Re: [AMBER] cpin file

From: Jason Swails <>
Date: Tue, 6 Oct 2015 12:14:09 -0400

On Tue, Oct 6, 2015 at 10:16 AM, Arjun Sharma <>

> Dear Jason,
> Could you please explain these terminologies in .cpin file from your
> explicit solvent CpHMD wikidot page and few other questions.
> STATEINF(0)%FIRST_CHARGE ? Is this total charge of the first residue ? I
> see 0, 95, 149 etc. in the list which didn’t make sense to me.

​It is the index of the first charge in the CHARGES array for that
residue. The number of charges that belong to that residue is going to be
natom * nstates, where natom is the number of atoms in that residue and
nstates is the number of defined titratable states for that residue.

STATEINF(0)%FIRST_STATE ? I see 0, 5 , 8 as values. Initially I thought
> its different states of a residue. But “8” ?

​This is an index into the STATENE array. Each residue has nstates
entries. So what this means is that the first residue has 5 states
(presumably deprotonated, then protonated on both oxygens in both the syn-
and anti- positions).

> STATENE=0,21.2486237123,21.2486237123,21.2486237123,21.2486237123…..what
> are these values ? Also why is first value zero ?

​The Delta G ref is calculated as Eref,1 - Eref,2. Since TI gives you a
single energy, what happens is that you give a 0 to the "deprotonated"
state and the full energy difference to the protonated state. So what this
essentially tells you is that the 4 protonated states have the same energy
(which they should), and the deprotonated state is 21.2486 kcal/mol *lower*
in energy (according to the pKa-adjusted TI energy).

> I calculated reference energies values for my model organic compound
> (CH3)3-C-COOH both in explicit and implicit solvent. Explicit solvation
> (IGB=0), I got -62.3070 kcal/mol (anti-) and -61.2687 kcal/mol (-syn). In
> Implicit solvation with O-radii = 1.5, I got -51.5209 kcal/mol (-anti) and
> -50.5030 kcal/mol (-syn) while changing O-radii to 1.3 gave me -59.8513
> kcal/mol (-anti) and -59.0642 (-syn). As you see values are not close for
> explicit solvation and implicit solvation (1.5 radii) and slightly closer
> with smaller radii (1.3). Does it makes sense, even though values of
> explicit and implicit (1.3 radii) don’t match ?

​I would have expected them to be slightly closer. But the syn- and anti-
reference energies should be identical (the difference in energies between
them should result in the appropriate population of syn- vs. anti-
protonation states that you see in experiment -- that's what the cpin file
you pasted shows).

> Also as you can see -anti is slightly favored over -syn. I’m not sure
> which one do I use as IGB2 value ? is it possible to use two values ?

​It's possible, but you should use the same value. Assign the same value
to all protonated states, then adjust that energy as needed to get the
correct pKa.​

In AMBER manual you specify that reference energy value calculated using TI
> is assigned to the unprotonated form and reference energy value to the
> protonated form is increased by adding pKaRTln(10) to that value. I’m not
> clear on this assigning part. Isn’t the reference energy the difference
> between two states ?

​It is pKa-adjusted. The TI energy gives you what the force field says is
the free energy difference between the two protonation states. However,
*experiment* tells you that the free energy difference is pKa*RT*ln(10), so
the reference energy is going to be the TI energy *minus* the
pKa*RT*ln(10), so that the force field sees the same free energy difference
as experiment. You just have to be sure that you keep your signs
straight. That's why the manual details the prescription "assign the
unprotonated form the reference energy and the protonated form increased by

> are reference energy and relative energy one and the same ?
> I ran explicit solvent CpHMD (same parameters as in TI calculations) at
> pH=pKa but I don’t see 50% protonation. It’s stuck at the starting residue
> state and there is no transition. I guess reference energy value is not
> right. But I don’t see anything wrong with the reference energy value
> calculated using sander TI method.

​You probably have a sign flipped somewhere. If this helps, using the TI
energy *exactly* as the reference energy results in a pKa of 0. So that
might be an easy place to start. Assign the TI energy as the reference
energy, then titrate around pH 0 to see if you get approximately pKa=0. If
not, try flipping the sign of the reference energy -- if you start seeing a
more reasonable pKa close to 0, then you obviously flipped the sign
somewhere by mistake. Once you have that working, you just add
RT*pKa*ln(10) to the protonated reference energy ( does this for
you when you specify "deprotonated" and the pKa of the titratable residue).

It takes awhile to really wrap your head around all of the signs and what's


Jason M. Swails
Rutgers University
Postdoctoral Researcher
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Received on Tue Oct 06 2015 - 09:30:03 PDT
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