Isomorphous replacement (MIR), cont’d
02-08-06
Lecture 8
Effect of heavy atoms on X-ray intensities
• How much difference should we expect by
the attachment of one or a few heavy
atoms?
• Crick and Magdoff (1956) estimated that:
– For centric reflections
– For acentric reflections
And we know that I is proportional to ∑fi2
Effect of heavy atoms on X-ray intensities
Attachment of heavy atom causes changes
in intensities
Native lysozyme
A Hg derivative
There is no change in reciprocal lattice spacing
Another example
Native phosphorylase
A Hg derivative
The problem of non-isomorphism
• Ideally, the attachment of heavy atoms only
substitutes disordered solvent molecules without
causing any structural changes to the
crystallized molecule.
• Non-isomorphism can be detected by
– Changes in unit cell dimensions
• Should be <0.5% along each cell axis
– Changes in intensities
• Orientational changes or conformational changes
• Ideal case: 10-20% difference throughout
different resolution shells
The problem of non-isomorphism – an example
Changes in molecule orientation result in changes in diffraction
intensities
Determination of heavy atom positions
• Isomorphous difference Patterson map will
give the Patterson function for the heavy
atom structure on half the scale.
– See the next slide for more information
• Heavy atom positions can be solved from
the isomorphous different Patterson map.
– See the slide after next for a simple example
Isomorphous difference Patterson
• Patterson function
• Isomorphous difference Patterson function
where
Δ|F|iso = |FPH| - |FP|
Isomorphous difference Patterson can be
used to solve heavy atom structure - a proof
Therefore,
An example: how to determine heavy atom positions
a
(-x2,y2) (x2,y2)
(-x1,y1) (x ,y )
1 1
y1-y2
2x1 2x2
b
A 2-fold axis along b axis
Space group symmetry: Pm
Two atoms in each asymmetric unit
We will have four peaks in each
asymmetric unit (shown in green)
(2x1, 0) (2x2, 0)
(x1-x2, y1-y2)
(x2-x1, y2-y1)
Use the Patterson map to
solve heavy atom positions
Patterson map symmetry:
P2/m
The origin along y axis is arbitrary in this Pm space group
Now FH can be calculated after the structure
of the heavy atoms are determined.
FH is the structure factor for the heavy atom
part of the derivative.
And we have:
FPH = FP + FH
Single isomorphous replacement
• For a centrosymmetric structure, phase angles are either
π or 0
– Phases can be determined for one heavy atom derivative
Assuming FP and FPH lie in the
same direction:
(a) If FPH> FP, then FP has the
same sign as FH
(b) If FPH< FP, then FP has the
opposite sign as FH
Only rarely does FPH and FP
have opposite signs,
resulting in:
(c) FP has the opposite sign as
FH
Single isomorphous replacement
•
For non-centrosymmetric structure
– Two phases are possible
FP + FH = FPH
(1) Draw a circle using radius FP
(2) Draw a vector FH with the arrowhead ending on
origin
(3) Use the starting point of FH as origin, draw a circle
using FPH as radius
(4) The two circles intercept at two points G and H,
giving two phase angles
Multiple isomorphous replacement
•
In theory, phases can be
unambiguously determined if
there are at least two heavy
atom derivatives.
– Circle FP and FPH1 intercept at
G and H
– Circle FP and FPH2 intercept at
H and L
– Therefore, the correct phase is
given by point H
•
In practice, measurement
errors and non-isomorphism
result in phase ambiguity and
lack of closure error.
Lack of closure error
If the FH + FP is not equal to FPH, then
ε is called the lack of closure error
The lack of closure error can be caused
by inaccuracy in data collection and nonisomorphism
During heavy atom refinement, our goal
is to minimize the lack of closure error by
adjusting heavy atom’s atomic
coordinates, occupany, and temperature
factor
Protein phase angles
To consider the lack of close error, a Gaussian
probability distribution is assumed for ε, and for
each reflection of each derivative:
α varies from 0 to 2π
N is the normalization factor
E is the root mean square of ε
If there are n derivatives, then
Phase probability curve
First example
Reflection j
One derivative only
Reflection j
Two derivatives
If there are n derivatives, then
Reflection k
The most probable phase and the best phase
αbest
αprobable
αprobable
αbest
The most probable phase is the phase with the highest
probability (this does not consider the shape of phase
probability curve)
The best phase, αbest, is used in calculating electron
density maps
The best phase and figure of merit
m is called the figure of merit
Figure of merit is a weighting factor
The value of m varies between
0 and 1. Phases with large m
values are more accurate.
Electron density calculation
Figure of merit m is like a weighting factor, the more
accurate the phases are, the larger the values of m are.
Electron density map
Protein coordinates are modeled based on
sequence and stereochemistry information