# Why only 6 water molecules are formed in the aerobic degradation of glucose?

I am studying the aerobic degradation of glucose and it seems that for every glucose molecule we should obtain $\ce{10H2O}$ molecules. However, it is known that we only obtain 6.

$\ce{C6H12O6 + 6O2 -> 6CO2 + 6H2O}$

(I am not going to focus in all products and reactants, but just in the important ones for the formation of water molecules)

First, in the glycolysis, for each molecule of glucose we obtain two water molecules, $\ce{2NADH+}$ and 2 pyruvate molecules. By the oxidation of two pyruvate molecules we obtain $\ce{2NADH+}$ and 2 acetyl Co-A molecules. So we are going to pass twice through the Krebs cycle, obtaining $\ce{6NADH+}$ and $\ce{2FADH2}$, and requiring 4 water molecules.

So, when we arrive to the electron transport chain, we have a negative balance of 2 water molecules, and we have $\ce{10NADH+}$ and $\ce{2FADH2}$. We have been told that for every of these molecules 2 electrons go to the electron transport chain, that means that a total of 24 electrons go to the system. The problem comes here:

$\ce{4e- + 4H+ + O2 = 2H2O}$

So, bearing in mind that we have $24e^-$, 12 water molecules should be formed, so at the end, we have gained 10 water molecules, but we know that the number of water molecules formed should be 6. So, clearly there is something wrong in my explanation. I would be very pleased if you could tell me what is wrong.

Your confusion comes entirely from this equation:

$\ce{C6H12O6 + 6O2 -> 6CO2 + 6H2O}$

This reaction is the combustion of glucose. This is not how glucose is oxidized in cells! Why so many biology texts and courses present this equation when introducing metabolism is beyond me.

Indeed, your tracking of water molecules is correct: starting with one glucose molecule, 2 waters are produced in glycolysis, 4 are consumed in the tricarboxylic acid cycle and 12 are produced during oxidation of NADH/QH2 (ie FADH2). This gives a net total of 10 produced.

Why is this different from the combustion of glucose? The answer lies in the oxygens introduced by inorganic phosphate during substrate level phosphorylation. Consider the balanced net reaction for the biological oxidation of glucose (simplified by ignoring ATP produced by oxidative phosphorylation and substituting ADP/ATP for GDP/GTP):

$\ce{C6H12O6 + 6O2 + 4ADP + 4P_i + 4H+ -> 6CO2 + 4ATP + 10H2O}$

In particular, consider the formation of ATP from ADP and Pi (HPO42-). In both substrate level phosphorylation reactions (catalyzed by GADPH/PGK in glycolysis and succinate-CoA ligase in the tricarboxylic acid cycle), inorganic phosphate nucleophilically attacks the activated carbonyl (thioester) of the substrate and is then transferred to ADP (to form ATP):

The oxygens of the original inorganic phosphate are coloured red. The key point is that an oxygen atom from HPO42- is transferred to the substrate. This oxygen is later removed in the form of carbon dioxide, via oxidative decarboxylation, during the conversion of pyruvate to acetyl-CoA and in the tricarboxylic acid cycle. This occurs four times for each glucose molecule entering glycolysis and is accompanied by the reduction of NAD+ to NADH. Given that NADH is used to reduce molecular oxygen during the electron transport chain:

$\ce{NADH + H+ + 1/2O2 -> NAD+ + H2O}$

...this explains where the four, apparently extra, water molecules come from when comparing the combustion of glucose with its biological oxidation.

• Good answer, but I would contest that the plain combustion reaction is actually very useful, even if it does not reflect the mechanism that cells use. The reason is that, if you balance your reaction involving substrate-level phosphorylation with ATP hydrolysis -- which always occurs alongside ATP synthesis in cells -- then you will get back the simple combustion equation exactly. And this must always be the case. So comparison with simple combustion helps us verify that we understand biochemistry correctly. – Roland Oct 10 '17 at 17:54
• @Roland Thanks! And good point. I find that it often causes a lot of confusion, but that may be more from poor explanation or comprehension. – canadianer Oct 10 '17 at 18:08
• @Roland, I am a high school student, I also had the same confusion, that why do so many textbooks compares combustion with respiration? in respiration 10 H20 are released, and I look upto the answer, but it seems like, i don't have enough knowledge to grasp it, can you explain it to me in simple terms? should I say that the other 4 H2O molecules are used somewhere else? – Arun Bhardwaj Feb 6 at 15:50
• @ArunBhardwaj, This is an old post, but as I mentioned in other comments, it's important to distinguish between (1) H2O derived from glucose oxidation and (2) H2O from ATP synthesis (respiration). Glucose oxidation in cells is always coupled with ATP hydrolysis to provide energy (for example, in muscle contraction). The ATP hydrolysis reaction is ATP + H2O --> ADP + H+ + Pi. So if you balance the the first reaction equation in this answer with 4 x ATP hydrolysis, the extra 4 ATP will cancel out and you get back the basic glucose oxidation reaction. So this is where the extra 4 ATP are used. – Roland Feb 7 at 9:11
• @ArunBhardwaj, (continued) The H2O formed during respiration is very different. Respiration is not even a pure chemical reaction, it's a physical process -- ATP synthase is actually more like a turbine! In respiration, the number of ATP formed by respiration is not fixed, but varies depending on conditions. It's usually somewhere between 2--3 ATP per NADH, but there are exceptions: some cells don't produce any ATP at all, but instead use all the energy in NADH to produce heat. So there is no single, exact chemical reaction formula for respiration. – Roland Feb 7 at 9:18

I think you are misconceiving the intent assumed within the statement "6 H20 generated from glycolysis". The number 6 is simply in relation to the number of carbons oxidized to CO2 within the TCA..which generates an electron at each oxidative occurance, in which the 6 electrons are then shuttled to OXPHOS requiring 3 O2 to form 6 H20. ... In short: Whenever they say 6 H20 are produced "aerobically" they are specifically referencing the simplified OXPHOS component (or simply: C6H12O6 + 6 O2 → 6 CO2 + 6 H2O).

Of note:

1. Stoichometry is relatively useless in real research. Don't overthink it. Metabolism is incredibly dynamic.

2. Real life H20 involvement within complete glucose oxidation is much more complex: Two turns of the TCA cycle generate: 4 CO2 from 2 acetyl CoA which requires 4+ H20 in total, but only has a net H20 loss of 4. Citrate synthase and fumerase both consume 2 H20 apiece to allow production of 4 CO2. The remaining H20 is used at aconitase, but no loss nor gain of H20 occurs in this reaction. Pyruvate dehydrogenase produces the other 2 CO2's making 6 CO2 in total, but does not require H20 at this step. ... OXPHOS activity: (following 2 turns of TCA) results in 48 total H20 produced between ATP synthase, cytochrome oxidase, and enolase - and being that 4H20 are consumed in the TCA, the net H20 production per glucose is actually 44 H20.

• Don't understand the comment that "Stoichometry is relatively useless in real research". I find it incredibly useful, and I like to think of myself as a real researcher. Admittedly, I've never had to carefully balance waters, but if you do it right, it works out. – Victor Chubukov Oct 9 '17 at 0:33
• This doesn't answer the question. The H2O produced in ATP synthesis is not relevant here, since those H2O molecules originate from the phosphate groups and are recovered during ATP hydrolysis. Also, ATP synthesis is not stoichiometrically fixed to glucose oxidation since chemiosmosis is a stochastic process, so it is not even possible to give an integer coefficient for H2O if you would involve ATP synthase. – Roland Oct 10 '17 at 20:16
• @VictorChubukov my apologies, my intention of saying "real" was not to contrast paid vs non-paid research, I meant it only as when looking at real-life biological systems, in vivo. Regarding the stoichometry, I was trying to imply that there can be multiple different reactions in the same place at the same time, that all produce the same product; and trying to break down those dynamics into an equation is impossible as of this moment; I guess I should of clarified what I meant as 'systems stoichometry'. Equation stoichometry itself is very usefull. Cheers. – Aaron43 Jan 19 '18 at 18:05

Two waters come in at each turn of the TCA cycle, one at citrate synthase and one at fumarase. Additionally, I don't think it's right to say that you get a net water from glycolysis (in fact if you look at what you wrote, the oxygens don't balance). I think what you're missing is you also get an ATP, and the water you do get at the 2PG->PEP step is really part of the net reaction ADP + phosphate -> ATP + water.

To really balance the waters you have to consider all the substrates and byproducts of glucose oxidation, including the ATP/GTP.

• Thank you very much. However, this only works if the ATP produced in the ATP-synthase does not need water to be formed. Why by this way (with ATP-synthase) the formation of ATP doesn't need water, but if is done in any other way, water is required? – maxbp Oct 7 '17 at 9:57
• Could it be that the formation of ATP by the ATP-synthase uses PO3 instead of HPO4 + H+? – maxbp Oct 7 '17 at 10:07
• Every phosphate fixation reaction will release a water molecule and every atp hydrolysis reaction will use a water molecule. That includes ATP synthase. – Victor Chubukov Oct 7 '17 at 13:19
• But then we have 40 water molecules, since in the electron transport chain 34 ATP are formed. – maxbp Oct 7 '17 at 15:20
• Sure. But then you have to write the equation as 1 glucose + 6 O2 + 34 PO4 + 34 ADP -> 6 CO2 + 34 ATP + 40 H2O. I think that balances. – Victor Chubukov Oct 7 '17 at 19:08