1. Experimental Phasing Andrew Howard ACA Summer School 22 July 2005

2. Experimental Phasing You can solve a structure with phases derived from experiments; it just may take some thinking. But the results will be statistically and esthetically satisfying.

3. Why don’t we always do this? Multiple experiments Sometimes requires specialized facilities Requires familiarity with a different set of software - so - We’ll often do difference Fouriers or molecular replacement even when we do have resources to do experimental phasing

4. Categories of Experimental Phasing Patterson methods Isomorphous replacement  Single isomorphous replacement  Multiple isomorphous replacement Anomalous diffraction  Multi-wavelength anomalous diffraction  Single-wavelenth anomalous diffraction  Optimized anomalous  ASIR / AMIR

5. General Concept Remember:  (r) = (1/V)  h F h exp(i  h ) exp(-2  i hr) We can measure F h We can’t trivially measure  h. So we seek an experimental probe that will enable us to estimate  h

6. Pattersons Calculate the following object: P(u) = (1/V 2 )  h |F h | 2 cos2  (hu) Note that h is a 3-vector in an integer- valued space, and u is a 3-vector in continuous space This allows for analysis of interatomic vectors, so if we have n atoms, we will find n(n-1)/2 peaks in the Patterson map in u.

7. Can we use this to solve structures? … sure, if n is moderate. Doesn’t require phase information directly! Whoopie! BUT If n=1000, n(n-1)/2 ~ 500000. Eech. So as a straight-ahead method for doing big molecular structures, this is a non-starter

8. Isomorphous replacement Relies on the fact that proteins and nucleic acids are almost entirely constructed from atoms with Z < 16, and mostly Z < 9. Scattering power for X-rays increases rapidly with Z Therefore if we have a small number of heavy atoms, our diffraction pattern will be significantly perturbed relative to the light- atom-only pattern

9. How does it work? Measure native data Measure data with heavy atom bound We rely on the fact that the Fourier transform is a linear transform:  (r) = (1/V)  h  (F h exp(i  h )) exp(-2  i hr) The inverse of that concept is applied to the problem we’re really trying to deal with.