proper or improper ncs?

From: Francis E Reyes
Date: 21 February 2012 13:47


Hi all

This structure has the following ncs (output via phenix.simple_ncs_from_pdb)
OPERATOR 1
CENTER:   18.3443  -55.4605   23.0986

ROTA 1:    1.0000    0.0000    0.0000
ROTA 2:    0.0000    1.0000    0.0000
ROTA 3:    0.0000    0.0000    1.0000
TRANS:     0.0000    0.0000    0.0000

OPERATOR 2
CENTER:   37.0405  -23.8676  -14.9388

ROTA 1:   -0.5444   -0.2202    0.8094
ROTA 2:    0.8330   -0.0278    0.5526
ROTA 3:   -0.0991    0.9751    0.1985
TRANS:    45.3456  -78.7231   53.0085


It looks two-foldish but I'm not sure if it's proper or improper. (I'm trying to rationalize the lack of peaks on the self rotation maps).


Any help would be appreciated.

F



---------------------------------------------
Francis E. Reyes

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From: Randy Read


Hi,

I'm sure there's a way to do this in CCP4, but using some little jiffy programs I've collected over the years...

That's a 133 degree rotation around an axis defined by the unit vector 0.290647, 0.624977, 0.724519, and there's a small screw component of about 2.4A along the axis, so it's an improper rotation.

On the other hand, if you combine a crystallographic symmetry operator with that transformation, you might still find that it's equivalent to a proper rotation.

Regards,

Randy Read

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From: Dirk Kostrewa 

Hi Francis,

a quick insertion of your rotation matrix into CONVROT (by Alexandre Urzhumtsev) gives the following equivalent polar rotation angles (definition as in X-PLOR/CNS or Rossmann, 1962):

phi,psi,kappa       :  111.86  128.68  133.38     291.86   51.32  226.62

For a proper two-fold, a kappa of 180 degrees would be expected. This looks (very) improper to me.

Best regards,

Dirk.

Am 21.02.12 14:47, schrieb Francis E Reyes:

--

*******************************************************
Dirk Kostrewa

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From: Ian Tickle 

Francis,

It's very easy to spot a 2-fold rotation or screw because the matrix
must be symmetric (or nearly so)**.  Your matrix very obviously is not
(i,e, A12 ne A21, A13 ne A31 etc).

** Proof:

 A rotation matrix is orthogonal, which implies inverse = transpose: A^-1 = A~.

A 2-fold rotation is proper which implies AA = I or A^-1 = A.

Take these together and you get A = A~ i.e. A is symmetric.

Surprising how many people aren't aware of this!

Cheers

-- Iann

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From: Randy Read


I've had a couple of requests for the jiffy program that interpreted the transformation, so here's the FORTRAN source code for decomp.f, for anyone who is interested.  It's very simple, so all you should need to do is something like:

gfortran decomp.f -o decomp

to get an executable (or replace gfortran by g77 or f77, as appropriate).  Put the matrix and vector into a file (say matvec.dat), then:

decomp < matvec.dat

It takes a matrix and vector as four lines on standard input, then decomposes the transformation into a rotation around an axis, a point on the axis, and a translation parallel to the axis.

Regards,

Randy

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From: Alexandre OURJOUMTSEV


Hi, everybody,

In addition to all comments, just as a pleasure to remind an old-fashion "no computer" way to calculate the rotation angle. The trace of a rotation matrix (the sum of its diagonal elements) is always equal to 2*cos(kappa)+1. Calculate it yourself for kappa = 180° and compare with the trace of the Francis's matrix.

ROTA 1:   -0.5444   -0.2202    0.8094
ROTA 2:    0.8330   -0.0278    0.5526
ROTA 3:   -0.0991    0.9751    0.1985

One can easily reject (still without a computer) other "crystallographic" angles.

Best regards,

Sasha Urzhumtsev