Punnett Square Calculator

Build Punnett squares for genetic crosses. See genotypic and phenotypic ratios instantly.

Starting values — Punnett Square Calculator

Dominance	Complete
Parent	1

Change any control above and the results recompute from these.

Loading interactive simulation...

Most families of four do not show the 3:1 ratio 🖖

A monohybrid cross predicts three dominant offspring to one recessive, and that number is a probability per child, not a quota per litter. Each child independently has a 3/4 chance of showing the dominant phenotype, so with four offspring the chance of landing exactly on 3:1 is 4 × (3/4)³ × (1/4) = 42.2%. In other words 57.8% of four-child families miss the textbook ratio entirely, and 31.6% show all four dominant with no recessive child at all. The ratio is real, but it only emerges across many crosses — which is precisely why Mendel counted thousands of pea plants rather than a handful, and why one unexpected family tells a breeder nothing.

A grid that counts every combination 🖖

A Punnett square is simply a table: list the alleles one parent can put into an egg along the top, the other parent's down the side, then fill each cell with the pair that meets there. Counting the cells gives the odds. Cross two Aa parents and you get the famous 3:1 ratio — three offspring showing the dominant trait for every one showing the recessive. Dihybrid crosses expand this to 9:3:3:1.

A cricket question that founded population genetics 🖖

Reginald Punnett, who popularized the square around 1905, was once puzzled about why dominant alleles didn't simply take over a population. He put the question to his cricket partner, the mathematician G. H. Hardy. Hardy's one-page reply in 1908 became the Hardy-Weinberg principle, the foundation of population genetics — showing that allele frequencies stay constant unless something disturbs them.

PUNNETT SQUARES — WHICH CROSS ARE YOU SETTING UP?

Which Punnett Case Are You In?

A Punnett square is one mechanism, but the number you get out depends entirely on which cross you fed in — and the two ratios it reports, genotypes and phenotypes, are not the same ratio. These six cases cover the classic monohybrid and what its 3:1 conceals, the same grid re-read under incomplete dominance, the cross you run when you cannot see a genotype, the cross that makes every offspring identical, and the two-locus versions of the first and third.

Monohybrid, complete dominance — the 3:1 everyone learns 1:2:1 → 3:1

Incomplete dominance — the same grid, a different ratio 1:2:1 → 1:2:1

Test cross — making a hidden allele visible 1:1 → 1:1

Homozygous parents — a uniform first generation 4/4 Aa → 4/4

Dihybrid — 9:3:3:1 is two monohybrids multiplied (3:1)² = 9:3:3:1

Dihybrid test cross — reading the gamete ratio directly 1:1:1:1 → 1:1:1:1

Monohybrid, complete dominance — the 3:1 everyone learns

What you know: Both parents heterozygous at one locus, and the dominant allele fully masks the recessive one.

Outcome: 1:2:1 → 3:1

Worked example: Aa × Aa fills four cells with AA, Aa, Aa and aa. The tool reports genotypes 1:2:1 and phenotypes 3:1 — 3/4 dominant, 1/4 recessive.

Open this case: Aa × Aa

Monohybrid, complete dominance — the 3:1 everyone learns. Four cells, three of them dominant in appearance. The two Aa cells and the one AA are indistinguishable by sight, so 1:2:1 reads as 3:1. Both parents heterozygous at one locus, and the dominant allele fully masks the recessive one. — Four cells, three of them dominant in appearance. The two Aa cells and the one AA are indistinguishable by sight, so 1:2:1 reads as 3:1.

Incomplete dominance — the same grid, a different ratio

What you know: The same heterozygous pair, but the heterozygote shows its own intermediate appearance rather than the dominant one.

Outcome: 1:2:1 → 1:2:1

Worked example: Aa × Aa again, with dominance set to incomplete. The four cells are unchanged, the genotype ratio is still 1:2:1 — and the phenotype ratio is now 1:2:1 too.

Open this case: Incomplete

Incomplete dominance — the same grid, a different ratio. The identical four cells. With the heterozygote showing its own phenotype, the phenotype ratio collapses onto the genotype ratio. The same heterozygous pair, but the heterozygote shows its own intermediate appearance rather than the dominant one. — The identical four cells. With the heterozygote showing its own phenotype, the phenotype ratio collapses onto the genotype ratio.

Test cross — making a hidden allele visible

What you know: One parent shows the dominant trait but its genotype is unknown; the other is homozygous recessive.

Outcome: 1:1 → 1:1

Worked example: Aa × aa gives two Aa and two aa: 1:1 in both genotypes and phenotypes. Had the unknown parent been AA, every offspring would have been dominant instead.

Open this case: Test cross

Test cross — making a hidden allele visible. Half the cells recessive. A single aa offspring is enough to prove the dominant-looking parent carried a. One parent shows the dominant trait but its genotype is unknown; the other is homozygous recessive. — Half the cells recessive. A single aa offspring is enough to prove the dominant-looking parent carried a.

Homozygous parents — a uniform first generation

What you know: Both parents homozygous, one for each allele.

Outcome: 4/4 Aa → 4/4

Worked example: AA × aa makes all four cells Aa: one genotype, one phenotype, 4/4 of the offspring looking like the dominant parent.

Open this case: AA × aa

Homozygous parents — a uniform first generation. One combination, four times over. The recessive allele is present in every offspring and visible in none. Both parents homozygous, one for each allele. — One combination, four times over. The recessive allele is present in every offspring and visible in none.

Dihybrid — 9:3:3:1 is two monohybrids multiplied

What you know: Both parents heterozygous at two loci that assort independently.

Outcome: (3:1)² = 9:3:3:1

Worked example: AaBb × AaBb needs 16 cells and produces 9 distinct genotypes, which collapse into four phenotype classes at 9:3:3:1.

Open this case: Dihybrid

Dihybrid — 9:3:3:1 is two monohybrids multiplied. Sixteen cells, nine genotypes, four appearances. The 9:3:3:1 is the product of two independent 3:1 crosses. Both parents heterozygous at two loci that assort independently. — Sixteen cells, nine genotypes, four appearances. The 9:3:3:1 is the product of two independent 3:1 crosses.

Dihybrid test cross — reading the gamete ratio directly

What you know: A parent heterozygous at two loci, crossed with one homozygous recessive at both.

Outcome: 1:1:1:1 → 1:1:1:1

Worked example: AaBb × aabb fills 16 cells that come out perfectly flat: 1:1:1:1 in both genotypes and phenotypes, 4/16 in each class.

Open this case: Dihybrid test cross

Dihybrid test cross — reading the gamete ratio directly. Four equal classes. Because one parent contributes a single gamete type, the offspring tally reads the other parent’s gametes. A parent heterozygous at two loci, crossed with one homozygous recessive at both. — Four equal classes. Because one parent contributes a single gamete type, the offspring tally reads the other parent’s gametes.

References (3)

The square, and the man it is named after: Edwards, A. W. F. (2012). "Reginald Crundall Punnett: First Arthur Balfour Professor of Genetics, Cambridge, 1912." Genetics 192(1), 3–13.
Cases 1, 4 and 5 — the original 3:1, the uniform F₁ and 9:3:3:1: Mendel, G. (1866). "Versuche über Pflanzen-Hybriden." Verhandlungen des naturforschenden Vereines in Brünn 4, 3–47. No DOI: the volume predates them, and the scans in circulation are of varying provenance, so none is linked here.
Case 6 — why a flat 1:1:1:1 is the test rather than the assumption: Griffiths, A. J. F. et al. Introduction to Genetic Analysis, 11th ed. W. H. Freeman, 2015 — testcrosses, gamete ratios, and the excess of parental classes that reveals linkage.

Problem solved in full

The 3:1 ratio in a real family of four 5 steps
Aa × Aa gives the 3:1 ratio everyone memorises. Work out why a real family of four almost never shows it — and what Mendel had to do about that.
1. Each parent contributes one allele, chosen at random from its two. Four equally likely combinations, and the square is just an organised way of listing them.
  
  $A a \times A a : gametes {A, a} \times {A, a} ⟹ AA : A a : aa = 1 : 2 : 1 = 25.0% : 50.0% : 25.0%$
2. Complete dominance collapses two of the four genotypes into one appearance. The genotype ratio and the phenotype ratio are different statements, and only the second is what you observe.
  
  $complete dominance \Rightarrow AA and A a look alike : dom : rec = 3 : 1 = 75.0% : 25.0%$
3. Now treat the offspring as independent trials. The chance of avoiding the recessive quarter every time is a power, and it decays slowly.
  
  $P (no recessive in n) = (\frac{3}{4})^{n} : n = 4 \Rightarrow 0.316, n = 11 \Rightarrow 0.042$
4. The dihybrid case is the same fact multiplied. Independent assortment means the two characters are independent events, so 9:3:3:1 is (3:1) times (3:1) — no new biology, just a product.
  
  $dihybrid = (3 : 1) \otimes (3 : 1) = 9 : 3 : 3 : 1 — independent assortment is a PRODUCT of two monohybrids$
5. Finally, what large numbers buy. Mendel's tall-versus-short counts sit within a fraction of a standard deviation of the expectation, which is what makes a ratio demonstrable at all.
  
  $χ^{2} = \sum \frac{( O - E ) ^{2}}{E} : 787 tall, 277 short of 1064 \Rightarrow E = 798, 266; χ^{2} = 0.61, p = 0.44$
Answer

Because a four-child family has a 32% chance of containing no recessive at all. The 3:1 is an expectation, not an outcome, and (3/4)⁴ = 0.316 says that nearly a third of such families look like perfect dominance with no hint of anything hiding. You need eleven offspring before the chance of seeing no recessive drops below 5%. This is exactly why Mendel counted peas rather than families: 1,064 plants in the tall/short cross, 787 to 277 against an expectation of 798 to 266, which is a χ² of 0.61 and about as good a fit as chance allows. Ronald Fisher later argued the fit across Mendel's experiments was too good, and the argument has run for a century — but the reason large numbers were needed in the first place is settled in the first line of this problem.

$3 : 1 phenotypic; (\frac{3}{4})^{4} = 32% of four-child families show no recessive at all$

By the 42Math team · Updated 2026-07-29

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

All Bio, Geo, Chem lessons →

Example problems

Aa × Aa - Aa × Aa with complete dominance: genotypes come out 1:2:1 but phenotypes 3:1, because AA and Aa look the same.
Dihybrid - AaBb × AaBb fills 16 cells and sorts them into four phenotype classes at 9:3:3:1.
Incomplete - The same 4-cell Aa × Aa grid, read with incomplete dominance: phenotypes become 1:2:1 because the heterozygote shows its own.
Test cross - Aa × aa gives 1:1 — the cross you run to find out whether a dominant-looking parent carries a recessive allele.
AA × aa - AA × aa makes every one of the 4 cells Aa: one genotype, one phenotype, and an F1 that looks like the dominant parent.
Dihybrid test cross - AaBb × aabb spreads 16 cells evenly at 1:1:1:1, so the offspring count reads the gamete ratio directly.