Question

In a large interbreeding population, 81% of the individuals are homozygous for a recessive character. In the absence of mutation or selection, what percentage of the next generation would be homozygous recessives?

Answer 1

In the absence of mutation or selection, and assuming random mating, the percentage of homozygous recessives in the next generation of a large interbreeding population would remain the same at 81%, due to Hardy-Weinberg equilibrium principles.

Step-by-step explanation:

When considering a population in which 81% of individuals are homozygous for a recessive trait, we can utilize principles of genetics to anticipate the percentages in subsequent generations. In the absence of mutation or selection, and assuming random mating, the population is likely to remain in Hardy-Weinberg equilibrium. Therefore, if 81% of the population is homozygous recessive (aa), the frequency of the recessive allele 'a' can be calculated as the square root of 0.81, which is q = 0.9. Since p + q = 1, the frequency of the dominant allele 'A' (p) would be 1 - q, which is 0.1. The expected distribution of genotypes can be calculated using the Hardy-Weinberg formula (p² + 2pq + q² = 1), giving us:

• p² = frequency of homozygous dominant (AA) individuals
• 2pq = frequency of heterozygous (Aa) individuals
• q² = frequency of homozygous recessive (aa) individuals, which again would be 81%.

According to this principle, in the next generation, the percentage of homozygous recessives would also be 81% because the allele frequencies in the population have not changed.