# Monohybrid and dihybrid inheritance

> CIE A-Level Biology · 9700
> Source: https://www.owlsprep.com/study/cie-9700-u17-monohybrid-and-dihybrid-inheritance/

This sub-topic covers Mendelian inheritance patterns for monohybrid and dihybrid crosses, how to construct Punnett squares, predict genotypic and phenotypic ratios, and apply core genetic principles to solve CIE A-Level exam questions.

**Prerequisites:** [Meiosis and gamete formation](https://www.owlsprep.com/study/cie-9700-u16-meiosis/); [Basic genetic terminology](https://www.owlsprep.com/study/cie-9700-u17-introduction-to-inheritance/)

## Learning objectives

- Distinguish between monohybrid and dihybrid inheritance patterns
- Construct Punnett squares for monohybrid and dihybrid crosses
- Predict accurate genotypic and phenotypic ratios from genetic crosses
- Explain the role of independent assortment in dihybrid inheritance
- Use test crosses to determine unknown genotypes of dominant phenotypes

## Monohybrid Inheritance

**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.

**Worked example:** 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?

1. 1. Identify parental genotypes: both parents are heterozygous
2. $$Tt \times Tt$$
3. 2. Determine possible gametes from each parent: each parent produces two gamete types
4. $$T \text{ and } t$$
5. 3. Construct a 2×2 Punnett square to combine gametes, resulting in offspring genotypes: 1 TT, 2 Tt, 1 tt
6. 4. Count phenotypes: all TT and Tt offspring are tall (dominant), only tt are dwarf (recessive).

*Conclusion:* The expected phenotypic ratio is 3 tall : 1 dwarf, and the genotypic ratio is 1 TT : 2 Tt : 1 tt.

> **Exam tip:** Always confirm if the question asks for genotypic or phenotypic ratio before writing your answer.

*Calculator:* forbidden

## Dihybrid Inheritance and Independent Assortment

**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.

> **info**
>
> The 9:3:3:1 phenotypic ratio is only expected for a cross between two double heterozygotes when genes are unlinked and dominance is complete.

**Worked example:** 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?

1. 1. Parental genotypes for both parents are:
2. $$RrYy$$
3. 2. By independent assortment, each parent produces 4 equal gamete types
4. $$RY, Ry, rY, ry$$
5. 3. Construct a 4×4 Punnett square to combine all gamete combinations, resulting in 16 total offspring genotypes.
6. 4. Count phenotypes by dominant/recessive rules:

*Conclusion:* The expected phenotypic ratio is 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green, simplified to 9:3:3:1.

*Calculator:* forbidden

## Test Crosses for Unknown Genotypes

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.

**Worked example:** 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?

1. 1. Dwarf plants are always homozygous recessive, with genotype:
2. $$tt$$
3. 2. 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.
4. 3. If the original tall plant were heterozygous (Tt), it produces T and t gametes in equal proportion. Crossing with tt gives:
5. $$50\% \ Tt \ (tall), 50\% \ tt \ (dwarf)$$

*Conclusion:* The observed ratio matches, so the original tall plant has genotype $Tt$ (heterozygous).

*Calculator:* forbidden

## Common pitfalls

- **Wrong:** Only listing 2 gamete types for a dihybrid double heterozygote, instead of 4.
  - Why it fails: This error comes from incorrectly assuming dominant alleles stay together, which is only true for linked genes.
  - Correct: Always sort alleles independently: each dominant allele can combine with either dominant or recessive allele of the second gene, giving 4 gamete types.
- **Wrong:** Claiming 9:3:3:1 is the expected ratio for all dihybrid crosses.
  - Why it fails: This ratio only applies to crosses between two double heterozygotes for unlinked genes, not all dihybrid crosses.
  - Correct: Always check the parental genotypes and confirm the genes are unlinked before stating the 9:3:3:1 ratio.
- **Wrong:** Confusing genotypic ratio with phenotypic ratio in exam answers.
  - Why it fails: CIE exam questions explicitly ask for one or the other, mixing them up results in lost marks even if working is correct.
  - Correct: Highlight or underline what the question asks for (genotype vs phenotype) before starting your working.
- **Wrong:** Forgetting to count all 16 boxes in a dihybrid Punnett square, leading to wrong ratio totals.
  - Why it fails: Rushing to complete the cross leads to miscounting, especially when multiple genotypes match the same phenotype.
  - Correct: Count all 16 squares first, then group them by phenotype to calculate the final ratio.

## Cheatsheet

| Cross Type | Parental Genotypes (unlinked) | Expected Phenotypic Ratio |
| --- | --- | --- |
| Monohybrid heterozygote cross | Aa × Aa | 3 dominant : 1 recessive |
| Monohybrid test cross | Aa × aa | 1 dominant : 1 recessive |
| Dihybrid double heterozygote cross | AaBb × AaBb | 9:3:3:1 |
| Dihybrid test cross (double heterozygote) | AaBb × aabb | 1:1:1:1 |

## 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.

- [Codominance and multiple alleles](https://www.owlsprep.com/study/cie-9700-u17-codominance-and-multiple-alleles/)
- [Sex linkage](https://www.owlsprep.com/study/cie-9700-u17-sex-linkage/)
- [Linkage and Crossing Over](https://www.owlsprep.com/study/cie-9700-u17-linkage-and-crossing-over/)

---

From [OwlsPrep](https://www.owlsprep.com) — free study guides for A-Level, IB, AP and IGCSE, written against the official syllabus. Canonical page: https://www.owlsprep.com/study/cie-9700-u17-monohybrid-and-dihybrid-inheritance/