From: yp sun (sunyeping_at_yahoo.com.cn)
Date: Sun Feb 24 2013 - 19:22:08 CST
Dear NAMD users,
I am now studying how to calculate binding free energy using NAMD. I find an answer in namd-I by LEWIN LI as following:
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)
I wonder how this method can be realized: how to take out the ligand from the protein-ligand complex and put the ligand somewhere fairly far away from the protein (using what software)? should the size of the water box be the same for calculating E(P-L) and E(P + L)? Should ions be added to neutralized both of the two systems?
Thanks in advance!
This archive was generated by hypermail 2.1.6 : Wed Dec 31 2014 - 23:20:56 CST