Question

Over the summer, someone left a large stock of food open and all the mice on campus feasted in Stimson. At the end of the summer all the mice return to one of the dorms on campus. As part of a lab exercise, mice are collected from three different Goucher dorms. This was done late in the spring semester (many generations of mice since August).

You make a hypothesis that all should have the same allele frequencies for gene A because they all came from one population in Stimson. From three different dorms, the genotypes of a large number of mice were determined. The gene has two alleles, A and a. The resulting genotype frequencies for each of the dorms are shown below, and they are not all the same. Indicate which dorm populations are at Hardy-Weinberg equilibrium. You must show how you got the answer for full credit.

                                                                                                                                    (12 pts)

Part I

Dorm	A/A	A/a	a/a	at equilibrium?
Lewis	0.36	0.48	0.16
Probst	0.64	0.32	0.04
Wagner	0.64	0.27	0.09

Answer 1

Answer : As per the given data, Lewis and Probst Dorms are at Hardy-Weinberg Equilibrium.

According to Hardy-Weinberg, the population which follow the equations, p+q = 1 and p2+2pq+q2=1 is in equilibrium,

where, p= allelic frequency of dominent character( Here p=A)

q= allelic frequency of recessive character(q=a)

p2= genotype frequency of homozygous dominent allele

q2 = genotype frequency of homozygous recessive allele

2pq = genotype frequency of heterozygous alleles

Lewis Dorm

In this dorm, p2=0.36, so p = 0.6

p+q = 1, so q = 1 - 0.6 = 0.4

Therefore q2  = 0.4 x 0.4 = 0.16

and 2pq = 2 x 0.4 x 0.6 = 0.48

Means A/A =0.36

A/a= 0.48

a/a=0.16

SO THIS DORM IS AT HARDEY-WEINBERG EQUILIBRIUM

Probst Dorm

In this dorm, p2=0.64, so p = 0.8

p+q = 1, so q = 1 - 0.8 = 0.2

Therefore q2  = 0.2 x 0.2 = 0.04

and 2pq = 2 x 0.8 x 0.2 = 0.32

Means A/A =0.64

A/a= 0.32

a/a=0.04

SO THIS DORM IS AT HARDEY-WEINBERG EQUILIBRIUM

Wagner Dorm

In order to be in HARDEY-WEINBERG EQUILIBRIUM, Wagner Dorm should have the following values

In this dorm, p2=0.64, so p = 0.8

p+q = 1, so q = 1 - 0.8 = 0.2

Therefore q2  = 0.2 x 0.2 = 0.04

and 2pq = 2 x 0.8 x 0.2 = 0.32

Means A/A =0.64

A/a= 0.32

a/a=0.04

but in the given data it is

A/A =0.64, A/a = 0.27 and a/a = 0.09

Since the given datas are not matching, this wagner dorm is not in HARDEY-WEINBERG EQUILIBRIUM