Although ATP is an important phosphate donor in biosynthetic reactions, the AMP

Question

Although ATP is an important phosphate donor in biosynthetic reactions, the AMP portion of the molecule is often produced in the cell along with the release of pyrophosphate. One important example is: R-COO^- + ATP + CoASH rightarrow R-CO-SCoA + AMP + Pyrophosphate Inorganic pyrophosphatase can hydrolyze pyrophosphate, yielding two molecules of inorganic phosphate (PO_4^3-): Pyrophosphate + H_2 O rightarrow 2 Phosphate Assume that the hydrolysis of R-CO-SCoA to R-COO^- + CoASH proceeds with a AG" of -30.8 kJ mol and that the hydrolysis of pyrophosphate yields a Delta G" of -33.0 kJ/mol. Calculate the K_eg at 25 degree C for the reaction (R 1) in the presence and absence of inorganic pyrophosphatase

Explanation / Answer

The hydrolysis reaction is

R-CO-SCoA + H2O --------> R-COO- + CoASH ……(1), G0’ = -30.8 kJ/mol

Pyrophosphate + H2O --------> 2 Phosphate …….(2), G0’ = -33.0 kJ/mol

ATP + H2O -------> AMP + Pyrophosphate …….(3), G0’ = -30.5 kJ/mol

The given reaction is

R-COO- + ATP + CoASH -------> R-CO-SCoA + AMP + Pyrophosphate ……(R1)

We first calculate in presence of inorganic pyrophosphatase enzyme.

Reaction (R1) is a combination of the reverse of reaction (1) and reaction (3). Since, we take the reverse of reaction (1), the free energy change for the reverse reaction is G0’ = -(-30.8 kJ/mol) = 30.8 kJ/mol

The overall free energy change for (R1) in presence of inorganic pyrophosphatase enzyme is

G0’rxn = (30.8 kJ/mol) + (-30.5 kJ/mol) = 0.3 kJ/mol (ans) (we simply add the free energy change for both the reactions).

Now, the equilibrium constant for the reaction is Keq’ (in presence of inorganic pyrophosphatase) at 25°C (= 273 + 25 = 298 K)

G0’rxn = -RTln(Keq’)

===> Keq’ = exp(-G0’/RT) = exp[-{(0.3 kJ/mol)/[(8.315 J/mol.K)(298 K)}] = exp(-[1.211*10-4]) = 0.9998 1.0 (ans).

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