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Am 20.10.2011 15:01, schrieb Mark Abraham:
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On 20/10/2011 7:25 PM, Vedat Durmaz wrote:
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thanks mark. i guess it took some time answering all these
questions. and i think you are right. trying to interpret each
computed energy term seperately in a physical manner is
senseless. especially since i'm not really deep inside the force
field & implementation stuff. however, with one of your
statements i cannot agree at all: <br>
<br>
imho, free energy diff's on the basis of two conformations and
their steady state distribution are independent from both the
energy and type of the intermediate state. it's only the
conversion's rate that does depend on it. <br>
</blockquote>
<br>
I'm not sure with which statement of mine you are disagreeing. </blockquote>
<br>
actually with this one: "Measuring the free energy difference with
simulations is hard because you cannot model the intermediate stages
of bond breaking and forming."<br>
<br>
<blockquote cite="mid:4EA01BAE.9090709@anu.edu.au" type="cite">The
free energy is a state function, and so the free energy difference
between E and Z isomers does not depend on the path taken for the
conversion. The difficulty is just that there is no path for a
cyclic alkane from E to Z in a normal MD force field.<br>
</blockquote>
and that's the reason for the disagreement. also imho, the path is
irrelevant. for a molecule given with a cis and a trans
configuration, respectively (same atoms, quasi-same topology), the
inner energy and entropy of both conformations (derived from
classical force field simulations) is enough in order to get the
free energy difference.<br>
<blockquote cite="mid:4EA01BAE.9090709@anu.edu.au" type="cite"> <br>
<blockquote cite="mid:4E9FDB12.4010207@zib.de" type="cite"> <br>
but the reason, why i refrained from extracting the potential
energy for a subset of my system was simply due to the
difficulties to find "detailed information" about how to play it
out although many gromacs users have been asking for the issue
over the last years. i needed nearly one full day to figure it
out even though it's a simple series of about 4 gromacs commands
only, given some md input & result files like md.xtc,
md.gro, complex.top, index.ndx:<br>
<br>
<code class="java plain">grompp -f mdSubset.mdp -c md.gro -p
complex.top -o mdSubsetTemp.tpr</code><br>
<div class="container" title="Hint: double-click to select code">
<div class="line number1 index0 alt2"><code class="java plain">echo
</code><code class="java string">"1"</code> <code
class="java plain">| tpbconv -s mdSubsetTemp.tpr -nsteps </code><code
class="java value">0</code> <code class="java plain">-n
index.ndx -o mdSubset.tpr</code></div>
</div>
<div class="container" title="Hint: double-click to select code">
<div class="line number1 index0 alt2"><code class="java plain">echo
</code><code class="java string">"1"</code> <code
class="java plain">| trjconv -s mdSubset.tpr -f md.xtc -o
mdSubset.xtc</code></div>
</div>
<code class="java plain">mdrun -s mdSubset.tpr -rerun
mdSubset.xtc -v -deffnm mdSubset</code><br>
<br>
where "1" stands for the group to be extracted (from the list of
groups in the index file) and "-nsteps 0" causes recomputation
of for the extracted subset at the given time steps only. i
mention it here so that other users looking for the details
might find them a little bit faster ... it's especially this
kind of "tiny tutorials" that i miss e. g. on the gromacs
website.<br>
</blockquote>
<br>
There's an art to trying to find detailed help. I know I've given
the advice to make subsets and use mdrun -rerun several times, but
I don't know how a newcomer is ever going to figure that out on
their own!<br>
</blockquote>
Your somehow right, but unfortunately, i'm no artist at all.
However, what's the reason for the availability of all the tutorials
and how-tos on the gromacs documentation site? i'll tell you: it
helps other researchers to reduce times of
searching/finding/figuring out. and besides, the information
retrieved from the tools' manpage or the manual is often far away
from being helpful!<br>
<br>
please, don't misunderstand me. i'm just saying: if there are
how-tos about how to plot data, perform calculations of pKa values
or diffusion constants, then the description of getting the
potEnergy of one of the systems components has an a fortiori right
to exist there or somewhere else at a central place within the
gromacs world! <br>
<br>
<blockquote cite="mid:4EA01BAE.9090709@anu.edu.au" type="cite"> <br>
BTW -nsteps 0 is not necessary. mdrun -rerun only computes on the
frames present in the input trajectory file. Also, if the index
group of interest is already present in the .tpr that ran the
simulation, then you can do the job with variations on only the
latter three commands.<br>
</blockquote>
<br>
thanks for the hints! <br>
<br>
<blockquote cite="mid:4EA01BAE.9090709@anu.edu.au" type="cite"> <br>
Mark<br>
</blockquote>
<br>
vedat<br>
<br>
<br>
<br>
ps: don't kill me now ...<br>
<blockquote cite="mid:4EA01BAE.9090709@anu.edu.au" type="cite"> <br>
<blockquote cite="mid:4E9FDB12.4010207@zib.de" type="cite"> <br>
thanks for listening and kind regards,<br>
<br>
vedat<br>
<br>
<br>
<blockquote cite="mid:4E9D9C57.4040104@anu.edu.au" type="cite">
<br>
<blockquote type="cite"> <br>
Q8 does anyone have an idea, how to perform the simulation
and on which energy terms to concentrate in order to get
reliable results? <br>
</blockquote>
<br>
You seem to be performing it OK, given that you've said very
little about any details... <br>
<br>
I think the problem is poorly constructed. You have some
experimental data that gives a general understanding of the
size of the free energy difference between the isomers. You
can't necessarily expect to reproduce that from (average)
potential energy differences between conformations of those
isomers. Measuring the free energy difference with simulations
is hard because you cannot model the intermediate stages of
bond breaking and forming. There are "alchemical" free energy
methods that could in principle treat this problem
effectively, but there will be some significant issues and you
are best doing your own homework there. <br>
<br>
Mark <br>
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