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

From: Jason Swails <>
Date: Wed, 19 Aug 2015 11:18:06 -0400

On Wed, 2015-08-19 at 10:43 -0400, Investigador QuĂ­mica 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!


Jason M. Swails
Rutgers University
Postdoctoral Researcher
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Received on Wed Aug 19 2015 - 08:30:04 PDT
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