In glycolysis, fructose 1,6-bisphosphate is converted to two products with a standard free-energy change ($\Delta$G ^' $\degree$) of 23.8 kJ/mol. Under what conditions (encountered in erythrocytes) will the free-energy change ($\Delta$G) be negative, enabling the reaction to proceed to products? A) The free-energy change will be negative if the concentrations of the two products are high relative to that of fructose 1,6-bisphosphate. B) The reaction will not go to the right spontaneously under any conditions because the $\Delta$G ^' $\degree$ is positive. C) Under standard conditions, enough energy is released to drive the reaction to the right. D) The free-energy change will be negative when there is a high concentration of fructose 1,6-bisphosphate relative to the concentration of products.

Question

Quizplus · Accepted Answer

According to the question, the standard free-energy change is positive, meaning the reaction is not spontaneous under standard conditions. For the reaction to proceed to products, the free-energy change needs to be negative. This can be achieved if the concentration of fructose 1,6-bisphosphate is high relative to the concentration of products because the reaction is more likely to proceed in the direction of lower concentration. Therefore, choice D is the best answer.

In Glycolysis, Fructose 1,6-Bisphosphate Is Converted to Two Products with a Standard