Trihybrid Cross Calculator - Punnett Square

How to use this calculator

1. Select the first parent’s genotype at each of the A, B, and C loci, then do the same for the second parent. Each pair of letters refers to one locus.
2. Read the Punnett square: first-parent gametes label the columns and second-parent gametes label the rows. Each cell joins one allele per locus from each parent.
3. Open the genotype probabilities to inspect all 27 classes, including zero-probability classes. Read the eight phenotype groups separately; each list totals 100%.
4. Choose percent or fraction beside a result to change its display scale. For example, 12.5 percent and 0.125 fraction represent the same probability.

Formula and method

For the assumptions behind these results, read the explanation below. You can also explore Punnett Square and Dihybrid Cross.

Read the letters one locus at a time

The three pairs in AaBbCc describe three loci, each with two possible alleles in this model. A gamete contains only one letter from each pair, such as AbC. Combining AbC with aBc produces AaBbCc: the calculator writes the two A-locus alleles together, followed by the B and C pairs.

A genotype records the allele pairs. A phenotype is the expressed category assigned by the dominance model. Here A– means AA or Aa, whereas aa means two recessive copies; the dash stands for either second allele. The same convention applies to B– and C–. Thus A–bbC– groups several genotypes that share dominant A and C phenotypes and a recessive B phenotype.

Why the square can be smaller than 8 × 8

Each heterozygous locus doubles the number of unique gamete types under independent assortment. AaBbCc therefore has eight, while aaBbCC has only two: aBC and abC. A parent homozygous at all three loci has one. The table removes repeated gamete labels and uses the number of unique types for its dimensions.

The largest grid contains 64 equally likely cells, from eight column gametes and eight row gametes. Smaller crosses use fewer cells. For aaBbCC × AabbCc, the first parent supplies two gamete types and the second supplies four, giving eight equally likely cells. Each cell represents 1/8 of the modeled outcomes, rather than 1/64.

Check the familiar ratio and individual genotypes

With two triple heterozygotes, each locus separately has genotype probabilities 1/4, 1/2, and 1/4. Multiplying the relevant three probabilities gives any combined genotype. For example, AABbcc uses 1/4 × 1/2 × 1/4 = 1/32, or 3.125%. This provides a quick independent check on a selected row.

Grouping the default cross by complete-dominance phenotype gives 27:9:9:9:3:3:3:1 in the displayed order. The three categories with two dominant phenotypes each receive 9/64; those with one each receive 3/64. These weights apply to AaBbCc × AaBbCc. Other parental choices change the distribution and can make some categories impossible.

There are 27 possible three-locus genotype classes because each locus has three genotype states. They are not all equally likely, and not all occur in every cross. Keeping zero rows visible makes comparisons between crosses easier. The genotype and phenotype lists are two summaries of the same offspring outcomes, so their totals should not be added together.

Know when the model applies

This calculation assumes diploid parents, three autosomal loci with two alleles each, equal segregation, independent assortment, and random union of gametes. It does not model linked-locus recombination, sex linkage, mutation, or differences in survival or transmission. Linked alleles may require information about their arrangement and recombination frequencies.

The phenotype groups additionally assume complete dominance at every locus, full expression, and no epistasis between loci. Epistasis can change phenotype grouping even when genotype segregation remains independent. Many human characteristics involve multiple genes and environmental effects, so these symbolic categories do not establish a personal prediction of appearance or health.