CALCULATE WITH CONFIDENCE

Genetics & Inheritance

Punnett Square & Inheritance Calculator

Build one-, two- or three-locus crosses from parental genotypes. Inspect gametes, every combination, genotype probabilities and the chosen phenotype model.

Biology · experimental measurements

Trace parental allele combinations into complete genotype and phenotype distributions.

Private calculations in your browser · explicit inputs and model boundaries
Example preview · One-locus crossFollow one allele from each parent at every locus

Locus A · Aa × Aa

Parent 1 gametes: A · a
Parent 2 gametes: A · a

A + AAA25% of combinations
A + aAa25% of combinations
a + AAa25% of combinations
a + aaa25% of combinations

Each small square lists four equally likely chromosome-copy combinations at one locus. Repeated letters represent different parental copies. Multiply locus probabilities only under the independent-assortment model; the complete gamete and offspring ledgers appear below.

  1. 1EnterProvide the known values
  2. 2CalculateResults update automatically
  3. 3VerifyReview the details and units
Try an example

Use AA, Aa or aa; add Bb and Cc pairs for up to three independent loci. Example: AaBb.

Use the same loci, in the same order. Genotypes are known model inputs, not inferred from appearance.

Used only for expected counts; the actual composition can differ.

Calculation result

—

Enter valid values to see the result.

Your entries are calculated in this browser and are not submitted to 365CALCS.COM.

Feedback

Understand the relationship

The reasoning behind the result

A square tracks transmission before it groups outcomes

A diploid parent contributes one of its two allele copies at a locus. Aa therefore contributes A or a with probability one half. AA contributes A with certainty, although the two chromosome copies are separate entries in a conventional square. The calculator combines identical gametes before constructing the complete cross ledger.

An Aa × Aa cross has four copy-level combinations: AA, Aa, aA and aa. The two middle combinations describe the same unordered genotype. Grouping them gives probabilities 1/4, 1/2 and 1/4, not four different genotypes.

Multiple loci add an independence assumption

P(AaBb) = P(Aa) × P(Bb), when loci assort independently

For AaBb, each parent's distinct gametes are AB, Ab, aB and ab. Their pairings form 16 combinations. A third heterozygous locus doubles the gametes again and expands the pairings to 64.

This multiplication is justified by the entered unlinked-locus model. Linked loci require haplotype phase and recombination information; entering extra allele pairs here does not model linkage. A test cross is represented by entering the homozygous recessive genotype as one parent. A backcross is represented by the actual chosen parental genotype.

Expression changes the phenotype groups, not segregation

Under complete dominance, AA and Aa share the A_ expression class and aa is separate. Under incomplete dominance, the heterozygote has its own intermediate class; under codominance, both allele contributions are expressed. These latter models still have the same genotype probabilities but different biological interpretations.

All entered loci use the selected model, and phenotype classes are combined without epistasis. Symbols are generic; the tool does not assign real colors, traits or medical meanings to them.

An expected count can be fractional

E[count in a class] = offspring count × class probability

Multiplying a probability by a hypothetical cohort size describes a mean over repeated model cohorts. It does not allocate a fixed number of actual offspring to each class. Use the multiple-offspring calculator for exact-count or range probabilities across repeated independent outcomes.

Follow the numbers

A double-heterozygote test cross

  1. Parent 1 is AaBb and parent 2 is aabb, with independent assortment and complete dominance.
  2. Parent 1 produces AB, Ab, aB and ab at 25% each; parent 2 produces only ab.
  3. The four offspring genotypes are AaBb, Aabb, aaBb and aabb, each at 25%.
  4. For 80 hypothetical offspring, each class has expected count 80 × 0.25 = 20.

The expected phenotype ratio is 1:1:1:1. Four equal expectations do not guarantee four equal observed counts.

Quick guide

How to use this calculator

  1. Enter both known genotypes with A/a, optionally B/b and C/c.
  2. Select a phenotype model that applies to every entered locus. The calculator does not infer dominance from the allele letters.
  3. Read each locus square, then inspect the full gamete, genotype and phenotype ledgers. Expected counts are probability-weighted averages.

Calculation method

Calculation and interpretation

Trace parental allele combinations into complete genotype and phenotype distributions.

P(gamete pair) = P(gamete from parent 1) × P(gamete from parent 2); P(genotype) = sum of matching pair probabilities; expected count = N × P.

Worked example

A double-heterozygote test cross

The expected phenotype ratio is 1:1:1:1. Four equal expectations do not guarantee four equal observed counts.

P(gamete pair) = P(gamete from parent 1) × P(gamete from parent 2); P(genotype) = sum of matching pair probabilities; expected count = N × P.

Supported inputs

Precision and limits

Educational inheritance model

Uses a diploid, two-allele model with equal segregation. It does not interpret a person's genetic test, assess disease or reproductive risk, establish parentage, or predict a real offspring's traits. Linkage, selection, incomplete penetrance, mutation, viability differences and environmental effects are excluded.

Supported genotype system

One to three loci, two alleles per locus, no epistasis, no linked haplotypes and one expression convention across all loci. ABO and other multiple-allele systems are not represented by A/a notation.

Continue calculating

Related calculators