Fasproblemet Tungmetallderivat (MIR, SIR) Anomalous dispersion Patterson kartor.

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Presentation transcript:

Fasproblemet Tungmetallderivat (MIR, SIR) Anomalous dispersion Patterson kartor

Vektorrepresentation av strukturfaktorer

SIR

Multiple isomorphous replacement

MIR

Friedels Lag (hkl = -h –k –l)

Anomalous dispersion

Patterson map Karta över vektorer mellan par av atomer För varje topp finns det två atomer, visar atomer relativt till varandra ej relativt till enhetscellen Varje atom bildar ett par (och vektor) med varje annan atom, dvs i en enhetscell med N atomer finns det N 2 vektorer. N self vectors och n(n-1) andra vektorer Intensitetet på toppen proportionell mot produkten av det ingående atomparet

Patterson in plane group p2 (0,0) a b SYMMETRY OPERATORS FOR PLANE GROUP P2 1) x,y 2) -x,-y

Patterson in plane group p2 (0,0) a b (0.1,0.2) SYMMETRY OPERATORS FOR PLANE GROUP P2 1) x,y 2) -x,-y

Patterson in plane group p2 (0,0) a b (0.1,0.2) (-0.1,-0.2) SYMMETRY OPERATORS FOR PLANE GROUP P2 1) x,y 2) -x,-y

Patterson in plane group p2 (0,0) a b (0.1,0.2) (-0.1,-0.2) SYMMETRY OPERATORS FOR PLANE GROUP P2 1) x,y 2) -x,-y

Patterson in plane group p2 (0,0) a b a b (0.1,0.2) (-0.1,-0.2) SYMMETRY OPERATORS FOR PLANE GROUP P2 1) x,y 2) -x,-y PATTERSON MAP 2D CRYSTAL

Patterson in plane group p2 (0,0) a b a b (0.1,0.2) (-0.1,-0.2) SYMMETRY OPERATORS FOR PLANE GROUP P2 1) x,y 2) -x,-y PATTERSON MAP 2D CRYSTAL

Patterson in plane group p2 (0,0) a b a b (0.1,0.2) (-0.1,-0.2) SYMMETRY OPERATORS FOR PLANE GROUP P2 1) x,y 2) -x,-y PATTERSON MAP 2D CRYSTAL What is the coordinate for the Patterson peak? Just take the difference between coordinates of the two happy faces. (x,y)-(-x,-y) or (0.1,0.2)-(-0.1,-0.2) so u=0.2, v=0.4

Patterson in plane group p2 (0,0) a b a b (0.1,0.2) (-0.1,-0.2) SYMMETRY OPERATORS FOR PLANE GROUP P2 1) x,y 2) -x,-y PATTERSON MAP 2D CRYSTAL What is the coordinate for the Patterson peak? Just take the difference between coordinates of the two happy faces. (x,y)-(-x,-y) or (0.1,0.2)-(-0.1,-0.2) so u=0.2, v=0.4 (0.2, 0.4)

Patterson in plane group p2 a (0,0) b PATTERSON MAP (0.2, 0.4) If you collected data on this crystal and calculated a Patterson map it would look like this.

Now I’m stuck in Patterson space. How do I get back to x,y, coordinates? a (0,0) b PATTERSON MAP (0.2, 0.4) Use our friends, the space group operators. The peaks positions correspond to vectors between smiley faces. SYMMETRY OPERATORS FOR PLANE GROUP P2 1) x,y 2) -x,-y x y -(-x –y) 2x 2y symop #1 symop #2

Now I’m stuck in Patterson space. How do I get back to x,y, coordinates? a (0,0) b PATTERSON MAP (0.2, 0.4) Use our friends, the space group operators. The peaks positions correspond to vectors between smiley faces. SYMMETRY OPERATORS FOR PLANE GROUP P2 1) x,y 2) -x,-y x y -(-x –y) 2x 2y symop #1 symop #2 set u=2x v=2y plug in Patterson values for 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 (0.2, 0.4) SYMMETRY OPERATORS FOR PLANE GROUP P2 1) x,y 2) -x,-y x y -(-x –y) 2x 2y symop #1 symop #2 set u=2x v=2y plug in Patterson values for u and v to get x and y. u=2x 0.2=2x 0.1=x v=2y 0.4=2y 0.2=y

Hurray!!!! SYMMETRY OPERATORS FOR PLANE GROUP P2 1) x,y 2) -x,-y x y -(-x –y) 2x 2y symop #1 symop #2 set u=2x v=2y plug in Patterson values for u and v to get x and y. u=2x 0.2=2x 0.1=x v=2y 0.4=2y 0.2=y HURRAY! we got back the coordinates of our smiley faces!!!! (0,0) a b (0.1,0.2)

Vektorrepresentation av strukturfaktorer

Patterson map