Some people are hypersensitive to the smell of asparagus, and can even detect a

Question

Some people are hypersensitive to the smell of asparagus, and can even detect a strong odor in the urine of a person who has recently eaten asparagus. this trait turns out to have a simple genetic basis. An indivigual with one or two copies of the A allele at the gene can smell asparagus in urine, whereas a person with two copies of the alternative "a" allele (aa genotypes) cannot. Assume that men and woman in the popultation hae the same allele frequencies at the asparagus-smelling gene and that marriage and children production are independent of the genotype at the gene. In the human population, 5% of alleles are A and 95% are a.

a. What is the probability that a randomly sampled individual from the population has two copies of the a allele?

b. What is the probability that both members of a randomly sampled married couple are aa at the asparagus-smelling gene?

c. What is the probability that both members of a randomly sampled married couple are heterozygotes at this locus ( meaning that each person has one allele A and one allele a)?

d) Consider the type of couple described in c. What is the probability that the first child of such a couple also has one A allele and one a allele (is a heterozygote)? Remember that the child must receive exactly one allele from each parent.

Explanation / Answer

Here we have

P(A) = 0.05, P(a) = 0.95

a:

The probability that both members of a randomly sampled married couple are aa at the asparagus-smelling gene is

P(aa) =P(a)P(a) = 0.95*0.95 = 0.9025

b:

Since each person is independent so the probability that both members of a randomly sampled married couple are aa at the asparagus-smelling gene is

P(both have aa) = P(aa)P(aa) = 0.9025*0.9025= 0.81450625

c:

P(Aa) = P(A)P(a) = 0.05*0.95 = 0.0475

So

P(both are heterozygotes) = P(Aa)P(Aa) = 0.0475*0.0475= 0.00225625

d:

Chid can recieve AA , aa or Aa froam parents. Possible sample space is S ={AA, aa, Aa, aA}

Child can receive Aa in two ways from parents, as Aa or aA. So the required probability is

2/4 = 0.5

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