How each cross is made
Every question starts from a hidden true model. Often it is the hypothesis you test; sometimes it is not (linked genes, a lethal allele, a sex-linked gene). The counts are drawn at random from that model, and a set is only used when χ² lands clearly on one side of the line: below 0.6 × the critical value or above 1.5 × it, so rounding never flips the decision. Every expected count is at least 5, and on the corn ear the kernels drawn are the counts in the key.
A worked dihybrid cross
In set P7QH22 (AP / intro college), 144 tarn snails from FfAa × FfAa are 81 striped and dark, 33 striped and pale, 21 unstriped and dark and 9 unstriped and pale. The 9 : 3 : 3 : 1 hypothesis expects 81, 27, 27 and 9, so χ² = 0 + 1.33 + 1.33 + 0 = 2.67. With 4 − 1 = 3 degrees of freedom the critical value is 7.81. 2.67 is less, so you fail to reject the null hypothesis: the counts are consistent with independent assortment. Open set P7QH22
What each level includes
High school: one-gene crosses (3 : 1, 1 : 1 test crosses and 1 : 2 : 1) with the ratio given and a one-gene corn ear; linked genes and Hardy-Weinberg start switched off but can be added. AP / intro college adds dihybrid crosses and test crosses, a sex-linkage check and map distance from a two-point test cross, as in AP Biology topics 5.3 and 5.4. A-level adds epistasis (9 : 3 : 4, 9 : 7 and 12 : 3 : 1) and multiple alleles from AQA 3.7.1, with linkage explained rather than mapped. College genetics adds 15 : 1 and three-point test crosses: gene order, map distances, coefficient of coincidence and interference.
Questions
How do you do a chi-square test on a genetic cross?
Work out each expected count from the ratio you are testing: total × fraction. For each class find (O − E)² ÷ E, then add them up to get χ². Compare χ² with the critical value for your degrees of freedom at p = 0.05. If χ² is bigger, reject the null hypothesis; if not, fail to reject it.
Why say “fail to reject” instead of “accept”?
A small χ² shows the counts are consistent with the hypothesis, not that it is true: another model might fit as well. The AP Biology course description asks students to state whether they reject or fail to reject the null hypothesis, so every key here uses those words.
How many degrees of freedom does the test have?
The number of classes minus one: 1 for 3 : 1, 3 for 9 : 3 : 3 : 1. For Hardy-Weinberg the AP sheet’s rule gives 3 − 1 = 2, while genetics texts subtract one more because p came from the same counts, giving 1. Every Hardy-Weinberg set here reaches the same decision either way.
What is the critical value at p = 0.05?
From the AP Biology equation sheet: 3.84 for 1 degree of freedom, 5.99 for 2, 7.81 for 3 and 9.49 for 4. The table beside the questions has the rest.
How do you find map distance from a test cross?
Add the recombinant offspring (the two smaller classes), divide by the total and multiply by 100. That recombination frequency is the distance in map units (cM), a fair estimate below about 30 cM. In a three-point cross, add the double crossovers to each region’s single crossovers.
What does p = 0.05 mean here?
If the null hypothesis were true, a χ² beyond the critical value would turn up by chance less than 5% of the time. So 0.05 is the risk of rejecting a hypothesis that is actually right. It is not the chance that the hypothesis is true.
Sources, checked October 25, 2026: AP Biology Course and Exam Description, effective Fall 2025 (PDF updated June 2026): science practice 5.C, topics 5.3 and 5.4, and the equation sheet’s chi-square table; AQA AS and A-level Biology 7401/7402 specification, version 1.5 (November 26, 2021, PDF), section 3.7.1 Inheritance; Wikipedia, Hardy-Weinberg principle, example chi-squared test (degrees of freedom for Hardy-Weinberg).