For a
dihybrid cross there are usually four phenotype categories, so there are 4 − 1 =
3 degrees of freedom, and the
critical value at p = 0.05 is
7.82.
Worked example: two
Drosophila heterozygous for two genes on different chromosomes are crossed. The expected ratio is 9 : 3 : 3 : 1. The 160 offspring are 94 wild type, 26 with vestigial wings only, 32 with ebony body only and 8 with both.
Null hypothesis: there is no significant difference between the observed numbers and those expected from a 9 : 3 : 3 : 1 ratio.
Expected numbers (160 ÷ 16 = 10 per part): 90, 30, 30, 10.
| Phenotype | O | E | (O − E)2 / E |
| wild type | 94 | 90 | 0.18 |
| vestigial | 26 | 30 | 0.53 |
| ebony | 32 | 30 | 0.13 |
| vestigial ebony | 8 | 10 | 0.40 |
χ2 = 0.18 + 0.53 + 0.13 + 0.40 = 1.24Conclusion: 1.24 is less than 7.82 at 3 degrees of freedom, so the probability that the difference is due to chance is greater than 0.05. Accept the null hypothesis: the results fit a 9 : 3 : 3 : 1 ratio, so the genes are assorting independently.
A dihybrid test cross (expected 1 : 1 : 1 : 1) is also tested with 3 degrees of freedom. If χ
2 is larger than 7.82, the null hypothesis is rejected. A common reason is that the two genes are linked on the same chromosome.