1 / 46

Interpreting difference Patterson Maps in Lab this week!

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

sylvia-edwards
Download Presentation

Interpreting difference Patterson Maps in Lab this week!

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 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) • HgCl2 • PCMBS • Hg(Acetate)2 • EuCl3 • GdCl3 • SmCl3 • 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.

  2. Patterson Review A Patterson synthesis is like a Fourier synthesis except for what two variables? Fourier synthesis r(xyz)=S |Fhkl| cos2p(hx+ky+lz -ahkl) hkl Patterson synthesis P(uvw)=SIhklcos2p(hu+kv+lw -0) hkl Patterson synthesis P(uvw)=S?hklcos2p(hu+kv+lw -?) hkl

  3. Hence, Patterson density map= electron density map convoluted with its inverted image. Patterson synthesis P(uvw)=SIhklcos2p(hu+kv+lw) Remembering Ihkl=Fhkl•Fhkl* And Friedel’s law Fhkl*= F-h-k-l P(uvw)=FourierTransform(Fhkl•F-h-k-l) P(uvw)=r(uvw) r (-u-v-w)

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

  5. a b 1 H H H H H H H H 3 2 Every Patterson peak corresponds to an inter-atomic vector • 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? Patterson Density Map single water molecule convoluted with its inverted image. Electron Density Map single water molecule in the unit cell

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

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

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

  9. b 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 What Plane group is this?

  10. b 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 Let’s consider only oxygen atoms 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.

  11. a (0,0) b 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?

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

  13. Patterson in plane group p2Lay down two copies of unit cell at origin, first copy with smile 1 at origin, second copy with simile 2 at origin. 2D CRYSTAL PATTERSON MAP (-0.2,-0.3) a (0,0) a (0,0) b b (0.2,0.3) What are the coordinates of this Patterson self peak? (a peak between atoms related by xtal sym) What is the length of the vector between faces? SYMMETRY OPERATORSFOR PLANE GROUP P2 1) x,y 2) -x,-y • 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

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

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

  16. Now I’m stuck in Patterson space. How do I get back to x,y coordinates? Remember the Patterson Peak positions (U,V) correspond to vectors between symmetry related smiley faces in the unit cell. That is, differences betrween our friends the space group operators. a (0,0) b (0.6, 0.4) (0.4, 0.6) SYMMETRY OPERATORSFOR PLANE GROUP P2 1) x,y 2) -x,-y PATTERSON MAP x , y -(-x, –y) 2x , 2y u=2x, v=2y symop #1 symop #2 plug in Patterson values for u and v to get x and y.

  17. Now I’m stuck in Patterson space. How do I get back to x,y, coordinates? SYMMETRY OPERATORSFOR PLANE GROUP P2 1) x,y 2) -x,-y a (0,0) b x y -(-x –y) 2x 2y symop #1 symop #2 (0.4, 0.6) set u=2x v=2y plug in Patterson values for u and v to get x and y. v=2y 0.6=2y 0.3=y u=2x 0.4=2x 0.2=x PATTERSON MAP

  18. Hurray!!!! SYMMETRY OPERATORSFOR PLANE GROUP P2 1) x,y 2) -x,-y a (0,0) b x y -(-x –y) 2x 2y (0.2,0.3) symop #1 symop #2 set u=2x v=2y plug in Patterson values for u and v to get x and y. 2D CRYSTAL v=2y 0.6=2y 0.3=y u=2x 0.4=2x 0.2=x HURRAY! we got back the coordinates of our smiley faces!!!!

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

  20. a (0,0) b Arbitrary choice of origin • Original origin choice • Coordinates x=0.2, y=0.3. (-0.2,-0.3) a (0,0) b (-0.3,-0.2) (0.2,0.3) (0.3,0.2) (0.8,0.7) (0.7,0.8) • New origin choice • Coordinates x=0.3, y=0.2. Related by 0.5, 0.5 (x,y) shift

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

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

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

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

  25. Harker section w=0 W=0 • X, Y, Z • -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.

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

  27. Resolving ambiguity in x,y,z • From w=0 Harker section x1=0.084, y1=0.133 • From v=1/2 Harker section, x2=0.416, z2=0.075 • Why doesn’t x agree between solutions? 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.

  28. Cheshire symmetry From w=0 Harker section xorig1=0.084, yorig1=0.133 From v=1/2 Harker section, xorig2=0.416, zorig2=0.075 • X, Y, Z • -X, -Y, Z • -X, Y, -Z • X, -Y, -Z • -X, -Y, -Z • X, Y, -Z • X, -Y, Z • -X, Y, Z • 1/2+X, Y, Z • 1/2-X, -Y, Z • 1/2-X, Y, -Z • 1/2+X, -Y, -Z • 1/2-X, -Y, -Z • 1/2+X, Y, -Z • 1/2+X, -Y, Z • 1/2-X, Y, Z • X,1/2+Y, Z • -X,1/2-Y, Z • -X,1/2+Y, -Z • X,1/2-Y, -Z • -X,1/2-Y, -Z • X,1/2+Y, -Z • X,1/2-Y, Z • -X,1/2+Y, Z • X, Y,1/2+Z • -X, -Y,1/2+Z • -X, Y,1/2-Z • X, -Y,1/2-Z • -X, -Y,1/2-Z • X, Y,1/2-Z • X, -Y,1/2+Z • -X, Y,1/2+Z • 1/2+X,1/2+Y, Z • 1/2-x,1/2-Y, Z • 1/2-X,1/2+Y, -Z • 1/2+X,1/2-Y, -Z • 1/2-X,1/2-Y, -Z • 1/2+X,1/2+Y, -Z • 1/2+X,1/2-Y, Z • 1/2-X,1/2+Y, Z • 1/2+X, Y,1/2+Z • 1/2-X, -Y,1/2+Z • 1/2-X, Y,1/2-Z • 1/2+X, -Y,1/2-Z • 1/2-X, -Y,1/2-Z • 1/2+X, Y,1/2-Z • 1/2+X, -Y,1/2+Z • 1/2-X, Y,1/2+Z • X,1/2+Y,1/2+Z • -X,1/2-Y,1/2+Z • -X,1/2+Y,1/2-Z • X,1/2-Y,1/2-Z • -X,1/2-Y,1/2-Z • X,1/2+Y,1/2-Z • X,1/2-Y,1/2+Z • -X,1/2+Y,1/2+Z • 1/2+X,1/2+Y,1/2+Z • 1/2-X,1/2-Y,1/2+Z • 1/2-X,1/2+Y,1/2-Z • 1/2+X,1/2-Y,1/2-Z • 1/2-X,1/2-Y,1/2-Z • 1/2+X,1/2+Y,1/2-Z • 1/2+X,1/2-Y,1/2+Z • 1/2-X,1/2+Y,1/2+Z Apply Cheshire symmetry operator #10 To x1 and y1 Xorig1=0.084 ½-xorig1=0.5-0.084 ½-xorig1=0.416 =xorig2 yorig1=0.133 -yorig1=-0.133=yorig2 Hence, Xorig2=0.416, yorig2=-0.133, zorig2=0.075

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

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

  31. 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

  32. 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

  33. 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

  34. 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

  35. 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

  36. 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

  37. 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

  38. From step 3 Xstep3= 0.02 ystep3=-0.70 zstep3=?.??? From step 4 Xstep4=-0.02 ystep4= ?.?? zstep4=-0.005 Clearly, Xstep3does not equalXstep4 . Use a Cheshire symmetry operator that transforms xstep3= 0.02 into xstep4=- 0.02. For example, let’s use: -x, -y, z And apply it to all coordinates in step 3. xstep3-transformed = - (+0.02) = -0.02 ystep3-transformed = - (- 0.70)= +0.70 Now xstep3-transformed = xstep4 And ystep3 has been transformed to a reference frame consistent with x and z from step 4. So we arrive at the following self-consistent x,y,z: Xstep4=-0.02, ystep3-transformed=0.70, zstep4=-0.005 Or simply, x=-0.02, y=0.70, z=-0.005 The x, y coordinate in step 3 describes one of the heavy atom positions in the unit cell. The x, z coordinate in step 4 describes a symmetry related copy. We can’t combine these coordinates directly. They don’t describe the same atom. Perhaps they even referred to different origins. How can we transform x, y from step 3 so it describes the same atom as x and z in step 4?

  39. From step 3 Xstep3= 0.02 ystep3=-0.70 zstep3=?.??? 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 From step 4 Xstep4=-0.02 ystep4= ?.?? zstep4=-0.005 Clearly, Xstep3does not equalXstep4 . Use a Cheshire symmetry operator that transforms xstep3= 0.02 into xstep4=- 0.02. For example, let’s use: -x, -y, z And apply it to all coordinates in step 3. xstep3-transformed = - (+0.02) = -0.02 ystep3-transformed = - (- 0.70)= +0.70 Now xstep3-transformed = xstep4 And ystep3 has been transformed to a reference frame consistent with x and z from step 4. So we arrive at the following self-consistent x,y,z: Xstep4=-0.02, ystep3-transformed=0.70, zstep4=-0.005 Or simply, x=-0.02, y=0.70, z=-0.005

  40. From step 3 Xstep3= 0.02 ystep3=-0.70 zstep3=?.??? 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 From step 4 Xstep4=-0.02 ystep4= ?.?? zstep4=-0.005 Clearly, Xstep3does not equalXstep4 . Use a Cheshire symmetry operator that transforms xstep3= 0.02 into xstep4=- 0.02. For example, let’s use: -x, -y, z And apply it to all coordinates in step 3. xstep3-transformed = - (+0.02) = -0.02 ystep3-transformed = - (- 0.70)= +0.70 Now xstep3-transformed = xstep4 And ystep3 has been transformed to a reference frame consistent with x and z from step 4. So we arrive at the following self-consistent x,y,z: Xstep4=-0.02, ystep3-transformed=0.70, zstep4=-0.005 Or simply, x=-0.02, y=0.70, z=-0.005

  41. From step 3 Xstep3= 0.02 ystep3=-0.70 zstep3=?.??? 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 From step 4 Xstep4=-0.02 ystep4= ?.?? zstep4=-0.005 Clearly, Xstep3does not equalXstep4 . Use a Cheshire symmetry operator that transforms xstep3= 0.02 into xstep4=- 0.02. For example, let’s use: -x, -y, z And apply it to all coordinates in step 3. xstep3-transformed = - (+0.02) = -0.02 ystep3-transformed = - (- 0.70)= +0.70 Now xstep3-transformed = xstep4 And ystep3 has been transformed to a reference frame consistent with x and z from step 4. So we arrive at the following self-consistent x,y,z: Xstep4=-0.02, ystep3-transformed=0.70, zstep4=-0.005 Or simply, x=-0.02, y=0.70, z=-0.005

  42. 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.

  43. 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.

More Related