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.
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.