[gmx-users] gmx distance lipid bilayer
xy21hb at 163.com
Wed Sep 30 18:13:54 CEST 2015
However, even if I choose B = C16 (last Carbon atom of sn-1 chain), it gives similar result.
Then I wonder how I could get the distance of my interest.
At 2015-09-30 23:20:28, "Justin Lemkul" <jalemkul at vt.edu> wrote:
>On 9/30/15 11:17 AM, xy21hb wrote:
>> Dear all,
>> I am using gmx distance to calculate the distance between the center of mass (c.o.m.) of the phosphorous group (PO4, 5 atoms)
>> and the c.o.m. of the lipid bilayer system in Justin's tutorial (a typical lipid DPPC bilayer system) by issuing, as shown on GROMACS website for gmx distance,
>> gmx distance -n index.ndx -select 'com of group "A" plus com of group "B"' -oxyz -oall -s md.tpr -f md.trr
>> where A = Phsphorous B = System.
>> The distance should be about the chain length of the lipid, so ~ 20 A
>> However, it gives ~ 0.1 nm. I just wonder why it is like that.
>Because your assumption is wrong. The bilayer is centered in the box, so the
>COM distance between the center of the system and the center of the P atoms is
>usually about the same position, so you get a value that is nearly zero.
>Justin A. Lemkul, Ph.D.
>Ruth L. Kirschstein NRSA Postdoctoral Fellow
>Department of Pharmaceutical Sciences
>School of Pharmacy
>Health Sciences Facility II, Room 629
>University of Maryland, Baltimore
>20 Penn St.
>Baltimore, MD 21201
>jalemkul at outerbanks.umaryland.edu | (410) 706-7441
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