Difference Patterson Tutorial Answer Page
PR 613: Protein Structure and Function -
Structural Biology I
October 10, 1997, "X-Ray Crystallography III -
Solving the Patterson Function"
© Charles Brenner, Ph.D.
Kimmel Cancer Institute
Thomas Jefferson University
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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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