calculations with eQuilibrator when solid compounds are involved

[Karel Olavarria]

Dear colleagues,

I have a “simple” question about eQuilibrator. If anyone can give me a short answer here, it will help me.
I am studying a metabolism where one metabolite is the solid state. Because the concepts of "concentration" or "activity" do not make sense in the case of solids, I am still capable to calculate the MDF for pathways where this metabolite participates. However, I realized that the obtained MDF result depends on the "concentration" value that I assign to this solid compound, even when I know that this conceptually not correct. My operative solution so far is to constrain the concentration of this solid compound to 1 M such that it does not have an impact on the calculation of Gibbs energy variation, but I am not sure if this is a correct approach. Could you recommend me some solution for this problem?

Thank you very much in advance,

Karel Olavarria

[Elad Noor]

Hi Karel,

It's not very easy to answer in general, but I will make a few assumptions and you can let me know if it's relevant or not in your case.

Solid compounds are not usually available for enzymes, so it's unlikely that your metabolite is an intermediate of the pathway. So, it is safe to assume it is the end product.

Maybe it's something that can precipitate (or a polymer).

I think the best option is to ignore the reaction where this metabolite becomes solid, and assume that it is practically irreversible.

E.g. if the pathway produces the monomer, which then is polymerized, I would just end the MDF analysis with the monomer. If it is some kind of precipitation process this is even more straightforward.

I hope this is helpful, if this does not apply to your case, I'll be happy to discuss further but I'll need more details probably.

Best, Elad

[Karel Olavarria]

Dear Elad,

 Thank you for your answer. 

 In the case I am studying, the solid is elemental sulfur (a substrate), and the pathway I am analyzing (sulfur disproportionation) is rather complex and with many unknown details.

 Elemental sulfur is simultaneously oxidized (to H2SO4) and reduced (to H2S). Oxidation occurs inside the cells and releases electrons that drive the extracellular reduction. ATP conservation is possible, and this conserved energy drives the CO2 fixation (outside of the scope of this discussion, but still interesting).

 I am representing the oxidation as follows:

 dimethyldisulfide + s_sulfanylglutathione <=> dimethyltrisulfide + glutathione

dimethyltrisulfide + 3 H2O + 2 NAD <=> 2 NADH + HSO3 + dimethyldisulfide
HSO3 + AMP + quinone <=> APS + quinol
APS + PPi <=> H2SO4 + ATP
2 NADH + 2 quinone + 2 ADP + 2 Pi <=> 2 NAD + 2 quinol + 2 ATP + 2 H2O

 If all these previous reactions proceed with a stoichiometric factor of 0.75, we will have the following net oxidation reaction:

 0.75 H2O + 1.5 ADP + 1.5 Pi + 0.75 PPi + 0.75 AMP + 2.25 quinone + 0.75 s_sulfanylglutathione <=> 2.25 ATP + 0.75 glutathione + 0.75 H2SO4 + 2.25 quinol

 On the other hand, the extracellular oxidation can be represented as:

 3 S + 3 H2O = HSO3 + 2 H2S     (stoichiometric factor 1)

HSO3 + 2 quinol + glutathione <=> s_sulfanylglutathione + 2 quinone + 3 H2O         (stoichiometric factor 1)

s_sulfanylglutathione + quinol <=> glutathione + H2S + quinone     (stoichiometric factor 0.25)

Thus, the net reduction process is:

 0.75 glutathione + 3 S + 2.25 quinol <=> 2.25 quinone + 0.75 s_sulfanylglutathione + 2.25 H2S

and the global disproportionation (dismutation) reaction is:

 3 S + 3 H2O + 0.75 H2O + 1.5 ADP + 1.5 Pi + 0.75 PPi + 0.75 AMP <=> 2.25 H2S + 0.75 H2SO4 + 2.25 ATP

 The mobilization of the elemental sulfur is not catalyzed by enzymes:

3 S + 3 H2O = sulfite + 2 H2S

This is consistent with your explanation ("Solid compounds are not usually available for enzymes"). However, even when the reaction is not catalyzed by an enzyme, I guess that thermodynamic laws constrain the feasibility space of this reaction and the feasibility space of the whole process. To tackle this problem, I am fixing the concentration of elemental sulfur to 1 M (the same treatment of H2O). Is that correct?

If this “operative” solution is not correct, what do you recommend?

Thank you very much in advance,

Karel Olavarria

[Robert Giessmann]

Hi there!

Exciting question! I guess that sulphur is dissolved in water, though very little ... I found a value of 1.9(±0.6) × 10−8 mole S8·kg−1 here: https://www.tandfonline.com/doi/pdf/10.1080/03086647808069875

Maybe you can check other sources, too. Or maybe you can find something with people who studied the reaction " 3 S + 3 H2O = sulfite + 2 H2S" in detail?

Best of luck / success with your project!
   Robert

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[Elad Noor]

Hi Karel and thanks Robert for your input,

Specifically for sulfur, we have an exception for having it in a solid phase. Elemental sulfur is actually one of the anchor points of the Gibbs formation energy definitions, so it will always have a formation energy of exactly 0 kJ/mol (in it's solid phase).

Fixing it to 1M is the correct way to go (it's not actually 1 M, it's just a "hack" for things that are not in solution).

I'm not sure about the biochemistry, but the reaction "3 S + 3 H2O = sulfite + 2 H2S" should be okay. If you think it is causing bugs, maybe you can just skip it and start directly with sulfite.

Best,

Elad