AMBER Archive (2003)

Subject: Re: charge neutralization in DNA

From: sychen (yuann_at_bioinfo.ndhu.edu.tw)
Date: Thu Mar 20 2003 - 10:52:38 CST

On Tue, 18 Feb 2003 14:27:20 -0800
"David A. Case" <case_at_scripps.edu> wrote:

> On Tue, Feb 18, 2003, Mahadevan Seetharaman wrote:
> >
> >
> > I have two positively charged N-groups in my DNA 10-mer (duplex). When I
> > do "addions unit Na+ 0", only 16 sodium ions are added (understandable,
> > since the net charge is -16, 18 phosphates and two N+). I think the system
> > should have 18 Na+ ions and 2 Cl- ions instead of 16 Na+ ions. If I want
> > to add 18 Na+ ions and 2 Cl- ions to this system, how would I do it, apart
> > from doing it manually.
> >
> You should be able to do the following:
>
> addIons x Na+ 18
> addIons x Cl- 2
>
> I don't know if you think this is doing it "manually" or not, but I don't
> think there is any simpler approach.
>
> ..good luck...dac

Right. So if I want to simulate DNA in vacuum, I have to modified the
partial chages of phosphorus and phosphate oxygens.
As the follow from all_nucleic94.lib,
          D(ATCG) D(ATCG)5 D(ATCG)3
P 1.1659 1.1659
O1P -0.7761 -0.7761
O2P -0.7761 -0.7761
O5' -0.4954 -0.6318 -0.4954
O3' -0.5232 -0.5232 -0.6549

base
net -1 -0.3079 -0.6921
chg

P group
net chg -1.4049 -1.155 -1.5366

Should I reduce the net charge on the phosphate group to
-0.4049, -0.8471, and -0.8445 respectively? And how to decide
the extent for reducing the charge on each atom?

Thank you for any ideas.

Best Regards,

sychen