Question

Question
The frequency of brown-eyed people in the population is 70%. Calculate the frequency of heterozygotes. The gene for brown eyes is dominant to the gene for blue eyes.

Answer 1

Answer:

Answer: Since q = 0.2, and p + q = 1, then p = 0.8 (80%). The frequency of heterozygous individuals. Answer: The frequency of heterozygous individuals is equal to 2pq. In this case, 2pq equals 0.32, which means that the frequency of individuals heterozygous for this gene is equal to 32% (i.e. 2 (0.8)(0.2) = 0.32).

Answer 2

$\bigstar\:\:\underline{\sf question :} \bigstar$

The frequency of brown-eyed people in the population is 70%. Calculate the frequency of heterozygotes. The gene for brown eyes is dominant to the gene for blue eyes.

$\sf\small\underline\red{Given:-}$

The frequency of brown-eyed people in the population is 70%.

$\sf\small\underline\red{To \: Find:-}$

Calculate the frequency of heterozygotes. The gene for brown eyes is dominant to the gene for blue eyes.

$\sf\small\underline\red{Solution:-}$

It is given that, the population frequency of dominant allele (p) is 70% Hence, the population frequency of recessive allele (q) will be 1 - 0.3 = 0.7.

$\sf\small{We \: know \: that \: p² + 2pq + q² = 1}$

where p² is frequency of homozygous dominant phenotype, 2pq is frequency of heterozygous dominant phenotype and q² is frequency of homozygous recessive phenotype.

$\sf\small{Hence, \: frequency \: of \: homozygous \: dominant \: phenotype \: (p² ) \: will \: be \: 0.7² = 0.49.}$

$\sf\small{Thus, \: the \: correct \: answer \: is \: '0.49.'}$