Question

This problem has been solved!See the answerHuman ABO blood groups are controlled by a single gene with three alleles, I^A I^B and i. The three alleles contribute to four blood group phenotypes, there are two other loci that control blood phenotypes, the MN blood group locus and the Rh factor locus. The MN blood group has two codominant allleles L^M and L^N. The Rh factor locus also has two alleles Rh+ and Rh-; Where Rh+ is dominant. The two phenotypes at the Rh factor locus are Rh+ and Rh-.Suppose in a population the frequency of the ABO allele I^A is .36 and I^B allele frequency is .26. In the same population the frequency of the L^M allele is .47 and the frequency of Rh+ is .62What is the frequency of the genotype I^A I^B L^M L^M Rh- Rh- if the population is in hardy weinberg equilibrium?

Answer 1

Answer: The correct answer would be 0.003

Explanation:

1. The probability of occurring of two independent events is equal to the product of probabilities of individual events.

For example, probability of event 1 =a and that of 2 = b.

Thus, the probability of both the events occurring together = a x b ....(1)

2. By Hardy-Weinberg equation, p + q = 1 .... (2)

and p² + q² + 2pq = 1 ..... (3)

The frequency of $I^{A}$ = 0.36 and that of $I^{B}$ = 0.26

Now, the frequency of  $L^{M}$ = 0.47

Thus, frequency of $L^{N}$ = 1 - 0.47 (by equation 2)

Thus, $L^{N}$ = 0.53

Similarly, the frequency of Rh- = 1 - 0.62 = 0.38 (by equation 2)

The frequency of individuals with genotype $I^{A}I^{B}L^{M}L^{M}Rh-Rh-$ = 0.36 x 0.26 x 0.47 x 0.47 x 0.38 x 0.8 ...(By equation 1)

Thus, the final answer would be 0.0029 ≅ 0.003