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A and B are 2 reactions.

If there is an excess availability of citrate, then which of the above reactions will prevail?

I know that citric acid is formed when oxaloacetic acid reacts with acetyl coA. But how does concentration of citric acid influence the above reactions

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This image is a bit misleading since both ways seem to be the reversion of each other, but in fact they are not (although the same metabolites are used).

Reaction A takes only place in gluconeogenesis which makes glucose from other metabolites (coming from the metabolism of fatty acids for example) and this pathway is only active when there is an abundance of energy present in the cell. Citrate activates the Fructose-1,6-bisphosphatase. The gluconeogenesis takes only place in liver and kidney.

Reaction B takes place in glycolysis and is the rate limiting for the whole pathway. It takes place when glucose is taken up and then used for the production of energy. The Phosphofructokinase (PFK) is inhibited by signals of excess energy as ATP (which is an allosteric inhibitor of PFK) as well as citrate and NADH/H+. While citrate is a weak allosteric inhibitor on its own, it acts as a strong inhibitor together with ATP. This makes sense since both molecules are indicators of free energy metabolites in the cell. See these two publications for details:

If there is an excess of AMP/ADP then the enzyme is activated and phosphorylates Fructose-1-phosphate.

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  • $\begingroup$ Although it is true that citrate can act as an inhibitor of PFK, I see to recall that the relevance of this in vivo is questionable. The key effect, not mentioned in the answer here, is that citrate is an activator of fructose 1,6 bisphosphatase, the enzyme for reaction A. Also worth mentioning that gluconeogenesis only takes place in liver and kidney. $\endgroup$ – Alan Boyd May 26 '14 at 16:57
  • $\begingroup$ @AlanBoyd It has both effects - and since they are adverse, this makes sense. PFK is activated by AMP/ADP which inhibits the Fructose-1,6-bisphosphatase. Thanks for mentioning. $\endgroup$ – Chris May 26 '14 at 18:44

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