From: Vani Krishna (vakri2002_at_yahoo.com)
Date: Thu Sep 02 2004 - 21:13:58 CDT
The method suggested by Lewyn is simple but I was
wondering how this method will work in case of two
strongly charged molecules?
and also if the binding energies are small, may be it
will be lost in the fluctuation part of the total
and also computaionally in charged systems, to
counteract the slow decay of charges, the box has to
be big and most of the computational time will be
wasted in simulating just water.
E(A+B+water) - E(A+water)-E(B+water) for binding
energy of AB
can reduce the computational cost, but there might
new problems of adding counterions for neutralising
the total charge. Then one has to seperately calculate
the effect of ions and the subtract them.. which seems
a bit painful.
Does anybody have some suggestions in this regard,
particularly charged systems?
Also is there a way to make NAMD output some 'intra'
and 'inter' energies of certain sets of atoms?
--- LEWYN LI <ll2150_at_columbia.edu> wrote:
> On Thu, 2 Sep 2004, ying xiong wrote:
> > Dear sir,
> > Could you please tell me whether NAMD program
> can calculate the bind
> > energy or binding free energy of protein and
> ligand ? If yes, how to
> > calculate it?
> > ying xiong
> > yxiong_at_mail.ccnu.edu.cn
> > 2004-09-02
> I presume by "binding (free) energy", you mean
> that change in
> (free) energy for the following reaction:
> protein (in water) + ligand (in water) ---->
> protein-ligand complex (in
> Here is my somewhat naive suggestion:
> I would try to compute the binding energy first by
> solvating the
> protein-ligand complex in a water box. You would
> need the x-ray or NMR
> structure of the protein-ligand complex, as well as
> a box of water
> molecules to do this. Once the system has
> equilibrated, you could just
> record the total energy. Call it E(P-L).
> Then you can take out the ligand from the
> protein-ligand complex
> and put the ligand somewhere fairly far away from
> the protein, in order to
> reduce any interaction between the protein and the
> ligand. Use a large
> water box if necessary. However, you should make
> sure that the ligand is
> still inside the water box. You can then
> re-equilibrate and re-compute
> the total energy. This value will now represent, in
> an approximate
> way, the total energy of the solvated protein and
> the solvated ligand.
> Call this E(P + L).
> You can estimate the binding energy simply as:
> E(binding) = E(P-L) - E(P + L)
> Of course, this estimate may or may not be close
> to experiments,
> because, in experiment, you usually determine the
> equilibrium constant and
> then calculate the binding FREE energy as dG =
> I think the binding free energy is more
> complicated to calculate,
> because you would need to determine the change in
> entropy of binding,
> which is a non-trivial task.
> Hope this helps.
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