Consider the following metabolic reaction:Succinyl-CoA + Acetoacetate - --> Acetoacetyl-CoA + Succinate (delta) G = - 1.25 kJ/molThe (delta) G for the hydrolysis of Succinyl-CoA is - 33.9 kJ/mol. What is the (delta) G for the hydrolysis of Acetoacetyl-CoA: Acetoacetyl-CoA - --> Acetoacetate + CoAa. - 35.2 kJ/molb. - 32.7 kJ/molc. + 32.7 kJ/mold. none of the above

Question

Biology

Cher

22 January, 06:58

Consider the following metabolic reaction:Succinyl-CoA + Acetoacetate - --> Acetoacetyl-CoA + Succinate (delta) G = - 1.25 kJ/molThe (delta) G for the hydrolysis of Succinyl-CoA is - 33.9 kJ/mol. What is the (delta) G for the hydrolysis of Acetoacetyl-CoA: Acetoacetyl-CoA - --> Acetoacetate + CoAa. - 35.2 kJ/molb. - 32.7 kJ/molc. + 32.7 kJ/mold. none of the above

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Sienna Strong · Answer 1

b. - 32.7 kJ/mol

Explanation:

For this metabolic reaction we have two coupled reactions taking place, which are:

(1) Acetoacetyl-CoA - --> Acetoacetate + CoA

(2) Succinyl-CoA + Acetoacetate - --> Acetoacetyl-CoA + Succinate

So the overall variaton of free energy (ΔG°) for the metabolic reaction will be the sum of the individual free energy variations of each of the coupled reactions:

ΔG°₁ + ΔG°₂ = ΔG°total.

In this case, the ΔG°total is the result of the Succinul-CoA hydrolysis, which is - 33.9 kJ/mol. And we already have the variation of free energy for reaction (2), which is ΔG°₂ = - 1.25 kJ/mol. So with this we can obtain the value for ΔG°₁, which corresponds to the variation of free energy for the hydrolysis of Acetoacetyl-CoA.

ΔG°₁ + (-1.25 kJ/mol) = ΔG°total.

ΔG°₁ = ΔG°total - ΔG°₂

ΔG°₁ = - 33.9 kJ/mol - (-1.25 kJ/mol)

ΔG°₁ = - 33.9 kJ/mol - (-1.25 kJ/mol)

ΔG°₁ = - 33.9 kJ/mol + 1.25 kJ/mol)

ΔG°₁ = - 32.65 kJ/mol → rounding decimals → - 32.7 kJ/mol