Consider the following metabolic reaction important in muscle and nerve cells:
ATP + creatine $\leftrightarrows$ phosphocreatine + ADP $\Delta$G°' = +12.6 kJ/mol
Under intracellular conditions,the $\Delta$G for the reaction,which is catalyzed by the enzyme creatine kinase,is ~0 kJ/mol.From this information we can conclude that
A)creatine kinase catalysis would not be necessary under intracellular conditions.
B)ATP has a greater phosphoryl group transfer potential compared to phosphocreatine.
C)at equilibrium,most intracellular creatine is phosphorylated.
D)the reaction operates close to equilibrium in cells.
E)None of the above is a correct conclusion.

Question

Quizplus · Accepted Answer

The fact that the standard free energy change (ΔG°') for the reaction is close to zero under intracellular conditions indicates that the reaction operates close to equilibrium in cells. This suggests that the concentrations of the reactants and products are roughly balanced, and therefore it may be important for creatine kinase to catalyze the reaction to ensure that ATP levels remain high and phosphocreatine levels remain low. However, it does not necessarily mean that most intracellular creatine is phosphorylated or that ATP has a greater phosphoryl group transfer potential than phosphocreatine.

Consider the Following Metabolic Reaction Important in Muscle and Nerve ⇆\leftrightarrows⇆

Consider the Following Metabolic Reaction Important in Muscle and Nerve $\leftrightarrows$