Question

The energy released is -151 kJ/mol. This energy can be converted into the synthesis of ATP at an efficiency of 0.7 (70% of the energy released is captured in ATP synthesis) and the oxidation of 1 succinate leads to the phosphorylation of 2 equivalents of ATP. Under these conditions, what is the maximum ratio of [ATP]/[ADP] attainable by oxidative phosphorylation when [Pi] = 1 mM? (Assume the free energy needed for ATP synthesis is +30.5 kJ/mol)

Answer 1

Question:

Answer:

Effiecny of energy conversion = 0.7 = 70%

Energy released = -151 KJ/mol

Energy available of ATP synthesis = 151 * 70/100 = 105.7 KJ/mol

Gibbs free energy for ATP synthesis = -RT ln (ATP/(ADP. Pi))

R = 8.314 J/mol = 0.008314 KJ/mol

T = 298 K (Standard temp)

Pi = 1mM = 0.001 M

3 molecules of ATP produced by 30.5*3=91.5 KJ/mol

So, only 3 molecules of ATP will be produced.

Gibbs free Energy available = -RT ln (ATP/ADP) * (1/Pi)

-105.7 = 0.008314 * 298 ln (3/ADP)*(1000/1)

- 42.66 = ln (3000/ADP)

-42.66 = ln 3000 – ln ADP

-42.66 = 8.006 – ln ADP

5.33 = ln ADP

ADP = 206.438 molecules

As only 3 molecules can be produced of ATP in the available energy.

So, Ratio of ATP/ADP = 3/206.438 = 0.015