Mendelian Genetics

Mendel found regular ratios in pea crosses and set out the laws of heredity. Dominance, segregation, and independent assortment explain monohybrid 3:1 and dihybrid 9:3:3:1 ratios. A Punnett square shows genotype and phenotype ratios at a glance. Switch the cross-type slider to compare monohybrid, test, and purebred crosses.

Mendel's Discovery

🌿 Laws of Heredity from Pea Plants

①Mendel cross-bred pea plants with 7 traits (height, flower color, seed shape, etc.)

②Found regular ratios (3:1)

③This was the start of genetics — "the laws of heredity"

Visualizing Punnett Squares

Cross type0

💡 Try changing the cross

①0: Aa × Aa → 3:1 (key monohybrid cross!)

②1: Aa × aa → 1:1 (test cross — checks genotype)

③2: AA × aa → all Aa (pure parents — F₁ all dominant)

Mendel's Laws

Law of Dominance

Aa → only dominant (A) shows

dominant masks recessive among allele pair

Law of Segregation

Aa × Aa → AA : Aa : aa = 1 : 2 : 1

alleles segregate during meiosis into separate gametes

Law of Independent Assortment

AaBb × AaBb → 9:3:3:1

genes on different chromosomes assort independently

Key Genetic Terms

Genetic Terms

ListCore Genetics Vocabulary

Allele

different genes for the same trait

A, a

Homozygous (purebred)

same allele × 2

AA, aa

Heterozygous (hybrid)

different alleles × 2

Genotype

genetic composition

AA, Aa, aa

Phenotype

observable trait

round seed, wrinkled seed

Test cross

cross with recessive purebred to check genotype

Aa × aa

Worked Examples

Example 1

Round seed (R) is dominant over wrinkled (r). A cross Rr × Rr yields 800 offspring. About how many are wrinkled (rr)?

Rr × Rr → phenotype ratio round : wrinkled = 3 : 1, so wrinkled is 1/4.

800 × 1/4 = 200.

▸ about 200

Law of segregation: Aa×Aa → 1/4 recessive homozygous (aa). Multiply the ratio (1/4) by the total.

Example 2

In a cross AaBb × AaBb, what is the probability of offspring recessive for both traits (aabb)? (genes independent)

Since the traits are independent, multiply probabilities. P(aa) = 1/4, P(bb) = 1/4.

1/4 × 1/4 = 1/16.

▸ 116

Law of independent assortment: find each trait’s probability separately and multiply. aabb = (1/4)(1/4) = 1/16.

Summary

Monohybrid

3 : 1

Aa × Aa

Dihybrid

9:3:3:1

AaBb × AaBb

CSAT-style

In a cross AaBb × AaBb, what fraction of offspring show both A and B as dominant (A_B₎? (genes independent)

①116

②316

③916

④34

⑤14

▸ ③ 916

In Aa×Aa, P(dominant A₎ = 3/4; in Bb×Bb, P(dominant B₎ = 3/4.

They are independent, so multiply: 3/4 × 3/4 = 9/16. (the 9 in 9:3:3:1)

🎯 Exam Points

①Segregation: Aa × Aa → genotype 1:2:1, phenotype 3:1

②Independent assortment: AaBb × AaBb → 9:3:3:1 (different chromosomes)

③Test cross: unknown × recessive purebred (aa) → infer from offspring

④Linkage: same chromosome → independent assortment fails

⑤Phenotype 3:1 → both parents are Aa (heterozygous)

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