Lecture 4 of the Molecular Genomics series. Reconstructing chromosomes from genotypes, and filling in what was never measured.
We have genotypes at many positions, and we know that nearby positions are inherited together because crossovers between them are rare. This lecture puts those two facts together. The result is the haplotype, which is the object that most modern genomic methods actually work with, and the procedure that exploits it, imputation, which is the main reason cheap chips are as useful as expensive ones.
A diploid animal carries two copies of each chromosome, one from each parent. A haplotype is the set of alleles carried on one of those copies, in order along the chromosome.
This is not what a genotype file tells you. A genotype records how many copies of each allele an animal has, but not how they are arranged between the two chromosomes. Consider an animal typed at three loci as
Here every locus is homozygous, so the arrangement is forced. One chromosome must carry A, b, C and so must the other. Written as haplotypes, the animal is AbC / AbC.
Now take an animal typed as
Only the first locus is heterozygous. One chromosome carries A and the other carries a, and since loci two and three are homozygous, both chromosomes must carry B and c. The haplotypes are ABc / aBc, again uniquely determined.
Haplotypes are determined unambiguously from a genotype whenever at most one locus is heterozygous. With two or more heterozygous loci the genotype is compatible with more than one arrangement, and additional information is required.
Take an animal typed as
Two loci are heterozygous. Both chromosomes carry a. The B and b must go to different chromosomes, and so must C and c, but which pairs with which is not determined. Two arrangements are compatible with the data:
With three heterozygous loci the problem grows. An animal typed Aa Bb Cc is compatible with four arrangements:
In general, an animal heterozygous at \(k\) loci is compatible with \(2^{k-1}\) haplotype pairs. The factor of one half arises because swapping the two chromosomes gives the same pair. At fifty thousand markers, of which perhaps a third are heterozygous, the number of arrangements is astronomical, and the problem cannot be solved one animal at a time.
The arrangements are not equally likely. Because recombination is rare over short distances, the haplotypes that actually exist in a population are a small subset of those that are combinatorially possible. Phasing works by finding the arrangement of each animal that is consistent with the haplotypes carried by everyone else.
Phasing is the process of assigning each allele to the paternal or maternal chromosome, that is, of reconstructing haplotypes from unphased genotypes. It is done computationally, by specialised software, on the whole dataset at once.
Two kinds of information are exploited, and their availability determines how well it works.
Family structure. If a sire, dam and offspring are all genotyped, most ambiguities resolve directly. At a locus where the offspring is heterozygous and the sire is homozygous, the allele the sire transmitted is known, so the other came from the dam. Trios are therefore much easier than unrelated individuals, and multi-generation pedigrees with genotypes in successive generations are easier still. Livestock populations are unusually favourable in this respect, because deep recorded pedigrees with many genotyped ancestors are normal.
Population haplotype sharing. Even among nominally unrelated animals, short stretches of chromosome are identical by descent from some shared ancestor in the past. The same haplotypes therefore recur across the population. Algorithms exploit this by dividing the chromosome into short windows, resolving the small number of haplotypes present in each window, and then stitching windows together into chromosome-length haplotypes.
Once haplotypes are known, missing genotypes can be filled in. Imputation is the process of making a statistically informed estimate of a genotype that was not observed. There are two distinct applications, and they differ in scale.
The second is the economically important one, but the first illustrates the principle.
Genotyping errors and failed calls are routine. Quality control removes markers and animals with too many, but scattered gaps remain, and most software will not accept missing values.
Suppose a haplotype is common in the population, say A-B-C-D-E. An animal is typed at these five loci and matches at four, with the middle one missing. If the surrounding four alleles identify the haplotype unambiguously, the missing genotype is not really unknown: the animal carries a chromosome segment that, in this population, essentially always carries C at that position.
The inference is strong precisely because recombination is rare over short distances. Filling in the gap is not guessing; it is reading the answer off the rest of the segment.
The commercially significant application follows the same logic on a larger scale.
The justification is identity by descent. Even in animals with no recorded relationship, short stretches of chromosome are identical copies of the same ancestral segment. The pattern of sharing across the population can be described, at least conceptually, by an unobserved genealogical tree, and the length of the shared segments depends on how long ago the common ancestor lived and therefore how many opportunities recombination has had to break them up.
A low-density chip costs a fraction of a high-density one. If high-density genotypes can be reconstructed computationally with high accuracy, then the cost per animal falls by that fraction while the information retained is nearly the same. Imputation converts a hardware cost into a software cost, and software scales.
This is why the choice of chip density, discussed in Lecture 2, is less consequential than it first appears. The usual design is a small high-density reference plus a large low-density target, imputed up.
Reported imputation accuracies are typically above 95%, and often above 99% for common variants in well-referenced populations. Several factors drive the number.
| Factor | Effect on accuracy |
|---|---|
| Size of the reference set | Larger is better; more haplotypes are represented |
| Relatedness of target to reference | Closer is much better; imputing across breeds is markedly harder |
| Density of the low-density chip | Higher density anchors the match more firmly |
| Extent of linkage disequilibrium | Longer-range LD makes segments easier to identify |
| Minor allele frequency | Rare variants are imputed much less accurately than common ones |
| Data quality | Errors in the reference propagate into every imputed animal |
Accuracy is assessed by masking: take animals that were genotyped at high density, delete the genotypes that a low-density chip would not have measured, impute them back, and compare with the truth. Simulation studies serve the same purpose under controlled conditions.
Accuracy measured as the proportion of genotypes correctly called is misleading when the minor allele is rare. If an allele has frequency 0.02, always guessing the common homozygote is right about 96% of the time while carrying no information at all. The correlation between imputed and true genotype is the honest measure, and it is the one that predicts how imputed data will behave in a subsequent analysis.
Counting arrangements. An animal is genotyped at six loci and is heterozygous at four of them. How many haplotype pairs are compatible with its genotype? Now suppose its sire is genotyped and is homozygous at three of those four loci. How many remain?
With \(k=4\) heterozygous loci there are \(2^{k-1} = 2^{3} = 8\) compatible pairs.
At each of the three loci where the sire is homozygous, the allele the sire transmitted is known with certainty, so the offspring's paternal allele at that locus is fixed and the maternal allele is whatever remains. Three of the four ambiguities are resolved, leaving one heterozygous locus with an unknown phase, that is \(2^{1-1} = 1\) arrangement. The phase is fully determined.
This is the quantitative version of why trios are so much easier than unrelated individuals, and why genotyping a few key sires improves phasing across the whole population rather than only for their own records.
When imputation fails. A breeding programme genotypes 5,000 local Ethiopian zebu cattle on a low-density chip and imputes to high density using a reference of 1,000 Holstein bulls. Imputation accuracy is reported as 91% of genotypes correctly called. Explain why this figure is both disappointing and optimistic.
Disappointing, because the reference and target are different populations. Imputation copies segments from reference haplotypes, so it can only reconstruct haplotypes that exist in the reference. Zebu haplotypes absent from Holstein cannot be recovered at all, and the algorithm will substitute the closest Holstein segment, which is wrong in a systematic rather than random way. Cross-breed imputation accuracy is characteristically much lower than within-breed.
Optimistic, because 91% correctly called is the wrong statistic. Much of that 91% is achieved on markers where one homozygote is common, where guessing the majority genotype is already right most of the time. The correlation between imputed and true genotype would be far lower, and it is the correlation that determines whether the imputed data are usable for association or prediction.
The remedy is a reference population drawn from the target population itself, even a small one. A few hundred locally genotyped animals typically outperform thousands of unrelated reference animals from another breed.
Segment length and ancestry. Two animals share a haplotype segment of 20 Mb. Two others share a segment of 1 Mb. Using the rule of thumb from Lecture 3, estimate how many generations back the common ancestor lies in each case, and state the assumption you are making.
A shared segment survives only if no crossover has fallen inside it on either lineage since the common ancestor. With \(g\) generations to the common ancestor there are about \(2g\) meioses in the path connecting the two animals. Using 1 cM per Mb, a segment of length \(L\) Mb is about \(L/100\) Morgans, so the expected number of crossovers within it is about \(2g \times L/100\).
Setting that expectation to about one gives \(g \approx 50/L\).
For \(L = 20\) Mb, \(g \approx 2.5\) generations: a very recent ancestor, a grandparent or great-grandparent. For \(L = 1\) Mb, \(g \approx 50\) generations: a distant ancestor, far beyond any recorded pedigree.
The assumptions are that recombination is uniform at 1 cM per Mb, which Lecture 3 showed is only an average, and that the segment length observed is typical rather than an extreme of a highly variable distribution. The relationship is nonetheless the basis for interpreting runs of homozygosity in Lecture 9: long segments mean recent ancestors.
Designing a reference. You have budget to genotype 200 animals at high density to serve as an imputation reference for a national population of 30,000. Which 200 would you choose, and why not simply take a random sample?
Choose the animals whose haplotypes are most widely represented in the population, which in a livestock population means the influential sires and their ancestors. A single widely used bull may contribute chromosome segments to thousands of descendants, so capturing his haplotypes at high density lets those segments be reconstructed in all of them.
A random sample is inefficient because it spends most of the budget on animals whose haplotypes are already represented, through their ancestors, by other members of the sample. The relevant quantity is not how many animals are in the reference but how much of the population's haplotype diversity they collectively carry.
Two refinements matter in practice. Selecting only elite sires biases the reference toward the selected part of the population and can under-represent haplotypes segregating in unselected animals, which matters if diversity, not prediction, is the goal. And in a population with several distinct subpopulations or breeds, each needs its own representation, since imputation does not transfer across them.
