What is the frequency of the dominant allele in this situation?

Ans: The frequency of the dominant allele in this population is 95.39%

Explanation:

To estimate the frequency of alleles in a population, use the Hardy-Weinberg equation. According to this equation:

p = frequency of dominant allele (say A)
q = frequency of recessive allele (say a)

For a population in genetic equilibrium:
p + q = 1 (The sum of the frequencies of both alleles is 100%.)

(p + q)2 = 1, therefore, p2 + 2pq + q2 = 1

The three terms of this binomial expansion indicate the frequencies of the three genotypes:

p2 = frequency of AA (homozygous dominant)
2pq = frequency of Aa (heterozygous)
q2 = frequency of aa (homozygous recessive)

The solution to your problem is as follows:

91% of the individuals are dominant for a character i.e. they have a dominant phenotype.

Here, 91% individuals have the genotype AA. Frequency of homozygous dominant (AA) in this problem is 91% = 91/100 = 0.91.

As mentioned earlier, homozygous dominant (AA) is designated as p2. Threfore, p2 = 0.91 and p = 0.9539 OR 95.39%.

Similarly, you can find the frequency of recessive allele (q) by substracting 0.9539 from 1 (since, p+q = 1). Therefore, frequency of the recessive allele will be 0.0461 OR 4.61%