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ADP has two phosphate groups, and can be hydrolysed to AMP in a reaction which involves a similar free energy change to that of hydrolysing ATP to ADP.

Why is the latter reaction, rather than the former reaction, coupled to energetically unfavourable reactions to provide an overall reaction with a negative change in Gibbs Free Energy?

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  • $\begingroup$ I have edited the question to change it from one of fact (that the poster could have checked) to one of evolutionary biochemistry, which I think is more interesting. $\endgroup$
    – David
    Commented Dec 30, 2018 at 16:25
  • $\begingroup$ Ah sorry, in my original question I wasn’t implying that ADP->AMP wasn’t as energetic as ATP->ADP, I was asking about the evolutionary biochem question to which you corrected it! Thanks! $\endgroup$
    – Sully Chen
    Commented Dec 31, 2018 at 10:19
  • $\begingroup$ The larger molecule (ATP) is easier to handle in biological systems. $\endgroup$ Commented Sep 26, 2020 at 8:40

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Summary

The original question was edited — with the approval of the poster — so that in brief it now asks:

Why is ATP hydrolysis to ADP, rather than ADP hydrolysis to AMP, the normal way in which cells drive reactions which alone involve positive changes in Gibbs Free Energy?

The scope of this is wider than that originally assumed in the answer from @user1136, but he has also contributed to this answer by his comments, which I have incorporated into my revision.

To my knowledge there is no generally-accepted provable answer to this question (as opposed to listing advantages for the system as it works today). What I shall do is present a hypothesis that could explain this situation. The fundamental idea† of this is that:

The contemporary use of ATP rather than ADP in energy transfer reactions evolved from a preference for NTPs, rather than NDPs, as the precursors of RNA synthesis. This preference was a result of the fact that the conversion of NTPs to NMPs and pyrophosphate could prevent the reversal of this (and other) synthetic processes in a way that conversions liberating phosphate could not.

AMP, ADP and ATP: Standard Free Energies of Hydrolysis

ATP and ADP

There is general agreement that the values for ∆G°′ (standard free energy change at 37˚C) for the hydrolysis of the terminal phosphate bond of ATP (β-γ) and ADP (α-β) are similar. (The following are from J.L. Jain, et al.:

ATP → ADP + Pi  –7.3 kcal/mol
ADP → AMP + Pi  –7.3 kcal/mol

Hence, there is no intrinsic energetic difference that might explain the preferential use of ATP over ADP as an ‘energy donor’ in biochemical processes, i.e. it is not evident why e.g.:

Glucose + ATP → Glucose 6-phosphate + ADP

and not

Glucose + ADP → Glucose 6-phosphate + AMP

And, as @user1136 mentions, there are examples where the energy of the hydrolysis of ADP to AMP can be used to ‘drive’ other reactions. He mentions polynucleotide phosphorylase, to which I would add adenylate kinase, as this is not a synthetic process and is pertinent to the energy metabolism of nucleotide phosphates:

ADP + ADP ⇄ AMP + ATP

The introduction of AMP into the discussion raises the question of the value of the ∆G°′ for the following reaction:

ATP → AMP + PPi

Many texts quote a value only slightly greater (ignoring the minus sign) than those for phosphate release (the one cited quotes –7.7 kcal/mol), but a 1995 review of the situation in the Journal of Biochemistry suggests that the difference is actually greater: –10.9 kcal/mol, as opposed to –7.8 kcal/mol for the hydrolysis of ATP to ADP and Pi.

The relevance of this is considered in the next section.

Biosynthetic processes and ATP

It seems to me that the example of polynucleotide phosphorylase is especially pertinent because it is a biosynthetic process in which the nucleotide monophosphate is actually incorporated into the product.

[AMP]n + ADP → [AMP]n+1 + Pi

However this ADP-driven non-template dependent synthesis of polynucleotides is not model for template-dependent RNA synthesis, one reason being that it can also operate in the reverse direction as a phosphorolytic 3ʹ to 5ʹ exoribonuclease reaction:

[AMP]n+1 + Pi → [AMP]n + ADP

Irreversibility is an important feature of template-dependent synthesis of RNA¶, and it is thought that the key to this is that fact that ATP is the substrate and is cleaved between the α and β phosphates liberating pyrophosphate rather than monophosphate.

[NMP]n + NTP → [NMP]n+1 + PPi

The reason generally proposed to explain why this reaction is irreversible is that in the cell pyrophosphatase hydrolyses the pyrophosphate to phosphate in an exothermic reaction. However the problem in evolutionary terms is that pyrophosphatases would have to evolve before this would work. However, if the ∆G°′ for pyrophosphate release has a greater absolute value than that for phosphate release, this in itself would affect the equilibrium of the reaction in a manner favourable to synthesis, albeit at the greater effective cost of an extra ATP. The later evolution of pyrophosphatases would have served to reinforce and enhance the irreversibility.

The Hypothesis

The hypothesis is that nucleoside tri-phosphates developed early in evolution to provide a means of making nucleic acid synthesis essentially irreversible by a mechanism that generated pyrophosphate. As the example of polynucleotide phosphorylase shows, nucleoside di-phosphates do not allow this.

A priori it would seem that there is no reason why ATP rather than ADP should be preferred for coupling biochemical reactions. However, if one accepts the argument that ATP synthesis was required for nucleic acid synthesis, there would seem no reason for having a set of reactions generating ADP for other purposes. The energy of hydrolysis of ATP to ADP in such reactions would, in any case, have to be utilized, otherwise it would represent a net loss. Why go any further?

It may have been that ATP was preceded in evolution by an an earlier ‘energy donor’. However I would suggest that once a mechanism evolved for synthesizing and using ATP (and other NTPs) for synthetic purposes, it was more efficient to use this for metabolic purposes, largely displacing any previous energy donor molecule.

† Footnote 1

I have touched on this very tangentially in my answer to a quite different question about ATP. The common idea, however, is that the purine and pyrimidine rings of NTPs serve no purpose in energy transfer, but are a relic of their requirement in RNA synthesis, where the NTPs are not merely providing energy, but are incorporated into the RNA product.

¶ Footnote 2

RNA and DNA synthesis are not the only synthetic processes in which this strategy for irreversibility is adopted. The tRNA amino-acylation reaction — essential for protein synthesis is as follows:

Amino Acid + ATP → Aminoacyl-AMP + PPi

And the first step in glycogen synthesis is the formation of UDP-glucose:

Glucose 1-phosphate + UTP → UDP-glucose + PPi

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  • $\begingroup$ @user1136 — Thanks for that — have downloaded the paper and will read it. Sounds the sort of thing that must have been discussed in TIBS at some stage. It is interesting in that it would provide less reversible phosphodiester bond formation without the hydrolysis of pyrophosphate, which one might suppose developed subsequently in evolution to lock things down. I'll revise my answer when I have a moment. $\endgroup$
    – David
    Commented Dec 31, 2018 at 9:29
  • $\begingroup$ This is amazing, exactly what I was wondering! $\endgroup$
    – Sully Chen
    Commented Dec 31, 2018 at 10:19
  • $\begingroup$ Just updated as promised. To try to make the argument easier to follow I relegated certain points to footnotes. Happy to be informed of any typos or mistakes. $\endgroup$
    – David
    Commented Jan 1, 2019 at 21:12
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It can, and a very famous example is polynucleotide phosphorylase, an enzyme of great historical importance in the elucidation of the genetic code.

That said, it is very uncommon, and polynucleotide phosphorylase is the only example that I know of (but other users may be able to provide additional examples).

There is certainly no thermodynamic reason why the 'high-energy' bond of ADP cannot be used as a source of free energy.

Polynucleotide phosphorylase was discovered in Ochoa's laboratory by Mairanna Grunberg-Manago. The story goes (from memory) that Ochoa was very dismissive when Grunberg-Manago told him that the enzyme uses dinucleotide phosphates, rather than trinucleotide phosphates, and he told her that her conclusion was impossible. He later realized his mistake and apologized, and the biosynthetic activity of the enzyme provided a simple and elegant method for making short RNAs that were used to unravel the genetic code.

(I do not have the source of the above story 'to hand' (as I am on holiday). Perhaps someone can help me with a more authentic account of the story? Otherwise I'll 'dig out' the original source in the coming days).

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  • $\begingroup$ Wow, that's really cool! If there is no thermodynamic reason, is it just by chance that life didn't really evolve to use the ADP->AMP pathway that much? $\endgroup$
    – Sully Chen
    Commented Dec 30, 2018 at 3:16
  • $\begingroup$ ATP can be hydrolyzed to AMP and pyrophosphate and the pyrophosphate is then subsequently hydrolyzed to phosphate. This is used by some enzymes to ensure the irreversibility of their reactions. $\endgroup$
    – canadianer
    Commented Dec 30, 2018 at 8:55
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    $\begingroup$ To be clear, I wasn’t critiquing your answer but rather adding some additional information that might be helpful to... someone. $\endgroup$
    – canadianer
    Commented Dec 30, 2018 at 12:38
  • $\begingroup$ @SullyChen — I don't think so. See my answer. BTW user1136, could you consider changing your name to something more memorable. $\endgroup$
    – David
    Commented Dec 30, 2018 at 16:30
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ATP -> ADP is easier to break than ADP -> AMP. I dont think i can explain why without invoking third-year chemistry notions of electron orbitals ...

Conceptually, the terminal phosphate link on the triphosphate is more unstable than the terminal one on the dipeptide, and so breaking this bond and liberating this bond energy to do work is easier, and the amount of energy is a small and useful amount. Breaking high energy bonds is harder, and if the energy released is in excess of what is need it can cause undesired side reactions.

Metabolism is step-wise to keep the unit of energy at appropriate level.

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    $\begingroup$ Please quote standard Gibbs Free Energies of hydrolysis to support your statement. And perhaps you would like to explain what dipeptides have to do with this. $\endgroup$
    – David
    Commented Dec 30, 2018 at 9:40

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