1. 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4

2. --2-fold symmetry axis
Chose a 2-fold axis as the origin.
How many molecules in the unit cell? asymmetric unit?
Draw an “o” at the upper left corner of the cell and make it the origin.
Estimate (x,y) coordinates of oxygen atom in fractions of a unit cell. Use the ruler provided. a o H Li
On the crystal, b
draw a unit cell
having this size
and shape.

3. 4 distinct two-fold axesb H Li

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

5. 4 Choices of origin Choice 1 Choice 2 Choice 3 Choice 4 b a H Li H Li

6. Which plane group? H Li b a

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

8. What are the coordinates of the other oxygen in the unit cell?Li
Symmetry operators in plane group p2
X, Y
-X,-Y a b
X2=0.80
Y2=0.80 1 2 3 4 1 2 3 4

9. Always allowed to add or subtract multiples of 1.0
X=0.2
Y=-0.8
X=1.8
Y=-0.2
X=1.2
X=0.8
X=-0.8
X=-.0.2 H Li b a
X=0.2
Y=0.2
X=1.8
Y=0.8
X=1.2
X=0.8
X=-0.8
X=-.0.2
X=0.2
Y=1.2
X=1.8
Y=1.8
X=1.2
X=0.8
X=-0.8
X=-.0.2

10. What would be the oxygen coordinates if we had drawn the unit cell with origin choice 2?Li 1 2 3 4
X=0.70
Y=0.20 1 2 3 4

11. Oxygen coordinates with origin choice 3?b H Li 1 2 3 4
X=0.30
Y=0.30 1 2 3 4

12. Oxygen coordinates with origin choice 4?b H Li 1 2 3 4
X=0.80
Y=0.30 1 2 3 4

13. Cheshire operators X , Y X+.5, Y X+.5, Y+.5 X , Y+.5 Choice 1 Choice 2

14. The 4 choices of origin are equally valid but once a choice is made, you must remain consistent.
X1=0.20
Y1=0.20
X1=0.70
Y1=0.20 H Li
X2=0.80
Y2=0.80
X2=0.30
Y2=0.80
Choice 3
Choice 4 H Li H Li
X1=0.30
Y1=0.30 Li H Li Li
X1=0.80
Y1=0.30 H H H H H Li Li Li
X2=0.70
Y2=0.70 H Li
X2=0.20
Y2=0.70

15. What did we learn?
There are multiple valid choices of origin for a unit cell.
The values of x,y,z for the atoms will depend on the choice of origin.
If a structure is solved independently by two crystallographers using different choices of origin, their coordinates will be related by a Cheshire opertor.
Adding 1 to x, y, or z, or any combination of x, y, and z is valid. It is just a unit cell translation.

16. Interpreting difference Patterson Maps in Lab this week!
Calculate an isomorphous difference Patterson Map (native-heavy atom). We collected 16 derivative data sets in lab (different heavy atoms at different concentrations)
Mersalyl, PCMBS
GdCl3
Did a heavy atom bind? How many?
What are the positions of the heavy atom sites?
Let’s review how heavy atom positions can be calculated from difference Patterson peaks.

17. 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

18. 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)

19. 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.

20. 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

21. 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

22. 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

23. 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

24. Calculating X,Y,Z coordinates from Patterson peak positions (U,V,W) Three Examples
Exceedingly simple 2D example
Straightforward-3D example, Pt derivative of polymerase b in space group P21212
Advanced 3D example, Hg derivative of proteinase K in space group P43212.

25. Plane group p2 b H

26. 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.

27. 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?

28. Symmetry operators in plane group p2
(-0.2,-0.3)
(0,0) x y
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. (-0.2, -0.3).
But, are these really the coordinates of the second face in the unit cell?
(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)

29. Patterson coordinates (U,V) are simply symop1-symop2.
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) x
(0,0) u y v
(0.2,0.3)
Patterson coordinates (U,V) are simply symop1-symop2.
Remember this bridge!
symop1= X , Y
-symop2= -(-X,-Y)
u=2X v=2Y
u=2(0.2) v=2(0.3)
u=0.4, v=0.6
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

30. Patterson in plane group p2
(-0.4, -0.6)
(-0.2,-0.3)
(0,0) x u
(0,0) y v
(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

31. Patterson in plane group p2
(0,0) u v
(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

32. 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) u v
(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.

33. 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) u v
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

34. Hurray!!!! 2D CRYSTAL SYMMETRY OPERATORS FOR PLANE GROUP P2 1) x,y
(0,0) u v
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!!!!

35. Devil’s advocate: What if I chose u,v= (0. 6,0. 4) instead of (0. 4,0
Devil’s advocate: What if I chose u,v= (0.6,0.4) instead of (0.4,0.6) to solve for smiley face (x,y)?
using Patterson (u,v) values 0.4, 0.6
to get x and y.
(0,0) u v
u=2x
0.4=2x
0.2=x
v=2y
0.6=2y
0.3=y
(0.6, 0.4)
(0.4, 0.6)
using Patterson (u,v) values 0.6, 0.4
to get x and y.
u=2x
0.6=2x
0.3=x
v=2y
0.4=2y
0.2=y
PATTERSON MAP
These two solutions do not give the same x,y? What is going on??????

36. Arbitrary choice of origin
(-0.2,-0.3)
(0,0) x y
(-0.3,-0.2)
(0.2,0.3)
(0,0) u v
(0.8,0.7)
(0.3,0.2)
(0.7,0.8)
Original origin choice
Coordinates x=0.2, y=0.3.
New origin choice
Coordinates x=0.3, y=0.2.
Related by 0.5, 0.5 (x,y) shift

37. Recap
Patterson maps are the convolution of the electron density of the unit cell with its inverted image.
The location of each peak in a Patterson map corresponds to the head of an inter-atomic vector with its tail at the origin.
There will be n2 Patterson peaks total, n peaks at the origin, n2-n peaks off the origin.
Peaks produced by atoms related by crystallographic symmetry operations are called self peaks.
There will be one self peak for every pairwise difference between symmetry operators in the crystal.
Written as equations, these differences relate the Patterson coordinates u,v,w to atom positions, x,y,z.
Different crystallographers may arrive at different, but equally valid values of x,y,z that are related by an arbitrary choice of origin or unit cell translation.

38. Polymerase b example, P21212
Difference Patterson map, native-Pt derivative.
Where do we expect to find self peaks?
Self peaks are produced by vectors between atoms related by crystallographic symmetry.
From international tables of crystallography, we find the following symmetry operators.
X, Y, Z
-X, -Y, Z
1/2-X,1/2+Y,-Z
1/2+X,1/2-Y,-Z
Everyone, write the equation for the location of the self peaks. 1-2, 1-3, and 1-4 Now!

39. Self Vectors X, Y, Z -X, -Y, Z X, Y, Z 1/2-X,1/2+Y,-Z -X, -Y, Z
u=2x, v=2y, w=0
1. X, Y, Z
3. ½-X,½+Y,-Z
u=2x-½,v=-½,w=2z
1. X, Y, Z
4. ½+X,½-Y,-Z
u=-½,v=2y-½,w=2z
Harker sections, w=0, v=1/2, u=1/2

40. Isomorphous difference Patterson map (Pt derivative)
W=0
V=1/2
U=1/2
X, Y, Z
-X, -Y, Z
u=2x, v=2y, w=0
1. X, Y, Z
3. ½-X,½+Y,-Z
u=2x-½,v=-½,w=2z
1. X, Y, Z
4. ½+X,½-Y,-Z
u=-½,v=2y-½,w=2z
Solve for x, y using w=0 Harker sect.

41. Harker section w=0 X, Y, Z -X, -Y, Z u=2x, v=2y, w=0 W=0 0.168=2x
Does z=0? No!
Solve for x, z using v=1/2 Harker sect.

42. Harker Section v=1/2 1. X, Y, Z 3. ½-X,½+Y,-Z u=2x-½,v=-½,w=2z V=1/2

43. Resolving ambiguity in x,y,z
From w=0 Harker section x1=0.084, y1=0.133
From v=1/2 Harker section, x2=0.416, z2=0.075
Why doesn’t x agree between solutions? They differ by an origin shift. Choose the proper shift to bring them into agreement.
What are the rules for origin shifts? Cheshire symmetry operators relate the different choices of origin. You can apply any of the Cheshire symmetry operators to convert from one origin choice to another.

44. Cheshire symmetry From w=0 Harker section xorig1=0.084, yorig1=0.133
From v=1/2 Harker section, xorig2=0.416, zorig2=0.075
X, Y, Z
-X, -Y, Z
-X, Y, -Z
X, -Y, -Z
-X, -Y, -Z
X, Y, -Z
X, -Y, Z
-X, Y, Z
1/2+X, Y, Z
1/2-X, -Y, Z
1/2-X, Y, -Z
1/2+X, -Y, -Z
1/2-X, -Y, -Z
1/2+X, Y, -Z
1/2+X, -Y, Z
1/2-X, Y, Z
X,1/2+Y, Z
-X,1/2-Y, Z
-X,1/2+Y, -Z
X,1/2-Y, -Z
-X,1/2-Y, -Z
X,1/2+Y, -Z
X,1/2-Y, Z
-X,1/2+Y, Z
X, Y,1/2+Z
-X, -Y,1/2+Z
-X, Y,1/2-Z
X, -Y,1/2-Z
-X, -Y,1/2-Z
X, Y,1/2-Z
X, -Y,1/2+Z
-X, Y,1/2+Z
1/2+X,1/2+Y, Z
1/2-x,1/2-Y, Z
1/2-X,1/2+Y, -Z
1/2+X,1/2-Y, -Z
1/2-X,1/2-Y, -Z
1/2+X,1/2+Y, -Z
1/2+X,1/2-Y, Z
1/2-X,1/2+Y, Z
1/2+X, Y,1/2+Z
1/2-X, -Y,1/2+Z
1/2-X, Y,1/2-Z
1/2+X, -Y,1/2-Z
1/2-X, -Y,1/2-Z
1/2+X, Y,1/2-Z
1/2+X, -Y,1/2+Z
1/2-X, Y,1/2+Z
X,1/2+Y,1/2+Z
-X,1/2-Y,1/2+Z
-X,1/2+Y,1/2-Z
X,1/2-Y,1/2-Z
-X,1/2-Y,1/2-Z
X,1/2+Y,1/2-Z
X,1/2-Y,1/2+Z
-X,1/2+Y,1/2+Z
1/2+X,1/2+Y,1/2+Z
1/2-X,1/2-Y,1/2+Z
1/2-X,1/2+Y,1/2-Z
1/2+X,1/2-Y,1/2-Z
1/2-X,1/2-Y,1/2-Z
1/2+X,1/2+Y,1/2-Z
1/2+X,1/2-Y,1/2+Z
1/2-X,1/2+Y,1/2+Z
Apply Cheshire symmetry operator #10
To x1 and y1
Xorig1=0.084
½-xorig1=
½-xorig1=0.416 =xorig2
yorig1=0.133
-yorig1=-0.133=yorig2
Hence,
Xorig2=0.416, yorig2=-0.133, zorig2=0.075

45. Advanced case,Proteinase K in space group P43212
Where are Harker sections?

48. -x - y ½+z - ( ½+y ½-x ¼+z) -½-x-y -½+x-y ¼
Symmetry operator 2 -Symmetry operator 4
-x y ½+z
- ( ½+y ½-x ¼+z)
-½-x-y -½+x-y ¼

49. -x - y ½+z - ( ½+y ½-x ¼+z) -½-x-y -½+x-y ¼
Symmetry operator 2 -Symmetry operator 4
-x y ½+z
- ( ½+y ½-x ¼+z)
-½-x-y -½+x-y ¼
Plug in u.
u=-½-x-y
0.18=-½-x-y
0.68=-x-y
Plug in v.
v=-½+x-y
0.22=-½+x-y
0.72=x-y
Add two equations and solve for y.
+(0.72= x-y)
1.40=-2y
-0.70=y
Plug y into first equation and solve for x.
0.68=-x-(-0.70)
0.02=x

50. -x - y ½+z - ( ½+y ½-x ¼+z) -½-x-y -½+x-y ¼
Symmetry operator 2 -Symmetry operator 4
-x y ½+z
- ( ½+y ½-x ¼+z)
-½-x-y -½+x-y ¼
Plug in u.
u=-½-x-y
0.18=-½-x-y
0.68=-x-y
Plug in v.
v=-½+x-y
0.22=-½+x-y
0.72=x-y
Add two equations and solve for y.
+(0.72= x-y)
1.40=-2y
-0.70=y
Plug y into first equation and solve for x.
0.68=-x-(-0.70)
0.02=x

51. -x - y ½+z - ( ½+y ½-x ¼+z) -½-x-y -½+x-y ¼
Symmetry operator 2 -Symmetry operator 4
-x y ½+z
- ( ½+y ½-x ¼+z)
-½-x-y -½+x-y ¼
Plug in u.
u=-½-x-y
0.18=-½-x-y
0.68=-x-y
Plug in v.
v=-½+x-y
0.22=-½+x-y
0.72=x-y
Add two equations and solve for y.
+(0.72= x-y)
1.40=-2y
-0.70=y
Plug y into first equation and solve for x.
0.68=-x-(-0.70)
0.02=x

52. -x - y ½+z - ( ½+y ½-x ¼+z) -½-x-y -½+x-y ¼
Symmetry operator 2 -Symmetry operator 4
-x y ½+z
- ( ½+y ½-x ¼+z)
-½-x-y -½+x-y ¼
Plug in u.
u=-½-x-y
0.18=-½-x-y
0.68=-x-y
Plug in v.
v=-½+x-y
0.22=-½+x-y
0.72=x-y
Add two equations and solve for y.
+(0.72= x-y)
1.40=-2y
-0.70=y
Plug y into first equation and solve for x.
0.68=-x-(-0.70)
0.02=x

54. ½-y ½+x ¾+z - ( -y -x ½-z) ½ ½+2x ¼+2z Symmetry operator 3
Plug in v.
v= ½+2x
0.48= ½+2x
-0.02=2x
-0.01=x
Plug in w.
w= ¼+2z
0.24= ¼+2z
-0.01=2z
-0.005=z

55. ½-y ½+x ¾+z - ( -y -x ½-z) ½ ½+2x ¼+2z Symmetry operator 3
Plug in v.
v= ½+2x
0.46= ½+2x
-0.04=2x
-0.02=x
Plug in w.
w= ¼+2z
0.24= ¼+2z
-0.01=2z
-0.005=z

56. ½-y ½+x ¾+z - ( -y -x ½-z) ½ ½+2x ¼+2z Symmetry operator 3
Plug in v.
v= ½+2x
0.46= ½+2x
-0.04=2x
-0.02=x
Plug in w.
w= ¼+2z
0.24= ¼+2z
-0.01=2z
-0.005=z

57. Xstep4=-0.02 ystep4= ?.?? zstep4=-0.005
From step 3
Xstep3= ystep3= zstep3=?.???
From step 4
Xstep4= ystep4= ?.?? zstep4=-0.005
Clearly, Xstep3 does not equal Xstep4 .
Use a Cheshire symmetry operator that transforms xstep3= 0.02 into xstep4=
For example, let’s use:
-x, -y, z
And apply it to all coordinates in step 3.
xstep3-transformed = - (+0.02) = -0.02
ystep3-transformed = - (- 0.70) = +0.70
Now xstep3-transformed = xstep4
And ystep3 has been transformed to a reference frame consistent with x and z from step 4. So we arrive at the following self-consistent x,y,z:
Xstep4=-0.02, ystep3-transformed=0.70, zstep4=-0.005
Or simply, x=-0.02, y=0.70, z=-0.005
The x, y coordinate in step 3 describes one of the heavy atom positions in the unit cell. The x, z coordinate in step 4 describes a symmetry related copy. We can’t combine these coordinates directly. They don’t describe the same atom. Perhaps they even
referred to different origins.
How can we transform x, y from step 3 so it describes
the same atom as x and z in step 4?

58. Xstep4=-0.02 ystep4= ?.?? zstep4=-0.005
From step 3
Xstep3= ystep3= zstep3=?.???
From step 4
Xstep4= ystep4= ?.?? zstep4=-0.005
Clearly, Xstep3 does not equal Xstep4 .
Use a Cheshire symmetry operator that transforms xstep3= 0.02 into xstep4=
For example, let’s use:
-x, -y, z
And apply it to all coordinates in step 3.
xstep3-transformed = - (+0.02) = -0.02
ystep3-transformed = - (- 0.70) = +0.70
Now xstep3-transformed = xstep4
And ystep3 has been transformed to a reference frame consistent with x and z from step 4. So we arrive at the following self-consistent x,y,z:
Xstep4=-0.02, ystep3-transformed=0.70, zstep4=-0.005
Or simply, x=-0.02, y=0.70, z=-0.005
Cheshire Symmetry Operators for space group P43212
X, Y, Z
-X, -Y, Z
-Y, X, 1/4+Z
Y, -X, 1/4+Z
Y, X, -Z
-Y, -X, -Z
X, -Y, 1/4-Z
-X, Y, 1/4-Z
1/2+X, 1/2+Y, Z
1/2-X, 1/2-Y, Z
1/2-Y, 1/2+X, 1/4+Z
1/2+Y, 1/2-X, 1/4+Z
1/2+Y, 1/2+X, -Z
1/2-Y, 1/2-X, -Z
1/2+X, 1/2-Y, 1/4-Z
1/2-X, 1/2+Y, 1/4-Z
X, Y, 1/2+Z
-X, -Y, 1/2+Z
-Y, X, 3/4+Z
Y, -X, 3/4+Z
Y, X, 1/2-Z
-Y, -X, 1/2-Z
X, -Y, 3/4-Z
-X, Y, 3/4-Z
1/2+X, 1/2+Y, 1/2+Z
1/2-X, 1/2-Y, 1/2+Z
1/2-Y, 1/2+X, 3/4+Z
1/2+Y, 1/2-X, 3/4+Z
1/2+Y, 1/2+X, 1/2-Z
1/2-Y, 1/2-X, 1/2-Z
1/2+X, 1/2-Y, 3/4-Z
1/2-X, 1/2+Y, 3/4-Z

59. Xstep4=-0.02 ystep4= ?.?? zstep4=-0.005
From step 3
Xstep3= ystep3= zstep3=?.???
From step 4
Xstep4= ystep4= ?.?? zstep4=-0.005
Clearly, Xstep3 does not equal Xstep4 .
Use a Cheshire symmetry operator that transforms xstep3= 0.02 into xstep4=
For example, let’s use:
-x, -y, z
And apply it to all coordinates in step 3.
xstep3-transformed = - (+0.02) = -0.02
ystep3-transformed = - (- 0.70) = +0.70
Now xstep3-transformed = xstep4
And ystep3 has been transformed to a reference frame consistent with x and z from step 4. So we arrive at the following self-consistent x,y,z:
Xstep4=-0.02, ystep3-transformed=0.70, zstep4=-0.005
Or simply, x=-0.02, y=0.70, z=-0.005
Cheshire Symmetry Operators for space group P43212
X, Y, Z
-X, -Y, Z
-Y, X, 1/4+Z
Y, -X, 1/4+Z
Y, X, -Z
-Y, -X, -Z
X, -Y, 1/4-Z
-X, Y, 1/4-Z
1/2+X, 1/2+Y, Z
1/2-X, 1/2-Y, Z
1/2-Y, 1/2+X, 1/4+Z
1/2+Y, 1/2-X, 1/4+Z
1/2+Y, 1/2+X, -Z
1/2-Y, 1/2-X, -Z
1/2+X, 1/2-Y, 1/4-Z
1/2-X, 1/2+Y, 1/4-Z
X, Y, 1/2+Z
-X, -Y, 1/2+Z
-Y, X, 3/4+Z
Y, -X, 3/4+Z
Y, X, 1/2-Z
-Y, -X, 1/2-Z
X, -Y, 3/4-Z
-X, Y, 3/4-Z
1/2+X, 1/2+Y, 1/2+Z
1/2-X, 1/2-Y, 1/2+Z
1/2-Y, 1/2+X, 3/4+Z
1/2+Y, 1/2-X, 3/4+Z
1/2+Y, 1/2+X, 1/2-Z
1/2-Y, 1/2-X, 1/2-Z
1/2+X, 1/2-Y, 3/4-Z
1/2-X, 1/2+Y, 3/4-Z

60. Xstep4=-0.02 ystep4= ?.?? zstep4=-0.005
From step 3
Xstep3= ystep3= zstep3=?.???
From step 4
Xstep4= ystep4= ?.?? zstep4=-0.005
Clearly, Xstep3 does not equal Xstep4 .
Use a Cheshire symmetry operator that transforms xstep3= 0.02 into xstep4=
For example, let’s use:
-x, -y, z
And apply it to all coordinates in step 3.
xstep3-transformed = - (+0.02) = -0.02
ystep3-transformed = - (- 0.70) = +0.70
Now xstep3-transformed = xstep4
And ystep3 has been transformed to a reference frame consistent with x and z from step 4. So we arrive at the following self-consistent x,y,z:
Xstep4=-0.02, ystep3-transformed=0.70, zstep4=-0.005
Or simply, x=-0.02, y=0.70, z=-0.005
Cheshire Symmetry Operators for space group P43212
X, Y, Z
-X, -Y, Z
-Y, X, 1/4+Z
Y, -X, 1/4+Z
Y, X, -Z
-Y, -X, -Z
X, -Y, 1/4-Z
-X, Y, 1/4-Z
1/2+X, 1/2+Y, Z
1/2-X, 1/2-Y, Z
1/2-Y, 1/2+X, 1/4+Z
1/2+Y, 1/2-X, 1/4+Z
1/2+Y, 1/2+X, -Z
1/2-Y, 1/2-X, -Z
1/2+X, 1/2-Y, 1/4-Z
1/2-X, 1/2+Y, 1/4-Z
X, Y, 1/2+Z
-X, -Y, 1/2+Z
-Y, X, 3/4+Z
Y, -X, 3/4+Z
Y, X, 1/2-Z
-Y, -X, 1/2-Z
X, -Y, 3/4-Z
-X, Y, 3/4-Z
1/2+X, 1/2+Y, 1/2+Z
1/2-X, 1/2-Y, 1/2+Z
1/2-Y, 1/2+X, 3/4+Z
1/2+Y, 1/2-X, 3/4+Z
1/2+Y, 1/2+X, 1/2-Z
1/2-Y, 1/2-X, 1/2-Z
1/2+X, 1/2-Y, 3/4-Z
1/2-X, 1/2+Y, 3/4-Z

61. Use x,y,z to predict the position of a non-Harker Patterson peak
x,y,z vs. –x,y,z ambiguity remains
In other words x=-0.02, y=0.70, z= or
x=+0.02, y=0.70, z= could be correct.
Both satisfy the difference vector equations for Harker sections
Only one is correct. 50/50 chance
Predict the position of a non Harker peak.
Use symop1-symop5
Plug in x,y,z solve for u,v,w.
Plug in –x,y,z solve for u,v,w
I have a non-Harker peak at u=0.28 v=0.28, w=0.0
The position of the non-Harker peak will be predicted by the correct heavy atom coordinate.

62. x y z -( y x -z) x-y -x+y 2z symmetry operator 1 -symmetry operator 5
u v w
First, plug in x=-0.02, y=0.70, z=-0.005
u=x-y = =-0.72
v=-x+y= = 0.72
w=2z=2*(-0.005)=-0.01
The numerical value of these co-ordinates falls outside the section we have drawn. Lets transform this uvw by Patterson symmetry u,-v,-w.
-0.72, 0.72,-0.01 becomes
-0.72,-0.72, 0.01 then add 1 to u and v
0.28, 0.28, 0.01 This corresponds to the peak shown u=0.28, v=0.28, w=0.01
Thus, x=-0.02, y=0.70, z= is correct. Hurray! We are finished!
In the case that the above test failed, we would change the sign of x.
(1) U, V, W (2)-U,-V, W (3) U, V,-W (4)-U,-V,-W
(5)-U, V, W (6) U,-V, W (7)-U, V,-W (8) U,-V,-W
(9)-V, U, W (10) V,-U, W (11)-V, U,-W (12) V,-U,-W
(13) V, U, W (14)-V,-U, W (15) V, U,-W (16)-V,-U,-W

63. Symmetry Operators are the Bridge between Atomic Coordinates and Patterson Peaks
symop #1
symop #2
x , y
-(-x, –y)
2x , 2y
u=2x, v=2y
(-0.2,-0.3)
(0,0) x
(0,0) u y v
(0.2,0.3)
(0.6, 0.4)
(0.4, 0.6)
SYMMETRY OPERATORS FOR PLANE GROUP P2
1) x,y ) -x,-y
PATTERSON MAP

64. Assignment Calculate an isomorphous difference Patterson map in lab.
Solve the positions of the heavy atom (x,y,z) from the peaks in the map (u,v,w).
follow the procedures in the handout
write neatly
check your answer
Next week hand in your calculation.
We will test the accuracy of your solution and use it to calculate phases and electron density.

65. Patterson space Crystal space
U=0.5
P43212 Symmetry operator difference 3-6
Calculate Y and Z
Calculate X and Y
W=0.25
Cheshire operator applied to Y and Z if two values of Y do not match
P43212 Symmetry operator difference 2-4
X,Y,Z referred to a common origin.
Check answer for peak off Harker section.
P43212 Symmetry operator difference 1-5
x y z
-( y x -z)
x-y -x+y 2z
u,v,w
If prediction lies outside Patterson asymmetric unit (0→0.5, 0→0.5,0→ 0.5) use
Patterson symmetry operators to find the symmetry equivalent peak in the
asymmetric unit. If the predicted peak is absent, then negate x value and re-calculate u,v,w. Predicted peak should be present if algebra is correct.

66. m230d_2011_scaled.mtz Intensity measurements (IHKL)
native-yen-b3.sca
prok_native_mucheng.sca
prok-gdcl3-matthew.sca
prok-gdcl3-maya.sca
prok_gdcl3_sean.sca
prok_iodide_jungreem.sca
prok_mersalyl_dan.sca
prok_mersalyl_dan_xds.sca
prok-pcmbs-1-5-stephen.sca
prok-pcmbs-andrew.sca
prok_pcmbs_cuiwen.sca
prok_pcmbs_pmsf_jose.sca
prok-pcmbs-pmsf-yazan.sca
prok_pcmbs_riki.sca
prok_pcmbs_sabai.sca
prok_pcmbs_smriti.sca
All data sets were appended into a spreadsheet. Each column contains IHKL of a different data set. Each row specifies a different HKL.
-using the CCP4 program CAD.
Intensity measurements were converted to structure factor
amplitudes (|FHKL|)
-using the CCP4 program TRUNCATE.
All data sets were scaled to a
reference native data set with the best statistics: prok-native-yen-b3
-using the CCP4 program SCALEIT.

67. Scale intensities by a constant (k) and resolution dependent exponential (B)
prok-native-yen
prok-gdcl3-matthew
H K L intensity sigma
H K L intensity sigma
comparison
-Probably first crystal is larger than the second.
-Multiply Saken’s data by k and B to put the data on the same scale.
/ = 2.65
/ = 1.65
/ = 3.07
/ = 2.45
/ = 0.70
/ = 1.83
/ = 1.24

68. -B*sin2q/l2 e

69. Initial intensity profile
low resolution high resolution.

70. Scaling constant K=0.883
intensity
low resolution high resolution.

71. after K= and B=-5.75
intensity
low resolution high resolution.

72. OVERALL FILE STATISTICS for resolution range 0.000 - 0.351
29 BOTH D D_pcmbs-stephen
30 BOTH Q SIGD_pcmbs-stephen
31 NONE F FP_pcmbs-andrew
32 NONE Q SIGFP_pcmbs-andrew
33 NONE D D_pcmbs-andrew
34 NONE Q SIGD_pcmbs-andrew
35 NONE F FP_pcmbs_cuiwen
36 NONE Q SIGFP_pcmbs_cuiwen
37 NONE D D_pcmbs_cuiwen
38 NONE Q SIGD_pcmbs_cuiwen
39 NONE F FP_pcmbs_pmsf_jose
40 NONE Q SIGFP_pcmbs_pmsf_jose
41 NONE D D_pcmbs_pmsf_jose
42 NONE Q SIGD_pcmbs_pmsf_jose
43 NONE F FP_pcmbs-pmsf-yazan
44 NONE Q SIGFP_pcmbs-pmsf-yazan
45 NONE D D_pcmbs-pmsf-yazan
46 NONE Q SIGD_pcmbs-pmsf-yazan
47 NONE F FP_pcmbs_riki
48 NONE Q SIGFP_pcmbs_riki
49 NONE D D_pcmbs_riki
50 NONE Q SIGD_pcmbs_riki
51 NONE F FP_pcmbs_sabai
52 NONE Q SIGFP_pcmbs_sabai
53 NONE D D_pcmbs_sabai
54 NONE Q SIGD_pcmbs_sabai
55 NONE F FP_pcmbs_smriti
56 NONE Q SIGFP_pcmbs_smriti
57 NONE D D_pcmbs_smriti
58 NONE Q SIGD_pcmbs_smriti
No. of reflections used in FILE STATISTICS
LIST OF REFLECTIONS
=================== ? ? ?
? ? ?
0.00
OVERALL FILE STATISTICS for resolution range
=======================
Col Sort Min Max Num % Mean Mean Resolution Type Column
num order Missing complete abs. Low High label
1 ASC H H
2 NONE H K
3 NONE H L
4 NONE I FreeR_flag
5 NONE F FP_native-yen
6 NONE Q SIGFP_native-yen
7 NONE F FP_native-mucheng
8 NONE Q SIGFP_native-mucheng
9 NONE F FP_gdcl3-matthew
10 NONE Q SIGFP_gdcl3-matthew
11 NONE F FP_gdcl3-maya
12 NONE Q SIGFP_gdcl3-maya
13 NONE D D_gdcl3-maya
14 NONE Q SIGD_gdcl3-maya
15 NONE F FP_gdcl3_sean
16 NONE Q SIGFP_gdcl3_sean
17 NONE D D_gdcl3_sean
18 NONE Q SIGD_gdcl3_sean
19 NONE F FP_iodide_jungreem
20 NONE Q SIGFP_iodide_jungreem
21 NONE D D_iodide_jungreem
22 NONE Q SIGD_iodide_jungreem
23 NONE F FP_mersalyl_dan
24 NONE Q SIGFP_mersalyl_dan
25 NONE D D_mersalyl_dan
26 NONE Q SIGD_mersalyl_dan
27 NONE F FP_pcmbs-stephen
28 NONE Q SIGFP_pcmbs-stephen