Rigid body refinement question
Dear phenixbb, What do I need to change in my .def file if I want to refine chain A-E of a structure as per the defaults but specify a single chain (in my case chain L) for just rigid refinement in all cycles? I had a look at the manual page (http://www.phenix-online.org/documentation/refinement.htm ) but didn't find it too clear. Thanks, Simon --------------------------------------------------------------- Dr Simon Kolstoe Laboratory for Protein Crystallography Centre for Amyloidosis & Acute Phase Proteins UCL Medical School Rowland Hill Street, London NW3 2PF Tel: 020 7433 2765 http://www.ucl.ac.uk/~rmhasek --------------------------------------------------------------
What do I need to change in my .def file if I want to refine chain A-E of a structure as per the defaults but specify a single chain (in my case chain L) for just rigid refinement in all cycles?
I think this should work: refinement.refine.strategy=rigid_body+individual_sites refinement.refine.sites.individual="chain A or chain B or chain C or chain D or chain E" refinement.refine.sites.rigid_body=chain L refinement.refine.mode=every_macro_cycle Ralf
Dear phenixbb I am refining a low resolution data (4.1 A). I first refined my data with CNS, then I am trying TLS refinement with Phenix. Although R-factors with bulk solvent and anisotropic scale are reducing, R-factor without them are increasing much. Start R-work = 0.3762, R-free = 0.4043 (no bulk solvent and anisotropic scale) Final R-work = 0.5019, R-free = 0.5204 (no bulk solvent and anisotropic scale) Start R-work = 0.2665, R-free = 0.3195 Final R-work = 0.2578, R-free = 0.3126 Why are R-factors with no bulk solvent and anisotropic scale increasing so much? Thanks, Shunsuke
Hi Shunsuke, at first glance, I don't see anything to worry about. It was a mistake to print this information at all: R-work = ..., R-free = ... (no bulk solvent and anisotropic scale) and I will probably remove it in one of the next versions. Your start values really are: Start R-work = 0.2665, R-free = 0.3195 and the final ones: Final R-work = 0.2578, R-free = 0.3126. Thanking some more... Actually, I don't understand why this is the case: Start R-work = 0.3762, R-free = 0.4043 (no bulk solvent and anisotropic scale) Final R-work = 0.5019, R-free = 0.5204 (no bulk solvent and anisotropic scale) - it is weird. If you send me (to my email address, not to the whole bb) the data and model then I will be able to provide some more diagnostics. Are you using a recent version of PHENIX ? Is it not a result of running TLS refinement first and then running a refinement without TLS - that would explain this, actually, since the ANISOU all will be converted to isotropic equivalents. Pavel. On 5/6/10 8:00 PM, Shunsuke Tagami wrote:
Dear phenixbb
I am refining a low resolution data (4.1 A). I first refined my data with CNS, then I am trying TLS refinement with Phenix. Although R-factors with bulk solvent and anisotropic scale are reducing, R-factor without them are increasing much.
Start R-work = 0.3762, R-free = 0.4043 (no bulk solvent and anisotropic scale) Final R-work = 0.5019, R-free = 0.5204 (no bulk solvent and anisotropic scale)
Start R-work = 0.2665, R-free = 0.3195 Final R-work = 0.2578, R-free = 0.3126
Why are R-factors with no bulk solvent and anisotropic scale increasing so much?
Thanks,
Shunsuke
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Dear Shunsuke, you indicated at the first line of you mail that you are working at a low resolution
I am refining a low resolution data (4.1 A).
where the bulk solvent contribution is essential ; at lower resolutions it becomes even dominant.
Start R-work = 0.3762, R-free = 0.4043 (no bulk solvent and anisotropic scale) Final R-work = 0.5019, R-free = 0.5204 (no bulk solvent and anisotropic scale)
Start R-work = 0.2665, R-free = 0.3195 Final R-work = 0.2578, R-free = 0.3126
Therefore it is normal that ignoring this component gives you a model with a very high R. So the real question is to be sure that the bulk solvent parameters have correct values (you did not give them). Looking at your total R-factors, I would believe it is OK. The R-factors perfectly correspond to the linear approximations : R = 0.091 ln(d) + 0.134 = 0.262 (you have 0.258) Rfree-R = 0.024 ln(d) + 0.020 = 0.054 (you have 0.055) (see Urzhumtsev, Afonine, Adams, 2009, Acta Cryst D65, 1283-1291) Best regards, Sacha
participants (5)
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Alexandre Urzhumtsev -
Pavel Afonine -
Ralf W. Grosse-Kunstleve -
Shunsuke Tagami -
Simon Kolstoe