Why does Aa × Aa give 3 : 1 and not 1 : 2 : 1?

The genotype ratio is 1 AA : 2 Aa : 1 aa. With complete dominance AA and Aa look the same, so three of the four cells show the dominant phenotype and one shows the recessive one.

Where does 9 : 3 : 3 : 1 come from?

Each gene independently gives ¾ dominant and ¼ recessive phenotype. Multiplying the two: ¾·¾ = 9/16 both dominant, ¾·¼ = 3/16 twice for the mixed classes, and ¼·¼ = 1/16 both recessive.

Why do my 100 simulated offspring not split exactly 75 : 25?

Each offspring is an independent random draw, so counts fluctuate around the expected value by roughly ±√(np(1−p)) ≈ ±4 for n = 100 and p = ¾. Click Simulate several times and watch the observed percentages approach the expected ones.

What is a test cross used for?

Crossing an individual with the dominant phenotype to a homozygous recessive (aa) reveals its genotype: AA × aa gives only dominant offspring, while Aa × aa gives a 1 : 1 split.

Punnett Square: Monohybrid & Dihybrid Crosses

Mendelian crosses: 2×2 / 4×4 Punnett grid, genotype and phenotype ratios (3:1, 9:3:3:1), incomplete dominance, 100 sampled offspring.

About this model

The simulator builds a Punnett square for one gene (monohybrid, 2×2 grid) or two genes (dihybrid, 4×4 grid). Each parent produces gametes that carry one allele per gene: a homozygote AA makes only A gametes, a heterozygote Aa makes A and a with equal probability, and in the dihybrid case AaBb makes AB, Ab, aB and ab, each with probability ¼ (independent assortment). Every cell of the grid is one equally likely gamete pairing, so the share of cells with a given genotype is its probability among offspring. With complete dominance the heterozygote shows the dominant phenotype, giving the classical 1 : 2 : 1 genotype ratio and 3 : 1 phenotype ratio for Aa × Aa, 1 : 1 for the test cross Aa × aa, and 9 : 3 : 3 : 1 for AaBb × AaBb. The incomplete-dominance toggle makes the heterozygote intermediate (red × white → pink flowers), so genotype and phenotype ratios coincide. The "Simulate 100 offspring" button draws random cells and accumulates observed counts next to the expected ones, showing that a real family of 4 or a sample of 100 rarely matches the ideal ratio exactly, while large samples converge to it. The model omits sex linkage, linked genes and recombination, codominance, epistasis, lethal alleles, mutation and any environmental influence on the phenotype; the eye-colour preset is a deliberately simplified one-gene toy, not real human genetics.

Who it's for: Grades 7–11 biology students meeting Mendel's laws, probability in heredity and sampling variation for the first time.

Key terms

Allele
Genotype and phenotype
Homozygous / heterozygous
Dominant and recessive
Incomplete dominance
Test cross
Independent assortment
Sampling error

How it works

Each parent makes gametes carrying one allele per gene (two kinds for a heterozygote). The Punnett square lists every gamete pairing with equal probability, so the fraction of cells with a genotype is its probability in the offspring. With complete dominance the heterozygote looks like the dominant homozygote; with incomplete dominance it is intermediate. Simulating offspring draws random cells, so observed counts scatter around the expected ratio — more draws make the percentages converge.

Key equations

Aa × Aa: ¼ AA + ½ Aa + ¼ aa → 3 dominant : 1 recessive

Aa × aa (test cross): ½ Aa + ½ aa → 1 : 1

AaBb × AaBb: (¾ A_ + ¼ aa)(¾ B_ + ¼ bb) → 9 : 3 : 3 : 1

P(genotype) = cells with that genotype / total cells

Frequently asked questions

Why does Aa × Aa give 3 : 1 and not 1 : 2 : 1?: The genotype ratio is 1 AA : 2 Aa : 1 aa. With complete dominance AA and Aa look the same, so three of the four cells show the dominant phenotype and one shows the recessive one.
Where does 9 : 3 : 3 : 1 come from?: Each gene independently gives ¾ dominant and ¼ recessive phenotype. Multiplying the two: ¾·¾ = 9/16 both dominant, ¾·¼ = 3/16 twice for the mixed classes, and ¼·¼ = 1/16 both recessive.
Why do my 100 simulated offspring not split exactly 75 : 25?: Each offspring is an independent random draw, so counts fluctuate around the expected value by roughly ±√(np(1−p)) ≈ ±4 for n = 100 and p = ¾. Click Simulate several times and watch the observed percentages approach the expected ones.
What is a test cross used for?: Crossing an individual with the dominant phenotype to a homozygous recessive (aa) reveals its genotype: AA × aa gives only dominant offspring, while Aa × aa gives a 1 : 1 split.

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About this model

How it works

Key equations

Frequently asked questions

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About this model

How it works

Key equations

Frequently asked questions