### Solving the Patterson Function

### © Charles Brenner, Ph.D.

for comments, clarifications, revisions or to help with illustrations, please email charles-brenner@uiowa.edu

Q. 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?

Six answers:

1. Large unit sizes of crystals create for small spot separations.

2. Proteins have more long range interactions than small molecules and thus have richer low-resolution data.

3. Protein crystals are generally less ordered than small molecule crystals and therefore diffract more weakly.

4. Small molecule crystals usually diffract to higher resolution.

5. Naturally occuring proteins never crystallize in centrosymmetric space groups.

6. Protein crystal diffraction patterns have a ring of diffuse water diffraction at 3.5 to 4 angstroms.

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Q. 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)?

A. In P2(1), heavy atom A at (.10, .01, .05) would have a symmetry mate at (.90, .51, .95) which would produce non-origin Patterson peaks at (.80, .50, .90) and (.20, .50, .10), the same vector peaks generated by heavy atom A at (.10, y, .05) for all values of y.

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Q. 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?

A.

Q. 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?

A. (.05, .01), (.95, .99), (.75, .89) and (.25, .11)

Q. Does that account for the Patterson map?

A. No.

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Q. 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?

A. The latter position is incompatible with the Patterson map, generating peaks at (.25, .89), (.75, .11), (.55, .01) and (.45, .99).

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Q. 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 it a unique solution?

A. We now know that A1 is at (.10, .24, .05) and A2 is at (.90, .74, .95). We know that B atoms are at (.35, y, .44) and (.65, y+ 1/2, .56). Call the first B atom B1 and the second B atom B2. If we examine the cross-peaks we see one that is (.25, .78, .39) and another that is (.25, .22, .39). We reason that these must correspond to B1-A1 and to A2-B2. If the first Patterson cross-peak, (.25, .78, .39), is B1-A1, then y = .02. However, if the second Patterson cross-peak, (.25, .22, .39) is B1-A1, then y =.46. According to this reasoning, there are two possible pairs of values for B peaks. They are either (.35, .02, .44) and (.65, .52, .56) or they are (.35, .46, .44) and (.35, .96, .56). These peaks have the same three-dimensional relationship to the A peaks we have chosen.

Nota bene. P1 and P2(1) Pattersons were chosen for this introductory tutorial because they are easy to solve. It should be appreciated that the degeneracy in the solutions observed herein are properties of these low symmetry space groups. Space groups that have symmetry elements in all three dimensions have unique Patterson solutions even when there is only a single heavy atom. In the future, we will add difference Patterson problems for space groups P2(1)2(1)2(1) and P3(1)21 that will take a bit more time to solve.

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