Question

Human males have one X-chromosome and one Y-chromosome, whereas females have two X- chromosomes, each chromosome being inherited from one parent. Hemophilia is a disease that exhibits X-chromosome-linked recessive inheritance, meaning that a male who inherits the gene variant that carries the disease on the X-chromosome is affected, whereas females that carry the aberrant gene in only one of her two X-chromosomes are not affected (because of the recessive character of the disease). The disease is generally fatal for women that inherit two such genes, an extremely rare event we can neglect for now.

Consider a woman who has an affected brother, which implies her mother must be a “carrier” of the disease with one good and one bad gene (males receive their only X-chromosome from their mothers). We are also told the father is not affected. Thus, the woman herself has a 50% chance of carrying the aberrant gene (daughters receive one X-chromosome from each parent, therefore she got the “good” gene from the father, but has a 50% chance of having gotten the “bad” gene from the mother). Let’s θ be the event “the woman carries the disease”, with θ=1 (0) meaning the woman carries (doesn’t carry) the disease. Given the information available, the prior probability p(θ=1) is 0.5.

a) Suppose the woman has two sons, neither of whom is affected. Let’s yi be the state of son i. yi=1 (0) means the son is affected (unaffected). Under the assumptions the sons are interchangeable and conditionally independent on θ, calculate the likelihoods p(y1=0, y2=0 | θ=1) and p(y1=0, y2=0 | θ=0). (Remember males receive one of their mother’s X-chromosomes). 


b) Calculate the posterior probability p(θ=1 | y1=0, y2=0). What does this probability represent? 


Answer 1