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

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 (-)|) 1 1 1 681.4 725.8 722.4 730.8 707.6 1 1 2 752.8 733.6 695.3 813.9 805.3 1 1 3 332.1 444.5 456.2 296.1 312.5 1 1 4 526.9 575.8 564.7 527.4 518.3 1 1 5 719.2 827.8 805.4 759.6 766.3 1 1 6 358.4 349.8 354.2 375.6 358.9 1 1 7 273.3 359.4 390.8 300.5 286.6 1 1 8 400.7 362.5 411.2 396.7 411.5 1 2 0 162.5 73.8 132.3 149.8 159.8

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

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)?

SIR Phasing FP=FP·Hg(+) - fHg(+) |FP | H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)| 9 2 1 486 586 611 536 499 Imaginary axis 500 |FP | 250 Real axis -500 -250 250 500 -250 -500

SIR Phasing FP=FP·Hg(+) - fHg(+) |FP | H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)| 9 2 1 486 586 611 536 499 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

SIR Phasing FP=FP·Hg(+) - fHg(+) |FP | H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)| 9 2 1 486 586 611 536 499 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

SIR Phasing FP=FP·Hg(+) - fHg(+) |FP | fHg(+) - H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)| 9 2 1 486 586 611 536 499 Imaginary axis 500 |FP | |fHg|=282 aHg=58° 250 282 58° fHg(+) Real axis -500 -250 250 500 - -250 -500

SIR Phasing FP=FP·Hg(+) - fHg(+) |FP | -fHg(+) |FP·Hg(+)| H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)| 9 2 1 486 586 611 536 499 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

SIR Phasing a) b) c) -fHg(+) |FP·Hg(+)| H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)| 9 2 1 486 586 611 536 499 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) 0 90 180 270 360 0 90 180 270 360 0 90 180 270 360

SIR Phasing 90 180 270 -fHg(+) |FP·Hg(+)| H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)| 9 2 1 486 586 611 536 499 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(+)| 0 90 180 270 360 90 180 270

Radius of circle is approximately |Fp| SIR Phasing H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)| 9 2 1 486 586 611 536 499 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|

best FP = |Fp|eia•P(a)da SIR Phasing H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)| 9 2 1 486 586 611 536 499 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(+)|

best FP = |Fp|eia•P(a)da SIR Phasing H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)| 9 2 1 486 586 611 536 499 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(+)|

Radius of circle is approximately |Fp| SIR Phasing H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)| 9 2 1 486 586 611 536 499 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|

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

RCullis = |FPH|-|FP+fH| |FPH| - |FP| SIR Phasing H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)| 9 2 1 486 586 611 536 499 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= 586-486 = 100

SIR Phasing 90 2.40 1.40 0.40 -0.40 180 270 Phasing Power = |fH| H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)| 9 2 1 486 586 611 536 499 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)

What Phasing Power is sufficient to solve the structure? SIR Phasing H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)| 9 2 1 486 586 611 536 499 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?

Include information from anomalous scattering SIRAS

SIRAS Phasing |FP | -fHg(+) |FP·Hg(+)| FP=FP·Hg(+) - fHg(+) H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)| 9 2 1 486 586 611 536 499 Imaginary axis 500 |FP | 250 Real axis -500 -250 250 500 -fHg(+) -250 |FP·Hg(+)| -500

FP=FP·Hg(-)* - fHg(-)* SIRAS Phasing FP=FP·Hg(+) - fHg(+) FP=FP·Hg(-)* - fHg(-)* H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)| 9 2 1 486 586 611 536 499 Imaginary axis 500 |FP | |fHg|=282 aHg*=48° 250 Real axis 282 48° fHg(-)* -500 -250 250 500 -fHg(+) -250 |FP·Hg(+)| -500

FP=FP·Hg(-)* - fHg(-)* SIRAS Phasing FP=FP·Hg(+) - fHg(+) FP=FP·Hg(-)* - fHg(-)* H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)| 9 2 1 486 586 611 536 499 Imaginary axis 500 |FP | 250 Real axis -500 -250 250 500 -fHg(-)* -fHg(+) -250 |FP·Hg(-)| |FP·Hg(+)| -500

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 0 90 180 270 360 Real axis -250 250 500 -fHg(-)* -500 -fHg(+) -250 |FP·Hg(-)| |FP·Hg(+)| -500

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

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

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?

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..

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 P41212. 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.

FP= FP·Gd(-)* - fGd(-)* MIRAS Phasing H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)| 9 2 1 486 586 611 536 499 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 9 2 1 281 27 53° 100 24 -114°

SIR SIRAS MIRAS

SIR SIRAS MIRAS

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

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

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

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

Enter Phasing Statistics in Spreadsheet

It’s time for Model Building

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.

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.

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

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

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.

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

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

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

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

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

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

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[1 +1 +1 +1 ]+10[-1 -1 -1 -1 ] F111=18[4 ] +10[-4 ] F111=72 -40 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 +1+0i +1+0i +1+0i ]+10[-1+0i +-1+0i +-1+0i +-1+0i] F111=18[1 +1 +1 +1 ]+10[-1 -1 -1 -1 ]

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

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[1 +1 +1 +1 ]+10[ 1 +1 +1 +1 ] F200= 18[4 ] +10[+4 ] F200= 72 +40 F200=112 |F200 | =112 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 +1+0i +1+0i +1+0i] +10[1+0i +1+0i +1+0i +1+0i] F200=18[1 +1 +1 +1 ] +10[+1 +1 +1 +1 ]

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

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[1 -1 +1 -1 ]+10[ 1 +1 -1 -1 ] F100= 18[0 ] +10[0 ] F100=0 +0 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 -1+0i +1+0i -1+0i] +10[1+0i +1+0i -1+0i -1+0i] F100=18[1 -1 +1 -1 ] +10[+1 +1 -1 -1 ]

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

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

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

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[1 +1 +1 +1 ]+10[-1 -1 -1 -1 ] F111=18[4 ] +10[-4 ] F111=72 -40 F111=32 |F111 | =32 and a=0° I=32*32=1024

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[1 +1 +1 +1 ]+18[-1 -1 -1 -1 ] F111=18[4 ] +18[-4 ] F111=72 -72 F111=0 |F111 | =0 and a=0° I=0*0=0

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.

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

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

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

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

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