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Solving Crystal Structures From Two-wavelength X-ray Powder .._1_

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Solving Crystal Structures From Two-wavelength X-ray Powder .._1_ Powered By Docstoc
					                                   Phasing
Today’s goal is to calculate phases (ap) for proteinase K using
  MIRAS method (PCMBS and GdCl3).
What experimental data do we need?
    1) from native crystal we measured |Fp| for native crystal
    2) from PCMBS derivative we measured |FPH1(h k l)|
    3) also from PCMBS derivative we measured |FPH1(-h-k-l)|
    4) FH1 for Hg atom of PCMBS
    5) from GdCl3 derivative we measured |FPH2(h k l)|
    6) also from GdCl3 derivative we measured |FPH2(-h-k-l)|
    7) FH2 for Gd atom of GdCl3


How many phasing triangles will we have for each structure factor if
  we use all the data sets ?
FP ( h k l) = FPH ( h k l) - FH ( h k l) is one type of phase triangle.
FP ( h k l) = FPH(-h-k-l) * - FH (-h-k-l)* is another type of phase triangle.
          Four Phase Relationships
                                  PCMBS
FP ( h k l) = FPH1 ( h k l) - FH1 ( h k l) isomorphous differences
FP ( h k l) = FPH1(-h-k-l)* - FH1 (-h-k-l)*   from Friedel mates (anomalous)




                                     GdCl3
FP ( h k l) = FPH2( h k l) - FH2 ( h k l)     isomorphous differences

FP ( h k l) = FPH2 (-h-k-l)* - FH2 (-h-k-l)* from Friedel mates (anomalous)
                  FP ( h k l) = FPH1 ( h k l) - FH1 (h k l)
                                   Imaginary axis
|Fp ( h k l) | = 1.8
                                            2



                                            1



                                                                                   Real axis

                       -2     -1                    1   2


                                            -1



                                            -2




                                                            Harker construction

                                                            1) |FP ( h k l) | –native measurement
                 FP ( h k l) = FPH1 ( h k l) - FH1 ( h k l)
                                 Imaginary axis
|Fp ( h k l) | = 1.8
|FH ( h k l) | = 1.4
aH (hkl)       = 45°                              |Fp |


                                                                            Real axis
          FH




                                                      Harker construction

                                                      1) |FP ( h k l) | –native measurement
                                                      2) FH (h k l) calculated from heavy
                                                      atom position.
                      FP ( h k l) = FPH1 ( h k l) - FH1 ( h k l)
                                        Imaginary axis
|FP ( hkl) | = 1.8
|FH1 ( hkl) | = 1.4
aH1 (hkl) = 45°                                           |Fp |
|FPH1 (hkl) | = 2.8
                                |FPH|
                                                                                        Real axis


                                             FH

                                                  |FPH|


                                                                  Harker construction
 Let’s look at the quality                                        1) |FP ( h k l) | –native measurement
                                                                  2) FH1 (h k l) calculated from heavy
 of the phasing statistics                                        atom position.
                                                                  3) |FP1(hkl)|–measured from
      up to this point.                                           derivative.
                                                       Which of the following
    SIR                                                graphs best represents
                                                        the phase probability
        Imaginary axis                                    distribution, P(a)?

                          |Fp |
                                              a)
                                                   0       90    180   270   360
|FPH|
                                  Real axis


             FH                               b)
                  |FPH|                            0       90   180    270   360



                                              c)

                                                   0       90   180    270   360
                                                   The phase probability
    SIR                                       distribution, P(a) is sometimes
                                                 shown as being wrapped
        Imaginary axis                          around the phasing circle.

                          |Fp |


                                                  0     90    180   270   360
|FPH|
                                  Real axis


             FH

                  |FPH|                                      90

                                                  180                0


                                                             270
                                               Which of the following is
      SIR                                       the best choice of Fp?

        Imaginary axis
                                                                  90



                          |Fp |                   a)       180          0


            90                                                   270



|FPH|
180                       0       Real axis
                                                                  90

             FH                                   b)       180          0


                  |FPH|                                          270
           270
                                                                 90
                                                  c)
                                                          180          0


                                                                 270

                                              Radius of circle is approximately |Fp|
      SIR
        Imaginary axis


                    |Fp |
                                    best F =         |Fp|eia•P(a)da
            90                                   a

|FPH|
180                         0   Real axis

                                                 Sum of
             FH
                                               probability
                  |FPH|
           270                                  weighted
                                               vectors Fp
                                            Usually shorter than Fp
      SIR
                best F =       |Fp|eia•P(a)da
       90                  a


180         0
                         Sum of
                       probability
      270               weighted
                       vectors Fp

 Best phase
      SIR        Which of the following is
                  the best approximation
                   to the Figure Of Merit
                 (FOM) for this reflection?

       90
                         a)    1.00
                         b)    2.00
180         0
                         c)    0.50
                         d)    -0.10
      270


                  FOM=|Fbest|/|FP|
                Radius of circle is approximately |Fp|
Which phase probability distribution would
 yield the most desirable Figure of Merit?


a)                      b)
                                     0
           90

                 0
                               270
                                     +
                                     +     90


     180                             180




           270                        0
                          c)
                               270


                                 +
                                  +        90


                                     180
    SIR                                          Which of the following is
                                                  the best approximation
        Imaginary axis
                                                 to the phasing power for
                                                      this reflection?


              |Fp |
                                                      a)   2.50
|FPH|                 Fbest                           b)   1.00
                                   Real axis

                      |Fp |
                                                      c)   0.50
            FH
                                                      d)   -0.50
                 |FPH|
                                     Phasing Power =               |FH|
                                                              Lack of closure


                                                |FH ( h k l) | = 1.4
                          Lack of closure = |FPH|-|FP+FH| = 0.5
                          (at the aP of Fbest)
    SIR                                        Which of the
        Imaginary axis                    following is the most
                                            desirable phasing
                                                 power?

              |Fp |
                                              a)   2.50
|FPH|                 Fbest                   b)   1.00
                              Real axis

                      |Fp |
                                              c)   0.50
            FH
                                              d)   -0.50
                 |FPH|
                               Phasing Power =             |FH|
                                                      Lack of closure




What Phasing Power is sufficient to solve the structure?        >1
|FP ( hkl) | = 1.8
                      SIR                        Which of the following is
                        Imaginary axis
|FPHg (hkl) | = 2.8
                                                    the RCullis for this
                                                       reflection?
                              |Fp |
                                                              a)   -0.5
                |FPH|                 Fbest                   b)   0.5
                                              Real axis

                                      |Fp |
                                                              c)   1.30
                            FH
                                                              d)   2.00
                                 |FPH|
                                                  RCullis =   Lack of closure

                                                          isomorphous difference

                                                 From previous page, LoC=0.5
                                              Isomorphous difference= |FPH| - |FP|
                                                                  1.0 = 2.8-1.8
SIRAS           Isomorphous differences
                 Anomalous differences
              Imaginary axis


                               FP ( h k l) = FPH(-h-k-l)* - FH (-h-k-l)*



                                                             Real axis



                                         We will calculate SIRAS phases
                                         using the PCMBS Hg site.

                                         FP –native measurement
                                         FH (hkl) and FH(-h-k-l) calculated
                                         from heavy atom position.
                                         FPH(hkl) and FPH(-h-k-l) –measured
                                         from derivative. Point to these on
                                         graph.


  To Resolve the phase ambiguity
SIRAS                          Isomorphous differences
                                Anomalous differences

  Imaginary axis




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




                                     0     90   180   270   360
SIRAS                          Isomorphous differences
                                Anomalous differences

  Imaginary axis



                               Which graph represents FOM of SIRAS phase?
                                   Which represents FOM of SIR phase?
                                         Which is better? Why?

                   Real axis



                                           90                90

                                     180         0     180         0


                                           270               270
Calculate phases and refine
  parameters (MLPhaRe)
    A Tale of Two Ambiguities
• We can solve both ambiguities in one
  experiment.
                Center of inversion ambiguity
                     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
Patterson map        coordinates (x,y,z→ -x,-y,-z)
    Choice of origin ambiguity
• I want to include the Gd data (derivative 2)
  in phase calculation.
• I can determine the Gd 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, fP, a protein
     electron density map could be calculated.
The amplitudes would be |FP|, the phases would be fP.
           r(x)=1/V*S|FP|e-2pi(hx+ky+lz-fP)




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-fP)

     What is the second heavy atom, Alex.
When the difference FPH2-FP is taken, the protein
  component is removed 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 by
   producing a very high peak when phases are
   calculated in the correct hand, but only noise when
   phases are calculated in the incorrect hand.
2) It will allow us to find the position of the second
   heavy atom and combine this data set into our
   phasing. Thus improving our phases.
               Phasing Procedures

1) Calculate phases for site x,y,z of Hg and run
   cross difference Fourier to find the Gd site. Note
   the height of the peak and Gd 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 Gd site. Compare the height
   of the peak with step 1.
3) Chose 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.
FH1(hkl)
FPH1(hkl)
FH1(-h-k-l)*
FPH1(-h-k-l)*
                              MIRAS
                               Imaginary axis

FH2(hkl)                                        Isomorphous
                                                   Deriv 2
FPH2(hkl)



                  Anomalous              Fp (hkl)
                    Deriv 1                                     Real axis




                         Isomorphous
                            Deriv 1




     GdCl3      FPH = FP+FH            for isomorphous differences
FH1(hkl)
FPH1(hkl)             GdCl3                FP ( h k l) = FPH1 (-h-k-l)* - FH1 (-h-k-l)*
FH1(-h-k-l)*
FPH1(-h-k-l)*
                                          Imaginary axis

FH2(hkl)                                                   Isomorphous
                                                              Deriv 2
FPH2(hkl)
FH2 (-h-k-l)*
FPH2(-h-k-l)*
                           Anomalous                Fp (hkl)
                             Deriv 1                                                 Real axis




                                    Isomorphous                    Anomalous
                                       Deriv 1                       Deriv 2




Harker Construction for MIRAS phasing (Multiple Isomorphous Replacement with Anomalous Scattering)
FH1(hkl)
FPH1(hkl)                                                   For opposite hand
FH1(-h-k-l)*
FPH1(-h-k-l)*                    Isomorphous
                                    Deriv 1       Imaginary axis

FH2(hkl)
FPH2(hkl)                                                                    Anomalous
                                                                               Deriv 2
FH2 (-h-k-l)*
FPH2(-h-k-l)*
                                                            Fp (hkl)
                                                                                                   Real axis




                Anomalous
                  Deriv 1
                                                                               Isomorphous
                                                                                  Deriv 2


 Note: in the presence of anomalous scattering, FH(HKL) ≠ FH(-H-K-L)* so if this opposite hand is the incorrect
hand, the phase angle (aopposite) will be significantly different than (-aoriginal). The electron density would not
                                just be inverted, but inverted and more noisy.
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
                    MIR phased map +
MIRAS phased map   Solvent Flattening +
                   Histogram Matching
                    MIR phased map +
MIRAS phased map   Solvent Flattening +
                   Histogram Matching
                           Density modification
                                                                   B) 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
Each grid point has a value r(x,y,z)




                     Y



                             X
                    Z
    represent r(x,y,z) in shades of gray




Y



       X
Z                    How many grid points have density 0?
                     How many grid points have density 1?
                     How many grid points have density x?
       ntotal=130 pixels total

r=0 n=60    r=1 n=6     r=2 n=17   r=3 n=5   r=4 n=1




r=5 n=2     r=6 n=12    r=7 n=2    r=8 n=4   r=9 n=1




r=10 n=25
   Histogram of electron density
         distribution P(r).


Number of
pixels with
density (r)




              0 1 2 3 4 5 6 7 8 9 10
                    Density (r)
                        Cumulative density distribution
                               function N(r).




                                                 Number of pixels with density < r
Number of pixels with
    density (r)




                        0 1 2 3 4 5 6 7 8 9 10
                                                                                     0 1 2 3 4 5 6 7 8 9 10
                               Density (r)
                                                                                     Cumulative density
                                        Histogram matching density
                                         distribution function N(r).




                                                                                  Number of pixels with density < r
Number of pixels with density < r




                                    Cumulative density                                                                Cumulative density
                                       observed                                                                             ideal
                                      Pixels with r=4 in original map will be modified to r= 2.5 in the histogram matched map.
HOMEWORK
      Barriers to combining phase
     information from 2 derivatives
1) Initial Phasing with PCMBS
   1) Calculate phases using coordinates you determined.
   2) Refine heavy atom coordinates
2) Find Gd site using Cross Difference Fourier map.
   1) Easier than Patterson methods.
   2) Want to combine PCMBS and Gd 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 Gd 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

				
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posted:9/2/2012
language:English
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