On Tue, Nov 27, 2012 at 12:03 PM, Subhani Bandara wrote:

I am using Dev1223 and need to generate LLG map from phenix.refine to see any additional anomalous scattering is there. In the output window of phenix.refine I added LLG map to map coefficients and checked write mtz file... option. I thought the the mtz file with LLG map will be written to the output folder of refinement where the other files are, but don't see any additional files generated other than usual files. Am I doing something wrong?

No, the LLG map coefficients should be part of the same MTZ file that all of the other map coefficients are written. At present I don't think they will be automatically opened in Coot when launched from the Phenix GUI (I will look into fixing this), nor will the "Auto-open MTZ" function in Coot recognize them. Instead, you need to select "Open MTZ, mmCIF, fcf, or phs" in Coot, pick the MTZ file, and once you get to the map setup dialog, choose "LLG" and "PHLLG" or whatever you decided to name the columns, and check "Is a difference map?". Randy's recommendation was to only look at peaks greater than the maximum negative value, so I would scroll up the contour level until all of the negative LLG density disappears. -Nat

On Tue, Nov 20, 2012 at 12:59 PM, Pavel Afonine wrote:

...

Also is there a way to calculate something like anomalous difference-difference map (Anomalous Fobs - Anomalous Fcalc) to identify any additional atoms bound at metal sites.

If you refine against anomalous data set (containing Fobs+ and Fobs-) then traditional anomalous difference map will be output by phenix.refine by default. Recent versions should have the ability to create LLG maps as well.

Which is *not* the same thing - the anomalous difference map Phenix outputs only shows the anomalous differences in F-obs, and not the difference between F-obs and F-calc. As a result, any weaker anomalous scatterers (like chlorine, etc.) may have their signal drowned out by the contribution of the metals. -Nat

Hi Carsten, some keywords have type "multiple" and some not. In your example, rigid_body has type multiple, that is you can specify as many of them as you want: rigid_body = chain A rigid_body = chain B rigid_body = chain C and that means you will be refining three rigid bodies. Same applied to tls, and many others. anisotropic is not multiple, so it can be only one. I recall I had reasons to make it this way when implementing it but it was long long time ago, so I forgot what it was. To see all attributes for all parameters: phenix.refine --show-defaults=all.all which will yield a long hard to digest list. Pavel On 11/20/12 6:25 PM, Schubert, Carsten [JRDUS] wrote:

Hello all,

I have a question about the parameter definition for refinement groups (hope I picked the right term here). How does the definition and handling of refinement groups vary from the "refine.sites.rigid_body" scope to the "refine.adp.anisotropic" scope? For instance:

refine { strategy = *individual_sites individual_sites_real_space *rigid_body \ *individual_adp group_adp tls *occupancies group_anomalous sites { individual = None torsion_angles = None rigid_body = chain A rigid_body = chain B } adp { individual { isotropic = None anisotropic = chain A anisotropic = chain B

}

would refine chains A and B as separate rigid groups, whereas only chain B would be refined anisotropically. Is this an intended behavior?

BTW "anisotropic = chain A or chain B" refines both chains as expected.

Cheers,

Carsten

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Hi Vincent, The output you show has the following meaning: The reindexing of the miller indices of (-k,-h,-l) of the P1 data gives you data in the space group "C 1 2 1 (x-y,x+y,z)". This is a special (no lattice translations) version of C2 that can be set to the standard setting by applying the transformation (x-y,x+y,z) to the cell. This will subsequently reindex your miller indices as well. The "C 1 2 1 (x-y,x+y,z) " is a full space group. The change of basis matrix stuck at the back of the standard symbol indicates which transformation is required to get it back to the traditional setting. INstead of specifying it in x,y,z, you can also make specify the COB on the unit cell basis vectors. This allows you to do fun things like P1 (2a,b,c) to describe the space group P1 with added operator (x+1/2,y,z), i.e. if you would incorrectly index your data. If you go here: http://cci.lbl.gov/cctbx/explore_symmetry.html and fill out the space group C 1 2 1 (x-y,x+y,z) you get this info List of symmetry operations: MatrixRotation-part typeAxis directionScrew/glide componentOrigin shiftx,y,z 1----y,-x,-z2[-1,1,0]0,0,00,0,0 i.e. your two-fold axis lies in the xy plane. I'll contact you off-list with more info. P On 21 November 2012 08:50, vincent Chaptal wrote:

Dear all,

I ran "phenix.explore_metric_**symmetry" with the command:

phenix.explore_metric_symmetry --unit_cell="97,97,225,78,78,**68" --space_group=P1

It output: ... ------------------------- Transforming point groups ------------------------- From P 1 to C 1 2 1 (x-y,x+y,z) using : * -k,-h,-l ...

I understand that going from P1 to C2, one needs to apply the transformation matrix (x-y, x+y,z) on the P1 cell to form the C2 cell, and (-k, -h, -l) on the reflections.

Naive question: why aren't the two matrices similar? The reciprocal space is the fourier transform of the real space; i was thinking that a reorientation matrix in the real space would be kept in the reciprocal space. My maths are not that good, and in P1 it is more complex than other space groups. Can someone tell me why the matrices are different?

Also, in C2 there is a 2-fold axis parallel to b, so reflections (h,k,l) are equivalent to (-h, k, -l). In P1, they are not. Applying the above transformation matrix on the reflections would give (hP1, kP1, lP1) transforms into (-kP1, -hP1, -lP1), and these are equivalent to (kP1, -hP1, lP1)? Is this correct?

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