Question

Because electron transport is coupled to ADP phosphorylation, it is theoretically possible to drive electron backwards in the chain, against the potential gradient, at the expense of ATP. This mechanism has been observed in some organisms. Consider a bacterium at 25 C that can transfer electrons from succinate to NAD+ by means of the transport

Succinate + NAD+ = Fumerate + NADH + H+

If the intracellular concentration of phosphate is constant at 10 mM, calculate the [ATP]/[ADP] ratio that would be required to maintain the steady-state ratio of [NADH]/[succinate] = 0.010. Assume that the intracellular concentrations of fumerate and NAD+ are equal.

Answer 1

Succinate + NAD+ = Fumerate + NADH + H+

ATP hydrolysis takes place during this reaction,

ATP = ADP + Pi

Given: Steady state reaction, phosphate is constant at 0.01 M, [NADH]/[succinate] = 0.01 M

Rate = [ADP] [Pi] = [FUMARATE] [NADH]  

            ATP              [NAD+] [SUCCINATE]

Hence: NAD+ = FUMARATE, and [NADH]    = 0.01 M

                                                  [SUCCINATE]

So, [ADP] [Pi] = 1 (ie, 0.01 divided by 0.01)

        ATP

Rearrange: [ATP] = 1
                 [ADP] [Pi]