Trihybrid Cross Calculator

Background.

A trihybrid cross follows three genes at once, and it is the point where drawing a Punnett square stops being sensible. Three heterozygous genes give each parent 2³ = 8 kinds of gamete, so the full grid is 8 × 8 = 64 boxes. Nobody fills in 64 boxes by hand. The standard method is the forked-line diagram, or more directly the product rule: work out the probability at each gene separately and multiply. That is how this calculator is built, and why its main question is 'what is the chance of this particular combination?' rather than 'draw me a grid'.

Pick a phenotype — dominant or recessive — at each of the three genes, enter both parents' genotypes, and the page returns the probability of exactly that combination, the same answer as an exact reduced fraction, and the combinatorics: how many distinct gametes each parent makes, how many cells the smallest equally-likely grid would need, and how many phenotype and genotype classes exist at all.

For the standard cross AaBbCc × AaBbCc the probability of being dominant at all three genes is (3/4)³ = 27/64 = 42.1875 %, and the probability of being recessive at all three is (1/4)³ = 1/64 = 1.5625 %. The eight classes form the series 27 : 9 : 9 : 9 : 3 : 3 : 3 : 1, which sums to 64. Beneath them sit 3³ = 27 genotype classes. Mendel derived all three of those series — 2ⁿ gametes, 3ⁿ genotypes, 4ⁿ combinations — for n differing characters in his 1866 paper, and tested them: his trifactorial experiment resolved 639 plants into exactly 27 classes.

The calculator enumerates all 64 cells internally rather than assuming the neat powers of 3/4, so it is equally correct for lopsided crosses. Enter AABbCc × AaBbcc, or aabbcc as the second parent for a three-point testcross, and the numbers adapt: a testcross flattens all eight classes to 12.5 % each, which is exactly the flat readout that makes three-point testcrosses the classical tool for gene mapping.

The assumption that carries all of this is independent assortment, and with three genes it is a stronger assumption than it looks. Two genes involve one pairwise relationship; three genes involve three, and every one of them has to be unlinked for 27:9:9:9:3:3:3:1 to hold. Genes that sit close together on the same chromosome are linked, and then the parental combinations turn up more often than the model says and the recombinant ones less often. Epistasis is the other common failure: the genes assort perfectly but one masks another's phenotype, merging visible classes. Reduced penetrance and lethal alleles move or delete classes too. A departure from the expected series in real progeny is usually biology, not arithmetic.

And the numbers on this page are exact ratios for an idealised cross, not a personal risk figure. They describe expected proportions over many offspring under a specific model; they do not predict any individual, and they should not be read as a clinical probability for a real family. Below the widget: the derivation, the forked-line method, the worked example against Mendel's own 639-plant experiment, and where the model stops being trustworthy.

What is trihybrid cross calculator?

A trihybrid cross is a mating tracked at three genes at once, written with three letters — AaBbCc × AaBbCc is the canonical case. Under Mendel's law of independent assortment, the allele passed on at each gene is chosen independently of the other two, so a parent heterozygous at all three produces 2³ = 8 gamete types (ABC, ABc, AbC, Abc, aBC, aBc, abC, abc) in equal numbers. Crossing eight gamete types against eight gives a 64-cell Punnett square in which every cell is equally likely.

Because 64 boxes are impractical, the trihybrid case is normally solved by the forked-line method or by the product rule. Both are the same idea: treat each gene as its own monohybrid cross, read off the probability of the phenotype you want at that gene, and multiply the three together. For a heterozygote × heterozygote cross the per-gene probabilities are 3/4 dominant and 1/4 recessive, so P(A_B_C_) = 3/4 × 3/4 × 3/4 = 27/64 and P(aabbcc) = 1/4 × 1/4 × 1/4 = 1/64. The eight combinations give the 27 : 9 : 9 : 9 : 3 : 3 : 3 : 1 series.

The underscore notation used throughout means 'either allele at that gene'. A_ covers AA and Aa alike, because complete dominance makes them indistinguishable; aa is written out in full because a recessive phenotype pins the genotype exactly. That asymmetry is why the triple-recessive class is the useful one experimentally — aabbcc is the only one of the eight phenotype classes that corresponds to a single genotype.

A three-point testcross is the trihybrid version of a testcross: the unknown individual is crossed to aabbcc, which contributes only recessive alleles, so every offspring's appearance reveals which gamete the unknown parent supplied. Under independent assortment the eight classes come out equal at 12.5 % each; the pattern of departure from that flat expectation is what classical three-point mapping uses to order three genes on a chromosome and measure the distances between them.

How to use this calculator.

1. 01Write each parent's genotype as six letters: characters 1–2 for gene 1, 3–4 for gene 2, 5–6 for gene 3. Uppercase is dominant, lowercase recessive, and the three genes must use three different letters.
2. 02Enter both parents using the same three genes in the same order. Order between the parents does not matter — swapping them cannot change any probability.
3. 03Set the three phenotype selectors to the combination you want. All eight combinations are valid, and a combination that cannot occur returns 0% with the fraction 0/1 rather than an error.
4. 04Read the headline probability and, for written work, the exact fraction beside it — 27/64 is exact where 42.19% is a rounded decimal.
5. 05Compare your class against the two anchors always shown: dominant at all three (the largest class) and recessive at all three (the smallest).
6. 06Use the combinatorics row to sanity-check the setup: distinct gametes should be 8 for a triple heterozygote, the smallest grid 64, genotype classes 27 and phenotype classes 8.
7. 07For a three-point testcross enter aabbcc as parent 2. Every class flattens to 12.5%, and any departure from that in real progeny is what you measure recombination frequencies from.
8. 08For two genes use the dihybrid cross calculator, which also gives expected counts for a chi-square test; for one gene use the Punnett square calculator, which adds incomplete dominance and codominance.

The formula.

Three loci, one rule. Segregation gives each gene a monohybrid probability; independent assortment lets you multiply across genes.

Step one, per gene. For each of the three genes, work out the chance of the phenotype you asked for exactly as you would for a single-gene cross. A heterozygote × heterozygote cross gives 3/4 dominant and 1/4 recessive; a cross with a homozygous dominant parent gives 1 and 0; a cross with two homozygous recessive parents gives 0 and 1.

Step two, multiply. Because the three genes assort independently,

P(class) = P₁ × P₂ × P₃

For AaBbCc × AaBbCc: P(A_B_C_) = ¾ × ¾ × ¾ = 27⁄64 = 42.1875 %, P(A_B_cc) = ¾ × ¾ × ¼ = 9⁄64 = 14.0625 %, P(A_bbcc) = ¾ × ¼ × ¼ = 3⁄64 = 4.6875 %, and P(aabbcc) = ¼ × ¼ × ¼ = 1⁄64 = 1.5625 %. Collecting all eight gives 27 : 9 : 9 : 9 : 3 : 3 : 3 : 1, summing to 64.

The combinatorics follow the same logic. For n genes at which a parent is heterozygous it makes 2ⁿ gamete types; a cross between two such parents fills 4ⁿ grid cells and produces 3ⁿ genotype classes and 2ⁿ phenotype classes under complete dominance. For n = 3 that is 8, 64, 27 and 8. Mendel stated all of these in his 1866 paper, and this calculator reports them so you can check your own setup.

Internally the code does not assume the powers of ¾ — it enumerates all 64 cells of the grid and counts. That matters for lopsided crosses such as AABbCc × AaBbcc, where the per-gene probabilities are not ¾ and ¼ and the tidy 27:9:9:9:3:3:3:1 does not appear. The test suite checks the enumeration against the per-gene product for both symmetric and asymmetric crosses, so the two routes are known to agree rather than assumed to.

ROUNDING STAGE. Rounding happens only at the final return, to ten decimal places, applied to values that are already exact rationals with denominator 64. There is no intermediate rounding and no threshold. The fraction output is reduced with an exact integer greatest common divisor computed from the raw counts, never from the rounded percentage — which is why it can report 27/64 and 1/8 exactly.

WHERE THE MODEL BREAKS. Independent assortment must hold for all three pairs of genes, not just on average. Linkage between any pair skews the series toward the parental combinations. Epistasis leaves the genotype proportions untouched but merges visible classes. Lethal alleles delete a class before it can be scored. Reduced penetrance puts individuals into the wrong visible class. In each case the counting is still right and the genotype-to-phenotype mapping, or the independence assumption, is what has failed.

INVALID DOMAIN. There is no singularity — the denominator is the constant 64, never a user value. The calculator rejects, field by field: a genotype that is not six letters, any gene whose two characters are different letters, two genes written with the same letter, a non-letter character, parents listing different genes or the same genes in a different order, and a phenotype selector that is neither dominant nor recessive. Asking for a class that cannot occur is not an error: it returns 0 % and the fraction 0/1, which is the correct answer.

A worked example.

Frequently asked questions.

References& sources.

1. [1]Abbott S. & Fairbanks D. J. (2016). Experiments on Plant Hybrids by Gregor Mendel. Genetics 204(2):407–422. doi:10.1534/genetics.116.195198. Peer-reviewed English translation of Mendel (1866). Source for the trifactorial experiment — 639 plants resolved into 27 classes, with the printed class averages of 10, 19, 43 and 78 — and for the 2ⁿ gamete / 3ⁿ genotype / 4ⁿ combination series. Open access. Retrieved 2026-07-29.
2. [2]OpenStax, Biology 2e, section 12.3 'Laws of Inheritance'. Rice University, 2018, CC BY 4.0. Source for the forked-line method — 'create rows equal to the number of genes being considered, and then segregate the alleles in each row on forked lines' — and for the trihybrid result 'the probability of F2 offspring having yellow, round, and tall traits is 3 × 3 × 3, or 27'. Free, openly licensed. Retrieved 2026-07-29.
3. [3]Biology LibreTexts, Map: Raven Biology 12th Edition, section 12.03 'Dihybrid Crosses and Mendel's Law of Independent Assortment'. Independent-publisher statement of the general multi-locus product rule — 'the probability of the desired genotype at the first locus multiplied by the probability of the desired genotype at the other loci' — worked through to a tetrahybrid case: '(1/4) × (1/4) × (1/4) × (1/4), we determine that 1/256 of the offspring will be quadruply homozygous recessive'. Free. Retrieved 2026-07-29.
4. [4]Reid J. B. & Ross J. J. (2011). Mendel's genes: toward a full molecular characterization. Genetics 189(1):3–10. doi:10.1534/genetics.111.132118. Source for the linkage caveat: Mendel's seven traits occupy only five of the seven pea linkage groups, with le and v about 12.6 map units apart, so 'an element of luck was involved with his choice of characters'. Open access via PMC. Retrieved 2026-07-29.
5. [5]MedlinePlus Genetics, US National Library of Medicine. Inheritance Patterns, page last updated 19 April 2021. Source for the classical single-gene inheritance definitions and for the position that most visible human traits are not governed by one or a few genes. Free. Retrieved 2026-07-29.