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Hi Qun

Anomalous scattering is a complex subject, I would advise you to read
some of the extensive literature on it (I learnt it from Blundell &
Johnson, which though in many ways is now very dated, still has a good
section on AS - if you have a copy see section 7.4 pp 170-5).  Also it's
covered extensively in International Tables and several other more
recent textbooks.

A quick explanation of AS is that the observed handedness of the
anomalous pairs (i.e. the observed difference between the intensities of
the reflection hkl and its inverted-beam partner -h-k-l) arises through
a combination of two factors: namely the handedness of the amino-acids
(i.e. an amino-acid and its symmetry inverse are non-superposable), and
the presence of one or more anomalously scattering atoms (usually at
least as heavy as sulphur).  The former results in the reflections hkl
and -h-k-l having phases of opposite sign but with equal contributions
to the scattered amplitude.  The latter always results in the same phase
lag of <= 90 deg in the amplitude contribution to hkl and -h-k-l from
the constellation of heavy atoms (of the same chemical element), but
again with equal contributions to their amplitudes.  The summation of
the two effects results in a breakdown of Friedel's law, i.e. hkl and
-h-k-l having different amplitudes (and intensities); thus the set of
intensities is handed (not superposable on its inverse).  If there were
no anomalous scatterers I(hkl) and I(-h-k-l) would always be equal
within experimental error and it would make absolutely no difference
whether the matrix inverted the co-ordinate system or not.  Otherwise
the matrix you use must be a rotation matrix (i.e. determinant = +1),
not an inversion matrix (determinant = -1), so that it maintains the
correct hand.

Looking at your values of averaged intensities I would say that these
are weighted averages, i.e. weighted by 1/sigma(I)^2.  Averaging in this
way is valid provided there is no significant anomalous scattering,
otherwise one should use the separate I(+) and I(-) for calculations.

BTW I would recommend CC'ing your responses to the BB, that way you'll
get a balanced view (I might be spouting complete nonsense for all you
know!).

HTH!

-- Ian

> -----Original Message-----
> From: Qun Wan [mailto:[EMAIL PROTECTED] 
> Sent: Tuesday, May 09, 2006 11:06 PM
> To: Ian Tickle
> Subject: RE: [ccp4bb]: permute unit cell indices
> Importance: High
> 
> Hi, Ian:
> Thanks for your information!
> I do use *-rej-anom.sca file for this comparison of isomorphism. My 
> crystal has unit cell 73.4 73.5 88.5 with space group P212121. 
> Therefore, I only need to swap a and b for comparison. I know the 
> answer that after the swap, to keep the matrix left-handed, I need to 
> use minus sign. I am not clear about the matrix. I do not think the 
> handedness is coming from the handedness of the amino acid, it is 
> from the difference of anomalous pairs. However, I am still confused 
> why where is handedness in the anomalous difference. Could you 
> explain that for me?
> Also, when I compare the -rej-anom.sca and -rej.sca like the 
> following,  I found some confusion. In anom.sca file, column 4 is the 
> intensity of hkl, column 5 is the sigma of hkl, colume 6 is the 
> intensity of
> -h-k-l, column 6 is the sigma of -h-k-l.  In sca file, I do not know 
> how the value of column 4 comes from, it is not the average of 
> intensities in anom.sca file of hkl and -h-k-l columns. Do you know 
> that?
> The -anom.sca file:
>      1
>   -985
>      72.774    74.053    88.494    90.000    90.000    90.000 
> p212121        
>    31   5   1       0      -1   166.7    91.7
>    31   5   2  1144.0   134.7  1090.0   103.9
>    31   5   3    46.3    84.2    79.9   118.5
>    31   5   4   369.1    95.8   354.1   158.3
> The sca fiel:
>    31   5   1   166.7    91.7
>    31   5   2  1110.2    82.3
>    31   5   3    57.6    68.7
>    31   5   4   365.1    82.0
> 
> 
> Your help is greatly appreciated!
> 
> 
> >Hi Qun
> >
> >Because the 3 axes in orthorhombic space groups are 
> non-equivalent but
> >in P212121 all are 2-fold screws, there are indeed 6 permutations for
> >assigning the axes (unless of course 2 of those are excluded from
> >consideration by some other information that you haven't mentioned).
> >These are klh and lhk in addition to the ones you mention.  For a
> >reference see International Tables Vol A, chap. 2, p. 21.  
> Normally the
> >standard permutation is decided by the order of the magnitudes of the
> >cell axis lengths, but presumably you are doing this because 
> all 3 cell
> >lengths are very nearly equal (within experimental error including
> >possible non-isomorphism), otherwise what you are proposing wouldn't
> >make sense.  If for example only 2 of the cell lengths were 
> nearly equal
> >this would reduce the number of feasible possibilities to 
> only 2 (e.g.
> >hkl and hlk if a!=b=c).  The minus signs in the matrix are 
> required to
> >maintain the same hand after transformation in the case that you have
> >measured anomalous differences.  Alternatively you could keep the
> >indices all positive and change the sign of the delta-anoms 
> in the cases
> >where the permutation is non-cyclic (i.e. hlk, lkh and khl).  If you
> >have no anomalous data then it makes no difference whether 
> you have the
> >minus signs or not, by Friedel's law the amplitudes of hkl and -h-k-l
> >are equal.
> >
> >HTH!
> >
> >-- Ian
> >
> >>  -----Original Message-----
> >>  From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
> >>  Behalf Of Qun Wan
> >>  Sent: Tuesday, May 02, 2006 10:14 PM
> >>  To: [email protected]
> >>  Subject: [ccp4bb]: permute unit cell indices
> >>
> >>  ***  For details on how to be removed from this list 
> visit the  ***
> >>  ***          CCP4 home page http://www.ccp4.ac.uk         ***
> >>
> >>
> >>  Hi,
> >>  I Have a question about permuting crystal indices. When I 
> compare two
> >>  crystal data sets to get crossR value to know how isomorphous they
> >>  are, I need to permute the indices of one of the crystal. 
> Both of my
> >>  crystals have space group P212121. I was told to permute 
> HKL to LKH,
> >>  HLK, and KHL. I do not have a clear mind the principle. 
> For example,
> >>  by whole arrangement, there will be 6 possibilities for the
> >>  permutation. Why only four exist here? The second question is more
> >>  important: if I want to change from HKL to KHL, the new 
> indices are
> >>  (0 -1 0    -1 0 0    0 0 -1), not (0 1 0  1 0 0   0 0 1). 
>  Why there
> >>  is minus sign in the indices?
> >>
> >>  Could anyone recommend me one paper or book to read to solve
> >>  my puzzle?
> >>
> >>  Your help is greatly appreciated!
> >  >
> >>
> >
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