[gmx-users] Lipid water simulation
p.c.kroon at rug.nl
Mon Mar 27 16:15:12 CEST 2017
On 27-03-17 03:55, Sheikh Imamul Hossain wrote:
> Hi all,
> I am trying to simulate 1024 dppc lipids with water. I have prepared my
> system using Charmm-Gui monolayer builder. Then I converted the atomistic
> system to coarse grained system using bacdward.py. The box size I got in
> the gro file was 18.246 18.38 23.7. Then I added CG water between the
> two dppc monolayers within the box 18.246 18.38 4.22. The I increased
> the box size to 18.246 18.38 50.00. The total number of CG water is
> 6144. I minimized the system then equilibrated the system three times with
> time steps 2fs, 10fs and 15fs. Then I did the production run for 2
> microsecond using 20fs time steps. The temperature 310K was used in surface
> tension coupling with surface tension 20mN/m per leaflet. I have few
> 1. After the first equilibration some (very few) CG water found in the air
> space of the box i.e in the top and bottom of the dppc monolayers. Which
> are supposed to be attached together with other water between the two
> monolayer. Is there any solution to prevent this?
If you're running in an NPT ensemble the pressure coupling should take
care of any air/vacuum bubbles. Depending on how far away from
equilibrium you start this may take a while though.
> 2. Is there any other option to build the CG monolayer system and adding CG
> water between the two monolayers? I used gmx solvate to add water.
You can probably also use the insane script to build monolayers (I'm not
sure though). If you have those you can stack them together (with some
space in between) using gmx editconf. You'll probably still need to
solvate it using gmx solvate.
> 3. I used 6 CG water per lipid. How many CG water can be added against the
> number of lipids?
Depends on what you want to simulate. If you're using Martini, 1 CG
water = 4 H2O molecules; so you can calculate the concentration of
lipids in water.
> After analyzing my system with the help of Bevan Lab DPPC tutorials I got
> the area per lipid 0.5845. I used the formula
> Area per lipid = (Box-X*Box-Y)/ # of lipid per leaflet
> Thanks in advance.
> Sincerely Your’s
> Sheikh Imamul Hossain
> PhD student at QUT
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