Because we know that 15% of the population does not have the allele we can assume that at least 85% of the population has at least on copy of the dominant allele.
Answer:[tex]61\\[/tex] %
Explanation:
As per HARDY-WEINBERG LAW
[tex]p^2 + 2pq + q^2 = 1 \\p + q = 1\\[/tex]
[tex]p = \\[/tex] represents the frequency of the dominant allele in the population
[tex]q =\\[/tex]represents the frequency of the recessive allele in the population
[tex]p^2 =\\[/tex]represents the percentage of homozygous dominant individuals
[tex]q^2 = \\[/tex]represents the percentage of homozygous recessive individuals
[tex]2pq =\\[/tex]represents the percentage of heterozygous individuals
Here the frequency of ‘tt” allele is [tex]0.15\\[/tex]
Thus, [tex]p^2 = 0.15\\p = \sqrt{0.15}\\p = 0.387\\[/tex]
Substituting this in below given equation, we get -
[tex]0.387 + q = 1\\q = 1 - 0.387 \\q = 0.613\\[/tex]
Frequency of T allele is represented by q.
Thus, frequency of T allele would be [tex]0.613\\= 61 %\\[/tex]%
