Most students can draw a Punnett square. Far fewer can tell you why the 9:3:3:1 ratio appears in a dihybrid cross, or what happens when you add a third gene into the mix. This post walks you through Mendelian inheritance from first principles — and includes a fully interactive simulator so you can run any cross yourself.
Gregor Mendel's pea plant experiments in the 1860s established two laws that form the bedrock of classical genetics. The Law of Segregation states that each organism carries two alleles per trait, and these separate during gamete formation so each gamete carries only one. The Law of Independent Assortment states that alleles at different loci sort into gametes independently of one another — producing all possible combinations with equal probability.
These laws hold whenever genes sit on different chromosomes (or are far enough apart on the same chromosome). When they do, the mathematics of inheritance is entirely predictable — and that predictability is exactly what we can simulate.
In CIE and Edexcel A Level questions, you're expected to state Mendel's laws explicitly before working a cross. One line suffices: "Alleles segregate during meiosis and assort independently, assuming the loci are on non-homologous chromosomes."
Each additional locus doubles the gamete types and squares the Punnett grid.
The simplest case: one gene with a dominant allele (A) and a recessive allele (a). Cross two heterozygotes (Aa × Aa) and you get the classic 1:2:1 genotypic ratio (AA : Aa : aa) and a 3:1 phenotypic ratio — three dominant phenotypes to every one recessive. Mendel observed this with seed shape: smooth (dominant) appeared roughly three times more often than wrinkled (recessive) in the F2 generation.
Add a second gene (B, controlling seed colour) and each parent now produces four gamete types: AB, Ab, aB, ab. The Punnett square grows to 4×4 = 16 squares. From AaBb × AaBb, the expected phenotypic ratio is 9:3:3:1 — nine plants showing both dominant traits, three showing dominant A only, three showing dominant B only, and one showing both recessives. This ratio is a direct consequence of independent assortment.
A third gene (C) produces eight gamete types per parent, giving an 8×8 = 64-square Punnett grid. Working this by hand is error-prone; the simulator below handles it instantly. The phenotypic ratio in an AaBbCc × AaBbCc cross is 27:9:9:9:3:3:3:1 across eight classes — but you can confirm this yourself in seconds using the tool.
| Cross type | Example (F1 × F1) | Gamete types | Grid size | Phenotypic classes |
|---|---|---|---|---|
| Monohybrid | Aa × Aa | 2 × 2 | 4 | 2 |
| Dihybrid | AaBb × AaBb | 4 × 4 | 16 | 4 |
| Trihybrid | AaBbCc × AaBbCc | 8 × 8 | 64 | 8 |
Select your cross type, enter any valid parental genotypes, and click Run cross. The simulator generates every offspring combination from the full Punnett grid and reports both genotypic and phenotypic frequencies. Try non-standard crosses: what does AA × aa look like? Or AaBB × aaBb?
The algorithm mirrors exactly what you would do by hand — just faster. For each parent, every possible gamete is generated by taking one allele from each locus pair. A heterozygote at two loci (AaBb) therefore produces four gametes: AB, Ab, aB and ab, each with equal probability. Every gamete from Parent 1 is then combined with every gamete from Parent 2, and alleles are sorted so the dominant (uppercase) allele is always listed first — this ensures genotype names are consistent regardless of which gamete contributed which allele.
Once all offspring genotypes are tallied, phenotype is determined by a simple rule: if a locus carries at least one uppercase allele, it expresses the dominant phenotype for that trait. Two lowercase alleles are needed to express the recessive phenotype. This is complete dominance — the simulator does not model co-dominance, incomplete dominance, or epistasis, all of which are worth exploring separately.
The 9:3:3:1 dihybrid ratio breaks down under epistasis — where one gene masks the expression of another. Labrador coat colour (B/b and E/e loci) produces a modified 9:3:4 ratio. This is A Level extension material and a common source of challenging exam questions.
A test cross uses a homozygous recessive parent to reveal an unknown genotype. Cross Aa × aa in the simulator — you should see a 1:1 ratio of heterozygotes to homozygous recessives, and a 1:1 phenotypic ratio of dominant to recessive. This is how Mendel inferred the existence of hidden recessive alleles before genetics was even a word.
Cross two true-breeding parents differing in both traits. Select Dihybrid, enter AABB and aabb. Every single offspring is AaBb — all heterozygous, all showing the dominant phenotype. Now cross those F1s with each other (AaBb × AaBb) to see the 9:3:3:1 emerge in the F2 generation. This is Mendel's original experiment, replicated in a second.
Select Trihybrid, enter AABbCc for both parents. Because the A locus is fixed (AA × AA always gives AA), the trihybrid reduces to an effective dihybrid at the B and C loci — you'll see only four phenotypic classes rather than eight, with ratios of 9:3:3:1. This is a classic trick question in A Level papers.
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