3
$\begingroup$

I having trouble understanding the equation of the cellular respiration. The thing that bothers me is the number of $\ce{H_2O}$ molecules. Generally, cellular respiration is written thus :

$\ce{C_6H_{12}O_6 + 6O_2 + 6H_2O -> 6CO_2 + 12H_2O}$

Yet studying the reaction on a molecular level, I realized that the water molecules just don't add up to be six! (on the left side) What I obtained from my textbook and Wikipedia is that (per 1 glucose molecule)

  • (Glycolysis) (in the reaction that 2-Phosphoglycerate(2PG) forms Phosphoenolpyruvate(PEP)) 2 molecules of $\ce{H_2O}$ are out of the reaction
  • (TCA cycle) (in the reaction that combines Oxaloacetate and Acetyl-CoA to form Citrate) 2 molecules of $\ce{H_2O}$ are into the reaction
  • (TCA cycle) (in the reaction that Citrate form Cis-Aconitate) 2 molecules of $\ce{H_2O}$ are out of the reaction
  • (TCA cycle) (in the reaction that Cis-Aconitate form D-Isocitrate) 2 molecules of $\ce{H_2O}$ are into the reaction
  • (TCA cycle) (in the reaction from Fumarate to Malate) 2 molecules of $\ce{H_2O}$ are into the reaction

These sums up to a net of (only) 2 molecules of $\ce{H_2O}$ into the process of glycolysis and TCA cycle... additional 6 molecules of $\ce{H_2O}$ comes from oxidative phosphorylation. Then, where can the remaining 4 molecules of $\ce{H_2O}$ possibly be?

$\endgroup$
  • 3
    $\begingroup$ Can you point to a source that uses the equation in that form? Everywhere that I have looked has no water on the LHS and 6 water on the RHS, and that is the equation that I'm familiar with. Incidentally in your accounting you missed the loss of 1 water at the aldolase step I believe. $\endgroup$ – Alan Boyd Mar 29 '14 at 9:11
  • $\begingroup$ My (possibly) outdated school textbook states the equation like what I wrote before... but I don't think it's trustworthy.(It's weird in many ways) I think the equation with only 6 water molecules is correct. So,(may I ask) is there a water molecule used in the aldolase step? two of them? $\endgroup$ – Chanhee Jeong Apr 20 '14 at 14:18
  • $\begingroup$ @AlanBoyd, I don't think H2O is involved in the aldolase reaction. Aldolase is a lyase, not a hydrolase; see chem.qmul.ac.uk/iubmb/enzyme/EC4/1/2/13.html $\endgroup$ – Roland Jul 25 '16 at 20:42
  • $\begingroup$ You might be interested in this anwer and references therein to the controversy on the role of water in the Krebs cycle, and in the discussion to this great question $\endgroup$ – user1136 Jul 29 '16 at 9:27
  • $\begingroup$ @AlanBoyd (and Chanhee Jeong) [comment 1-3] This ref, for example, gives the equation for the complete oxidation of glucose as the OP has stated it. I think this paper gets the stoichiometry of water right, and is a very good paper, but also contains [IMO] the following clanger: _ $\endgroup$ – user1136 Jul 30 '16 at 15:46
3
$\begingroup$

First thing to make clear is that net $6 \, \ce{H_2O}$ go out of the reaction.($12 \,\ce{H_2O}-6\,\ce{H_2O}$)

Let me tell you my calculation, you should then be able to figure out what went wrong.

For the Left hand Side, $6\,\ce{H_2O}$ are accounted here :

  • $2\,\ce{H_2O}$ go in conversion of 2-Phosphoglycertae to phosphoenolpyruvate.
  • $2\,\ce{H_2O}$ in TCA from conversion of Oxoloacetate to Citrat by combining with Acetyl CoA.(One for each Acetyl CoA molecule)
  • The aconitase -cis-aconitase-isocitrate water molecules cancel out. Equal no go in and out.
  • $2\,\ce{H_2O}$ in TCA from conversion of fumarate to Malate.

For the right hnd side, if we calculate the no. of NADH formed = 10 and assume that each participates in respiratory chain, we have $10\,\ce{H_2O}$ from there. The other $2$ will be from the $2$ FADH produced.

So we have 6 on the LHS and 12 on the RHS. Hope this makes sense !

$\endgroup$
  • 1
    $\begingroup$ I'm afraid that per a NADH or a FADH2, these yield only 2 electrons, reducing only 1/2 water molecule. So, only 6H2O produced at the respiratory chain $\endgroup$ – Chanhee Jeong Apr 20 '14 at 14:22
  • $\begingroup$ @ChanheeJeong, No, 2 electrons reduce 1/2 O2 molecule, so forms 1 H2O. So this answer is correct I think. $\endgroup$ – Roland Jul 25 '16 at 21:15
  • $\begingroup$ The reduction of one 02 to one H2O is a four-electron reduction. The reduction of one O2 to one H2O2 (hydrogen peroxide) is a two-electron reduction. $\endgroup$ – user1136 Jul 31 '16 at 14:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.