Hi, unless I'm missing something, I think there is no need to use insertion code... All you need is to: 1) edit your PDB file to add altloc identifiers to individual components, 2) make sure the whole thing has the same residue number and chain id (example: see residue number 22 here http://www.rcsb.org/pdb/files/1EJG.pdb) 3) either go by the automatic grouping of occupancies (=do nothing, run phenix.refine), or instruct phenix.refine how it should combine individual conformers using occupancies { constrained_group { selection = .. } } as described here: http://phenix-online.org/documentation/refinement.htm#anch20 If still confused, then send me a PDB file with the extract of residues in question and tell me how they are expected to be grouped and I will send you back an example. Pavel. On 3/21/11 8:48 AM, Nathaniel Echols wrote:

...

I am working on the refinement of a structure about 2.8A, which the substrate ligand has an odd conformation. A blob of negative difference density about 4.8 sigma between the one portion of the ligand and the other. If I break down the ligand into two parts, the negative density disappears, however, one part of the ligands does not fit well. We suspect a mixture of conformations.

Any suggestions would be appreciated on how to setup refinement for ligands with alternative conformation. In theory it's just like setting up alternate conformations for

On Mon, Mar 21, 2011 at 6:35 AM, wrote: protein residues. However, it sounds like you have something similar to a partially hydrolyzed ATP, where conformation "A" is ATP, and conformation "B" is ADP (perhaps in combination with phosphate). There is nothing in phenix.refine that prevents you from doing this - unfortunately, it probably requires editing the PDB file manually to create the alternate conformations and rename the altloc (the column to the left of the residue name). For the other half of the hydrolyzed ligand, if you decide to model this explicitly, I'd recommend using the insertion code (the column to the right of the residue number), and leaving the residue number the same. Thus for ATP+ADP+Pi, you would end up with:

ATP = conformer A, residue 1 ADP = conformer B, residue 1 Pi = conformer B, residue 1P

Messy, but it should work.

-Nat _______________________________________________ phenixbb mailing list [email protected] http://phenix-online.org/mailman/listinfo/phenixbb