Symmetry operator 2 Symmetry operator 4 x y

Symmetry operator 3 -Symmetry operator 6 ½-y ½+x ¾+z - ( -y -x ½-z)

From step 3 Xstep 3= 0. 02 ystep 3=-0. 70 zstep 3=? . ?

Use x, y, z to predict the position of a non-Harker Patterson peak •

x y z symmetry operator 1 -symmetry operator 5 -( y x -z) u

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

Symmetry operator 2 -Symmetry operator 4 -x -y - ( ½+y ½-x -½-x-y -½+x-y ½+z ¼+z) ¼ Plug in u. u=-½-x-y 0. 18=-½-x-y 0. 68=-x-y Plug in v. v=-½+x-y 0. 22=-½+x-y 0. 72=x-y Add two equations and solve for y. 0. 68=-x-y +(0. 72= x-y) 1. 40=-2 y -0. 70=y Plug y into first equation and solve for x. 0. 68=-x-y 0. 68=-x-(-0. 70) 0. 02=x

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

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

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

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

Use x, y, z to predict the position of a non-Harker Patterson peak • x, y, z vs. –x, y, z ambiguity remains In other words x=-0. 02, y=0. 70, z=-0. 005 or x=+0. 02, y=0. 70, z=-0. 005 could be correct. • • Both satisfy the difference vector equations for Harker sections Only one is correct. 50/50 chance Predict the position of a non Harker peak. Use symop 1 -symop 5 Plug in x, y, z solve for u, v, w. Plug in –x, y, z solve for u, v, w I have a non-Harker peak at u=0. 28 v=0. 28, w=0. 0 The position of the non-Harker peak will be predicted by the correct heavy atom coordinate.

x y z symmetry operator 1 -symmetry operator 5 -( y x -z) u v w x-y -x+y 2 z First, plug in x=-0. 02, y=0. 70, z=-0. 005 u=x-y = -0. 02 -0. 70 =-0. 72 v=-x+y= +0. 02+0. 70= 0. 72 w=2 z=2*(-0. 005)=-0. 01 The numerical value of these coordinates falls outside the section we have drawn. Lets transform this uvw by Patterson symmetry v, -u, -w. -0. 72, -0. 01 becomes -0. 72, 0. 01 then add 1 to u and v 0. 28, 0. 01 This corresponds to the peak shown u=0. 28, v=0. 28, w=0. 01 Thus, x=-0. 02, y=0. 70, z=-0. 005 is correct. Hurray! We are finished! In the case that the above test failed, we would change the sign of x.