Chi-squared
2
( )
analysis
Objective
• To analyse the results of genetic crosses by applying the chi-square test
Purpose of the chi-squared (2)test
• It is a statistical test used to test the significance of deviations/differences
between numbers observed in an experiment or investigation and the numbers
expected from a given hypothesis.
• The measure of the deviation is called chi-squared.
• This statistical test can be used only on raw counts; it cannot be used on
proportions, percentages and measurements (e.g. length, time, weight, volume).
Performing the chi-squared (2)test
• The 2 test is particularly valuable in genetics to determine whether offspring
phenotype ratios fit expected Mendelian ratios.
• For example, when Mendel self-crossed heterozygous tall plants, he obtained
787 tall and 277 dwarf plants. On the basis of his law of segregation, the
expected ratio is 3:1.
• We can now perform the 2 test to determine the significance of the difference
between the observed and expected values following a series of steps.
• Step 1: Write the null hypothesis which states:
There is no significant difference between the observed and expected results
OR
The difference between the observed and expected results is due to chance alone.
Performing the chi-squared (2)test
• Step 2: Calculate the expected results
• Step 3: Calculate the 2 value using the formula
• The information for steps 2 and 3 are usually placed in a table.
Performing the chi-squared (2)test
Phenotypic ratio
O
E
O-E
(O-E)2
(O-E)2/E
Tall
3
Dwarf
1
Total
787
(¾ x 1064) = 798
(787-798) = -11
277
(¼ x 1064) = 266
(277-266) = 11
1064
1064
-
(-11)2 = 121
(121/798) = 0.152
(11)2 = 121
(121/266) = 0.455
0.607
NB: The total value
for O and E must be
exactly the same.
2 = 0.607
Step 4: Determine the degrees of freedom (d.f.)
d.f. represents the number of ways in which the observed classes are free to vary.
Degrees of freedom = n-1 where n is the number of different expected phenotypes.
In this example the d.f is 1.
Performing the chi-squared (2)test
• Step 5: Determine the probability associated with the calculated 2 value for the degrees of
freedom.
This requires the use of a chi-square table.
In this example, the 2 value (0.607) lies between a probability of 0.50 and 0.30.
This means that the difference we observed can be expected 30-50% of the times the
experiment is carried .
Performing the chi-squared (2)test
• If the probability is greater than 5% or 0.05 then accept the null hypothesis which states that
there is no significant difference between the observed and expected results.
• If the probability is less than 5% or 0.05 then there is a significant deviation from the
expected results and the null hypothesis is rejected. Therefore, some factor other than
chance is responsible for the observed values being different from the expected.
The 5% probability critical values
• If the value of 2 is less than the 5%
critical value for the d.f, then
accept the null hypothesis.
• If the value of 2 is greater than the
5% critical value for the d.f, then
reject the null hypothesis.
Question
• In the garden pea, yellow cotyledon colour is dominant to green and inflated pod
shape is dominant to the constricted form. A cross between two heterozygous
yellow, inflated pod shape peas produced the following progeny:
193 green, inflated
184 yellow, constricted
556 yellow, inflated
61 green, constricted
Perform a chi-square test to determine if the results are due to chance alone.
Phenotypic
ratio
O
E
O-E
(O-E)2
(O-E)2/E
Yellow,
inflated
Yellow,
constricted
Green,
inflated
Green,
constricted
Total
556
184
193
61
994
Phenotypic
ratio
O
E
O-E
(O-E)2
(O-E)2/E
Yellow,
inflated
Yellow,
constricted
Green,
inflated
Green,
constricted
9
3
3
1
556
559.125
-3.125
9.766
0.017
184
186.375
-2.375
5.641
0.030
193
186.375
6.625
43.891
0.235
61
62.125
-1.125
1.266
0.020
2 = 0.302
d.f. = 3
Probability is greater than 0.95 or 90%
Accept the null hypothesis that the results are due to chance alone.
Total
994
994
0.302
Questions
1. A genetics engineer was attempting to cross a tiger and a cheetah. She predicted a
phenotypic outcome of the traits she was observing to be in the following ratio 4 stripes
only: 3 spots only: 9 both stripes and spots. When the cross was performed and she
counted the individuals she found 50 stripes only, 41 with spots only and 85 with both.
According to the chi-square test, did she get the predicted outcome?
2. In corn, purple kernels are dominant over yellow kernels and full kernels are dominant
over shrunken kernels. A corn plant heterozygous for purple and full kernels is crossed
with a plant having yellow and shrunken kernels. The following progeny are obtained:
• Purple, full – 112
• Purple, shrunken – 103
• Yellow, full – 91
• Yellow, shrunken – 94
Test your genetic hypothesis with a chi-square test.
3. A plant with purple flowers was crossed with a plant with white flowers. All the plants in
the F1 generation had purple flowers. The F1 plants were self-fertilized to produce the F2
generation with 105 purple flowers and 45 white flowers. Perform a chi-square test on this
data.