[phenixbb] TLS question

Anastassis Perrakis a.perrakis at nki.nl
Tue Jan 27 07:20:06 PST 2009

On 27 Jan 2009, at 16:10, Jeffrey Chao wrote:

> hello all-
> i apologize if this question has previously been posted, but i was not
> able to track down an answer by google.  i am refining a 2.75A
> structure that has three copies of the protein in the asu with phenix
> 1.4-3 and using NCS with each chain as an NCS group.  after molecular
> replacement, i set all of the b-factors to the wilson b-factor and
> then performed four different refinements:
> grouped b-factor (1 group per residue)	: Rwork 22.6 	Rfree 27.0
> individual b-factor					: Rwork 21.4	Rfree 26.6
> grouped b + tls (						: Rwork 21.9 Rfree 25.2
> individual + tls						: Rwork 21.2 Rfree 24.8
> i defined the tls groups with the following:
> adp {
>     tls = chain A
>     tls = chain B
>     tls = chain C
>   }
> after grouped b-factor the atoms of each residue all have the same b-
> factor as i expected. after grouped b-factor+tls, however, each atom
> of a residue has a unique b-factor, which i did not expect.

This has to do with the way that phenix output the Bs (another  
Phenix outputs the total B, including the contribution to TLS.
A simple way to think about it, is that since the TLS tensor has an  
its contribution to each different atom is space is different. Thus,
even if atoms have a grouped B factor (same for the group) the TLS for  
chain will contribute different to each atom and the output total B  
will be
different (but not dramatically for atoms close to each other)

> i realize
> that this confusion is likely caused by me not fully understanding how
> tls refinement is implemented.  if someone could explain this to me, i
> will be able to sleep better.  thank you for your time, jeff

bed time reading:
"Use of TLS parameters to model anisotropic displacements in  
macromolecular refinement"


> Jeffrey Chao, PhD
> 1300 Morris Park Ave
> ASB-Golding 601
> Bronx NY 10461
> jchao at aecom.yu.edu
> 718-430-8597
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