# How to calculate Oxygen consumption?

I am looking into Glycolysis and TCA cycle. I have learnt that oxygen is the final electron acceptor.

So, let's say we know the amount of NADH and FADH2 produced under some circumstance related to the Glycolysis and TCA cycle.

How would we calculate the amount of oxygen consumption level?

• (+1) One method, at one time almost universal in biochemistry labs, it to use a Clarke oxygen electrode attached to a (y-time) chart recorder. This allows continuous monitoring of the oxygen concentration in a stirred chamber, from which the rate of oxygen consumption may be calculated. A common setup is to measure the oxygen consumption of mitochondria isolated from liver or some other source under various experimental conditions. – user1136 Nov 2 '16 at 21:48
• If you are interested in Glycolysis and the TCA cycle, it might be of interest to take a look at the (often overlooked?) Pasteur Effect: in many cell types oxygen inhibits respiration, But quite possibly you know all about this (and its explanation) – user1136 Nov 2 '16 at 21:56

Assuming that you know the amount of NADH and FADH$_2$ produced by glycolysis and TCA cycle, you can determine the reducing equivalent produced: for each NADH or FADH$_2$ produced, two reducing equivalents are produced. It means that the re-oxidation of NADH and FADH$_2$ will release 2 electrons. Here are the corresponding half-reactions:

$\textrm{NADH} = \textrm{NAD}^+ + \textrm{H}^+ + 2~\textrm{e}^-$

$\textrm{FADH}_2 = \textrm{FAD} + 2~\textrm{H}^+ + 2~\textrm{e}^-$

Knowing the total reducing equivalent produced, divide it by four in order to obtain dioxygen consumption: O$_2$ is reduced into 2 H$_2$O by accepting four electrons. Here is the corresponding half-reaction:

$\textrm{O}_2 + 4~\textrm{e}^- + 4~\textrm{H}^+ = 2~\textrm{H}_2\textrm{O}$

Of course this only applies to aerobic organisms: some microbes are able to use other electron acceptors than O2.

• This is of course correct, though I think it might help to just write the reaction: 2 NADH + O2 -> 2 NAD+ + 2 H2O – Victor Chubukov Mar 3 '17 at 22:17
• The half-reaction for oxygen reduction is not balanced, missing 4 H+. – Roland Mar 4 '17 at 16:53