Question

if ATP + H20 --> ADP +Pi and is -30.5 kJ/mole

and removing a phosphate from X to give X +Phosphate (with water) is -25.5 kJ/mole. what is overall change in free energy to add a P to X using ATP

+56
	+4.5
	-56
	-4.5

Answer 1

Ans. Ans. Given, the standard free energy change of phosphate hydrolysis for-

            ATP + H₂O ------> ADP + Pi           ; dG⁰ = -30.5 kJ/ mol            - Reaction 1

            X-P + H₂O ---------> X + Pi            ; dG⁰ = -25.5 kJ/ mol            - Reaction 2

#. Phosphorylation of X

            X + Pi ------> X-P               - Reaction 3

Phosphorylation (addition of a phosphate group) is the reverse of hydrolysis of a phosphate group.

That is, reaction 3 is the reverse of reaction 2.

When a reaction is reversed, the sign of standard free energy change is also reversed.

So,

            X + Pi ------> X-P               ; dG⁰ = +25.5 kJ/mol                     - Reverse of Rxn 2

#. The phosphorylation of X (Rxn 3) can be coupled to ATP hydrolysis as follow-

            X + Pi -----------> X-P                                 ; dG⁰ = +25.5 kJ/mol           

            ATP + H₂O ------> ADP + Pi                       ; dG⁰ = -30.5 kJ/ mol

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Net coupled Rxn: X + ATP + H₂O ----> X-P + ADP   ; dG⁰_net = ?

The dG⁰ for the net coupled reaction is given by-

            dG⁰_net = dG⁰ of Rxn 3 + dG⁰ of Rxn 1

            Or, dG⁰_net = 25.5 kJ/mol + (-30.5 kJ/mol) = -5.0 kJ/mol

Hence, coupling phosphorylation of X with ATP hydrolysis gives dG⁰ = -5.0 kJ/mol

# None of the options provided are equal to the calculated value of -5.0 kJ/mol.

Note: Though option D. can be treated as correct as being closest to the calculated, it’s advised to recheck the typo errors, if any.