I do a fair bit of biochemistry, and one of the things I've taken on faith up until now is that ATP is stable because it has a high kinetic barrier for falling apart despite the spontaneity of its decomposition. I understand that it does, but what about the transition state of ATP hydrolysis in an aqueous solution is so high energy? For that matter, what is the transition state of spontaneous ATP hydrolysis?


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This is not exactly an answer making strong use of references etc. but it does give a good hint. The most ‘energetic’ bonds in ATP are the phosphoric acid anhydride ones i.e. the $\ce{P-O-P}$, which, to hydrolyse ATP, must be attacked by water. The most logical path of attack is one according to the $\mathrm{S_N2}$ mechanism, where a lone pair of water attacks the rear side of a phosphorus-oxygen bond and liberates some phosphate species. ($\mathrm{S_N1}$ can be ruled out, since it would result in a phosphocation which is surrounded by three electron-withdrawing oxygens making the cation highly unstable.)

The problem in this mechanism is physiological pH. It is safe to say that the entire ATP molecule is at least a dianion, if not even a tri- or tetraanion. Now water’s lone pair, i.e. its partially most negatively charged area must approach this already highly negative molecule — but that would result in the repulsion of same charges leading to a high activation barrier.

This would be circumvented by protonation, and indeed even at equilibrium there is a certain probability that a higher number of protons attaches to ATP, making it less negative. This probably helps to account for the slow but steady decay of ATP.

  • $\begingroup$ facepalms Should have realized this. XP Thank you! $\endgroup$ Jan 27, 2016 at 2:08

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