Chi-Square Test in Genetics: 3:1 and 9:3:3:1 Worked Examples

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Genetics-I · A-level, AP Biology and introductory university study
A genetic ratio is a probability model, not a promise that every small sample will match it exactly. Learn to test a 3:1 or 9:3:3:1 prediction, show the arithmetic and write a defensible conclusion.

Original teaching diagram: expected phenotype counts under a 3:1 model.

What does a chi-square test tell you in genetics?

Pearson’s chi-square goodness-of-fit test compares observed phenotype counts with counts predicted by a stated model. It asks whether the discrepancy is unusually large relative to sampling variation under that model. It does not identify the biological cause of a discrepancy.

Start with the cross and its assumptions. For Aa × Aa, complete dominance and equal survival give phenotype probabilities of 3/4 dominant and 1/4 recessive. A dihybrid AaBb × AaBb cross predicts 9:3:3:1 only under suitable assumptions, including independent assortment and complete dominance at both loci. A different cross needs a different prediction.

Expected count: E = total offspring × predicted proportion
Test statistic: χ² = Σ[(O − E)²/E]
Degrees of freedom: df = categories − 1, when probabilities are fixed in advance.

Calculate one contribution for every category and add them. Use counts, not percentages. Keep full precision during intermediate steps and round the final statistic.

Worked example 1: a 3:1 ratio

Suppose a class scores 160 offspring: 118 dominant and 42 recessive. These are invented teaching data. The null hypothesis, H₀, states that population phenotype probabilities are 0.75 and 0.25. The alternative states that they differ from that prediction. Choose the significance level, α = 0.05, before examining the result.

Phenotype	Observed O	Expected E	(O − E)²/E
Dominant	118	160 × 3/4 = 120	4/120 = 0.0333
Recessive	42	160 × 1/4 = 40	4/40 = 0.1000
Total	160	160	χ² = 0.1333

Two categories give df = 1. At α = 0.05, the critical value is approximately 3.841. Because 0.1333 is smaller, fail to reject H₀. Write: These data do not provide evidence of a departure from the predicted 3:1 phenotype ratio at the 5% significance level.

The same absolute difference of two offspring contributes more in the smaller category because the formula divides by the expectation. Adding absolute differences alone is not a chi-square calculation.

Worked example 2: a 9:3:3:1 ratio

For 320 offspring, suppose the four counts are 176, 64, 62 and 18. Keep the categories in the same order as the prediction: A_B_, A_bb, aaB_, aabb. An underscore means either allele may occupy that position without changing the phenotype in this model.

Category	O	Probability	E	Contribution
A_B_	176	9/16	180	0.0889
A_bb	64	3/16	60	0.2667
aaB_	62	3/16	60	0.0667
aabb	18	1/16	20	0.2000

Adding unrounded contributions gives χ² ≈ 0.6222. Here df = 3, and the 5% critical value is approximately 7.815. Fail to reject the model. Compatibility with the prediction does not by itself establish independent assortment.

Check assumptions before interpreting the result

Independent observations: score each offspring once. Counting the same organism repeatedly does not increase the independent sample size.
Exclusive categories: every observation belongs in one category; settle ambiguous scoring rules beforehand.
A justified model: do not try many ratios and report the best fit as though it were predicted in advance.
Adequate expectations: a common introductory rule requires every expected count to be at least five. Small expectations call for an appropriate exact or simulation-based method.
Correct degrees of freedom: estimating model parameters from the same data may change df; the simple k − 1 rule has a context.

What if the result is significant?

Consider a separate sample of 100 offspring with counts 90 and 10. Under 3:1, expected counts are 75 and 25. The statistic is 225/75 + 225/25 = 12. With df = 1, it exceeds 3.841, so reject that model at α = 0.05. Investigate phenotype scoring, the assumed parental cross, differential survival and other possible explanations. The test alone does not demonstrate linkage.

Frequently asked questions

Does a large p-value prove the genetic ratio?

No. It means the data do not supply sufficient evidence against the tested model at your chosen threshold. A small sample can have limited power to detect departures.

What does the p-value actually mean?

It is the probability, under the null model and its assumptions, of a discrepancy at least as extreme as the observed one. It is not the probability that the null hypothesis is true.

Can I combine small phenotype categories?

Do not combine categories merely to obtain a convenient result. Categories have biological meaning. Use a suitable exact or simulation approach when the approximation is unreliable.

Exam answer checklist

State the hypotheses; show expected counts; calculate all contributions; give df and the decision threshold; then conclude in biological context. Use “fail to reject” instead of “proved correct.” Follow the rounding and tables supplied by your course.