Since a crystallographic 3-fold generates the trimer (of dimers) then A1-B2 can be the ASU in this case.
Just generate the symmetry mates with the A1-B1 dimer and then make a new PDB of A1-B2 then generate the symmetry mates of A1-B2 to see that the lattice is complete.
I compulsively made and attached an illustration to show what I mean.
James
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On Oct 20, 2011, at 3:40 AM, Kayashree M wrote: Thank you Sir for the suggestions.
Hi,
If, in your case, no possible asymmetric unit can contain A1-B2, then
you deposit A1-B1 (or I suppose A2-B2...) and indicate to the PDB (like
placing cards in the header cards) the operator to be used (and the
subunit it applies to) in order to generate the most likely biological
dimer. Normally the PDB can take care of that.
The protein looks like the letter "C", So in one of the trimer it is arranged
as "C" while in the other trimer (stacked) it is arranged like "inverted C", So
The dimer A1-B1 and A2-B2 are same while A1-B1 and A1-B2 are different.
Thanking you
With Regards
Kavya
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