Trihybrid Cross Calculator (64-box Punnett square, AaBbCc phenotype ratio)

Understanding the trihybrid cross calculator — concept and principle

A trihybrid cross follows three genes at once, each with a dominant and a recessive allele. This calculator works out the full 64-box Punnett square for any combination of parent genotypes and reports the phenotype ratio, the number of gamete types, how many different genotypes and phenotypes appear and the fraction of offspring in each of the eight phenotype classes.

Each box is one of the 64 equally likely offspring; 27 show all three dominant traits.

How it is calculated

Each parent makes 2 × 2 × 2 = 8 gamete types when heterozygous for all three genes.

For each gene, P(dominant phenotype) comes from the single-gene cross: Aa × Aa gives 3/4 A_ and 1/4 aa; AA × anything gives 4/4 A_; aa × aa gives 0/4.

P(phenotype) = P(gene A class) × P(gene B class) × P(gene C class), written over 64.

"A_" means at least one dominant A allele; "aa" is homozygous recessive.
The ratio lists the non-zero classes in the order A_B_C_, A_B_cc, A_bbC_, aaB_C_, A_bbcc, aaB_cc, aabbC_, aabbcc, reduced by their common factor.

The principle

Mendel's law of segregation says each parent passes one of its two alleles at random, and the law of independent assortment says genes on different chromosomes are inherited independently. Independence means probabilities multiply. So instead of filling all 64 boxes, you can solve three small one-gene crosses and multiply. For a heterozygote cross, each gene gives 3 dominant to 1 recessive, and (3 + 1)³ expands to 27 + 9 + 9 + 9 + 3 + 3 + 3 + 1 = 64.

Worked example

Default inputs: AaBbCc × AaBbCc.

Gamete types: 2³ = 8 for each parent, so 8 × 8 = 64 boxes.
Each gene: 3/4 dominant, 1/4 recessive.
A_B_C_ = 3/4 × 3/4 × 3/4 = 27/64 (42.19 %).
Two dominant and one recessive: 3/4 × 3/4 × 1/4 = 9/64 (14.06 %) each for A_B_cc, A_bbC_ and aaB_C_.
One dominant: 3/64 (4.69 %) each for A_bbcc, aaB_cc and aabbC_; aabbcc = 1/64 (1.56 %).
Ratio 27 : 9 : 9 : 9 : 3 : 3 : 3 : 1, with 27 genotypes (3³) and 8 phenotypes.

How to read the result

The ratio is an expected probability, not a guarantee. In a litter or a small sample the counts will vary by chance; a chi-square test can show whether observed numbers fit the expected ratio. If one parent is homozygous for a gene, some classes disappear and the ratio becomes simpler. Large departures from the expected ratio may suggest linkage, lethal genotypes or incomplete dominance.

Common mistakes

Treating the ratio as exact counts in a small family.
Applying it to genes on the same chromosome (linked genes).
Assuming dominance when the trait shows incomplete or co-dominance.

Keep in mind: The model assumes complete dominance, unlinked genes and equally viable offspring. Real inheritance, especially in humans, is often more complex and is not predictable from this model alone.

Frequently asked questions

Why are there 64 boxes?

Each parent makes 8 gamete types, and 8 × 8 = 64 equally likely combinations.

How many genotypes come from AaBbCc × AaBbCc?

27, because each gene has three genotypes (AA, Aa, aa) and 3³ = 27.

What changes if the genes are linked?

Parental allele combinations appear more often than independent assortment predicts, so the ratio shifts.

Sources: OpenStax, Biology 2e, Chapter 12 "Mendel's Experiments and Heredity"

Checked 9 Nov 2026. Mendel’s laws of segregation and independent assortment; product rule for independent genes (standard genetics, e.g. OpenStax Biology 2e, ch. 12 “Mendel’s Experiments and Heredity”).

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