Trihybrid Cross Calculator

Calculate offspring probabilities for trihybrid crosses using multiplication rule.

Gene A cross

Gene B cross

Gene C cross

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Why Multiplication Replaces the Grid

A trihybrid Punnett square has 64 boxes and a tetrahybrid has 256. The product rule handles any number of independently assorting genes in one line, because independent events multiply.

P(combined) = P(A) × P(B) × P(C)

Heterozygous genes	Gametes	Grid boxes	Phenotype classes	Genotype classes
1	2	4	2	3
2	4	16	4	9
3	8	64	8	27
4	16	256	16	81

Note the pattern: 2ⁿ phenotype classes and 3ⁿ genotype classes for n heterozygous genes. The classic 27:9:9:9:3:3:3:1 trihybrid ratio is simply (3:1) applied three times and multiplied out.

Working Any Combination

The rule handles partial dominance patterns just as easily. For A_B_cc from AaBbCc × AaBbCc: (3/4) × (3/4) × (1/4) = 9/64. There is no need to identify which of the 64 boxes qualify.

For "at least one dominant across all three", it is often easier to calculate the complement — the probability of the unwanted outcome subtracted from one.

Target	Calculation	Result
A_B_C_	(3/4)³	27/64 = 42.2%
aabbcc	(1/4)³	1/64 = 1.6%
A_B_cc	(3/4)(3/4)(1/4)	9/64 = 14.1%
Exactly one recessive trait	3 × (3/4)²(1/4)	27/64 = 42.2%

When Independence Fails

The product rule requires genes to assort independently, meaning they sit on different chromosomes or far apart on the same one. Linked genes violate it, producing excess parental combinations.

With three genes, a testcross to a fully recessive parent is the standard diagnostic: independent assortment gives eight equally frequent classes, while linkage produces two common parental classes and two rare double-crossover classes. The rarest pair identifies which gene lies in the middle.

Worked Examples

Example 1: AaxAa, BbxBb, CcxCc (all het x het)

Each: 3/4 dom, 1/4 rec

Result: A_B_C_ = 27/64 = 42.2%

Classic 27:9:9:9:3:3:3:1 ratio

Example 2: AAxAa, BBxBb, CCxCc

Each: 100%, 75%, 75%

Result: All dom = 1×0.75×0.75=56.25%

Modified ratios from homozygous parents

Example 3: Calculating a specific combination

A_B_cc from a full trihybrid cross

Result: (3/4)(3/4)(1/4) = 9/64

No grid required. Each gene contributes independently, so the probabilities simply multiply.

Example 4: Detecting linkage in a three-point cross

Testcross giving unequal class frequencies

Result: Genes are linked

Independent assortment predicts eight equal classes. Two common and two rare classes indicate linkage, and the rarest pair identifies the middle gene.

Common Mistakes

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Drawing a 64-box grid

Multiplying single-gene probabilities gives the same answer in one line and extends to any number of genes.

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Applying the product rule to linked genes

Independent assortment is required. Linked genes produce an excess of parental combinations, and a testcross reveals it.

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Confusing phenotype and genotype class counts

Three heterozygous genes give 8 phenotype classes but 27 genotype classes. The distinction matters when counting expected categories.

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Expecting exact ratios in small samples

The 27:9:9:9:3:3:3:1 ratio needs several hundred offspring before observed counts approach expectation.

Frequently Asked Questions

Multiplication rule basis?

For independently assorting genes (on different chromosomes or far apart), P(A and B) = P(A) × P(B). Each gene segregates independently per Mendel's second law.

Number of phenotypic classes?

n independently assorting heterozygous genes → 2^n phenotypic classes. 3 genes → 8 classes. Ratio coefficients follow (3+1)^n = 4^n = 64 total combinations for n=3.

How many phenotype classes does a trihybrid cross give?

Eight, since each gene contributes two possibilities: 2³. There are 27 genotype classes, from 3³.

Why can I multiply probabilities?

Because independently assorting genes are independent events. The probability of a combined outcome is the product of the individual probabilities.

What is the trihybrid phenotypic ratio?

27:9:9:9:3:3:3:1, which is simply the 3:1 monohybrid ratio applied three times and multiplied out.

When does the product rule fail?

When genes are linked. A testcross showing unequal class frequencies indicates linkage rather than independent assortment.