Protein/Membrane system size

This page collects notes on how to decide how to initially setup a membrane protein simulation. Please do not follow these notes blindly — they are more guidelines than actual rules.

Once you've decided on box dimensions and number of lipids you can use any of the Membrane protein insertion methods to set up the system.

System size

• Decide on the bilayer composition. Have you got force field parameters for the lipids?
• System size:
  • rule of thumb 1: have at least 2-3 lipid layers between protein and periodic box boundary in the x-y plane to ensure(?) membrane-like behaviour; more may be necessary: you will need to monitor the hydrophobic mismatch and make the system big enough so that the bilayer at the box edges has its natural thickness.
  • rule of thumb 2: have at least 1-1.5nm of water between protein and z-boundary (so that Coulomb forces are effectively screened)
  • Especially for coarse-grained self-assembly simulations you also need to take into account the lipid-to-water ratio and its affect on membrane structure.
• Add ions at eg 100 mM (+counter ions) for additional Coulomb screening
• Remember that run time scales like <math>N \log N</math> (or even <math>N^2</math>) and <math>N</math> scales with the volume, i.e. <math>L^3</math>! Thus: keep your system small and pretty (ie the smallest system that still behaves like a big system... invariably you are trading size-artifacts versus speed and thus sampling)

Number of lipids

Estimate the cross sectional area of the protein, eg using the radius of gyration

<math>A_{\mathrm{prot}} = \pi R_G</math>

and with the approximate area per lipid <math>A_l = 64</math> Å the number of lipids in two leaflets is for a box with dimensions <math>L_x</math> and <math>L_y</math>

<math>N = \frac{2(L_x L_y - A_\mathrm{prot})}{A_l}</math>.

If you are simulating a mixed bilayer, eg POPE:POPG 4:1 then you will need <math>4N/5</math> POPE lipids and <math>N/5</math> POPG ones.

Number of ions

Once the system is solvated you can calculate the numbers of ions <math>N</math> needed for a given concentration <math>c</math> (in addition to any counter ions):

<math>c = \frac{n}{V} = \frac{N/N_A}{V}</math>

(where <math>N_A = 6.02214179 \times 10^{23} \mathrm{mol}^{-1}</math> is the Avogadro constant).

The problem is to get the water volume of the inhomogenous system, consisting of membrane and protein with the water. An approximation is to use the volume of a water molecule at standard conditions, <math>v_w = 30 \mathrm{\AA}^3</math> and calculate <math>V = N_w v_w</math> from the number of waters in the system, <math>N_w</math>. The equations become even simpler if we use the standard concentration of water, <math>c_w = n_w/(N_w v_w) = 55.5\, \mathrm{mol/l}</math>.

Using all this we get

<math>N_+ = \frac{c_+}{V} = \frac{c_+}{c_w} N_w</math>

for the (monovalent) cations, and correspondingly for monovalent anions

<math>N_- = N_+</math>

Example

For a concentration of 100 mM = 0.1 M and 8000 water molecules we will need to add

<math>N_+ = N_- = \frac{0.1\,\mathrm{M}}{55.5\,\mathrm{M}} \times 8000 = 14</math>

ions of each kind (eg 14 sodium and 14 chloride and whatever counterions are required to make the system charge neutral).