Chi-squared testAQA A-Level Biology: Revision notes
Section 1
Why we use the chi-squared test
In genetic crosses, observed numbers rarely match expected ratios exactly, because fertilisation is random. The chi-squared (χ²) test is a statistical test used to compare the goodness of fit of observed phenotypic ratios with the expected ratios. It tells you whether the differences between observed and expected numbers are significant or due to chance.
Use it for categorical data in whole numbers (counts, not percentages) with a large enough sample.
Section 2
Null hypothesis and the formula
State a null hypothesis: there is no significant difference between the observed and expected numbers (any difference is due to chance).
The formula is χ² = Σ (O − E)² / E, where O is the observed number and E the expected number in each class.
Steps: (1) work out expected numbers by dividing the total by the ratio parts; (2) for each class find (O − E)² / E; (3) add up the values.
Example: 295 round and 105 wrinkled, expected 300 and 100: 25/300 + 25/100 = 0.083 + 0.25 = 0.33.
Use the actual counts, never percentages. Always use the expected numbers, not the expected ratio, as E.
Section 3
Degrees of freedom and critical values
The degrees of freedom equal the number of classes − 1. For a monohybrid cross with two phenotypes the value is 1. For a dihybrid cross with four phenotypes it is 3.
Compare your calculated χ² with the critical value at p = 0.05 (a 5% probability that the difference is due to chance). Common critical values at 5%: 1 df = 3.84, 2 df = 5.99, 3 df = 7.82.
Section 4
Drawing conclusions
- If χ² is less than the critical value: the difference is not significant (p > 0.05). Accept the null hypothesis: any difference is due to chance, and the data fit the expected ratio.
- If χ² is greater than the critical value: the difference is significant (p < 0.05). Reject the null hypothesis: the difference is unlikely to be due to chance, so the expected ratio is not correct.
Example: 40.4 > 7.82 shows the genes are not assorting independently, suggesting linkage. Only a statistical test can show whether a difference is significant; it does not explain why.
Write the comparison and conclusion together: 'χ² (value) is greater/less than the critical value (value), so ...'.
Section 5
Using the test in genetics
Use χ² to test expected ratios such as 3 : 1, 1 : 2 : 1, 9 : 3 : 3 : 1, 1 : 1 : 1 : 1 (a test cross), or epistatic ratios such as 9 : 3 : 4. A result that fits a 9 : 3 : 3 : 1 ratio supports independent assortment. A significant result for a test cross suggests the genes may be linked, or that other factors such as selection are involved.
Larger samples give more reliable results. If expected numbers are very small (below 5), the test is not reliable.
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Chi-squared test
- A student crosses two heterozygous pea plants and expects a 3 : 1 ratio of round to wrinkled seeds in the offspring. She counts 400 seeds: 295 are round and 105 are wrinkled. She uses a chi-squared (χ²) test to compare her observed numbers with the numbers expected.Calculate χ² for her results. Use the formula χ² = Σ (O − E)² / E and give your answer to two decimal places.2 marks
- A researcher crosses two fruit flies that are heterozygous for body colour and wing length. He expects the offspring to show a 9 : 3 : 3 : 1 phenotypic ratio. Among 160 offspring he counts 79 grey normal-winged, 40 grey vestigial-winged, 29 ebony normal-winged and 12 ebony vestigial-winged flies. He calculates χ² = 5.11. The critical value of χ² at p = 0.05 with 3 degrees of freedom is 7.82.State and explain the conclusion the researcher should draw from his result.2 marks
- A student test crosses a fly heterozygous for two genes with a fly that is homozygous recessive for both genes. If the genes assort independently, a 1 : 1 : 1 : 1 ratio of four phenotypes is expected. Among 160 offspring there are 62 of the first phenotype, 58 of the second, 22 of the third and 18 of the fourth. The critical value of χ² at p = 0.05 with 3 degrees of freedom is 7.82.State a null hypothesis for this investigation and explain how the number of degrees of freedom is found.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).