SSX-PepT Se: Difference between revisions
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mkdir <attendee_name>_SSX (e.g. Greta_SSX) | mkdir <attendee_name>_SSX (e.g. Greta_SSX) | ||
cd <attendee_name>_SSX | cd <attendee_name>_SSX | ||
find ../tutorial_data/PepT_SE_145/ -name xtal* > | find ../tutorial_data/PepT_SE_145/ -name xtal* > XSCALE.INP | ||
sed -i s/^/INPUT_FILE=/ | sed -i s/^/INPUT_FILE=/ XSCALE.INP | ||
paste this into the header: | paste this into the header: | ||
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PRINT_CORRELATIONS=FALSE ! this is not yet in the official BUILT=20171111 | PRINT_CORRELATIONS=FALSE ! this is not yet in the official BUILT=20171111 | ||
WFAC1=1 | WFAC1=1 | ||
Alternatively, solve the structure with only cluster 1 ! | |||
== solve structure with [[ccp4com:SHELX C/D/E|SHELX C/D/E]] == | == solve structure with [[ccp4com:SHELX C/D/E|SHELX C/D/E]] == | ||
(use [[ccp4com:hkl2map|hkl2map]])! | (use [[ccp4com:hkl2map|hkl2map]])! There are 458 residues, out of which 18 are SeMet, in PepT. | ||
== Notes == | |||
other option: run xscale_isocluster with the option -clu 1 and sort by strength, take the upper ones and solve the structure. | other option: run xscale_isocluster with the option -clu 1 and sort by strength, take the upper ones and solve the structure. | ||
Or | |||
run xscale_isocluster with the option -clu 1 and sort by strength*cos(angle), take the upper ones and solve the structure. | |||
That would be | |||
awk '/INPUT_FILE/{ one = $1 ; getline ; print one " " $7*cos($8*3.14159/180) " " $8;getline}' XSCALE.1.INP |grep INPUT_FILE >temp | |||
awk '/INPUT_FILE/{ one = $1 ; getline ; print one " " $7*cos($8*3.14159/180) " " $8;getline}' XSCALE.4.INP |grep INPUT_FILE >>temp | |||
sort -nr -k2 temp >temp1 |
Latest revision as of 11:57, 29 November 2017
This requires 145 XDS_ASCII.HKL files collected at X06SA (SLS).
generating XSCALE.INP for scaling
mkdir <attendee_name>_SSX (e.g. Greta_SSX) cd <attendee_name>_SSX find ../tutorial_data/PepT_SE_145/ -name xtal* > XSCALE.INP sed -i s/^/INPUT_FILE=/ XSCALE.INP
paste this into the header:
UNIT_CELL_CONSTANTS= 104.056 110.896 110.369 90 90 90 SPACE_GROUP_NUMBER=20 MINIMUM_I/SIGMA=1 OUTPUT_FILE=all_145.hkl SAVE_CORRECTION_IMAGES=FALSE PRINT_CORRELATIONS=FALSE ! this is not yet in the official BUILT=20171111 WFAC1=1
scale everything together
rm xscale.inp xscale_par
-->have a look at XSCALE.LP
run xscale_isocluster
xscale_isocluster -dim 3 -i all_145.hkl | tee XSCALE_ISOCLUSTER.LP
--> check output on screen (which is saved on XSCALE_ISOCLUSTER.LP). The program says that we obtain 4 clusters. Check with coot:
coot iso.pdb
We don't really see the clusters as separate when looking at iso.pdb in coot. So let us just take the 4 clusters that xscale_isocluster detects as an opportunity to use groups of data sets for further work, without implying any deeper meaning of the clusters.
look at the 4 clusters
load iso.1.pdb into coot. Also load iso.2.pdb , iso.3.pdb, iso.4.pdb and use the Display Manager for switching the clusters on/off
decide which clusters one would take for structure solution
Here: 1. and 4. cluster
cp XSCALE.INP XSCALE_old.INP cp XSCALE.LP XSCALE_old.LP awk '/INPUT_FILE/{ one = $1 ; getline ; print one " " $7 " " $8;getline}' XSCALE.1.INP |grep 'INPUT_FILE' >temp awk '/INPUT_FILE/{ one = $1 ; getline ; print one " " $7 " " $8;getline}' XSCALE.4.INP |grep 'INPUT_FILE'>>temp sort -nr -k2 temp >temp1
--> look at temp1 , combined clusters ,sorted by strength --> remove lowest (until strength of 0.6)
awk '{print $1}' temp1 > XSCALE.INP
paste this into header:
UNIT_CELL_CONSTANTS= 104.056 110.896 110.369 90 90 90 SPACE_GROUP_NUMBER=20 MINIMUM_I/SIGMA=1 OUTPUT_FILE=1+4.hkl SAVE_CORRECTION_IMAGES=FALSE PRINT_CORRELATIONS=FALSE ! this is not yet in the official BUILT=20171111 WFAC1=1
Alternatively, solve the structure with only cluster 1 !
solve structure with SHELX C/D/E
(use hkl2map)! There are 458 residues, out of which 18 are SeMet, in PepT.
Notes
other option: run xscale_isocluster with the option -clu 1 and sort by strength, take the upper ones and solve the structure. Or
run xscale_isocluster with the option -clu 1 and sort by strength*cos(angle), take the upper ones and solve the structure. That would be
awk '/INPUT_FILE/{ one = $1 ; getline ; print one " " $7*cos($8*3.14159/180) " " $8;getline}' XSCALE.1.INP |grep INPUT_FILE >temp awk '/INPUT_FILE/{ one = $1 ; getline ; print one " " $7*cos($8*3.14159/180) " " $8;getline}' XSCALE.4.INP |grep INPUT_FILE >>temp sort -nr -k2 temp >temp1