Chi-Squared Test

You have carried out a genetic cross and counted the offspring. The numbers look roughly like a 9:3:3:1 ratio — but not exactly. Is the deviation just chance, or is it real and telling you that Mendel's law does not apply (because of linkage or epistasis, say)? The chi-squared (χ²) test is the statistical tool that lets you answer this question quantitatively. OCR A-Level Biology A specification module 6.1.2(g) requires you to use the chi-squared test to analyse results from genetic crosses, including calculating the value, using degrees of freedom and critical values, and interpreting the outcome.

Key Definitions:

• Null hypothesis (H₀) — there is no significant difference between observed and expected values; any difference is due to chance.
• Observed (O) — the actual number in each category.
• Expected (E) — the number predicted by the null hypothesis.
• Degrees of freedom (df) — the number of categories minus 1.
• Critical value — the tabulated χ² value above which the null hypothesis is rejected at a chosen probability (usually p = 0.05).
• p-value — the probability that the observed deviation (or a more extreme one) would occur by chance alone if H₀ were true.

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When to Use the Chi-Squared Test

Chi-squared is used for categorical data (counts, not measurements) arranged in discrete classes. In genetics, typical applications are:

• Monohybrid crosses (3:1 expected).
• Dihybrid crosses (9:3:3:1 expected).
• Sex-linked crosses.
• Detecting deviations that suggest linkage or epistasis.

You should not use chi-squared on percentages or on data with very small expected counts (conventionally, each expected value should be at least 5).

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The Procedure

Step 1: State the null hypothesis

H₀: the observed offspring numbers fit the expected Mendelian ratio (e.g. 9:3:3:1). Any difference is due to chance.

Step 2: Calculate the expected values

Multiply the total number of offspring by the proportion expected in each class. For a 9:3:3:1 ratio with 160 offspring, expected values are 90, 30, 30, 10.

Step 3: Apply the formula

$\chi^2 = \sum \dfrac{(O - E)^2}{E}$χ2=∑E(O−E)2​

For each category, compute (O − E)², divide by E, and sum across all categories.

Step 4: Work out degrees of freedom

$\text{df} = n - 1$df=n−1

where n is the number of categories. For a 9:3:3:1 cross with 4 categories, df = 3. For a 3:1 monohybrid, df = 1.

Step 5: Compare to the critical value

Look up the critical value for df and p = 0.05 (the usual significance threshold in biology).

• If χ² calculated < critical value → difference is not significant → accept the null hypothesis.
• If χ² calculated ≥ critical value → difference is significant → reject the null hypothesis.

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Critical Values Table (p = 0.05)

df	Critical value (p = 0.05)
1	3.84
2	5.99
3	7.82
4	9.49
5	11.07

For a 9:3:3:1 dihybrid cross, the value you need is 7.82. For a 3:1 monohybrid, it is 3.84.

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Worked Example 1: Monohybrid Cross

A geneticist crosses two heterozygous pea plants (Tt × Tt) and counts 200 offspring: 142 tall and 58 short.

Expected (3:1 ratio)

• Tall: 200 × 3/4 = 150
• Short: 200 × 1/4 = 50

Calculate χ²

Class	O	E	O − E	(O − E)²	(O − E)²/E
Tall	142	150	−8	64	0.427
Short	58	50	8	64	1.280

χ² = 0.427 + 1.280 = 1.71

Interpret

df = 2 − 1 = 1. Critical value = 3.84. Since 1.71 < 3.84, the deviation is not significant. Accept H₀: the data are consistent with a 3:1 ratio.

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Worked Example 2: Dihybrid Cross

Cross two heterozygous flies (RrYy × RrYy). Out of 160 offspring observed:

• Round yellow: 95
• Round green: 28
• Wrinkled yellow: 25
• Wrinkled green: 12

Expected (9:3:3:1 ratio)

• Round yellow: 160 × 9/16 = 90
• Round green: 160 × 3/16 = 30
• Wrinkled yellow: 160 × 3/16 = 30
• Wrinkled green: 160 × 1/16 = 10

Calculate χ²

Class	O	E	O − E	(O − E)²	(O − E)²/E
Round yellow	95	90	5	25	0.278
Round green	28	30	−2	4	0.133
Wrinkled yellow	25	30	−5	25	0.833
Wrinkled green	12	10	2	4	0.400

χ² = 0.278 + 0.133 + 0.833 + 0.400 = 1.644

Interpret

df = 4 − 1 = 3. Critical value = 7.82. Since 1.644 < 7.82, the deviation is not significant. Accept H₀: the data fit a 9:3:3:1 ratio.

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Worked Example 3: Detecting Linkage

A dihybrid test cross (AaBb × aabb) in fruit flies is expected to give a 1:1:1:1 ratio. Out of 400 offspring:

• AB (grey long): 170
• ab (black vestigial): 160
• Ab (grey vestigial): 35
• aB (black long): 35

Expected (1:1:1:1)

• Each class: 100

Calculate χ²