A population is in Hardy-Weinberg equilibrium. If the frequency of a r...
Explanation:1. Understanding Hardy-Weinberg Equilibrium:Hardy-Weinberg equilibrium is a principle in population genetics that describes the genetic equilibrium in a population under certain conditions. It states that the frequencies of alleles and genotypes in a population remain constant from generation to generation in the absence of other influences.
The equation for Hardy-Weinberg equilibrium is:
p^2 + 2pq + q^2 = 1
Where:
- p^2 represents the frequency of the homozygous dominant genotype (AA).
- 2pq represents the frequency of the heterozygous genotype (Aa).
- q^2 represents the frequency of the homozygous recessive genotype (aa).
- p represents the frequency of the dominant allele (A).
- q represents the frequency of the recessive allele (a).
2. Given Information:In the given question, the frequency of the recessive allele (q) is mentioned as 0.30. Therefore, we can calculate the frequency of the dominant allele (p) using the equation:
p = 1 - q
p = 1 - 0.30
p = 0.70
3. Calculating the Percentage of Individuals with Dominant Trait:To calculate the percentage of individuals with the dominant trait, we need to consider the frequency of the homozygous dominant genotype (AA) and the frequency of the heterozygous genotype (Aa).
The frequency of the homozygous dominant genotype (AA) can be calculated using p^2:
Frequency of AA = p^2
Frequency of AA = (0.70)^2
Frequency of AA = 0.49
Similarly, the frequency of the heterozygous genotype (Aa) can be calculated using 2pq:
Frequency of Aa = 2pq
Frequency of Aa = 2 * 0.70 * 0.30
Frequency of Aa = 0.42
Therefore, the percentage of individuals with the dominant trait (AA and Aa) can be calculated as:
Percentage of individuals with dominant trait = (Frequency of AA + Frequency of Aa) * 100
Percentage of individuals with dominant trait = (0.49 + 0.42) * 100
Percentage of individuals with dominant trait = 91%
So, the correct answer is 91%.