# Homework Help: Atp synthase and gibbs free energy

1. Dec 6, 2007

### jessawells

Hi,

I'm trying to answer a question that i'm stuck on. The question is as follows:

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If the Gibbs free energy change (deltaG) for ATP hydrolysis in a cell is -57 kJ/mol and the free energy change for transporting a proton from the cytoplasm/inter-membrane space into the mitochondrial matrix is -21.5 kJ/mol

a) What is the minimum number of c-subunits that an ATP synthase needs in order to synthesize rather than hydrolyze ATP.
b) If the ATP synthase that you described in part (a) had its OSCP subunit removed, would it still synthesize ATP in the conditions described above? Why or why not?

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For a), I put that the minimum number of c-subunits is 3 because since atp hydrolysis has a deltaG of -57 kJ/mol, then atp synthesis must have deltaG = +57 kJ/mol. We need the total free energy from transporting protons to be greater in magnitude than the deltaG of atp synthesis - eg. the overall process must be negative. so -21.5 * 3 = -64.5 kJ/mol, and the overall free energy change (from both the proton transport and atp synthesis) is -7.5 kj/mol. Does my reasoning make sense?

i'm not sure how to answer part b). As far as I know, the OSCP subunit anchors the F1 so that it does not rotate along with the gamma subunit, but I'm not sure how removing it would affect ATP synthesis. any help is much appreciated!

2. Mar 3, 2008

### Feniouk

Hi jessawells,

Your logic behind the calculations on thermodynamics is fine. But ATP synthase has 3 catalytic sites, and 3 ATP molecules are synthethised per 360 degree revolution of the rotor subunits.
So the number of c-subunits in the c-oligomer should be calculated for THREE ATP molecules, not one. This yields in your case -57x3 = -171 kJ/mol, corresponding to at least 8 c-subunits (-21.5x8 = -172 kJ/mol)

Loss of OSCP will uncouple the enzyme, so ATP synthesis will be inhibited.

Hope it helps.

I also suggest you to take a look at http://www.atpsynthase.info
There is an ATP synthase FAQ with thermodynamics explained.

Regards,

Boris.