Reparameterizing EEF1 for a new VDW cutoff - 08/03/05 02:35 PM

Hi All,

This question is probably for an expert in EEF1. I need to use a larger VDW cutoff in my simulations, while using EEF1. The problem is this, the EEF1 solvation free energies were parameterized with a 9 angstom VDW cutoff, so if I change the cutoff I need to reparameterize the parameters used in the solvation free energie calculations.

This is how I reparameterized them:

1) I recalculated the VDW energy correction from 16 angstroms (my cutoff) to Infinity (~0.04 kcal/mol)

2) Subtracted 0.16 kcal/mol from G,ref, G,free, and H,ref in solvpar.inp, since Lazaridis had added 0.20 initially to G,ref and H,ref.

3) I did not subtract 0.16 from atoms that had G,ref and G,free = 0, or from NC2, NH3 or OC.

Note, the reason I subtracted the value from G,free also is because I assume this will keep the buried atoms at a solvation free energy of 0.

Did I carry out this reparameterization correctly?

Thanks,

Ed

PS - I see the alpha values are a function of G,free, but I assume these are not hard-coded, and will be calculated with the new G,free values.

This question is probably for an expert in EEF1. I need to use a larger VDW cutoff in my simulations, while using EEF1. The problem is this, the EEF1 solvation free energies were parameterized with a 9 angstom VDW cutoff, so if I change the cutoff I need to reparameterize the parameters used in the solvation free energie calculations.

This is how I reparameterized them:

1) I recalculated the VDW energy correction from 16 angstroms (my cutoff) to Infinity (~0.04 kcal/mol)

2) Subtracted 0.16 kcal/mol from G,ref, G,free, and H,ref in solvpar.inp, since Lazaridis had added 0.20 initially to G,ref and H,ref.

3) I did not subtract 0.16 from atoms that had G,ref and G,free = 0, or from NC2, NH3 or OC.

Note, the reason I subtracted the value from G,free also is because I assume this will keep the buried atoms at a solvation free energy of 0.

Did I carry out this reparameterization correctly?

Thanks,

Ed

PS - I see the alpha values are a function of G,free, but I assume these are not hard-coded, and will be calculated with the new G,free values.