Monohybrid and dihybrid inheritance
CIE A-Level BiologyΒ· Unit 17: InheritanceΒ· 25 min read
1. Monohybrid Inheritanceβ β ββββ± 8 min
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Monohybrid Inheritance
The inheritance of a single gene that controls one specific characteristic, with one allele inherited from each parent.
Example:
Stem height in pea plants, controlled by one gene with two alleles: tall (T) and dwarf (t).
Monohybrid inheritance follows Mendel's law of segregation: alleles separate during gamete formation, so each gamete carries only one allele for each gene. Dominant alleles are expressed in both homozygous and heterozygous genotypes, while recessive alleles are only expressed when homozygous.
In pea plants, tall stem (T) is dominant to dwarf stem (t). Cross two heterozygous tall plants. What is the expected phenotypic ratio of offspring?
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- Identify parental genotypes: both parents are heterozygous
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- Determine possible gametes from each parent: each parent produces two gamete types
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- Construct a 2Γ2 Punnett square to combine gametes, resulting in offspring genotypes: 1 TT, 2 Tt, 1 tt
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- Count phenotypes: all TT and Tt offspring are tall (dominant), only tt are dwarf (recessive).
Exam tip:
Always confirm if the question asks for genotypic or phenotypic ratio before writing your answer.
2. Dihybrid Inheritance and Independent Assortmentβ β β βββ± 10 min
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Dihybrid Inheritance
The inheritance of two unlinked genes (located on separate homologous chromosomes) that each control a different characteristic.
Example:
Seed shape and seed colour in pea plants, controlled by two separate genes.
For unlinked genes, alleles assort independently during meiosis I: the allele inherited for one gene does not affect the allele inherited for the other. A double heterozygote (one dominant and one recessive allele for each gene) produces 4 different gamete types in equal proportion.
In pea plants, round seeds (R) are dominant to wrinkled (r), and yellow seeds (Y) are dominant to green (y). Cross two double heterozygous round yellow plants. What is the expected phenotypic ratio?
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- Parental genotypes for both parents are:
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- By independent assortment, each parent produces 4 equal gamete types
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- Construct a 4Γ4 Punnett square to combine all gamete combinations, resulting in 16 total offspring genotypes.
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- Count phenotypes by dominant/recessive rules:
3. Test Crosses for Unknown Genotypesβ β β βββ± 7 min
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Individuals with a dominant phenotype can be either homozygous dominant or heterozygous, so their genotype cannot be determined from phenotype alone. A test cross crosses the unknown individual with a homozygous recessive individual, since homozygous recessive individuals always produce one type of gamete carrying the recessive allele.
A tall pea plant of unknown genotype is test crossed with a dwarf plant. 50% of offspring are tall and 50% are dwarf. What is the genotype of the original tall plant?
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- Dwarf plants are always homozygous recessive, with genotype:
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- If the original tall plant were homozygous dominant (TT), all offspring would inherit one T from the parent and one t from the dwarf parent, resulting in all Tt (tall) offspring. This does not match the result.
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- If the original tall plant were heterozygous (Tt), it produces T and t gametes in equal proportion. Crossing with tt gives:
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4. Common Pitfalls
Wrong move:
Only listing 2 gamete types for a dihybrid double heterozygote, instead of 4.
Why:
This error comes from incorrectly assuming dominant alleles stay together, which is only true for linked genes.
Correct move:
Always sort alleles independently: each dominant allele can combine with either dominant or recessive allele of the second gene, giving 4 gamete types.
Wrong move:
Claiming 9:3:3:1 is the expected ratio for all dihybrid crosses.
Why:
This ratio only applies to crosses between two double heterozygotes for unlinked genes, not all dihybrid crosses.
Correct move:
Always check the parental genotypes and confirm the genes are unlinked before stating the 9:3:3:1 ratio.
Wrong move:
Confusing genotypic ratio with phenotypic ratio in exam answers.
Why:
CIE exam questions explicitly ask for one or the other, mixing them up results in lost marks even if working is correct.
Correct move:
Highlight or underline what the question asks for (genotype vs phenotype) before starting your working.
Wrong move:
Forgetting to count all 16 boxes in a dihybrid Punnett square, leading to wrong ratio totals.
Why:
Rushing to complete the cross leads to miscounting, especially when multiple genotypes match the same phenotype.
Correct move:
Count all 16 squares first, then group them by phenotype to calculate the final ratio.
5. Quick Reference Cheatsheet
Cross Type | Parental Genotypes (unlinked) | Expected Phenotypic Ratio |
|---|---|---|
Monohybrid heterozygote cross | Aa Γ Aa | 3 dominant : 1 recessive |
Monohybrid test cross | Aa Γ aa | 1 dominant : 1 recessive |
Dihybrid double heterozygote cross | AaBb Γ AaBb | 9:3:3:1 |
Dihybrid test cross (double heterozygote) | AaBb Γ aabb | 1:1:1:1 |
6. Frequently Asked
Why is 9:3:3:1 only for unlinked genes?
Linked genes are on the same chromosome and do not assort independently, so they do not produce this expected phenotypic ratio.
What is the purpose of a test cross?
A test cross crosses a dominant phenotype individual of unknown genotype with a homozygous recessive individual. Offspring phenotypes reveal the unknown genotype.
When this came up on past exams
AI-estimated based on syllabus patterns β cross-check with official past papers for accuracy. Use only as revision-focus signals.
- 2022 Β· 22
Dihybrid cross ratio prediction
- 2023 Β· 12
Monohybrid test cross interpretation
- 2024 Β· 21
Linked vs unlinked dihybrid comparison
Going deeper
What's Next
Mastering monohybrid and dihybrid inheritance is the foundation for all advanced genetics topics in CIE A-Level Biology. The core skills of gamete identification, Punnett square construction, and ratio prediction you practiced here transfer directly to more complex inheritance patterns. This topic regularly forms the basis of extended response questions on Paper 2, so it is critical to be confident with ratio calculations and test cross interpretation before moving on. Understanding how independent assortment works for unlinked genes also helps you recognise deviations from expected ratios when genes are linked or interact via epistasis.