Mendelian Ratios Calculators
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Monohybrid Cross (Aa × Aa)
F2 genotypes: 1/4 AA : 2/4 Aa : 1/4 aa = 1:2:1. F2 phenotypes (with complete dominance): 3/4 dominant (AA + Aa) : 1/4 recessive (aa) = 3:1. Punnett square: 2×2 grid. Example: Tall (TT) × Dwarf (tt) → F1 all Tt (tall) → F1 × F1 → 3 Tall : 1 Dwarf.
Dihybrid Cross (AaBb × AaBb)
F2 genotypes: 16 combinations in a 4×4 Punnett square. F2 phenotypes (complete dominance, independent assortment): 9/16 A_B_ : 3/16 A_bb : 3/16 aaB_ : 1/16 aabb = 9:3:3:1. Shortcut: multiply individual monohybrid probabilities: P(A_B_) = 3/4 × 3/4 = 9/16.
Modified Ratios
- Incomplete dominance: 1:2:1 (e.g., snapdragon red × white → pink)
- Codominance: 1:2:1 (both alleles expressed, e.g., blood type M/N)
- Dominant epistasis: 12:3:1
- Recessive epistasis: 9:3:4
- Duplicate dominant: 15:1
- Complementary genes: 9:7
Chi-Square Test
Test observed ratios: χ² = Σ(O−E)²/E. df = (classes − 1). For 3:1: df=1; for 9:3:3:1: df=3. Critical value (df=1, p=0.05) = 3.841. If χ² < critical value: data consistent with Mendelian ratio.
Glossary
Frequently Asked Questions
Mendelian ratios are the characteristic phenotype frequencies produced in genetic crosses, predicted by Gregor Mendel's laws (1865). They arise from the equal probability of each allele being transmitted to offspring (law of segregation) and from alleles at different loci assort independently into gametes (law of independent assortment). Key ratios: F2 monohybrid (Aa × Aa): 3 dominant : 1 recessive. F2 dihybrid (AaBb × AaBb): 9:3:3:1. Testcross (Aa × aa): 1:1 (dominant : recessive). These ratios assume: Mendelian inheritance; complete dominance; no linkage between genes being studied.
Cross: Aa × Aa (heterozygous × heterozygous). Step 1: Determine gametes. Each parent makes A and a gametes, each at 50% frequency. Step 2: Fill 2×2 Punnett square: A and a across top and left. Step 3: 4 genotype combinations: AA (1/4); Aa (2/4); Aa (2/4); aa (1/4) → 1 AA : 2 Aa : 1 aa. Step 4: Phenotypes with complete dominance: AA and Aa both show dominant phenotype → 3/4 dominant : 1/4 recessive = 3:1. Probability approach (faster): P(A_) = P(AA) + P(Aa) = (1/4) + (2/4) = 3/4. P(aa) = 1/4. Result: 3:1.
Dihybrid cross: AaBb × AaBb. Step 1: Gametes of each parent: AB, Ab, aB, ab (each 1/4). Step 2: 4×4 Punnett square: 4 gamete types across the top and left. Step 3: Fill 16 cells. Step 4: Group by phenotype (with complete dominance): A_B_: any box with at least one A and at least one B → count = 9. A_bb: at least one A, two b → count = 3. aaB_: two a, at least one B → count = 3. aabb: two a, two b → count = 1. Ratio: 9:3:3:1. Shortcut: instead of 16-cell Punnett square, multiply individual locus probabilities: P(A_) = 3/4; P(B_) = 3/4; P(A_B_) = 9/16.
Incomplete dominance: the heterozygote (Aa) has an intermediate phenotype — neither allele is fully dominant. Example: Antirrhinum (snapdragon) flowers: RR = red; rr = white; Rr = pink (intermediate). Cross F1 × F1 (Rr × Rr): F2 genotype ratio still 1:2:1 (same as always — genotype ratios don't change). But F2 phenotype ratio: 1/4 red (RR) : 2/4 pink (Rr) : 1/4 white (rr) = 1:2:1. The phenotype ratio now reflects the genotype ratio because each genotype has a distinct phenotype. Compare: complete dominance (Aa looks like AA) → 3:1 phenotype ratio. Incomplete dominance (Aa looks intermediate) → 1:2:1 phenotype ratio. Codominance: both alleles fully expressed (e.g., AB blood type, MN blood group) → also 1:2:1 phenotype ratio but with 3 distinct phenotype classes.