SAMPLE PROBLEMS
I have provided the full solution for problem 1 and the answer for
problem 2.
Solution: Problem 1
Population 1: 210 MM, 86 Mm, and 8 mm
210+86+8 = 304 individuals sampled
Each has 2 alleles, so 3042=608 total alleles
Let allele M = p and allele m = q
Allele frequencies:
p=(2102)+86
p=506/608
p=0.832
q=(82)+86
q=102/608
q=0.168
p=0.832, q = 0.168
Expected Genotypic Frequencies
p2+2pq+q2
=0.8322+2(0.8320.168)+0.1682
=0.693+0.279+0.028
Expected individuals of each genotype
=0.693304=210.6 MM
=0.279304=84.9Mm
=0.028304=8.6mm
MM
Mm
mm
Observed
210
86
8
Expected
210.6
84.9
8.6
Use Chi-square (2) to test if expected probabilities are equal to the observed.
2 
(observed  expected ) 2 (observed  expected ) 2 (observed  expected ) 2


expected
expected
expected
2 
(210  210.6) 2 (86  84.9) 2 (8  8.6) 2


210.6
84.9
8.6
2 = 0.05
Contrast with value for chi-square from a table (with 1 degree of freedom and a 0.05
probability) = 3.81
Since 0.05 is less than 3.81, we accept the null hypothesis, that the observed population is
in Hardy-Weinberg Equilibrium and does not differ from the expected values.
Problem 2.
260 MM, 42 Mm, and 68 mm
Allele frequencies
p=0.759
q=0.241
Genotypic frequencies
p2=0.577
2pq=0.365
q2=0.058
Expected Genotypes
MM = 213.4
Mm = 135.2
Mm = 21.4
2 = 175.8
175.8 > 3.8
Reject null hypothesis of HWE. The observed distribution of genotypes in this
population does not fit HWE and some evolutionary force is acting upon it.