Re: [AMBER] Equilibration phase and RMSD

From: Carlos Simmerling <carlos.simmerling.gmail.com>
Date: Thu, 11 Sep 2014 11:05:56 -0400

thanks, it wasn't clear since you said it could form a homodimer or
heterodimer. if you're simulating the entire system that is in the pdb file
it should be reasonably stable, at least for a while. Of course it needs
careful equilibration. We normally do this in many steps. when you say that
the RMSD increases linearly, can you tell us what the input file is for the
step when it increases, and what the values of the RMSD are during that
step?

On Thu, Sep 11, 2014 at 10:56 AM, Ibrahim Said <saidibrahim569.gmail.com>
wrote:

> Thank you all
> My protein is a pdb file and it has x-ray coordinates and its resolution is
> 2.10 A. The forcefield is ff12SB and TIP3P water model is the solvent. I
> hope I can find a help. I will start to change the forcefield may it is the
> cause as I read in internet.
>
> Ibrahim
>
>
> On Thu, Sep 11, 2014 at 2:43 PM, Carlos Simmerling <
> carlos.simmerling.gmail.com> wrote:
>
> > Can you tell us more about the initial structure? Where did it come from?
> > Is there anything unusual about your setup?
> > On Sep 10, 2014 4:22 PM, "Ibrahim Said" <saidibrahim569.gmail.com>
> wrote:
> >
> > > Dear Amber users,
> > > I am using a a subunit of a protein can form a homodimer or
> heterodimer.
> > I
> > > want to characterize the flexibility of its N-terminal and C-terminal
> > > domains.
> > > My problem is The RMSD during equilibration phase increases approx.
> > > linearly, which may indicate unfolding of the protein . I ran
> > equilibration
> > > for 1 ns. Please, any one has sugestions.
> > > The potocol of heating is
> > > &cntrl
> > > imin = 0, irest = 0,
> > > ntb = 1, ntx = 1,
> > > cut = 10, ntr = 0,
> > > ntc = 2, ntf = 2,
> > > tempi= 0.0, temp0 = 298.0,
> > > ntt = 3, gamma_ln = 1.0, ntp = 0,
> > > nstlim = 100000, dt = 0.002,
> > > ntpr = 500, ntwr = 500,
> > > ntwx = 500,
> > > /
> > > Keep solute fixed with weak restraint
> > > 500.0
> > > RES 1 181
> > > END
> > > END
> > >
> > > The protocol for equilibration is
> > > &cntrl
> > > imin = 0, irest = 1,
> > > ntb = 2, ntx = 5,
> > > cut = 10, ntr = 0,
> > > ntc = 2, ntf = 2,
> > > temp0 = 298.0,
> > > ntt = 3, gamma_ln = 1.0, ntp = 1,
> > > nstlim = 100000, dt = 0.002,
> > > ntpr = 500, ntwr = 500,
> > > ntwx = 500,
> > > &END
> > >
> > > All the best
> > > Ibrahim
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Received on Thu Sep 11 2014 - 08:30:02 PDT
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