Trihybrid Cross Probabilities

Biology • Mendelian Genetics

Written by STEM Calculators Team Published January 8, 2026 Updated February 24, 2026

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Trihybrid Cross Probabilities (3 genes, complete dominance)

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Independent assortment for loci A/a, B/b, C/c. Computed by independence (no 8×8 grid).

Enter genotypes like AaBbCc. For each locus: dominant phenotype is A_ (AA or Aa), recessive is aa.

Parent 1 genotype

Accepted: AaBbCc, “Aa Bb Cc”, etc. Must contain exactly 2 alleles per locus.

Parent 2 genotype

Output focus

Specific patterns use A_/aa, B_/bb, C_/cc.

Show probability tree visualization Show bar charts

Gametes (per locus)

Shown after calculation.

Parent 1

Parent 2

Results

Expected phenotype ratio (simplified)

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Order: 0 recessive, then 1 recessive (3 classes), then 2 recessive (3 classes), then 3 recessive.

Specific phenotype probability

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Shown even in “Full table” mode (uses current pattern selection).

Phenotype probabilities table

Phenotype class	Recessive traits	Probability	Percent	Ratio count (out of 64)

Batch mode (CSV paste/upload)

Paste CSV with columns: parent1,parent2 (header optional). Uses the current specific phenotype selection.

CSV data

Upload CSV file

Upload fills the textarea automatically.

Batch results

#	Parent 1	Parent 2	Ratio	Pattern	P(pattern)

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Trihybrid Cross Probabilities

A trihybrid cross tracks inheritance at three independent genes: A/a, B/b, and C/c, assuming complete dominance at each locus (A dominates a, B dominates b, C dominates c). Instead of building an 8×8 Punnett square, probabilities are computed using independence (the multiplication rule).

Key assumptions

Complete dominance: A_ means AA or Aa (dominant phenotype), aa is recessive.
Independent assortment: loci A, B, and C segregate independently (unlinked).
Mendelian segregation: each parent contributes one allele per gene via gametes.

If loci are linked (on the same chromosome and close together), independence may not hold. In that case, predicted ratios can deviate from these results.

Step 1: Gametes at one locus

For a single gene (for example A/a), the genotype determines gamete allele probabilities:

\[ \begin{aligned} \text{AA} &\Rightarrow P(A)=1,\; P(a)=0 \\ \text{Aa} &\Rightarrow P(A)=\frac{1}{2},\; P(a)=\frac{1}{2} \\ \text{aa} &\Rightarrow P(A)=0,\; P(a)=1 \end{aligned} \]

Step 2: Phenotype probability at one locus

Under complete dominance, the recessive phenotype occurs only for the homozygous recessive genotype. Using the probability that each parent contributes the recessive allele:

\[ \begin{aligned} P(\text{aa}) &= P(a)_{\text{P1}} \cdot P(a)_{\text{P2}} \\ P(\text{A\_}) &= 1 - P(\text{aa}) \end{aligned} \]

The same logic applies at B/b and C/c: \(P(\text{bb})=P(b)_{\text{P1}}\cdot P(b)_{\text{P2}}\), \(P(\text{B\_})=1-P(\text{bb})\), and similarly for C/c.

Step 3: Independence across loci

With independent assortment, the probability of a combined phenotype class is the product of the locus-level phenotype probabilities. For example, for the class A_B_cc (dominant at A and B, recessive at C):

\[ P(\text{A\_B\_cc}) = P(\text{A\_}) \cdot P(\text{B\_}) \cdot P(\text{cc}) \]

The 8 phenotype classes

Because each locus has two phenotype states (dominant “_” or recessive), there are \(2^3=8\) phenotype classes:

A_B_C_
A_B_cc
A_bbC_
aaB_C_
A_bbcc
aaB_cc
aabbC_
aabbcc

Why “out of 64” is natural here

If both parents are heterozygous at a locus (e.g., Aa × Aa), the dominant:recessive phenotype probability is \(3:1\) at that locus. Across three loci, the combined denominator becomes:

\[ 4^3 = 64 \]

That is why the classic trihybrid heterozygous cross AaBbCc × AaBbCc yields the famous phenotype ratio:

\[ 27:9:9:9:3:3:3:1 \]

The calculator also simplifies ratios by dividing all “counts out of 64” by their greatest common divisor.

Distribution by number of recessive traits

Another intuitive summary groups offspring by how many recessive phenotypes they show: 0, 1, 2, or 3 recessive traits. This helps visualize how “rare” multiple recessive traits become when each locus independently favors the dominant phenotype.

How to interpret the visuals

Probability tree (A → B → C): each branch multiplies probabilities to reach a leaf class.
8-bar phenotype chart: compares probabilities of all phenotype classes.
Recessive-count distribution: summarizes outcomes by 0/1/2/3 recessive traits.

Tip: Use “Probability of a specific phenotype pattern” to answer questions like: “What is the probability of showing recessive phenotype only at C?” (A_B_cc), or “What is the probability of being recessive at all three?” (aabbcc).

Frequently Asked Questions

How does the trihybrid cross calculator compute probabilities without an 8x8 Punnett square?

It computes phenotype probabilities at each locus (A, B, and C) from the parent genotypes, then multiplies them using independence. For example, P(A_B_cc) = P(A_) x P(B_) x P(cc).

What does A_ mean in trihybrid cross results?

A_ means the dominant phenotype at the A locus, which occurs for AA or Aa. The recessive phenotype occurs only for aa.

Why does the calculator report a ratio out of 64?

When each locus is a heterozygous cross like Aa x Aa, outcomes at one locus have denominator 4, so across three independent loci the denominator becomes 4^3 = 64. The tool converts probabilities to counts out of 64 and simplifies the ratio.

How do I compute the probability of a specific phenotype pattern like aabbcc?

Switch to the specific pattern option and choose aa at A, bb at B, and cc at C, then click Calculate. The tool multiplies the three recessive probabilities to return P(aabbcc).

How do I use batch mode for many genotype pairs?

Paste or upload a CSV with columns parent1 and parent2 containing genotype strings like AaBbCc. The batch output includes the simplified ratio for the current pattern selection and the corresponding P(pattern), and it can be copied or downloaded as CSV.

Trihybrid Cross Probabilities (3 genes, complete dominance)

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