Re: [AMBER] the better box size for calculating host-guest interaction energies

From: Investigador Química <>
Date: Thu, 20 Aug 2015 10:09:12 -0400

Dear Jason, thank you for your kind and clear explanation.
You are right. For the three isolated and solvated systems generated using
"solvateoct TIP3PBOX 11" we have:

            Box (x=y=z) triangulated 3-points waters sum of charges

H-G 64,477 6598 -0.99950000
H 51,248 3254 -0.16460000
G 37,612 1321 -0.03280000

My problem is how can I run the simulations with the water counts exactly
matched between the bound and unbound simulations?

In AMBER tutorial 21 the following values are used for solvatebox structure TIP3PBOX 16.50 iso solvatebox guest TIP3PBOX 13.16 iso solvatebox host TIP3PBOX 10.18 iso solvatebox b2host TIP3PBOX 9.91 iso

and manually they removed waters over 1500.

In my case I do'nt know how to choose the appropriated number of waters or
the numbers of the
TIP3PBOX Nr? iso

Could you please help me?

Thank you for your help and time!

Best regards

2015-08-19 11:18 GMT-04:00 Jason Swails <>:

> On Wed, 2015-08-19 at 10:43 -0400, Investigador Química wrote:
> > Dear Jason, thank you for your kind explanation. I recognize my first
> > question was not fortunate. Please let me to explain my question.
> > If I have the Etot for a host H, for a guest G and for the complex H-G, I
> > can calculate the interaction energy as E= E(complex H-G) - E(H) - E(G),
> > isn't?.
> Loosely speaking, yes. However, there are countless ways to do this
> wrong. For example, if the sum of the host and guest systems do not
> have exactly the same number (and *kind*) of atoms and molecules as the
> bound complex, then this energy difference is completely arbitrary (and
> meaningless). After all, energy is extensive, so you can tune the
> interaction energy to be any value you want by simply adding solvent to
> either the bound or unbound species and not to the other.
> There are many ways to compute the interaction energies (MM/PBSA is one
> example, and the linear response theory/linear interaction energy is
> another). Whichever you use must make sure that the only energies that
> are left after taking the difference are the interactions between the
> atoms in the host and the atoms in the guest.
> > If I get the energies for the three entities using solvateoct and
> > TIP3PBOX from 8 to 11 each time (for the three entities each time) I get
> > different interaction energies. Should the interaction energy best value
> be
> > the lower one?
> No. This is known as the variational principle and is common in QM
> calculations (i.e., a method is variational if a lower energy *ensures*
> a better approximation of the true wavefunction). Force fields are
> *certainly* not variational, and even variational QM Hamiltonians won't
> be variational in an interaction energy calculation like this.
> > How can I know that, if I have no experimental values to
> > compare with?
> You can't. You have to use your judgement and rationalize which model
> for computing the interaction energy is the best.
> > For instance these were the values I got when running 3 ns
> > of MD equilibration according to AMBER tutorial 1 section 5:
> 3 ns is a very short simulation.
> > box 8: E= -11067 kcal/mol
> > box 9: E= -10363 kcal/mol
> > box 10: E= -8276 kcal/mol
> > box 11: E= -6485 kcal/mol
> These are huge differences. You haven't described exactly how you are
> computing the interaction energies, but this looks to me like the fewer
> particles you use, the lower your energy gets. This can happen if you
> have 3 separate systems (H-G, H, G) each independently solvated and you
> take the energy of each one and subtract them. With an 11 A buffer, H+G
> will have a *lot* more water-water interactions between the two systems
> than the HG system compared with, say, an 8 A buffer. Which means that
> the total interaction energy will *appear* lower, because your
> difference includes more interactions than *just* those between your
> host and guest.
> Possible differences arising from box size effects will be related to
> periodic images interacting with each other, different conformational
> ensembles caused by either incomplete sampling or periodicity artifacts,
> and some net charge effects if you have a non-neutral unit cell with
> periodic boundary conditions. However, I wouldn't expect any of those
> effects (excluding incomplete sampling) to be larger than ~10 kcal/mol,
> let alone up to 5000!
> HTH,
> Jason
> --
> Jason M. Swails
> BioMaPS,
> Rutgers University
> Postdoctoral Researcher
> _______________________________________________
> AMBER mailing list

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Investigación Facultad de Química
Universidad de Santiago de Chile
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Received on Thu Aug 20 2015 - 07:30:04 PDT
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