The presence of a peak in the fluorescence doesn't guarantee low
enough standard deviations in intensities to ensure 'measurable'
anomalous differences.
How well did your crystal diffract?
The problem reported earlier is most likely a bug (which I hope can be
reproduced).
Cheers
Peter
2009/8/11, [email protected]
Hi, I had similar problems recently, I did a wavelength scan of my crystal at the synchrotron (selenium edge) and collected one dataset at the peak wavelength.The scan gives me a distinct peak. However when I processed the data and evaluated in xtriage it says that there is no anomolous signal. I am unable to explain this? any advice? Shya
Uber Odd
I reprocessed it and it seems to have resolved itself.
Unfortunately though there seems to be no real significant anomalous differences in this dataset. Back to the drawing board.
FR
On Aug 11, 2009, at 8:54 AM, Francis E Reyes wrote:
1.4-87
FR
On Aug 11, 2009, at 8:53 AM, Peter Zwart wrote:
which version are you using?
P
2009/8/11 Francis E Reyes
: Hi all
I ran xtriage on my dataset to 4.5.here's the relevant section pertaining to anomalous signal
Analyses of anomalous differences
Table of measurability as a function of resolution
The measurability is defined as the fraction of Bijvoet related intensity differences for which |delta_I|/sigma_delta_I > 3.0 min[I(+)/sigma_I(+), I(-)/sigma_I(-)] > 3.0 holds. The measurability provides an intuitive feeling of the quality of the data, as it is related to the number of reliable Bijvoet differences. When the data are processed properly and the standard deviations have been estimated accurately, values larger than 0.05 are encouraging. Note that this analyses relies on the correctness of the estimated standard deviations of the intensities.
unused: - 90.6596 [ 0/1 ] bin 1: 90.6596 - 9.2614 [1371/1382] 0.0375 bin 2: 9.2614 - 7.3520 [1369/1376] 0.0033 bin 3: 7.3520 - 6.4230 [1380/1388] 0.0000 bin 4: 6.4230 - 5.8358 [1353/1359] 0.0016 bin 5: 5.8358 - 5.4176 [1367/1371] 0.0000 bin 6: 5.4176 - 5.0982 [1361/1369] 0.0000 bin 7: 5.0982 - 4.8429 [1378/1382] 0.0000 bin 8: 4.8429 - 4.6321 [1351/1359] 0.0000 bin 9: 4.6321 - 4.4538 [1383/1389] 0.0000 bin 10: 4.4538 - 4.3001 [1390/1393] 0.0000 unused: 4.3001 - [ 0/0 ]
The full resolution range seems to contain a useful ammount of anomalous signal. Depending on your specific substructure, you could use all the data available for the location of the heavy atoms, or cut the resolution to speed up the search.
If values larger than 0.05 are encouraging why is it telling me that the entire resolution contains a useful amount of anomalous signal? Did I process this incorrectly?
Thanks
FR
--------------------------------------------- Francis Reyes M.Sc. 215 UCB University of Colorado at Boulder
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