Question

question 6 of 25
a fly has two alleles for the color of its eyes. the green allele is recessive, and is represented by q. the blue allele is dominant, and is represented by p. if 44 of 100 organisms are green, what is p?
homozygous dominant + heterozygous + homozygous recessive = 1
$p^{2}+2pq + q^{2}=1$
a. 0.56
b. 0.34
c. 0.66
d. 0.44

Explanation:

Step1: Find the frequency of the recessive phenotype

The frequency of the recessive phenotype (green - eyed flies) is \( q^{2}=\frac{44}{100} = 0.44\).

Step2: Calculate \( q\)

Take the square - root of \( q^{2}\). So \( q=\sqrt{0.44}\approx0.66\).

Step3: Use the Hardy - Weinberg equation \(p + q=1\)

Since \(p + q = 1\), then \(p=1 - q\). Substitute \(q\approx0.34\) into the equation, \(p = 1-0.34 = 0.66\).

Answer:

C. 0.66