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In the Krebs cycle, where do the hydrogens and electrons that NAD+ and FAD accept come from? It seems that citric acid only loses two hydrogens because it starts out with eight hydrogens and then becomes oxaloacetic acid, which has four hydrogens.

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  • $\begingroup$ where have you looked for answers? have you tried google - this is a relatively straightforward question that should be easily found... try googling "where do electrons from krebs cycle come from" $\endgroup$ – Vance L Albaugh Jul 28 '16 at 14:02
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    $\begingroup$ Yes, I have googled this, discussed the question with friends, and thought about it myself and I have not found an answer. That's why I'm asking on stack exchange. $\endgroup$ – user3163829 Jul 28 '16 at 14:26
  • $\begingroup$ [1-2]. In the Krebs cycle, all of the electrons passed to both FAD and NAD come from carbon-hydrogen and carbon-carbon bonds. None come from water, although this seems to be quite a common misconception, and this ref, in a major academic journal, even goes so far as to say that the Krebs cycle should be considered a water-splitting process! $\endgroup$ – user1136 Jul 28 '16 at 14:29
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    $\begingroup$ [2-2] You may be interested in this answer, and the references therein. Great question, btw. I suppose the lesson is: Don't believe everything you read on Google! $\endgroup$ – user1136 Jul 28 '16 at 14:30
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    $\begingroup$ Very interesting question. Clearly this is far from obvious, with conflicting statements in the literature. I don't know if one should call it "splitting water", but the TCA cycle does take up 3 oxygen atoms from H2O (one via phosphate) which end up in CO2. Given that, it's not clear to me where the electrons of H2O end up, especially considering that electrons in the carboxyl groups are delocalized ... @TomD, were you able to find the missing references you mention in your other answer? $\endgroup$ – Roland Jul 28 '16 at 17:51
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This is a question that has been a source of disagreement between some quite senior biochemists, so, although I am no angel, I waited a little to find an elusive volume of a journal before rushing in. At the outset it seemed to me that:

  1. What was important in the oxidation of acetyl-CoA in the tricarboxylic acid (TCA) cycle was the source of electrons, not the source of hydrogen atoms. Oxidation is the removal of electrons.
  2. The source of the electrons must be the carbohydrate being oxidized. It could not come from anywhere else (such as water).

This point was made in a comment (not an answer) by TomD, who pointed to his answer to a different but related SE Biology question. In that answer he refers to two ‘letters to the editor’ on the subject in TIBS vol.6, p.6 (1981), which unfortunately are not available in the online version of the journal. I recalled them from my own university library’s archives and have made them available here as a text pdf, although in the meantime TomD posted a scan himself. I recommend reading the letters carefully to consider their arguments, but quote extracts from each here to show why I personally tend towards their view.

D.E. Atkinson

Atkinson’s key point, as I understand it, is that although the H may come from water, this is only as a result of ionization — the water is not oxidized, so cannot provide electrons:

“Electrons cannot be tagged and traced, but the stoichiometry is clear: in the oxidation of an acetyl group to 2CO2, eight electrons are lost. Of these, six are used in the reduction of NAD+ to NADH and two in the reduction of the flavin of succinate dehydrogenase. Water is not oxidized; thus no electrons are supplied from water to be: ‘raised to the level of NADH2’ or for any other role. When water ionizes, the electrons all remain associated with the OH ion; whatever the proton may do, it cannot reduce anything.”

B. Herreros and J. Garcia-Sancho

These authors emphasize that it is the C–C and C–H bonds that are providing the electrons:

“As pointed out by Losada the carbon compounds themselves are the only source of reducing power in the TCA cycle, not water. The difficulty in realizing it arises from the contabilization of the reducing equivalents as pairs of H rather than as electron pairs. It is true that one molecule of glucose can not supply 12 pairs of H, but it can certainly supply 12 pairs of electrons…

… In the TCA cycle, the electrons shared in the C–C and the C–H bonds are the source of reducing power; they are transfered first to pyridine nucleotide (and flavin nucleotide) and then to the vicinity of oxygen through the respiratory chain.”

Towards the end of his letter, Atkinson writes “I have found it a useful pedagogical device to ask students to work through respiration (glycolysis plus the citrate cycle), keeping careful account of water and proton so that their equations will sum to the proper balanced overall equation.” I am afraid I haven’t provided that in this answer, limiting myself to surveying the wood rather than examining each tree.

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The two electrons and a proton of NADH comes from a H$^-$ ion of a Krebs cycle intermediate during Isocitrate dehydrogenase reaction and Malate dehydrogenase reaction.

In the following reaction, Isocitrate is oxidised to Oxalosuccinate and in the process two Hydrogens are lost one as H$^-$ and the other as H+. This H$^-$ is accepted by NAD+ to form NADH.

Note: Oxalosuccinate is an intermediate formed during oxidative decarboxylation of Isocitrate.

1st

Now coming to FADH$_2$ $-$

During oxidation of Succinate to Fumarate, Succinate looses a H$^-$ and a H+ ion which constitute the two protons and electrons of FADH$_2$ . Here's an image that illustrate the mechanism:

2nd

The alpha-ketoglutarate dehydrogenase complex reaction is similar to pyruvate dehydrogenase complex reaction(PDC). PDC reaction is an oxidative decarboxylation reaction of Pyruvate in which the 2e$^-$ of NADH are derived from the electron shared between COO$^2-$ and C=O groups of Pyruvate. Image 1

The Hygrogen (H+) of NADH is either derived from the H+ of CoA-SH or from the medium. There's a series of complex reactions which need to be studied to fully understand the source of H of this NADH.

So you can understand that the 2 e$^-$ of NADH produced in Alpha-ketoglutarate dehydrogenase complex reaction are the electrons between COO$^2-$ and C=O groups of Alpha-ketoglutarate and the H is either from the medium or the H+ of CoA-SH.

Reference: Biochemistry by Berg

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  • $\begingroup$ The malate dehydrogenase step is similar to ICD? Whats the intermediate there? But isn't the other reaction, the a-ketoglutarate step similar to pyruvate dehydrogenase? The former is a C-H bond, while the latter is a C-C bond. Correct me if wrong! Sanjukta Ghosh @TomD ? $\endgroup$ – Polisetty Aug 13 '16 at 16:35
  • $\begingroup$ H- ions? Do you have a real source for that? $\endgroup$ – Chris Aug 13 '16 at 17:20
  • $\begingroup$ @Chis I have deduced it from the diagram that I have shared from Biochemistry by Berg. They haven't mentioned H- ion, but that it what the picture says. Referring to the first diagram, H- and H+ must leave the Isocitrate to form Alpha-keto acid, oxalosuccinate. $\endgroup$ – Tyto alba Aug 13 '16 at 17:42
  • $\begingroup$ @David I'd like to hear your views on the concept behind my answer. Do you find it correct? $\endgroup$ – Tyto alba Aug 14 '16 at 18:06
  • $\begingroup$ @Polisetty Yes you are right MD step is similar to ICD step. Thanks for pointing out, I was in a hurry so I had messed up, but now I've corrected it. The PD step is also similar to AKD step. $\endgroup$ – Tyto alba Aug 14 '16 at 19:09

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