[Chimera-users] building on functional groups

[Elaine Meng]

Hi Jesse,
I see I was wrong in previously suggesting you could add all the hydrogens first; sorry about that.  Lysine nitrogens are automatically assigned a formal positive, tetrahedral atom type.  So the first step of building should be to select that nitrogen and use the “modify structure” section of Build Structure tool to change it from 4 bonds tetrahedral to 3 bonds trigonal. In my testing just now, I specified a new name for the residue (via the options in Build Structure) because I thought it might not work to just leave it as LYS.  Next select one of the two hydrogens on that nitrogen and change it to C with 3 bonds trigonal.  Next select one of the two hydrogens on that carbon and change it to O with 1 bond, select the other hydrogen and change it to C with 4 bonds tetrahedral.  

Now the atom types will have been automatically adjusted as appropriate for an acetylated lysine.  Now you can use addh on the whole structure, and any subsequent charge calculations should be fine.  In my test the modified residue was named UNK and its net charge was 0.

(You can see atom types by selecting the atom(s) and using menu: Actions… Label… IDATM type.  The label won’t update automatically if the type is changed, you would have to re-label to see a difference.  The types are documented here: <http://www.rbvi.ucsf.edu/chimera/docs/UsersGuide/idatm.html> )

Best,
Elaine
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Elaine C. Meng, Ph.D.                       
UCSF Computer Graphics Lab (Chimera team) and Babbitt Lab
Department of Pharmaceutical Chemistry
University of California, San Francisco

On Oct 22, 2014, at 5:00 AM, JESSE JAYNES <jjsqrd at bellsouth.net> wrote:

> Thank you Elaine.  I appreciate it.  
> 
> I have been able to acetylate the lysine residues but now I am having difficulty in eliminating the charge.  I noticed that each hydrogen on a normal epsilon amino group of lysine is assigned +0.34 charge.  It would obviously be less than +1 and when I try and assign it +0 it basically is not capable of conducting the calculation.  I will continue to read and work on it.
> 
> Thanks,
> Jesse