NADH reduces disulfide bond (section bank question)

[Vicodin_]

Hi I am really confused about this question in section bank C/P...

A protein contains 4 disulfide bonds. In order to break these bonds the researchers added a minimum of :

A.2 moles of NADH for each mole of protein
B.4 moles of NADH for each mole of protein
C.2 moles of NAD+ for each mole of protein
D:4 moles of NAD+ for each mole of protein

The answer is B. I got the answer right but I don't understand why it's 4 moles of NADH instead of 8 moles, because you would need to add 2 H for each disulfide bond.

Thanks!

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[bobeanie95]

During oxidation of cysteine residues, 2 electrons are removed to form a single bond. The reduction of NAD+ involves 2 electrons to make NADH. Therefore, one mole of NADH would reduce one bond (donation of 2 electrons).

 

[aldol16]

Draw out a basic mechanism. NADH is basically a hydride donor. A hydride is basically a lone pair of electrons (plus a hydrogen nucleus). So you would have a disulfide bond where the hydride attacks one sulfur, kicking out the other sulfur as an anion, which then picks up a proton from solution. So overall, you reduced the disulfide by 2 electrons and picked up a hydride and proton while doing it. The key point here is that in redox, you care about electrons (or hydrides), not about how many H atoms you add. You can always pick up protons in solution. So follow the electrons and they'll lead you to the right answer in redox.