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Bonded length for two positive charged headgroup
- htaoyu
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9 years 2 months ago #4462
by htaoyu
Bonded length for two positive charged headgroup was created by htaoyu
Hi all,
I am trying to set up a simulation for a lipid-like molecule using the dry martini force. In my molecule, the two head groups are two Q0 beads with +1 charges. I am curious which bond length should I use for these two beads. In the dry_martini_v2.1_lipids.tip file, I saw two different bond lengths for the headgroups, i.e. 0.45 nm and 0.37 nm. For the (+1)-(-1) charged zwitterionic headgroup, a 0.45nm bond length is used. However, I did not find any example for the (+1)-(+1) type headgroup.
Thanks,
Hongtao
I am trying to set up a simulation for a lipid-like molecule using the dry martini force. In my molecule, the two head groups are two Q0 beads with +1 charges. I am curious which bond length should I use for these two beads. In the dry_martini_v2.1_lipids.tip file, I saw two different bond lengths for the headgroups, i.e. 0.45 nm and 0.37 nm. For the (+1)-(-1) charged zwitterionic headgroup, a 0.45nm bond length is used. However, I did not find any example for the (+1)-(+1) type headgroup.
Thanks,
Hongtao
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- jaakko
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9 years 1 month ago #4473
by jaakko
Replied by jaakko on topic Bonded length for two positive charged headgroup
Hi Hongtao,
have you simulated your molecule atomistically or using standard Martini? If you have you can just fit the bond length so that the observed bond length distributions in simulations match. Without knowing what the maximum of the actual bond length distributions should be it's hard to give a guess what the equilibrium bond length needs to be. But you can see from the values in the file that for a headgroup without charge-charge interactions the value is 0.45 nm and with attractive charge-charge it's 0.37 nm. With repulsive charge-charge interactions it's likely larger than either of those but that again depends on the exact structure of your molecule.
- Jaakko
have you simulated your molecule atomistically or using standard Martini? If you have you can just fit the bond length so that the observed bond length distributions in simulations match. Without knowing what the maximum of the actual bond length distributions should be it's hard to give a guess what the equilibrium bond length needs to be. But you can see from the values in the file that for a headgroup without charge-charge interactions the value is 0.45 nm and with attractive charge-charge it's 0.37 nm. With repulsive charge-charge interactions it's likely larger than either of those but that again depends on the exact structure of your molecule.
- Jaakko
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