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

From: Adrian Roitberg <>
Date: Thu, 20 Aug 2015 12:41:52 -0400

I am also confused as to why your TOTAL charge is not the same as the
sum of H and G charges... Unless you are changing the number of ions,
which would make it very hard to compute what you are trying to compute.


On 8/20/15 10:09 AM, Investigador Química wrote:
> 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

Dr. Adrian E. Roitberg
Department of Chemistry
University of Florida
AMBER mailing list
Received on Thu Aug 20 2015 - 10:00:04 PDT
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