Dihybrid Cross Calculator
Cross two parents for two genes and see each parent's gametes, the 16-box Punnett square, and every genotype and phenotype with its probability. Set incomplete dominance, codominance or a gene interaction such as 9:3:4, and test the offspring you counted against the expected ratio.
Calculator
Set up the cross
Type both parents, such as RrYy × rryy, or set them below.
Parent 1
AaBb
Parent 2
AaBb
Each gene controls its own trait. Choose an interaction when both act on one trait (epistasis).
Letters and names (optional)
Name the traits, and the lists and the copied answer use the names: round, yellow instead of R_Y_.
Phenotype ratio
9 : 3 : 3 : 1
Of every 16 offspring, on average 9 are A_B_, 3 A_bb, 3 aaB_ and 1 aabb.
- Genotype ratio
- 1:2:1:2:4:2:1:2:1
- Genotypes and phenotypes
- 9 and 4
- Gametes of parent 1 (AaBb)
- AB, Ab, aB, ab, 1/4 each
- Gametes of parent 2 (AaBb)
- AB, Ab, aB, ab, 1/4 each
Punnett square
AaBb × AaBb, 16 boxes
| AB | Ab | aB | ab | |
|---|---|---|---|---|
| AB | AABBphenotype 1 | AABbphenotype 1 | AaBBphenotype 1 | AaBbphenotype 1 |
| Ab | AABbphenotype 1 | AAbbphenotype 2 | AaBbphenotype 1 | Aabbphenotype 2 |
| aB | AaBBphenotype 1 | AaBbphenotype 1 | aaBBphenotype 3 | aaBbphenotype 3 |
| ab | AaBbphenotype 1 | Aabbphenotype 2 | aaBbphenotype 3 | aabbphenotype 4 |
Phenotypes
Tap one to find it in the square
Genotypes
9 genotypes
Test your counts
Counted the offspring of a real cross? Enter them to test them against 9:3:3:1 with chi-square.
Enter how many offspring fell into each class, 0 for none, to test them against 9:3:3:1.
Worked example
Mendel's peas: round yellow × round yellow, both RrYy
- 01Each parent makes four gametes, RY, Ry, rY and ry, a quarter each: 4 × 4 = 16 boxes
- 02Phenotypes: 9 R_Y_ round yellow, 3 R_yy round green, 3 rrY_ wrinkled yellow, 1 rryy wrinkled green
- 03Mendel counted 315, 108, 101 and 32, 556 in all; 9:3:3:1 expects 312.75, 104.25, 104.25 and 34.75
- 04χ² = 0.016 + 0.135 + 0.101 + 0.218 = 0.470, with df = 3, so p = 0.925
The ratio is 9:3:3:1, and Mendel's counts fit it: a χ² of 0.470 is far below 7.815, the value needed to reject the ratio at the 0.05 level.
How it works
How to do a dihybrid cross
Start with each parent's gametes. A gamete takes one allele of each gene, and when the genes assort independently every combination is equally likely: AaBb makes AB, Ab, aB and ab, a quarter each. The FOIL order (first, outer, inner, last) lists them without missing one.
Draw the square: one parent's gametes down the side, the other's across the top, and fill each box with the alleles of its row and its column. Each of the 16 boxes is a sixteenth of the offspring. Count the boxes of each genotype, then group the genotypes that look alike into phenotypes.
Or skip the square and multiply. The two genes are independent, so Aa × Aa gives ¾ A_ and Bb × Bb gives ¾ B_, and A_B_ is ¾ × ¾ = 9/16. When two genes act on one trait, the boxes stay the same but some classes look alike and merge: 9:3:4, 12:3:1, 9:7, 15:1, 13:3 or 9:6:1.
To test real offspring, compare the counts with the ratio by chi-square. A p above 0.05 means the counts fit: a deviation that size often happens by chance. Below 0.05, look for linked genes, a genotype that dies early, offspring put in the wrong class, or genes that interact.
Questions
Frequently asked
What is the phenotype ratio of a dihybrid cross?
9:3:3:1, when both parents are heterozygous for both genes (AaBb × AaBb) and each gene is completely dominant: 9 in 16 show both dominant traits, 3 and 3 show one of them, and 1 shows both recessive traits. The genotype ratio is 1:2:1:2:4:2:1:2:1, nine genotypes in the 16 boxes.
How many gametes does AaBb make?
Four kinds, AB, Ab, aB and ab, a quarter each, because each gamete takes one allele of each gene and the genes assort independently. AaBB makes two kinds, AB and aB, and AABB only one.
What is a dihybrid test cross?
A cross with an individual recessive for both genes, aabb. All its gametes carry ab, so the offspring show which gametes the other parent made: AaBb gives 1:1:1:1, AaBB gives 1:1, and AABB gives offspring that all look alike.
What do ratios like 9:3:4, 12:3:1 and 9:7 mean?
That two genes act on one trait, which is called epistasis. The 16 boxes are the same as for 9:3:3:1, but some classes look alike. In recessive epistasis the 3 and the 1 merge into 4 (Labrador coat colour); in dominant epistasis the 9 and a 3 merge into 12 (summer squash colour); with complementary genes everything but the 9 looks the same, 9:7 (sweet pea flowers).
How do I use chi-square on a dihybrid cross?
Count the offspring in each phenotype class, and work out the counts the ratio predicts for the same total. Add up (observed − expected)² ÷ expected over the classes, and compare the sum with 7.815, the critical value for 3 degrees of freedom at the 0.05 level. Below it, the counts fit the ratio: Mendel's 315, 108, 101 and 32 give 0.470.
Why don't my results match 9:3:3:1?
Chance first: small crosses stray from the ratio, which the chi-square test allows for. Beyond chance, the genes may be linked, one genotype may die before it is counted, one gene may be incompletely dominant (3:1:6:2:3:1 instead), or the two genes may act on the same trait.
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