1. r(xyz)=S |Fhkl| cos2p(hx+ky+lz -ahkl)
Wednesday: compute an experimental electron density map of proteinase K
Fourier synthesis
r(xyz)=S |Fhkl| cos2p(hx+ky+lz -ahkl)
hkl

2. Combine data from 3 Crystals How many intensities have we measured to phase each reflection?
Native PCMBS (Hg) GdCl3
Jeannette Sean Greg
H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)|)

3. How many circles to intersect in the complex plane?
Which phasing equations are applicable to our data?
PCMBS (Hg)
Yes No
FP(hkl)= FPHg (hkl) fHg(hkl) X
Isomorphous
Anomalous
Dispersive
FP(hkl)= FPHg(-h-k-l)* - fHg(-h-k-l)* X
FP(hkl)= FPHg(hkl)ln - fHg(hkl)ln X
For GdCl3 (Gd)
Yes No
FP(hkl)= FPGd (hkl) fGd(hkl) X
Isomorphous
Anomalous
Dispersive
FP(hkl)= FPGd(-h-k-l)* - fGd(-h-k-l)* X
How many circles to intersect in the complex plane? X
FP(hkl)= FPGd(hkl)ln - fGd(hkl)ln

4. FP(hkl)=FPHg (hkl) - fHg(hkl) Isomorphous Differences
Begin with SIR Phasing with PCMBS (Hg)
FP(hkl)=FPHg (hkl) - fHg(hkl)
Isomorphous Differences
We measured |FP| and |FPHg|.
How do we graph these quantities in complex plane?
How did we obtain the coordinates of Hg?
How do we obtain fHg(hkl)?

5. SIR Phasing FP=FP·Hg(+) - fHg(+) |FP |
H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)|
Imaginary axis
500
|FP |
250
Real axis
-500
-250
250
500
-250
-500

6. SIR Phasing FP=FP·Hg(+) - fHg(+) |FP |
H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)|
Imaginary axis
500
Hg coordinates (x,y,z) from Patterson =(0.197, 0.755, 0.935)
|FP |
250
Real axis
-500
-250
250
500
-250
-500
f’ and f” are anomalous scattering corrections specific for wavelength used.
fHg is a real number proportional to the number of e- in Hg

7. SIR Phasing FP=FP·Hg(+) - fHg(+) |FP |
H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)|
Imaginary axis
500
Hg coordinates (x,y,z) from Patterson =(0.197, 0.755, 0.935)
|FP |
fHg=(fHg+f’+if)[e2pi*(h(x)+k(y)+l(z))+
e2pi*(h(-x)+k(-y)+l(½+z))+
e2pi*(h(½-y)+k(½+x)+l(¾+z)+
e2pi*(h(½+y)+k(½-x)+l(¼+z)+
e2pi*(h(½-x)+k(½+y)+l(¾-z)+
e2pi*(h(½+x)+k(½-y)+l(¼-z)+
e2pi*(h(y)+k(x)+l(-z)+
e2pi*(h(-y)+k(-x)+l(½-z)]
250
Real axis
-500
-250
250
500
-250
-500
f’ and f” are anomalous scattering corrections specific for wavelength used.
fHg is a real number proportional to the number of e- in Hg

8. SIR Phasing FP=FP·Hg(+) - fHg(+) |FP | fHg(+) -
H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)|
Imaginary axis
500
|FP |
|fHg|=282
aHg=58°
250
282
58°
fHg(+)
Real axis
-500
-250
250
500 -
-250
-500

9. SIR Phasing FP=FP·Hg(+) - fHg(+) |FP | -fHg(+) |FP·Hg(+)|
H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)|
Imaginary axis
500
|FP |
250
Real axis
-500
-250
250
500
-fHg(+)
-250
Let’s look at the quality of the phasing statistics up to this point.
|FP·Hg(+)|
-500

10. SIR Phasing a) b) c) -fHg(+) |FP·Hg(+)|
H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)|
Which of the following graphs best represents the phase probability distribution, P(a)?
500
-500
-250
250
Imaginary axis
|FP |
-fHg(+)
Real axis
|FP·Hg(+)| a) b) c)

11. SIR Phasing 90 180 270 -fHg(+) |FP·Hg(+)|
H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)|
The phase probability distribution, P(a) is sometimes shown as being wrapped around the phasing circle.
500
-500
-250
250
Imaginary axis
|FP |
-fHg(+)
Real axis
|FP·Hg(+)| 90
180
270

12. Radius of circle is approximately |Fp|
SIR Phasing
H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)|
Which of the following is the best choice of Fp?
Imaginary axis 90 90
180
270
500 a) b) c)
|FP |
180
270
Real axis 90
-500
500
180
270
-500 90
|FP·Hg(+)|
180
270
Radius of circle is approximately |Fp|

13. best FP = |Fp|eia•P(a)da
SIR Phasing
H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)|
Imaginary axis
best FP = |Fp|eia•P(a)da a
500 90
|FP |
180
Real axis
Sum of probability weighted vectors Fp
Usually shorter than Fp
-500
500
270
-500
|FP·Hg(+)|

14. best FP = |Fp|eia•P(a)da
SIR Phasing
H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)|
Imaginary axis
best FP = |Fp|eia•P(a)da a
500 90
|Fbest |
180
Real axis
Sum of probability weighted vectors Fp
Usually shorter than Fp
-500
500
270
-500
|FP·Hg(+)|

15. Radius of circle is approximately |Fp|
SIR Phasing
H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)|
Which of the following is the best approximation to the Figure Of Merit (FOM) for this reflection?
Imaginary axis
500 90
1.00
2.00
0.20
-0.20
|Fbest |
180
Real axis
-500
500
270
-500
|FP·Hg(+)|
FOM=|Fbest|/|FP|
Radius of circle is approximately |Fp|

16. SIR Phasing
Which phase probability distribution would yield the most desirable Figure of Merit? c)
500 90
|Fbest | a) b)
180
-500
500 + + +
270 90
270 90
270
180
180
-500

17. RCullis = |FPH|-|FP+fH| |FPH| - |FP|
SIR Phasing
H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)|
Which of the following
is the RCullis for this reflection?
Imaginary axis
500 90
-0.5
0.5
1.30
2.00
|Fbest |
180
Real axis
-500
500
270
RCullis = |FPH|-|FP+fH| |FPH| - |FP|
-500
|FP·Hg(+)|
|FPH|-|FP+fH| = Lack of closure = 200
|FPH|-|FP| = Isomorphous difference= = 100

18. SIR Phasing 90 2.40 1.40 0.40 -0.40 180 270 Phasing Power = |fH|
H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)|
Which of the following is the best approximation to the phasing power for this reflection?
Imaginary axis
500 90
2.40
1.40
0.40
-0.40
|Fbest |
180
Real axis
-500
500
-fHg(+)
270
Phasing Power = |fH|
Lack of closure
-500
|FP·Hg(+)|
|fH ( h k l) | = 282
|FPH|-|FP+fH| = Lack of closure = 200
(at the aP of Fbest)

19. What Phasing Power is sufficient to solve the structure?
SIR Phasing
H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)|
Which of the following is the most desirable phasing power?
Imaginary axis
500 90
2.40
1.40
0.40
-0.40
|Fbest |
180
Real axis
-500
500
-fHg(+)
270
Phasing Power = |fH|
Lack of closure
-500
|FP·Hg(+)|
>1
What Phasing Power is sufficient to solve the structure?

20. Include information from anomalous scattering
SIRAS

21. SIRAS Phasing |FP | -fHg(+) |FP·Hg(+)|
FP=FP·Hg(+) - fHg(+)
H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)|
Imaginary axis
500
|FP |
250
Real axis
-500
-250
250
500
-fHg(+)
-250
|FP·Hg(+)|
-500

22. FP=FP·Hg(-)* - fHg(-)*
SIRAS Phasing
FP=FP·Hg(+) - fHg(+)
FP=FP·Hg(-)* - fHg(-)*
H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)|
Imaginary axis
500
|FP |
|fHg|=282
aHg*=48°
250
Real axis
282
48°
fHg(-)*
-500
-250
250
500
-fHg(+)
-250
|FP·Hg(+)|
-500

23. FP=FP·Hg(-)* - fHg(-)*
SIRAS Phasing
FP=FP·Hg(+) - fHg(+)
FP=FP·Hg(-)* - fHg(-)*
H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)|
Imaginary axis
500
|FP |
250
Real axis
-500
-250
250
500
-fHg(-)*
-fHg(+)
-250
|FP·Hg(-)|
|FP·Hg(+)|
-500

24. Isomorphous differences Anomalous differences
SIRAS
Isomorphous differences
Anomalous differences
Imaginary axis
500
|FP |
Which P(a) corresponds to SIR?
Which P(a) corresponds to SIRAS?
250
Real axis
-250
250
500
-fHg(-)*
-500
-fHg(+)
-250
|FP·Hg(-)|
|FP·Hg(+)|
-500

25. and now, its time for… ?
A Tale
Of Two
Ambiguities ? ? ?

26. Two Ambiguities to Resolve
What are the coordinates of Gd?
Patterson methods offer only 1:48 chance of choosing an origin for Gd that is consistent with PCMBS.
Which is the correct hand of space group?
Measured intensities do not distinguish between P41212 and P43212.
Patterson map

27. A Cross difference Fourier resolves both ambiguities
Consider an electron density map calculated using amplitudes |FP|, and newly determined phases aP.
r(xyz) =1/V*S|FP|e-2pi(hx+ky+lz-aP)
If we replace the coefficients with |FPGd-FP|, the result will be an electron density map corresponding to what structural feature?

28. r(x)=1/V*S|FPGd-FP|e-2pi(hx-aP)
It's the Gd atom, Alex.
The difference FPGd-FP cancels the protein contribution and we are left with only the contribution from the Gd atom.
This cross difference Fourier will help us in two ways:
It will give us the coordinates x,y,z for the Gd atom.
-The coordinates of Gd will be referred to an origin consistent with Hg (because the phases aP were calculated using the Hg site.
2) It will resolve the handedness ambiguity
-The correct hand of the space group will produce a Gd peak height higher than the incorrect space group..

29. Phasing Procedures
Calculate aP using Hg x,y,z. Use these phases in a cross difference Fourier to find the Gd x,y,z.
-Note the height of the peak and Gd coordinates.
Negate x,y,z of Hg and invert the space group from P43212 to P Calculate a second set of phases and calculate a 2nd cross difference Fourier to find the Gd site.
-Compare the height of the peak with step 1.
Choose the handedness which produces the highest peak for Gd. Use the corresponding hand of space group and Hg, and Gd coordinates to make a combined set of phases aP.

30. FP= FP·Gd(-)* - fGd(-)*
MIRAS Phasing
H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)|
Imaginary axis
500
FP= FP·Gd(+) - fGd(+)
FP= FP·Gd(-)* - fGd(-)*
250
Real axis
-500
-250
250
500
-250
-500
H K L fH+f’ f”(-) aH fH+f’ f” aH
° °

31. SIR SIRAS MIRAS

32. SIR SIRAS MIRAS

33. Density modification A) Solvent flattening.
Calculate an electron density map.
If r<threshold, -> solvent
If r>threshold -> protein
Build a mask
Set density value in solvent region to a constant (low).
Transform flattened map to structure factors
Combine modified phases with original phases.
Iterate

34. Histogram matching Calculate an electron density map.
Calculate the electron density distribution. It’s a histogram. How many grid points on map have an electron density falling between 0.2 and 0.3 etc?
Compare this histogram with ideal protein electron density map.
Modify electron density to resemble an ideal distribution.
Number of times a particular electron density value is observed.
Electron density value

35. MIR phased map + Solvent Flattening + Histogram Matching
MIRAS phased map

36. MIR phased map + Solvent Flattening + Histogram Matching
MIRAS phased map

37. Enter Phasing Statistics in Spreadsheet

38. It’s time for Model Building

39. Problem Set 5b
Calculate |F|, a, and I for reflections 1,1,1, and 2,0,0, and 1,0,0, of a NaCl crystal assuming that:
a) The magnitude of the atomic scattering factor, f, for an ion is equal to its number of electrons.
b) I=F•F*, where F* is the complex conjugate of F.

40. Z=0 Z=½ Cl Cl Cl Na Na Na Cl Cl Cl Na Na Na (0,½,0) (½,½,0) (0,½,½)
(½,½,½) Cl Cl Cl Na Na Na
(0,0,0)
(½,0,0)
(0,0,½)
(½,0,½) y x z
Cl-: 0,0,0; ½,½,0; ½,0,½; 0,½,½.
Na+: 0,½,0; ½,0,0; 0,0,½; ½,½,½.

41. Face Centered Cubic Na Cl Na Cl Na Cl y x z

42. Structure Factor Equation
Fhkl=Sfje2pi(hx+ky+lz) j

43. Structure Factor Equation
Fhkl=Sfje2pi(hx+ky+lz) j
A sum over all atoms in the unit cell:
from j=1 to j=N.
N=total atoms in unit cell.

44. Structure Factor Equation
Cl- (0,0,0) Cl- (½,½,0) Cl- (0,½,½) Cl- (½,0,½) Na+(0,0,½) Na+(0,½,0) Na+(½,0,0) Na+(½,½,½)
Fhkl=f1e2pi(hx1+ky1+lz1)
+f2e2pi(hx2+ky2+lz2)
+f3e2pi(hx3+ky3+lz3)
+f4e2pi(hx4+ky4+lz4)
+f5 e2pi(hx5+ky5+lz5)
+f6e2pi(hx6+ky6+lz6)
+f7e2pi(hx7+ky7+lz7)
+f8e2pi(hx8+ky8+lz8) N
Fhkl=Sfje2pi(hx+ky+lz) j

45. Collect the form factors
Cl- (0,0,0) Cl- (½,½,0) Cl- (0,½,½) Cl- (½,0,½) Na+(0,0,½) Na+(0,½,0) Na+(½,0,0) Na+(½,½,½)
Fhkl=fCl- e2pi(hx1+ky1+lz1)
+fCl-e2pi(hx2+ky2+lz2)
+fCl-e2pi(hx3+ky3+lz3)
+fCl-e2pi(hx4+ky4+lz4)
+fNa+ e2pi(hx5+ky5+lz5)
+ fNa+ e2pi(hx6+ky6+lz6)
+ fNa+ e2pi(hx7+ky7+lz7)
+fNa+e2pi(hx8+ky8+lz8)
Collect the form factors

46. Collect the form factors
Cl- (0,0,0) Cl- (½,½,0) Cl- (0,½,½) Cl- (½,0,½) Na+(0,0,½) Na+(0,½,0) Na+(½,0,0) Na+(½,½,½)
Fhkl=fCl- [e2pi(hx1+ky1+lz1)
+e2pi(hx2+ky2+lz2)
+e2pi(hx3+ky3+lz3)
+e2pi(hx4+ky4+lz4)]
+fNa+ [e2pi(hx5+ky5+lz5)
+ e2pi(hx6+ky6+lz6)
+ e2pi(hx7+ky7+lz7)
+e2pi(hx8+ky8+lz8)]
Collect the form factors

47. F111
Cl- (0,0,0) Cl- (½,½,0) Cl- (0,½,½) Cl- (½,0,½) Na+(0,0,½) Na+(0,½,0) Na+(½,0,0) Na+(½,½,½)
Fhkl=fCl- [e2pi(hx1+ky1+lz1)+e2pi(hx2+ky2+lz2)+e2pi(hx3+ky3+lz3)+e2pi(hx4+ky4+lz4)]
+fNa+[e2pi(hx5+ky5+lz5)+e2pi(hx6+ky6+lz6)+e2pi(hx7+ky7+lz7)+e2pi(hx8+ky8+lz8)]
Substitute h,k,l

48. F111
Cl- (0,0,0) Cl- (½,½,0) Cl- (0,½,½) Cl- (½,0,½) Na+(0,0,½) Na+(0,½,0) Na+(½,0,0) Na+(½,½,½)
F111=fCl- [e2pi(1x1+1y1+1z1)+e2pi(1x2+1y2+1z2)+e2pi(1x3+1y3+1z3)+e2pi(1x4+1y4+1z4)]
+fNa+[e2pi(1x5+1y5+1z5)+e2pi(1x6+1y6+1z6)+e2pi(1x7+1y7+1z7)+e2pi(1x8+1y8+1z8)]
Substitute x,y,z

49. Unit vector in complex plane with direction given by exponent
F111
Cl- (0,0,0) Cl- (½,½,0) Cl- (0,½,½) Cl- (½,0,½) Na+(0,0,½) Na+(0,½,0) Na+(½,0,0) Na+(½,½,½)
F111=fCl-[e2pi(1•0+1•0+1•0)+e2pi(1•½+1•½+1•0)+e2pi(1•0+1•½+1•½)+e2pi(1•½+1•0+1•½)]
+fNa+[e2pi(1•0+1•0+1•½)+e2pi(1•0+1•½+1•0)+e2pi(1•½+1•0+1•0)+e2pi(1•½+1•½+1•½)]
Unit vector in complex plane with direction given by exponent
Substitute x,y,z

50. F111
Cl- (0,0,0) Cl- (½,½,0) Cl- (0,½,½) Cl- (½,0,½) Na+(0,0,½) Na+(0,½,0) Na+(½,0,0) Na+(½,½,½)
F111=fCl-[e2pi(hx1+ky1+lz1) +e2pi(hx2+ky2+lz2) +e2pi(hx3+ky3+lz3) +e2pi(hx4+ky4+lz4) ] +fNa+[e2pi(hx5+ky5+lz5) +e2pi(hx6+ky6+lz6) +e2pi(hx7+ky7+lz7) +e2pi(hx8+ky8+lz8)]
F111=18[e2pi(1*0+1*0+1*0) +e2pi(1*½+1*½+1*0) +e2pi(1*0+1*½+1*½) +e2pi(1*½+1*0+1*½) ]+10[e2pi(1*0+1*0+1*½) +e2pi(1*0+1*½+1*0) +e2pi(1*½+1*0+1*0) +e2pi(1*½+1*½+1*½)]
F111=18[e2pi(0) e2pi(1) e2pi(1) e2pi(1) ]+10[e2pi(½) e2pi(½) e2pi(½) e2pi(3/2) ]
F111=18[ ]+10[ ]
F111=18[4 ] [-4 ]
F111=
F111=32
|F111 | =32 and a=0°
I=32*32=1024
F111=18[cos(0)+isin(0) +cos(2p)+isin(2p) +cos(2p)+isin(2p) +cos(2p)+isin(2p)]+10[cos(p)+isin(p) +cos(p)+isin(p) +cos(p)+isin(p) +cos(p)+isin(p)]
F111=18[1+0i i i i ]+10[-1+0i i i i]
F111=18[ ]+10[ ]

51. How are the ions arranged w.r.t. the 1,1,1 Bragg Planes?Na Cl Cl Na Na Cl Cl Na Cl Cl Na Cl Na Na Cl Na Na Cl Cl Na y x z

52. F200 Xx Xx Xx Xx Xx Xx Xx Xx
Cl- (0,0,0) Cl- (½,½,0) Cl- (0,½,½) Cl- (½,0,½) Na+(0,0,½) Na+(0,½,0) Na+(½,0,0) Na+(½,½,½)
F200= 18[e2pi(2*0+0*0+0*0) +e2pi(2*½+0*½+0*0) +e2pi(2*0+0*½+0*½) +e2pi(2*½+0*0+0*½) ]+10[e2pi(2*0+0*0+0*½) +e2pi(2*0+0*½+0*0) +e2pi(2*½+0*0+0*0) +e2pi(2*½+0*½+0*½)]
F200= 18[e2pi(0) e2pi(1) e2pi(0) e2pi(1) ]+10[e2pi(0) e2pi(0) e2pi(1) e2pi(1) ]
F200= 18[ ]+10[ ]
F200= 18[4 ] [+4 ]
F200=
F200=112
|F200 | = a=0
I = 122*122 = 12,544
F200=18[cos(0)+isin(0) +cos(2p)+isin(2p) +cos(0)+isin(0) +cos(2p)+isin(2p)]+10[cos(2p)+isin(2p) +cos(0)+isin(0) +cos(2p)+isin(2p) +cos(2p)+isin(2p)]
F200=18[1+0i i i i] [1+0i i i i]
F200=18[ ] [ ]

53. How are the ions arranged w.r.t. the 2,0,0 Bragg Planes?Na Cl Na Cl Na Cl y x z

54. F100 Xx Xx Xx Xx Xx Xx Xx Xx
Cl- (0,0,0) Cl- (½,½,0) Cl- (0,½,½) Cl- (½,0,½) Na+(0,0,½) Na+(0,½,0) Na+(½,0,0) Na+(½,½,½)
F100= 18[e2pi(1*0+0*0+0*0) +e2pi(1*½+0*½+0*0) +e2pi(1*0+0*½+0*½) +e2pi(1*½+0*0+0*½) ] +10[e2pi(1*0+0*0+0*½) +e2pi(1*0+0*½+0*0) +e2pi(1*½+0*0+0*0) +e2pi(1*½+0*½+0*½) ]
F100= 18[e2pi(0) e2pi(½) e2pi(0) e2pi(½) ]+10[e2pi(0) e2pi(0) e2pi(½) e2pi(½) ]
F100= 18[ ]+10[ ]
F100= 18[0 ] [0 ]
F100=
F100=0
|F100 | =0 a=0
I = 0*0 = 0
F100=18[cos(0)+isin(0) +cos(p)+isin(p) +cos(0)+isin(0) +cos(p)+isin(p)]+10[cos(0)+isin(0) +cos(0)+isin(0) +cos(p)+isin(p) +cos(p)+isin(p)]
F100=18[1+0i i i i] [1+0i i i i]
F100=18[ ] [ ]

55. How are the ions arranged w.r.t. the 1,0,0 Bragg Planes?Na Cl Na Cl Na Cl y x z

56. Na Cl Cl Na Na Cl Cl Na Cl Cl Na Cl Na Na Cl Na Na Cl Cl Na
Why is the intensity of the 1,1,1 reflection of the KCl crystal very weak compared to that of NaCl, even though these salts have the same face centered cubic structure? Na Cl Cl Na Na Cl Cl Na Cl Cl Na Cl Na Na Cl Na Na Cl Cl Na y x z

57. Cl Na Cl Cl K K Cl K Cl K Cl K Cl K Cl K Cl K Cl Cl K K Cl K Cl K Cl
Why is the intensity of the 1,1,1 reflection of the KCl crystal very weak compared to that of NaCl, even though these salts have the same face centered cubic structure? Cl Na Cl Cl K K Cl K Cl K Cl K Cl K Cl K Cl K Cl Cl K K Cl K Cl K Cl y x z

58. Why is the intensity of the 1,1,1 reflection of the KCl crystal very weak compared to that of NaCl, even though these salts have the same face centered cubic structure? X
F111
Cl- (0,0,0) Cl- (½,½,0) Cl- (0,½,½) Cl- (½,0,½) Na+(0,0,½) Na+(0,½,0) Na+(½,0,0) Na+(½,½,½)
F111=fCl-[e2pi(hx1+ky1+lz1) +e2pi(hx2+ky2+lz2) +e2pi(hx3+ky3+lz3) +e2pi(hx4+ky4+lz4) ] +fNa+[e2pi(hx5+ky5+lz5) +e2pi(hx6+ky6+lz6) +e2pi(hx7+ky7+lz7) +e2pi(hx8+ky8+lz8)]
F111=18[e2pi(1*0+1*0+1*0) +e2pi(1*½+1*½+1*0) +e2pi(1*0+1*½+1*½) +e2pi(1*½+1*0+1*½) ]+10[e2pi(1*0+1*0+1*½) +e2pi(1*0+1*½+1*0) +e2pi(1*½+1*0+1*0) +e2pi(1*½+1*½+1*½)]
F111=18[e2pi(0) e2pi(1) e2pi(1) e2pi(1) ]+10[e2pi(½) e2pi(½) e2pi(½) e2pi(3/2) ]
F111=18[ ]+10[ ]
F111=18[4 ] [-4 ]
F111=
F111=32
|F111 | =32 and a=0°
I=32*32=1024

59. Why is the intensity of the 1,1,1 reflection of the KCl crystal very weak compared to that of NaCl, even though these salts have the same face centered cubic structure?
F111
Cl- (0,0,0) Cl- (½,½,0) Cl- (0,½,½) Cl- (½,0,½) Na+(0,0,½) Na+(0,½,0) Na+(½,0,0) Na+(½,½,½)
F111=fCl-[e2pi(hx1+ky1+lz1) +e2pi(hx2+ky2+lz2) +e2pi(hx3+ky3+lz3) +e2pi(hx4+ky4+lz4) ] +fNa+[e2pi(hx5+ky5+lz5) +e2pi(hx6+ky6+lz6) +e2pi(hx7+ky7+lz7) +e2pi(hx8+ky8+lz8)]
F111=18[e2pi(1*0+1*0+1*0) +e2pi(1*½+1*½+1*0) +e2pi(1*0+1*½+1*½) +e2pi(1*½+1*0+1*½) ]+18[e2pi(1*0+1*0+1*½) +e2pi(1*0+1*½+1*0) +e2pi(1*½+1*0+1*0) +e2pi(1*½+1*½+1*½)]
F111=18[e2pi(0) e2pi(1) e2pi(1) e2pi(1) ]+18[e2pi(½) e2pi(½) e2pi(½) e2pi(3/2) ]
F111=18[ ]+18[ ]
F111=18[4 ] [-4 ]
F111=
F111=0
|F111 | =0 and a=0°
I=0*0=0

60. Problem set 4 Show structure factors are real numbers when the motif is centrosymmetric.
Fh=Sfj[cos(2phx)+isin(2phx)]+fj[cos(2ph(-x))+isin(2ph(-x))]
collect fj terms:
Fh=Sfj[cos(2phx)+cos(-2phx) +isin(2phx)+isin(-2phx)]
Substitute the following identities:
cos(-f)=cos(f)
sin(-f)=-sin(f)
Fh=Sfj[cos(2phx)+cos(2phx) +isin(2phx)-isin(2phx)]
Add terms:
Fh=Sfj*(2cos(2phx))
Imaginary terms cancel. It’s real.

61. best F = |Fp|eia•P(a)da
SIR 90
180
270
best F = |Fp|eia•P(a)da a
Sum of probability weighted vectors Fp
Best phase

62. Ambiguity in handedness?
Santa Monica
UCLA ?
LAX
Structures recovered from Patterson
Structure
Patterson

63. Ambiguity of Origin: example in 2 dimensions
Choice 1 b a
X=0.80 Y=0.30
X=0.20 Y=0.70

64. The Zn2+ ion can be specified by any of these 8 coordinates.
Choice 1
Choice 2 b a b a
X=0.30 Y=0.30
X=0.80 Y=0.30
X=0.70 Y=0.70
X=0.20 Y=0.70
Choice 3
Choice 4 b a b a
X=0.20 Y=0.20
X=0.70 Y=0.20
X=0.30 Y=0.80
X=0.80 Y=0.80

65. Barriers to combining phase information from 2 derivatives
Initial Phasing with PCMBS
Calculate phases using coordinates you determined.
Refine heavy atom coordinates
Find Eu site using Cross Difference Fourier map.
Easier than Patterson methods.
Want to combine PCMBS and Eu to make MIRAS phases.
Determine handedness (P43212 or P41212 ?)
Repeat calculation above, but in P41212.
Compare map features with P43212 map to determine handedness.
Combine PCMBS and Eu sites (use correct hand of space group) for improved phases.
Density modification (solvent flattening & histogram matching)
Improves Phases
View electron density map