Today: compute the experimental electron density

map of proteinase K

Fourier synthesisr(xyz)=S |Fhkl| cos2p(hx+ky+lz -ahkl)

hkl

3 Crystals5 Measured Quantities

H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)|)1 1 1 681.4 725.8 722.4 730.8 707.61 1 2 752.8 733.6 695.3 813.9 805.31 1 3 332.1 444.5 456.2 296.1 312.51 1 4 526.9 575.8 564.7 527.4 518.31 1 5 719.2 827.8 805.4 759.6 766.31 1 6 358.4 349.8 354.2 375.6 358.91 1 7 273.3 359.4 390.8 300.5 286.61 1 8 400.7 362.5 411.2 396.7 411.51 2 0 162.5 73.8 132.3 149.8 159.8

Native PCMBS (Hg) EuCl3 (Eu)

Measureable differences

2) Anomalous Differences (i.e. differences between Friedel pairs):

3) Dispersive differences (differences due to changing the

wavelength):

FP(hkl)=FPH (hkl)

- fH(hkl)

1) Isomorphous differences(between native and derivative)

FP(hkl)=FPH (hkl) ln

- fH(hkl) ln

FP(hkl)=FPH(-h-k-l)

* - fH(-h-k-l)*

3 Crystals5 Measured Quantities

H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)|1 1 1 681.4 725.8 722.4 730.8 707.61 1 2 752.8 733.6 695.3 813.9 805.31 1 3 332.1 444.5 456.2 296.1 312.51 1 4 526.9 575.8 564.7 527.4 518.31 1 5 719.2 827.8 805.4 759.6 766.31 1 6 358.4 349.8 354.2 375.6 358.91 1 7 273.3 359.4 390.8 300.5 286.61 1 8 400.7 362.5 411.2 396.7 411.51 2 0 162.5 73.8 132.3 149.8 159.8

Native PCMBS (Hg) EuCl3 (Eu)

Vector equations for this experiment

Isomorphous and

Anomalous Differences

Isomorphous and

Anomalous Differences

FP(hkl)=FPHg (hkl)

- fHg(hkl)

FP(hkl)=FPHg(-h-k-l)

* - fHg(-h-k-l)*

FP(hkl)=FPEu (hkl)

- fEu(hkl)

FP(hkl)=FPEu(-h-k-l)

* - fEu(-h-k-l)*

PCM

BS

(Hg)

For E

uCl 3

(Eu)

Vector equations for this experiment

Isomorphous Differences FP(hkl)=FPHg (hkl)

- fHg(hkl)

We have collecting data on the native and derivative crystals.We know the coordinates of Hg.How many unknown quantities remain?

SIR Phasing

500

500

-500

-250

-500

250-250

250

H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)|

9 2 1 486 586 611 536 499

Real axis

Imaginary axis

|FP |

FP=FP·Hg(+) - fHg(+)

SIR Phasing

500

500

-500

-250

-500

250-250

250

H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)|

9 2 1 486 586 611 536 499 Imaginary axis

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)]

|FP |

FP=FP·Hg(+) - fHg(+)

Real axis

f’ and f” are anomalous scattering corrections specific for wavelength used.

x,y,z are Hg coordinates from Patterson map (0.197, 0.755, 0.935)

fHg is a real number proportional to the number of e- in Hg

SIR Phasing

500

500

-500

-250

-500

250-250

250

H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)|

9 2 1 486 586 611 536 499 Imaginary axis

No!fHg=fHg[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)]+f’+if”

|FP |

FP=FP·Hg(+) - fHg(+)

Real axis

fHg ≠ |FP|-|FPH (+)|

f’ and f” are anomalous scattering corrections specific for wavelength used.

x,y,z are Hg coordinates from Patterson map (0.197, 0.755, 0.935)

fHg is a real number proportional to the number of e- in Hg

SIR Phasing

500

500

-500

-250

-500

250-250

250

H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)|

9 2 1 486 586 611 536 499

|fHg|=282 aHg=58°

Imaginary axis

|FP |

FP=FP·Hg(+) - fHg(+)

-

Real axis

282

58°fHg(+)

SIR Phasing

500

500

-500

-250

-500

250-250

250

H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)|

9 2 1 486 586 611 536 499 Imaginary axis

|FP |

FP=FP·Hg(+) - fHg(+)

-fHg(+)

Real axis

|FP·Hg(+)|Let’s look at the quality of the phasing statistics

up to this point.

SIR Phasing H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)|

9 2 1 486 586 611 536 499

0 90 180 270 360

0 90 180 270 360

0 90 180 270 360

Which of the following graphs best represents the phase probability

distribution, P(a)?

a)

b)

c)

500

500

-500

-250

-500

250-250

250

Imaginary axis

|FP |

-fHg(+)Real axis

|FP·Hg(+)|

SIR Phasing H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)|

9 2 1 486 586 611 536 499

500

500

-500

-250

-500

250-250

250

Imaginary axis

|FP |

-fHg(+)Real axis

|FP·Hg(+)|

0 90 180 270 360

The phase probability distribution, P(a) is sometimes

shown as being wrapped around the phasing circle.

90

0180

270

SIR Phasing H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)|

9 2 1 486 586 611 536 499Which of the following is the best choice of Fp?

a)

b)

c)

Radius of circle is approximately |Fp|

500

500

-500

-500

Imaginary axis

|FP |

Real axis

|FP·Hg(+)|

90

0180

270

90

0180

270

90

0180

270

90

0180

270

500

SIR Phasing H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)|

9 2 1 486 586 611 536 499

500

-500

-500

Imaginary axis

|FP |

Real axis

|FP·Hg(+)|

best FP = |Fp|eia•P(a)daa

Sum of probability weighted vectors Fp

Usually shorter than Fp

0

90

180

270

500

SIR Phasing H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)|

9 2 1 486 586 611 536 499

500

-500

-500

Imaginary axis

|Fbest | Real axis

|FP·Hg(+)|

best FP = |Fp|eia•P(a)daa

Sum of probability weighted vectors Fp

Usually shorter than Fp

0

90

180

270

90

0180

270

SIR

Sum of probability weighted vectors Fp

Best phase

best F = |Fp|eia•P(a)daa

90

0180

270

SIR

a) 1.00b) 2.00c) 0.50d) -0.10

Which of the following is the best approximation to the Figure Of Merit

(FOM) for this reflection?

FOM=|Fbest|/|FP|Radius of circle is approximately |Fp|

Which phase probability distribution would yield the most desirable Figure of Merit?

0

90

180

270 ++

0

90

180

270 ++

a) b)

c)

90

0180

270

Fbest|FPH|

|FPH|

Imaginary axis

SIR

Real axis

|Fp | a) 2.50b) 1.00c) 0.50d) -0.50

Which of the following is the best approximation

to the phasing power for this reflection?

Lack of closure = |FPH|-|FP+FH| = 0.5(at the aP of Fbest)

|Fp |fH

|fH ( h k l) | = 1.4

Phasing Power = |fH| Lack of closure

Fbest|FPH|

|FPH|

Imaginary axis

SIR

Real axis

|Fp | a) 2.50b) 1.00c) 0.50d) -0.50

Which of the following is the most

desirable phasing power?

|Fp |fH

What Phasing Power is sufficient to solve the structure? >1

Phasing Power = |fH| Lack of closure

fH

SIR

a) -0.5b) 0.5c) 1.30d) 2.00

Which of the following is the RCullis for this

reflection?

RCullis = Lack of closure isomorphous difference

From previous page, LoC=0.5Isomorphous difference= |FPH| - |FP|

1.0 = 2.8-1.8

|FP ( hkl) | = 1.8|FP•Hg (hkl) | = 2.8

Fbest

Imaginary axis

Real axis

|Fp |

|Fp |

|FPH|

|FPH|

SIRAS Phasing

500

500

-500

-250

-500

250-250

250

H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)|

9 2 1 486 586 611 536 499 Imaginary axis

|FP |

FP=FP·Hg(-)* - fHg(-)*

Real axis

|fHg|=282 aHg*=48°

282

48°

fHg(-)*

|FP·Hg(+)|

-fHg(+)

SIRAS Phasing

500

500

-500

-250

-500

250-250

250

H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)|

9 2 1 486 586 611 536 499 Imaginary axis

|FP |

FP=FP·Hg(-)* - fHg(-)*

Real axis

-fHg(-)*

|FP·Hg(+)||FP·Hg(-)|

-fHg(+)

Isomorphous differencesAnomalous differencesSIRAS

0 90 180 270 360

Which P(a) corresponds to SIR?Which P(a) corresponds to SIRAS?

Remember, because the position of Hg was determined using a Patterson map there is an ambiguity in handedness.

The Patterson map has an additional center of symmetry not present in the real crystal. Therefore, both the site x,y,z and -x,-y,-z are equally consistent with Patterson peaks.

Handedness can be resolved by calculating both electron density maps and choosing the map which contains structural features of real proteins (L-amino acids, right handed a-helices).

If anomalous data is included, then one map will appear significantly better than the other.

Note: Inversion of the space group symmetry (P43212 →P41212) accompanies inversion of the coordinates (x,y,z→ -x,-y,-z)

Center of inversion ambiguity

Patterson map

Choice of origin ambiguity

• I want to include the Eu data (derivative 2) in phase calculation.

• I can determine the Eu site x,y,z coordinates using a difference Patterson map.

• But, how can I guarantee the set of coordinates I obtain are referred to the same origin as Hg (derivative 1)?

• Do I have to try all 48 possibilities?

Use a Cross difference Fourier to resolve the handedness ambiguity

With newly calculated protein phases, aP, a protein electron density map could be calculated.

The amplitudes would be |FP|, the phases would be aP. r(xyz)=1/V*S|FP|e-2pi(hx+ky+lz-a

P)

Answer: If we replace the coefficients with |FPH2-FP|, the result is an electron density map corresponding to this structural feature.

r(x)=1/V*S|FPH2-FP|e-2pi(hx-aP

)

What is the second heavy atom, Alex.When the difference FPH2-FP is taken, the protein contribution

is cancelled and we are left with only the contribution from the second heavy atom.

This cross difference Fourier will help us in two ways:1) It will resolve the handedness ambiguity

-high peak in difference map calculated with aP in correct hand-only noise in difference map calculated with aP in incorrect

hand.2) It will improve our electron density map of the protein

-identify the position of the 2nd heavy atom -include 2 new vector equations for Eu (more accurate aP)

Phasing Procedures

1) Calculate phases for site x,y,z of Hg and run cross difference Fourier to find the Eu site. -Note the height of the peak and Eu coordinates.

2) Negate x,y,z of Hg and invert the space group from P43212 to P41212. Calculate a second set of phases and run a second cross difference Fourier to find the Eu site. -Compare the height of the peak with step 1.

3) Chose the handedness which produces the highest peak for Eu. Use the corresponding hand of space group and Hg, and Eu coordinates to make a combined set of phases.

|FP | = 486 ± 2

MIRAS Phasing

500

500

-500

-250

-500

FP=FP·Hg(+) - fHg(+)

250-250

250

H K L |FP| |FPH (+)| |FPH (-)| |FPH (+)| |FPH (-)|

9 2 1 486 586 611 536 499

H K L fH+f’ f”(-) aH fH+f’ f” aH9 2 1 281 27 53° 100 24 -114°

FP=FP·Hg(-)* - fHg(-)*

Real axis

Imaginary axis

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

MIR phased map + Solvent Flattening + Histogram Matching

MIRAS phased map

MIR phased map + Solvent Flattening + Histogram Matching

MIRAS phased map

Density modificationB) 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

HOMEWORK

Barriers to combining phase information from 2 derivatives

1) Initial Phasing with PCMBS1) Calculate phases using coordinates you determined.2) Refine heavy atom coordinates

2) Find Eu site using Cross Difference Fourier map.1) Easier than Patterson methods. 2) Want to combine PCMBS and Eu to make MIRAS phases.

3) Determine handedness (P43212 or P41212 ?)1) Repeat calculation above, but in P41212.2) Compare map features with P43212 map to determine

handedness. 4) Combine PCMBS and Eu sites (use correct hand of

space group) for improved phases.5) Density modification (solvent flattening & histogram

matching)1) Improves Phases

6) View electron density map