Comments on: NADH.H+ Reoxidation during Anaerobic Conditions

By: fateh

fateh — Wed, 22 Jul 2009 21:36:00 +0000

The answer here might be option (G)

By: Matthew

Matthew — Fri, 10 Apr 2009 08:27:30 +0000

I think I may have phrased my question wrong. What I was asking was why “NADH + H+” was now the terminology rather than what I previously thought: simply “NADH”. Why the additional H+? The whole equation works when only NADH is used.

By: biochemistryquestions

biochemistryquestions — Thu, 09 Apr 2009 19:34:32 +0000

Hi, Matthew!
Thanks for sharing your doubts with me.
(I assume that you are very young since your call it a nightmare…happily you have not found real nightmares yet!)

Lets try to answer you:

Let’s balance some global equations:

This is the very global equation:
C6 H12 O6 (glucose) + 6 O2 —–> 6 CO2 + 6 H2O (you can see 12 Hydrogens at the left and 12 at the right)

Let’s see if it is true, buy analizing two big steps:

Aerobic glycolysis:
C6 H12 O6 +O2——-> 2 (C3 H4 O3) (pyruvic) + 2 H2O

From Pyruvic to CO2 -includes pyruvic decarboxylation and krebs cyle-(you can balance and check)

2 (C3 H4 O3)pyruvic + 5 O2 —> 6 CO2 + 4 H2O

(If you add both reactions you obtain the more global reaction written above)

So, you can see that there are 12 hydrogens at the left and 12 hydrogens at the right.

This 12 hydrogens are included in 6 molecules of water, since they go, from the reduced cofactors formed in these processes, to the respiratory chain.

By: Matthew

Matthew — Wed, 08 Apr 2009 15:29:33 +0000

But hang on…..

in all those reactions assuming only 1 glucose molecule is used therefor only 12 hydrogen atoms are available for the whole reaction. If this system of NADH + H+ is used, I find a total of 22 hydrogen atoms would be required. What am I missing? where are the extra H+ ions coming from?

Ie. Glycolysis: Glucose ==> 2x Pyruvate 4H removed (2x NADH + H+)
Link Reaction: 2x Pyruvate ==> 2x Acetyl CoA 4H removed (2x NADH + H+)
Krebs Cycle:2x Oxaloacetate + Acetyl ==> Citrate ==>…==> Oxaloacetate(2x (3x NADH + H+)
+ (FADH2) = 18H)

18H + 2H + 2H = 22H

Please explain where the 10 extra H+ ions are coming from cos this is a nightmare. The entire reaction is limited to 12H as far as I can see as no more are added anywhere.

By: biochemistryquestions

biochemistryquestions — Sat, 07 Mar 2009 22:44:13 +0000

The right form of writing NAD in the oxidized form is NAD+ since this chemical structure has a positive charge (like the N in NH4+); in a reduction process,
NAD+ + H2 —> A Hydrogen with two electrons are bound to the NAD+, so it becomes NADH
As you see, the only “part” of the H2 that results bound to the NAD+ is H: and that is why we should write the reduced form as NADH
but since a proton H+ of the original H2 is released to the environment, then it is used to write the reduced form of NAD+ as NADH. H+, indicating also the proton (H+) that has been released.
Thanks for your question.

By: TAREK

TAREK — Sat, 07 Mar 2009 21:17:07 +0000

why NAD & NADH+H+ are written as it, and what is the different?