AMBER Archive (2006)

Subject: Re: AMBER: For different ligand conformation

From: David Mobley (dmobley_at_gmail.com)
Date: Mon Jul 31 2006 - 17:29:40 CDT

Pravas,

You may need multiple starting structures. Depending on how different
the conformations are, interconversion times can be very long (on a
system I study, I've found timescales in excess of 5 ns for a rotation
of an aromatic ring by 60^o, and never (>>5ns) seen flip of a phenyl
ring around its axis of symmetry. MD tends not to work very well at
sampling such things in tight binding sites because energy barriers
are high.

Replica exchange may help somewhat, or you can just keep multiple
docking poses that are significantly different and use these for
starting orientations (let me know if you want a preprint of a paper
where I do this sort of thing for binding free energy calculations; it
should be in J Chem Phys in the next few weeks).

Thanks,
David

On 7/31/06, pkb bioinfo <pkstruct_at_gmail.com> wrote:
>
> Dear AMBER users,
>
> I am simulating a ligand in the protein.Experimentally it has been proven
> that the ligand stays in the active site
> in two different conformation and the catalysis give two different products.
>
> I did docking in Autodock3 and hoped during simulation two different
> conformation will be apparent.
> From the simulation one conformation is somwhow clear but the other
> conformation is not visible.
>
> So please guide me any procedure in which I will be able increase the
> flexibility of the ligand and I will be able to see
> the other conformation.These are the follwing AMBER input files for
> periodic boundary simulation.
>
> Initial mnimisation with cartesian restraints on solute
> &cntrl
> imin=1,maxcyc=200,
> ntpr=5,
> ntr=1,
> &end
> Group input for restarined atoms
> 100.0
> RES 1 500
> END
> END
>
> Minimization of the entire molecule
> &cntrl
> imin=1,maxcyc=200,
> ntpr=5,
> &end
>
> Heating up the system equilibration stage 1
> &cntrl
> nstlim=5000, dt=0.002, ntx=1, irest=0, ntpr=500, ntwr=5000, ntwx=5000,
> tempi =100.0, temp0=300.0, ntt=1, tautp=2.0, ig=209858,
> ntb=1, ntp=0,
> ntc=2, ntf=2,
> nrespa=2,
> &end
>
>
> Constant pressure constant temperature equilibration stage 2
> &cntrl
> nstlim=500000, dt=0.002, ntx=5, irest=1, ntpr=500, ntwr=500, ntwx=500,
> temp0=300.0, ntt=1, tautp=2.0,
> ntb=2, ntp=1, ntc=2, ntf=2,
> nrespa=1,
> &end
>
> Thanking you in advance
>
> Regards
> Pravas
>
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