Dear Noguchi, If I understand correctly, you have one protein with 8 identical repeats and a few additional residues at the N-terminus that you cannot see in the electron density map. This means, that the N-terminal repeat is different from the other repeats and theoretically, you should submit the complete molecule in the lower-symmetry space group, including data processed at lower symmetry.
However, since the N-terminal residues are not visible, I suspect that they are not involved in any crystal contacts and that your protein molecules have random orientations within the 8-fold internal symmetry, averaging-out any differences between the different repeats. In this case, I would submit the partial molecule in the high-symmetry space group, but it will probably be something the pdb has not foreseen and you may have to do some explanation. Concerning the lower Rfree in your lower symmetry space group: With non-crystallographic symmetry, you have to carefully select your free reflections (including all NCS mates), otherwise your free reflections will not be free, but coupled via NCS to your working set reflections, which I suspect will be the case for you. Best, Herman -----Ursprüngliche Nachricht----- Von: CCP4 bulletin board [mailto:CCP4BB@JISCMAIL.AC.UK] Im Auftrag von Hiroki Noguchi Gesendet: Montag, 27. November 2017 13:30 An: CCP4BB@JISCMAIL.AC.UK Betreff: [EXTERNAL] [ccp4bb] Spacegroup for a symmetric monomeric protein Dear All, We have a rather confusion situation currently with our designer proteins. They are highly symmetric and consist of identical amino acid repeats in tandem that are organized in 3D around a central (rotation axis). Our first series of proteins like this was the Pizza (3ww9) which crystallized in P212121. However for other proteins we have now a bit of a problem. After crystallization of our new proteins some of them crystallized in such a way that we can solve the structure with such a symmetry that only a fragment of the protein is present. For instance we only see 2 repeats in the asymmetric unit but in fact in the monomeric structure there should be 8. The 8fold symmetric protein can be reconstructed from crystal symmetry. My question is now. Should we solve and deposit this crystal structure to the most symmetrical smallest asymmetric unit resulting in an incomplete protein in the asymmetric unit or should we solve the structure in a lower space group such that the asymmetric unit contains a full length monomeric protein. The protein has additional few amino acids at N-terminal but we can not see these regions on electronic density map. When comparing the two R free we see that the lowest spacegroup has a slightly better (4%) R free value than the higher spacegroup with an incomplete protein in the asymmetric unit. The PDB document (https://urldefense.proofpoint.com/v2/url?u=https-3A__cdn.rcsb.org_wwpdb_docs_documentation_annotation_wwPDB-2DA-2D2017Aug02-2DV2.8.pdf&d=DwIBaQ&c=Dbf9zoswcQ-CRvvI7VX5j3HvibIuT3ZiarcKl5qtMPo&r=HK-CY_tL8CLLA93vdywyu3qI70R4H8oHzZyRHMQu1AQ&m=1Y5lSTX28-d9UR3haS7N8B16DvGd8reLwqeZwMRY3yg&s=mUPJXE8ewFueTn3Jfbg-QgxnjguFB5i3CjJvQz1v08Q&e= ) is described, "The asymmetric unit may be one molecule or one subunit of a multimeric protein, but it can also be more than one." on page 34. So should we go for the highest symmetry as observed from the diffraction data, or go for a lower symmetry agreeing with a full protein. Thank you for you help. Kindest Regards, Hiroki noguchi KU LEUVEN, Belgium
