1
$\begingroup$

We have recently postulated that ATP generated by substrate-level phosphorylation may have a different phosphorylation potential to the ATP generated by oxidative phosphorylation. This could be because ATP generated via ATP synthase is released into a more basic environment than the cytoplasm.

My question is: Has anybody measured the phosphorylation potential of ATP during the growth of E. coli under anaerobic conditions, and, if so, how does it compare with the phosphorylation potential of ATP during the growth of E. coli under aerobic conditions?

$\endgroup$
9
  • 2
    $\begingroup$ Please tell us the results of your own research to try to answer this question. $\endgroup$
    – David
    Commented Feb 3, 2022 at 20:18
  • 1
    $\begingroup$ @David my cynical side says this question is purely to raise the profile of the OPs paper linked in the question. $\endgroup$
    – bob1
    Commented Feb 3, 2022 at 22:44
  • 1
    $\begingroup$ @bob1 — One has to take questions of this sort at face value, and respond by making sure they fulfil the usual prerequisites. Anyone trying to publicize their own published science on this list — especially on this type of subject — is wasting their time, in my opinion, as I think they are addressing the wrong audience. But that’s their problem. $\endgroup$
    – David
    Commented Feb 3, 2022 at 22:57
  • 1
    $\begingroup$ Do you have an open-access version of your publication? $\endgroup$
    – user338907
    Commented Feb 3, 2022 at 23:04
  • 1
    $\begingroup$ @MansiEl-Mansi You are probably the best person to answer your own question. This site is for biologists as well as general public but there are only a handful of experts who study different fields of biology. You ask a question about a very specific topic (that too asking for literature evidence for something). You have better chances of finding an answer yourself than relying on users here, because this is your research topic. Did you check out google scholar? $\endgroup$
    – WYSIWYG
    Commented Feb 8, 2022 at 12:14

1 Answer 1

2
$\begingroup$

Based on the way this question is phrased, the answer is trivially no. No matter how you make an ATP molecule, if you measure hydrolysis in a consistent manner there will be no difference.

A fundamental axiom in chemistry is that identical molecules are indistinguishable from each other. Specifically, I mean that the entropy and enthalpy are State Functions - they only depend on the current state of the system and not how the system got there. This is best summed up with Hess' Law.

The only things that could change the free energy of hydrolysis of an ATP molecule are:

  1. A change in the isotopic makeup of the ATP molecule
  2. A change in the environment in which the hydrolysis is occuring

The first is very, very unlikely to be dependent on aerobic vs. anaerobic respiration. I mean I guess the isotopic makeup of the oxygen could be ever-so-slightly different or something, but I would be shocked if it was. Even then it almost certainly wouldn't make much of a difference in energy of hydrolysis.

The second, is actually very likely to occur. Changes in temperature will always affect any reaction for which the change in entropy is non-negligible (remember $ \Delta G = \Delta H - T \Delta S $). As you suggest, changes in pH would certainly effect a measured change in energy due to a change in the abundance of the various protonated forms of ATP (and keep in mind hydronium if you're considering spontaneous hydrolysis). Indeed, almost any change in the environment could affect the measured energy.

But, importantly, these are not due to differences in the actual ATP molecule. These are due to differences in the environment that the reaction is taking place in, not the environment it was produced in. If you take two ATP molecules and place them in an identical environment they will always have the same energies (isotopic differences excluded). Any evidence contradicting this is either wrong or a suggestion of a fundamental paradigm shift upending years of basic physics and chemistry.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .