Free Punnett Square Calculator (2×2 and 4×4, with every step shown)

What a Punnett square is actually doing

Every body cell carries two alleles for a gene — one from each parent. Meiosis separates them, so each sperm or egg carries exactly one allele per gene. A Punnett square is a bookkeeping grid for every way those gametes can meet at fertilisation.

This is why the axes of the square hold gametes, not parent genotypes. If you write "Aa" across the top of a 2×2, you have mislabelled the grid. The top of an Aa parent's axis reads A in one column and a in the other, because those are the two gametes that parent can make.

Each box is one equally likely fertilisation. Count the boxes and you have your probabilities. The genotype is the pair of letters in the box; the phenotype is what that combination looks like, which is set by dominance: one uppercase allele is enough to show the dominant trait, and only a pair of lowercase alleles shows the recessive one.

• Uppercase = dominant allele (A). Lowercase = recessive allele (a).
• AA is homozygous dominant, Aa is heterozygous, aa is homozygous recessive.
• AA and Aa look identical — that is the whole point of dominance.
• Aa and aA are the same genotype. Write the dominant allele first so you do not double-count.

Worked example: Aa × Aa, the 3:1 cross

Both parents are heterozygous. Each makes two kinds of gamete: A and a. That gives a 2×2 grid with four boxes.

Fill each box by taking the gamete from its row and the gamete from its column: A+A = AA, A+a = Aa, a+A = Aa, a+a = aa.

Now count. Genotypes: 1 AA, 2 Aa, 1 aa — a 1:2:1 genotype ratio. Phenotypes: three boxes contain at least one A and one box is aa — a 3:1 phenotype ratio.

Both ratios describe the same four boxes. Which one the question wants is the difference between a mark and no mark, so read the wording: "genotypic ratio" means the letters, "phenotypic ratio" means the appearance. This is also the cross behind the classic carrier problem — two heterozygous parents have a 1-in-4 (25%) chance of a homozygous recessive child, even though neither parent shows the recessive trait.

Dihybrid crosses: FOIL, 16 boxes, and 9:3:3:1

With two genes, each parent makes four gametes instead of two, so the grid becomes 4×4 with 16 boxes. Listing the gametes is where most errors start, and FOIL is the fastest reliable method.

Treat AaBb as (A + a)(B + b) and multiply out: First A×B = AB, Outer A×b = Ab, Inner a×B = aB, Last a×b = ab. Every gamete carries exactly one allele from gene A and exactly one from gene B — never two A alleles, never zero B alleles.

Cross AaBb × AaBb and the 16 boxes sort into four phenotypes: 9 A_B_, 3 A_bb, 3 aaB_, 1 aabb. That famous 9:3:3:1 is not a fact to memorise, it is a product. Gene A on its own gives 3:1 dominant to recessive, gene B on its own gives 3:1, and because the genes assort independently you multiply the fractions: 3/4 × 3/4 = 9/16, 3/4 × 1/4 = 3/16, and so on.

The genotype ratio for the same cross is 1:2:1:2:4:2:1:2:1 across AABB, AABb, AAbb, AaBB, AaBb, Aabb, aaBB, aaBb, aabb — nine genotypes collapsing into four appearances.

A homozygous parent still gets a 4×4. AABB makes gametes AB, AB, AB, AB — all four identical. Keep all four on the axis rather than collapsing to one, because the box counts have to stay out of 16 for the ratios to come out right.

Mistakes that cost marks

Almost every wrong Punnett square answer comes from one of a short list of slips. None of them are about the biology being hard.

• Putting parent genotypes on the axes instead of gametes. The axis of an Aa parent is A and a, in separate rows or columns.
• Answering with the genotype ratio when the question asked for the phenotype ratio. Aa × Aa is 1:2:1 genotypically and 3:1 phenotypically — different numbers, same square.
• Writing dihybrid genotypes grouped by chromosome instead of by gene: ABab instead of AaBb. Keep both alleles of one gene next to each other or the columns will not line up.
• Counting Aa and aA as two different genotypes. They are one. Write the dominant allele first every time and the tally becomes obvious.
• Building gametes wrong in a dihybrid — AA and Ab are not valid gametes from AaBb. One allele per gene, always.
• Treating the ratio as a guarantee. A 3:1 ratio does not mean three of your next four offspring show the dominant trait; it is the probability per offspring.
• Using a plain Punnett square where the genetics do not fit: incomplete dominance and codominance change the phenotype ratio to 1:2:1, linked genes break the 9:3:3:1 split entirely, and sex-linked traits need X and Y notation rather than a letter pair.

How to check the answer before you hand it in

Four fast checks catch nearly every arithmetic error, and you can run them without redoing the square.

• Your box counts must add up to the number of boxes: 4 for a monohybrid, 16 for a dihybrid. If your genotype counts sum to 15, you skipped a box.
• Your percentages must add up to 100%. In a 16-box grid each box is 6.25%, so 9 boxes is 56.25%.
• For a dihybrid, solve each gene separately as its own 2×2 and multiply the fractions — the forked-line method. If gene A gives 3/4 dominant and gene B gives 1/2 dominant, then the double-dominant phenotype must be 3/4 × 1/2 = 3/8 = 6/16. If your 4×4 disagrees, you filled a box wrong.
• No offspring can carry an allele that neither parent has. If a B appears in your grid and neither parent genotype contains B, you copied something across incorrectly.
• Sanity-check the direction of the answer: a cross with a homozygous dominant parent (AA × anything) can never produce a recessive-looking offspring, because every box inherits at least one A.