Graphical calculations of the expectation of Identity by Descent

    Consider a single locus A in two individuals 1 & 2. Designate the four alleles at this locus in these two individuals as A1 & A2, and A3 & A4, respectively. [We do not care whether the four alleles are similar or dissimilar by allelic state (DNA sequence)]. Individuals 3 and 4 are full sibs, and inherit one allele from each parent. Individual 5 is their offspring. What is the probability that 5 will inherit two alleles that are identical by descent (IDB) at this locus?

    A1 passes from 1 to 3 with a probability of 1/2, AND to 4 with a probability of 1/2. Given that it has passed to both offspring, it then passes from 3 to 5 with a probability of 1/2, AND from 4 to 5 with a probability of 1/2. Thus, the joint probability that A1 has passed from 1 to 5 through both 3 and 4 is (1/2)4 = 1/16.

    The same calculation can be repeated for alleles
A2, A3 & A4. Then, the combined probability that 5 has two alleles identical by descent at this locus, that is, has two copies of A1 OR A2 OR A3 OR A4 , is 1/16 + 1/16 + 1/16 + 1/16 = 1/4.

    Since this calculation applies for every locus individually, the total fraction of loci that are identical by descent is also 1/4, which is the Inbreeding Coefficient for full-sibs. A small sample of loci may depart from expectation, but as the sample of the genome increases, the observed fraction converges on expectation because of the Central Limit Theorem
.

    HOMEWORK:
The genetic consequences of first-cousin marriages (e.g., between Charles Darwin and Emma Wedgwood) are often of interest. Extend the graphical method above to show that the inbreeding coefficient for the children of such marriages is 1/16.


Text material © 2024 by Steven M. Carr