## Re: [Apbs-users] Electrostatic Free energy of folding in water (per residue)

 Re: [Apbs-users] Electrostatic Free energy of folding in water (per residue) From: Steve Edgcomb - 2006-10-05 18:20:35 ```Hi, Thank you again for responding. I am double checking the parameters and hopefully that is the problem =20= with the direct calculation. However, I still don't understand how the thermodynamic cycle removes =20= the self energies. I add Fv and then subtract it. I also subtract Uv and then I add it. I think this is the same as the direct calculation: i.e. Direct calcualtion: DG,elec,aq=3DFaq-Uaq Thermodynaimic cycle: Fv-Uv+Faq-Fv-Uaq+Uv=3DFaq-Uaq + 0 =3D Direct =20 calculation How is the cycle different than the direct calculation? At what point do the self-energies cancel? Regardless of the values of Fv and Uv, where do I subtract a term =20 that I don't add back? Thanks again with your help. Steve > >> I am still missing something. >> Lets see if this diagram helps: >> >> Folded State in water (Faq) >> Unfolded State in water (Uaq) - residue from the pqr file without >> changing its position >> Folded State Vacuum (Fv) >> Unfolded State Vacuum (Uv) >> >> DG,elec,v >> Uv <--> Fv >> DGu,solv | | || DGf,solv >> Uaq <--> Faq >> DG,elec,aq >> >> DGf,solv =3D Faq-Fv (where values for each state are from APBS) >> DG,elec,v =3D Fv-Uv >> DGu,solv =3D Uaq-Uv > > This seems reasonable. > >> I think DG,elec,aq should be on the order of ~1kcal/(mol res). >> Calculating directly gives values that are approximately 100 times >> too large in magnitude. >> =46rom your e-mail, I think you are saying: >> >> DG,elec,v+DGf,solv-DGu,solv > > Yes, this is how I would calculate it. > >> But this gives the same value because the vacuum terms cancel. >> (i.e. Fv-Uv+Faq-Fv-Uaq+Uv=3DFaq-Uaq) > > Be careful about this cancellation. Coulombic interactions aren't > reproduced accurately by PB solvers due to grid artifacts and the > inclusion of self interactions. Removal of these self-interactions > through the free energy cycle I mentioned is *almost always* > necessary for accurate answers. The symptom of failing to remove > these interactions are very large energies. > >> It seems I am misunderstanding something obvious, but I don=92t know >> what. >> >> Oddly, >> DGf,solv-DGu,solv >> is only ~10 x too large in magnitude. > > This quantity should calculated accurately by APBS. How do your > parameters (radii, surface definitions, dielectric constants, etc.) > compare to the paper you're benchmarking against? > > Thanks, > > Nathan > ```

 [Apbs-users] Electrostatic Free energy of folding in water (per residue) From: Steve Edgcomb - 2006-09-28 20:12:16 ```Hi, I am trying to calculate the Electrostatic Free energy of folding in =20 water (DGelec) for a protein on a per residue basis. =46rom the =20 literature (Elcock A. JMB (2001) 312 885-896) I expect values on the =20 scale of 4 KJ /(mol res) (1 kcal/(mol/res)). With APBS I am getting =20 values that are about 100 times too large. The pdb converted to pqr with pdb2pqr and the PARSE option is used as =20= the native state (N). The APBS-calculated energy for each side-chain =20= atom is extracted with a perl script from the output of =91calcenergy =20= comps=92 and added together to get per-residue energy. The unfolded =20 state (U) is modeled as the individual side chain with coordinates =20 from the pqr without the rest of the protein. Again, the per atom =20 energies are totaled to get the per side chain energy. I thought I could calculate the folding energy directly from the =20 difference of the calculated energy for each state in water. DGelec =3D DGN(aq)-DGU(aq). =46rom the Balanol-PKA tutorial, it appeared to me that this would =20 work as long as the spacing and coordinates are the same in the =20 calculations for both states (i.e. that should take care of the self =20 energies). This appears not to work by ~100 to KJ/(mol res). I am including the in file for both the Native and Unfolded states. =20 The Unfolded in file is run on each individual amino acid. I have =20 run out of ideas about why my values are too large and I would be =20 grateful for helpful input. Thanks, Steve In file for native state: read mol pqr last_1.pdb.pqr end elec name 1 mg-auto dime 129 129 97 cglen 82.9787 75.6908 64.4589 fglen 68.8110 64.5240 57.9170 cgcent mol 1 fgcent mol 1 mol 1 lpbe bcfl sdh ion 1 0.1 2 ion -1 0.1 2 pdie 12 sdie 78.540 srfm smol chgm spl2 sdens 10.00 srad 1.40 swin 0.30 temp 298.15 gamma 0.105 calcenergy comps calcforce no end print energy 1 end quit In file for unfolded state (run individually for each side chain): read mol pqr last_1.pdb.pqr mol pqr last_1_res_1.pqr end elec name 1 mg-auto dime 129 129 97 cglen 82.9787 75.6908 64.4589 fglen 68.8110 64.5240 57.9170 cgcent mol 1 fgcent mol 1 mol 1 lpbe bcfl sdh ion 1 0.1 2 ion -1 0.1 2 pdie 12 sdie 78.540 srfm smol chgm spl2 sdens 10.00 srad 1.40 swin 0.30 temp 298.15 gamma 0.105 calcenergy no calcforce no end elec name 2 mg-auto dime 129 129 97 cglen 82.9787 75.6908 64.4589 fglen 68.8110 64.5240 57.9170 cgcent mol 1 fgcent mol 1 mol 2 lpbe bcfl sdh ion 1 0.1 2 ion -1 0.1 2 pdie 78.540 sdie 78.540 srfm smol chgm spl2 sdens 10.00 srad 1.40 swin 0.30 temp 298.15 gamma 0.105 calcenergy comps calcforce no end print energy 2 end quit ```
 Re: [Apbs-users] Electrostatic Free energy of folding in water (per residue) From: Nathan Baker - 2006-10-02 00:13:30 ```Hello -- Based on the input files you've included, it looks as if you're not =20 correctly removing self-energies. =46rom your e-mail, it seems you're =20= interested in the difference in electrostatic energy between two =20 configurations. Since the configurations are different, you can't =20 rely on cancellation of self-energies between the two calculations. =20 Instead, what you want to do is calculate solvation energies for each =20= of the configurations and then take the difference of these solvation =20= energies. This difference (the change in solvation energy due to un/=20 folding) can be added to the difference in Coulombic energy =20 (electrostatic energy in a homogeneous dielectric) change to give the =20= total electrostatic energy change. Does that make sense? Thanks, Nathan On Sep 28, 2006, at 3:11 PM, Steve Edgcomb wrote: > Hi, > I am trying to calculate the Electrostatic Free energy of folding in > water (DGelec) for a protein on a per residue basis. =46rom the > literature (Elcock A. JMB (2001) 312 885-896) I expect values on the > scale of 4 KJ /(mol res) (1 kcal/(mol/res)). With APBS I am getting > values that are about 100 times too large. > > The pdb converted to pqr with pdb2pqr and the PARSE option is used as > the native state (N). The APBS-calculated energy for each side-chain > atom is extracted with a perl script from the output of =91calcenergy > comps=92 and added together to get per-residue energy. The unfolded > state (U) is modeled as the individual side chain with coordinates > from the pqr without the rest of the protein. Again, the per atom > energies are totaled to get the per side chain energy. > > I thought I could calculate the folding energy directly from the > difference of the calculated energy for each state in water. > DGelec =3D DGN(aq)-DGU(aq). > > =46rom the Balanol-PKA tutorial, it appeared to me that this would > work as long as the spacing and coordinates are the same in the > calculations for both states (i.e. that should take care of the self > energies). This appears not to work by ~100 to KJ/(mol res). > > I am including the in file for both the Native and Unfolded states. > The Unfolded in file is run on each individual amino acid. I have > run out of ideas about why my values are too large and I would be > grateful for helpful input. > > Thanks, > Steve > In file for native state: > > read > mol pqr last_1.pdb.pqr > end > elec name 1 > mg-auto > dime 129 129 97 > cglen 82.9787 75.6908 64.4589 > fglen 68.8110 64.5240 57.9170 > cgcent mol 1 > fgcent mol 1 > mol 1 > lpbe > bcfl sdh > ion 1 0.1 2 > ion -1 0.1 2 > pdie 12 > sdie 78.540 > srfm smol > chgm spl2 > sdens 10.00 > srad 1.40 > swin 0.30 > temp 298.15 > gamma 0.105 > calcenergy comps > calcforce no > end > print energy 1 end > quit > > In file for unfolded state (run individually for each side chain): > > read > mol pqr last_1.pdb.pqr > mol pqr last_1_res_1.pqr > end > elec name 1 > mg-auto > dime 129 129 97 > cglen 82.9787 75.6908 64.4589 > fglen 68.8110 64.5240 57.9170 > cgcent mol 1 > fgcent mol 1 > mol 1 > lpbe > bcfl sdh > ion 1 0.1 2 > ion -1 0.1 2 > pdie 12 > sdie 78.540 > srfm smol > chgm spl2 > sdens 10.00 > srad 1.40 > swin 0.30 > temp 298.15 > gamma 0.105 > calcenergy no > calcforce no > end > elec name 2 > mg-auto > dime 129 129 97 > cglen 82.9787 75.6908 64.4589 > fglen 68.8110 64.5240 57.9170 > cgcent mol 1 > fgcent mol 1 > mol 2 > lpbe > bcfl sdh > ion 1 0.1 2 > ion -1 0.1 2 > pdie 78.540 > sdie 78.540 > srfm smol > chgm spl2 > sdens 10.00 > srad 1.40 > swin 0.30 > temp 298.15 > gamma 0.105 > calcenergy comps > calcforce no > end > print energy 2 end > quit > > ----------------------------------------------------------------------=20= > --- > Take Surveys. Earn Cash. Influence the Future of IT > Join SourceForge.net's Techsay panel and you'll get the chance to =20 > share your > opinions on IT & business topics through brief surveys -- and earn =20 > cash > http://www.techsay.com/default.php?=20 > page=3Djoin.php&p=3Dsourceforge&CID=3DDEVDEV > _______________________________________________ > apbs-users mailing list > apbs-users@... > https://lists.sourceforge.net/lists/listinfo/apbs-users > -- Assistant Professor, Dept. of Biochemistry and Molecular Biophysics Center for Computational Biology, Washington University in St. Louis Web: http://cholla.wustl.edu/ ```
 Re: [Apbs-users] Electrostatic Free energy of folding in water (per residue) From: Steve Edgcomb - 2006-10-02 19:53:43 ```I am still missing something. Lets see if this diagram helps: Folded State in water (Faq) Unfolded State in water (Uaq) - residue from the pqr file without =20 changing its position Folded State Vacuum (Fv) Unfolded State Vacuum (Uv) DG,elec,v Uv <--> Fv DGu,solv | | || DGf,solv Uaq <--> Faq DG,elec,aq DGf,solv =3D Faq-Fv (where values for each state are from APBS) DG,elec,v =3D Fv-Uv DGu,solv =3D Uaq-Uv I think DG,elec,aq should be on the order of ~1kcal/(mol res). Calculating directly gives values that are approximately 100 times =20 too large in magnitude. =46rom your e-mail, I think you are saying: DG,elec,v+DGf,solv-DGu,solv But this gives the same value because the vacuum terms cancel. (i.e. Fv-Uv+Faq-Fv-Uaq+Uv=3DFaq-Uaq) It seems I am misunderstanding something obvious, but I don=92t know = what. Oddly, DGf,solv-DGu,solv is only ~10 x too large in magnitude. Thanks again, Steve I am including the .in files for the vacuum calculations. The aq .in =20= files have not changed. In file for protein in vacuum: read mol pqr last_1.pdb.pqr end elec name 1 mg-auto dime 129 129 97 cglen 82.9787 75.6908 64.4589 fglen 68.8110 64.5240 57.9170 cgcent mol 1 fgcent mol 1 mol 1 lpbe bcfl sdh pdie 2 sdie 2 srfm smol chgm spl2 sdens 10.00 srad 1.40 swin 0.30 temp 298.15 gamma 0.105 calcenergy comps calcforce no end print energy 1 end quit in file for residue in vacuum: read mol pqr last_1.pdb.pqr mol pqr last_1_res_1.pqr end elec name 1 mg-auto dime 129 129 97 cglen 82.9787 75.6908 64.4589 fglen 68.8110 64.5240 57.9170 cgcent mol 1 fgcent mol 1 mol 1 lpbe bcfl sdh pdie 2 sdie 2 srfm smol chgm spl2 sdens 10.00 srad 1.40 swin 0.30 temp 298.15 gamma 0.105 calcenergy no calcforce no end elec name 2 mg-auto dime 129 129 97 cglen 82.9787 75.6908 64.4589 fglen 68.8110 64.5240 57.9170 cgcent mol 1 fgcent mol 1 mol 2 lpbe bcfl sdh pdie 2 sdie 2 srfm smol chgm spl2 sdens 10.00 srad 1.40 swin 0.30 temp 298.15 gamma 0.105 calcenergy comps calcforce no end print energy 1 end quit On Oct 1, 2006, at 5:13 PM, Nathan Baker wrote: > Hello -- > > Based on the input files you've included, it looks as if you're not > correctly removing self-energies. =46rom your e-mail, it seems you're > interested in the difference in electrostatic energy between two > configurations. Since the configurations are different, you can't > rely on cancellation of self-energies between the two calculations. > Instead, what you want to do is calculate solvation energies for each > of the configurations and then take the difference of these solvation > energies. This difference (the change in solvation energy due to un/ > folding) can be added to the difference in Coulombic energy > (electrostatic energy in a homogeneous dielectric) change to give the > total electrostatic energy change. > > Does that make sense? > > Thanks, > > Nathan > ```
 Re: [Apbs-users] Electrostatic Free energy of folding in water (per residue) From: Nathan Baker - 2006-10-05 11:00:22 ```Hi Steve -- > I am still missing something. > Lets see if this diagram helps: > > Folded State in water (Faq) > Unfolded State in water (Uaq) - residue from the pqr file without =20 > changing its position > Folded State Vacuum (Fv) > Unfolded State Vacuum (Uv) > > DG,elec,v > Uv <--> Fv > DGu,solv | | || DGf,solv > Uaq <--> Faq > DG,elec,aq > > DGf,solv =3D Faq-Fv (where values for each state are from APBS) > DG,elec,v =3D Fv-Uv > DGu,solv =3D Uaq-Uv This seems reasonable. > I think DG,elec,aq should be on the order of ~1kcal/(mol res). > Calculating directly gives values that are approximately 100 times =20 > too large in magnitude. > =46rom your e-mail, I think you are saying: > > DG,elec,v+DGf,solv-DGu,solv Yes, this is how I would calculate it. > But this gives the same value because the vacuum terms cancel. > (i.e. Fv-Uv+Faq-Fv-Uaq+Uv=3DFaq-Uaq) Be careful about this cancellation. Coulombic interactions aren't =20 reproduced accurately by PB solvers due to grid artifacts and the =20 inclusion of self interactions. Removal of these self-interactions =20 through the free energy cycle I mentioned is *almost always* =20 necessary for accurate answers. The symptom of failing to remove =20 these interactions are very large energies. > It seems I am misunderstanding something obvious, but I don=92t know =20= > what. > > Oddly, > DGf,solv-DGu,solv > is only ~10 x too large in magnitude. This quantity should calculated accurately by APBS. How do your =20 parameters (radii, surface definitions, dielectric constants, etc.) =20 compare to the paper you're benchmarking against? Thanks, Nathan > Thanks again, > Steve > > I am including the .in files for the vacuum calculations. The =20 > aq .in files have not changed. > > In file for protein in vacuum: > read > mol pqr last_1.pdb.pqr > end > elec name 1 > mg-auto > dime 129 129 97 > cglen 82.9787 75.6908 64.4589 > fglen 68.8110 64.5240 57.9170 > cgcent mol 1 > fgcent mol 1 > mol 1 > lpbe > bcfl sdh > pdie 2 > sdie 2 > srfm smol > chgm spl2 > sdens 10.00 > srad 1.40 > swin 0.30 > temp 298.15 > gamma 0.105 > calcenergy comps > calcforce no > end > print energy 1 end > quit > > > in file for residue in vacuum: > read > mol pqr last_1.pdb.pqr > mol pqr last_1_res_1.pqr > end > elec name 1 > mg-auto > dime 129 129 97 > cglen 82.9787 75.6908 64.4589 > fglen 68.8110 64.5240 57.9170 > cgcent mol 1 > fgcent mol 1 > mol 1 > lpbe > bcfl sdh > pdie 2 > sdie 2 > srfm smol > chgm spl2 > sdens 10.00 > srad 1.40 > swin 0.30 > temp 298.15 > gamma 0.105 > calcenergy no > calcforce no > end > elec name 2 > mg-auto > dime 129 129 97 > cglen 82.9787 75.6908 64.4589 > fglen 68.8110 64.5240 57.9170 > cgcent mol 1 > fgcent mol 1 > mol 2 > lpbe > bcfl sdh > pdie 2 > sdie 2 > srfm smol > chgm spl2 > sdens 10.00 > srad 1.40 > swin 0.30 > temp 298.15 > gamma 0.105 > calcenergy comps > calcforce no > end > print energy 1 end > quit > > On Oct 1, 2006, at 5:13 PM, Nathan Baker wrote: > >> Hello -- >> >> Based on the input files you've included, it looks as if you're not >> correctly removing self-energies. =46rom your e-mail, it seems = you're >> interested in the difference in electrostatic energy between two >> configurations. Since the configurations are different, you can't >> rely on cancellation of self-energies between the two calculations. >> Instead, what you want to do is calculate solvation energies for each >> of the configurations and then take the difference of these solvation >> energies. This difference (the change in solvation energy due to un/ >> folding) can be added to the difference in Coulombic energy >> (electrostatic energy in a homogeneous dielectric) change to give the >> total electrostatic energy change. >> >> Does that make sense? >> >> Thanks, >> >> Nathan >> > -- Assistant Professor, Dept. of Biochemistry and Molecular Biophysics Center for Computational Biology, Washington University in St. Louis Web: http://cholla.wustl.edu/ ```
 Re: [Apbs-users] Electrostatic Free energy of folding in water (per residue) From: Steve Edgcomb - 2006-10-05 18:20:35 ```Hi, Thank you again for responding. I am double checking the parameters and hopefully that is the problem =20= with the direct calculation. However, I still don't understand how the thermodynamic cycle removes =20= the self energies. I add Fv and then subtract it. I also subtract Uv and then I add it. I think this is the same as the direct calculation: i.e. Direct calcualtion: DG,elec,aq=3DFaq-Uaq Thermodynaimic cycle: Fv-Uv+Faq-Fv-Uaq+Uv=3DFaq-Uaq + 0 =3D Direct =20 calculation How is the cycle different than the direct calculation? At what point do the self-energies cancel? Regardless of the values of Fv and Uv, where do I subtract a term =20 that I don't add back? Thanks again with your help. Steve > >> I am still missing something. >> Lets see if this diagram helps: >> >> Folded State in water (Faq) >> Unfolded State in water (Uaq) - residue from the pqr file without >> changing its position >> Folded State Vacuum (Fv) >> Unfolded State Vacuum (Uv) >> >> DG,elec,v >> Uv <--> Fv >> DGu,solv | | || DGf,solv >> Uaq <--> Faq >> DG,elec,aq >> >> DGf,solv =3D Faq-Fv (where values for each state are from APBS) >> DG,elec,v =3D Fv-Uv >> DGu,solv =3D Uaq-Uv > > This seems reasonable. > >> I think DG,elec,aq should be on the order of ~1kcal/(mol res). >> Calculating directly gives values that are approximately 100 times >> too large in magnitude. >> =46rom your e-mail, I think you are saying: >> >> DG,elec,v+DGf,solv-DGu,solv > > Yes, this is how I would calculate it. > >> But this gives the same value because the vacuum terms cancel. >> (i.e. Fv-Uv+Faq-Fv-Uaq+Uv=3DFaq-Uaq) > > Be careful about this cancellation. Coulombic interactions aren't > reproduced accurately by PB solvers due to grid artifacts and the > inclusion of self interactions. Removal of these self-interactions > through the free energy cycle I mentioned is *almost always* > necessary for accurate answers. The symptom of failing to remove > these interactions are very large energies. > >> It seems I am misunderstanding something obvious, but I don=92t know >> what. >> >> Oddly, >> DGf,solv-DGu,solv >> is only ~10 x too large in magnitude. > > This quantity should calculated accurately by APBS. How do your > parameters (radii, surface definitions, dielectric constants, etc.) > compare to the paper you're benchmarking against? > > Thanks, > > Nathan > ```
 Re: [Apbs-users] Electrostatic Free energy of folding in water (per residue) From: Nathan Baker - 2006-10-07 13:17:16 ```Hi Steve -- > Thank you again for responding. > I am double checking the parameters and hopefully that is the =20 > problem with the direct calculation. > However, I still don't understand how the thermodynamic cycle =20 > removes the self energies. > I add Fv and then subtract it. I also subtract Uv and then I add it. > I think this is the same as the direct calculation: > > i.e. > Direct calcualtion: DG,elec,aq=3DFaq-Uaq > > Thermodynaimic cycle: Fv-Uv+Faq-Fv-Uaq+Uv=3DFaq-Uaq + 0 =3D Direct =20= > calculation The solvation energy of the folded state should be calculated as DGFs =3D Faq(APBS) - Fv(APBS) where Faq(APBS) and Fv(APBS) are the electrostatics energies obtained =20= from APBS PB calculations on the folded state. The solvation energy =20 of the unfolded state should be calculated as DGUs =3D Uaq(APBS) - Uv(APBS) where Uaq(APBS) and Uv(APBS) are the electrostatics energies obtained =20= from APBS PB calculations on the folded state. The change in =20 solvation energy upon folding is then DDGs =3D DGFs - DGUs =3D Faq(APBS) - Fv(APBS) - Uaq(APBS) + Uv(APBS) This calculation also serves to eliminate self energies in the folded =20= and unfolded states. This DDGs needs to be added to the change in Coulombic energy upon =20 folding DGc =3D Fc(coul) - Uc(coul) where Fc(coul) and Uc(coul) are the *analytic* Coulombic energies of =20 the molecule in the folded and unfolded states. These are calculated =20= in a homogeneous dielectric, usually chosen as 1 or the protein =20 interior dielectric. The net free energy change upon folding is DDG =3D DDGs + DGc =3D Faq(APBS) - Fv(APBS) - Uaq(APBS) + Uv(APBS) + Fc=20= (coul) - Uc(coul) Since Fv(APBS) does not equal Fc(coul) and Uv(APBS) does not equal Uc=20 (coul), there is no cancellation terms. Does this make more sense? Thanks, Nathan > > How is the cycle different than the direct calculation? > > At what point do the self-energies cancel? > > Regardless of the values of Fv and Uv, where do I subtract a term =20 > that I don't add back? > > Thanks again with your help. > > Steve > > > > > >> >>> I am still missing something. >>> Lets see if this diagram helps: >>> >>> Folded State in water (Faq) >>> Unfolded State in water (Uaq) - residue from the pqr file without >>> changing its position >>> Folded State Vacuum (Fv) >>> Unfolded State Vacuum (Uv) >>> >>> DG,elec,v >>> Uv <--> Fv >>> DGu,solv | | || DGf,solv >>> Uaq <--> Faq >>> DG,elec,aq >>> >>> DGf,solv =3D Faq-Fv (where values for each state are from APBS) >>> DG,elec,v =3D Fv-Uv >>> DGu,solv =3D Uaq-Uv >> >> This seems reasonable. >> >>> I think DG,elec,aq should be on the order of ~1kcal/(mol res). >>> Calculating directly gives values that are approximately 100 times >>> too large in magnitude. >>> =46rom your e-mail, I think you are saying: >>> >>> DG,elec,v+DGf,solv-DGu,solv >> >> Yes, this is how I would calculate it. >> >>> But this gives the same value because the vacuum terms cancel. >>> (i.e. Fv-Uv+Faq-Fv-Uaq+Uv=3DFaq-Uaq) >> >> Be careful about this cancellation. Coulombic interactions aren't >> reproduced accurately by PB solvers due to grid artifacts and the >> inclusion of self interactions. Removal of these self-interactions >> through the free energy cycle I mentioned is *almost always* >> necessary for accurate answers. The symptom of failing to remove >> these interactions are very large energies. >> >>> It seems I am misunderstanding something obvious, but I don=92t know >>> what. >>> >>> Oddly, >>> DGf,solv-DGu,solv >>> is only ~10 x too large in magnitude. >> >> This quantity should calculated accurately by APBS. How do your >> parameters (radii, surface definitions, dielectric constants, etc.) >> compare to the paper you're benchmarking against? >> >> Thanks, >> >> Nathan >> > -- Assistant Professor, Dept. of Biochemistry and Molecular Biophysics Center for Computational Biology, Washington University in St. Louis Web: http://cholla.wustl.edu/ ```