Answer:
q = √0.16 which = 0.4
q² = frequency of homozygous recessive = 0.16
p = 0.6
p² = frequency of homozygous dominant = p X p = 0.6 X 0.6 = 0.36
2pq = frequency of heterozygous dominant = 0.48
Explanation:
In this question it is shown that brown hair is dominant over blond hair.
The total population of people with both brown hair and blond hair = 200
Brown-haired = homozygous brown (BB) + heterozygous brown (Bb) = 168
Using the Hardy Weinberg equation we can calculate frequencies.
This equation is written as p² + 2pq + q² = 1
Blond hair (bb) = (Total population) 200 - 168 (Brown haired) = 32
Percent of Blond hair of total population = 16% or 0.16 which is equal to q²
q² = frequency of homozygous recessive = 0.16
q = √0.16 which = 0.4
Sum of both frequencies = 100% = p + q = 1
so p = 1 - q
p = 1 - 0.4
p = 0.6
p² = frequency of homozygous dominant = p X p = 0.6 X 0.6 = 0.36
2pq = Frequency of heterozygous brown hair
2pq = 2(0.6)(0.4) = 0.48
2pq = frequency of heterozygous dominant = 0.48
