Answer:
the population has a Hardy-Weinberg equilibrium
Explanation:
According to the case and the information given, the genotype frequencies are:
EE (p2) = 0.61
Ee (2pq) = 0.26
ee (q2) = 0.13
To obtain the allele frequencies we can perform the following equation
Frequency of E = p = p2 + 1/2 (2pq) = 0.61 + 1/2 (0.26) = 0.61+ 0.13 = 0.74
replacing the data we obtain:
Frequency of e = q = 1- p = 1 - 0.74 = 0.26
Regarding the expected frequencies of the genotype, taking into account that there is a Hardy-Weinberg equilibrium:
EE (p2) = (0.74) 2 = 0.547
Ee (2pq) = 2 × (0.74) (0.26) = 0.384
ee (q2) = (0.26) 2 = 0.067
We know that: (p² + 2pq + q² = 1)
replacing with the obtained data:
0.547 + 0.384 + 0.067 = 0.998 ~ 1
finally we can affirm that the population has a Hardy-Weinberg equilibrium
