a)A population of lab rats recently escaped into the sewer system of Washington,
ID: 57511 • Letter: A
Question
a)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)
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? Let freq (e)=p and freq (e)=q. (Tip first calculate p and q from the observed, and provided, phenotypic frequencies)
freq EE =
freq Ee =
freq ee =
b) Suppose that the phenotypes of the original adult population in question 8 were not equally likely to survive and reproduce (selection). Tip: the original adult population is the one under selection, not the one in HW eq, so which frequencies do you use?
The fitness coefficients are WEE = 1 WEe = 0.6 Wee = 0.2
What would the phenotypic frequencies be after selection?
Freq EE after selection =
Freq Ee after selection =
Freq ee after selection =
c) What would be the allelic frequencies after selection?
freq E (p)
=
freq e (q)=
61% normal eyes
(EE)
26% cross-eyed
(Ee)
13% blind
(ee)
Explanation / Answer
a)
Based on the given data, the genotypic frequencies are as follows:
Allele frequencies:
Expected genotype frequencies (assuming Hardy-Weinberg):
Thus, (p² + 2pq + q² = 1)
0.547 + 0.384 + 0.067 = 0.998 ~ 1
Thus, the population is under HW equilibrium.
b)
Form the given data, after selection:
Allele frequencies:
Expected genotype frequencies (assuming Hardy-Weinberg after selection):
Thus, the modified Hardy-Weinberg formulate is (p2w11 + 2pqw12 + q2w22):
1.69 + 0.78 + 0.09 = 2.46
Thus, the population is not under HW equilibrium.
c)
Allele frequencies:
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