On 2013-03-31 12:21, Kavyashree M wrote:
Thank you Sir
Note that the free energy differences are rigorously correct only if the
array of the cells in the grid correspond units of phase space with the
same volume. This is close to impossible to achieve, but the plots may
still give insight.
Regards
kavya
On Sun, Mar 31, 2013 at 11:58 AM, bipin singh <[email protected]> wrote:
g_sham calculates free energy landscapes by computing the joint
probability distribution from the two dimensional plane constructed
using two quantities (in your case it will be rmsd and radius of
gyration). Conformations sampled during the simulation were projected
on this two dimensional plane, and the number of points occupied by
each cell was counted. The grid cell containing the maximum number of
points is then assigned as the reference cell, with a free energy
value of zero. Free energies for all the other cells were assigned
with respect to this reference cell using the following equation:
ΔG = -kbT ln P(x,y)/Pmin
P(x,y) is the estimate of probability density function obtained from a
histogram of MD data and Pmin is the maximum of the probability
density function. Kb is the Boltzmann constant, and T is the
temperature corresponding to each simulation.
On Sun, Mar 31, 2013 at 10:35 AM, Kavyashree M <[email protected]> wrote:
Dear users,
Can someone kindly explain how g_sham calculates
the free energy landscape of given two quantities say,
rmsd and radius of gyration.
Any references are welcome.
Thank you
with Regards
Kavya
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Thanks and Regards,
Bipin Singh
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