if a population following hardy weinberg principle the frequency of dominant allele equal to 81 recessive allele equal to 19 calculate the percentage of heterozygous allele
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Answer:
1. The frequency of two alleles in a gene pool is 0.19 (A) and 0.81(a). Assume that the population
is in Hardy-Weinberg equilibrium.
(a) Calculate the percentage of heterozygous individuals in the population.
According to the Hardy-Weinberg Equilibrium equation, heterozygotes are represented by the 2pq
term. Therefore, the number of heterozygous individuals (Aa) is equal to 2pq which equals
2 0.19 × 0.81 = 0.31 or 31%
(b) Calculate the percentage of homozygous recessives in the population.
The homozygous recessive individuals (aa) are represented by the q
2
term in the H-W equilibrium
equation which equals 0.81 0.81 = 0.66 or 66%
2. An allele W, for white wool, is dominant over allele w, for black wool. In a sample of 900 sheep,
891 are white and 9 are black. Calculate the allelic frequencies within this population, assuming
that the population is in H-W equilibrium.
The allelic frequency of w is represented by the q term and the allelic frequency W is represented by the
p term. To calculate the value of q, realize that qq or q
2
represents the homozygous recessive individuals
or the black sheep in this case. Since there are 9 black sheep, the frequency of black sheep
=
# individuals =
9
= 0.01 , thus ww = q
2 = 0.01
total individuals 900
q = = = 0.1
Additionally, p + q = 1 thus p = 1 – q or p = 1 – 0.1 or 0.9 p = W = 0.9 and q = w =0.1
3. In a population that is in Hardy-Weinberg equilibrium, the frequency of the recessive
Explanation:
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