Question

In glycolysis, fructose-1,6-bisphosphate is converted to two products with a standard free-energy change (ΔG°') of 23.8 kJ/mol. Under what conditions (encountered in a normal cell) will the free-energy change (ΔG) be negative, enabling the reaction to proceed to the right?

A) Under standard conditions, enough energy is released to drive the reaction to the right.
B) If the concentrations of the two products are high relative to that of fructose-1,6-bisphosphate.
C) The reaction will not go to the right spontaneously under any conditions because ΔG°' is positive.
D) When there is a high concentration of fructose-1,6-bisphosphate relative to the concentration of products.
E) When there is a high concentration of products relative to the concentration of fructose-1,6-bisphosphate.

Answer 1

Option D

Step-by-step explanation:

This reaction can be promoted by high amounts of substrate, which also follows the LeChatlier principle.