An individual with one or two copies of the A allele at the gene (AA or Aa genot

Question

An individual with one or two copies of the A allele at the gene (AA or Aa genotypes) can smell the apparatus in urine, whereas a person with two copies of the alternative "a" allele (aa genotypes) cannot. Assume that men and women in the population have the same allele frequencies at the asparagus- smelling gene and that marriage and child 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 (man and woman) are aa at the asparagus- smelling gene?

C) What is the probability that both members of a randomly sampled married couple ( man and woman) 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 part C. What is the probability that the first child of such a couple had one A allele and one a allele? Remember that the child must receive exactly one allele from each parent.

Explanation / Answer

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

A. Probability that a randomly selected individual has two copies of a allele ( This probability will be same for male & female since allele frequency for both the geender in same)
= P( genotype aa)
= P(a) * P(a)
= 0.95 * 0.95
= 0.9025

B. Probability that both members of a randomly sampled married couple are aa
= P (man is aa & woman is aa)
= P(man is aa) * P(woman is aa) ( Since both the events are independent)
= 0.9025*0.9025 (From A)
=0.835

C. Probability that both members of a randomly sampled married couple are Aa
= P (man is Aa & woman is Aa)
= P(man is Aa) * P(woman is Aa) ( Since both the events are independent)
= P(A)*P(a)*P(A)*P(a)
=0.95*0.05*0.95*0.05
=0.0023

D. Probability that the first chind is Aa given that both the parents are Aa
= P(child gets A from father & a from mother) + P(child gets a from father & A from mother)
= P(child gets A from father ) * P(child gets a from mother) + P(child gets a from father)*P(child gets A from mother)
=0.5*0.5 +0.5*0.5
=0.5

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