1. It is amazing to consider that in 1933 it was possible to publish a paper in Science "demonstrating" that because crystalline pepsin did not diffract X-rays, it was clear that proteins are amorphous and merely inert carriers of enzyme activity, the physical basis for which could not yet be physically studied. Bernal and Crowfoot's Nature paper the following year, demonstrating diffraction by crystalline pepsin set the stage, just 63 years ago, to account for biology's most fascinating chemistry with physics.
(How were Bernal and Crowfoot able to observe diffraction when others had failed? Recognizing that pepsin crystals lose their optical birefringence when dried, they mounted their crystals in sealed capillary tubes for data collection. How do the large unit cell sizes of protein crystals and their hydration make protein crystal diffraction patterns different than small molecule diffraction patterns? Are there any additional differences between proteins and small molecules that would alter their respective diffraction patterns? answer page)
2. The physics of X-ray diffraction are no longer considered to be interesting by physicists. As you have learned, space group theory was so firmly established by 1935 that all possible space groups were canonized in the International Tables. It would be hard to expect biologists to keep up with the latest physical theories that are en vogue in current physics just as it would be hard for a working physicist today to keep up with the latest function of p53. However, the amount of physics that you have to learn to appreciate X-ray diffraction is roughly the equivalent of the basic biology of the double helix that you would expect a physicist to grasp. It will "not escape your notice" that much of our basic biology is indebted to physical methods. Let's review a few additional facts of physics so that we can proceed to new material.
Heavy atom methods make it possible to get out of this conundrum. If you can attach a heavy atom to a unique location or locations on your macromolecule of interest without changing the structure or symmetry of your macromolecule and without destroying the ability of the crystal to diffract, you can use the Patterson function to solve the position of the heavy atom.
If you understand the answers to this outstanding question, you will
become a good student of X-ray crystallography.
With a single solved heavy atom derivative, you are still "underdetermined" in terms of solving the problem of the protein's phases--in the 360 degrees of possible phases, the two possible phases for each reflection are equally likely. Additional phase information provided, for example, by a second derivative can break phase ambiguity and get you in the business of interpreting an electron density map rather than trying to get phases to generate an initial map. Precisely how the observed native and derivative amplitudes (Fp and Fph) and the calculated heavy atom phases (a h) are used to calculate two possible native phases (a p) is not prohibitively difficult to understand but it is not the subject of today's lecture.
In 1954, Perutz and co-workers calculated a difference Patterson (Fph - Fp)2 with the amplitudes of a mercury-labeled hemoglobin crystal and the amplitudes of an isomorphous native hemoglobin crystal. Here, the scatter of the light atoms is mathematically removed (leaving noise, of course) so that the difference Patterson map ought to show simply the vectors between heavy atoms.
The Patterson map contains equal peaks corresponding to the vector from atom A to atom B and the vector from atom B to atom A. You should be able to appreciate how this fact makes Patterson maps centrosymmetric. You can also see that the vector from atom A in one cell to atom A in neighboring (or any other) cells will always fall on the origin.
If you were presented with a Patterson map that had peaks at (.25, .78, .39) and (.75, .22, .61), you would report that one heavy atom was at (.25, .78, .39) with respect to the other. The conventional way to do this would be to assign one heavy atom to the origin (0, 0, 0) and the other to (.25, .78, .39). If you take our first example at face value, for two heavy atoms to have positions of (.10, .24. .05) and (.35, .02, .44) in P1, it is likely that some work was done already to assign a different atom to position (0, 0, 0).
Crystallographers reading this tutorial will recognize the self-restraint being exercised in not launching into a discussion of difference Fourier methods!
| atom ID | operator | position | vector | Patterson site |
| A | 1 | (.10, .24. .05) | A2-A1 | (.80, .50, .90) |
| A | 2 | (.90, .74, .95) | A1-A2 | (.20, .50, .10) |
| operator 1= | (x, y, z) | |||
| operator 2= | (-x, y +1/2, -z) | |||
| vector 1-2= | (2x, 1/2, 2z) | |||
| vector 2-1= | (-2x, 1/2, -2z) |
Notice that the symmetry operator that takes all values of y and translates them half a cell along y makes it such that all non-origin peaks generated by crystallographic symmetry in P2(1) Patterson maps are at the v = .50 section. These "Harker" sections are always the first sections examined in Patterson solutions. We now take the symmetry operations of P2(1) from the International Tables, (x, y, z) and (-x, y + 1/2, -z), and derive general algebraic expressions of the Patterson vectors. Subtracting operator 2 from operator 1, we can say that in Patterson space, there will be a peak at (u, v, w) = (2x, .50, 2z). Subtraction in the opposite direction yields the centrosymmetric peak, (u, v, w) = (-2x, .50, -2z). Because 2x = +/- .20, x could equal .10, .90, .40 or .60. Initially, the z solution is also 4-fold degenerate at z = .05, .95, .45 or .55.
It should not surprise you that x could equal .10 or .90. In fact, if a heavy atom in P2(1) falls on x = .10, symmetry generates the atom's mate at .90. If the heavy atom falls at x = .10, can z still equal .05, .95, .45 or .55? The answer is no. If one is taking x to be .10, then one has assigned a particular vector equation [either (2x, 1/2, 2z) or (-2x, 1/2, -2z)] to a particular peak. If you assign the first equation to the peak at (.20, .50, .10), you will conclude that x must equal .10 or .60 because 2x = .20. You are have now "taken the position" that 2z (and not -2z) must equal .10 and you will find that only two z solutions, .05 and .55, are compatible with z = .10.
Because peaks on the v = 1/2 Harker section are independent of y, y can have any value. Can you convince yourself that heavy atom A at (.10, .01, .05) or atom A at (.10, .33, .05) would generate the same Patterson peaks as atom A at (.10, .24. .05)? answer page
Use the four tables printed below as sets of alternative two-dimensional unit cells to account for the P2(1) Patterson results. Label the horizontal axes x and the vertical axes z (both ascending from origins in lower left corners). Mark the position of the heavy atoms A in the top table at positions x = .10, z = .05 and x = .90, z = .95 in each of the twelve unit cells. In the second table, mark A atoms at x = .10, z = .55 and x = .90, z = .45. In the third table, mark A atoms at x = .40, z = .45 and x = .60, z = .55. In the last table, mark the last remaining correct solution. Can you see that each of these representations are equivalent descriptions of the same physical structure and produce the same pair of interatomic vectors? By marking one of the sets of unit cells in another color with an incorrect solution, such as x = .10, z = .45, can you see how this is not the same physical arrangement? answer page Choice of one of the correct unit cells over the other is arbitrary and simply fixes the origin of the unit cell with respect to a heavy atom. Because we began this discussion by stating that we had a heavy atom at (.10, .24, .05) we will stick with those coordinates, even though three other x, z combinations would be acceptable and fixing y at 0 would be more conventional.
NOTE: These four tables should appear as 3 by 4 arrays of identically
sized unit cells. You may need to use the lastest
browser
for this to appear correctly.
| atom ID | operator | position | vector | Patterson site |
| A | 1 | (.10, .24. .05) | B1-A1 | (.25, .78, .39) |
| B | 1 | (.35, .02, .44) | A2-A1 | (.80, .50, .90) |
| A | 2 | (.90, .74, .95) | B2-A1 | (.55, .28, .51) |
| B | 2 | (.65, .52, .56) | A2-B1 | (.55, .72, .51) |
| B2-B1 | (.30, .50, .12) | |||
| operator 1= | (x, y, z) | B2-A2 | (.75, .78, .61) | |
| operator 2= | (-x, y +1/2, -z) | A1-B1 | (.75, .22, .61) | |
| A1-A2 | (.20, .50, .10) | |||
| vector 1-2= | (2x, 1/2, 2z) | A1-B2 | (.45, .72. .49) | |
| vector 2-1= | (-2x, 1/2, -2z) | B1-A2 | (.45, .28, .49) | |
| B1-B2 | (.70, .50, .88) | |||
| A2-B2 | (.25, .22, .39) |
The unique solutions for x and z from the B-derived Harker peaks at (.30, .50, .12) and (.70, .50, .88) are x = .15, z = .06 or .56 and x = .35, z = .44 or .94. If there were no heavy atom A, then either of these solutions would be equivalent and serve to fix the origin as above. However, the assignment of heavy atom A to (.10, .24. .05) should make some of these solutions for B correct and some incorrect.
Let's examine the consequences of hypothesizing that heavy atom B is at x = .15 to see how far it will take us in explaining all of the Patterson peaks. If B has an x value of .15, its symmetry mate will be located at x = .85. Knowing that A atoms fall at x = .10 and .90, it is satisfying to find Patterson cross-peaks at u = .25, and .75 because these could represent vectors of (.10 - .85) and (.85 - .10). The absence of Patterson sites at u = .05 and .95 is troubling, however. Can you provide the set of Patterson cross-peaks for heavy atom A in its correct positions and heavy atom B at x = .15, z = .06 and x = .85, z = .94? Does that account for the Patterson map? answer page
If heavy atom B is not at x = .15, it must be at .35. Can it be
at both x = .35, z = .44 and x = .35, z = .94 or must it be at one or the
other? answer
page Given that the y value of atom A was assigned to .24 in
the solution (.10, .24. .05), can you now derive the y position of your
B atoms? Is this a unique solution? answer
page