Re: [Apbs-users] Electrostatic Free energy of folding in water (per residue)

[Nathan B. <nat...@ma...>]

Hi Steve --

> I am still missing something.
> Lets see if this diagram helps:
>
> Folded State in water (Faq)
> Unfolded State in water (Uaq) - residue from the pqr file without =20
> changing its position
> Folded State Vacuum (Fv)
> Unfolded State Vacuum (Uv)
>
>                        DG,elec,v
>                     Uv  <--> Fv
>    DGu,solv   | |            ||  DGf,solv
>                      Uaq <--> Faq
>                         DG,elec,aq
>
> DGf,solv =3D Faq-Fv (where values for each state are from APBS)
> DG,elec,v =3D Fv-Uv
> DGu,solv =3D Uaq-Uv

This seems reasonable.

> I think DG,elec,aq should be on the order of ~1kcal/(mol res).
> Calculating directly gives values that are approximately 100 times =20
> too large in magnitude.
> =46rom your e-mail, I think you are saying:
>
> DG,elec,v+DGf,solv-DGu,solv

Yes, this is how I would calculate it.

> But this gives the same value because the vacuum terms cancel.
> (i.e.  Fv-Uv+Faq-Fv-Uaq+Uv=3DFaq-Uaq)

Be careful about this cancellation.  Coulombic interactions aren't =20
reproduced accurately by PB solvers due to grid artifacts and the =20
inclusion of self interactions.  Removal of these self-interactions =20
through the free energy cycle I mentioned is *almost always* =20
necessary for accurate answers.  The symptom of failing to remove =20
these interactions are very large energies.

> It seems I am misunderstanding something obvious, but I don=92t know =20=

> what.
>
> Oddly,
> DGf,solv-DGu,solv
> is only ~10 x too large in magnitude.

This quantity should calculated accurately by APBS.  How do your =20
parameters (radii, surface definitions, dielectric constants, etc.) =20
compare to the paper you're benchmarking against?

Thanks,

Nathan

> Thanks again,
> Steve
>
> I am including the .in files for the vacuum calculations.  The =20
> aq .in files have not changed.
>
> In file for protein in vacuum:
> read
>     mol pqr last_1.pdb.pqr
> end
> elec name 1
>     mg-auto
>     dime 129 129 97
>     cglen 82.9787 75.6908 64.4589
>     fglen 68.8110 64.5240 57.9170
>     cgcent mol 1
>     fgcent mol 1
>     mol 1
>     lpbe
>     bcfl sdh
>     pdie 2
>     sdie 2
>     srfm smol
>     chgm spl2
>     sdens 10.00
>     srad 1.40
>     swin 0.30
>     temp 298.15
>     gamma 0.105
>     calcenergy comps
>     calcforce no
> end
> print energy 1 end
> quit
>
>
> in file for residue in vacuum:
> read
>     mol pqr last_1.pdb.pqr
>     mol pqr last_1_res_1.pqr
> end
> elec name 1
>     mg-auto
>     dime 129 129 97
>     cglen 82.9787 75.6908 64.4589
>     fglen 68.8110 64.5240 57.9170
>     cgcent mol 1
>     fgcent mol 1
>     mol 1
>     lpbe
>     bcfl sdh
>     pdie 2
>     sdie 2
>     srfm smol
>     chgm spl2
>     sdens 10.00
>     srad 1.40
>     swin 0.30
>     temp 298.15
>     gamma 0.105
>     calcenergy no
>     calcforce no
> end
> elec name 2
>     mg-auto
>     dime 129 129 97
>     cglen 82.9787 75.6908 64.4589
>     fglen 68.8110 64.5240 57.9170
>     cgcent mol 1
>     fgcent mol 1
>     mol 2
>     lpbe
>     bcfl sdh
>     pdie 2
>     sdie 2
>     srfm smol
>     chgm spl2
>     sdens 10.00
>     srad 1.40
>     swin 0.30
>     temp 298.15
>     gamma 0.105
>     calcenergy comps
>     calcforce no
> end
> print energy 1 end
> quit
>
> On Oct 1, 2006, at 5:13 PM, Nathan Baker wrote:
>
>> Hello --
>>
>> Based on the input files you've included, it looks as if you're not
>> correctly removing self-energies.  =46rom your e-mail, it seems =
you're
>> interested in the difference in electrostatic energy between two
>> configurations.  Since the configurations are different, you can't
>> rely on cancellation of self-energies between the two calculations.
>> Instead, what you want to do is calculate solvation energies for each
>> of the configurations and then take the difference of these solvation
>> energies.  This difference (the change in solvation energy due to un/
>> folding) can be added to the difference in Coulombic energy
>> (electrostatic energy in a homogeneous dielectric) change to give the
>> total electrostatic energy change.
>>
>> Does that make sense?
>>
>> Thanks,
>>
>> Nathan
>>
>

--
Assistant Professor, Dept. of Biochemistry and Molecular Biophysics
Center for Computational Biology, Washington University in St. Louis
Web: http://cholla.wustl.edu/