Chi-squared test on dihybrid cross dataSpec D3.2.21
In short
A chi-squared test on data from a dihybrid cross tests whether the difference between observed and expected numbers is statistically significant. The null hypothesis is that there is no significant difference, for example that the genes are unlinked. If the calculated value exceeds the critical value at p = 0.05, the null hypothesis is rejected.
- State the null hypothesis (H₀): there is no significant difference between observed and expected results, for example the genes assort independently (9:3:3:1). The alternative hypothesis (H₁) is that there is a significant difference, for example the genes are linked.
- Calculate expected numbers (E) from the ratio and the total.
- Calculate χ² = Σ (O − E)² ÷ E for all classes.
- Degrees of freedom = number of classes − 1 (3 for four phenotype classes).
- Compare χ² with the critical value at p = 0.05 (7.815 for 3 degrees of freedom).
- If χ² is greater than the critical value, the difference is statistically significant: reject H₀. If it is smaller, the difference is not significant: accept (do not reject) H₀, because the difference can be explained by chance.
Testing a 9:3:3:1 ratio
An F2 from a dihybrid cross gave 300 round yellow, 108 round green, 92 wrinkled yellow and 20 wrinkled green seeds (total 520). Use a chi-squared test to decide whether this fits 9:3:3:1.
- Expected: 9/16 × 520 = 292.5; 3/16 × 520 = 97.5; 97.5; 1/16 × 520 = 32.5.
- Round yellow: (300 − 292.5)² ÷ 292.5 = 0.19.
- Round green: (108 − 97.5)² ÷ 97.5 = 1.13.
- Wrinkled yellow: (92 − 97.5)² ÷ 97.5 = 0.31.
- Wrinkled green: (20 − 32.5)² ÷ 32.5 = 4.81.
- χ² = 0.19 + 1.13 + 0.31 + 4.81 = 6.44, with 3 degrees of freedom.
Answer: 6.44 < 7.815, so the difference is not significant at p = 0.05: accept H₀; the results fit a 9:3:3:1 ratio.
Nature of science: the F2 is a sample representing all possible offspring. The test asks whether the sample's deviation from expected is larger than chance alone would often produce.
Quick check
Why do AA and Aa individuals have the same phenotype?
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One dominant allele produces enough functional protein for the trait to appear.
Give the genotype of a person with blood group O.
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ii.
What phenotype does a heterozygote show in incomplete dominance?
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An intermediate phenotype, e.g. pink four o'clock flowers.
State the outlier rule for a box-and-whisker plot.
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More than 1.5 × IQR above Q3 or below Q1.
HL only What ratio does a dihybrid test cross with unlinked genes give?
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1:1:1:1.
Written and checked against the IB Biology HL specification · Updated October 2026