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

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

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 $H_2O$ are out of the reaction
  • (TCA cycle) (in the reaction that combines Oxaloacetate and Acetyl-CoA to form Citrate) 2 molecules of $H_2O$ are into the reaction
  • (TCA cycle) (in the reaction that Citrate form Cis-Aconitate) 2 molecules of $H_2O$ are out of the reaction
  • (TCA cycle) (in the reaction that Cis-Aconitate form D-Isocitrate) 2 molecules of $H_2O$ are into the reaction
  • (TCA cycle) (in the reaction from Fumarate to Malate) 2 molecules of $H_2O$ are into the reaction

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

Thanks in advance

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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. –  Alan Boyd Mar 29 '14 at 9:11
My (possibly) outdated school textbook states the equation like what I wrote before...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? –  Chanhee Jeong Apr 20 '14 at 14:18

1 Answer 1

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

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

For the Left hand Side, $6$ $H_2O$ are accounted here :

  • $2$ $H_2O$ go in conversion of 2-Phosphoglycertae to phosphoenolpyruvate.
  • $2$ $H_2O$ 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$$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$$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 !

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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 –  Chanhee Jeong Apr 20 '14 at 14:22

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