Hardy-Weinberg: Within a population of birds, 21 percent of them display the recessive phenotype for a particular trait. What are the allele frequencies for this trait assuming the population is in genetic equilibrium?
I'm absolutely confused on this ;l
No worries! I can explain the concept of the Hardy-Weinberg equilibrium and guide you through the process of calculating allele frequencies.
The Hardy-Weinberg equilibrium is a principle in population genetics that describes the relationship between allele and genotype frequencies in an idealized, non-evolving population. According to the Hardy-Weinberg equilibrium, the allele frequencies within a population will remain constant across generations if certain conditions are met.
The conditions required for genetic equilibrium are as follows:
1. No mutations occur.
2. The population is large (to avoid genetic drift).
3. Mating is random (no sexual selection).
4. There is no migration (no gene flow).
5. Natural selection does not affect the trait in question.
Now, let's move on to calculating allele frequencies using the information provided.
The equation to determine the allele frequencies in the Hardy-Weinberg equilibrium is as follows:
p^2 + 2pq + q^2 = 1
Here,
- p represents the frequency of one allele (let's call it A)
- q represents the frequency of the other allele (let's call it a)
Given that 21% of the birds display the recessive phenotype (aa), we can conclude that q^2 = 0.21. Remember, q^2 represents the frequency of the aa genotype.
Now, to find the value of q, we can take the square root of 0.21:
q = √(0.21) ≈ 0.458
Since p + q = 1, we can find the value for p by subtracting q from 1:
p = 1 - q = 1 - 0.458 ≈ 0.542
So, the allele frequencies for this trait assuming the population is in genetic equilibrium are approximately p = 0.542 (frequency of the dominant allele) and q = 0.458 (frequency of the recessive allele).