Interpreting difference Patterson Maps in Lab today! Calculate an isomorphous difference Patterson Map (native-heavy atom). We collected 8 derivative data.

Interpreting difference Patterson Maps in Lab today! Calculate an isomorphous difference Patterson Map (native-heavy atom). We collected 8 derivative data sets in lab –3 PCMBS –4 EuCl 3 –1 GdCl 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 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

Difference Patterson synthesis P(uvw)=  F PH(hkl) -F P (hkl) | 2 cos2  (hu+kv+lw -  ) Difference Patterson We use isomorphous differences as coefficients, so the features in the map correspond to heavy atoms. hkl Contributions of protein atoms cancel out. Much simplified.

Patterson map= electron density map convoluted with its inverted image. P(uvw)=  (xyz)  (-x-y-z) Peak positions correspond to interatomic vectors in the unit cell. Imagine a vector connecting every pair of heavy atoms. Translate each vector so one atom of the pair is at the origin. Peak is located at position of other atom in the pair. Both forward and reverse vectors included (centrosymmetric). If n atoms in unit cell, then n 2 peaks in Patterson.

Pt derivative DNA polymerase  (1) x, y, z (2) -x, -y, z (3) -x + 1/2, y + 1/2, -z (4) x + 1/2, -y + 1/2, -z x z y P2 1 2 1 2

Contains 2 and 2 1 axes (1) x, y, z (reference) (2) -x, -y, z (2-fold around z) (3) -x + 1/2, y + 1/2, -z (2 1 around y) (4) x + 1/2, -y + 1/2, -z (2 1 around x) x z y

One symop per asymmetric unit (1) x, y, z (2) -x, -y, z (3) -x + 1/2, y + 1/2, -z (4) x + 1/2, -y + 1/2, -z x z y

|F PH -F P | gives Pt contribution (1) x, y, z (2) -x, -y, z (3) -x + 1/2, y + 1/2, -z (4) x + 1/2, -y + 1/2, -z x z y These x,y,z coordinates represent the solution we would like to obtain from interpreting the difference Patterson map

Difference Patterson contains peaks between every pair of Pt. (1) x, y, z (2) -x, -y, z (3) -x + 1/2, y + 1/2, -z (4) x + 1/2, -y + 1/2, -z x z y 16 vectors, every pairwise difference between symops. (1) x, y, z (2) -x, -y, z (3) -x + 1/2, y + 1/2, -z (4) x + 1/2, -y + 1/2, -z

Each vector translated to origin 16 vectors, every pairwise difference between symops. (1) x, y, z (2) -x, -y, z (3) -x + 1/2, y + 1/2, -z (4) x + 1/2, -y + 1/2, -z x z y (1) x, y, z (2) -x, -y, z (3) -x + 1/2, y + 1/2, -z (4) x + 1/2, -y + 1/2, -z

Assign difference vector to peak 16 vectors, every pairwise difference between symops. u w y (1) x, y, z (2) -x, -y, z (3) -x + 1/2, y + 1/2, -z (4) x + 1/2, -y + 1/2, -z

Assign difference vector to peak 16 vectors, every pairwise difference between symops. u w y (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 (2) -x, -y, z (3) -x + 1/2, y + 1/2, -z (4) x + 1/2, -y + 1/2, -z x z y

Assign difference vector to peak 16 vectors, every pairwise difference between symops. (1) x, y, z (4) x + 1/2, -y + 1/2, -z u w y (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 (2) -x, -y, z (3) -x + 1/2, y + 1/2, -z (4) x + 1/2, -y + 1/2, -z x z y -(1)(x, y, z ½, -2y+1/2,-2z u=1/2

Harker sections 1.Planes perpendicular to a rotation or screw axis. 2.It occurs when one of the coordinates of the difference vector is a constant. 3.They offer clues to assigning a Patterson peak to the correct difference vector equation.

In what planes do we expect difference Patterson peaks in P2 1 2 1 2? Name ops u v w Andrew 1-2 2x 2y 0 David 1-3 2x-1/2 -1/2 2z Jessica 1-4 -1/2 2y-1/2 2z Hanyoung 2-3 -1/2 -2y-1/2 2z Jeff 2-4 -2x-1/2 -1/2 2z John 3-4 -2x 2y 0 Scott 2-1 -2x -2y 0 Steve 3-1 -2x+1/2 1/2 -2z 4-1 1/2 -2y+1/2 -2z 3-2 1/2 2y+1/2 -2z 4-2 2x+1/2 1/2 -2z 4-3 2x -2y 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

In what planes do we expect difference Patterson peaks in P2 1 2 1 2? a)w=0 b)w=1/2 c) v=1/2 d) u=1/2 (1)x, y, z (2) -x, -y, z (3) -x + 1/2, y + 1/2, -z (4) x + 1/2, -y + 1/2, -z

Harker sections in P2 1 2 1 2 (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 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

Which of these difference vectors is likely to correspond to the difference Patterson peak shown here? W=0 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 a) b) c) d) They are all equally likely.

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

What is the value of z? 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 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=2x, v=2y, w=0 0.168=2x 0.084=x 0.266=2y 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=2x- ½,v=-½,w=2z 0.333=2x-1/2 0.833=2x 0.416=x 0.150=2z 0.075=z

What are the coordinates x,y,z for the heavy atom? V=1/2 0.333=2x-1/2 0.833=2x 0.416=x 0.150=2z 0.075=z 0.168=2x 0.084=x 0.266=2y 0.133=y W=0 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

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 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 orig1 and y orig1 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

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

Symmetry operator 2 -Symmetry operator 4 - 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. 0.68=-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

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

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

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?

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

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.

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

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 map that they were assigned. 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.

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.

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

m230d_2015_scaled2.mtz H K L FreeR_flag FP_native-jeannette SIGFP_native-jeannette FP_native-john SIGFP_native-john FP_native-aj SIGFP_native-aj FP_native-joshua SIGFP_native-joshua FP_native-mimi SIGFP_native-mimi FP_native-wenyang SIGFP_native-wenyang FP_eucl3-beccah SIGFP_eucl3-beccah D_eucl3-beccah SIGD_eucl3-beccah FP_eucl3-jessica SIGFP_eucl3-jessica D_eucl3-jessica SIGD_eucl3-jessica Up to 72 columns All data sets were entered in a spreadsheet. Each column label a different measured quantity. Each row specifies a different HKL. - using the CCP4 program CAD. 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 -0.66 3.67 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 -0.73 3.31 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 -5.45 4.89 Etc. for thousands of reflections

m230d_2015_scaled2.mtz 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-jeannette - using the CCP4 program SCALEIT.

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-yen 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-gdcl3-matthew 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 first crystal is larger than the second. -Multiply Saken’s data by k and B to put the data on the same scale.

e -B*sin 2  / 2

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

Which of the following corresponds to a native Patterson map of NNQQNY? a) b) c) d)

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)110 2 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 (|F PH | - |F P |) 2 What will be the only features in the map? a)Vectors between protein atoms b)Vectors between heavy atoms 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)Vectors between protein atoms b)Vectors between heavy atoms 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? (0.8, 0.0, 0.3) (0.2, 0.5, 0.7) a)0.6,-0.5,-0.4 b)-0.6,0.5,0.4 c)a and b d)No such peak exists

Which of the following would correspond to a difference Patterson peak? (0.8, 0.0, 0.3) (0.2, 0.5, 0.7) a)0.6,-0.5,-0.4 b)-0.6,0.5,0.4 c)a and b d)No such peak exists (0.8, 0.0, 0.3) -(0.2, 0.5, 0.7) (0.6,-0.5,-0.4) u v w (0.2, 0.5, 0.7) -(0.8, 0.0, 0.3) (-0.6, 0.5, 0.4) u v 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? (x, y, z) (-x,y+1/2,-z) a)-2x,1/2,-2z b)2x,-1/2,2z c)a and b d)No such peak exists

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

Choice of origin ambiguity Each is a valid choice of cell. Show two choices of origin in NNQQNY The coordinates of the heavy atoms would be different in each. Each of these choices of unit cell boundaries would produce the same Patterson map.(same set of interatomic vectors regardless of how cell boundaries are drawn). May arrive at either of these definitions of unit cell depending on choice of peak used.

How many heavy atoms are expected in the unit cell with space group P2 1 2 1 2 if there is only one heavy atom in the asymmetric unit, ? a)4 b)4 2 c)Not enough information given (1)x, y, z (2) -x, -y, z (3) -x + 1/2, y + 1/2, -z (4) x + 1/2, -y + 1/2, -z

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”.

Non Harker peaks u v w = 0.4404 0.4404 0.1304 -u -v w = -0.4404 -0.4404 0.1304 u v -w = 0.4404 0.4404 -0.1304 -u -v -w = -0.4404 -0.4404 -0.1304 -u v w = -0.4404 0.4404 0.1304 u -v w = 0.4404 -0.4404 0.1304 -u v -w = -0.4404 0.4404 -0.1304 u -v -w = 0.4404 -0.4404 -0.1304 -v u w = -0.4404 0.4404 0.1304 v -u w = 0.4404 -0.4404 0.1304 -v u -w = -0.4404 0.4404 -0.1304 v -u -w = 0.4404 -0.4404 -0.1304 v u w = 0.4404 0.4404 0.1304 -v -u w = -0.4404 -0.4404 0.1304 v u -w = 0.4404 0.4404 -0.1304 -v -u -w = -0.4404 -0.4404 -0.1304

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

Patterson map to coordinates a b (0,0) a b Plane group p2 Symmetry operators are 1) x, y 2)-x,-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). (2)= -x, -y (1)=-( x, y) ---------- u=-2x v=-2y (0.4, 0.4) (0.6, 0.6)

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 P21212 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 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.

Coordinates to Patterson map Plane group p2 Symmetry operators are x,y -x, -y How many atoms in unit cell? In asymmetric unit? a b (0,0) a b (x,y) (-x+1,-y+1) 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) (-x,-y)

Coordinates to Patterson map a b (0,0) a b (x,y) (-x,-y) -2x, -2y (1)= x, y (2)=-(-x,-y) ---------- u=2x v=2y (2)= -x, -y (1)=-( x, y) ---------- u=-2x v=-2y Plane group p2 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?

Coordinates to Patterson map Plane group p2 Symmetry operators are 1) x, y 2)-x,-y a b (0,0) a b (x,y) (-x,-y) -2x, -2y 2x, 2y If x=0.3, y=0.8 What will be the coordinates of the Patterson peaks? (1)= x, y (2)=-(-x,-y) ---------- u=2x v=2y (2)= -x, -y (1)=-( x, y) ---------- u=-2x v=-2y

a b (0,0) a b (x,y) (-x,-y) -2x, -2y 2x, 2y -2x, -2y 2x, 2y -2x, -2y 2x, 2y -2x, -2y 2x, 2y Plane group p2 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? (1)= x, y (2)=-(-x,-y) ---------- u=2x v=2y u=2*0.3 v=2*0.8 u=0.6 v=1.6 (2)= -x, -y (1)=-( x, y) ---------- u=-2x v=-2y u=-2*0.3 v=-2*0.8 u=-0.6 v=-1.6 Coordinates to Patterson map (-0.6,-1.6) (-0.6, 1.6)

a b (0,0) a b (x,y) (-x,-y) -2x, -2y 2x, 2y -2x, -2y 2x, 2y -2x, -2y 2x, 2y -2x, -2y 2x, 2y Plane group p2 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? (1)= x, y (2)=-(-x,-y) ---------- u=2x v=2y u=2*0.3 v=2*0.8 u=0.6 v=1.6 u=0.6 v=0.6 (2)= -x, -y (1)=-( x, y) ---------- u=-2x v=-2y u=-2*0.3 v=-2*0.8 u=-0.6 v=-1.6 u= 0.4 v= 0.4 (0.4, 0.4) (0.6, 0.6) (-0.6,-1.6) -0.6, 1.6) Coordinates to Patterson map

Patterson map to coordinates a b (0,0) a b Plane group p2 Symmetry operators are 1) x, y 2)-x,-y What are coordinates for heavy atom? (2)= -x, -y (1)=-( x, y) ---------- u=-2x v=-2y (0.4, 0.4) (0.6, 0.6)

Patterson map to coordinates a b (0,0) a b (2)= -x, -y (1)=-( x, y) ---------- u=-2x v=-2y 0.4=-2x -0.2=x 0.4=-2y -0.2=y (0.4, 0.4) 0.8=x 0.8=y x=0.3, y=0.8  not the same sites as was used to generate Patterson map 

Patterson map to coordinates a b (0,0) a b Plane group p2 Symmetry operators are 1) x, y 2)-x,-y (2)= -x, -y (1)=-( x, y) ---------- u=-2x v=-2y 0.6=-2x -0.3=x 0.6=-2y -0.3=y (0.6, 0.6) 0.7=x 0.7=y x=0.3, y=0.8  not the same sites as was used to generate Patterson map 

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

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

b a It’s fine! (x=0.2, y=0.2) corresponds to origin choice 4. Choice 1 Choice 2 Choice 3 Choice 4 b a b a b a X=0.80 Y=0.30 X=0.30 Y=0.30 X=0.30 Y=0.80 X=0.80 Y=0.80 X=0.20 Y=0.70 X=0.70 Y=0.70 X=0.70 Y=0.20 X=0.20 Y=0.20

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  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!

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

These peaks are sorted into descending order of height, the top 50 are selected for output The number of symmetry related peaks rejected for being too close to the map edge is 24 Peaks related by symmetry are assigned the same site number Order No. Site Height/Rms Grid Fractional coordinates Orthogonal coordinates 1 1 1 253.87 0 0 0 0.0000 0.0000 0.0000 0.00 0.00 0.00 2 79 55 14.14 65 52 96 0.4959 0.3967 0.5000 33.64 26.91 50.92 3 67 45 13.14 6 6 71 0.0469 0.0469 0.3708 3.18 3.18 37.76 4 68 46 12.72 66 0 73 0.5000 0.0000 0.3790 33.92 0.00 38.59 5 59 40 12.71 60 8 48 0.4525 0.0580 0.2495 30.69 3.94 25.41 6 60 40 12.71 8 60 48 0.0580 0.4525 0.2495 3.94 30.69 25.41 7 43 30 12.08 66 14 23 0.5000 0.1051 0.1214 33.92 7.13 12.36 8 45 31 11.29 58 58 25 0.4421 0.4421 0.1288 29.99 29.99 13.12 9 29 10 6.67 0 46 0 0.0000 0.3478 0.0000 0.00 23.59 0.00

1 2 34 1 2 34 Use this edge to measure x Use this edge to measure y 1 2 34 1 2 34 Use this edge to measure x Use this edge to measure y 1 2 34 1 2 34 Use this edge to measure x Use this edge to measure y 1 2 34 1 2 34 Use this edge to measure x Use this edge to measure y 1 2 34 1 2 34 Use this edge to measure x Use this edge to measure y 1 2 34 1 2 34 Use this edge to measure x Use this edge to measure y

1) Draw this symbol on all the 2-fold symmetry axes you see. 2) Chose one 2-fold axis as the origin. 3) 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. b a Crystal Structure of NNQQNY Peptide from Sup35 Prion having this size and shape, with Draw a unit cell corners on 2-folds (0,0)

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 1 Choice 2 Choice 3 Choice 4 b a b a b a b a

Which plane group? b a

X=0.80 What are the coordinates of the red zinc using origin choice 1? Y=0.30 1 2 34 1 2 34 b a

What are the coordinates of the red zinc with origin choice 2? a X=0.30 Y=0.30 b a 1 2 34 1 2 3 4 X = +0.3 Y = +0.3

What are the coordinates of the red zinc with origin choice 3? a X=0.30 Y=0.80 b a 1 2 34 1 2 34 X = +0.3 Y = +0.8

What are the coordinates of the red zinc with origin choice 4? a X=0.80 Y=0.80 b a 1 2 34 1 2 34 X = +0.8 Y = +0.8

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

b a What are the coordinates of the 2 nd Zn ion in the unit cell? Choice 1 Choice 2 Choice 3 Choice 4 b a b a b a X=0.80 Y=0.30 X=0.30 Y=0.30 X=0.30 Y=0.80 X=0.80 Y=0.80

X=0.20 What are the coordinates of the purple zinc using origin choice 1? Y=0.70 1 2 34 b a 1 2 34 X 2 =-0.80 Y 2 =-0.30 Symmetry operators in plane group p2 X, Y -X,-Y X=0.80 Y=0.30

X = +0.8 Y = +0.3 Always allowed to add or subtract multiples of 1.0 X = +0.2 Y = +0.7 X = -0.8 Y = -0.3 b a

b a The 4 choices of origin are equally valid but once a choice is made, you must remain consistent. Choice 1 Choice 2 Choice 3 Choice 4 b a b a b a X=0.80 Y=0.30 X=0.30 Y=0.30 X=0.30 Y=0.80 X=0.80 Y=0.80 X=0.20 Y=0.70 X=0.70 Y=0.70 X=0.70 Y=0.20 X=0.20 Y=0.20

Dear Sung Chul, I apologize for the lack of clarity. Both x,y,z and -x,y,z can satisfy the vector equations in step 6 because both the Harker sections used thus far contain mirror symmetry that is consistent with either sign of "x". In other words, the peaks in these Harker sections are related by Patterson symmetry that makes them consistent with either sign of "x" However, there are other difference vector equations in this space group (for example symop 1 - symop 5) that are not reflected by these mirror planes. These can be used to discriminate which sign of "x" is correct. Only one of these choices of the sign of "x" will be able to predict correctly Patterson peaks specified by this difference vector. My thought is that one sign of "x" is consistent with space group P43212, and the other sign of "x" is consistent with space group P41212. Mike On 01/08/2015 06:21 PM, 하성철 wrote: > > Dear Michael R. Sawaya, > > I have learned a lot basic concept and practical methods using presentation materials of the lectures for CHEM M230D Course. THey are prepared well so that anyone can understand the contents easily even without the explanation by a lecturer. > > I have one question about content of the presentation ppt file for the lectures for CHEM M230D Course. In the ppt file or on the web page for “Difference Patterson Maps and Determination of Heavy Atom Sites”, I don’t know why both x,y,z and -x,y,z can satisfy the difference vector equations in the step 6. Could you explain the reason? > > Best regards, > > SungChul Ha >

Interpreting difference Patterson Maps in Lab today! Calculate an isomorphous difference Patterson Map (native-heavy atom). We collected 8 derivative data.

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Interpreting difference Patterson Maps in Lab today! Calculate an isomorphous difference Patterson Map (native-heavy atom). We collected 8 derivative data.

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Presentation on theme: "Interpreting difference Patterson Maps in Lab today! Calculate an isomorphous difference Patterson Map (native-heavy atom). We collected 8 derivative data."— Presentation transcript:

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