1. 1 2 34 1 2 34 1 2 34 1 2 34 1 2 34 1 2 34 1 2 34 1 2 34

2. b a 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. H Li H H H H H H H H H H H H H H H H H

3. b a Two-folds H Li H H H H H H H H H H H H H H H H H

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

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

6. b a Which plane group? H Li H H H H H H H H H H H H H H H H H

7. X=0.2 What are the coordinates of the oxygen using origin choice 1? b a H Li H H H H H H H H H H H H H H H H H X=0.20 Y=0.20 1 2 34 1 2 34

8. H Li H H H H H H H H H H H H H H H H H b a X=0.20 Y=0.20 What are the coordinates of the other oxygen in the unit cell? X=0.80 Y=0.80 Symmetry operators in plane group p2 X, Y -X,-Y 1 2 34 1 2 34

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

10. H Li H H H H H H H H H H H H H H H H H What would be the oxygen coordinates if we had drawn the unit cell with origin choice 2? b a 1 2 34 X=0.70 Y=0.20 1 2 34

11. Oxygen coordinates with origin choice 3? X=0.30 Y=0.30 H Li H H H H H H H H H H H H H H H H H b a 1 2 34 1 2 34

12. H H H H H H H H H H H H H H H H H H b a Oxygen coordinates with origin choice 4? 1 2 34 X=0.80 Y=0.30 1 2 34

13. Cheshire operators X, Y X+.5, Y X+.5, Y+.5 X, Y+.5 X=0.20 Y=0.20 X=0.70 Y=0.20 X=0.30 Y=0.30 X=0.80 Y=0.30 Choice 1 Choice 2 Choice 3 Choice 4

14. The 4 choices of origin are equally valid but once a choice is made, you must remain consistent. H H Li H H H H H H H H H H H H H H H H b a H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H Choice 1 Choice 2 Choice 3 Choice 4 X 1 =0.20 Y 1 =0.20 X 1 =0.70 Y 1 =0.20 X 1 =0.80 Y 1 =0.30 X 1 =0.30 Y 1 =0.30 X 2 =0.80 Y 2 =0.80 X 2 =0.30 Y 2 =0.80 H Li H X 2 =0.70 Y 2 =0.70 X 2 =0.20 Y 2 =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. To compare atomic coordinates between structures solved with different origins, a Cheshire operator must be applied. 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) –HgCl 2 Hg(Acetate) 2 PCMBS –PIP –EuCl 3 GdCl 3 SmCl 3 How many heavy atom sites per asymmetric unit, if any? 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 synthesis P(uvw)=  ? hkl cos2  (hu+kv+lw -  ) hkl Patterson synthesis P(uvw)=  I hkl cos2  (hu+kv+lw -  ) hkl Patterson Review A Patterson synthesis is like a Fourier synthesis except for what two variables? Fourier synthesis  (xyz)=  |F hkl | cos2  (hx+ky+lz -  hkl ) hkl

18. Hence, Patterson density map= electron density map convoluted with its inverted image. Patterson synthesis P(uvw)=  I hkl cos2  (hu+kv+lw) Remembering I hkl =F hkl F hkl * And Friedel’s law F hkl *= F -h-k-l P(uvw)=FourierTransform(F hklF -h-k-l ) P(uvw)=  (uvw)  (-u-v-w)

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

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

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

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

24. Calculating X,Y,Z coordinates from Patterson peak positions (U,V,W) Three Examples 1.Exceedingly simple 2D example 2.Straightforward-3D example, Pt derivative of polymerase  in space group P2 1 2 1 2 3.Advanced 3D example, Hg derivative of proteinase K in space group P4 3 2 1 2.

25. Plane group p2 H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H b

26. Let’s consider only oxygen atoms H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H b 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? 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? (0,0) a b

28. Symmetry operators in plane group p2 (0,0) a b (0.2,0.3) (-0.2,-0.3) SYMMETRY OPERATORS FOR PLANE GROUP P2 1) x,y 2) -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. 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? Yes! Equivalent by unit cell translation. (-0.2+1.0, -0.3+1.0)=(0.8, 0.7)

29. 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. (0,0) a b a b (0.2,0.3) (-0.2,-0.3) SYMMETRY OPERATORS FOR PLANE GROUP P2 1) x,y 2) -x,-y PATTERSON MAP 2D CRYSTAL 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? 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

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

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

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

34. Hurray!!!! SYMMETRY OPERATORS FOR PLANE GROUP P2 1) x,y 2) -x,-y x y -(-x –y) 2x 2y symop #1 symop #2 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 HURRAY! we got back the coordinates of our smiley faces!!!! (0,0) a b (0.2,0.3) 2D CRYSTAL

35. 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)? a (0,0) b PATTERSON MAP (0.4, 0.6) (0.6, 0.4) using Patterson (u,v) values 0.4, 0.6 to get x and y. u=2x 0.4=2x 0.2=x v=2y 0.6=2y 0.3=y u=2x 0.6=2x 0.3=x v=2y 0.4=2y 0.2=y using Patterson (u,v) values 0.6, 0.4 to get x and y. These two solutions do not give the same x,y? What is going on??????

36. Arbitrary choice of origin Original origin choice Coordinates x=0.2, y=0.3. (0,0) a b (0.2,0.3) (-0.2,-0.3) (0.8,0.7) (0.3,0.2) (-0.3,-0.2) (0.7,0.8) (0,0) a b 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 n 2 Patterson peaks total, n peaks at the origin, n 2 -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  example, P2 1 2 1 2 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. 1. X, Y, Z 2. -X, -Y, Z 3. 1/2-X,1/2+Y,-Z 4. 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 1. X, Y, Z 2. -X, -Y, Z 3. 1/2-X,1/2+Y,-Z 4. 1/2+X,1/2-Y,-Z 1. X, Y, Z 2.-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 1. X, Y, Z 2.-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 W=0 1. X, Y, Z 2.-X, -Y, Z u=2x, v=2y, w=0 0.168=2x 0.084=x 0.266=2y 0.133=y Does z=0? No! Solve for x, z using v=1/2 Harker sect.

42. Harker Section v=1/2 V=1/2 1. X, Y, Z 3. ½-X,½+Y,-Z u=2x- ½,v=-½,w=2z 0.333=2x-1/2 0.833=2x 0.416=x 0.150=2z 0.075=z

43. Resolving ambiguity in x,y,z From w=0 Harker section x 1 =0.084, y 1 =0.133 From v=1/2 Harker section, x 2 =0.416, z 2 =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 1. X, Y, Z 2. -X, -Y, Z 3. -X, Y, -Z 4. X, -Y, -Z 5. -X, -Y, -Z 6. X, Y, -Z 7. X, -Y, Z 8. -X, Y, Z 9.1/2+X, Y, Z 10.1/2-X, -Y, Z 11.1/2-X, Y, -Z 12.1/2+X, -Y, -Z 13.1/2-X, -Y, -Z 14.1/2+X, Y, -Z 15.1/2+X, -Y, Z 16.1/2-X, Y, Z 17. X,1/2+Y, Z 18. -X,1/2-Y, Z 19. -X,1/2+Y, -Z 20. X,1/2-Y, -Z 21. -X,1/2-Y, -Z 22. X,1/2+Y, -Z 23. X,1/2-Y, Z 24. -X,1/2+Y, Z 25. X, Y,1/2+Z 26. -X, -Y,1/2+Z 27. -X, Y,1/2-Z 28. X, -Y,1/2-Z 29. -X, -Y,1/2-Z 30. X, Y,1/2-Z 31. X, -Y,1/2+Z 32. -X, Y,1/2+Z 33.1/2+X,1/2+Y, Z 34.1/2-x,1/2-Y, Z 35.1/2-X,1/2+Y, -Z 36.1/2+X,1/2-Y, -Z 37.1/2-X,1/2-Y, -Z 38.1/2+X,1/2+Y, -Z 39.1/2+X,1/2-Y, Z 40.1/2-X,1/2+Y, Z 41.1/2+X, Y,1/2+Z 42.1/2-X, -Y,1/2+Z 43.1/2-X, Y,1/2-Z 44.1/2+X, -Y,1/2-Z 45.1/2-X, -Y,1/2-Z 46.1/2+X, Y,1/2-Z 47.1/2+X, -Y,1/2+Z 48.1/2-X, Y,1/2+Z 49. X,1/2+Y,1/2+Z 50. -X,1/2-Y,1/2+Z 51. -X,1/2+Y,1/2-Z 52. X,1/2-Y,1/2-Z 53. -X,1/2-Y,1/2-Z 54. X,1/2+Y,1/2-Z 55. X,1/2-Y,1/2+Z 56. -X,1/2+Y,1/2+Z 57.1/2+X,1/2+Y,1/2+Z 58.1/2-X,1/2-Y,1/2+Z 59.1/2-X,1/2+Y,1/2-Z 60.1/2+X,1/2-Y,1/2-Z 61.1/2-X,1/2-Y,1/2-Z 62.1/2+X,1/2+Y,1/2-Z 63.1/2+X,1/2-Y,1/2+Z 64.1/2-X,1/2+Y,1/2+Z From w=0 Harker section x orig1 =0.084, y orig1 =0.133 From v=1/2 Harker section, x orig2 =0.416, z orig2 =0.075 Apply Cheshire symmetry operator #10 To x 1 and y 1 X orig1 =0.084 ½-x orig1 =0.5-0.084 ½-x orig1 =0.416 =x orig2 y orig1 =0.133 -y orig1 =-0.133=y orig2 Hence, X orig2 =0.416, y orig2 =-0.133, z orig2 =0.075

45. Advanced case,Proteinase K in space group P4 3 2 1 2 Where are Harker sections?

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

49. 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. 0.68=-x-y +(0.72= x-y) 1.40=-2y -0.70=y Plug y into first equation and solve for x. 0.68=-x-y 0.68=-x-(-0.70) 0.02=x

50. 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. 0.68=-x-y +(0.72= x-y) 1.40=-2y -0.70=y Plug y into first equation and solve for x. 0.68=-x-y 0.68=-x-(-0.70) 0.02=x

51. 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. 0.68=-x-y +(0.72= x-y) 1.40=-2y -0.70=y Plug y into first equation and solve for x. 0.68=-x-y 0.68=-x-(-0.70) 0.02=x

52. 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. 0.68=-x-y +(0.72= x-y) 1.40=-2y -0.70=y Plug y into first equation and solve for x. 0.68=-x-y 0.68=-x-(-0.70) 0.02=x

54. Symmetry operator 3 -Symmetry operator 6 ½-y ½+x ¾+z - ( -y -x ½-z) ½ ½+2x ¼+2z 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. Symmetry operator 3 -Symmetry operator 6 ½-y ½+x ¾+z - ( -y -x ½-z) ½ ½+2x ¼+2z 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. Symmetry operator 3 -Symmetry operator 6 ½-y ½+x ¾+z - ( -y -x ½-z) ½ ½+2x ¼+2z 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. From step 3 X step3 = 0.02 y step3 =-0.70 z step3 =?.??? From step 4 X step4 =-0.02 y step4 = ?.?? z step4 =-0.005 Clearly, X step3 does not equal X step4. Use a Cheshire symmetry operator that transforms x step3 = 0.02 into x step4 =- 0.02. For example, let’s use: -x, -y, z And apply it to all coordinates in step 3. x step3-transformed = - (+0.02) = -0.02 y step3-transformed = - (- 0.70) = +0.70 Now x step3-transformed = x step4 And y step3 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: X step4 =-0.02, y step3-transformed =0.70, z step4 =-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. From step 3 X step3 = 0.02 y step3 =-0.70 z step3 =?.??? From step 4 X step4 =-0.02 y step4 = ?.?? z step4 =-0.005 Clearly, X step3 does not equal X step4. Use a Cheshire symmetry operator that transforms x step3 = 0.02 into x step4 =- 0.02. For example, let’s use: -x, -y, z And apply it to all coordinates in step 3. x step3-transformed = - (+0.02) = -0.02 y step3-transformed = - (- 0.70) = +0.70 Now x step3-transformed = x step4 And y step3 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: X step4 =-0.02, y step3-transformed =0.70, z step4 =-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. From step 3 X step3 = 0.02 y step3 =-0.70 z step3 =?.??? From step 4 X step4 =-0.02 y step4 = ?.?? z step4 =-0.005 Clearly, X step3 does not equal X step4. Use a Cheshire symmetry operator that transforms x step3 = 0.02 into x step4 =- 0.02. For example, let’s use: -x, -y, z And apply it to all coordinates in step 3. x step3-transformed = - (+0.02) = -0.02 y step3-transformed = - (- 0.70) = +0.70 Now x step3-transformed = x step4 And y step3 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: X step4 =-0.02, y step3-transformed =0.70, z step4 =-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. From step 3 X step3 = 0.02 y step3 =-0.70 z step3 =?.??? From step 4 X step4 =-0.02 y step4 = ?.?? z step4 =-0.005 Clearly, X step3 does not equal X step4. Use a Cheshire symmetry operator that transforms x step3 = 0.02 into x step4 =- 0.02. For example, let’s use: -x, -y, z And apply it to all coordinates in step 3. x step3-transformed = - (+0.02) = -0.02 y step3-transformed = - (- 0.70) = +0.70 Now x step3-transformed = x step4 And y step3 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: X step4 =-0.02, y step3-transformed =0.70, z step4 =-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=-0.005 or x=+0.02, y=0.70, z=-0.005 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 = -0.02-0.70 =-0.72 v=-x+y= +0.02+0.70= 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=-0.005 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 u (0,0) v PATTERSON MAP (0.4, 0.6) x, y -(-x, –y) 2x, 2y u=2x, v=2y symop #1 symop #2 (0.6, 0.4) (0,0) x y (0.2,0.3) (-0.2,-0.3) SYMMETRY OPERATORS FOR PLANE GROUP P2 1) x,y 2) -x,-y

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. Calculate Y and Z Calculate X and Y X,Y,Z referred to a common origin. 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. Crystal space U=0.5 W=0.25 Cheshire operator applied to Y and Z if two values of Y do not match P43212 Symmetry operator difference 1-5 x y z -( y x -z) x-y -x+y 2z u,v,w Check answer for peak off Harker section. P43212 Symmetry operator difference 3-6 P43212 Symmetry operator difference 2-4 Patterson space

66. m230d_2010_scaled.mtz prok-eu-yanran.sca prok-gd-1-10-david.sca prok-gd-1-1-keith.sca prok-gd-1-1-shewit.sca prok-gd-1-4-wei.sca prok-hgac-1-4-ralph.sca prok-hgcl2-1-1-huimin.sca prok-hgcl2-travis.sca prok-native-amanda.sca prok-native-natalie.sca prok-native-saken.sca prok-pcmbs-1-1-alex.sca prok-pcmbs-1-1-toby.sca prok-pcmbs-alexsin.sca prok-pip-1-1-annie.sca prok-pip-1-1-lauren.sca prok-pip-1-4-christine.sca prok-pmsf-angelica.sca prok-pmsf-camellia.sca prok-pmsf-catherine.sca prok-pmsf-yuewei.sca prok-sm-1-1-bill.sca prok-sm-1-4-xinghong.sca Intensity measurements (I HKL ) All data sets were appended into a spreadsheet. Each column contains I HKL of a different data set. Each row specifies a different HKL. - using the CCP4 program CAD. Intensity measurements were converted to structure factor amplitudes (|F HKL |) - using the CCP4 program TRUNCATE. All data sets were scaled to a reference native data set with the best statistics: prok-native-natalie - using the CCP4 program SCALEIT.

67. Scale intensities by a constant (k) and resolution dependent exponential (B) H K L intensity sigma 1 0 10 106894.0 1698.0 1 0 11 41331.5 702.3 1 0 12 76203.2 1339.0 1 0 13 28113.5 513.6 1 0 14 6418.2 238.7 1 0 15 45946.4 882.7 1 0 16 26543.8 555.6 prok-native-natalie 106894.0 / 40258.7 = 2.65 41331.5 / 25033.2 = 1.65 76203.2 / 24803.6 = 3.07 28113.5 / 11486.3 = 2.45 6418.2 / 9180.5 = 0.70 45946.4 / 25038.8 = 1.83 26543.8 / 21334.6 = 1.24 prok-native-saken H K L intensity sigma 1 0 10 40258.7 1222.9 1 0 11 25033.2 799.8 1 0 12 24803.6 771.5 1 0 13 11486.3 423.9 1 0 14 9180.5 353.6 1 0 15 25038.8 783.0 1 0 16 21334.6 686.4 comparison -Probably Natalie used a larger crystal than Saken. -Multiply Saken’s data by k and B to put the data on the same scale.

68. OVERALL FILE STATISTICS for resolution range 0.000 - 0.399 ======================= Col Sort Min Max Num % Mean Mean Resolution Type Column num order Missing complete abs. Low High label 1 ASC 0 42 0 100.00 22.2 22.2 56.53 1.60 H H 2 NONE 0 30 0 100.00 9.1 9.1 56.53 1.60 H K 3 NONE 0 63 0 100.00 23.8 23.8 56.53 1.60 H L 4 NONE 0.0 19.0 0 100.00 9.48 9.48 56.53 1.60 I FreeR_flag 5 NONE 6.2 2336.4 1623 94.96 232.23 232.23 40.76 1.60 F FP_native-natalie 6 NONE 0.8 34.0 1623 94.96 3.59 3.59 40.76 1.60 Q SIGFP_native-natalie 7 NONE 6.5 1420.2 5392 83.26 248.69 248.69 33.97 1.70 F FP_native-amanda 8 NONE 0.6 56.2 5392 83.26 3.51 3.51 33.97 1.70 Q SIGFP_native-amanda 9 NONE 1.6 2486.3 9128 71.66 278.06 278.06 48.05 1.79 F FP_native-saken 10 NONE 1.0 56.3 9128 71.66 5.84 5.84 48.05 1.79 Q SIGFP_native-saken 11 NONE 2.0 2428.1 5239 83.74 258.80 258.80 56.53 1.69 F FP_Eu-yanran 12 NONE 1.5 117.4 5239 83.74 11.53 11.53 56.53 1.69 Q SIGFP_Eu-yanran 13 NONE -109.8 175.0 5280 83.61 -2.52 15.74 56.53 1.69 D D_Eu-yanran 14 NONE 0.0 80.8 5280 83.61 18.37 18.37 56.53 1.69 Q SIGD_Eu-yanran 15 NONE 4.1 2334.3 5370 83.33 254.01 254.01 56.53 1.70 F FP_Gd-david 16 NONE 0.7 37.2 5370 83.33 3.57 3.57 56.53 1.70 Q SIGFP_Gd-david 17 NONE -52.0 47.8 5415 83.19 -0.99 8.20 56.53 1.70 D D_Gd-david 18 NONE 0.0 28.4 5415 83.19 5.75 5.75 56.53 1.70 Q SIGD_Gd-david 19 NONE 13.7 2166.6 14898 53.75 307.13 307.13 48.05 1.91 F FP_Gd-keith 20 NONE 2.3 178.1 14898 53.75 23.20 23.20 48.05 1.91 Q SIGFP_Gd-keith 21 NONE -234.6 266.1 15375 52.27 -0.08 22.57 48.05 1.91 D D_Gd-keith 22 NONE 0.0 287.9 15375 52.27 35.67 35.67 48.05 1.91 Q SIGD_Gd-keith 23 NONE 7.0 2361.1 5566 82.72 253.74 253.74 56.53 1.70 F FP_Gd-shewit 24 NONE 1.6 161.6 5566 82.72 11.67 11.67 56.53 1.70 Q SIGFP_Gd-shewit 25 NONE -128.6 103.2 5651 82.46 -1.60 17.22 56.53 1.70 D D_Gd-shewit 26 NONE 0.0 102.4 5651 82.46 18.70 18.70 56.53 1.70 Q SIGD_Gd-shewit 27 NONE 3.7 2404.7 480 98.51 233.25 233.25 56.53 1.60 F FP_Gd-wei 28 NONE 0.8 19.5 480 98.51 3.33 3.33 56.53 1.60 Q SIGFP_Gd-wei 29 NONE -77.1 64.6 601 98.13 0.31 12.27 56.53 1.60 D D_Gd-wei 30 NONE 0.0 32.2 601 98.13 5.45 5.45 56.53 1.60 Q SIGD_Gd-wei 31 NONE 7.0 2175.4 5797 82.00 256.78 256.78 56.53 1.71 F FP_HgAc-ralph 32 NONE 0.7 76.2 5797 82.00 6.24 6.24 56.53 1.71 Q SIGFP_HgAc-ralph 33 NONE -66.7 57.1 5912 81.65 0.35 6.39 56.53 1.71 D D_HgAc-ralph 34 NONE 0.0 55.2 5912 81.65 10.02 10.02 56.53 1.71 Q SIGD_HgAc-ralph 35 NONE 4.1 2025.3 5506 82.91 263.03 263.03 56.53 1.70 F FP_HgCl2-huimin 36 NONE 1.4 96.4 5506 82.91 7.62 7.62 56.53 1.70 Q SIGFP_HgCl2-huimin 37 NONE -96.9 125.9 6249 80.60 1.60 9.71 56.53 1.70 D D_HgCl2-huimin 38 NONE 0.0 98.4 6249 80.60 12.11 12.11 56.53 1.70 Q SIGD_HgCl2-huimin 39 NONE 9.1 2020.0 8928 72.28 254.72 254.72 40.76 1.69 F FP_HgCl2-travis 40 NONE 1.8 147.6 8928 72.28 11.15 11.15 40.76 1.69 Q SIGFP_HgCl2-travis 41 NONE -140.7 137.7 10384 67.76 -3.31 13.56 40.76 1.69 D D_HgCl2-travis 42 NONE 0.0 166.3 10384 67.76 17.25 17.25 40.76 1.69 Q SIGD_HgCl2-travis 43 NONE 6.1 2276.3 5412 83.20 254.27 254.27 56.53 1.70 F FP_PCMBS-alexj 44 NONE 0.8 49.5 5412 83.20 3.78 3.78 56.53 1.70 Q SIGFP_PCMBS-alexj 45 NONE -75.3 72.9 5413 83.20 0.23 7.72 56.53 1.70 D D_PCMBS-alexj 46 NONE 0.0 33.8 5413 83.20 5.98 5.98 56.53 1.70 Q SIGD_PCMBS-alexj 47 NONE 0.7 2275.5 5067 84.27 251.80 251.80 56.53 1.69 F FP_PCMBS-alexs 48 NONE 0.5 50.7 5067 84.27 3.90 3.90 56.53 1.69 Q SIGFP_PCMBS-alexs 49 NONE -71.6 78.1 5319 83.49 0.60 9.27 56.53 1.69 D D_PCMBS-alexs 50 NONE 0.0 34.1 5319 83.49 6.16 6.16 56.53 1.69 Q SIGD_PCMBS-alexs 51 NONE 2.2 2246.6 337 98.95 233.26 233.26 48.05 1.60 F FP_PCMBS-toby 52 NONE 1.1 31.4 337 98.95 3.04 3.04 48.05 1.60 Q SIGFP_PCMBS-toby 53 NONE -63.7 68.4 458 98.58 -0.68 7.52 48.05 1.60 D D_PCMBS-toby 54 NONE 0.0 27.6 458 98.58 4.86 4.86 48.05 1.60 Q SIGD_PCMBS-toby 55 NONE 6.2 2265.1 5371 83.33 247.65 247.65 48.05 1.70 F FP_PIP-annie AND SO ON……………. No. of reflections used in FILE STATISTICS 32212 LIST OF REFLECTIONS =================== 0 0 4 2.00 1816.34 26.07 ? ? 1882.86 56.33 1194.04 117.40 0.00 0.00 1499.12 26.90 0.00 0.00 ? ? ? ? 1188.25 43.78 0.00 0.00 1544.44 19.48 0.00 0.00 1445.92 74.58 0.00 0.00 1601.80 51.68 0.00 0.00 1733.87 88.65 0.00 0.00 1782.74 49.52 0.00 0.00 ? ? ? ? ? ? ? ? ? ? ? ? ? ? 1549.52 63.20 0.00 0.00 ? ? ? ? ? ? 1954.73 56.31 ? ? ? ?