Question

The frequency of an allele in a population of manatees is 0.15. If the population is at Hardy–Weinberg for this locus, what number of 600 individuals should be homozygous for this allele?

Answer 1

13.5

Step-by-step explanation:

In Hardy-Weinberg Equation, there are two equations necessary to solve a Hardy-Weinberg Equilibrium question:

p + q = 1 --------------------- (1)

p² + 2pq + q² = 1 ---------- (2)

p is the frequency of the dominant allele.

q is the frequency of the recessive allele.

p² is the frequency of individuals with the homozygous dominant genotype.

2pq is the frequency of individuals with the heterozygous genotype.

q² is the frequency of individuals with the homozygous recessive genotype.

Now, from the question... Given that;

The frequency of an allele in a population of manatees is 0.15

Let the frequency be 'p' which is equal to 0.15

If the population is at Hardy–Weinberg for this locus, what number of 600 individuals should be homozygous for this allele?

For Homozygous allele frequency i.e. For p²; we will have

p² = 0.15x 0.15 = 0.0225

Since p² = 0.0225, the number of 600 individuals that should be homozygous for this allele will be 600 × 0.0225 = 13.5

Answer 2

13.5

Step-by-step explanation:

Hardy-Weinberg principle states that allele and genotype frequencies in a population in the absence of mutation, genetic drift and other evolutionary influences

if f (r) = t then f (rr) = t², f(R) = u, the f(RR) = u², f(Rr) = 2tu where t and u are frequencies

the number of homozygous for this = 0.15² × 600 = 13.5