Interpreting difference Patterson Maps in Lab this week! Calculate an isomorphous difference Patterson Map (native-heavy atom) for each derivative data.

Interpreting difference Patterson Maps in Lab this week! Calculate an isomorphous difference Patterson Map (native-heavy atom) for each derivative data set. We collected12 derivative data sets in lab (different heavy atoms at different concentrations) –HgCl 2 –PCMBS –Hg(Acetate) 2 –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.

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

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)

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.

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

Patterson maps are more complicated 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

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

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.

What Plane group is this? 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

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.

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

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)

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

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)

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)

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 betrween 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.

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

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

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??????

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

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.

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

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.

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.

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

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? Arbitrary origin choice can bring them into agreement. What are the rules for origin shifts? 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 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

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 coordinates 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.

Interpreting difference Patterson Maps in Lab this week! Calculate an isomorphous difference Patterson Map (native-heavy atom) for each derivative data.

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Interpreting difference Patterson Maps in Lab this week! Calculate an isomorphous difference Patterson Map (native-heavy atom) for each derivative data.

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Presentation on theme: "Interpreting difference Patterson Maps in Lab this week! Calculate an isomorphous difference Patterson Map (native-heavy atom) for each derivative data."— Presentation transcript:

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