In Lab on Tuesday at noon We will

In Lab on Tuesday at noon • We will determine what are the two data sets that produce the strongest isomorphous difference Patterson peaks. • We have several native data sets and several Hg, Eu, and Gd derivative data sets. • A different pair will be assigned to each person. • Each person will calculate a difference Patterson between their assigned pair of derivative and a native. • We will record the height of the highest peak on the u=0. 5 section for each map. • The native-derivative pair with the highest peak will be used by the whole class. • We will all interpret the difference Patterson map to calculate coordinates x, y, z for the heavy atom.

Which of the following is true of Patterson maps? a)Every peak in the map corresponds to a vector between atoms in the unit cell b)It is always centrosymmetric c)It has the same unit cell parameters as the crystal d)It can be computed without knowing phases. e)All of the above

Which of the following are true of Patterson maps? a)Every peak in the map corresponds to a vector between atoms in the unit cell b)It is always centrosymmetric c)It has the same unit cell parameters as the crystal d)It can be computed without knowing phases. e)All of the above

Lesson: native Patterson maps offer some structural information but are too complex to offer an interpretation of the atomic coordinates.

If there are 110 atoms in the unit cell, how many peaks in Patterson map? a a)1102 c b)110 c) One peak for each molecule d)None of the above

Lesson: there are n 2 peaks in a Patterson map. n = number of atoms in unit cell n peaks are at the origin n 2 -n peaks off origin

If I calculate an isomorphous difference Patterson map with coefficients (|FPH| - |FP|)2 What will be the only features in the map? a c a a)Vectors between protein atoms b)Vectors between heavy atoms c c) No features expected d) Either B or C

If I calculate an isomorphous difference Patterson map with coefficients |FPH| - |FP| What will be the only features in the map? a c a a)Vectors between protein atoms b)Vectors between heavy atoms c c) No features expected d) Either B or C

Lesson: difference Patterson maps have fewer features than ordinary Patterson maps and so can be more easily interpreted.

Which of the following would correspond to a difference Patterson peak? a)0. 6, -0. 5, -0. 4 a c (0. 8, 0. 0, 0. 3) b)-0. 6, 0. 5, 0. 4 (0. 2, 0. 5, 0. 7) c)a and b d)No such peak exists

Which of the following would correspond to a difference Patterson peak? a)0. 6, -0. 5, -0. 4 a c (0. 8, 0. 0, 0. 3) -(0. 2, 0. 5, 0. 7) (0. 6, -0. 5, -0. 4) u (0. 8, 0. 0, 0. 3) b)-0. 6, 0. 5, 0. 4 v w (0. 2, 0. 5, 0. 7) -(0. 8, 0. 0, 0. 3) (-0. 6, 0. 5, 0. 4) u (0. 2, 0. 5, 0. 7) v c)a and b d)No such peak exists w

Lesson: difference Patterson peaks correspond to vectors between atoms in the unit cell.

Which of the following would correspond to a difference Patterson peak? a)-2 x, 1/2, -2 z a c (-x, y+1/2, -z) (x, y, z) b)2 x, -1/2, 2 z c)a and b d)No such peak exists

Which of the following would correspond to a difference Patterson peak? a)-2 x, 1/2, -2 z a c (-x, y+1/2, -z) -( x, y, z) (-2 x, 1/2, -2 z) u (x, y, z) b)2 x, -1/2, 2 z c)a and b v w ( x, y , z) -(-x, y+1/2, -z) ( 2 x, -1/2, 2 z) u v d)No such peak exists w

Lesson: Even if you don’t know the heavy atom coordinates, you can still write an equation describing the position of the peaks they would produce in the difference Patterson map as long as you have what information?

What information do you need to write the equations for Patterson peaks? a) The unit cell parameters b)The space group c)The symmetry operators d) B or C

Lesson: symmetry operators are the bridge between atomic coordinates in the crystal and Patterson peaks a (-x, y+1/2, -z) -( x, y, z) (x, y, z) c , -2 x (-2 x, 1/2, -2 z) u v , 2 x c w y -2 x, -2 a (u, v, w) , -2 x (-x, y+1/2, -z) , 2 x Crystal 2 y -2 y 2 y , 2 x -2 y 2 y Difference Patterson Map

How many heavy atoms are expected in the unit cell with space group P 21212 if there is only one heavy atom in the asymmetric unit, ? a)4 (1)x, y, z (2) -x, -y, z (3) -x + 1/2, y + 1/2, -z (4) x + 1/2, -y + 1/2, -z b)42 c)Not enough information given

Lesson: The number of atoms in the unit cell is an integer multiple of the number of symmetry operators. In this case, the integer multiple was specified as “ 1”.

In what planes do we expect difference Patterson peaks in P 21212? a)w=0 (1)x, y, z (2) -x, -y, z (3) -x + 1/2, y + 1/2, -z (4) x + 1/2, -y + 1/2, -z b)w=1/2 c) v=1/2 d) u=1/2

In what planes do we expect difference Patterson peaks in P 21212? 1. X, Y, Z 2. -X, -Y, Z u=2 x, v=2 y, w=0 (1)x, y, z (2) -x, -y, z (3) -x + 1/2, y + 1/2, -z (4) x + 1/2, -y + 1/2, -z 1. X, Y, Z 3. ½-X, ½+Y, -Z u=2 x-½, v=-½, w=2 z 1. X, Y, Z 4. ½+X, ½-Y, -Z u=-½, v=2 y-½, w=2 z

Lesson: Vectors between symmetry-related atoms often lie on planes. We call these planes Harker sections W=0 V=1/2 U=1/2

Which of these difference vectors is likely to correspond to the difference Patterson peak shown here? W=0 a) 1. X, Y, Z 2. -X, -Y, Z u=2 x, v=2 y, w=0 b) 1. X, Y, Z 3. ½-X, ½+Y, -Z u=2 x-½, v=-½, w=2 z c) 1. X, Y, Z 4. ½+X, ½-Y, -Z u=-½, v=2 y-½, w=2 z d) They are all equally likely.

Harker section w=0 W=0 1. X, Y, Z 2. -X, -Y, Z u=2 x, v=2 y, w=0 0. 168=2 x 0. 084=x 0. 266=2 y 0. 133=y

What is the value of z? W=0 1. X, Y, Z 2. -X, -Y, Z u=2 x, v=2 y, w=0 0. 168=2 x 0. 084=x 0. 266=2 y 0. 133=y a)Zero b) x/y c) not specified by this Harker section.

What is the value of z? W=0 1. X, Y, Z 2. -X, -Y, Z u=2 x, v=2 y, w=0 0. 168=2 x 0. 084=x 0. 266=2 y 0. 133=y a)Zero b) x/y c) not specified by this Harker section. How can we determine the z coordinate?

Harker Section v=1/2 V=1/2 1. X, Y, Z 3. ½-X, ½+Y, -Z u=2 x-½, v=-½, w=2 z 0. 333=2 x-1/2 0. 833=2 x 0. 416=x 0. 150=2 z 0. 075=z

What are the coordinates x, y, z for the heavy atom? V=1/2 W=0 0. 168=2 x 0. 084=x 0. 266=2 y 0. 133=y 0. 333=2 x-1/2 0. 833=2 x 0. 416=x a) x=0. 084, y=0. 133, z=0. 075 b) x=0. 416, y=0. 133, z=0. 075 c) None of the above 0. 150=2 z 0. 075=z

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.

Cheshire symmetry From w=0 Harker section xorig 1=0. 084, yorig 1=0. 133 From v=1/2 Harker section, xorig 2=0. 416, zorig 2=0. 075 1. 2. 3. 4. 5. 6. 7. 8. X, -X, 9. 10. 11. 12. 13. 14. 15. 16. 1/2+X, 1/2 -X, 1/2+X, 1/2 -X, Y, -Y, -Y, Y, Z Z -Z -Z Z Z 33. 34. 35. 36. 37. 38. 39. 40. 1/2+X, 1/2+Y, 1/2 -x, 1/2 -Y, 1/2 -X, 1/2+Y, 1/2+X, 1/2 -Y, 1/2 -X, 1/2 -Y, 1/2+X, 1/2+Y, 1/2+X, 1/2 -Y, 1/2 -X, 1/2+Y, 1/2+X, 1/2 -X, 1/2+X, 1/2 -X, Z Z -Z -Z Z Z Y, -Y, -Y, Y, Z Z -Z -Z Z Z 41. 42. 43. 44. 45. 46. 47. 48. 17. 18. 19. 20. 21. 22. 23. 24. X, 1/2+Y, -X, 1/2 -Y, -X, 1/2+Y, X, 1/2 -Y, -X, 1/2 -Y, X, 1/2+Y, X, 1/2 -Y, -X, 1/2+Y, Z Z -Z -Z Z Z 49. 50. 51. 52. 53. 54. 55. 56. 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 25. 26. 27. 28. 29. 30. 31. 32. X, -X, Y, 1/2+Z -Y, 1/2+Z Y, 1/2 -Z -Y, 1/2+Z 57. 58. 59. 60. 61. 62. 63. 64. 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 -Y, 1/2+Z Y, 1/2 -Z -Y, 1/2+Z Apply Cheshire symmetry operator #10 To xorig 1 and yorig 1 Xorig 1=0. 084 ½-xorig 1=0. 5 -0. 084 ½-xorig 1=0. 416 =xorig 2 yorig 1=0. 133 -yorig 1=-0. 133=yorig 2 Hence, Xorig 2=0. 416, yorig 2=-0. 133, zorig 2=0. 075

Advanced case, Proteinase K in space group P 43212 • Where are Harker sections?

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

Symmetry operator 2 -Symmetry operator 4 -x -y - ( ½+y ½-x -½-x-y -½+x-y ½+z ¼+z) 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=-2 y -0. 70=y Plug y into first equation and solve for x. 0. 68=-x-y 0. 68=-x-(-0. 70) 0. 02=x

Symmetry operator 3 -Symmetry operator 6 ½-y ½+x ¾+z - ( -y -x ½-z) ½ ½+2 x ¼+2 z Plug in v. v= ½+2 x 0. 48= ½+2 x -0. 02=2 x -0. 01=x Plug in w. w= ¼+2 z 0. 24= ¼+2 z -0. 01=2 z -0. 005=z

Symmetry operator 3 -Symmetry operator 6 ½-y ½+x ¾+z - ( -y -x ½-z) ½ ½+2 x ¼+2 z Plug in v. v= ½+2 x 0. 46= ½+2 x -0. 04=2 x -0. 02=x Plug in w. w= ¼+2 z 0. 24= ¼+2 z -0. 01=2 z -0. 005=z

From step 3 Xstep 3= 0. 02 ystep 3=-0. 70 zstep 3=? . ? ? ? From step 4 Xstep 4=-0. 02 ystep 4= ? . ? ? zstep 4=-0. 005 Clearly, Xstep 3 does not equal Xstep 4. Use a Cheshire symmetry operator that transforms xstep 3= 0. 02 into xstep 4=- 0. 02. The x, y coordinate in step 3 describes one of the heavy atom For example, positions in thelet’s unit use: cell. The x, z coordinate in step 4 describes a -x, -y, copy. z symmetry related We can’t combine these coordinates And apply it todon’t all coordinates stepatom. 3. directly. They describe the in same Perhaps they even xreferred = - (+0. 02) to different origins. = -0. 02 step 3 -transformed ystep 3 -transformed = - (- 0. 70) = +0. 70 Now xstep 3 -transformed = xx, How can we transform y from step 3 so it describes step 4 And ystep 3 atom has been transformed to a reference the same as x and z in step 4? frame consistent with x and z from step 4. So we arrive at the following self-consistent x, y, z: Xstep 4=-0. 02, ystep 3 -transformed=0. 70, zstep 4=-0. 005 Or simply, x=-0. 02, y=0. 70, z=-0. 005

From step 3 Xstep 3= 0. 02 ystep 3=-0. 70 zstep 3=? . ? ? ? From step 4 Xstep 4=-0. 02 ystep 4= ? . ? ? zstep 4=-0. 005 Clearly, Xstep 3 does not equal Xstep 4. Use a Cheshire symmetry operator that transforms xstep 3= 0. 02 into xstep 4=- 0. 02. For example, let’s use: -x, -y, z And apply it to all coordinates in step 3. xstep 3 -transformed = - (+0. 02) = -0. 02 ystep 3 -transformed = - (- 0. 70) = +0. 70 Now xstep 3 -transformed = xstep 4 And ystep 3 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: Xstep 4=-0. 02, ystep 3 -transformed=0. 70, zstep 4=-0. 005 Or simply, x=-0. 02, y=0. 70, z=-0. 005 Cheshire Symmetry Operators for space group P 43212 X, -Y, Y, Y, -Y, X, -X, Y, Z -Y, Z X, 1/4+Z -X, 1/4+Z X, -Z -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, Y, Y, -Y, X, -X, Y, 1/2+Z -Y, 1/2+Z X, 3/4+Z -X, 3/4+Z X, 1/2 -Z -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

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 symop 1 -symop 5 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.

x y z symmetry operator 1 -symmetry operator 5 -( y x -z) u v w x-y -x+y 2 z 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=2 z=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. 01 becomes -0. 72, 0. 01 then add 1 to u and v 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! (1) U, (5)-U, (9)-V, (13) V, V, U, U, W (2)-U, -V, W (6) U, -V, W (10) V, -U, W (14)-V, -U, W (3) U, V, -W (4)-U, -V, -W W (7)-U, V, -W (8) U, -V, -W W (11)-V, U, -W (12) V, -U, -W W (15) V, U, -W (16)-V, -U, -W In the case that the above test failed, we would change the sign of x.

Assignment • 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 • Friday, hand in your calculation. • We will test the accuracy of your solution and use it to calculate phases and electron density.

Patterson space U=0. 5 Crystal space P 43212 Symmetry operator difference 3 -6 25 W=0. metry Sym P 43212 or diff operat erence 2 -4 Calculate Y and Z Calculate X and Y ak r pe o f wer ion. s n 1 -5 k a sect ce c n e e er Ch arker diff r H o t ra off pe yo u, v, w P 12 432 Sym tr me z -z) x x 2 z y y + ( -x x-y y If prediction lies outside Patterson asymmetric unit (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 recalculate u, v, w. Predicted peak should be present if algebra is correct. Cheshire operator applied to Y and Z if two values of Y do not match X, Y, Z referred to a common origin.

m 230 d_2015_scaled 2. mtz 3 2 64 10. 00 130. 30 2. 20 174. 93 2. 12 150. 51 4. 01 144. 96 2. 04 103. 41 4. 04 164. 00 2. 62 126. 54 1. 31 6. 82 2. 62 149. 88 1. 83 3 2 65 9. 00 175. 48 1. 66 191. 37 2. 00 197. 89 3. 23 177. 46 1. 75 159. 61 2. 69 202. 44 2. 22 180. 36 1. 13 5. 15 2. 26 170. 56 1. 66 3 2 66 17. 00 110. 19 2. 60 129. 09 2. 69 141. 68 4. 54 121. 00 2. 29 82. 97 5. 16 165. 87 2. 24 97. 76 1. 65 4. 36 3. 30 103. 57 2. 45 Etc. for thousands of reflections -0. 66 -0. 73 -5. 45 Up to 72 columns SIGD_eucl 3 -jessica SIGFP_eucl 3 -jessica SIGD_eucl 3 -beccah SIGFP_eucl 3 -beccah SIGFP_native-wenyang SIGFP_native-mimi SIGFP_native-joshua SIGFP_native-aj SIGFP_native-john SIGFP_native-jeannette Free. R_flag L K H All data sets were entered in a spreadsheet. Each column label a different measured quantity. Each row specifies a different HKL. -using the CCP 4 program CAD. 3. 67 3. 31 4. 89

m 230 d_2015_scaled 2. mtz Intensity measurements were converted to structure factor amplitudes (|FHKL|) -using the CCP 4 program TRUNCATE. All data sets were scaled to a reference native data set with the best statistics: prok-native-jeannette -using the CCP 4 program SCALEIT.

Scale intensities by a constant (k) and resolution dependent exponential (B) prok-native-yen H 1 1 1 1 K 0 0 0 0 prok-gdcl 3 -matthew L intensity sigma 10 106894. 0 1698. 0 11 41331. 5 702. 3 12 76203. 2 1339. 0 13 28113. 5 513. 6 14 6418. 2 238. 7 15 45946. 4 882. 7 16 26543. 8 555. 6 H 1 1 1 1 K 0 0 0 0 L 10 11 12 13 14 15 16 intensity sigma 40258. 7 1222. 9 25033. 2 799. 8 24803. 6 771. 5 11486. 3 423. 9 9180. 5 353. 6 25038. 8 783. 0 21334. 6 686. 4 comparison 106894. 0 41331. 5 76203. 2 28113. 5 6418. 2 45946. 4 26543. 8 / / / / 40258. 7 25033. 2 24803. 6 11486. 3 9180. 5 25038. 8 21334. 6 = = = = 2. 65 1. 65 3. 07 2. 45 0. 70 1. 83 1. 24 -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 q/l 2 -B*sin e

Non Harker peaks u -u -u u -v v v -v v -v u -u w w -w -w = = = = 0. 4404 -0. 4404 0. 4404 -0. 4404 0. 1304 -0. 1304

Symmetry Operators are the Bridge between Atomic Coordinates and Patterson Peaks symop #1 symop #2 (-0. 2, -0. 3) x, y -(-x, –y) 2 x , 2 y u=2 x, v=2 y (0, 0) x y (0, 0) u v (0. 2, 0. 3) (0. 6, 0. 4) (0. 4, 0. 6) SYMMETRY OPERATORS FOR PLANE GROUP P 2 1) x, y 2) -x, -y PATTERSON MAP

Patterson map to coordinates (0, 0) a (0. 4, 0. 4) b Plane group p 2 Symmetry operators are 1) x, y 2)-x, -y b (0. 6, 0. 6) (2)= -x, -y (1)=-( x, y) -----u=-2 x v=-2 y Use the Patterson map on the right to calculate coordinates (x, y) for heavy atom. Draw a circle on unit cell on left at (x, y).

Interpreting difference Patterson Maps in Lab today! • Calculate an isomorphous difference Patterson Map (native-heavy atom). We collected 6 derivative data sets in lab – 3 PCMBS – 3 Eu. Cl 3 • 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.

Patterson Review A Patterson synthesis is like a Fourier synthesis except for what two variables? Fourier synthesis r(xyz)=S |Fhkl| cos 2 p(hx+ky+lz -ahkl) hkl Patterson synthesis P(uvw)=S I? hkl -? ) hkl cos 2 p(hu+kv+lw -0) hkl

Hence, Patterson density map= electron density map convoluted with its inverted image. Patterson synthesis P(uvw)=S Ihkl cos 2 p(hu+kv+lw) Remembering Ihkl=Fhkl • Fhkl* And Friedel’s law Fhkl*= F-h-k-l P(uvw)=Fourier. Transform(Fhkl • F-h-k-l) P(uvw)=r(xyz) r (-x-y-z)

Significance? P(uvw)=r(xyz) r (-x-y-z) The Patterson map contains a peak for every interatomic vector in the unit cell. If n atoms in unit cell, then n 2 peaks in Patterson. The peaks are located at the head of the interatomic vector when its tail is placed at the origin. So, n of these peaks on the origin. If the coefficients are |FPH - FP|2, the interatomic vectors are between only heavy atoms in unit cell. Vectors involving protein atoms cancel out. Much simplified.

For 2015 • Too many details about how to solve for x, y, z. Didn’t understand what was the motive for solving the Patterson. • Show a movie of a protein-heavy atom complex. Maybe in P 21212 like pol b. • Make protein a blob, rotate in movie to show the symmetry, then make protein disappear, leaving only heavy atoms. • Draw vectors between the heavy atoms and label with vector equations • Show that some vectors have a pre-defined coordinate that depends on symmetry operator. Leads to Harker section. • Emphasize that the symmetry operator must be known in order to back calculate the Patterson to coordinates.

Calculating X, Y, Z coordinates from Patterson peak positions (U, V, W) Three Examples 1. Exceedingly simple 2 D example 2. Straightforward-3 D example, Pt derivative of polymerase b in space group P 21212 3. Advanced 3 D example, Hg derivative of proteinase K in space group P 43212.

Coordinates to Patterson map (-x, -y) (0, 0) a (-x+1, -y+1) b b (x, y) Plane group p 2 Symmetry operators are x, y -x, -y How many atoms in unit cell? In asymmetric unit? How many peaks will be in the Patterson map? (n 2) How many peaks at the origin? (n) How many non-origin peaks? (n 2 -n)

Coordinates to Patterson map y a (0, 0) b , -2 x (0, 0) -2 -2 x, a (-x, -y) y -2 b (x, y) Plane group p 2 Symmetry operators are 1) x, y 2)-x, -y If x=0. 3, y=0. 8 What will be the coordinates of the Patterson peaks? (2)= -x, -y (1)=-( x, y) -----u=-2 x v=-2 y (1)= x, y (2)=-(-x, -y) -----u=2 x v=2 y

Coordinates to Patterson map y (0, 0) b , -2 x a (0, 0) (-x, -y) y -2 , 2 x -2 -2 x, a 2 y b (x, y) Plane group p 2 Symmetry operators are 1) x, y 2)-x, -y If x=0. 3, y=0. 8 What will be the coordinates of the Patterson peaks? (2)= -x, -y (1)=-( x, y) -----u=-2 x v=-2 y (1)= x, y (2)=-(-x, -y) -----u=2 x v=2 y

Coordinates to Patterson map (-0. 6, -1. 6) y (0, 0) b , -2 x a (0, 0) (-x, -y) y -2 , 2 x -2 -2 x, -2 a 2 y b y -2 x, y , 2 x 2 y -2 x, -2 (x, y) , 2 x 2 y y -2 x, -2 , 2 x 2 y (-0. 6, 1. 6) Plane group p 2 Symmetry operators are 1) x, y 2)-x, -y If x=0. 3, y=0. 8 What will be the coordinates of the Patterson peaks? (2)= -x, -y (1)=-( x, y) -----u=-2 x v=-2 y u=-2*0. 3 v=-2*0. 8 u=-0. 6 v=-1. 6 (1)= x, y (2)=-(-x, -y) -----u=2 x v=2 y u=2*0. 3 v=2*0. 8 u=0. 6 v=1. 6

Coordinates to Patterson map (-0. 6, -1. 6) y (0, 0) b , -2 x a (0, 0) (-x, -y) y -2 , 2 x -2 x, -2 y -2 x, -2 a 2 y (0. 4, 0. 4) , 2 y 2 x b -2 y , , -2 x (0. 6, 0. 6) -2 x (x, y) , 2 x 2 y -0. 6, 1. 6) Plane group p 2 Symmetry operators are 1) x, y 2)-x, -y If x=0. 3, y=0. 8 What will be the coordinates of the Patterson peaks? (2)= -x, -y (1)=-( x, y) -----u=-2 x v=-2 y u=-2*0. 3 v=-2*0. 8 u=-0. 6 v=-1. 6 u= 0. 4 v= 0. 4 (1)= x, y (2)=-(-x, -y) -----u=2 x v=2 y u=2*0. 3 v=2*0. 8 u=0. 6 v=1. 6 u=0. 6 v=0. 6

Patterson map to coordinates (0, 0) a (0. 4, 0. 4) b Plane group p 2 Symmetry operators are 1) x, y 2)-x, -y b (0. 6, 0. 6) (2)= -x, -y (1)=-( x, y) -----u=-2 x v=-2 y What are coordinates for heavy atom?

Patterson map to coordinates (0, 0) a (0. 4, 0. 4) b b x=0. 3, y=0. 8 not the same sites as was used to generate Patterson map (2)= -x, -y (1)=-( x, y) -----u=-2 x v=-2 y 0. 4=-2 x -0. 2=x 0. 4=-2 y -0. 2=y 0. 8=x 0. 8=y

Patterson map to coordinates (0, 0) a (0, 0) b b Plane group p 2 Symmetry operators are 1) x, y 2)-x, -y x=0. 3, y=0. 8 not the same sites as was used to generate Patterson map (2)= -x, -y (1)=-( x, y) -----u=-2 x v=-2 y 0. 6=-2 x -0. 3=x 0. 6=-2 y -0. 3=y 0. 7=x 0. 7=y a (0. 6, 0. 6)

NNQQNY structure Example from NNQQNY (PDB ID code 1 yjo) Zn ion bind between N and C-termini

It’s fine! (x=0. 2, y=0. 2) corresponds to origin choice 4. Choice 2 Choice 1 a a b X=0. 30 Y=0. 30 b X=0. 80 Y=0. 30 X=0. 70 Y=0. 70 X=0. 20 Y=0. 70 Choice 3 Choice 4 a b a X=0. 70 Y=0. 20 X=0. 30 Y=0. 80 b X=0. 20 Y=0. 20 X=0. 80 Y=0. 80

Cheshire operators • • X , Y X+. 5, Y+. 5 X=0. 70 Y=0. 70 Choice 2 X=0. 30 Y=0. 30 Choice 2 X=0. 20 Y=0. 20 X=0. 80 Y=0. 80 Choice 4

Recap • Where n is the number of atoms in the unit cell, there will be n 2 Patterson peaks total, n peaks at the origin, n 2 -n peaks off the origin. • That is, there will be one peak for every pairwise difference between symmetry operators in the crystal. • Written as equations, these differences relate the Patterson peak coordinates u, v, w to atomic coordinates, 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.

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. • Adding 1 to x, y, or z, or any combination of x, y, and z is valid. It is just a unit cell translation. • If a structure is solved independently by two crystallographers using different choices of origin, their coordinates will be related by a Cheshire operator.

Polymerase b example, P 21212 • • 1. 2. 3. 4. • 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!

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

24 50 are selected for output The number of symmetry related peaks rejected for being too close to the map edge is These peaks are sorted into descending order of height, the top Peaks related by symmetry are assigned the same site number 5 4 3 2 1 43 60 59 68 67 79 1 10 31 30 40 40 46 45 55 1 6. 67 11. 29 12. 08 12. 71 12. 72 13. 14 14. 14 253. 87 0 58 66 8 60 66 6 65 0 46 58 14 60 8 0 6 52 0 0 25 23 48 48 73 71 96 0 0. 0000 0. 4421 0. 5000 0. 0580 0. 4525 0. 5000 0. 0469 0. 4959 0. 0000 0. 3478 0. 4421 0. 1051 0. 4525 0. 0580 0. 0000 0. 0469 0. 3967 0. 0000 0. 1288 0. 1214 0. 2495 0. 3790 0. 3708 0. 5000 0. 00 29. 99 33. 92 3. 94 30. 69 33. 92 3. 18 33. 64 0. 00 23. 59 29. 99 7. 13 30. 69 3. 94 0. 00 3. 18 26. 91 0. 00 13. 12 12. 36 25. 41 38. 59 37. 76 50. 92 0. 00 Orthogonal coordinates 6 45 Fractional coordinates 7 29 Grid 8 Order No. Site Height/Rms 9

1 Use this edge to measure y 2 2 2 4 Use this edge to measure y 3 3 3 4 4 2 4 2 1 3 4 1 Use this edge to measure x 2 3 4 Use this edge to measure y 1 4 Use this edge to measure x 2 3 1 3 2 Use this edge to measure x 2 4 1 1 3 4 Use this edge to measure y Use this edge to measure x 2 4 1 3 3 Use this edge to measure x 2 2 1 1 Use this edge to measure x

Crystal Structure of NNQQNY Peptide from Sup 35 Prion 1) Draw this symbol on all the 2 -fold symmetry axes you see. 2) Chose one 2 -fold axis as the origin. 3) (0, 0) a Draw a unit cell b having this size and shape, with corners on 2 -folds 4) How many NNQQNY molecules in the unit cell? asymmetric unit? 5) Label the upper left corner of the cell with “(0, 0)” thus designating it as the origin. label the horizontal axis “a”. Label the other axis “b” 6) Measure x and y distances from the origin to one of the zinc ions (sphere) located within the unit cell boundaries. The ruler provided measures in fractions of a unit cell. One side of the ruler is for measuring x, the other is for y. Round off answers to nearest 0. 10.

two-fold axes

4 distinct sets of two-fold axes

Unit Cell Choice 1 Choice 2 Choice 3 Choice 4

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

Which plane group? a b

What are the coordinates of the red zinc using origin choice 1? X=0. 80 Y=0. 30 a b 1 2 3 4 1 2

What are the coordinates of the red zinc with origin choice 2? a X=0. 30 Y=0. 30 a b X = +0. 3 Y = +0. 3 1 1 2 3 4 2

What are the coordinates of the red zinc with origin choice 3? a X=0. 30 Y=0. 80 a b 1 X = +0. 3 Y = +0. 8 1 2 3 4 2 3

What are the coordinates of the red zinc with origin choice 4? a X=0. 80 Y=0. 80 a b 1 X = +0. 8 Y = +0. 8 1 2 3 4 2 3

Cheshire operators • • X , Y X+. 5, Y+. 5 X=0. 80 Y=0. 30 Choice 1 X=0. 30 Y=0. 30 Choice 2 X=0. 30 Y=0. 80 Choice 3 X=0. 80 Y=0. 80 Choice 4

What are the coordinates of the 2 nd Zn ion in the unit cell? Choice 2 Choice 1 a a b X=0. 30 Y=0. 30 b X=0. 80 Y=0. 30 Choice 3 Choice 4 a a b b X=0. 30 Y=0. 80 X=0. 80 Y=0. 80

What are the coordinates of the purple zinc using origin choice 1? X=0. 80 Y=0. 30 Symmetry operators in plane group p 2 X, Y -X, -Y a b X 2=-0. 80 Y 2=-0. 30 1 1 2 3 4 X=0. 20 Y=0. 70 2

Always allowed to add or subtract multiples of 1. 0 X = -0. 8 Y = -0. 3 a b X = +0. 8 Y = +0. 3 X = +0. 2 Y = +0. 7

The 4 choices of origin are equally valid but once a choice is made, you must remain consistent. Choice 2 Choice 1 a a b X=0. 30 Y=0. 30 b X=0. 80 Y=0. 30 X=0. 70 Y=0. 70 X=0. 20 Y=0. 70 Choice 3 Choice 4 a b a X=0. 70 Y=0. 20 X=0. 30 Y=0. 80 b X=0. 20 Y=0. 20 X=0. 80 Y=0. 80