Question

A population of lab rats recently escaped into the sewer system of Washington, D.C. A brave researcher sampled this population and estimated the following phenotypic frequencies, which were caused by a single gene with two alleles: E and e. 61% normal eyes (EE) 26% cross-eyed (Ee) 13% blind (ee) a) If this population were to reach a Hardy-Weinberg equilibrium (HWE) in one generation, what would the phenotypic frequencies be in that new HWE generation

Answer 1

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