Two-Gene Cross Calculators

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A two-gene cross (dihybrid cross) follows the inheritance of two independently assorting gene loci simultaneously. When two heterozygous parents (AaBb × AaBb) cross, they produce four types of gametes (AB, Ab, aB, ab) in equal proportions, and the 4×4 Punnett square yields 16 genotype combinations. Under complete dominance and independent assortment (Mendel's second law), the F2 phenotype ratio is 9:3:3:1 (9 dominant both : 3 dominant A only : 3 dominant B only : 1 recessive both). Deviations from this ratio indicate gene interaction (epistasis) or linkage.

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Dihybrid Cross Setup

Parents: AaBb × AaBb. Gametes (each parent): AB, Ab, aB, ab — each at 1/4 frequency. 4×4 Punnett square → 16 genotype combinations → group by phenotype.

9:3:3:1 Phenotype Ratio

A_B_ (both dominant): 9/16. A_bb (A dominant, b recessive): 3/16. aaB_ (a recessive, B dominant): 3/16. aabb (both recessive): 1/16. Probability shortcut: multiply single-gene probabilities. P(A_B_) = 3/4 × 3/4 = 9/16; P(A_bb) = 3/4 × 1/4 = 3/16; etc.

Modified Ratios from Epistasis

  • Dominant epistasis: 12:3:1 (A_ masks B)
  • Recessive epistasis: 9:3:4 (aa masks B)
  • Duplicate dominant: 15:1 (A_ or B_ → same phenotype)
  • Complementary: 9:7 (A and B both needed)

Chi-Square Test

Test observed ratio against 9:3:3:1: χ² = Σ(O−E)²/E; df = 3. Critical χ² (df=3, p=0.05) = 7.815. If χ² > 7.815: reject 9:3:3:1 → investigate alternative ratios.

Glossary

Dihybrid Cross
A genetic cross following two independently assorting loci (AaBb × AaBb); produces 16 genotype combinations; F2 phenotype ratio = 9:3:3:1 under complete dominance and independent assortment.
9:3:3:1 Ratio
The F2 phenotype ratio from a dihybrid cross: 9 A_B_ : 3 A_bb : 3 aaB_ : 1 aabb; calculated by P(A_) × P(B_) product rule; modified by epistasis (12:3:1; 9:3:4; 9:7; 15:1).
Epistasis
Gene interaction where one gene masks or modifies expression of another; produces modified dihybrid ratios (12:3:1; 9:3:4; 9:7; 15:1); all total 16 genotypic combinations but grouped differently.

Frequently Asked Questions

A dihybrid cross follows two independently assorting gene loci simultaneously. Standard cross: AaBb × AaBb. Each parent produces 4 gamete types: AB, Ab, aB, ab (each 1/4). The 4×4 Punnett square gives 16 combinations → grouped by phenotype: A_B_ (any A dominant, any B dominant): 9/16. A_bb: 3/16. aaB_: 3/16. aabb: 1/16. Ratio: 9:3:3:1. This requires: both genes show complete dominance; genes are on different chromosomes (or far apart on the same one) → independent assortment (Mendel's second law).

The product rule: for independently assorting genes, the probability of a combined outcome = product of individual probabilities. From a monohybrid Aa × Aa cross: P(A_) = 3/4; P(aa) = 1/4. For two genes: P(A_B_) = P(A_) × P(B_) = 3/4 × 3/4 = 9/16. P(A_bb) = P(A_) × P(bb) = 3/4 × 1/4 = 3/16. P(aaB_) = 1/4 × 3/4 = 3/16. P(aabb) = 1/4 × 1/4 = 1/16. This avoids drawing the full 16-cell Punnett square and is faster for exam problems. Always verify: 9/16 + 3/16 + 3/16 + 1/16 = 16/16 = 1 ✓.

Modified ratios arise when the two genes interact (epistasis): Dominant epistasis (12:3:1): presence of A_ masks effect of B. Example: A_ produces yellow pigment regardless of B; aaB_ = purple; aabb = white. Recessive epistasis (9:3:4): aa genotype masks B expression. Example: aa prevents pigment production; B has no color to express → aaB_ and aabb both look the same (no color). Complementary genes (9:7): both A and B required for phenotype. Neither alone is sufficient. Duplicate dominant (15:1): A_ or B_ alone is sufficient → only aabb shows recessive phenotype. All modified ratios still total 16 cells; they are different groupings of the 9:3:3:1 genotype classes.

Chi-square goodness of fit test: H₀ = ratio is 9:3:3:1 (independent assortment, complete dominance). For total n offspring: Expected counts: E(A_B_) = n × 9/16; E(A_bb) = n × 3/16; E(aaB_) = n × 3/16; E(aabb) = n × 1/16. Calculate: χ² = Σ(O−E)²/E for all 4 classes. Degrees of freedom = 4 − 1 = 3. Critical value: χ² (df=3, p=0.05) = 7.815. Decision: χ² < 7.815 → data consistent with 9:3:3:1; χ² > 7.815 → significantly different → investigate alternative ratios (epistasis, linkage, or non-Mendelian inheritance).