Question

The coenzyme NADP+ is the terminal electron acceptor in chloroplasts, according the reaction

2 H2O + 2 NADP+ → 2 NADPH + 2 H+ + O2

NADP+ + H+ + 2 e- → NADPH E˚'= -0.315 V

O2 + 4 H+ + 4 e- → 2 H2O E˚'= 0.816 V

Given the information above, calculate the equilibrium constant for this reaction at 25˚C and pH 7.0.

How can the chloroplast overcome this unfavorable equilibrium?

Answer 1

The equations for the half cell reactions are as follows:

NADP++H++2e-NADPH Eo'= -0.315 V ----(i) O2+4H++4e-2H2O Eo'= 0.816 V -----(ii)

Multiplying (i) by 2 and then subtracting (ii) from (i) we get,

2NADP++2H2O 2NADPH+2H++O2 ;   

(i) We know the relation between standard potential and Keq as folllows;

,

(ii) So  

As is positive and is negative, therefore the reaction is non-spontaneous and unfavorable. So to overcome the unfavorable equilibrium, the chloroplast must lower the pH by generating more H+ concentration, i.e to favor the forward reaction. This can be achieved by increasing the concentration of reactant like water and NADP+. Chloroplast transfer more water molecules to the thylakoid lumen and induce splitting of them so that protons produced accumulate within the lumen of thylakoid. Another modification done by chloroplast is to generate more protons that are transfered from stroma to lumen via NADP reductase enzyme which reduces NADP+ to NADPH+H+. All these factors increases protons content in the lumen resulting in low pH which enables to create a proton gradient across the thylakoid membrane necessary for ATP synthesis.