[Chimera-users] sigma and contour level

Elaine Meng meng at cgl.ucsf.edu
Wed Jul 2 09:56:00 PDT 2008


Hi Hazel,
We are glad you like Chimera!

There is a "Volume Mean, SD, RMS" tool (under Tools in the Volume  
Viewer dialog, among other places) that reports statistics for a  
density map and allows calculation of what contour level corresponds  
to a particular value of sigma.  There are a few more details in this  
previous message:

http://www.cgl.ucsf.edu/pipermail/chimera-users/2008-May/002651.html

At least when this message was written, we were not sure whether the  
official definition of sigma is standard deviations from the mean or  
standard deviations from zero.  Perhaps it depends on the prior  
processing of the map.  Either way, however, the conversion can be  
performed using the values reported by "Volume Mean, SD, RMS."

I hope this helps,
Elaine
-----
Elaine C. Meng, Ph.D.                          meng at cgl.ucsf.edu
UCSF Computer Graphics Lab and Babbitt Lab
Department of Pharmaceutical Chemistry
University of California, San Francisco
                      http://www.cgl.ucsf.edu/home/meng/index.html




On Jul 2, 2008, at 8:01 AM, Levy, Hazel C. wrote:

> Hi,
> I am a postdoctoral researcher and an enthusiastic Chimera user.   
> We are trying to finish up a manuscript to submit tomorrow to JSB  
> and we have been frantically trying to figure out how to convert  
> from Chimera's "contour level" to the more conventional "sigma  
> level" for density map viewing at different noise levels.
>
> We understand that the higher the contour level the smaller the map  
> volume displayed (less noisy), but we are not quit sure how to  
> report this as a sigma level.
>
> Each map that we are comparing has a different contour range  
> assigned, I am wondering if sigma level corresponds to some  
> proportion of the maximum volume.
>
> Please let us know how to do this conversion or please point us in  
> the direction toward the person that can help us.
>
> Thanks you so much...this information will settle a long-standing  
> mystery for us.
> Hazel Levy
>
>



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