Consider a population of 10,000 turtles. They can be striped or spotted. The striped allele is dominant, and the spotted is recessive. Assign the variables for homozygous dominant, recessive, and heterozygous forms. Bonus- If 36 turtles are homozygous recessive, calculate the frequency of SS (homozygous dominant) form.
SS (homozygous dominant)
ss (homozygous recessive)
Ss (heterozygous)
Frequency of ss (homozygous recessive) form = 36/10,000 = 0.0036 or 0.36%.
Since ss represents the recessive phenotype, it must be aa. Therefore, p^2 (frequency of SS) + 2pq (frequency of Ss) + q^2 (frequency of ss) = 1
We know q^2 (frequency of ss) = 0.0036.
Therefore, p^2 + 2pq = 1 - 0.0036
p^2 + 2pq = 0.9964
We also know that S and s are the only two alleles, so p + q = 1.
We can use this to solve for p:
p + q = 1
p = 1 - q
Substituting into the equation above:
(1-q)^2 + 2(1-q)q = 0.9964
1 - 2q + q^2 + 2q - 2q^2 = 0.9964
-q^2 + 2q - 0.0036 = 0
q^2 - 2q + 0.0036 = 0
Using the quadratic formula:
q = (2 ± √(4 - 4(1)(0.0036))) / (2(1))
q = (2 ± 0.12) / 2
q = 1 or q = 0.0036
Since q cannot be 1 (otherwise there would be no S alleles in the population), the only valid solution is:
q = 0.0036
p = 1 - q = 0.9964
Therefore, the frequency of SS (homozygous dominant) form is p^2 = 0.9964^2 = 0.9928 or 99.28%.