A biologist found that 20% of a population of bats are brown, which is recessive. assuming the population shows Hardy-Weinberg equilibrium, determine the genotype and the allele frequencies of the population?
To determine the genotype and allele frequencies in a population showing Hardy-Weinberg equilibrium, we can use the Hardy-Weinberg equation. In a population with two alleles, A and a, the equation 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)
Given that 20% of the bat population is brown (recessive), we can assume that q^2 = 0.20. Therefore, q = sqrt(0.20) ≈ 0.447.
To calculate p, we subtract q from 1 since p + q = 1. Thus, p = 1 - 0.447 ≈ 0.553.
Now, we can calculate p^2 and 2pq:
- p^2 = (0.553)^2 ≈ 0.306
- 2pq = 2 * 0.553 * 0.447 ≈ 0.494
The genotype and allele frequencies in the population are as follows:
- Genotype frequencies: AA ≈ 0.306, Aa ≈ 0.494, aa ≈ 0.20
- Allele frequencies: Frequency of A ≈ 0.553, Frequency of a ≈ 0.447
To determine the genotype and allele frequencies of a population under Hardy-Weinberg equilibrium, we can use the following equations:
1. Genotype Frequencies:
- P² + 2PQ + Q² = 1, where P is the frequency of the dominant allele (B) and Q is the frequency of the recessive allele (b).
- P² represents the frequency of homozygous dominant individuals (BB), 2PQ represents the frequency of heterozygous individuals (Bb), and Q² represents the frequency of homozygous recessive individuals (bb).
2. Allele Frequencies:
- P + Q = 1, where P is the frequency of the dominant allele (B) and Q is the frequency of the recessive allele (b).
Given that 20% of the bat population is brown (recessive phenotype, bb), we can infer that the frequency of the recessive allele (Q) is √0.2 = 0.447.
Next, we can determine the frequency of the dominant allele (P):
- Since P + Q = 1, we can substitute the value of Q, leading to P + 0.447 = 1.
- Solving for P, we find P = 1 - 0.447 = 0.553.
Now, let's calculate the genotype frequencies:
- P² = (0.553)² = 0.306 or 30.6% (frequency of BB)
- Q² = (0.447)² = 0.200 or 20% (frequency of bb)
- 2PQ = 2 * 0.553 * 0.447 = 0.494 or 49.4% (frequency of Bb)
Therefore, under Hardy-Weinberg equilibrium, the genotype frequencies are approximately 30.6% BB (homozygous dominant), 20% bb (homozygous recessive), and 49.4% Bb (heterozygous). The allele frequencies are approximately 55.3% B (dominant allele) and 44.7% b (recessive allele).