Fourier transform

Fourier transform

Combine phases from the cat and amplitude from the duck

Phase problem Four methods to solve the phase problem Heavy-atom derivative Molecular replacement Isomorphous replacement Multiwavelength anomalous dispersion (MAD phasing)

Fasproblemet Tungmetallderivat (MIR, SIR) Anomalous dispersion Patterson kartor

© 2006 Academic Press

© 2006 Academic Press

© 2006 Academic Press

© 2006 Academic Press

© 2006 Academic Press

© 2006 Academic Press

© 2006 Academic Press

© 2006 Academic Press

© 2006 Academic Press

© 2006 Academic Press

Friedels Lag (hkl = -h –k –l)

Anomalous dispersion

Anomalous dispersion

Draw the location of all 2-fold symmetry axes using the ellipse symbol. Draw a unit cell. How many molecules in the unit cell? asymmetric unit? Estimate (x,y) coordinates of oxygen atom in fractions of a unit cell. Use the ruler provided. a b H Li

Two-folds a b H Li

Unit Cell a b H Li Choice 1 Choice 2 Choice 3 Choice 4

4 Choices of origin Choice 1 Choice 2 Choice 3 Choice 4 b a H Li H Li

What are the coordinates of the oxygen using origin choice 1? b Li Li Li 1 2 3 4 H H H H H H Li Li Li Li Li Li H H H H H H Li Li Li 1 2 3 4 Li Li Li X=0.20 Y=0.20 X=0.2 H H H H H H Li Li Li

What are the coordinates of the other oxygen in the unit cell? b H Li X=0.20 Y=0.20 1 2 3 4 X=0.80 Y=0.80 1 2 3 4 Symmetry operators in plane group p2 X, Y -X,-Y

Always allowed to add or subtract multiples of 1.0 X=0.2 Y=0.2 X=1.8 Y=0.8 X=1.2 X=0.8 X=2.2 X=2.8 b H Li

Patterson Review Fourier synthesis A Patterson synthesis is like a Fourier synthesis except for what two variables? Fourier synthesis r(xyz)=S |Fhkl| cos2p(hx+ky+lz -ahkl) hkl Patterson synthesis P(uvw)=S Ihkl cos2p(hu+kv+lw -0) hkl Patterson synthesis P(uvw)=S ?hkl cos2p(hu+kv+lw -?) hkl

Hence, Patterson density map= electron density map convoluted with its inverted image. Patterson synthesis P(uvw)=S Ihkl cos2p(hu+kv+lw) Remembering Ihkl=Fhkl•Fhkl* And Friedel’s law Fhkl*= F-h-k-l P(uvw)=FourierTransform(Fhkl•F-h-k-l) P(uvw)=r(uvw) r (-u-v-w)

Significance? P(uvw)=r(uvw) r (-u-v-w) The Patterson map contains a peak for every interatomic vector in the unit cell. The peaks are located at the head of the interatomic vector when its tail is placed at the origin.

Electron Density vs. Patterson Density b a b Lay down n copies of the unit cell at the origin, where n=number of atoms in unit cell. H H H 1 2 3 H For copy n, atom n is placed at the origin. A Patterson peak appears under each atom. Patterson Density Map single water molecule convoluted with its inverted image. Electron Density Map single water molecule in the unit cell

Every Patterson peak corresponds to an inter-atomic vector b 3 sets of peaks: Length O-H Where? Length H-H Length zero How many peaks superimposed at origin? How many non-origin peaks? H 1 2 3 H Patterson Density Map single water molecule convoluted with its inverted image. Electron Density Map single water molecule in the unit cell

Patterson maps are more complicated to interpret than electron density maps. Imagine the complexity of a Patterson map of a protein a b H Unit cell repeats fill out rest of cell with peaks Patterson Density Map single water molecule convoluted with its inverted image. Electron Density Map single water molecule in the unit cell

Patterson maps have an additional center of symmetry b plane group pm H plane group p2mm H Patterson Density Map single water molecule convoluted with its inverted image. Electron Density Map single water molecule in the unit cell

Calculating X,Y,Z coordinates from Patterson peak positions (U,V,W) A simple 2D example using a water molecule

Plane group p2 b H

Let’s consider only oxygen atoms b H Analogous to a difference Patterson map where we subtract out the contribution of the protein atoms, leaving only the heavy atom contribution. Leaves us with a Patterson containing only self vectors (vectors between equivalent atoms related by crystal symmetry). Unlike previous example.

How many faces? (0,0) a b In unit cell? In asymmetric unit? How many peaks will be in the Patterson map? How many peaks at the origin? How many non-origin peaks?

Symmetry operators in plane group p2 Coordinates of one smiley face are given as 0.2, 0.3. Coordinates of other smiley faces are related by symmetry operators for p2 plane group. For example, symmetry operators of plane group p2 tell us that if there is an atom at (0.2, 0.3), there is a symmetry related atom at (-x,-y) = (-0.2, -0.3). But, are these really the coordinates of the second face in the unit cell? (-0.2,-0.3) (0,0) a b (0.2,0.3) SYMMETRY OPERATORS FOR PLANE GROUP P2 1) x,y 2) -x,-y Yes! Equivalent by unit cell translation. (-0.2+1.0, -0.3+1.0)=(0.8, 0.7)

Patterson in plane group p2 Lay down two copies of unit cell at origin, first copy with smile 1 at origin, second copy with simile 2 at origin. 2D CRYSTAL PATTERSON MAP (-0.2,-0.3) (0,0) a (0,0) a b b (0.2,0.3) What are the coordinates of this Patterson self peak? (a peak between atoms related by xtal sym) What is the length of the vector between faces? SYMMETRY OPERATORS FOR PLANE GROUP P2 1) x,y 2) -x,-y Patterson coordinates (U,V) are simply symop1-symop2. Remember this bridge! symop1 X , Y = 0.2, 0.3 symop2 -(-X,-Y) = 0.2, 0.3 2X, 2Y = 0.4, 0.6 = u, v

Patterson in plane group p2 (-0.4, -0.6) (-0.2,-0.3) (0,0) a a (0,0) b b (0.2,0.3) (0.6, 0.4) (0.4, 0.6) SYMMETRY OPERATORS FOR PLANE GROUP P2 1) x,y 2) -x,-y 2D CRYSTAL PATTERSON MAP

Patterson in plane group p2 (0,0) a b (0.6, 0.4) (0.4, 0.6) If you collected data on this crystal and calculated a Patterson map it would look like this. PATTERSON MAP

Now I’m stuck in Patterson space. How do I get back to x,y coordinates? Remember the Patterson Peak positions (U,V) correspond to vectors between symmetry related smiley faces in the unit cell. That is, differences between our friends the space group operators. (0,0) a b (0.6, 0.4) (0.4, 0.6) SYMMETRY OPERATORS FOR PLANE GROUP P2 1) x,y 2) -x,-y PATTERSON MAP x , y -(-x, –y) 2x , 2y u=2x, v=2y symop #1 symop #2 plug in Patterson values for u and v to get x and y.

Now I’m stuck in Patterson space. How do I get back to x,y, coordinates? SYMMETRY OPERATORS FOR PLANE GROUP P2 1) x,y 2) -x,-y (0,0) a b x y -(-x –y) 2x 2y symop #1 symop #2 (0.4, 0.6) set u=2x v=2y plug in Patterson values for u and v to get x and y. u=2x 0.4=2x 0.2=x v=2y 0.6=2y 0.3=y PATTERSON MAP

Hurray!!!! 2D CRYSTAL SYMMETRY OPERATORS FOR PLANE GROUP P2 1) x,y (0,0) a b x y -(-x –y) 2x 2y symop #1 symop #2 (0.2,0.3) set u=2x v=2y plug in Patterson values for u and v to get x and y. 2D CRYSTAL u=2x 0.4=2x 0.2=x v=2y 0.6=2y 0.3=y HURRAY! we got back the coordinates of our smiley faces!!!!