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Learn9700Paper 4 · A Level Structured QuestionsInheritanceMonohybrid and dihybrid inheritance
Inheritance · Subtopic 2 of 2

Monohybrid and Dihybrid Inheritance

Monohybrid: one gene. Two heterozygotes (Aa × Aa) give 3 : 1 dominant : recessive. Codominance: both alleles expressed (e.g. Iᴬ Iᴮ blood group AB). Multiple alleles: more than two alleles in the population (Iᴬ, Iᴮ, Iᴼ). Sex linkage: gene on the X chromosome, so males (XY) show recessive traits more often (e.g. haemophilia). Dihybrid: two genes on different chromosomes; AaBb × AaBb gives 9 : 3 : 3 : 1; a test cross with aabb gives 1 : 1 : 1 : 1. Autosomal linkage: genes on the same chromosome are inherited together, giving more parental types and fewer recombinants than expected. Epistasis: one gene affects the expression of another, modifying ratios (e.g. 9 : 3 : 4). The chi-squared test χ2=∑(O−E)2E\chi^2=\sum\frac{(O-E)^2}{E} checks whether differences from expected ratios are due to chance.

Cambridge 9700 Paper 4Edexcel YBI11 WBI11AQA 7402 Paper 2≈ 55 minRevision 1 · updated 2026-11-02
Section 1 of 14

Learning objectives

  • Draw genetic diagrams for monohybrid and dihybrid crosses.
  • Explain codominance, multiple alleles and sex linkage.
  • Explain linkage and epistasis.
  • Use the chi-squared test.
Pause and think

Why are colour-blind men much more common than colour-blind women?

Section 2 of 14

Core idea

Start hereMendel's 9 : 3 : 3 : 1

Mendel crossed peas for years and found the same pattern again and again.

The big idea

Unlinked genes assort independently, giving predictable ratios. Linkage, epistasis and chance change what we observe; chi-squared tests whether chance explains it.

Dihybrid cross AaBb × AaBb
  1. gametes: AB, Ab, aB, ab\text{gametes: AB, Ab, aB, ab}
    Each parent, equal frequencies.
  2. 4×4=16 combinations4\times4=16\ \text{combinations}
    Punnett square.
  3. 9 A_B_:3 A_bb:3 aaB_:1 aabb9\,A\_B\_:3\,A\_bb:3\,aaB\_:1\,aabb
    Group by phenotype.
Worked exampleChi-squared

O: 90, 30, 28, 12. E: 90, 30, 30, 10.

Goal
χ² and conclusion.
Strategy
Σ(O−E)²/E; compare with 7.82.
  1. 0+0+430+410=0.530+0+\frac{4}{30}+\frac{4}{10}=0.53
    Sum terms.
  2. 0.53<7.820.53<7.82
    3 df, p = 0.05.
Answerχ² = 0.53; no significant difference.
✓ CheckObserved close to expected ✓.

Summary

  • 3 : 1, 9 : 3 : 3 : 1.
  • Sex linkage, codominance.
  • χ² test.
Key words
Allele
Alternative form of a gene.
Epistasis
Interaction where one gene affects another's expression.
HistoryMendel's peas. Gregor Mendel's dihybrid crosses of round/wrinkled and yellow/green peas gave 315 : 108 : 101 : 32, very close to 9 : 3 : 3 : 1.
Fun factLabrador colours. Labrador coat colour is a classic case of epistasis: a second gene decides whether pigment is deposited at all, giving yellow dogs.
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