Apparent twinning in P 1 21 1

From: Yuri Pompeu
Date: 27 September 2011 19:16
Hello everyone,
I have a 2.3A data set that could be scaled in C 2 2 21  and P 1 21 1
Intensity statistics tests indicate twinning (pseudo-merohedral h,-k,-h-l in P 1 21 1)
I find a good MR solution and when I try to refine it with the twin law I get fairly good maps and decent Rs 21-28%. I can see features tha were not in the search model
Which leads me to think that this a valid solution. The one thing that bothers me however is the fact that my beta angle in P 1 21 1 is 104 (not close to 90) and that the geometry gets worse after refinement?
Any suggestions?
cheers

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From: Nat Echols

I've seen this before - the conventions for the C2221 and P21 unit cells are very different, so even if beta=104 in P21, the equivalent C2221 cell can still have all angles equal to 90.  And you can definitely have pseudo-merohedral twinning in these circumstances (I did too - PDB ID 3ori).

The problem with geometry is a separate issue - probably the automatic weighting not working properly, or an improper fixed weight.

-Nat

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From: Linda Schuldt
Dear Yuri,

in a monoclinic space group an orthorhombic lattice metric can be
simulated when one of the following conditions is fulfilled:
i) a = c [e.g. in Wittmann & Rudolph (2007) Acta Cryst. D63, 744-749],
ii) the beta angle is close to 90° [e.g. in Larsen et al. (2002) Acta
Cryst. D58, 2055-2059 ] or
iii) c cos beta is about -a/2 [e.g. in Declercq & Evrard, (2002) Acta
Cryst. D57, 1829-1835]. The a and b axes of the orthorhombic cell are
identical to the monoclinic a and c axes, respectively. The length of the
orthorhombic b-axis can also be calculated by "c(monoclinic) cos(beta-90°)
= 1/2b(orthorhomic)".

I would assume that you have the case iii with a quite high twin fraction.
If I recall correctly, Declercq and Evrard have a nice figure in their
paper showing the geometric relationship. If not, let me know and I can
sent you a figure.

Good luck!
Linda


Yuri Pompeu schrieb:

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From: Yuri Pompeu
These papers describe something similar to what I see.
Acta Cryst. (2001). D57, 1829-1835
Acta Cryst. (2009). D65, 388-392

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From: Eleanor Dodson
You might like to look at this..

It tries to explain likely twinning possibilities in P21.

If you get C2222 and P21, then probably a~=c - then Beta can have any value.

 C222 axes are then always possible with  a* +c*  , a*-c*, b*  all having angles ~ 90

Without twinning you wont get 222 symmetry though. Pointless helps here.



Eleanor


 http://www.ccp4.ac.uk/dist/html/twinning.html

i) a = c [e.g. in Wittmann&  Rudolph (2007) Acta Cryst. D63, 744-749], iii) c cos beta is about -a/2 [e.g. in Declercq&  Evrard, (2002) Acta

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From: Yuri Pompeu
After I ran DETWIN with the estimated 0.46 alpha, my completeness for the detwinned data is now down to 54%!!!
Is this normal behavior? (I am guessing yes since the lower symmetry untwinned dat is P1 21 1)

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From: Garib N Murshudov

Why do you detwin? It would not be normal procedure if you have twinning. Molecular replacement programs usually do not have much problem with twinned data and refinement programs can deal with them more or less properly.

when you detwin then errors are increased (As far as I remember proportional to 1/(1-2alpha), if alpha is 0.46 then errors will increase more than 10 times). Moreover it is very likely that twin and pseudo rotation are close to each other and estimated twin fractions may not be accurate.

regards

Garib

Garib N Murshudov

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From: Phil Jeffrey
Yuri,

Detwinning relies on having both twin-related reflections present to calculate either/both of the the de-twinned data values.  Therefore it magnifies incompleteness depending on where your missing data is with respect to the twin operator.

I'd recommend against trying to do this with a twin fraction close to 0.5.  From the DETWIN docs:

Itrue(h1) = ((1-tf)*iTw(h1) -tf*iTw(h2)) / (1-2tf)

i.e. tf = twin fraction, so 1/(1-2tf) becomes a large number and it's multiplying a weighted term of the form: (iTw(h1) - iTw(h2)) which becomes a very small number as the twin fraction approaches 0.5.  The latter difference can easily be less than sigma(I), and so the signal/noise of your data plummets.

Better to use REFMAC and phenix.refine's abilities to compensate for the twin fraction directly in refinement and leave your data as it is.

Phil Jeffrey
Princeton

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From: Yuri
I am refining in phenix twin law h,-k,-h-l.
Without twinning I was around 0.36 Rfree and 0.25 with twinning.
I am noticing however that my Rgap keeps increasing slowly... (little concerned now its at 8% - 0.18-0.26 - to 2.4A)
Maps look decent for 2.4A I have a lot of clashes however some are just bad waters though. phenix does not do real space when twinning is enabled.
Any ideas here?

thank you much

--
Yuri Pompeu

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From: Yuri

I am now  realizing that! I have been refining with twin law enabled, its been somewhat succesful.
I have been using phenix and now getting ready to try refmac...
regards

--  Yuri Pompeu