How this instrument works
A dihybrid cross tracks two genes at once across a mating, rather than the single gene a simpler monohybrid cross follows. Mendel's law of independent assortment states that, absent gene linkage, the alleles for one gene segregate into gametes independently of the alleles for another gene — so the combined outcome across two genes can be found by treating each gene's own 3:1 or similar segregation separately and then combining the probabilities.
This instrument reports its result as a ratio of phenotype classes rather than a single number, because a dihybrid cross does not reduce to one figure — it produces up to four distinct phenotype classes (dominant-dominant, dominant-recessive, recessive-dominant, recessive-recessive) in proportions that depend on both parents' genotypes. The notation uses an underscore, as in A_B_, to mean 'dominant phenotype at this gene, regardless of which allele pair produced it' — since AA and Aa both look dominant, the underscore stands in for 'either allele.'
Select a genotype for each parent from three options: AaBb (heterozygous at both genes), AABB (homozygous dominant at both), or aabb (homozygous recessive at both). The classic 9:3:3:1 ratio only appears from a heterozygous-by-heterozygous cross (AaBb × AaBb); crossing a fully homozygous parent with another genotype instead collapses the outcome to a simpler ratio, which the instrument reports exactly as it works out for whichever pair you select.
- Select a genotype for Parent 1 genotype from the three offered combinations: AaBb, AABB, or aabb.
- Select a genotype for Parent 2 genotype the same way — parents can be the same genotype or different.
- Read Offspring phenotype-class ratio — a ratio string like '9 A_B_ : 3 A_bb : 3 aaB_ : 1 aabb', not a single number.
- Remember the underscore convention: A_ means dominant phenotype at that gene (from AA or Aa), while a lowercase pair like aa or bb written out fully means the recessive phenotype at that gene.
- This models two independently assorting genes; genes linked close together on the same chromosome are inherited together more often, which skews a real cross away from these ratios.
Worked example — AaBb x AaBb, the classic dihybrid cross
Set Parent 1 genotype to AaBb and Parent 2 genotype to AaBb — both parents heterozygous at both genes, the textbook dihybrid self-cross. Offspring phenotype-class ratio reads 9 A_B_ : 3 A_bb : 3 aaB_ : 1 aabb.
That ratio comes directly from multiplying each gene's independent 3:1 segregation: 9/16 of offspring show the dominant phenotype at both genes (A_B_), 3/16 show dominant at the first gene and recessive at the second (A_bb), another 3/16 show the reverse (aaB_), and 1/16 show recessive at both (aabb) — the 9:3:3:1 proportions that gave Mendel's independent assortment its first quantitative confirmation.
Questions
Why is the result a ratio string like '9 A_B_ : 3 A_bb : 3 aaB_ : 1 aabb' instead of a single number?
Because a dihybrid cross doesn't have one answer — it distributes offspring across up to four distinct phenotype classes at once, and the interesting result is how those classes compare to each other, not any single class's share alone. Reporting the full ratio, rather than one number, is what actually answers 'what phenotypes should I expect, and in what proportions.'
What does the underscore in 'A_B_' mean?
It stands for 'either allele' at that position, because AA and Aa look identical from the outside — both show the dominant phenotype. A_B_ therefore means 'dominant phenotype at gene A and dominant phenotype at gene B,' covering the AABB, AaBB, AABb and AaBb genotypes together, since a Punnett square tracks genotypes but a phenotype ratio only cares about which trait is visible.
What happens if I cross AABB with aabb instead of two heterozygotes?
Every offspring inherits one dominant allele of each gene from the AABB parent and one recessive allele of each gene from the aabb parent, so all offspring end up AaBb — heterozygous at both genes and showing the dominant phenotype at both. The ratio for that cross is simply 1 A_B_, since there is only one phenotype class rather than four; this is the textbook 'all F1 offspring are double heterozygotes' result.
What ratio results from a testcross, AaBb x aabb?
An even 1 A_B_ : 1 A_bb : 1 aaB_ : 1 aabb — each of the four phenotype classes appears in equal proportion. This specific cross, a double heterozygote against a fully homozygous recessive parent, is the classic 'testcross' geneticists use in practice: since the aabb parent can only contribute recessive alleles, the offspring's phenotypes directly reveal which gametes the heterozygous parent produced, which is also why any deviation from this even 1:1:1:1 split in a real testcross is the standard evidence for gene linkage.
What is 'independent assortment' and why does it matter here?
Independent assortment is Mendel's second law: the allele pair for one gene segregates into gametes without influencing how the allele pair for a different gene segregates, provided the two genes are on different chromosomes or far enough apart on the same one. It is what allows this calculator to compute each gene's ratio separately and then simply multiply the probabilities together — without it, the four phenotype classes would not appear in these clean, predictable proportions.
Does this calculator account for gene linkage?
No — it assumes the two genes assort independently, which is the standard textbook dihybrid-cross model. Genes located close together on the same chromosome are often inherited together more frequently than independent assortment predicts, a phenomenon called linkage, which skews real observed ratios away from 9:3:3:1 or 1:1:1:1; detecting and measuring that skew is itself the classic evidence biologists use to establish that two genes are linked.