Hi Garib I think that was the first thing we tried but it gave very poor geometry for the ACE and also it's giving a VDW outlier for the ACE C - MET N bond. So it looks like it's not recognising the link.
Cheers -- Ian On 14 February 2014 11:51, Garib Murshudov <ga...@mrc-lmb.cam.ac.uk> wrote: > Hi Ian > > dod you try without link and standard ACE atom naming. Refmac should be > able to deal N-terminal activation and few other things. At least it was > the intention when it was written. Bugs may have been (self)introduced to > prevent this from happening. If it is so then I would like to know. > > Regards > Garib > > > On 13 Feb 2014, at 22:57, Ian Tickle <ianj...@gmail.com> wrote: > > > All, I'm having problems refining a structure with an N-terminal > acetylated MET residue. I'm trying it with both Refmac & Buster. Buster > works fine & gives perfect planar geometry for the ACE-MET linkage. Refmac > gives a pyramidal acetyl group after refinement which to my eyes is wrong > (sp2 C atom?). > > I have this line in my input PDB: > > LINKR C ACE A 0 N MET A 1 > ACE_C-N > > which as I understand it should solve the problem. However, looking at > the CIF entry for the ACE_C-N link I see restraints defined for bonds, > angles & torsion angles but not for the CC(=O)N plane. So the problem > seems to be that the planar restraints for this link group are missing - or > are they defined elsewhere? Anyway I added planar restraints to the > ACE_C-N link entry & it solves the problem, at least for regularisation - I > still have the same problem with refinement. Refmac in regularisation mode > now gives the correct (planar) geometry for the ACE-MET linkage. I'm just > puzzled why no-one has noticed this, after all post-translational > acetylation is surely not that uncommon (according to Wikipedia > 80% of > human proteins are N-term acetylated!). > > Further, looking at the entry for ACE I see: > > ACE ACE 'ACETYL GROUP ' non-polymer > 7 3 . > > ACE O O O 0.000 0.000 0.000 0.000 > ACE C C C1 0.000 -1.044 -0.606 0.000 > ACE H H H 0.000 -1.978 -0.069 0.000 > ACE CH3 C CH3 0.000 -1.041 -2.113 0.000 > ACE H3 H H 0.000 -0.541 -2.464 0.865 > ACE H2 H H 0.000 -2.038 -2.468 0.000 > ACE H1 H H 0.000 -0.540 -2.464 -0.864 > > Where did the extra H atom (3rd atom) come from? Acetyl is CH3C=O: the > extra H atom would make it acetaldehyde which of course has nothing > whatsoever to do with acetylation! Is this the reason for the lack of > planar link restraints (though that wouldn't explain why the other link > restraints are present)? > > Any insights appreciated! > > Cheers > > -- Ian > > > Dr Garib N Murshudov > MRC-LMB > Francis Crick Avenue > Cambridge > CB2 0QH UK > Web http://www.mrc-lmb.cam.ac.uk, > http://www2.mrc-lmb.cam.ac.uk/groups/murshudov/ > > > >
