NADH + H⁺ has more reducing power than FADH₂, therefore: A) oxidation of NADH will contribute to form a larger $\Delta$p and more ATP B) oxidation of FADH₂ by respiratory ETS contributes less to $\Delta$p formation, and as a consequence fewer ATP can be made C) the NADH + H⁺/NAD⁺ redox pair has a more negative E°' value than the FAD/FADH₂ pair D) all of the above E) none of the above

Question

Quizplus · Accepted Answer

NADH has a higher reducing power than FADH2, which means it can donate electrons more readily. This results in a larger proton gradient (Δp) when NADH is oxidized, leading to the production of more ATP compared to when FADH2 is oxidized. The more negative E°' value of the NADH/NAD+ redox pair compared to the FAD/FADH2 pair reflects this higher reducing power. Therefore, all the statements A, B, and C are correct, making D the correct choice.

NADH + H+ Has More Reducing Power Than FADH2, Therefore Δ\DeltaΔ

NADH + H⁺ Has More Reducing Power Than FADH₂, Therefore $\Delta$