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Punnett square calculator

The square of a mono or dihybrid cross: genotype and phenotype ratios, from 3:1 to 9:3:3:1.

Parameters

The cross
Gene A

Square

4 equally likely cells
Monohybrid · complete dominanceParent 2AaAaParent 1AaAaAAAaAaaaAAAaaa1/42/41/4dominant 3/4recessive 1/4

AAAaaa

Meiosis
gametes A, a and A, a → 4 equally likely cells
Count
genotypes 1 AA : 2 Aa : 1 aa
Regrouping
A masks a: dominant 3/4, recessive 1/4
Note

The 9:3:3:1 is (3:1) × (3:1) and assumes independent assortment.

Result

continuous computation
Offspring
Aa × Aa3 : 1phenotypes, dominant then recessive
Dominant homozygotes
1/4
Heterozygotes
2/4
Recessive homozygotes
1/4
Dominant phenotype
3/4
Recessive phenotype
1/4
Gametes crossed
2 × 2

Calculation, assumptions and sources

OverviewPunnett square of a cross: the gamete grid, the genotype (1:2:1) and phenotype (3:1) ratios of the monohybrid, or the four classes of the dihybrid (9:3:3:1).

Punnett square of a cross: the gamete grid, the genotype (1:2:1) and phenotype (3:1) ratios of the monohybrid, or the four classes of the dihybrid (9:3:3:1). The ratios are per-offspring probabilities, under complete dominance and, for the dihybrid, with independent genes.

Method & formulasgametes: one allele of each pair, in equal shares
gametes: one allele of each pair, in equal sharesAa × Aa → 1 AA : 2 Aa : 1 aa → 3 dominant : 1 recessivedihybrid: (3:1) × (3:1) = 9:3:3:1

Each cell, the product of two gametes, has the same probability; counting cells gives the ratios. The independent dihybrid is the product of two monohybrids.

Gamete
the reproductive cell, carrying a single allele of each gene once meiosis has separated the pairs.
Test cross
a cross with a recessive homozygote (aa), whose offspring reveal the genotype of the tested parent.
F1, F2
Mendel’s generations. The F1 of an AA × aa is uniform (all Aa); in the F2 (F1 × F1), the recessive reappears in one quarter.
Independent assortment
two genes transmitted independently, the condition of the 9:3:3:1, met for different chromosomes.
Validity domainThe model assumes one autosomal two-allele gene, complete dominance, independent genes in a dihybrid and large offspring counts.

The model assumes one autosomal two-allele gene, complete dominance, independent genes in a dihybrid and large offspring counts. It does not cover linked genes (distorted ratios, the basis of mapping), codominance and incomplete dominance (three phenotypes), sex-linked genes (ratios depending on the cross direction) or incomplete penetrance. In those cases the genotypic square stays right; only its phenotypic reading changes.

Per-offspring probabilitiesA litter of four from an Aa × Aa has only about a 42% chance of being exactly 3-1, and can be 4-0.

A litter of four from an Aa × Aa has only about a 42% chance of being exactly 3-1, and can be 4-0. The binomial distribution describes litters; the chi-square test judges whether observed counts are compatible with the expected ratios.

Reading the resultIn a test cross (unknown dominant × aa), half-recessive offspring point to an Aa parent; with no recessive at all, the parent is AA, unless the litter is too small to conclude.

In a test cross (unknown dominant × aa), half-recessive offspring point to an Aa parent; with no recessive at all, the parent is AA, unless the litter is too small to conclude.