In a certain population, the dominant phenotype of a certain trait occurs 91% of the time. What is the frequency of the dominant allele?

Question

Maya Anderson · Accepted Answer

Frequency of dominant allele is 7% and that of recessive allele is 3%.

Just 9% of organisms are homozygous for the recessive allele, expressing the recessive trait, whereas 91% of organisms exhibit dominant phenotype, which includes both homozygous and heterozygous organisms with 2/1 dominant allele. This population would have a 3% frequency of the recessive allele (3x3=9) and a 7% frequency of the dominant allele (1-3=7), according to Hardy Weinberg's law. This is due to the fact that the sum of homozygous recessive plus all dominant organisms equals 100% and the frequency of dominant allele plus recessive alleles equals ONE.

Lydia Adams · Answer

undefined If the dominant phenotype occurs 91% of the time, then the frequency of the dominant allele can be calculated using the Hardy-Weinberg equation. Let p be the frequency of the dominant allele. Since the dominant phenotype occurs 91% of the time, the frequency of individuals with at least one dominant allele (genotypes AA and Aa) is 91%. Therefore, the frequency of individuals with the homozygous dominant genotype (AA) is $ p^2 $ and the frequency of individuals with the heterozygous genotype (Aa) is $ 2pq $, where q represents the frequency of the recessive allele.

Given that the dominant phenotype occurs 91% of the time, the frequency of individuals with the recessive phenotype (aa) is 9%.

Using this information, we can set up the equation:

$$ p^2 + 2pq + q^2 = 1 $$

Solving for p:

$$ p^2 + 2pq = 0.91 $$
$$ p^2 + 2p(1-p) = 0.91 $$
$$ p^2 + 2p - 2p^2 = 0.91 $$
$$ -p^2 + 2p + 0.91 = 0 $$
$$ p^2 - 2p - 0.91 = 0 $$

Using the quadratic formula, we find:

$$ p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

Where:
- a = 1
- b = -2
- c = -0.91

Plugging in the values:

$$ p = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-0.91)}}{2(1)} $$
$$ p = \frac{2 \pm \sqrt{4 + 3.64}}{2} $$
$$ p = \frac{2 \pm \sqrt{7.64}}{2} $$
$$ p = \frac{2 \pm 2.76}{2} $$

Since we're dealing with a frequency, we choose the positive root:

$$ p = \frac{2 + 2.76}{2} $$
$$ p = \frac{4.76}{2} $$
$$ p = 2.38 $$

Therefore, the frequency of the dominant allele (p) is 0.238.