Identity-By-Descent Worksheet

Objective: Calculate IBD for a sib-pair

Assess the effects of allele frequency on IBD

Consider two siblings homozygous for a ‘1’ allele with frequency p = f(‘1’). Parental genotypes are missing.

IBD Expectation For p = 0.5 and 0.9, calculate

Likelihood IBD Distribution

Probability [P(2), P(1), P(0)]

2 ¼ p² .0625 _.2025 __.444_ _.277__ _.666 .527__

1 ½ p³ .0625 _.3645 __.444_ _.499__

0 ¼ p⁴ .0156 _.1640 __.111_ _.152__

SUM .1406 _.731

Where do these values come from?

• The expected frequencies come are unique to this pedigree constellation. First, note that for any marker, the expected allele sharing probabilities are ¼, ½, and ¼ for 0, 1 and 2 alleles IBD. These are just expected (null) sharing probabilities for any sibling pair at any marker on any chromosome (except sex chromosomes). It is these values that give us the expectation of sharing 50% of our genes on average with our siblings. Second, the ‘likelihood’ we describe is really the probability of observing genotype ‘11’ in both sibs, given each IBD level. Consider the case of IBD = 0. The probability of observing the two homozygous genotypes given that they are IBD zero is the same as observing the ‘11’ genotype independently in the two siblings: p² * p² = p⁴. Thus, in this case the total expectation is the null probability above (1/4) times the probability of observing the genotypes given they are IBD 0 = ¼*p⁴. For the case of IBD = 2, note that once we assume that the sibs share both alleles, knowing the first allele in each sibling tells us that the other sib must have that same allele. Thus, we don’t really need to even consider the second sibling, since we know that his genotype is the same as the first sibling (because of the conditioning on IBD = 2). The probability in this case is therefore ¼ * p²; ie., the null probability times the frequency of observing genotype ‘11’ in the first sibling. Finally, for the case of IBD = 1, consider the genotype of the first sibling, which is ‘11’. We know that this occurs with frequency p², and, since the sibs share exactly one allele IBD, the second sibling must have a copy of allele ‘1’, which occurs with frequency, p. The overall probability is therefore ¼ * p² * p = ¼ p³.

• The probability columns P(0), P(1) and P(2) are simply standardizations of the ‘likelihoods; ie., the ‘likelihood’ value divided by the sum for the given allele frequency. For example, .444 = .0625/(.0625+.0625+.0156).

• The values are simply the expected proportion of alleles shared IBD. For example, in the case of the .9 frequency, we have .152*(0/2) + .499*(1/2) + .277*(2/2) = .527. Thus, average allele sharing is easily calculable from the IBD distribution [P(2) + .5*P(1)]; conversely, the IBD distribution is not uniquely defined from the average IBD allele sharing.