What Is the Chi-Squared Test Used for in Genetics?
Learn how the chi-squared test checks observed genetic ratios against expected offspring numbers, with a worked AQA A Level Biology example and exam tip.
The chi-squared test is used in genetics to decide whether differences between observed offspring numbers and numbers expected from a genetic cross are likely to have arisen by chance. For AQA A Level Biology, it tests how well observed phenotypic frequencies fit a predicted genetic ratio.
How the test works
First, use the predicted ratio to calculate the expected number in each phenotype category. A phenotype is an observable characteristic, such as whether an offspring shows a dominant or recessive trait. To calculate an expected number, multiply the total number of offspring by that category’s fraction of the predicted ratio. The expected numbers must add up to the total observed.
Then compare the expected numbers with the observations using:
χ2=∑(O−E)2E\chi^2=\sum\frac{(O-E)^2}{E}χ2=∑E(O−E)2Here, OOO is the observed number, EEE is the expected number, and ∑\sum∑ means add the value for every category.
The null hypothesis states that any difference between observed and expected numbers is due to chance. Compare the calculated χ2\chi^2χ2 with the critical value at p=0.05p=0.05p=0.05. The degrees of freedom equal the number of categories minus 111. If the calculated value is greater than or equal to the critical value, reject the null hypothesis. The difference is statistically significant.
Worked example
Suppose a monohybrid cross predicts a 3:13:13:1 ratio of dominant to recessive phenotypes. Among 100100100 offspring, you observe 787878 dominant and 222222 recessive. The expected numbers are 757575 and 252525:
χ2=(78−75)275+(22−25)225=0.48\chi^2=\frac{(78-75)^2}{75}+\frac{(22-25)^2}{25}=0.48χ2=75(78−75)2+25(22−25)2=0.48There are 222 categories, so the degrees of freedom are:
2−1=12-1=12−1=1At p=0.05p=0.05p=0.05, the critical value is 3.843.843.84. Since 0.48<3.840.48<3.840.48<3.84, do not reject the null hypothesis. The difference could reasonably be due to chance, so the results are consistent with the predicted ratio.
Exam mistake to avoid
A non-significant result does not prove that the predicted inheritance ratio is correct. It only means there is insufficient evidence to reject the null hypothesis. Also show every calculation, use the correct degrees of freedom, and compare the values in the correct direction.
