2. Fourier transform

3. Fourier transform

4. Combine phases from the cat and amplitude from the duck

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

6. Fasproblemet Tungmetallderivat (MIR, SIR) Anomalous dispersion
Patterson kartor

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18. Friedels Lag (hkl = -h –k –l)

19. Anomalous dispersion

20. Anomalous dispersion

21. 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

22. Two-folds a b H Li

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

24. 4 Choices of origin Choice 1 Choice 2 Choice 3 Choice 4 b a H Li H Li

25. 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

26. 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

27. 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

28. 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

29. 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)

30. 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.

31. Electron Density vs. Patterson Densityb 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

32. Every Patterson peak corresponds to an inter-atomic vectorb
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

33. 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

34. Patterson maps have an additional center of symmetryb
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

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

36. Plane group p2 b H

37. Let’s consider only oxygen atomsb 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.

38. 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?

39. 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.
( , )=(0.8, 0.7)

40. 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

41. 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

42. 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

43. 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.

44. 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

45. 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!!!!