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PPT – Interpreting difference Patterson Maps in Lab this week! PowerPoint presentation | free to view - id: 1c19e0-ZDc1Z

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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?
- Lets review how heavy atom positions can be

calculated from difference Patterson peaks.

Patterson Review

- A Patterson synthesis is like a Fourier synthesis

except for what two variables?

Fourier synthesis r(xyz)S Fhkl cos2p(hxkylz

-ahkl) hkl

Patterson synthesis P(uvw)S ?hkl cos2p(hukvlw

-?) hkl

Patterson synthesis P(uvw)S Ihkl cos2p(hukvlw

-0) hkl

Hence, Patterson density map electron density

map convoluted with its inverted image.

- Patterson synthesis
- P(uvw)S Ihkl cos2p(hukvlw)
- Remembering IhklFhklFhkl
- And Friedels law Fhkl F-h-k-l

P(uvw)FourierTransform(FhklF-h-k-l) - P(uvw)r(uvw) r (-u-v-w)

Electron Density vs. Patterson Density

Lay down n copies of the unit cell at the origin,

where nnumber of atoms in unit cell.

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

- 3 sets of peaks
- Length O-H
- Where?
- Length H-H
- Where?
- Length zero
- Where?
- How many peaks superimposed at origin?
- How many non-origin peaks?

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 than electron

density maps. Imagine the complexity of a

Patterson map of a protein

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

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.

What Plane group is this?

Lets consider only oxygen atoms

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?

Symmetry operators in plane group p2

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?

(-0.2,-0.3)

a

(0,0)

b

(0.2,0.3)

SYMMETRY OPERATORSFOR PLANE GROUP P2 1) x,y 2)

-x,-y

Yes! Equivalent by unit cell translation. (-0.21

.0, -0.31.0)(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.

2D CRYSTAL

PATTERSON MAP

(-0.2,-0.3)

a

(0,0)

a

(0,0)

b

b

(0.2,0.3)

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 OPERATORSFOR PLANE GROUP P2 1) x,y 2)

-x,-y

- 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

(-0.4, -0.6)

(-0.2,-0.3)

a

(0,0)

b

(0.2,0.3)

(0.6, 0.4)

(0.4, 0.6)

SYMMETRY OPERATORSFOR PLANE GROUP P2 1) x,y 2)

-x,-y

PATTERSON MAP

2D CRYSTAL

Patterson in plane group p2

a

(0,0)

b

(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 Im 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 betrween our friends the space group

operators.

a

(0,0)

b

(0.6, 0.4)

(0.4, 0.6)

SYMMETRY OPERATORSFOR PLANE GROUP P2 1) x,y 2)

-x,-y

PATTERSON MAP

x , y -(-x, y) 2x , 2y

u2x, v2y

symop 1 symop 2

plug in Patterson values for u and v to get x and

y.

Now Im stuck in Patterson space. How do I get

back to x,y, coordinates?

SYMMETRY OPERATORSFOR PLANE GROUP P2 1) x,y 2)

-x,-y

a

(0,0)

b

x y -(-x y) 2x 2y

symop 1 symop 2

(0.4, 0.6)

set u2x v2y plug in Patterson values for u

and v to get x and y.

v2y 0.62y 0.3y

u2x 0.42x 0.2x

PATTERSON MAP

Hurray!!!!

SYMMETRY OPERATORSFOR PLANE GROUP P2 1) x,y 2)

-x,-y

a

(0,0)

b

x y -(-x y) 2x 2y

symop 1 symop 2

(0.2,0.3)

set u2x v2y plug in Patterson values for u

and v to get x and y.

2D CRYSTAL

v2y 0.62y 0.3y

u2x 0.42x 0.2x

HURRAY! we got back the coordinates of our

smiley faces!!!!

Devils 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)

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

and y.

b

v2y 0.62y 0.3y

u2x 0.42x 0.2x

(0.6, 0.4)

(0.4, 0.6)

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

and y.

v2y 0.42y 0.2y

PATTERSON MAP

u2x 0.62x 0.3x

These two solutions do not give the same x,y?

What is going on??????

Arbitrary choice of origin

(-0.2,-0.3)

a

(0,0)

b

(-0.3,-0.2)

(0.2,0.3)

(0.3,0.2)

(0.8,0.7)

(0.7,0.8)

- Original origin choice
- Coordinates x0.2, y0.3.

- New origin choice
- Coordinates x0.3, y0.2.

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,1/2Y,-Z
- 1/2X,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
- 1/2-X,1/2Y,-Z
- 1/2X,1/2-Y,-Z

- X, Y, Z
- -X, -Y, Z
- u2x, v2y, w0

1. X, Y, Z 3. ½-X,½Y,-Z u2x-½,v-½,w2z

1. X, Y, Z 4. ½X,½-Y,-Z u-½,v2y-½,w2z

Harker sections, w0, v1/2, u1/2

Isomorphous difference Patterson map (Pt

derivative)

- X, Y, Z
- -X, -Y, Z
- u2x, v2y, w0

1. X, Y, Z 3. ½-X,½Y,-Z u2x-½,v-½,w2z

1. X, Y, Z 4. ½X,½-Y,-Z u-½,v2y-½,w2z

Solve for x, y using w0 Harker sect.

Harker section w0

W0

- X, Y, Z
- -X, -Y, Z
- u2x, v2y, w0

0.1682x 0.084x

0.2662y 0.133y

Does z0? No!

Solve for x, z using v1/2 Harker sect.

Harker Section v1/2

1. X, Y, Z 3. ½-X,½Y,-Z u2x-½,v-½,w2z

V1/2

0.3332x-1/2 0.8332x 0.416x

0.1502z 0.075z

Resolving ambiguity in x,y,z

- From w0 Harker section x10.084, y10.133
- From v1/2 Harker section, x20.416, z20.075
- Why doesnt 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

From w0 Harker section xorig10.084,

yorig10.133 From v1/2 Harker section,

xorig20.416, zorig20.075

- X, Y, Z
- -X, -Y, Z
- -X, Y, -Z
- X, -Y, -Z
- -X, -Y, -Z
- X, Y, -Z
- X, -Y, Z
- -X, Y, Z
- 1/2X, Y, Z
- 1/2-X, -Y, Z
- 1/2-X, Y, -Z
- 1/2X, -Y, -Z
- 1/2-X, -Y, -Z
- 1/2X, Y, -Z
- 1/2X, -Y, Z
- 1/2-X, Y, Z
- X,1/2Y, Z

- 1/2X,1/2Y, Z
- 1/2-x,1/2-Y, Z
- 1/2-X,1/2Y, -Z
- 1/2X,1/2-Y, -Z
- 1/2-X,1/2-Y, -Z
- 1/2X,1/2Y, -Z
- 1/2X,1/2-Y, Z
- 1/2-X,1/2Y, Z
- 1/2X, Y,1/2Z
- 1/2-X, -Y,1/2Z
- 1/2-X, Y,1/2-Z
- 1/2X, -Y,1/2-Z
- 1/2-X, -Y,1/2-Z
- 1/2X, Y,1/2-Z
- 1/2X, -Y,1/2Z
- 1/2-X, Y,1/2Z
- X,1/2Y,1/2Z

Apply Cheshire symmetry operator 10 To x1 and y1

Xorig10.084 ½-xorig10.5-0.084 ½-xorig10.416

xorig2 yorig10.133 -yorig1-0.133yorig2

Hence, Xorig20.416, yorig2-0.133, zorig20.075

Advanced case,Proteinase K in space group P43212

- Where are Harker sections?

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Symmetry operator 2-Symmetry operator 4

-x - y ½z - ( ½y

½-x ¼z) -½-x-y -½x-y ¼

Symmetry operator 2-Symmetry operator 4

-x - y ½z - ( ½y

½-x ¼z) -½-x-y -½x-y ¼

Plug in u. u-½-x-y 0.18-½-x-y 0.68-x-

y Plug in v.

v-½x-y 0.22-½x-y

0.72x-y Add two equations and solve for y.

0.68-x-y (0.72 x-y) 1.40-2y

-0.70y Plug y into first equation and solve for

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

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Symmetry operator 3 -Symmetry operator 6

½-y ½x ¾z - ( -y -x

½-z) ½ ½2x ¼2z

Plug in v. v ½2x 0.48

½2x -0.022x -0.01x Plug

in w. w ¼2z 0.24

¼2z -0.012z -0.005z

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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,

lets 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-transformed0.70, zstep4-0.005 Or

simply, x-0.02, y0.70, 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 cant combine these coordinates

directly. They dont 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?

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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, y0.70, z-0.005 or
- x0.02, y0.70, 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 u0.28 v0.28, w0.0
- The position of the non-Harker peak will be

predicted by the correct heavy atom coordinate.

x y z -( y x -z) x-y

-xy 2z

symmetry operator 1 -symmetry operator 5 u

v w

First, plug in x-0.02, y0.70, z-0.005 ux-y

-0.02-0.70 -0.72 v-xy 0.020.70

0.72 w2z2(-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 u0.28, v0.28, w0.01 Thus, x-0.02,

y0.70, z-0.005 is correct. Hurray! We are

finished! In the case that the above test

failed, we would change the sign of x.