From: Jérôme Hénin (jhenin_at_ifr88.cnrs-mrs.fr)
Date: Wed Dec 15 2010 - 09:31:20 CST
Dear Ajasja,
> Regarding the presentation of a 5D space. It is indeed a problem. My
> current<http://www.ijs.si/ijs/dept/epr/ajasja/mails/mtssl/5D.png>solution
> is to draw all the 2D projections as a density plot (here I'm
> showing the histogram of angles during the simulation, but a 5D PMF could
> be presented in the same manner). But this is not very useful, except
> perhaps for spotting differences between two such sets of projections.
I find the 2d projections useful to detect correlated degrees of
freedom, but that only applies once there is sufficient sampling. For
example, in your graphs, chi3 and chi5 seem mostly uncorrelated.
> Regarding the overlap of the sidechain with the backbone. If I
> understood correctly then all the bins along the given reaction coordinate
> are supposed to be evenly sampled. But If the backbone is fixed and if using
> 15 degree bins, then such
> conformations<http://www.ijs.si/ijs/dept/epr/ajasja/mails/mtssl/overlapMTSSL.png>are
> inevitable (otherwise the sampling is not uniform).
> The backbone is green and the label is shown in red. The angles are -22,
> 123, -80, -87, 95 degrees.
> Inside the width of the bin such overlap can not be avoided. So does this
> mean that such bins are simply not sampled?
That is correct. In fact, the theoretical outcome of uniform sampling
may become unattainable when diffusion is very slow, the RC range is
too large, or the dimension is greater than one. In any case, really
forbidden areas (e.g. major steric clashes) remain forbidden - if the
bias ever becomes strong enough to overcome those, the simulation is
likely to crash.
> I have knowledge of (1D) relaxed dihedral
> scans<http://www.ijs.si/ijs/dept/epr/ajasja/mails/mtssl/mtssl_dihedral_potentials.png>around
> each single bond. With the added complication, that the Chi5 scan is
> very dependent on the value of Chi3. For the Chi3 angle there are two
> distinct minima, so it should perhaps be possible to split this in to two
> (or few) 4D runs.
>
> I haven't really thought about choosing any other collective variables. The
> projections of the CA-O could be interesting, but as pointed out by the above
> reference <http://pubs.acs.org/doi/full/10.1021/ct100413b> "simple, yet
> hand-picked, collective variables, as a set of chosen distances, turn out to
> miss some relevant features of the free energy surface. " Since I don't yet
> know what the relevant features of my landscape are, this could be a
> problem.
If I were you, I would start playing around with 2d surfaces and see
what can be resolved. Of course, the major question is what
information exactly you want to extract from these simulations.
> This is not directly connected, but it has been buging me anyway:
> Could a 1D relaxed dihedral scan be done using NAMD's CV module? (if the CV
> module is active during minimization this could be done using harmonic
> restraints and SMD.
Interesting that you should ask now. The code update that makes this
possible went into the public CVS just yesterday. Feel free to test
it. Here is an example input for a relaxed scan (assuming you have
defined a dihedral angle colvar named 'chi'):
harmonic {
colvars chi
centers -70
targetCenters 290
targetNumSteps 10000
targetNumStages 36
forceConstant 0.5
}
This will run a full 360 degree scan, starting from -70 in 10 degree
intervals, running 10000 steps of minimization at each stage. Don't
forget to subtract the restraint energy (filed under MISC in the NAMD
output) from the total potential energy to get the real force field
energy.
> A relaxed dihedral scan is basically the change in potential energy (dU)
> with regard to the chosen dihedral angle. By running an ABF on the same
> dihedral one obtains the change in free energy (dG).
> So is the difference between the two equal to TdS? (Using G = U + pV - TS
> and neglecting the pV term and any electrical work at constant temperature)
You are also assuming that the change in average energy is mirrored by
the change in minimum energy.
Cheers,
Jerome
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