# Reaction for G-3-P oxidizing to 1,3-BPG

1. May 28, 2008

### teddy1975

okay, so here's another question: the reaction for G-3-P oxidizing to 1,3-BPG is:

I understand that the hydride ion leaves G-3-P and donates two electrons and one proton, which neutralizes the charge on NAD+ and gives NADH. But it is only giving one 'hydrogen', while the product side of the reaction shows another one (the hydrogen proton). Where does this hydronium ion come from? Why is it in the reaction? I read somewhere that it is pulled out of 'solution', but what does that mean?

2. May 30, 2008

### MichaelXY

I am no biologist but I think it is Cytochromes in the mitochondrial inner membrane.

3. May 31, 2008

### epenguin

Just think first of a neutral substrate being oxidised to a neutral product, like alcohol -> aldehyde. $$H^-$$ is transferred to NAD+ giving you the neutral ring NADH as you have explained. What would that leave the alcohol? A positively charged molecule with unpaired electron that can't exist. Write down the structure and you will find yourself predicting that and electron pair folds in to make a double bond C=O leaving a free proton so you have accounted for your proton production. In our case, oxidation of an aldehyde, the impossible positively charged subtrate molecule after the H- has left takes on an OH-, so finishes up neutral acid molecule RCOOH. Absorbtion of an OH- is equivalent to production of a H+. Alternatively you can think of the substrate abstracting an OH- from an H2O molecule leaving an H+.

(Then just as a complication the acid molecule which is formally RCOOH, at physiological pH's will be practically totally dissociated into RCOO- + H+.)

Note here we are not too concerned with exactly how it happens, the real steps or 'mechanism', just with the overall results or starting and end products, the 'stoichiometry'. The reaction is equivalent to transfer of H2 to NAD+, or reduction of the NAD, by an H2 molecule; H2 is equivalent to H- + H+.

Last edited: May 31, 2008
4. Jun 2, 2008

### epenguin

Sorry, just delete the "with unpaired electron" above.
Try and predict with 'bow and arrow' chemistry what happens when the H- leaves.