Chi-Squared Test - Monohybrid Crosses

For a monohybrid cross there are normally two phenotype categories, so there is 2 − 1 = 1 degree of freedom, and the critical value at p = 0.05 is 3.84.Worked example: pea plants heterozygous for height (Tt) are crossed. The offspring are 160 tall and 40 short, a total of 200. Is this consistent with a 3 : 1 ratio?Null hypothesis: there is no significant difference between the observed numbers and the numbers expected from a 3 : 1 ratio.Expected numbers: tall = ¾ × 200 = 150; short = ¼ × 200 = 50.Calculation:
Tall: (160 − 150)² / 150 = 100 / 150 = 0.67
Short: (40 − 50)² / 50 = 100 / 50 = 2.00χ² = 0.67 + 2.00 = 2.67Conclusion: 2.67 is less than the critical value of 3.84 at 1 degree of freedom. The probability that the difference is due to chance is greater than 0.05, so the null hypothesis is accepted. The results fit a 3 : 1 ratio, supporting the idea that tall is dominant and both parents were Tt.The same method tests a 1 : 1 ratio from a test cross, or a 1 : 2 : 1 ratio for codominance (which has three categories and so 2 degrees of freedom, critical value 5.99).Tips: always use the actual counts, check that the expected numbers add up to the total, and state the conclusion in terms of the null hypothesis and the probability.