1. 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 between their assigned pair of derivative and a native.
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.

2. Which of the following is true of Patterson maps?
Every peak in the map corresponds to a vector between atoms in the unit cell
It is always centrosymmetric
It has the same unit cell parameters as the crystal
It can be computed without knowing phases.
All of the above

3. Which of the following are true of Patterson maps?
Every peak in the map corresponds to a vector between atoms in the unit cell
It is always centrosymmetric
It has the same unit cell parameters as the crystal
It can be computed without knowing phases.
All of the above

4. Lesson: native Patterson maps offer some structural information but are too complex to offer an interpretation of the atomic coordinates.

5. If there are 110 atoms in the unit cell, how many peaks in Patterson map?
b)110
c) One peak for each molecule
d)None of the above a c

6. If there are 110 atoms in the unit cell, how many peaks in Patterson map?
b)110
c) One peak for each molecule
d)None of the above a c

7. Lesson: there are n2 peaks in a Patterson map
Lesson: there are n2 peaks in a Patterson map. n = number of atoms in unit cell n peaks are at the origin n2-n peaks off origin

8. a)Vectors between protein atoms
If I calculate an isomorphous difference Patterson map with coefficients (|FPH| - |FP|)2 What will be the only features in the map? a
a)Vectors between protein atoms
b)Vectors between heavy atoms
c) No features expected
d) Either B or C c a c

9. If I calculate an isomorphous difference Patterson map with coefficients |FPH| - |FP| What will be the only features in the map? a
a)Vectors between protein atoms
b)Vectors between heavy atoms
c) No features expected
d) Either B or C c a c

10. Lesson: difference Patterson maps have fewer features than ordinary Patterson maps and so can be more easily interpreted.

11. Which of the following would correspond to a difference Patterson peak?
b)-0.6,0.5,0.4
c)a and b
d)No such peak exists a
(0.8, 0.0, 0.3) c
(0.2, 0.5, 0.7)

12. 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
(0.6,-0.5,-0.4) a
u v w
(0.8, 0.0, 0.3) c
(0.2, 0.5, 0.7)
-(0.8, 0.0, 0.3)
(-0.6, 0.5, 0.4)
(0.2, 0.5, 0.7)
u v w

13. Lesson: difference Patterson peaks correspond to vectors between atoms in the unit cell.

14. Which of the following would correspond to a difference Patterson peak?
a)-2x,1/2,-2z
b)2x,-1/2,2z
c)a and b
d)No such peak exists a
(x, y, z) c
(-x,y+1/2,-z)

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

16. 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?

17. 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

18. 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

19. Lesson: symmetry operators are the bridge between atomic coordinates in the crystal and Patterson peaks
-2x, -2y
2x, 2y
-2x, -2y a a
(-x, y+1/2, -z)
-( x, y, z)
(x, y, z)
(-2x, 1/2, -2z)
(u,v,w) c c
u v w
2x, 2y
-2x, -2y
2x, 2y
(-x,y+1/2,-z)
Difference Patterson Map
Crystal

20. c)Not enough information given
How many heavy atoms are expected in the unit cell with space group P21212 if there is only one heavy atom in the asymmetric unit, ?
a)4
b)42
c)Not enough information given
x, y, z
(2) -x, -y, z
(3) -x + 1/2, y + 1/2, -z
(4) x + 1/2, -y + 1/2, -z

21. c)Not enough information given
How many heavy atoms are expected in the unit cell with space group P21212 if there is only one heavy atom in the asymmetric unit, ?
a)4
b)42
c)Not enough information given
x, y, z
(2) -x, -y, z
(3) -x + 1/2, y + 1/2, -z
(4) x + 1/2, -y + 1/2, -z

22. 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”.

23. In what planes do we expect difference Patterson peaks in P21212?
a)w=0
b)w=1/2
c) v=1/2
d) u=1/2
x, y, z
(2) -x, -y, z
(3) -x + 1/2, y + 1/2, -z
(4) x + 1/2, -y + 1/2, -z

24. In what planes do we expect difference Patterson peaks in P21212?
X, Y, Z
-X, -Y, Z
u=2x, v=2y, w=0
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
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

25. W=0
V=1/2
U=1/2
Lesson: Vectors between symmetry-related atoms often lie on planes. We call these planes Harker sections

26. Which of these difference vectors is likely to correspond to the difference Patterson peak shown here?
W=0
X, Y, Z
-X, -Y, Z
u=2x, v=2y, w=0 a) b) c)
d) They are all equally likely.
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

27. Which of these difference vectors is likely to correspond to the difference Patterson peak shown here?
W=0
X, Y, Z
-X, -Y, Z
u=2x, v=2y, w=0 a) b) c)
d) They are all equally likely.
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

28. Harker section w=0 X, Y, Z -X, -Y, Z u=2x, v=2y, w=0 W=0 0.168=2x

29. What is the value of z? a)Zero b) x/y
X, Y, Z
-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.

30. What is the value of z? a)Zero b) x/y
X, Y, Z
-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?

31. 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

32. What are the coordinates x,y,z for the heavy atom?
None of the above

33. What are the coordinates x,y,z for the heavy atom?
None of the above

34. 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.

35. 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 xorig1 and yorig1
Xorig1=0.084
½-xorig1=
½-xorig1=0.416 =xorig2
yorig1=0.133
-yorig1=-0.133=yorig2
Hence,
Xorig2=0.416, yorig2=-0.133, zorig2=0.075

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

39. -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

40. -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

41. -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

42. -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

43. -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

45. 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

46. 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

47. 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

48. Xstep4=-0.02 ystep4= ?.?? zstep4=-0.005
From step 3
Xstep3= ystep3= zstep3=?.???
From step 4
Xstep4= ystep4= ?.?? zstep4=-0.005
Clearly, Xstep3 does not equal Xstep4 .
Use a Cheshire symmetry operator that transforms xstep3= 0.02 into xstep4=
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?

49. Xstep4=-0.02 ystep4= ?.?? zstep4=-0.005
From step 3
Xstep3= ystep3= zstep3=?.???
From step 4
Xstep4= ystep4= ?.?? zstep4=-0.005
Clearly, Xstep3 does not equal Xstep4 .
Use a Cheshire symmetry operator that transforms xstep3= 0.02 into xstep4=
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

50. Xstep4=-0.02 ystep4= ?.?? zstep4=-0.005
From step 3
Xstep3= ystep3= zstep3=?.???
From step 4
Xstep4= ystep4= ?.?? zstep4=-0.005
Clearly, Xstep3 does not equal Xstep4 .
Use a Cheshire symmetry operator that transforms xstep3= 0.02 into xstep4=
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

51. Xstep4=-0.02 ystep4= ?.?? zstep4=-0.005
From step 3
Xstep3= ystep3= zstep3=?.???
From step 4
Xstep4= ystep4= ?.?? zstep4=-0.005
Clearly, Xstep3 does not equal Xstep4 .
Use a Cheshire symmetry operator that transforms xstep3= 0.02 into xstep4=
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

52. 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= or
x=+0.02, y=0.70, z= 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.

53. 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 = =-0.72
v=-x+y= = 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= 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

54. 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.

55. 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.

56. m230d_2015_scaled2.mtz All data sets were entered in a spreadsheet.
SIGFP_native-jeannette
SIGFP_native-wenyang
SIGFP_native-joshua
SIGFP_eucl3-beccah
SIGFP_eucl3-jessica
FP_native-jeannette
FP_native-wenyang
SIGD_eucl3-beccah
SIGD_eucl3-jessica
SIGFP_native-john
SIGFP_native-mimi
FP_native-joshua
FP_eucl3-beccah
FP_eucl3-jessica
Up to 72 columns
SIGFP_native-aj
FP_native-john
FP_native-mimi
D_eucl3-beccah
D_eucl3-jessica
FP_native-aj
FreeR_flag H K L
m230d_2015_scaled2.mtz
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.
Etc. for thousands of reflections

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

58. Scale intensities by a constant (k) and resolution dependent exponential (B)
prok-native-yen
prok-gdcl3-matthew
H K L intensity sigma
H K L intensity sigma
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.
/ = 2.65
/ = 1.65
/ = 3.07
/ = 2.45
/ = 0.70
/ = 1.83
/ = 1.24

59. -B*sin2q/l2 e

60. Non Harker peaks
u v w =
-u -v w =
u v w =
-u -v w =
-u v w =
u v w =
-u v w =
u v w =
-v u w =
v u w =
-v u w =
v u w =
v u w =
-v u w =
v u w =
-v u w =

61. 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 ) -x,-y
PATTERSON MAP

62. Patterson map to coordinates
(0,0) a
(0,0) a
(0.4, 0.4) b b
(0.6, 0.6)
(2)= -x, -y
(1)=-( x, y)
u=-2x v=-2y
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).

63. Interpreting difference Patterson Maps in Lab today!
Calculate an isomorphous difference Patterson Map (native-heavy atom). We collected 6 derivative data sets in lab
3 PCMBS
3 EuCl3
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.

64. 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

65. 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(xyz) r (-x-y-z)

66. Significance? P(uvw)=r(xyz) r (-x-y-z)
The Patterson map contains a peak for every interatomic vector in the unit cell.
If n atoms in unit cell, then n2 peaks in Patterson.
The peaks are located at the head of the interatomic vector when its tail is placed at the origin. So, n of these peaks on the origin.
If the coefficients are |FPH - FP|2, the interatomic vectors are between only heavy atoms in unit cell. Vectors involving protein atoms cancel out. Much simplified.

67. 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.

68. 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.

69. Coordinates to Patterson map
(-x,-y)
(0,0) a
(0,0) a
(-x+1,-y+1) b b
(x,y)
Plane group p2
Symmetry operators are
x,y
-x, -y
How many atoms in unit cell?
In asymmetric unit?
How many peaks will be in the Patterson map? (n2)
How many peaks at the origin? (n)
How many non-origin peaks? (n2-n)

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

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

72. Coordinates to Patterson map
(-0.6,-1.6)
-2x, -2y
2x, 2y
-2x, -2y
(0,0) a
(0,0) a
(-x,-y)
2x, 2y
-2x, -2y b b
-2x, -2y
2x, 2y
-2x, -2y
2x, 2y
(x,y)
(-0.6, 1.6)
Plane group p2
Symmetry operators are
1) x, y
2)-x,-y
(2)= -x, -y
(1)=-( x, y)
u=-2x v=-2y
u=-2*0.3 v=-2*0.8
u= v=-1.6
(1)= x, y
(2)=-(-x,-y)
u=2x v=2y u=2*0.3 v=2*0.8
u=0.6 v=1.6
If x=0.3, y=0.8
What will be the coordinates of the Patterson peaks?

73. Coordinates to Patterson map
(-0.6,-1.6)
-2x, -2y
2x, 2y
-2x, -2y
(0,0) a
(0,0) a
(-x,-y)
2x, 2y
(0.4, 0.4)
-2x, -2y b b
-2x, -2y
2x, 2y
-2x, -2y
2x, 2y
(0.6, 0.6)
(x,y)
-0.6, 1.6)
Plane group p2
Symmetry operators are
1) x, y
2)-x,-y
(2)= -x, -y
(1)=-( x, y)
u=-2x v=-2y
u=-2*0.3 v=-2*0.8
u= v=-1.6
u= v= 0.4
(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
If x=0.3, y=0.8
What will be the coordinates of the Patterson peaks?

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

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

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

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

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

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

80. Cheshire operators X , Y X+.5, Y X+.5, Y+.5 X , Y+.5 Choice 2 Choice 2

81. Recap
Where n is the number of atoms in the unit cell, there will be n2 Patterson peaks total, n peaks at the origin, n2-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.

82. 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.

83. 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!

84. 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

86. 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

87. 1 2 3 4
Use this edge to measure x 1 2 3 4
Use this edge to measure y 1 2 3 4
Use this edge to measure x 1 2 3 4
Use this edge to measure y 1 2 3 4
Use this edge to measure x 1 2 3 4
Use this edge to measure y 1 2 3 4
Use this edge to measure x 1 2 3 4
Use this edge to measure y 1 2 3 4
Use this edge to measure x 1 2 3 4
Use this edge to measure y 1 2 3 4
Use this edge to measure x 1 2 3 4
Use this edge to measure y

88. Crystal Structure of NNQQNY Peptide from Sup35 Prion
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. a
(0,0)
Draw a unit cell b
having this size
and shape, with
corners on 2-folds

89. two-fold axes

90. 4 distinct sets of two-fold axes

91. Unit Cell
Choice 1
Choice 2
Choice 3
Choice 4

92. 4 Choices of origin Choice 1 Choice 2 a a b b Choice 3 Choice 4 b a b

93. Which plane group? b a

94. What are the coordinates of the red zinc using origin choice 1?
X=0.80
Y=0.30 b a 1 2 3 4 1 2 3 4

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

96. What are the coordinates of the red zinc with origin choice 3?
X=0.30
Y=0.80 b a
X = +0.3
Y = +0.8 1 2 3 4 1 2 3 4

97. What are the coordinates of the red zinc with origin choice 4?
X=0.80
Y=0.80 b a
X = +0.8
Y = +0.8 1 2 3 4 1 2 3 4

98. Cheshire operators X , Y X+.5, Y X+.5, Y+.5 X , Y+.5 Choice 1 Choice 2

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

100. What are the coordinates of the purple zinc using origin choice 1?
X=0.80
Y=0.30
Symmetry operators in plane group p2
X, Y
-X,-Y b a
X2=-0.80
Y2=-0.30 1 2 3 4
X=0.20
Y=0.70 1 2 3 4

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

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