AMBER Archive (2005)

Subject: RE: AMBER: E of amber calculations

From: Ross Walker (ross_at_rosswalker.co.uk)
Date: Thu Dec 15 2005 - 10:23:46 CST

Dear Young,

For implicit solvent calculations such comparisons can be made fairly easily
as the energy represents that of just the solute. However, you need to run a
sufficiently long simulation (much longer than in the tutorials) and compare
the average potential energy. The energy of a single snapshot is not very
meaningful but the average over a section of the trajectory is. You also
need to ensure that you are comparing Potential Energy and not total energy.
Although for well equilibrated systems at the same temperature that show the
same degree of fluctuations in the temperature you can compare average total
energies and recover the same information as comparing the potential
energies.

For explicit solvent simulations such as that in the tutorial things are
much more difficult due to the huge number of degrees of freedom introduced
by the solvent. In this case the energy of the solute is masked by
fluctuations in the energy of the solvent. This can take a long time to
reach equilibrium so a large amount of sampling is required. For example, in
the case of the tutorial the solute structure undergoes a huge changing
converting from A form to B form. Such a change can displace a large amount
of water, displacing it from equilibrium. The water will then re-adjust to
the new solute structure. Such a readjustment can take quite a while to
occur and so to make comparisons you would either have to stop the explicit
solvent simulations, remove the solute and run some implicit solvent
simulations, or allow sufficient simulation time for the system to
re-equilibrate. Both methods have their disadvantages.

The real take home message is that the simulations in the tutorials are
typically too short to obtain meaningful ensemble averages from.

I hope this helps.
All the best
Ross

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> -----Original Message-----
> From: owner-amber_at_scripps.edu
> [mailto:owner-amber_at_scripps.edu] On Behalf Of YoungJin Cho
> Sent: Wednesday, December 14, 2005 09:35
> To: amber_at_scripps.edu
> Subject: AMBER: E of amber calculations
>
> Hi,
> I'd better ask than being confused longer.
> My question is about energy in amber calculations.
> In the DNA tutorial, It shows the polyA-polyT decamer.
> In comparison with starting B and A form DNA (the tutorials shows how
> it changes from A to B in a water box), my question is from
> .out files.
> At the beginning, from the fist minimazation (A form), energy changes
> from -2.8014E+04 to -4.6001E+04 while B form changes from -2.4745E+04
> to -3.8989E+04 in a water box.
> First, does it mean it has a B form DNA starting structure with
> somewhat higher energy? and the first minimized B form structure is
> less stable than A from mininized structure?
>
> My question is about how we can mention the stability of the molecule
> in terms of energy in this case.
> Same comparison can go on with tutorial's data.
> Minimization2: (B) -3.9276 -> -4.1777E+04
> (A) -4.6577 -> -4.9411E+04
> heating Dynamincs (B) -44677.2451 -> -32447.7077
> (A) -52921.4514 -> -37836.6256
> Dynamics2 (B) -32534.5748 -> -33533.4505
> (A) -38138 -> -39319
>
> What I understood is in considering A, the starting energy is getting
> lower while it changed into B form DNA. Anyway, my ultimate
> question is
> if there is a way that we can simply compare stability/ energy of
> current conformation of the molecule.
>
> with big thanks in advance,
>
> Young Jin
>
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