On 7 Mar 2018, at 20:27, Daniel M. Himmel, Ph. D. <[log in to unmask]> wrote:

Rajesh,

Ca2+ usually is coordinated by five atoms, whereas Mg2+ usually is

coordinated by six.  I've seen that a lot in structures that I've determined.

-Daniel

On Wed, Mar 7, 2018 at 2:14 PM, Eleanor Dodson <[log in to unmask]> wrote:

You just calculate a anom map - it doesnt need you to dselect Ca or Mg or anything else.. 

Judge the peak heights by the signal you get over any S sites in the model. The relative signal you expect depends on your wavelength of course.

Eleanor

On 7 March 2018 at 19:09, Rajesh Kumar <[log in to unmask]> wrote:

Hi Ivan,

https://csgid.org/csgid/metal_sites/

suggest that Mg++ is the correct metal. Ca++ is not acceptable.

I am going to validate it by calculating anomalous peak for Ca++ too. 

Thank you

Rajesh 

On Wed, Mar 7, 2018 at 1:06 PM, Ivan Shabalin <[log in to unmask]u> wrote:

Hi, Rajesh,

It does look like Ca to me. It has around 2.4 A distances to waters and quite many electrons (18), which correspond to rather wide and high density peak in the middle.

Mg has distances at 2.07, and 10 electrons (should be a smaller peak).

Na has 2.41 distances to waters, 10 electrons (should be a smaller peak).

After you place a metal and refine it, try using CheckMyMetal to validate it:

https://csgid.org/csgid/metal_sites/


It takes into account preferable coordination geometries, coordination bond distances, and compares ADPs to average ADP for the surrounding atoms.

If the dataset is very good, and the wavelength of X-ray is rather long, you might even get anomalous signal for Ca:

http://skuld.bmsc.washington.edu/scatter/AS_periodic.html

Ivan



With best regards,
Ivan Shabalin, Ph.D.
Research Scientist,
Department of Molecular Physiology and Biological Physics,
University of Virginia,
1340 Jefferson Park Avenue, Pinn Hall,Room 4223,
Charlottesville, VA 22908

On 03/06/2018 05:19 PM, Rajesh Kumar wrote:

Dear All,

Have you had experience with this kind of density? I am wandering what this could be?

Thank you very much for the help.

-Rajesh

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