1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4
--2-fold symmetry axis Chose a 2-fold axis as the origin. How many molecules in the unit cell? asymmetric unit? Draw an “o” at the upper left corner of the cell and make it the origin. Estimate (x,y) coordinates of oxygen atom in fractions of a unit cell. Use the ruler provided. a o H Li On the crystal, b draw a unit cell having this size and shape.
4 distinct two-fold axes b H Li
Unit Cell a b H Li Choice 1 Choice 2 Choice 3 Choice 4
4 Choices of origin Choice 1 Choice 2 Choice 3 Choice 4 b a H Li H Li
Which plane group? H Li b a
What are the coordinates of the oxygen using origin choice 1? Li Li Li H H H H H H Li a Li Li b Li Li Li H 1 2 3 4 H H H H H Li Li Li Li Li Li Y=0.2 X=0.2 H H H H H H Li Li Li 1 2 3 4
What are the coordinates of the other oxygen in the unit cell? Li Symmetry operators in plane group p2 X, Y -X,-Y a b X2=0.80 Y2=0.80 1 2 3 4 1 2 3 4
Always allowed to add or subtract multiples of 1.0 X=0.2 Y=-0.8 X=1.8 Y=-0.2 X=1.2 X=0.8 X=-0.8 X=-.0.2 H Li b a X=0.2 Y=0.2 X=1.8 Y=0.8 X=1.2 X=0.8 X=-0.8 X=-.0.2 X=0.2 Y=1.2 X=1.8 Y=1.8 X=1.2 X=0.8 X=-0.8 X=-.0.2
What would be the oxygen coordinates if we had drawn the unit cell with origin choice 2? Li 1 2 3 4 X=0.70 Y=0.20 1 2 3 4
Oxygen coordinates with origin choice 3? b H Li 1 2 3 4 X=0.30 Y=0.30 1 2 3 4
Oxygen coordinates with origin choice 4? b H Li 1 2 3 4 X=0.80 Y=0.30 1 2 3 4
Cheshire operators X , Y X+.5, Y X+.5, Y+.5 X , Y+.5 Choice 1 Choice 2
The 4 choices of origin are equally valid but once a choice is made, you must remain consistent. X1=0.20 Y1=0.20 X1=0.70 Y1=0.20 H Li X2=0.80 Y2=0.80 X2=0.30 Y2=0.80 Choice 3 Choice 4 H Li H Li X1=0.30 Y1=0.30 Li H Li Li X1=0.80 Y1=0.30 H H H H H Li Li Li X2=0.70 Y2=0.70 H Li X2=0.20 Y2=0.70
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. If a structure is solved independently by two crystallographers using different choices of origin, their coordinates will be related by a Cheshire opertor. Adding 1 to x, y, or z, or any combination of x, y, and z is valid. It is just a unit cell translation.
Interpreting difference Patterson Maps in Lab this week! Calculate an isomorphous difference Patterson Map (native-heavy atom). We collected 16 derivative data sets in lab (different heavy atoms at different concentrations) Mersalyl, PCMBS GdCl3 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 Fourier synthesis 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)=S Ihkl cos2p(hu+kv+lw -0) hkl Patterson synthesis P(uvw)=S ?hkl cos2p(hu+kv+lw -?) hkl
Hence, Patterson density map= electron density map convoluted with its inverted image. Patterson synthesis P(uvw)=S Ihkl cos2p(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)
Significance? P(uvw)=r(uvw) r (-u-v-w) The Patterson map contains a peak for every interatomic vector in the unit cell. The peaks are located at the head of the interatomic vector when its tail is placed at the origin.
Electron Density vs. Patterson Density b a b Lay down n copies of the unit cell at the origin, where n=number of atoms in unit cell. H H H 1 2 3 H 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
Every Patterson peak corresponds to an inter-atomic vector b 3 sets of peaks: Length O-H Where? Length H-H Length zero How many peaks superimposed at origin? How many non-origin peaks? H 1 2 3 H Patterson Density Map single water molecule convoluted with its inverted image. Electron Density Map single water molecule in the unit cell
Patterson maps are more complicated to interpret than electron density maps. Imagine the complexity of a Patterson map of a protein a b H 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
Patterson maps have an additional center of symmetry b plane group pm H plane group p2mm H Patterson Density Map single water molecule convoluted with its inverted image. Electron Density Map single water molecule in the unit cell
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.
Plane group p2 b H
Let’s consider only oxygen atoms b H 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? (0,0) a b 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?
Symmetry operators in plane group p2 (-0.2,-0.3) (0,0) 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. (-0.2, -0.3). But, are these really the coordinates of the second face in the unit cell? (0.2,0.3) SYMMETRY OPERATORS FOR 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)
Patterson coordinates (U,V) are simply symop1-symop2. 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. 2D CRYSTAL PATTERSON MAP (-0.2,-0.3) (0,0) x (0,0) u y v (0.2,0.3) Patterson coordinates (U,V) are simply symop1-symop2. Remember this bridge! symop1= X , Y -symop2= -(-X,-Y) u=2X v=2Y u=2(0.2) v=2(0.3) u=0.4, v=0.6 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 OPERATORS FOR PLANE GROUP P2 1) x,y 2) -x,-y
Patterson in plane group p2 (-0.4, -0.6) (-0.2,-0.3) (0,0) x u (0,0) y v (0.2,0.3) (0.6, 0.4) (0.4, 0.6) SYMMETRY OPERATORS FOR PLANE GROUP P2 1) x,y 2) -x,-y 2D CRYSTAL PATTERSON MAP
Patterson in plane group p2 (0,0) u v (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
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 between our friends the space group operators. (0,0) u v (0.6, 0.4) (0.4, 0.6) SYMMETRY OPERATORS FOR 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.
Now I’m stuck in Patterson space. How do I get back to x,y, coordinates? SYMMETRY OPERATORS FOR PLANE GROUP P2 1) x,y 2) -x,-y (0,0) u v 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. u=2x 0.4=2x 0.2=x v=2y 0.6=2y 0.3=y PATTERSON MAP
Hurray!!!! 2D CRYSTAL SYMMETRY OPERATORS FOR PLANE GROUP P2 1) x,y (0,0) u v x y -(-x –y) 2x 2y symop #1 symop #2 (0.2,0.3) set u=2x v=2y plug in Patterson values for u and v to get x and y. 2D CRYSTAL 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!!!!
Devil’s advocate: What if I chose u,v= (0. 6,0. 4) instead of (0. 4,0 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)? using Patterson (u,v) values 0.4, 0.6 to get x and y. (0,0) u v u=2x 0.4=2x 0.2=x v=2y 0.6=2y 0.3=y (0.6, 0.4) (0.4, 0.6) using Patterson (u,v) values 0.6, 0.4 to get x and y. u=2x 0.6=2x 0.3=x v=2y 0.4=2y 0.2=y PATTERSON MAP These two solutions do not give the same x,y? What is going on??????
Arbitrary choice of origin (-0.2,-0.3) (0,0) x y (-0.3,-0.2) (0.2,0.3) (0,0) u v (0.8,0.7) (0.3,0.2) (0.7,0.8) Original origin choice Coordinates x=0.2, y=0.3. 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 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.
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!
Self Vectors X, Y, Z -X, -Y, Z X, Y, Z 1/2-X,1/2+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
Isomorphous difference Patterson map (Pt derivative) W=0 V=1/2 U=1/2 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.
Harker section w=0 X, Y, Z -X, -Y, Z u=2x, v=2y, w=0 W=0 0.168=2x Does z=0? No! Solve for x, z using v=1/2 Harker sect.
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
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? 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 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
Advanced case,Proteinase K in space group P43212 Where are Harker sections?
-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
-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. +(0.72= x-y) 1.40=-2y -0.70=y Plug y into first equation and solve for x. 0.68=-x-(-0.70) 0.02=x
-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. +(0.72= x-y) 1.40=-2y -0.70=y Plug y into first equation and solve for x. 0.68=-x-(-0.70) 0.02=x
-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. +(0.72= x-y) 1.40=-2y -0.70=y Plug y into first equation and solve for x. 0.68=-x-(-0.70) 0.02=x
-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. +(0.72= x-y) 1.40=-2y -0.70=y Plug y into first equation and solve for x. 0.68=-x-(-0.70) 0.02=x
½-y ½+x ¾+z - ( -y -x ½-z) ½ ½+2x ¼+2z Symmetry operator 3 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
½-y ½+x ¾+z - ( -y -x ½-z) ½ ½+2x ¼+2z Symmetry operator 3 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
½-y ½+x ¾+z - ( -y -x ½-z) ½ ½+2x ¼+2z Symmetry operator 3 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
Xstep4=-0.02 ystep4= ?.?? zstep4=-0.005 From step 3 Xstep3= 0.02 ystep3=-0.70 zstep3=?.??? From step 4 Xstep4=-0.02 ystep4= ?.?? zstep4=-0.005 Clearly, Xstep3 does not equal Xstep4 . 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?
Xstep4=-0.02 ystep4= ?.?? zstep4=-0.005 From step 3 Xstep3= 0.02 ystep3=-0.70 zstep3=?.??? From step 4 Xstep4=-0.02 ystep4= ?.?? zstep4=-0.005 Clearly, Xstep3 does not equal Xstep4 . 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 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
Xstep4=-0.02 ystep4= ?.?? zstep4=-0.005 From step 3 Xstep3= 0.02 ystep3=-0.70 zstep3=?.??? From step 4 Xstep4=-0.02 ystep4= ?.?? zstep4=-0.005 Clearly, Xstep3 does not equal Xstep4 . 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 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
Xstep4=-0.02 ystep4= ?.?? zstep4=-0.005 From step 3 Xstep3= 0.02 ystep3=-0.70 zstep3=?.??? From step 4 Xstep4=-0.02 ystep4= ?.?? zstep4=-0.005 Clearly, Xstep3 does not equal Xstep4 . 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 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
Symmetry Operators are the Bridge between Atomic Coordinates and Patterson Peaks symop #1 symop #2 x , y -(-x, –y) 2x , 2y u=2x, v=2y (-0.2,-0.3) (0,0) x (0,0) u y v (0.2,0.3) (0.6, 0.4) (0.4, 0.6) SYMMETRY OPERATORS FOR PLANE GROUP P2 1) x,y 2) -x,-y PATTERSON MAP
Assignment Calculate an isomorphous difference Patterson map in lab. Solve the positions of the heavy atom (x,y,z) from the peaks in the map (u,v,w). follow the procedures in the handout write neatly check your answer Next week hand in your calculation. We will test the accuracy of your solution and use it to calculate phases and electron density.
Patterson space Crystal space U=0.5 P43212 Symmetry operator difference 3-6 Calculate Y and Z Calculate X and Y W=0.25 Cheshire operator applied to Y and Z if two values of Y do not match P43212 Symmetry operator difference 2-4 X,Y,Z referred to a common origin. Check answer for peak off Harker section. P43212 Symmetry operator difference 1-5 x y z -( y x -z) x-y -x+y 2z u,v,w 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.
m230d_2011_scaled.mtz Intensity measurements (IHKL) native-yen-b3.sca prok_native_mucheng.sca prok-gdcl3-matthew.sca prok-gdcl3-maya.sca prok_gdcl3_sean.sca prok_iodide_jungreem.sca prok_mersalyl_dan.sca prok_mersalyl_dan_xds.sca prok-pcmbs-1-5-stephen.sca prok-pcmbs-andrew.sca prok_pcmbs_cuiwen.sca prok_pcmbs_pmsf_jose.sca prok-pcmbs-pmsf-yazan.sca prok_pcmbs_riki.sca prok_pcmbs_sabai.sca prok_pcmbs_smriti.sca All data sets were appended into a spreadsheet. Each column contains IHKL of a different data set. Each row specifies a different HKL. -using the CCP4 program CAD. Intensity measurements were converted to structure factor amplitudes (|FHKL|) -using the CCP4 program TRUNCATE. All data sets were scaled to a reference native data set with the best statistics: prok-native-yen-b3 -using the CCP4 program SCALEIT.
Scale intensities by a constant (k) and resolution dependent exponential (B) prok-native-yen prok-gdcl3-matthew 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 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. 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
-B*sin2q/l2 e
Initial intensity profile low resolution high resolution.
Scaling constant K=0.883 intensity low resolution high resolution.
after K=1.1329 and B=-5.75 intensity low resolution high resolution.
OVERALL FILE STATISTICS for resolution range 0.000 - 0.351 29 BOTH 0.0 0.0 23592 12.15 0.00 0.00 56.46 1.70 D D_pcmbs-stephen 30 BOTH 0.0 0.0 23592 12.15 0.00 0.00 56.46 1.70 Q SIGD_pcmbs-stephen 31 NONE 1.1 1428.1 230 99.14 292.43 292.43 56.46 1.70 F FP_pcmbs-andrew 32 NONE 0.7 65.0 230 99.14 6.64 6.64 56.46 1.70 Q SIGFP_pcmbs-andrew 33 NONE -94.7 91.4 369 98.63 -0.57 10.26 56.46 1.70 D D_pcmbs-andrew 34 NONE 0.0 73.5 369 98.63 10.57 10.57 56.46 1.70 Q SIGD_pcmbs-andrew 35 NONE 1.0 1921.9 309 98.85 287.65 287.65 56.46 1.70 F FP_pcmbs_cuiwen 36 NONE 0.7 61.3 309 98.85 7.28 7.28 56.46 1.70 Q SIGFP_pcmbs_cuiwen 37 NONE -91.0 89.9 358 98.67 0.99 10.27 56.46 1.70 D D_pcmbs_cuiwen 38 NONE 0.0 93.4 358 98.67 11.59 11.59 56.46 1.70 Q SIGD_pcmbs_cuiwen 39 NONE 7.2 1534.3 89 99.67 285.42 285.42 40.68 1.70 F FP_pcmbs_pmsf_jose 40 NONE 1.2 139.5 89 99.67 10.39 10.39 40.68 1.70 Q SIGFP_pcmbs_pmsf_jose 41 NONE -153.0 111.3 174 99.35 -0.68 13.99 40.68 1.70 D D_pcmbs_pmsf_jose 42 NONE 0.0 146.2 174 99.35 16.49 16.49 40.68 1.70 Q SIGD_pcmbs_pmsf_jose 43 NONE 0.8 1290.2 424 98.42 287.16 287.16 48.02 1.70 F FP_pcmbs-pmsf-yazan 44 NONE 0.6 47.9 424 98.42 6.93 6.93 48.02 1.70 Q SIGFP_pcmbs-pmsf-yazan 45 NONE -115.5 95.1 765 97.15 -1.24 9.48 48.02 1.70 D D_pcmbs-pmsf-yazan 46 NONE 0.0 55.5 765 97.15 10.98 10.98 48.02 1.70 Q SIGD_pcmbs-pmsf-yazan 47 NONE 6.1 1563.7 10388 61.32 332.65 332.65 43.41 2.00 F FP_pcmbs_riki 48 NONE 0.8 86.0 10388 61.32 7.75 7.75 43.41 2.00 Q SIGFP_pcmbs_riki 49 NONE -128.4 124.3 10427 61.17 0.24 8.85 43.41 2.00 D D_pcmbs_riki 50 NONE 0.0 88.8 10427 61.17 11.72 11.72 43.41 2.00 Q SIGD_pcmbs_riki 51 NONE 9.0 1270.0 543 97.98 287.74 287.74 48.02 1.70 F FP_pcmbs_sabai 52 NONE 0.6 52.9 543 97.98 6.08 6.08 48.02 1.70 Q SIGFP_pcmbs_sabai 53 NONE -83.4 102.5 808 96.99 0.43 7.94 48.02 1.70 D D_pcmbs_sabai 54 NONE 0.0 61.2 808 96.99 9.45 9.45 48.02 1.70 Q SIGD_pcmbs_sabai 55 NONE 8.5 1452.5 1240 95.38 288.38 288.38 48.02 1.70 F FP_pcmbs_smriti 56 NONE 0.6 81.1 1240 95.38 6.07 6.07 48.02 1.70 Q SIGFP_pcmbs_smriti 57 NONE -89.0 102.4 1732 93.55 -0.06 8.88 48.02 1.70 D D_pcmbs_smriti 58 NONE 0.0 94.3 1732 93.55 9.11 9.11 48.02 1.70 Q SIGD_pcmbs_smriti No. of reflections used in FILE STATISTICS 26855 LIST OF REFLECTIONS =================== 0 0 12 17.00 186.54 5.69 278.24 11.23 ? ? 116.84 2.90 0.00 0.00 163.16 3.25 0.00 0.00 118.72 4.84 0.00 0.00 84.13 0.99 0.00 0.00 171.07 5.60 0.00 0.00 259.16 11.71 0.00 0.00 245.47 4.71 0.00 0.00 159.70 8.95 0.00 0.00 264.64 7.11 0.00 0.00 142.73 5.05 0.00 0.00 ? ? ? ? 194.52 5.51 0.00 0.00 OVERALL FILE STATISTICS for resolution range 0.000 - 0.351 ======================= Col Sort Min Max Num % Mean Mean Resolution Type Column num order Missing complete abs. Low High label 1 ASC 0 39 0 100.00 20.9 20.9 56.46 1.70 H H 2 NONE 0 28 0 100.00 8.6 8.6 56.46 1.70 H K 3 NONE 0 59 0 100.00 22.3 22.3 56.46 1.70 H L 4 NONE 0.0 19.0 0 100.00 9.47 9.47 56.46 1.70 I FreeR_flag 5 NONE 0.6 1060.8 885 96.70 291.57 291.57 56.46 1.70 F FP_native-yen 6 NONE 0.5 37.5 885 96.70 3.36 3.36 56.46 1.70 Q SIGFP_native-yen 7 NONE 1.0 941.8 5734 78.65 286.40 286.40 56.46 1.79 F FP_native-mucheng 8 NONE 0.6 87.0 5734 78.65 17.38 17.38 56.46 1.79 Q SIGFP_native-mucheng 9 NONE 9.8 1454.3 4366 83.74 305.42 305.42 56.46 1.80 F FP_gdcl3-matthew 10 NONE 1.5 105.6 4366 83.74 15.07 15.07 56.46 1.80 Q SIGFP_gdcl3-matthew 11 NONE 0.9 1500.2 62 99.77 292.26 292.26 56.46 1.70 F FP_gdcl3-maya 12 NONE 0.6 42.9 62 99.77 4.95 4.95 56.46 1.70 Q SIGFP_gdcl3-maya 13 NONE -95.4 97.7 173 99.36 0.12 16.32 56.46 1.70 D D_gdcl3-maya 14 NONE 0.0 52.7 173 99.36 7.98 7.98 56.46 1.70 Q SIGD_gdcl3-maya 15 NONE 5.1 1352.8 986 96.33 293.14 293.14 48.02 1.70 F FP_gdcl3_sean 16 NONE 0.2 42.2 986 96.33 3.55 3.55 48.02 1.70 Q SIGFP_gdcl3_sean 17 NONE -40.7 51.1 1341 95.01 -0.06 4.22 48.02 1.70 D D_gdcl3_sean 18 NONE 0.0 49.3 1341 95.01 5.44 5.44 48.02 1.70 Q SIGD_gdcl3_sean 19 NONE 12.8 1623.3 491 98.17 291.95 291.95 43.41 1.70 F FP_iodide_jungreem 20 NONE 0.9 126.6 491 98.17 8.78 8.78 43.41 1.70 Q SIGFP_iodide_jungreem 21 NONE -152.8 101.7 533 98.02 0.79 8.40 43.41 1.70 D D_iodide_jungreem 22 NONE 0.0 135.1 533 98.02 13.98 13.98 43.41 1.70 Q SIGD_iodide_jungreem 23 NONE 4.5 1958.6 1325 95.07 308.60 308.60 19.78 1.72 F FP_mersalyl_dan 24 NONE 0.3 19.5 1325 95.07 1.60 1.60 19.78 1.72 Q SIGFP_mersalyl_dan 25 NONE -26.0 27.7 1360 94.94 0.01 3.03 19.78 1.72 D D_mersalyl_dan 26 NONE 0.0 29.1 1360 94.94 2.65 2.65 19.78 1.72 Q SIGD_mersalyl_dan 27 NONE 12.2 1488.7 2170 91.92 290.41 290.41 56.46 1.70 F FP_pcmbs-stephen 28 NONE 1.0 119.8 2170 91.92 14.94 14.94 56.46 1.70 Q SIGFP_pcmbs-stephen