Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 45 / 1 / ii / Solution

Label the two paternal ancestral marker copies

P_{1}, P_{2}

and the two maternal copies

M_{1}, M_{2}

. Treat these as four distinct ancestral copies in an outbred pedigree; equal observed alleles need not imply identity by descent. By Mendelian segregation, each child independently receives either paternal copy with probability

1/2

. Thus the indicator

I_{P}

that the two full siblings inherit the same paternal copy satisfies

P (I_{P} = 1) = P (P_{1}, P_{1}) + P (P_{2}, P_{2}) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} .

The same argument gives

P (I_{M} = 1) = 1/2

for the maternal copy. Paternal and maternal meioses are independent, so

I_{P}

and

I_{M}

are independent. Consequently the number

J = I_{P} + I_{M}

of copies shared identical by descent has a binomial distribution, giving

P (J = 0) = \frac{1}{4}, P (J = 1) = \frac{1}{2}, P (J = 2) = \frac{1}{4} .

This is Mendelian IBD sharing of full siblings. It describes an ordinary, unselected sib pair: conditioning on a shared disease phenotype can change the distribution if there is genetic linkage to its causal locus.

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