Consider the net reaction leading to formation of fructose 6-phosphate from glucose and ATP.

ΔG° for the reaction equals:

A. + 3.96 kcal/mol, and Keq for the reaction is greater than 1.

B. - 3.60 kcal/mol, and Keq for the reaction is greater than 1.

C. - 3.60 kcal/mol, and Keq for the reaction is less than 1.

D. + 3.96 kcal/mol, and Keq for the reaction is less than 1.

The answer is B

Explanation:

This question can be better understood with the knowledge of the glycolytic pathway as fructose 6-phosphate is not directly formed from glucose and ATP.

During glycolysis (breakdown of glucose), glucose is first of all phosphorylated (in the presence of ATP) to form glucose 6-phosphate (during the process of phosphorylation, glucose combines with one phosphate group from ATP to form glucose 6-phosphate, ADP is then released subsequently). Glucose 6-phosphate then undergoes isomerization to form fructose 6-phosphate.

The sum of ΔG° for the formation of glucose 6-phosphate and ADP from glucose and ATP is - 4000 cal/mol while the sum for the ΔG° for the formation of fructose 6-phosphate from glucose 6-phosphate is + 400 cal/mol. The ΔG° for the formation of fructose 6-phosphate from glucose and ATP is:

-4000 + 400 = - 3,600 cal/mol ⇔ - 3.6 kcal/mol

When ΔG° is negative for any reaction, it means the reaction proceeds spontaneously under standard conditions and that at equilibrium, more products are formed than the reactants. The Keq in such cases are greater than 1 as shown in the relationship between ΔG° and Keq below:

ΔG = - InKeq

NOTE:

ΔG° = Gibb's free energy

Keq = Equilibrium constant