Agreed, PISA is the way to go,and that is what RCSB would have used if the author had not already volunteered the information that the biomolecule is a dimer. But i also agree with Randy Read that it is educational to look and see for yourself. Also empowering to know that you _can_ always look for yourself instead of relying on a highly sophisticated application to tell you. And if one day the PISA server is down and you need a quick answer. . .
On 06/07/2016 05:13 PM, Eleanor Dodson wrote:
> Well PISA would tell you if A and C make crystallographic dimers..
> E
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> On 7 June 2016 at 21:47, Edward A. Berry <[log in to unmask] <mailto:[log in to unmask]>> wrote:
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> On 06/07/2016 04:20 PM, Edward A. Berry wrote:
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> On 06/07/2016 03:00 PM, Randy Read wrote:
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> I like Eleanor’s suggestion of using PISA to find the assemblies automatically. That’s probably the best thing to do first.
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> However, there’s an alternative I like that uses your brain’s pattern-matching abilities. In coot, view your molecules as C-alpha traces and display a generous number symmetry copies, also as C-alpha. Center on an atom in one of the monomers (either use Draw->Go to atom, or middle click followed by the “p” key so that coot really knows you’re centered on that atom). Then just repeatedly hit the “o” key, which goes to the same view of the next NCS-related monomer. As you’re running through them, keep an eye on the neighbours, to see if any relationship among neighbours always recurs. This can also be a useful way to get a hint of missed symmetry, because then there will be views that should have a different packing environment but nonetheless look the same!
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> More direct- find a residue in the b-d interface. Go to same residue of A. Show sym atoms. You will see a sym mate of C making the same interface with A as you saw between B and D when you selected your residue. middle-click on an atom in that model, (verify that it is indeed chain C), and read the symmetry operator from the status bar below the graphics. Use this to fill in the template that the annotator provided you: "A+ moving C with symmetry operation -x,y,1-z= dimer3"
> (or am I missing something?)
> eab
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> Or perhaps A and C make crystallographic dimers, with their own sym-mates. In that case you have 4 examples of the dimer, despite only 6 monomers in the AU. "A+ moving A with symmetry operation -x,y,1-z= dimer3, C+ moving C with symmetry operation -x,y,1-z= dimer4"
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> Or, much less likely I think, you actually have 2 dimers and 2 monomers.
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