Name:_____________________
Hardy-Weinberg Assignment
Objective: This assignment will give you a chance to apply your knowledge of HardyWeinberg principle and allele frequencies to a real world dataset.
Background: Rock pocket mice live in rocky habitats in the American southwest. Coat
color in rock pocket mice is largely controlled by a single gene (melanocortin-1-receptor,
Mc1r) that has two alleles. Two color morphs (Figure 1) are present within the species
such that
DD = dark morph
Dd = dark morph
dd = light morph
Figure 1 – Dark (left) and light (right) morphs of the rock pocket mouse.
Key Concept Questions
1. What is the Hardy-Weinberg (HW) principle?
a. Describe it and give the formula, making sure to define each symbol.
p2 + 2 pq + q2 = 1
p and q =allele frequencies for a population
The HW principle serves as a null hypothesis model for evolution at a
genetic locus. If observed genotype frequencies do not match expected
genotype frequencies, then the population is evolving.
b. What are the five assumptions of the Hardy-Weinberg principle?
1.
2.
3.
4.
5.
no natural selection
no genetic drift or random allele frequency changes
no gene flow
no mutation
random mating
c. What does the HW principle test?
The HW principle tests if the observed genotype frequencies match
expected genotype frequencies.
d. What can you say about a population if observed genotype frequencies
differ from expected genotype frequencies?
The population in evolving.
Application Questions
2. Sampling of the rock pocket mouse populations in Arizona produced 53
homozygous dark, 49 heterozygous dark, and 76 light mice. Fill in the table by
calculating the genotype frequencies.
Genotype
Number of Mice
Genotype Frequency
DD
53
0.298
Dd
49
0.275
Dd
76
0.427
3. What are the allele frequencies for D and d given the information from Question
2?
D = (53 x 2 + 49)/356 = 0.44
d = (76 x 2 + 49)/356 = 0.56
4. Given the answer to Question 3, what are the expected genotype frequencies in
this rock pocket mouse population?
p=D
q=d
p2 = 0.442 = 0.194
q2 = 0.562 = 0.313
2pq = 2 x 0.44 x 0.56 = 0.493
5. Fill in the table below using your answers from Questions 2 and 4.
Genotype
Expected Frequency
Observed Frequency
DD
0.194
0.298
Dd
0.313
0.275
dd
0.493
0.427
a. Are the observed and expected genotypic frequencies the same or
different? What does this suggest about the population?
The observed and expected frequencies are different. This means that the
HW equilibrium is not upheld implying the population is evolving and that
one of the assumptions of HW is being broken.
b. Which frequencies appear to be the most different?
The heterozygote frequency and homozygote dominant frequencies appear
to be the most different
6. In Arizona, where this population occurs, there are areas that contain light rocks
and those that contain dark rocks. Explain how this information and Figure 2
relate to your answer in Question 5. (hint: what assumption of HW might be
violated in this population?)
Could this information help
explain why some frequencies
are more different than others?
Figure 2 shows that light mice
and dark mice are not well
camouflaged on dark and light
rocks, respectively. Natural
selection could be acting on the
alleles by predation.
Heterozygotes could be less
abundant because emigration is
selected against.
Figure 2 – Dark and light morphs on light (left) and
dark (right) rock types.