Question

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Using the free energies of hydrolysis given for reactions A and B, calculate a combined favorable free energy for the 2 equations given and the overall equations for the combined reaction. Write the new reaction. (5 pt)

Donuts-P + H2O donuts + P -15 kJ/mole

ATP + H2O ADP + P -30.5 kJ/mole

Answer 1

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

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

            Donut-P + H₂O ----> Donut + Pi ; dG⁰ = -15.0 kJ/ mol            - Reaction 2

#. Phosphorylation of donut (thermodynamically favorable reaction)

            Donut + Pi ------> Donut-P + H₂O                     - 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,

            Donut + Pi ------> Donut-P ; dG⁰ = +15.0 kJ/mol                 - Reverse of Rxn 2

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

            Donut + Pi ------> Donut-P + H₂O                                   ; dG⁰ = +15.0 kJ/mol           

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

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Net coupled Rxn: Donut + ATP ----> Donut-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 = 15.0 kJ/mol + (-30.5 kJ/mol) = -15.5 kJ/mol

Hence, coupling phosphorylation of Donut with ATP hydrolysis gives dG⁰ = -15.5 kJ/mol