# Re: Solvation free energy difference of alchemical transformation of K+ into Na+

From: Hannes Loeffler (Hannes.Loeffler_at_stfc.ac.uk)
Date: Tue Jan 10 2017 - 06:15:02 CST

It would be useful it you kept the discussion in one thread.

Your attachment suggests that you plot λ vs <U> but your really need
<∂U/∂λ>. Is that what it is? How have you extracted this?

Maybe you send all relevant input files (control file, parmfile,
coordinates/alchfile) and output files (alchOutFile) so that somebody on
the list can take a look.

On Tue, 10 Jan 2017 16:39:01 +0530
Abhishek Kumar Singh <bo13m1002_at_iith.ac.in> wrote:

> Hello,
>
> I have alchemically transformed K+ to Na+ at the center of water
> boxes (40, 60 and 80 A edge length respectively) using NAMD software
> and CHARMM36 forcefield. A harmonic spring of force constant = 10
> kcal/mol.A^-2 has been applied throughout the transformation. Temp
> and Pressure has been kept at 310K and 1 atm respectively.
> Free energy of alchemical transformation has been calculated using
> thermodynamic integaration. The derivative has been computed as
> (∂G/∂λ) = ⟨ U(λ=1)- U)(λ=0) ⟩λ, where λ is the coupling parameter
> such that λ=1 and λ=0 are K+ and Na+ vdw parameters respectively.
> Free energy derivatives has been computed for 11 λvdw values ( 1.0,
> 0.9, 0.8, 0.7, 0.6, 0.5, 0.4, 0.3, 0.2, 0.1 and 0.0). The area under
> the plot (derivate vs λ) gives me = -23.40, -23.46 and -23.4
> ​8​
> kcal/mol for the box sizes of 40A, 60A and 80A respectively. I am
> attaching alongwith the derivate vsλ values for 40A water box.
>
> The standard CHARMM36 prm file actually states that the value should
> be -18.6 kcal/mol.
> So what can be reason for this difference. Is there any problem in
> this procedure ?
> Your suggestions are most welcome.
>
> Sincerely
> Abhishek​

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