Using the table of critical values:
1. Set your α level. Conventionally, α = 0.05.
The α level represents a cut-off value for an acceptable amount of deviation. For example, if
α = 0.05, this represents a 5% chance of seeing these observed frequencies.
2. Calculate your Χ2 using the formula
3. Determine the degrees of freedom (df), and locate the corresponding row in the table.
4. Within the row, locate where your Χ2 lies (your exact value likely won’t be in the table, so you’ll
have to find the range where it would be located). Then, follow this line upwards to find the
associated range of p-values.
a. If p > α  accept the null hypothesis
There is a greater than 5% chance of obtaining the observed frequencies due to normal random
variation.
b. If p < α  reject the null; accept the alternative hypothesis
There is a smaller than 5% chance of obtaining the observed frequencies due to normal random
variation. Therefore, the observed frequencies are likely not consistent with the expected
distribution.
Applying Chi-Square Analysis
to Genetics Problems
So much of Mendelian genetics involves ratios. See the table below for the characteristic ratios we have seen.
Table 3. Characteristic phenotypic and genotypic ratios associated with monohybrid and dihybrid crosses.
F2 ratio
Phenotypic
Genotypic
Monohybrid cross
3:1
1:2:1
Dihybrid cross
9:3:3:1
------We can use the chi-square test to compare observed frequencies of phenotypic and genotypic classes to the frequencies
that would be expected, based on Mendelian ratios.
Recall that these ratios are based on characters that assort independently (are not linked). One application of
the chi-square test to genetics can therefore be to indicate whether two characters are linked or not.
Example 1
In flowers, red petals (A) are dominant to white petals (a). You think that a particular plant is
heterozygous (Aa), and to test this theory you perform a test cross. You obtain 120 progeny,
55 of which are red, and 65 of which are white.
a) In this test cross, what is the genotype of the second plant? aa
b) What are the expected proportions of red to white plants in the progeny?
Red: 50%
c)
White: 50%
What are the expected frequencies of red to white plants?
Red: 50% x 120 = 60
White: 60
d) Define a null hypothesis and an alternative hypothesis:
H0: The observed data follows a 1:1 distribution
HA: The observed data does not follow a 1:1 distribution
e) Define a significance level: α = _0.05___
f)
Fill in the chi-square calculation table:
Class
O
E
(O-E)2
red
55
60
25
white
65
60
25
Total =
g)
(O-E)2/E
0.417
0.417
X2 =
0.834
How many degrees of freedom are in your data? _1_________
h) Compare your test value, X2, to the table of critical values. What is the P value for your X2?
__0.1-0.5 (or 10%-50%)_____________
i)
Do you reject or accept the null hypothesis, H0?
j)
Draw a general conclusion based on your test:
□ accept
□ reject
□ The observed frequencies match the expected distribution
□ The observed frequencies do not match the expected distribution
k)
Draw a specific conclusion based on your test:
□ The red plant is heterozygous Aa
□ The red plant is homozygous AA
Example 2. A plant geneticist has two pure lines, one with purple petals, and one with blue. She hypothesizes that
the phenotypic difference is due to two alleles of one gene. To test this idea, she aims to look for a 3:1 ratio in the F 2.
She crosses the lines and finds that all F1 progeny are purple. The F1 plants are selfed, and 400 F2 plants are obtained.
Of these, 320 are purple and 80 are blue. Use the chi-square test to determine if these results fit her hypothesis.
α= __0.05_____ H0: The observed data follows a 3:1 distribution
HA: The observed data does not follow a 3:1 distribution
Class
O
E
(O-E)2
(O-E)2/E
Purple
320
300
400
1.33
80
100
400
4.00
400
400
Blue
df= _1______
Total = X2=
5.33
P value: 0.01-0.025________  reject the null hypothesis
accept the alternative hypothesis
General Conclusion: The data does not follow a 3:1 distribution
Specific Conclusion: Flower colour is not due to two alleles of one gene
Example 3. Thomas Hunt Morgan studied the inheritance of characters in fruit flies, Drosophila melanogaster.
He mated fruit flies with the following phenotypes/genotypes:
Parent
Phenotype
Gray body/Normal wings
Genotype
Gg Ww
Black body/Vestigial wings
gg ww
He obtained 2300 progeny with the following genotypes and frequencies:
Gg Ww 965
Gg ww 206
gg ww 944
gg Ww 185
Use the chi-square test to determine whether these results support the hypothesis that the genes for body colour and
wing type assort independently.
α= 0.05
H0: The observed data follows a 1:1:1:1 distribution (based on Punnett square prediction)
HA: The observed data doesn’t follow a 1:1:1:1
Class
O
E
(O-E)2
(O-E)2/E
GgWw
965
575
152100
264.522
ggww
944
575
136161
236.802
Ggww
206
575
136161
236.802
ggWw
185
575
152100
264.522
Total = X2=
1002.648
2300
df= 3
P value: <<0.005

reject the null hypothesis
accept the alternative hypothesis
General Conclusion: The observed data doesn’t follow a 1:1:1:1
Specific Conclusion: The genes for body colour and wing type do not assort independently