Crystallography & Symmetry
patterns derived from their groups
Concept

Heavy atom — where it appears

An atom of high atomic number among many light ones, whose scattering is strong enough to be found on its own. Locating a few of them first, from a difference map, is how a structure too large to solve directly is phased.

Named by 9 essays across one field — each of them below, with the objects they name alongside it.

How many reflections the centre test needs. The error rate of two tests for a centre of symmetry against the number of reflections used, measured on 60 centrosymmetric and 60 non-centrosymmetric structures at each point. The moment test — the one in every textbook, comparing ⟨|E|² − 1⟩ to its two theoretical values — reaches one error in twenty at 160 reflections and one in a hundred at 320. A likelihood ratio, which uses each reflection's own value instead of one average, reaches the same at 40 and 80. The gap is the price of summarising a distribution by its mean, and it is about a factor of four.

How many reflections it takes to know there is a centre

The test for a centre of symmetry compares one average of the intensities against two theoretical values a quarter apart. Whether that is a measurement depends on how many reflections went into the average, and the only honest way to find out is to run the test on structures whose answer is already known and count the mistakes.

diffraction · Statistics The same structure, mapped from intensities and from differences. Left, the ordinary Patterson map of the structure: 14762 interatomic vectors, a continuous field of overlapping peaks, and the two vectors between the anomalous scatterers — circled — nowhere among its strongest. Right, the map from squared Bijvoet differences over the same reflections: two peaks after the origin, and they are those two vectors. The difference map is 7381 times smaller a problem to read.

A map of the atoms that break the law

Feed a Patterson synthesis the differences between the two halves of each Friedel pair instead of the intensities, and the map that comes back holds the vectors between the anomalous scatterers and nothing else. Two atoms among a hundred and twenty-two: 14,762 vectors become two.

diffraction · The Patterson function Which cells the flipping solver finds. For four cells of twenty atoms — all carbon, a quarter sulphur, one sulphur, one lead — the share of random starting phase sets that charge flipping takes to the right structure within 250 cycles, against the threshold below which density has its sign reversed. Each point pools several random arrangements of atoms with several starts each, and the atoms fall off with angle as real atoms do. The cells with sulphur are solved more often than carbon alone, the quarter-sulphur cell most of all, each at a threshold near one standard deviation; the lead cell is never solved at any threshold.

The heavy atom the flipping solver cannot see past

Sayre's identity degrades as a cell's atoms become unequal, most of all for a quarter of heavy atoms three times the weight of the rest, and the obvious prediction is that a solver built on the same sparseness should struggle there too. It does not. Charge flipping solves a cell of carbon with sulphur in it more often than carbon alone, and never solves a cell with one lead atom — because a threshold stated in standard deviations belongs to whichever atom owns the map's variance.

diffraction · Direct methods Three difference maps of one substructure, and the map they approximate. Patterson syntheses over 1,681 reflections of a structure with 120 light atoms and two anomalous scatterers, from squared Bijvoet differences at one wavelength, squared dispersive differences between two wavelengths, and the two normalised and summed; the fourth panel is the Patterson of the anomalous atoms' own intensities. Circles mark the two vectors between the anomalous atoms. Each map's correlation with the exact one is printed beneath it: Bijvoet 0.775, dispersive 0.838, combined 0.999.

A second wavelength completes the map

The Bijvoet difference carries the anomalous substructure's intensity times the sine squared of a phase; the dispersive difference between two wavelengths carries it times the cosine squared. Normalised and added, the two make the substructure's own Patterson, correlation 0.999. It needs no fewer reflections than either half, because the threshold was never the average. What the dispersive half brings is a way of failing the Bijvoet half cannot have: two datasets, and a scale between them.

diffraction · The Patterson function Success of the flipping solver against how far off the subtracted lead is. The share of forty starts on the cell of nineteen carbons and one lead that the flipping solver solves, with the lead's density subtracted at a position displaced from the true one by up to sixteen grid steps, half the cell. The rate does not fall with the offset: the points scatter without a trend about the pooled rate of 40%, the band marking two binomial standard deviations, and the original solver, with nothing subtracted, solves none. Refining the position from the map does no better than holding it.

The lead is wherever it is taken out

Charge flipping solves a cell of nineteen carbons and one lead once the lead's own density is taken out before each flip and put back after. The obvious worry was precision: a lead subtracted a fraction of a grid step from where it really is leaves a residual larger than any carbon. The residual is there, larger than a carbon at a tenth of a step. It does not matter. Subtracted two steps away, or half the cell away, the lead still lets the solver succeed at the same rate, because the solver builds the whole structure round wherever it is put.

diffraction · Direct methods A scale error between two wavelengths has a sign. The contrast of the substructure's vectors in a dispersive difference Patterson, the weaker of its two peaks over the strongest other peak, as the second dataset is scaled wrongly, with no metal atom and with two metal atoms that do not scatter anomalously. An error that weakens the second dataset loses the map at 3.7% with no metal; one that strengthens it at 50%. The dashed line is where the substructure's peak is predicted to reach zero, a quarter of the fractional change in f′ between the wavelengths. The metals shorten the strong side and hardly move the weak one.

A scale error between two wavelengths has a sign

A dispersive difference compares two datasets, so a wrong scale factor between them adds the whole structure to it. On a structure whose own Patterson is a haze, the substructure's vectors survived a scale error of fifty per cent, and heavy atoms that do not scatter anomalously were expected to take over much sooner. They do, but only in one direction. Scale the second dataset the other way and the substructure is erased at four per cent whatever else is in the crystal, because the error's linear term cancels the signal. Scaling the datasets onto each other is itself an error of that sign.

diffraction · The Patterson function Four flipping rules on four cells. The best success rate of the flipping solver, pooled over five arrangements and eight starts, on four cells: twenty carbons, a quarter sulphur, one sulphur among twenty and one lead among twenty. Four rules for what is flipped: below a number of standard deviations of the map, below a height set by the lightest atom's weight, the same height above the map's mean, and the original rule with the known heavy atom's density removed first. On the lead cell only the last solves anything, 55% against nothing; on both sulphur cells it raises the rate to near certainty; on carbon alone it is the original rule.

The lead taken out solves what no threshold could

Charge flipping never solved a cell of nineteen carbons and one lead, and two repairs were proposed: a threshold set by the composition rather than by the map's spread, and the lead's own density removed before each flip. The threshold does nothing for the lead cell at any setting, measured from zero or from the map's mean. Removing the lead takes it from nought to fifty-five per cent, and lifts both cells with sulphur in them to near certainty. The trouble was never where the threshold sat.

diffraction · Direct methods Success of the flipping solver against the error in the vector between two subtracted leads. The share of 80 starts on a cell of eighteen carbons and two leads that the flipping solver solves, with both leads' density subtracted, the first at its true place and the second displaced from its true place by an error in a random direction. The rate holds to a quarter of a grid step (21, 19, 18, 11, 0, 0, 0 successes at errors of 0, 0.1, 0.25, 0.5, 1, 2, 4 steps), falls by half at half a step, and is zero from one step on. With one lead the place did not matter at all; with two, the vector between them does.

Two leads need their vector to half a step

With one heavy atom in a cell with no symmetry, charge flipping did not care where the atom was subtracted: it rebuilt the structure round whatever place it was given. With two heavy atoms that freedom runs out, as predicted, because no translation can put both under wrong references at once. Subtract one of them a whole grid step from where it belongs and the solver never succeeds. But the tolerance is five times looser than the estimate it was predicted from: up to a quarter of a step the rate does not move, at half a step it halves, and the residual the solver absorbs is three carbons' worth, not one.

diffraction · Direct methods Fitting the outer shells makes it worse, and correcting the fit recovers it. The substructure's weaker peak as a share of its height with no scaling, on data with no error, with the scale error each scaling builds in, for five falloffs and three scalings. Fitting everything loses 42% with no falloff and less as the falloff grows. Fitting only the outer shells loses more, and at the steepest falloffs builds in a large error of the opposite sign. The fit corrected by the ratio the composition expects recovers the whole peak at every falloff, within a few per cent.

Fit the scale, then put the lean back

Scaling two wavelengths to each other builds in an error of the cancelling sign, because their means differ for a physical reason. The proposal was to fit against the part the edge does not touch, at high resolution. It does the opposite: f₀ falls with angle and f′ does not, so the outer shells lean furthest and fitting them costs more. Taking the scale from the composition alone reproduces the lean as well. What works is fitting as usual and then putting back the difference the composition expects, and that recovers the whole peak.

diffraction · The Patterson function

Named alongside it

The objects these essays reach for when they reach for this one.

Phase problemStructure factorAnomalous scatteringDirect methodsHeavy atom methodThe Patterson functionElectron densityResolutionFriedel lawNormalised structure factorOrigin choiceBias

All concepts