maybe- but strain is relative to the "relaxed" state, and you need an
energy for that (since it won't be zero in the MM function).
On Wed, Mar 19, 2014 at 1:14 PM, Jio M <jiomm.yahoo.com> wrote:
> Dear Carlos
>
> That was nice suggestion indeed.
>
> thinking over this, can we see this like in othrewise way: If I consider a
> polymer chain and do MM energy on each resdiue the polymer made of it can
> give me hot spot where I have most strain along the chain length.
>
>
>
>
> On Wednesday, March 19, 2014 4:43 PM, Carlos Simmerling <
> carlos.simmerling.gmail.com> wrote:
>
> I interpret "strain" as being the difference in energy between the knitted
> form and the isolated polymers (how much intramolecular energy penalty they
> pay in order to interact with each other). You can't really just do MM on
> the knitted form to determine this, because the un-knitted form does not
> have zero energy. In your case, getting the average energy of the free
> polymers might be difficult - but it's the only way I know to estimate
> internal strain.
>
>
>
> On Wed, Mar 19, 2014 at 12:14 PM, Jio M <jiomm.yahoo.com> wrote:
>
> Hi All
> >
> >I have polymer chains (4 chains) knitted together and done explicit
> solvent simulation. I am interested to know how much each chain has
> strain/energy (may be not even a correct term to use). I can calculate
> single point energy (MM energy; using imin=5) of backbone but would it be
> sufficient (if not satisfactory) to represent such behaviour?
> >Also I am not sure if I can just use atoms as there is residue based
> decomposition of energy or is there any other method that can help in such
> analysis.
> >
> >thanks
> >Jiom
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Received on Wed Mar 19 2014 - 12:00:03 PDT
