Question

If a natural population of 60 individuals is in Hardy-Weinberg equilibrium for a gene with two alleles B and b, with the gene frequency of allele B of 0.7, the genotype frequency of Bb will be:

• A. ( 0.21 )
• B. ( 0.42 )
• C. ( 0.49 )
• D. ( 0.56 )

Answer 1

The Correct Option is B

According to the Hardy-Weinberg equilibrium, if there are two alleles for a gene in a population, say B and b, with frequencies \( p \) and \( q \) respectively, then \( p + q = 1 \). The genotype frequencies are given by \( p^2 \) (for BB), \( 2pq \) (for Bb), and \( q^2 \) (for bb). Given that the frequency of allele B ((p)) is 0.7. Since (p + q = 1), the frequency of allele b ((q)) is (1 - p = 1 - 0.7 = 0.3). The genotype frequency of heterozygotes (Bb) is given by (2pq). Substituting the values of (p) and (q): Frequency of Bb = \( 2 \times 0.7 \times 0.3 = 2 \times 0.21 = 0.42 \). The population size (60 individuals) is irrelevant for calculating the genotype frequency in Hardy-Weinberg equilibrium, as it describes the proportions of genotypes in the population.