D3.2.21 (HL)—Chi-squared test

The chi-squared test compares observed and expected genetic results to judge whether differences could be due to chance in inheritance problems.

Syllabus
First assessment 2025
Objective
D3.2.21
Level
HL

Exam analysis

Chance of appearing2%of analysed past papers
Latest appearanceNovember 2021
Most common paperPaper1
Typical marks1–2

Recent exam appearances

November 2021Paper1 ["HL"] · TZ035[ 1 ]D3.2.21 (HL)—Chi-squared test
November 2019Paper2 ["HL"] · TZ02(c)[ 2 ]D3.2.21 (HL)—Chi-squared test

Coverage 2019–2021 · Updated 16 Jul 2026

Chi-Squared Begins with a Genetic Model and Expected Counts

HL only
Statement Meaning in a dihybrid goodness-of-fit test
null hypothesis, H₀ observed departures from the expected genetic ratio are due to chance sampling
alternative hypothesis, H₁ the departures are too large for that chance model; another explanation is needed

Convert the ratio to expected counts before calculating χ². For a total of 556 offspring and a 9:3:3:1 model: multiply 556 by 9/16, 3/16, 3/16 and 1/16 to obtain 312.75, 104.25, 104.25 and 34.75.

Categories must be mutually exclusive, observations should be independent and expected counts should be large enough for the approximation. The expected ratio must come from a biological model chosen before inspecting the deviations.

Chi-squared tests whether counts fit a specified expectation; it does not identify the true alternative mechanism when they do not fit.

Chi-Squared Adds the Scaled Departure of Every Category

HL only

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

Category O E (O − E)² / E
both dominant 315 312.75 0.016
A dominant only 108 104.25 0.135
B dominant only 101 104.25 0.101
both recessive 32 34.75 0.218
sum 556 556 χ² = 0.470

For four mutually exclusive categories with one fixed total and no parameters estimated from the data, degrees of freedom = categories − 1 = 3.

Use counts, not percentages, inside the formula. Keep expected values unrounded during calculation, and include every category contribution before comparing with a critical value.

A chi-squared table compares observed and expected Drosophila phenotype counts, sums category contributions to 0.47 and compares the result with the p equals 0.05 critical value for three degrees of freedom.

Compare Chi-Squared with the Critical Value, Then State Only What the Test Supports

HL only
At p = 0.05 Statistical decision Genetic conclusion
χ² ≤ critical value fail to reject H₀ deviations are not significant; data are consistent with the expected ratio
χ² > critical value reject H₀ deviations are significant; the stated ratio does not adequately explain the data

For df = 3, the p = 0.05 critical value is 7.815. Because 0.470 < 7.815, fail to reject H₀: the observed Drosophila counts are consistent with a 9:3:3:1 expectation.

Do not say the test proves H₀ or proves independent assortment. A non-significant result means the observed departure is small enough to be explained by sampling under the model.

A significant result directs further investigation. Possible causes include linkage, differential survival, non-random fertilization, scoring error or an incorrect dominance model; chi-squared alone does not choose among them.

Inheritance Models Are Built from Chromosomes and Tested with Offspring Counts

HL only

meiosis predicts gamete combinations → chromosome loci determine whether genes assort independently or travel together → crossing over creates reciprocal recombinants → test-cross counts reveal parental phase and recombinant frequency → chi-squared tests whether departures from a chosen ratio exceed chance expectation

Offspring pattern First model to examine
9:3:3:1 F2 two unlinked genes with complete dominance
1:1:1:1 test cross four equally frequent gametes from an unlinked double heterozygote
two large and two small reciprocal test-cross classes linked loci with crossing over
significant χ² departure assumptions, linkage, viability, fertilization or scoring require review

Keep prediction and evidence separate: define the genetic model first, calculate expected counts from it, compare observations with the expectation, and change the model only when chromosome location or statistical evidence requires it.

Chi-squared test

HL only

[ 2 ]

The chi-squared value was calculated as shown. Deduce, with reasons, whether the observed ratio differed significantly from the expected Mendelian ratio.

c2=Σ( Observed − Expected )2 Expected =1002.6c^{2}=\Sigma \frac{(\text { Observed }- \text { Expected })^{2}}{\text { Expected }}=1002.6
Probability
Degrees of freedom0.9950.9750.200.100.050.0250.020.010.0050.0020.001
10.000040.0011.6422.7063.8415.0245.4126.6357.8799.55010.828
20.0100.0513.2194.6055.9917.3787.8249.21010.59712.42913.816
30.0720.2164.6426.2517.8159.3489.83711.34512.83814.79616.266
40.2070.4845.9897.7799.48811.14311.66813.27714.86016.92418.467
50.4120.8317.2899.23611.07012.83313.38815.08616.75018.90720.515
60.6761.2378.55810.64512.59214.44915.03316.81218.54820.79122.458
70.9891.6909.80312.01714.06716.01316.62218.47520.27822.60124.322

Concept essentials

  • Chi-squared compares observed counts with expected counts.
  • Expected values must come from a stated genetic hypothesis or ratio.
  • A larger chi-squared value means a poorer fit to the expected ratio.