Question

We will use the Hardy Weinberg assumption were:

p = Dominant allele

q = Recessive allele

p + q = 1

p2 = frequency of homozygous dominant

q2 = frequency of homozygous recessive

2pq = frequency of heterozygous

Recessive frequency of recessive genotype = 1% = 0.01

Calculate the frequency of the recessive allele (q) = √0.010= 0.1

Calculate the frequency of the dominant allele

p = 1-q = 0.9

Calculate the frequency of homozygous dominant (p2)

p2 = 0.92 = 0.81

Calculate the frequency of heterozygous

2pq = 2 x 0.01 x 0.81 = 0.162

Now, to determine the number of each genotype in the population, multiply each frequency by the total number of the population (1000)

Number of recessive individuals = 1000x0.01 = 10

Number of homozygous dominant individuals = 1000 x 0.81= 810

Number of heterozygous individuals = 810 – 10 =180

The correct answer is 810 W/W, 180 W/w, 10 w/w

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