Monohybrid Cross Calculators

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A monohybrid cross follows the inheritance of a single gene with two alleles. In a cross between two heterozygous parents (Aa × Aa), the F2 generation produces a 3:1 phenotype ratio (3 dominant : 1 recessive) when there is complete dominance. This ratio is predicted by Mendel's law of segregation: each parent produces two types of gametes (A and a) in equal proportions, and these combine randomly in a 2×2 Punnett square to give genotype frequencies 1 AA : 2 Aa : 1 aa. Testcrosses (Aa × aa) produce a 1:1 ratio and are used to determine the genotype of an individual with the dominant phenotype.

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Aa × Aa Cross (F1 × F1)

Gametes: each Aa parent produces A (50%) and a (50%) gametes. Punnett square (2×2): AA, Aa, Aa, aa. Genotype ratio: 1 AA : 2 Aa : 1 aa. Phenotype ratio (complete dominance): 3 dominant (AA + Aa) : 1 recessive (aa) = 3:1. Probabilities: P(A_) = 3/4; P(aa) = 1/4.

Testcross (Aa × aa)

Cross an unknown individual with a homozygous recessive (aa): If unknown is AA: all offspring dominant (AA × aa → all Aa). If unknown is Aa: half dominant, half recessive (Aa × aa → 1/2 Aa : 1/2 aa → 1:1). Used to determine genotype of dominant-phenotype individual.

P, F1, and F2 Generations

P (parental): TT × tt → F1: all Tt (tall). F1 × F1 (Tt × Tt) → F2: 3 Tall : 1 Dwarf. Mendel's pea plant example. Classic representation of law of segregation.

Chi-Square Test

Test observed F2 against 3:1: χ² = (O_dom − E_dom)²/E_dom + (O_rec − E_rec)²/E_rec. df = 1. Critical value (p=0.05, df=1) = 3.841. If χ² < 3.841: data consistent with 3:1.

Glossary

Monohybrid Cross
A genetic cross following one gene (Aa × Aa); F2 genotype ratio 1:2:1 (AA:Aa:aa); F2 phenotype ratio 3:1 (dominant:recessive) with complete dominance; Mendel's law of segregation.
Testcross
Cross of unknown genotype (dominant phenotype) × homozygous recessive (aa); all dominant offspring → unknown is AA; 1:1 dominant:recessive offspring → unknown is Aa; determines genotype.
Law of Segregation
Mendel's first law: the two alleles of a gene separate during gamete formation; each gamete receives only one allele; produces 1:2:1 genotype and 3:1 phenotype ratios in F2 from Aa×Aa.

Frequently Asked Questions

A monohybrid cross follows the inheritance of one gene. Standard F1 × F1 cross (Aa × Aa): gametes from each parent: A (50%) and a (50%). Punnett square: 4 equally likely combinations: AA (25%), Aa (50%), aa (25%). Genotype ratio: 1:2:1. Phenotype ratio (with complete dominance — A dominant over a): AA and Aa are both dominant phenotype = 3/4 dominant. aa = 1/4 recessive. Phenotype ratio = 3:1. Mendel's original data: tall × dwarf → F1 all tall → F2: 787 tall : 277 dwarf ≈ 2.84:1 ≈ 3:1.

For Aa × Aa: Step 1: Write gametes of parent 1 across the top: A, a. Step 2: Write gametes of parent 2 down the left side: A, a. Step 3: Fill each cell by combining the column header + row header. Top-left: AA; top-right: Aa; bottom-left: Aa; bottom-right: aa. Step 4: Count genotypes: 1 AA, 2 Aa, 1 aa (ratio 1:2:1). Step 5: Phenotypes (if A is dominant): AA and Aa are phenotypically identical = 3 dominant. aa = 1 recessive. Ratio = 3:1. For Aa × aa (testcross): gametes top: A, a; gametes left: a, a. Cells: Aa, aa, Aa, aa → 1 Aa : 1 aa → 1:1 phenotype ratio.

A testcross crosses an individual of unknown genotype (showing dominant phenotype) with a homozygous recessive individual (aa). The genotype of the aa parent is known; the genotype of the other parent is unknown (could be AA or Aa). Predictions: If unknown = AA: all offspring are Aa (dominant phenotype) → 100% dominant in offspring. If unknown = Aa: half offspring Aa (dominant), half aa (recessive) → 1:1 phenotype ratio. By observing offspring phenotypes, you can infer the unknown parent's genotype. Why testcross and not self-cross: for diploid organisms; testcross immediately reveals the recessive allele in the unknown parent because the recessive aa parent contributes only a alleles.

Chi-square goodness-of-fit: H₀ = ratio is 3:1. Total offspring = N. Expected: E_dom = 3N/4; E_rec = N/4. χ² = (O_dom − E_dom)²/E_dom + (O_rec − E_rec)²/E_rec. df = number of phenotypic classes − 1 = 2 − 1 = 1. Critical value: χ²₀.₀₅(df=1) = 3.841. Example: 90 total; 65 dominant, 25 recessive: E_dom = 67.5; E_rec = 22.5. χ² = (65−67.5)²/67.5 + (25−22.5)²/22.5 = 6.25/67.5 + 6.25/22.5 = 0.093 + 0.278 = 0.371. 0.371 < 3.841 → fail to reject H₀ → data consistent with 3:1 ratio.