I was wondering how the reaction catalysed by adenylate kinase and its equilibrium of approximately 1 leads to [AMP] changing greatly with a small change in [ATP]. The below information was provided but I fail to see how the percentage changes were achieved. If anyone could point me in the right direction for a mathematical/equilibrium calculation as to how these percentages might come to be I would be incredibly grateful.

$\ce{2ADP <=> ATP + AMP}$

The equilibrium constant for this reaction is given by $K_\mathrm{eq} = \dfrac{[\ce{AMP}] [\ce{ATP}]}{[\ce{ADP}]^2}$ and has a value of near 1.

Therefore $[\ce{AMP}]$ ~ $\dfrac{[\ce{ADP}]^2}{[\ce{ATP}]}$

As [ADP] rises during muscle contraction, it is rapidly converted into AMP and ATP, and because [AMP] concentrations are only ~2% of [ATP] a 10% decrease in [ATP] will result in a ~400% increase in [AMP].


2 Answers 2


[AMP] concentrations are only ~2% of [ATP]

If the product of AMP and ATP concentration is equal to the square of the ADP concentration, the ADP concentration is $\sqrt{0.02}$ of the ATP concentration, so 0.14 of the ATP concentration. In a well-energized cell, therefor, you could have concentrations of 5 mM, 0.7 mM, and 0.1 mM for ATP, ADP and AMP, respectively, for example.

If you use up (that is hydrolyze to ADP) 10 % of the ATP, the new ATP concentration would be 4.5 mM and the new ADP concentration would be 1.2 mM. With the concentration of AMP still at 0.1 mM, the system would be out of equilibrium:

$Q = \frac{\pu{4.5 mM} \cdot \pu{0.1 mM}}{\pu{1.2 mM}^2} = 0.28$

So, some of the ADP will react to form AMP and ATP. The approximate concentrations at equilibrium will be 0.21 mM, 0.97 mM and 4.4 mM for ATP, ADP and AMP, respectively. So the AMP concentration more than doubles when 10% of the ATP is hydrolyzed to ADP.

I'm not sure why the claim is an increase by 400%. In this example, I set the ATP concentration to 5 mM. This was arbitrary but does not affect the result because the problem scales linearly (e.g. if you have the system at equilibrium and dilute everything by a factor 10, it is still at equilibrium).


I believe what's being said here is that if you have a cell and it contains 1000 molecules of ATP, then there will be ~20 molecules of AMP (2% of 1000). If 100 molecules of ATP (10% of 1000) are used up they form 100 ADP. Using the equation you provided, you then get 50 AMP and 50 ATP from those 100 ADP. If there were only ~20 AMP to begin with, and you now have 20+50 = 70 AMP, then you have increased the amount of AMP by 70/20*100 = 350%. So not quite 400% (hence the ~400%) but close enough.

As a reference, you can see this paper where, they state discuss the ratio of ATP to AMP and ADP. I find it easier to think in terms of ratios instead of percentages for this case. From the paper:

Although produced in a few other reactions, the major cellular source of AMP appears to be the reaction catalysed by adenylate kinases, enzymes that catalyse the interconversion of the three adenine nucleotides (figure 1b). There is little change in free energy in this reaction, so it is readily reversible with an equilibrium constant close to 1 such that, at equilibrium, [ATP].[AMP]/[ADP]2 will be ≈1. If this reaction is indeed at or near equilibrium (which appears to be the case in most eukaryotic cells), and cells maintain an ATP : ADP ratio of 10 : 1 as discussed above, it follows that the ratio of ATP : AMP will be 100 : 1. In a cell with a fully charged ‘battery', the concentration of AMP will, therefore, be 100-fold lower than that of ATP and 10-fold lower than that of ADP.


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