I really have no view providing that if you include the whole side
chain you must set the atom occupancy to 0.00 forthe invisible atoms..
Eleanor
For someone who doesn't have a view, "you must" is a pretty strong
phrase ;)
I believe we can all agree on the fact that the side chain atoms are
in the crystallized material, so one could argue that they should be
included in the final model. However, the density for them is so weak
that they are deemed 'not visible', so one could argue they should
not be included in the final model.
If one chooses to set the occupancy to less than 1.0, wouldn't one
have to include other locations for the atoms in question, so that
the sum of the occupancies equals 1.0? Anything else doesn't make
physical sense, because we are talking about covalently bound atoms.
Because the atoms are in the crystal, one solution is to refine them
with occupancy = 1.0 and let the B factor reflect the disorder. That,
to me, seems to be the approach that makes the most physical sense.
Leaving atoms out of the model is, IMHO, the least appropriate
solution, because it does not make physical sense. If one can 'see'
the C-alpha atom, the C-beta atom has to be there as well. One can
always contour at very low levels; at some level there will be some
density, so what criteria should we use to decide whether an atom
should be included in the model or not? I think this would require
much more deliberation and consensus-finding from the community than
simply letting the B factor reflect the degree of disorder. One could
include some kind of map correlation coefficient, as discussed in
other contexts, but non-crystallographers will likely have more
problems with that than with a simple B-factor, which most people
have heard of by now.
Of course, other solutions are chosen for entire disordered residues,
such as at termini or in loops. Those are indeed usually left out of
the model, but it is again not quite clear what the (objective)
criteria are for that.
Best - MM
Nicholas Noinaj wrote:
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Hi,
i would like to get opinions on whether or not one removes side-
chain atoms where there is no density. for example, if one can
only observe density up to the beta-carbon for lysine (say at > 0.5
sigma), does one leave the lysine side chain intact, knowing it
must be disordered, or does one terminate at the beta-carbon,
making the coordinates reflect what is actually observed in the
density.
It seems both approaches are published and people seem to have
conflicting opinions on the topic. It would be nice to come to
some concensus, possibly clear up the issue for us newbies.
Thanks in advance for all feedback!
Cheers,
NIck
________________________________________
Nicholas Noinaj
University of Kentucky College of Medicine
Department of Molecular and Cellular Biochemistry
The Center for Structural Biology
Biomedical Biological Sciences Research Building, Rm 236
741 S. Limestone
Lexington, Ky 40536
Lab: 859-323-8183
Cell: 859-893-4789
Home: 859-228-0978
[EMAIL PROTECTED]
noinaj.com
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