I don't want to confuse things further, but as a PS to Ian's answer that clearly tells you how to get Z..
You should be aware that a crystal might also have non-crystallographic symmetry. Ian's answer is right for Z, but as you also mentioned monomers I thought I should mention that if the a.u. comprises a multimer then you would have to multiply. i.e. symops will give the number of copies of the asymmetric unit in the unit cell, this might not be the number of monomers. Peter On 2 August 2012 10:12, Ian Tickle <ianj...@gmail.com> wrote: > Hi Careina > > The obvious answer is to look it up in International Tables vol. A. > If you don't have access to that you can look up the text file > $CLIBD/syminfo.lib in the CCP4 distribution and work it out from > there. Fpr a given space group you need to count the number of > 'symop' lines. That's the number of asymmetric units in the primitive > cell. If you want the number in a centred cell, also count the number > of 'cenop' lines: that's the number of centring translations, so to > get the total no of a.u.'s in the centred cell you would multiply > these two numbers. > > I'm afraid there isn't an easier way, for example you can't get it > straightforwardly from the space group symbol. > > HTH > > -- Ian > > On 2 August 2012 09:37, Careina Edgooms <careinaedgo...@yahoo.com> wrote: > > Dear ccp4 > > > > I ask a very fundamental question because I have not had formal training > in > > this and I would like to understand. > > How can I obtain the multiplicity (z) from the space group? So for > example > > if the space group is P222 how do I know that there are 4 monomers in the > > unit cell? Or if it is P422 then there is 8? I am only concerning myself > > with a primitive lattice for now because I am sure the others are more > > complicated. > > thanks > > Careina >
