OU Crystallography Lab

Department of Chemistry & Biochemistry
Chemical Crystallography Laboratory

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Structure Solution

A crystal structure is considered solved when the phases of enough reflections are known well enough to reveal most if not all of the atoms in the unique part of the unit cell. There are several ways to solve the crystal structure of a small molecule compound. The technique most commonly used today is called direct methods. Another series of common methods still in use are based on the Patterson function. Most Patterson methods are significantly aided by the presence of one or more heavy atoms in the structure. In addition to these methods, difficult small-molecule structures are sometimes solved either by superposition maps or by rotation and translation functions, that are both very specialized extensions of the Patterson method.

Solving protein crystal structures are usually more challenging. If the crystal structure of a similarly sized protein, having similar cell parameters and space group symmetry, is already known, then the phases from the previously known structure may be used as a starting point to solve the current problem. This method is known as molecular replacement. Sometimes the solutions of heavy atom compounds are soaked into crystals in hopes that the heavy atoms will settle in a few specific locations of the crystals. The locations of the heavy atoms are determined and these phases are used to bootstrap the structure of the protein. Usually this method requires that two or more such heavy atom derivatives be prepared giving this method the name multiple isomorphous replacement. At synchrotrons, the wavelength can be tuned to enhance the anomalous scattering of a given atom type (e.g. Se that has replaced S in methionines), and anonalous scattering is used to enhance the phasing power of one or more heavy atom derivatives. The methods to solve protein crystal structures are beyond the scope of these notes; however, a good introduction to these methods is presented at Bernhard Rupp’s site in the Phasing Techniques section.

Table of Contents

Direct Methods

H. Schenk has prepared an introduction to Direct Methods.

Currently nearly all small-molecule crystal structures are solved by direct methods. Direct methods use probability relationships to assign phases to a small subset of the data. From modified electron density maps based on this subset data it is possible to extract enough of the atom positions to consider that the structure is solved.

The essential core of direct methods assumes that there is information about the phases contained in the structure factor amplitudes. Two general features of crystal structures have led to the development of a broad range of mathematical relations among the structure factor phases based upon knowing the values of the amplitudes.

  1. The electron density of the correct model must be ≥ 0 throughout the unique volume of the unit cell (positive electron density condition).
  2. The structure is composed of discrete atoms (discrete atom condition).

Positive electron density condition -- If totally random phases are input into the electron density function, then it is unlikely that the resulting electron density would have ρ(r) ≥ 0 for all r. This criterion of positive electron density led Karle and Hauptman in 1950 to derive numerous inequalities of determinants that relate the phase angles of different structure factors to one another. They later received a Nobel prize for their pioneering work on phase determination in crystallography.

Discrete atom condition -- In 1952, three researchers (Sayre, 1952; Cochran, 1952; and Zachariasen, 1952) independently arrived at an important relation involving the phases using the discrete atom condition. Sayre’s derivation is the most clear and is given below.

For structures with well resolved, equal atoms, Sayre observed that the functions ρ(r) and ρ2(r) are quite similar and show maxima at the same positions.

The Fourier transform of ρ(r) is (1/V)Fh. The structure factor Fh is related to the positions of the atoms by:

Fh = Σ fh exp[2πi(h · rj)]

where the summation j runs over the N data. For equal atom structures this becomes:

Fh = fh Σ exp[2πi(h · rj)]

A similar expression can be written for the Fourier transform of ρ2(r)

Gh = gh Σ exp(2πih · rj)

where gh is the scattering factor of the squared atom.

It can be shown using convolutions that

T[g(x)*h(x)]=T[g(x)] * T[h(x)].

Thus the transform of ρ2(r), that is equal to (1/V)Gh, is also equal to (1/V)Fh * (1/V)Fh. Since Fh is a discrete function defined only at the points of the reciprocal lattice, the convolution becomes a summation:

Gh = (1/V) Σ Fk Fh-k

where the summation is over the k peaks. From the expressions for Fh and Gh above then

Fh = (fh / gh) Gh = θh Gh

Fh = (θh / V) Σ Fk Fh-k

which is Sayre’s equation. Multiplying both sides of this last expression by F-h gives

|Fh|2 = (θh / V) Σ |FhFk Fh-k| exp[i(φ-h + φk + φh-k)]

where the summation is over the k reflections. Peaks with large values of |Fh| will also have large values for |Fh|2 and presumably at least some terms in the right hand expression with large values for |Fk| and |Fh-k|. For these terms in the right hand summation it follows that

φ-h + φk + φh-k = 0

For structures in centrosymmetric space groups this expression becomes

S(-h) · S(k) · S(h-k) = +

where S(h) stands for the sign of reflection h. In the last two expressions the = indicate only approximate equalities. The probability that the above expressions are true increases with increasing values for |Fh|, |Fk|, and |Fh-k|.

Direct methods work with structure factors that have been modified to behave as if they denote the scattering from point atoms located at the same positions as the original atoms. The modified structure factors, called E values or normalized structure factors, are calculated as shown below.

Ehkl2 = Fhkl2 / (εfj2)

where fj = fjo exp(-B sin2θ / λ2) is the scattering factor for the jth atom and ε is an integer, 1 or greater, that corrects for the fact that some classes of reflections have expectation values that are less than ∑ fj2 by an integer amount. Note that the phases associated with the Ehkl values are the same as the phases associated with the Fhkl values.

It has been found that only a fraction of the whole data set are needed to correctly identify an initial model of the structure. The data with the strongest E values contain the most information about the locations of the atoms. Because of these facts, about 10% of the whole data set having the strongest E values are chosen to carry out direct methods.

Structure Invariants

For a given basis system describing the unit cell, let rj be the positional vector of the jth atom. From this reference system the structure factor would be

Fh = ∑ fj exp(2πi h · rj)

where the summation j runs over the N atoms in the cell. Consider the effect on the structure factor if the origin is shifted by q. With this shift of the origin the positional vector of the jth atom becomes rj’ = rj - q and the structure factor expression becomes

Fh’ = ∑ fj exp[2πi h · rj’)] =
fj exp(2πi h · (rj - q) =
Fh exp(-2πi h · q)

From this relation, it is clear that the structure factor modulus does not change with an origin shift, and that the phase value changes according to

φh’ = φh - 2πi h · q

Thus the structure factor amplitudes are said to be structure invariants because they are not changed with a shift or translation in the unit cell origin.

Do any structure factors exist whose phase would not change with an origin shift? From the above expression only the phase of F000 = ∑ fj is invariant to any origin translation.

Are there functions involving the phases that would remain invariant of any origin translation? Consider the product

Fh1 Fh2 ... Fhn = |Fh1 Fh2 ... Fhn| exp[i(φ1 + φ2 + ... + φn)]

According to the phase relation above the product of structure factors transformed by an origin shift would become

Fh1Fh2’ ... Fhn’ =
Fh1 Fh2 ... Fhn exp[-2πi (h1 + h2 + ... + hn) · q]

This suggests that the product of structure factors would be invariant to an origin shift if

h1 + h2 + ... + hn = 0

Products of structure factors that satisfy the last expression are called structure invariants, since their values do not depend on the origin, and therefore depend only on the structure (Hauptman & Karle, 1953; Giacovazo, 1998). The phase relations developed by Sayre is also a structure invariant.

Several structure invariants include:

Current direct methods programs operate successfully the vast majority of time. Modern direct methods combine a variety of methods to generate phases usually starting from a random phase set and then refining the phases with an annealing step and with some tangent refinement and extension. The tangent refinement formula is shown below.

tan(φhkl) = {∑κ(H,K) sin[φ(K) + φ(H - K)]} / {∑κ(H,K) cos[φ(K) + φ(H - K)]}

Patterson Methods

In 1935, A. L. Patterson published a classic paper on the utility of a Fourier map that uses |F|2 as coefficients and phase angles all assumed to be 0°.(Patterson, 1935) He demonstrated that such a map gave peaks corresponding to all vectors between any given pair of points.

P(u,v,w) = ∑ Fhkl2cos[2π(hu+kv+lw)]

The peak heights were found to be proportional to Zi·Zj. Thus a point at uvw in a Patterson map indicates that there are atoms in the crystal at (x1y1z1) and at (x2y2z2) such that

u = x1 - x2,      v = y1 - y2,      w = z1 - z2

For a crystal structure with N atoms, there will be N2 peaks in the Patterson map. N of these peaks will be peaks of zero length corresponding to the vector of given point to itself. The remaining N2 - N peaks are distributed throughout the cell. Since the cell of the Patterson function is the same size as the cell of the crystal, the Patterson function is much more densely packed than the corresponding electron density map. This higher density of peaks causes many peaks in the Patterson map to be overlapped. The greater intrinsic breadth of Patterson peaks accentuates this overlap. Vectors that are connected between the atom sites, each with a definite width, cause the vector peaks to be as broad as the sum of the widths of the two atom peaks.

Consider a structure with only one heavy atom, say an iodine atom with Z = 53 and the remaining atoms no heavier than an oxygen with Z = 8. Then the Zlight * Zlight peaks would have intensities roughly proportional to 64. The Zlight * Zheavy peaks would have intensites roughly proportional to 424 and the Zheavy * Zheavy peaks would have intensities roughly proportional to 2809. Thus the vectors that have at least one end on a heavy atom are relatively easy to identify in the Patterson map.

Because of the many overlaps and the broad peaks, Patterson maps tend to be an almost featureless distribution of peak density. To remedy this problem the |F|2 values are typically modified to make the scattering appear to come from point atoms. In modern programs this sharpening of the map is often accomplished by using E2 values instead of |F|2.

A less serious problem with Patterson maps is the very large peak at the origin. Patterson showed that it was possible to subtract the origin peak from the map by subtracting the average value of |F|2 from the coefficients.

|F|2origin removed = |F|2 - ∑ fj2

where fj is the scattering factor for the jth atom. For a Patterson function sharpened to resemble scattering from point atoms, the fj values become equal to Zj. E2 values are calculated to have an average value of 1. So Patterson maps calculated using E2 - 1 as coefficients will have the large peak at the origin removed.

Patterson maps have the same symmetry as that of the corresponding Laue group for the sample. Patterson maps are centrosymmetric because the vectors between the atoms can point both directions. Although symmetry elements of the crystal’s space group do not necessarily appear as such in the Patterson map, they do leave their traces in the form of higher concentrations of peaks. The locations to look for these stronger peaks are called Harker lines and planes (Harker, 1936). These regions of the Patterson map correspond to the vectors between atoms related by the space group symmetry of the structure. For example in P21 the symmetry operators are x, y, z and -x, ½+y, -z. In the Patterson map there should be a concentration of vectors at 2x, ½, 2z in the v = ½ plane. This method of locating vectors can be successfully applied to finding a heavy atom in a crystal structure.

First, consider the relative peak heights of vectors that are in a map. If the crystal structure only contains light (Z < 10) atoms then the peaks will all be of the same height, Zlight * Zlight’ except for accidental overlaps and a slight build up in the corresponding Harker lines or planes. If however, there is a single heavy atom in the compound, then the peaks between the heavy atom and the light atoms will have heights proportional to Zlight * Zheavy, and peaks between symmetry-related heavy atoms in the cell will be proportional to Zheavy * Zheavy. Thus the peaks corresponding to the heavy atom should stand out in comparison to the rest of the Patterson map peaks.

The vectors between symmetry-related peaks are obtained from the symmetry operators of the space group. A table is created with the symmetry operators listed along the left side column and along the top row. The vectors are determined by subtracting the left side operator from the top row operator. An example of this process is illustrated for the space group P21/c.

Table 1. Harker Lines and Planes for P21/c.
  x, y, z     -x, -y, -z     -x, ½+y, ½-z     x, ½-y, ½+z
x, y, z     0, 0, 0     -2x, -2y, -2z     -2x, ½, ½-2z     0, ½-2y, ½
-x, -y, -z     2x, 2y, 2z     0, 0, 0     0, ½+2y, ½     2x, ½, ½+2z
-x, ½+y, ½-z     2x, ½, ½+2z     0, ½-2y, ½     0, 0, 0     2x, -2y, 2z
x, ½-y, ½+z     0, ½+2y, ½     -2x, ½, ½-2z     -2x, 2y, -2z     0, 0, 0

The Harker line for this space group is at u = 0 and w = ½; and the Harker plane for this space group is at v = ½. To make sure that a heavy atom is properly located from the peaks in the Harker line (from the v coordinate locate y) and the Harker plane (from u and w locate x and z), check that there is also a vector in the Patterson corresponding to 2x, 2y, 2z.

For the a structure with only one heavy atom in the asymmetric unif of this example space group, the highest peak at u, ½, w in the Patterson map would be equated to the expression 2x, ½, ½+2z. Thus the x and z coordinates for the heavy atom would be at x = u/2, z = (w - ½)/2. The highest peak in the 0, v, ½ line in the map should correspond to the expression 0, ½+2y, ½ giving y = (v - ½)/2. Then verify these coordinates by looking in the Patterson map for a strong peak with u, v, w coordinates of u = 2x, v = 2y, and w = 2z.

Use these x, y, z coordinates for the heavy atom to calculate a simple Fourier map. From this new Fourier map it is usually possible to locate most if not all of the remaining atoms in the crystal structure.

Dr. Donald Ward has prepared tables of Harker lines and planes for all space groups and published these tables in Patterson Peaks. A brief teaching edition of this book is available on the web at http://www2.chemistry.msu.edu/staff/ward/PattPeaks/brief.shtml.

A Patterson map shows vectors between all atoms creating a large number of images of the molecule. One way to think about all of these images is to consider moving each atom of the structure to the origin and then plotting all vectors between atoms. If the Patterson function could be manipulated to leave only one image of the structure, then the structure would be solved. This type of manipulation is performed by a technique known as Patterson superposition. The steps in performing a superposition calculation are discussed below.

A heavy atom vector is located in the Patterson map. One copy of the Patterson map is shifted in space by this vector amount and for every point in the map, the intensities of the shifted map are superimposed on the intensities of an unshifted Patterson map. The minimum intensity between these two maps is then stored as a shifted map. A new high intensity peak is chosen from the shifted map and a copy of the original Patterson map is moved by this vector amount and the intensities of the shifted map are compared with the intensies of this moved Patterson. Again a minimum function is used in the comparison process. With each shifting and comparison step, the new map contains many fewer peaks. Usually 3-4 such superpositions are needed to reveal a single image of the structure. At this point, the symmetry elements of the space group are located in the map and the positions of the atoms (the remaining peaks in the map) are shifted to put the space group’s symmetry elements in the conventional position(s).

There are a few details about the superposition method that should be mentioned. First the original Patterson map should be calculated on a fine grid so that each point represents ≤ 0.25 Å separation. Both the original Patterson and all shifted maps should be calculated for the entire unit cell. Finally, the superposition program should properly interpolate the intensites from the map based on the shift vectors. At this time, there are no commercially available superposition programs.

Molecular Replacement

The molecular replacement method was first developed by Rossmann and Blow (1963, 1964).8 This method utilizes a structurally-similar model as represented in a known crystal structure or chemical model. Parts of the known model that are believed to not be similar to the problem structure are removed. The model is then rotated in all unique directions of a Patterson map and translated within the unit cell of the Patterson map to attempt to locate a fit with the model.


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