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Clarification on why vacuum simulation needed for hydration free energy #1310

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zanebeckwith opened this issue Dec 12, 2023 · 2 comments
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@zanebeckwith
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Hi, thanks for the great project!

I was a bit surprised to see that the examples here (and in the yank-examples repo) for hydration free energy include not one thermodynamic leg (decoupling the solute from the solvent), but two (also decoupling in vacuum). I was hoping you could help me understand.

Is this because Yank decouples (i.e. turns off only intermolecular interactions) the sterics but annihilates (i.e. turns off inter- and intra-molecular interactions) the electrostatics? And so the vacuum simulation is needed to, effectively, turn back on the intramolecular Coulombic interaction?

Is my understanding correct that, if the openmmtools.AlchemicalRegion were created here with annihilate_electrostatics=False, then there would be no need for the vacuum simulation at all? Or is there something else I'm missing?

Sorry for the ignorant question! Thanks for the help!

@Lnaden
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Lnaden commented Dec 12, 2023

The short answer is that both free energy transformations are needed to get the complete "free energy of hydration." You do get a free energy difference from the "solvent phase" (decoupling solute from solvent), but that transform only moves the solute into a state where its technically in a box, but that box is still in pressurized, volumetric space with periodic copies of itself still being computed. A true, full hydration free energy takes the free energy from the "solute fully by itself with nothing around it and unconstrained" (vacuum) and insert it into "solute surrounded by solvent."

Its true if you only Decouple Electrostatics, then solute-solute type interactions are still left on. If you Annihilate Electrostatics, the solute-solute electrostatics are turned off; however, that also isn't the true vacuum state as the solute needs to have all of its intra-molecular interactions on to be accurate.

The long answer is here on the theory site. It only covers the Binding free energy example, but all the theory still applies as you can think of a "receptor in solvent" as just a "very complicated solvent" for this purposes. Or in reverse, a hydration free energy is just a binding free energy with a very simple receptor. Its not a perfect analogy, but for high-order theory, it translates well enough.

@zanebeckwith
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Thanks @Lnaden, I appreciate you taking the time to help! And sorry for the long delay on my response!

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