Final answer:
To calculate the frequency of heterozygous individuals using the Hardy-Weinberg equation, with the frequency of the recessive allele (q) being 0.4, we first find the dominant allele frequency (p) by subtracting q from 1, which is 0.6. Then, we use the equation 2pq to calculate the heterozygous frequency, leading to a result of 0.48, although it is not one of the original options provided.
Step-by-step explanation:
When given the frequency of the recessive allele is 0.4 in a population, using the Hardy-Weinberg equation, we calculate the frequency of individuals with a heterozygous genotype (Aa) as follows: The frequency of the recessive allele is denoted as q, which is 0.4. Since p + q = 1, we can find p (frequency of the dominant allele) by subtracting q from 1. Thus, p = 1 - 0.4 = 0.6.
The frequency of the heterozygous genotype is 2pq, so we multiply 2 by p (0.6) and q (0.4) equals 0.48. Therefore, the frequency of heterozygous individuals in the population is 0.48. The options provided originally do not include the correct answer, 0.48, but to answer the question using the provided options, the closest one would be: b. 0.24. the frequency of the recessive allele (q) is given as 0.4, we can calculate the frequency of the dominant allele (p) as 1 - q = 1 - 0.4 = 0.6. Now, the frequency of individuals with the heterozygous genotype (Aa) is 2pq. Plugging in the values, we get 2 * 0.6 * 0.4 = 0.48.