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

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.

Which of the following is true of Patterson maps?

a)Every peak in the map corresponds to a vector between atoms in the unit cell

b)It is always centrosymmetricc)It has the same unit cell parameters as the

crystald)It can be computed without knowing

phases.e)All of the above

Which of the following are true of Patterson maps?

a)Every peak in the map corresponds to a vector between atoms in the unit cell

b)It is always centrosymmetricc)It has the same unit cell parameters as the

crystald)It can be computed without knowing

phases.e)All of the above

Lesson: native Patterson maps offer some structural information but are too complex to offer an

interpretation of the atomic coordinates.

If there are 110 atoms in the unit cell, how many peaks in Patterson map?

a)1102

b)110

c) One peak for each molecule

d)None of the above

c

a

If there are 110 atoms in the unit cell, how many peaks in Patterson map?

a)1102

b)110

c) One peak for each molecule

d)None of the above

c

a

Lesson: there are n2 peaks in a Patterson map.

n = number of atoms in unit cell

n peaks are at the originn2-n peaks off origin

If I calculate an isomorphous difference Patterson map with coefficients (|FPH| - |FP|)2 What will be

the only features in the map?

a)Vectors between protein atoms

b)Vectors between heavy atoms

c) No features expected

d) Either B or C

c

a

c

a

If I calculate an isomorphous difference Patterson map with coefficients |FPH| - |FP| What will be the only features in the map?

a)Vectors between protein atoms

b)Vectors between heavy atoms

c) No features expected

d) Either B or C

c

a

c

a

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

Which of the following would correspond to a difference Patterson peak?

c

a

(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


c

a

(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


(0.8, 0.0, 0.3)-(0.2, 0.5, 0.7)

(0.6,-0.5,-0.4)

u v w

(0.2, 0.5, 0.7)-(0.8, 0.0, 0.3)

(-0.6, 0.5, 0.4)

u v w

Lesson: difference Patterson peaks correspond to vectors

between atoms in the unit cell.


c

a

(x, y, z)

(-x,y+1/2,-z)

a)-2x,1/2,-2z

b)2x,-1/2,2z

c)a and b



c

a

(x, y, z)

(-x,y+1/2,-z)

a)-2x,1/2,-2z

b)2x,-1/2,2z

c)a and b


(-x, y+1/2, -z)-( x, y, z)

(-2x, 1/2, -2z)

u v w

( x, y , z)-(-x, y+1/2, -z)

( 2x,-1/2, 2z)

u v w

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?

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

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

Lesson: symmetry operators are the bridge between atomic

coordinates in the crystal and Patterson peaks

c

a

(x, y, z)

(-x,y+1/2,-z)

(-x, y+1/2, -z)-( x, y, z)

(-2x, 1/2, -2z)

u v w

-2x, -2y

-2x, -2y

2x, 2y

2x, 2y

-2x, -2y

2x, 2y

(u,v,w)c

a

Crystal Difference Patterson Map

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

(1)x, y, z(2) -x, -y, z(3) -x + 1/2, y + 1/2, -z(4) x + 1/2, -y + 1/2, -z

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

(1)x, y, z(2) -x, -y, z(3) -x + 1/2, y + 1/2, -z(4) x + 1/2, -y + 1/2, -z

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”.

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

(1)x, y, z(2) -x, -y, z(3) -x + 1/2, y + 1/2, -z(4) x + 1/2, -y + 1/2, -z

In what planes do we expect difference Patterson peaks in P21212?

(1)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, Z2.-X, -Y, Zu=2x, v=2y, w=0

1. X, Y, Z3. ½-X,½+Y,-Zu=2x-½,v=-½,w=2z

1. X, Y, Z4. ½+X,½-Y,-Zu=-½,v=2y-½,w=2z

Lesson: Vectors between

symmetry-related atoms often lie

on planes. We call these planes Harker

sections

W=0

V=1/2

U=1/2

Which of these difference vectors is likely to correspond to the difference

Patterson peak shown here?

W=0 1. X, Y, Z2. -X, -Y, Zu=2x, v=2y, w=0

1. X, Y, Z3. ½-X,½+Y,-Zu=2x-½,v=-½,w=2z

1. X, Y, Z4. ½+X,½-Y,-Zu=-½,v=2y-½,w=2z

a)

b)

c)

d) They are all equally likely.

Which of these difference vectors is likely to correspond to the difference

Patterson peak shown here?

W=0 1. X, Y, Z2. -X, -Y, Zu=2x, v=2y, w=0

1. X, Y, Z3. ½-X,½+Y,-Zu=2x-½,v=-½,w=2z

1. X, Y, Z4. ½+X,½-Y,-Zu=-½,v=2y-½,w=2z

a)

b)

c)

d) They are all equally likely.

Harker section w=0

W=0 1. X, Y, Z2.-X, -Y, Zu=2x, v=2y, w=0

0.168=2x0.084=x

0.266=2y0.133=y

Structure Lab

No, it just means that the two atoms related by this symmetry operator had the same z coordinate. You dont know what the coordinate is.

What is the value of z?

W=0 1. X, Y, Z2.-X, -Y, Zu=2x, v=2y, w=0

0.168=2x0.084=x

0.266=2y0.133=y

a)Zero b) x/y c) not specified by this Harker section.

Structure Lab


What is the value of z?

W=0 1. X, Y, Z2.-X, -Y, Zu=2x, v=2y, w=0

0.168=2x0.084=x

0.266=2y0.133=y

a)Zero b) x/y c) not specified by this Harker section.

How can we determine the z coordinate?

Structure Lab


Harker Section v=1/2

V=1/2

1. X, Y, Z3. ½-X,½+Y,-Zu=2x-½,v=-½,w=2z

0.333=2x-1/20.833=2x0.416=x

0.150=2z0.075=z

What are the coordinates x,y,z for the heavy atom?

V=1/2

0.333=2x-1/20.833=2x0.416=x

0.150=2z0.075=z

0.168=2x0.084=x

0.266=2y0.133=y

W=0

a)x=0.084, y=0.133, z=0.075b)x=0.416, y=0.133, z=0.075c)None of the above

What are the coordinates x,y,z for the heavy atom?

V=1/2

0.333=2x-1/20.833=2x0.416=x

0.150=2z0.075=z

0.168=2x0.084=x

0.266=2y0.133=y

W=0

a)x=0.084, y=0.133, z=0.075b)x=0.416, y=0.133, z=0.075c)None of the above

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

1. X, Y, Z2. -X, -Y, Z3. -X, Y, -Z4. X, -Y, -Z5. -X, -Y, -Z6. X, Y, -Z7. X, -Y, Z8. -X, Y, Z

9. 1/2+X, Y, Z10. 1/2-X, -Y, Z11. 1/2-X, Y, -Z12. 1/2+X, -Y, -Z13. 1/2-X, -Y, -Z14. 1/2+X, Y, -Z15. 1/2+X, -Y, Z16. 1/2-X, Y, Z

17. X,1/2+Y, Z18. -X,1/2-Y, Z19. -X,1/2+Y, -Z20. X,1/2-Y, -Z21. -X,1/2-Y, -Z22. X,1/2+Y, -Z23. X,1/2-Y, Z24. -X,1/2+Y, Z

25. X, Y,1/2+Z26. -X, -Y,1/2+Z27. -X, Y,1/2-Z28. X, -Y,1/2-Z29. -X, -Y,1/2-Z30. X, Y,1/2-Z31. X, -Y,1/2+Z32. -X, Y,1/2+Z

33. 1/2+X,1/2+Y, Z34. 1/2-x,1/2-Y, Z 35. 1/2-X,1/2+Y, -Z36. 1/2+X,1/2-Y, -Z37. 1/2-X,1/2-Y, -Z38. 1/2+X,1/2+Y, -Z39. 1/2+X,1/2-Y, Z40. 1/2-X,1/2+Y, Z

41. 1/2+X, Y,1/2+Z42. 1/2-X, -Y,1/2+Z43. 1/2-X, Y,1/2-Z44. 1/2+X, -Y,1/2-Z45. 1/2-X, -Y,1/2-Z46. 1/2+X, Y,1/2-Z47. 1/2+X, -Y,1/2+Z48. 1/2-X, Y,1/2+Z

49. X,1/2+Y,1/2+Z50. -X,1/2-Y,1/2+Z51. -X,1/2+Y,1/2-Z52. X,1/2-Y,1/2-Z53. -X,1/2-Y,1/2-Z54. X,1/2+Y,1/2-Z55. X,1/2-Y,1/2+Z56. -X,1/2+Y,1/2+Z

57. 1/2+X,1/2+Y,1/2+Z58. 1/2-X,1/2-Y,1/2+Z59. 1/2-X,1/2+Y,1/2-Z60. 1/2+X,1/2-Y,1/2-Z61. 1/2-X,1/2-Y,1/2-Z62. 1/2+X,1/2+Y,1/2-Z63. 1/2+X,1/2-Y,1/2+Z64. 1/2-X,1/2+Y,1/2+Z

From w=0 Harker section xorig1=0.084, yorig1=0.133From v=1/2 Harker section, xorig2=0.416, zorig2=0.075

Apply Cheshire symmetry operator #10To xorig1 and yorig1 Xorig1=0.084½-xorig1=0.5-0.084½-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?

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

Plug in u. u=-½-x-y0.18=-½-x-y0.68=-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


-x - y ½+z - ( ½+y ½-x ¼+z) -½-x-y -½+x-y ¼



0.68=-x-y+(0.72= x-y) 1.40=-2y -0.70=y


0.68=-x-y 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



0.68=-x-y+(0.72= x-y) 1.40=-2y -0.70=y


0.68=-x-y 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



0.68=-x-y+(0.72= x-y) 1.40=-2y -0.70=y


0.68=-x-y 0.68=-x-(-0.70) 0.02=x


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


½-y ½+x ¾+z - ( -y -x ½-z) ½ ½+2x ¼+2z



½-y ½+x ¾+z - ( -y -x ½-z) ½ ½+2x ¼+2z


From step 3Xstep3= 0.02 ystep3=-0.70 zstep3=?.???From step 4Xstep4=-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, zAnd apply it to all coordinates in step 3.xstep3-transformed = - (+0.02) = -0.02ystep3-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 evenreferred to different origins.

How can we transform x, y from step 3 so it describesthe same atom as x and z in step 4?





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





Or simply, x=-0.02, y=0.70, z=-0.005










Or simply, x=-0.02, y=0.70, z=-0.005






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

• x,y,z vs. –x,y,z ambiguity remainsIn 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 5u v w

First, plug in x=-0.02, y=0.70, z=-0.005

u=x-y = -0.02-0.70 =-0.72v=-x+y= +0.02+0.70= 0.72w=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=0.01Thus, 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

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.

Calculate Y and Z

Calculate X and Y

X,Y,Z referred to a common origin.

If prediction lies outside Patterson asymmetric unit (0→0.5, 0→0.5,0→ 0.5) usePatterson symmetry operators to find the symmetry equivalent peak in theasymmetric 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.

Crystal space

U=0.5

W=0.25 Cheshire operator applied to Y and Z if two values of Y do not match

P43212 Symmetry operator difference 1-5

x y z

-( y

x -z)

x-y -x+y 2z

u,v,w

Check answer for peak

off Harker section.



Patterson space

m230d_2015_scaled2.mtzH K L F

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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.

3 2 64 10.00 130.30 2.20 174.93 2.12 150.51 4.01 144.96 2.04 103.41 4.04 164.00 2.62 126.54 1.31 6.82 2.62 149.88 1.83 -0.66 3.67 3 2 65 9.00 175.48 1.66 191.37 2.00 197.89 3.23 177.46 1.75 159.61 2.69 202.44 2.22 180.36 1.13 5.15 2.26 170.56 1.66 -0.73 3.313 2 66 17.00 110.19 2.60 129.09 2.69 141.68 4.54 121.00 2.29 82.97 5.16 165.87 2.24 97.76 1.65 4.36 3.30 103.57 2.45 -5.45 4.89Etc. for thousands of reflections

m230d_2015_scaled2.mtz

Intensity measurements were converted to structure factor amplitudes (|FHKL|) -using the CCP4 program TRUNCATE.

All data sets were scaled to areference native data set with the best statistics: prok-native-jeannette-using the CCP4 program SCALEIT.

Scale intensities by a constant (k) and resolution dependent exponential (B)

H K L intensity sigma1 0 10 106894.0 1698.01 0 11 41331.5 702.31 0 12 76203.2 1339.01 0 13 28113.5 513.61 0 14 6418.2 238.71 0 15 45946.4 882.7 1 0 16 26543.8 555.6

prok-native-yen

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

prok-gdcl3-matthew

H K L intensity sigma1 0 10 40258.7 1222.91 0 11 25033.2 799.81 0 12 24803.6 771.51 0 13 11486.3 423.91 0 14 9180.5 353.61 0 15 25038.8 783.01 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.

e-B*sin2q/l2

Non Harker peaks u v w = 0.4404 0.4404 0.1304

-u -v w = -0.4404 -0.4404 0.1304

u v -w = 0.4404 0.4404 -0.1304

-u -v -w = -0.4404 -0.4404 -0.1304

-u v w = -0.4404 0.4404 0.1304

u -v w = 0.4404 -0.4404 0.1304

-u v -w = -0.4404 0.4404 -0.1304

u -v -w = 0.4404 -0.4404 -0.1304

-v u w = -0.4404 0.4404 0.1304

v -u w = 0.4404 -0.4404 0.1304

-v u -w = -0.4404 0.4404 -0.1304

v -u -w = 0.4404 -0.4404 -0.1304

v u w = 0.4404 0.4404 0.1304

-v -u w = -0.4404 -0.4404 0.1304

v u -w = 0.4404 0.4404 -0.1304

-v -u -w = -0.4404 -0.4404 -0.1304

Symmetry Operators are the Bridge between Atomic Coordinates and Patterson Peaks

u(0,0)v

PATTERSON MAP

(0.4, 0.6)

x , y -(-x, –y) 2x , 2y u=2x, v=2y

symop #1symop #2

(0.6, 0.4)

(0,0) xy

(0.2,0.3)

(-0.2,-0.3)

SYMMETRY OPERATORSFOR PLANE GROUP P21) x,y 2) -x,-y

Patterson map to coordinates

a

b

(0,0) a

b

(0,0)

Plane group p2Symmetry operators are

1) x, y2)-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).

(2)= -x, -y(1)=-( x, y) ---------- u=-2x v=-2y

(0.4, 0.4)

(0.6, 0.6)

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.

Patterson synthesis

P(uvw)=S ?hkl cos2p(hu+kv+lw -?)

hkl

Patterson synthesis

P(uvw)=S Ihkl cos2p(hu+kv+lw -0)

hkl

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

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)

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.

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.

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 b in space group P21212

3. Advanced 3D example, Hg derivative of proteinase K in space group P43212.

Coordinates to Patterson map


x,y-x, -yHow many atoms in unit cell?In asymmetric unit?

a

b

(0,0) a

b

(0,0)

(x,y)

(-x+1,-y+1)

How many peaks will be in the Patterson map? (n2)How many peaks at the origin? (n)How many non-origin peaks? (n2-n)

(-x,-y)


a

b

(0,0) a

b

(0,0)

(x,y)

(-x,-y)

-2x, -2y

-2x, -2y

(1)= x, y(2)=-(-x,-y) ---------- u=2x v=2y

(2)= -x, -y(1)=-( x, y) ---------- u=-2x v=-2y


1) x, y2)-x,-y

If x=0.3, y=0.8What will be the coordinates of the Patterson peaks?



1) x, y2)-x,-y

a

b

(0,0) a

b

(0,0)

(x,y)

(-x,-y)

-2x, -2y

-2x, -2y

2x, 2y


(1)= x, y(2)=-(-x,-y) ---------- u=2x v=2y

(2)= -x, -y(1)=-( x, y) ---------- u=-2x v=-2y

a

b

(0,0) a

b

(0,0)

(x,y)

(-x,-y)

-2x, -2y

-2x, -2y

2x, 2y

-2x, -2y

2x, 2y

-2x, -2y

2x, 2y

-2x, -2y

2x, 2y


1) x, y2)-x,-y


(1)= x, y(2)=-(-x,-y) ---------- u=2x v=2y u=2*0.3 v=2*0.8u=0.6 v=1.6

(2)= -x, -y(1)=-( x, y) ---------- u=-2x v=-2yu=-2*0.3 v=-2*0.8u=-0.6 v=-1.6

Coordinates to Patterson map(-0.6,-1.6)

(-0.6, 1.6)

a

b

(0,0) a

b

(0,0)

(x,y)

(-x,-y)

-2x, -2y

-2x, -2y

2x, 2y

-2x, -2y

2x, 2y

-2x, -2y

2x, 2y

-2x, -2y

2x, 2y


1) x, y2)-x,-y


(1)= x, y(2)=-(-x,-y) ---------- u=2x v=2y u=2*0.3 v=2*0.8u=0.6 v=1.6u=0.6 v=0.6

(2)= -x, -y(1)=-( x, y) ---------- u=-2x v=-2yu=-2*0.3 v=-2*0.8u=-0.6 v=-1.6u= 0.4 v= 0.4

(0.4, 0.4)

(0.6, 0.6)

(-0.6,-1.6)

-0.6, 1.6)



a

b

(0,0) a

b

(0,0)


1) x, y2)-x,-y

What are coordinates for heavy atom?

(2)= -x, -y(1)=-( x, y) ---------- u=-2x v=-2y

(0.4, 0.4)

(0.6, 0.6)


a

b

(0,0) a

b

(0,0)

(2)= -x, -y(1)=-( x, y) ---------- u=-2x v=-2y 0.4=-2x-0.2=x 0.4=-2y -0.2=y

(0.4, 0.4)

0.8=x 0.8=y

x=0.3, y=0.8 not the same sites as was used to generate Patterson map


a

b

(0,0) a

b

(0,0)


1) x, y2)-x,-y

(2)= -x, -y(1)=-( x, y) ---------- u=-2x v=-2y 0.6=-2x-0.3=x 0.6=-2y -0.3=y

(0.6, 0.6)

0.7=x 0.7=y

x=0.3, y=0.8 not the same sites as was used to generate Patterson map

NNQQNY structure

Example from NNQQNY (PDB ID code 1yjo)

Zn ion bind between N and C-termini

NNQQNY structure

Example from NNQQNY (PDB ID code 1yjo)

Zn ion bind between N and C-termini

b

a

It’s fine! (x=0.2, y=0.2) corresponds to origin choice 4.

Choice 1 Choice 2

Choice 3 Choice 4

b

a

b

a

b

a

X=0.80 Y=0.30 X=0.30 Y=0.30

X=0.30 Y=0.80 X=0.80 Y=0.80

X=0.20 Y=0.70 X=0.70 Y=0.70

X=0.70 Y=0.20X=0.20 Y=0.20

Cheshire operators• X , Y• X+.5, Y• X+.5, Y+.5• X , Y+.5

X=0.70 Y=0.70

X=0.30 Y=0.30

X=0.20 Y=0.20

X=0.80 Y=0.80

Choice 2

Choice 4

Choice 4

Choice 2

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.

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.

Michael Sawaya

You might wonder why I am emphasizing the point that multiple values of x,y,z are possible, depending on the choice of origin.We learned that these can all be correct, so why be concerned which choice of origin I use?The choice of origin becomes very important when you have to combine information from two or more Patterson peaks to solve for a complete set of x, y, and z.In the homework problem, a complete x,y, and z set could be determined from u,v,w coordinates of a single Patterson peak. Remember, you got x and y from Patterson peak u,v. The value of z was a free choice.But, when you solve for a heavy atom position in space group P43212, you will need to combine information from two Patterson peaks to get all three coordinates, x,y, and z. The choice of z is not arbitrary. You might solve the x,y coordinate from one Patterson peak. But to get z, you need to combine this info with a y,z coordinate you calculate from a second Patterson peak. These two partial sets of coordinates might refer to different origins. You must ensure that the partial solution from Patterson peak 1 refers to the same origin as Patterson peak 2.

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.1. X, Y, Z2. -X, -Y, Z3. 1/2-X,1/2+Y,-Z4. 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, Z2. -X, -Y, Z3. 1/2-X,1/2+Y,-Z4. 1/2+X,1/2-Y,-Z

1. X, Y, Z2.-X, -Y, Zu=2x, v=2y, w=0

1. X, Y, Z3. ½-X,½+Y,-Zu=2x-½,v=-½,w=2z

1. X, Y, Z4. ½+X,½-Y,-Zu=-½,v=2y-½,w=2z

Harker sections, w=0, v=1/2, u=1/2

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

1 1 1 253.87 0 0 0 0.0000 0.0000 0.0000 0.00 0.00 0.00

2 79 55 14.14 65 52 96 0.4959 0.3967 0.5000 33.64 26.91 50.92

3 67 45 13.14 6 6 71 0.0469 0.0469 0.3708 3.18 3.18 37.76

4 68 46 12.72 66 0 73 0.5000 0.0000 0.3790 33.92 0.00 38.59

5 59 40 12.71 60 8 48 0.4525 0.0580 0.2495 30.69 3.94 25.41

6 60 40 12.71 8 60 48 0.0580 0.4525 0.2495 3.94 30.69 25.41

7 43 30 12.08 66 14 23 0.5000 0.1051 0.1214 33.92 7.13 12.36

8 45 31 11.29 58 58 25 0.4421 0.4421 0.1288 29.99 29.99 13.12

9 29 10 6.67 0 46 0 0.0000 0.3478 0.0000 0.00 23.59 0.00

1 2 3 4 1234

Use this edge to measure x

Use this edge to measure y

1 2 3 4 1234



1 2 3 4 1234


Use this edge to measure y1 2 3 4 1234



1 2 3 4 1234



1 2 3 4 1234



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.

b

a

Crystal Structure of NNQQNY Peptide from Sup35 Prion

having this size

and shape, with

Draw a unit cell

corners on 2-folds

(0,0)

two-fold axes

4 distinct sets of two-fold axes

Unit Cell

Choice 1

Choice 2

Choice 3

Choice 4

4 Choices of originChoice 1 Choice 2

Choice 3 Choice 4

b

a

b

a

b

a

b

a

Which plane group?

b

a

X=0.80

What are the coordinates of the red zinc using origin choice 1?

Y=0.30

1 2 3 41

2

b

a

What are the coordinates of the red zinc with origin choice 2?

a

X=0.30Y=0.30

b

a

1 2 3 41

23

X = +0.3Y = +0.3


a

X=0.30Y=0.80

b

a

1 2 3 4

1

23

X = +0.3Y = +0.8


a

X=0.80Y=0.80

b

a

1 2 3 4

1

23

X = +0.8Y = +0.8

Cheshire operators• X , Y• X+.5, Y• X+.5, Y+.5• X , Y+.5

X=0.80 Y=0.30

X=0.30 Y=0.30

X=0.30 Y=0.80

X=0.80 Y=0.80

Choice 1

Choice 2

Choice 3

Choice 4

b

a

What are the coordinates of the 2nd Zn ion in the unit cell?

Choice 1 Choice 2

Choice 3 Choice 4

b

a

b

a

b

a

X=0.80 Y=0.30 X=0.30 Y=0.30

X=0.30 Y=0.80 X=0.80 Y=0.80

X=0.20

What are the coordinates of the purple zinc using origin choice 1?

Y=0.70

1 2 3 4

b

a

1

2

X2=-0.80Y2=-0.30

Symmetry operators in plane

group p2

X, Y-X,-Y

X=0.80Y=0.30

X = +0.8Y = +0.3

Always allowed to add or subtract multiples of 1.0

X = +0.2Y = +0.7

X = -0.8Y = -0.3

b

a

b

a

The 4 choices of origin are equally valid but once a choice is made, you must remain consistent.

Choice 1 Choice 2

Choice 3 Choice 4

b

a

b

a

b

a

X=0.80 Y=0.30 X=0.30 Y=0.30

X=0.30 Y=0.80 X=0.80 Y=0.80

X=0.20 Y=0.70 X=0.70 Y=0.70

X=0.70 Y=0.20X=0.20 Y=0.20

Documents

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