Well, as my weblet was used here, maybe I should explain.

 

This is actually not so easy to understand, and I have seen many students 
having problems here.

 

First, generating symmetry mates in the vicinity of a given copy does not mean 
they are aligned with the generating operator’s axis (which I believe is what 
you want to show).

A program that does that right needs to expand the cell around the object and 
also consider the translational neighbors. Afaik, Pymol does it right.

 

The second problem is, that you need to understand the generator 6sub5. What 
does it do?

If you read the definition of a screw axis (I might suggest a certain textbook, 
p 221):

An operator N(s) rotates the motif by 360/N degrees and shifts it along the 
operator axis by a fraction of s/N, N times.

The moment s>1, you will eventually end up outside the unit cell, and you need 
to translate the generated copies back into

the unit cell. That is what these listed operators (equipoint transformations) 
mean.

 

Example:

 

say 6sub5, first generation: rotate by 60, shift by 5/6 = translation is 0.833 
on z. Check. 

Next, again take this copy, rotate by 60, shift by 5/6 = already 0.833, then 
another 0.833 = 1.666 

aha! outside, so we translate back -1.0 :  0.666 Check.

 

the rest you can figure out easily.

 

The important thing to realize and remember is that when you change the axis to 
its enantiomorphic mate, say 6sub5 -> 6sub1, you

get at the first sight similar looking operators, but the handedness of the 
screw operation is reversed. Some programs invert the

coordinates automatically for you if you know only the point group during 
structure solution but are unsure about the space group.

Our you just try them all (in MR).

 

here the operator set for 6sub 1

 

| 6 equipoint transformations (X)'=[R]*(X)+(T) :                              | 

 
+-----------------------------------------------------------------------------+ 

 | R(11) R(12) R(13) R(21) R(22) R(23) R(31) R(32) R(33)   T(1)   T(2)   T(3)  
| 

 | 1.0   0.0   0.0   0.0   1.0   0.0   0.0   0.0   1.0    0.000  0.000  0.000  
| 

 | 1.0  -1.0   0.0   1.0   0.0   0.0   0.0   0.0   1.0    0.000  0.000  0.167  
| 

 | 0.0  -1.0   0.0   1.0  -1.0   0.0   0.0   0.0   1.0    0.000  0.000  0.333  
| 

 |-1.0   0.0   0.0   0.0  -1.0   0.0   0.0   0.0   1.0    0.000  0.000  0.500  
| 

 |-1.0   1.0   0.0  -1.0   0.0   0.0   0.0   0.0   1.0    0.000  0.000  0.667  
| 

 | 0.0   1.0   0.0  -1.0   1.0   0.0   0.0   0.0   1.0    0.000  0.000  0.833  
| 

 
+-----------------------------------------------------------------------------+ 

 

see the difference?

 

An example for 31 vs 32 is given here

http://www.ruppweb.org/Garland/gallery/Ch5/pages/Biomolecular_Crystallography_Fig_5-33.htm

http://www.ruppweb.org/Garland/gallery/Ch5/thumbnails/Biomolecular_Crystallography_Fig_5-34.jpg

 

So your question what to do depends on what you want to show, and that is not 
quite clear. But you should

be able to figure it out by now.

 

BR

 

 

From: CCP4 bulletin board [mailto:[email protected]] On Behalf Of Smith Lee
Sent: Friday, January 20, 2017 6:05 PM
To: [email protected]
Subject: [ccp4bb] on how to use pdbset to get the symmetry mate coordinate

 

Dear All,

 

I have a pdb with P 65 space group. 

 

6 equipoint transformations (X)'=[R]*(X)+(T) :                              | 
 
+-----------------------------------------------------------------------------+ 
 | R(11) R(12) R(13) R(21) R(22) R(23) R(31) R(32) R(33)   T(1)   T(2)   T(3)  
| 
 | 1.0   0.0   0.0   0.0   1.0   0.0   0.0   0.0   1.0    0.000  0.000  0.000  
| 
 | 1.0  -1.0   0.0   1.0   0.0   0.0   0.0   0.0   1.0    0.000  0.000  0.833  
| 
 | 0.0  -1.0   0.0   1.0  -1.0   0.0   0.0   0.0   1.0    0.000  0.000  0.667  
| 
 |-1.0   0.0   0.0   0.0  -1.0   0.0   0.0   0.0   1.0    0.000  0.000  0.500  
| 
 |-1.0   1.0   0.0  -1.0   0.0   0.0   0.0   0.0   1.0    0.000  0.000  0.333  
| 
 | 0.0   1.0   0.0  -1.0   1.0   0.0   0.0   0.0   1.0    0.000  0.000  0.167  |
 
6 equipoint transformations decoded into atom coordinate format :           | 
 
+-----------------------------------------------------------------------------+ 
 |  X   ,  Y   ,  Z                                                            
| 
 |  X -Y,  X   ,5/6 +Z                                                         
| 
 | -Y   ,  X -Y,2/3 +Z                                                         
| 
 | -X   , -Y   ,1/2 +Z                                                         
| 
 | -X +Y, -X   ,1/3 +Z                                                         
| 
 |  Y   , -X +Y,1/6 +Z     
Will you please show me step by step (what exactly to input in the black in the 
graphical pdbset interface) by pdbset how can I get the symmetry mate pdb, for 
example how to get the pdb for 1 of 6 equipoint transformation X -Y,  X   ,5/6 
+Z?
 
I am looking forward to getting a reply from you.
 
Smith

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