How to do a trihybrid cross without a Punnett square

Table of contents

1. Step 1: Split it into three monohybrid crosses
2. Step 2: Multiply for “and” (the product rule)
3. Step 3: Add for “or” (the sum rule)
4. The forked-line (branching) method
5. Counting shortcuts
6. How to find the genotypic ratio of a trihybrid cross
7. When the parents aren’t identical
8. Common mistakes
9. Practice
10. Related
11. Sources

Level: High school and introductory college · You’ll need: the dihybrid cross and its product-rule shortcut

A trihybrid cross follows three genes at once. Two fully heterozygous parents (AaBbCc × AaBbCc) each make 8 kinds of gamete, so the Punnett square has 8 × 8 = 64 boxes. Drawing that is slow and error-prone.

You don’t need to. Because the three genes assort independently, a trihybrid cross is really three monohybrid crosses happening at the same time. Solve each one and multiply.

Assumptions: complete dominance at each gene and independent assortment (the genes are on different chromosomes or far apart on the same one). Linked genes need recombination frequencies instead.

Step 1: Split it into three monohybrid crosses

For AaBbCc × AaBbCc:

Gene	Cross	Genotypes	Phenotypes
A	Aa × Aa	¼ AA, ½ Aa, ¼ aa	¾ dominant, ¼ recessive
B	Bb × Bb	¼ BB, ½ Bb, ¼ bb	¾ dominant, ¼ recessive
C	Cc × Cc	¼ CC, ½ Cc, ¼ cc	¾ dominant, ¼ recessive

Step 2: Multiply for “and” (the product rule)

The chance of a particular combination is the product of the three single-gene chances.

• P(AaBbCc) = ½ × ½ × ½ = 1/8 (8 of the 64 boxes)
• P(aabbcc) = ¼ × ¼ × ¼ = 1/64
• P(AABbcc) = ¼ × ½ × ¼ = 1/32
• P(dominant for all three traits) = ¾ × ¾ × ¾ = 27/64

Step 3: Add for “or” (the sum rule)

When several different combinations answer the question, work out each one and add them.

What fraction of offspring show exactly two dominant traits? There are three ways: recessive for A only, for B only, or for C only. Each has probability ¾ × ¾ × ¼ = 9/64. So the answer is 9/64 + 9/64 + 9/64 = 27/64.

The forked-line (branching) method

The forked-line method draws the multiplication as a tree, which is handy when you need every phenotype class. Branch on gene A, then B, then C, and multiply along each path:

A gene       B gene       C gene       Phenotype     Probability
             ┌ ¾ B_ ──┬── ¾ C_  →  A_B_C_      27/64
     ┌ ¾ A_ ─┤        └── ¼ cc  →  A_B_cc       9/64
     │       └ ¼ bb ──┬── ¾ C_  →  A_bbC_       9/64
     │                └── ¼ cc  →  A_bbcc       3/64
─────┤
     │       ┌ ¾ B_ ──┬── ¾ C_  →  aaB_C_       9/64
     └ ¼ aa ─┤        └── ¼ cc  →  aaB_cc       3/64
             └ ¼ bb ──┬── ¾ C_  →  aabbC_       3/64
                      └── ¼ cc  →  aabbcc       1/64

Collected together, that’s the trihybrid phenotypic ratio 27 : 9 : 9 : 9 : 3 : 3 : 3 : 1 (eight phenotype classes adding up to 64). The trihybrid calculator lists the same classes in a fixed trait order, so its 9s and 3s can appear interleaved.

Counting shortcuts

For n genes, each heterozygous in both parents:

What you’re counting	Formula	Trihybrid (n = 3)
Gamete types per parent	2ⁿ	8
Boxes in the Punnett square	4ⁿ	64
Different genotypes	3ⁿ	27
Different phenotypes (complete dominance)	2ⁿ	8

To list a parent’s gametes without a square, branch the same way: A or a, then B or b, then C or c, giving ABC, ABc, AbC, Abc, aBC, aBc, abC, abc.

How to find the genotypic ratio of a trihybrid cross

Each gene’s genotypic ratio is 1 : 2 : 1, so the trihybrid ratio is (1 : 2 : 1)³, with 27 genotypes. Writing all 27 in a row is unwieldy. It’s easier to group them by how many genes are heterozygous:

Heterozygous genes	Example	Number of genotypes	Boxes each	Boxes in total
None	AABBCC, aabbcc	8	1	8
One	AaBBCC	12	2	24
Two	AaBbCC	6	4	24
All three	AaBbCc	1	8	8

The total is 8 + 24 + 24 + 8 = 64, so every box is accounted for. To get any single genotype’s share, use the product rule: each homozygous gene contributes ¼ and each heterozygous gene contributes ½.

When the parents aren’t identical

The method works for any pair of parents. Take AaBbCc × aaBbcc:

Gene	Cross	Result
A	Aa × aa	½ Aa, ½ aa
B	Bb × Bb	¼ BB, ½ Bb, ¼ bb
C	Cc × cc	½ Cc, ½ cc

• P(aaBbcc) = ½ × ½ × ½ = 1/8
• P(dominant for all three) = ½ × ¾ × ½ = 3/16

Common mistakes

1. Adding when you should multiply. “And” (all three conditions at once) means multiply.
2. Forgetting the sum rule. “Exactly two dominant” has three routes. Add all of them.
3. Using ¾ for a genotype. ¾ is a phenotype probability (A_). A specific genotype uses that gene’s genotype fraction (¼ or ½ in an Aa × Aa cross).
4. Applying it to linked genes. The product rule only works if the genes assort independently.

Practice

1. For AaBbCc × AaBbCc, what is the probability of AaBBcc?
2. For the same cross, what fraction of offspring show all three recessive traits?
3. For AaBbCc × AaBbcc, what fraction show the dominant phenotype for all three traits?

Answers

1. ½ × ¼ × ¼ = 1/32
2. ¼ × ¼ × ¼ = 1/64
3. ¾ × ¾ × ½ = 9/32. The C gene is Cc × cc, so only ½ show the dominant C trait.

Check any of these in the trihybrid cross calculator. It shows the full 64-box square and every genotype’s probability.

Related

• Trihybrid cross calculator
• Dihybrid cross explained simply
• Genotypic and phenotypic ratios as percentages

Sources

• OpenStax, Biology 2e, 12.3 Laws of Inheritance (forked-line and probability methods, the gamete and genotype counts in Table 12.5, the 27:9:9:9:3:3:3:1 trihybrid ratio). The 4ⁿ row and general-n formulas follow from the same arithmetic.

Share this article

• Copy link