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you have 6 [presumably] identical copies. how you choose to average them
is somewhat arbitrary.
1) you could just do 2-fold averaging within each dimer
2) you could do 3-fold averaging between your various dimers
3) you could do 6-fold averaging of your monomers.
from your description, i think perhaps you are doing #1.
i am not entirely sure whether you are doing it correctly, your masking
choice would require proper symmetry within dimers, and from your
description i cannot tell whether that is the case.
or perhaps you are trying to do #1 and #2, in which case you really need
to check the DM output statistics on overlap removal, because the
algorithm is deliberately designed to prevent this.
#3 should give you your best phase refinement/extension, assuming
you can identify the boundary for a monomer, and this is the one i will
you only need a mask for one monomer, say A1.
then you need to determine operators from A1 to each of the
A1 -> A1 (the identity matrix)
A1 -> A2
A1 -> B1
A1 -> B2
A1 -> C1
A1 -> C2
On Tue, 16 Nov 1999, Jan Abendroth wrote:
> Dear all,
> once again a question concerning a slightly complicated case of dm and
> The situation:
> Within the the asymmetric unit of my crystal there are six copies of the
> protomer grouped as three dimers A, B and C.
> I could determine the following symmetry operators:
> inter-dimer: A -> B
> B -> C
> intra-dimer: A1 -> A2 (from subunit 1 of dimer A to subunit 2 of dimer
> B1 -> B2
> C1 -> C2
> All symmetry operators are 2-folds with a transition-component. None of
> them falls on a crystallographic axis.
> For the symmetry operations A -> B, A -> B and A1 -> A2 I use A as mask,
> for B1 -> B2 i use B as mask and for C1 -> C2 C as mask.
> The question:
> Are these symmetry operations enough to cover the whole symmetry?
> Or do I have to specify the operations A -> C, A1 -> B1, A1 -> B2 ,...
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unless they are mothers with infants, or doctors with patients, or lovers.
- Ursula K. Le Guin
David J. Schuller
modern man in a post-modern world
University of California-Irvine
- From: Jan Abendroth <Jan.Abendroth@uni-koeln.de>