Question

Synthesis of the activated form of acetate (acetyl‑CoA) is carried out in an ATP‑dependent process. Acetate+CoA+ATP⟶Acetyl−CoA+AMP+PPi Acetate+CoA+ATP⟶Acetyl−CoA+AMP+PPi The Δ????′∘ΔG′∘ for the hydrolysis of acetyl‑CoA to acetate and CoA is −32.2 kJ/mol−32.2 kJ/mol , and that for hydrolysis of ATP to AMP and PPiPPi is −30.5 kJ/mol−30.5 kJ/mol . Calculate ?G’o for the ATP –dependent synthesis of acetyl-CoA.

Answer 1

Answer:

$\Delta G^{o}$ of the reaction is 1.7 kJ/mol.

Hydrolysis of the pyrophosphate shifts the equilibrium of the reaction to the right, making the formation of acetyl-CoA energetically more favorable.

Explanation:

The synthesis of acetyl CoA is as follows.

Form the given data,

$Acetyl\rightarrow CoA \rightarrow Acetate+ CoA; \Delta G_{1}^{o}=-32.2\,kJ.mol^{-1}$

$ATP\rightarrow AMP+PP_{i};\Delta G_{2}^{o}=-30.5kJ.mol^{-1}$

Total the above two steps we get,

$Acetate+CoA+ATP\rightarrow Acetyl-CoA+AMP+PP_{i}$

$\Delta G^{o}=32.2+(-30.5)kJ.mol^{-1}=1.7kJ.mol^{-1}$

Therefore, $\Delta G^{o}$ of the reaction is 1.7 kJ/mol.

According to the Lechatelier's principle, a system at equilibrium always shows a tendency to regain the state of equilibrium., if equilibrium disturbed by an external factor only.

The enzyme inorganic pyrophosphotase catalyze the reaction.

The enzyme removes the ppi , one of the product in acetyl -CoA  reaction.

This would disturb the equilibrium therefore, system try to regain the state of equilibrium. And the equilibrium shifted to the right side.

Therefore, Hydrolysis of the pyrophosphate shifts the equilibrium of the reaction to the right, making the formation of acetyl-CoA energetically more favorable.