Test an inheritance null with chi-square reasoning

Mechanism before a label

An inheritance model predicts category probabilities before the offspring are counted. Multiply those probabilities by the total count to obtain expected frequencies. Pearson chi-square adds (observed−expected)^2/expected across categories; each category must be counted once.

The approximate chi-square reference requires independent observations, mutually exclusive categories and adequately large expected counts. For a fully specified four-category ratio, degrees of freedom are 4−1=3. Estimating parameters from the same data can change degrees of freedom; related offspring or pooled biased samples can violate independence.

A large statistic is evidence of disagreement with the specified null under its assumptions. Failure to reject does not prove the inheritance model, and rejection does not uniquely establish linkage. Misclassification, unequal survival or an incorrect dominance assumption can also cause disagreement.

Predict, compare and calculate

The following record is an original synthetic teaching dataset. Its assigned conditions are not a claim that an actual organism was measured. Write your prediction before revealing the changed condition.

Observed categories	Value (offspring)
AB	90
A_bb	30
aaB_	30
aabb	10

The predeclared null is 9:3:3:1 with N=160; these equal expected counts.

Your prediction before comparison

This scratch prediction is not saved automatically. Copy it to the lesson notebook to retain it.

Evidence conditionReference comparisonChanged condition / follow-up

Test an inheritance null with chi-square reasoning

Supplied synthetic evidence: The predeclared null is 9:3:3:1 with N=160; these equal expected counts.

Observed categories: reference condition

Quantity	Value	Unit
A_B_	90	offspring
A_bb	30	offspring
aaB_	30	offspring
aabb	10	offspring

Task: If the fourth expected count were 1, what would you check before using the same reference distribution?

This deliberately perfect packet is not evidence that real observations must match exactly.

Compare with the conditional interpretation

The chi-square statistic is zero for this constructed exact match.

The changed packet contributes 10²/90+(-10)²/30+0+0=4.444…. Because 4.444<7.815, fail to reject at the stipulated 5% level. Do not report a p-value that has not been calculated.

Case index for this investigation

0. Track homologs and chromatids through meiosis
1. Explain crossing over and assortment without mixing mechanisms
2. Derive cross probabilities and distinguish expectation from outcome
3. Use a testcross to estimate recombination and retain limits
4. Test an inheritance null with chi-square reasoning
5. Distinguish dominance, codominance and multiple alleles
6. Interpret chromosome-linked and cytoplasmic inheritance
7. Separate genotype, environment and their interaction

No numerical confidence, laboratory credit or independent mastery is inferred from opening a supplied case. Save your first prediction and comparison in the notebook.

Reset this unsaved comparison

Change the counts, not the prespecified model

Original synthetic practice. Start the case, predict a result, then edit one class. These scratch values are not saved or graded. The four categories must remain mutually exclusive; repeated counts are not new replicates.

Start / reset this calculation

Worked reasoning

The changed packet contributes 10²/90+(-10)²/30+0+0=4.444…. Because 4.444<7.815, fail to reject at the stipulated 5% level. Do not report a p-value that has not been calculated.

The chi-square statistic is zero for this constructed exact match.

Apply to a new case

If the fourth expected count were 1, what would you check before using the same reference distribution?

Compare your reasoning with the worked response

The small expected count may make the chi-square approximation unreliable. Use a justified exact or simulation-based method with the prespecified model, or redesign sampling; do not merge categories merely to obtain a preferred decision.

This public response supplies feedback, not a score for independent reasoning. Record which part of your prediction changed and why.

Evidence and investigation boundary

This deliberately perfect packet is not evidence that real observations must match exactly. The critical-value decision assumes the null and approximation conditions; no proof of independence follows.

For a supervised investigation, predeclare the question, biological unit of replication, treatment/comparison assignment, quantities and units, exclusion rules, graph and uncertainty method. Keep the raw records and the first prediction separate from later revisions. A simulated case does not supply laboratory hours or establish AP authorization. No physical manipulation is required by this lesson.

Source and continuation

Framework topics: 5.3. Current AP Biology framework; supporting scientific reference. Sources checked 30 October 2026; all teaching wording, questions and numbers here are original. Return to AP Biology to compare this mechanism with another scale of organization.