In a population the proportion of homozygotes in 18 times, the proportion of homozygous recessive. Calculate the frequency of the 2 alleles?
To calculate the frequency of the two alleles, we need to use the Hardy-Weinberg equation.
In the Hardy-Weinberg equation, the frequency of homozygotes (p^2) is equal to the proportion of homozygotes (18 times) divided by the total population size. The frequency of homozygous recessive individuals (q^2) is equal to the proportion of homozygous recessive individuals divided by the total population size.
Let's denote the frequency of the dominant allele as 'p' and the frequency of the recessive allele as 'q'. According to the Hardy-Weinberg equation, p^2 + 2pq + q^2 = 1, where 1 represents the total frequency of alleles in the population.
In our case, the proportion of homozygotes (18 times) is equal to p^2, and the proportion of homozygous recessive individuals is equal to q^2. Let's assume the total population size is 'N'.
Therefore, we have the following equations:
p^2 = 18/N (Equation 1)
q^2 = ? (To calculate)
p + q = 1 (Equation 3)
Substituting Equation 1 into Equation 3, we get:
p + √(18/N) = 1
Simplifying the equation, we find:
p = 1 - √(18/N)
Substituting this value of p into Equation 3, we get:
1 - √(18/N) + q = 1
Solving for q, we find:
q = √(18/N)
Now, we can calculate the frequency of the two alleles:
The frequency of the dominant allele (p) is equal to 1 - √(18/N).
The frequency of the recessive allele (q) is equal to √(18/N).