Question

Inherited traits in humans are determined by materials in specific locations on chromosomes. Each normal human has 23 pairs of chromosomes, one chromosome in each pair coming from each parent. A gene is a portion of each chromosome in a pair. The material which makes up a gene on the pair of chromosome comes in forms called alleles. Each distinct combination of alleles is called a genotype. Consider a gene that has two alleles A and a. Suppose that the gene exhibits itself through a trait (such as hair colour or blood-type) with two versions. We call A dominant and a recessive if individuals with genotypes AA and Aa have the same version of the trait and the individuals with genotype aa have the other version. The two versions of the trait are called phenotypes. We shall call the phenotype exhibited by individuals with genotypes AA and Aa the dominant trait, and the other trait will be called the recessive trait.

Assume that the allele A is dominant, that individuals mate independently of genotype and that the genotypes AA, Aa and aa occur in the population with probabilites 1/4, 1/2, and 1/4 respectively. We are going to observe an individual whose parents are not available, and we shall observe the phenotype of this individual. 1

(a) List all six possible parental genotypes of this individual. You may use symbol such as (AA, Aa), remember that (AA, Aa) = (Aa, AA)

(b) Let B1, B2, . . . , B6 represent the six events in (a), find P(Bi) for all i = 1, . . . 6. A probability like P(Bi) is often called the prior probability.

(c) Let E be the event that the observed individual has the dominant trait. Find P(E|Bi) for i = 1, . . . , 6.

(d) Find the probability that the observed individual will have the dominant trait.

(e) Find the posterior probability P(Bi |E) for i = 1, . . . , 6. Note: you may summarize your result as follows:

Answer 1