Question

Martian microbe 4 has a new type of lactate dehydrogenase that is a membrane-bound enzyme that oxidizes lactate to pyruvate, donating 2 electrons to ubiquinone and pumping out 4 protons to the periplasm per lactate. What is the Delta p that can be generated by this enzyme, given an E_h of -185 mV for pyruvate/lactate, and +100 mV for ubiquinone/ubiquinol?

Answer 1

The energy released on electron transfer between substrate and product redox pools is equal to the energy required to pump the protons (Ripple etal. 2012).

Lactate+ ubiquinone à pyruvate+ ubiquinol

∆G = -2 ( E_h^{ubiquinone/ubiquinol} - E_h^{pyruvate/lactate}) + n_p∆P

E_h^{ubiquinone/ubiquinol} and E_h^{pyruvate/lactate} are the redox potentials of the ubiquinone/ubiquinol and pyruvate/lactate. n_p is the number of protons pumped per two electrons transferred. ∆P is the proton motive force

The factor of 2 accounts for the 2-electron transfer.

ΔG is zero when the enzyme operates at equilibrium and the flux is zero. Thus,

n_p∆P = 2 (E_h^{ubiquinone/ubiquinol} - E_h^{pyruvate/lactate})

∆P = 2 (E_h^{ubiquinone/ubiquinol} - E_h^{pyruvate/lactate})/ n_p

∆P = 2 (100-(-185))/4

∆P = 142.5 mV