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 <[email protected]> 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/



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