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)

– 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.

Patterson synthesis

P(uvw)=?hkl cos2(hu+kv+lw -)

hkl

Patterson synthesis

P(uvw)=Ihkl cos2(hu+kv+lw -)

hkl

Patterson Review

A Patterson synthesis is like a Fourier synthesis except for what two variables?

Fourier synthesis

(xyz)=|Fhkl| cos2(hx+ky+lz -hkl)

hkl

Hence, Patterson density map= electron density map convoluted

with its inverted image.

Patterson synthesis

P(uvw)=Ihkl cos2(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)=(uvw) (-u-v-w)

a

b

Electron Density vs. Patterson Density

H H

H H

H HH H

Electron Density Mapsingle water molecule in the unit cell

Patterson Density Mapsingle water molecule convoluted with its inverted image.

a

bLay down n copies of the unit cell at the origin, where n=number of atoms in unit cell.

1

23

For copy n, atom n is placed at the origin.A Patterson peak appears under each atom.

Every Patterson peak corresponds to an inter-atomic vector

H H



a

b

3 sets of peaks:

Length O-HWhere?

Length H-HWhere?

Length zero Where?

How many peaks superimposed at origin?

How many non-origin peaks?

1

23H H

H HH H

Patterson maps are more complicated than electron density maps. Imagine the complexity of a Patterson map of a protein



a

b

Unit cell repeats fill out rest of cell with peaksH H

H H

Patterson maps have an additional center of symmetry



a

b

H H H H

H H H H

plane group pm plane group p2mm

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 in space group P21212

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

What Plane group is this?

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

HH

H

H

H

H

H

H

HH

H

b

Let’s consider only oxygen atoms

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

HH

H

H

H

H

H

H

HH

H

b

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?

•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?

(0,0) ab

Symmetry operators in plane group p2

(0,0) ab

(0.2,0.3)

(-0.2,-0.3)

SYMMETRY OPERATORSFOR PLANE GROUP P21) x,y2) -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.

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?

Yes! Equivalent by unit cell translation.(-0.2+1.0, -0.3+1.0)=(0.8, 0.7)

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.

(0,0) ab

a(0,0)b

(0.2,0.3)

(-0.2,-0.3)


PATTERSON MAP2D CRYSTAL

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?

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

Patterson in plane group p2

(0,0) ab

a(0,0)b

(0.2,0.3)

(-0.2,-0.3)


PATTERSON MAP2D CRYSTAL

(0.4, 0.6)

(-0.4, -0.6)

(0.6, 0.4)

Patterson in plane group p2

a(0,0)b

PATTERSON MAP

(0.4, 0.6)

If you collected data on this crystaland calculated a Patterson mapit would look like this.

(0.6, 0.4)

Now I’m stuck in Patterson space. How do I get back to x,y coordinates?

a(0,0)b

PATTERSON MAP

(0.4, 0.6)

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.


x , y -(-x, –y) 2x , 2y u=2x, v=2y

symop #1symop #2

(0.6, 0.4)

plug in Patterson valuesfor u and v to get x and y.

Now I’m stuck in Patterson space. How do I get back to x,y, coordinates?

a(0,0)b

PATTERSON MAP

(0.4, 0.6)


x y-(-x –y)2x 2y

symop #1symop #2

set u=2x v=2y plug in Patterson valuesfor u and v to get x and y.

u=2x0.4=2x0.2=x

v=2y0.6=2y0.3=y

Hurray!!!!


x y-(-x –y)2x 2y

symop #1symop #2

set u=2x v=2y plug in Patterson valuesfor u and v to get x and y.

u=2x0.4=2x0.2=x

v=2y0.6=2y0.3=y

HURRAY! we got back the coordinatesof our smiley faces!!!!

(0,0) ab

(0.2,0.3)

2D CRYSTAL

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)b

PATTERSON MAP

(0.4, 0.6)

(0.6, 0.4)

using Patterson (u,v) values 0.4, 0.6to get x and y.

u=2x0.4=2x0.2=x

v=2y0.6=2y0.3=y

u=2x0.6=2x0.3=x

v=2y0.4=2y0.2=y

using Patterson (u,v) values 0.6, 0.4to get x and y.

These two solutions do not give the same x,y? What is going on??????

Arbitrary choice of origin

• Original origin choice• Coordinates x=0.2, y=0.3.

(0,0) ab

(0.2,0.3)

(-0.2,-0.3)

(0.8,0.7) (0.3,0.2)

(-0.3,-0.2)

(0.7,0.8)

(0,0) ab

• 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 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

Isomorphous difference Patterson map (Pt derivative)

W=0

V=1/2

U=1/2

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

Solve for x, y using w=0 Harker sect.

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

Does z=0? No!

Solve for x, z using v=1/2 Harker sect.

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

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.

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 x1 and y1 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


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

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

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?

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

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 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 v0.28, 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.

Interpreting difference Patterson Maps in Lab this week!

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