphate (F2,6BP).

In cellular level, during glycolysis, the reactions catalyzed by hexokinase, phosphofructokinase, and pyruvate kinase are virtually irreversible; hence, these enzymes would be expected to have regulatory as well as catalytic role.