Question

in addition to the abo blood groups, humans have an mn blood type system in which the m and n alleles are codominant to one another. a group of scientists found that in one population the frequency of the m allele was 0.82, and the frequency of the n allele was 0.18. assuming that the population is in hardy-weinberg equilibrium, which of the following is the expected frequency of individuals with the genotype mn? a 0.148 b 0.295 c 0.640 d 0.672

Explanation:

Step1: Recall Hardy-Weinberg formula for heterozygotes

For a gene with two alleles (let frequency of \( M \) be \( p \) and \( N \) be \( q \)), the frequency of heterozygous genotype \( MN \) is given by \( 2pq \) (since \( M \) can come from one parent and \( N \) from the other, or vice versa).
Given \( p = 0.82 \) (frequency of \( M \) allele) and \( q = 0.18 \) (frequency of \( N \) allele).

Step2: Calculate \( 2pq \)

Substitute \( p = 0.82 \) and \( q = 0.18 \) into the formula \( 2pq \).
\[
2pq = 2\times0.82\times0.18
\]
First, calculate \( 0.82\times0.18 = 0.1476 \).
Then, multiply by 2: \( 2\times0.1476 = 0.2952 \approx 0.295 \)

Answer:

B. 0.295