Lost track of the hydrogen in step 3 of the Krebs cycle

Question

I'm trying to follow the Krebs cycle step by step, accounting for the number of hydrogens in each molecule before and after reactions.

I've run into a problem starting from step 3.

Isocitrate has five hydrogens. It undergoes oxidisation to Alpha-Ketoglutarate, with NAD+ being reduced to NADH/H+.

I would say this process entails the loss of two hydrogens from Isocitrate, and the gain of those hydrogens in NADH/H+.

What I don't understand (and this lack of understanding cascades through steps 4 and 5) is that Alpha Ketoglutarate has four hydrogens.

My arithmetic says that there should be only three.

Where does the fourth hydrogen come from?

Note: If I start from Step 6, my accounting method works, all the way round to Step 3.

Step 6: Succinate (four hydrogens) is oxidised to Fumarate (two hydrogens), reducing FAD to FADH2.

Step 7: Fumarate (two hydrogens) is reduced to Malate (four hydrogens), by the addition of H2O, gaining two hydrogens.

Step 8: Malate (four hydrogens) is oxidised to Oxaloacetate (two hydrogens), reducing NAD+ to NADH/H.

Etc.

I came across a similar question at https://biology.stackexchange.com/questions/93775/what-exactly-happens-to-hydrogen-atoms-in-step-4-of-citric-acid-cycle but as this was about step 4, and for me the problem occurs one step earlier.

Answer 1

Your problem stems from the statement

"I would say this process entails the loss of two hydrogens from Isocitrate, and the gain of those hydrogens in NADH/H+."

This statement is incorrect. In isocitrate, the carboxylic acid functional groups are deprotonated, as would be the case at neutral pH. The oxidation reaction thus involves loss of only $\ce{H-}$ to NAD. There is no loss of $\ce{H+}$, because that has already been removed.

If you start with isocitric acid, which is isocitrate in its neutral form with $\ce{-CO2H}$ instead of $\ce{-CO2-}$, then you will need to remove $\ce{H-}$ and $\ce{H+}$ to accomplish the reaction and your above statement will be correct.

The textbook image you showed takes a confusing approach of implicitly associating an $\ce{H+}$ with isocitrate even though it is not actually on the molecule, so that "NADH/H+" is indicated as a product even though only NADH has actually been produced in the reaction shown.