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Grade 10-11 (age 15-17)

high school Mendelian Genetics

Mendel crossed pea plants by the thousands and counted the traits of their offspring, finding that traits do not blend but reappear in fixed ratios. The laws of dominance, segregation and independent assortment explain why crossing one pair of traits gives 3 : 1 and crossing two pairs gives 9 : 3 : 3 : 1. A Punnett square, with each parent's gametes written along the top and the side and each box filled with a pairing, shows the genotype and phenotype ratios of the offspring at a glance. In the diagram you can compare the Punnett squares and resulting ratios for a cross between hybrids, a test cross and a cross between purebreds.

Mendel's Discovery

🌿 Patterns Found in Pea Crosses
①Mendel chose seven clearly contrasting pea traits, such as height, flower color and seed shape, and crossed plants over several generations
②The F₁ hybrids from crossing purebreds showed only one trait, and in the F₂ from crossing the F₁ plants the two traits reappeared in about a 3 : 1 ratio
③From this he concluded that the factors deciding a trait (now called genes) come in pairs and separate when gametes form
④Genes on the same chromosome, however, travel together (linkage), so the law of independent assortment does not hold for them

Punnett Squares

Aa × Aa (hybrid cross)
💡 Comparing the Three Crosses
①Aa × Aa: genotypes AA : Aa : aa = 1 : 2 : 1 and phenotypes dominant : recessive = 3 : 1
②Aa × aa (test cross): offspring are Aa and aa in a 1 : 1 ratio, which shows the dominant parent was Aa
③AA × aa: all offspring are Aa and show only the dominant trait
④Each box of a Punnett square is one offspring formed when two gametes meet, so the ratio of the four boxes is the ratio of the offspring's probabilities

Mendel's Laws of Inheritance

Law of Dominance
Aa → only the dominant trait (A) shows
crossing purebreds with contrasting traits gives F₁ hybrids that show only the dominant trait
Law of Segregation
Aa × Aa → AA : Aa : aa = 1 : 2 : 1
during meiosis the two alleles of a pair separate and go into different gametes
Law of Independent Assortment
AaBb × AaBb → 9 : 3 : 3 : 1
two pairs of alleles on different chromosomes separate independently of each other

Genetic Terms to Know

Genetic Terms

Basic Genetics Vocabulary
Allele
different forms of a gene for the same trait, found at the same place on homologous chromosomes
A, a
Homozygous (purebred)
has two identical alleles
AA, aa
Heterozygous (hybrid)
has two different alleles
Aa
Genotype
the combination of alleles an individual carries
AA, Aa, aa
Phenotype
the trait that shows on the outside
round seed, wrinkled seed
Test cross
crossing with a recessive purebred (aa) to find the genotype of a dominant individual
Aa × aa

Worked Examples

Example 1
Round seeds (R) are dominant over wrinkled seeds (r). A cross of Rr × Rr gives 800 offspring. About how many have wrinkled seeds (rr)?
1
For Rr × Rr the phenotype ratio is round : wrinkled = 3 : 1, so wrinkled seeds make up 1/4 of the total.
2
800 × 1/4 = 200
▸ about 200
One of the four boxes in the Punnett square is rr, so the probability is 1/4. The actual count is estimated by multiplying the probability by the total, and the larger the total, the closer the ratio comes to 1/4.
Example 2
In the cross AaBb × AaBb, what is the probability of an offspring that is recessive for both traits (aabb)? (The two genes are on different chromosomes.)
1
The two genes separate independently, so find the probability for each trait and multiply. The probability of aa is 1/4, and so is the probability of bb.
2
1/4 × 1/4 = 1/16
▸ 116
The final 1 in 9 : 3 : 3 : 1 is exactly aabb. You could draw both traits in one 16-box square, but finding 3 : 1 for each trait and multiplying is quicker.

Summary

Phenotype Ratio of a Monohybrid Cross
3 : 1
Aa × Aa
Phenotype Ratio of a Dihybrid Cross
9 : 3 : 3 : 1
AaBb × AaBb
CSAT-style problem, adapted
In the cross AaBb × AaBb, what fraction of the offspring show the dominant trait for both A and B (AB)? (The two genes are on different chromosomes.)
①116
②316
③916
④34
⑤14
▸ ③ 916
1
In Aa × Aa the probability of the dominant trait (A) is 3/4, and in Bb × Bb the probability of the dominant trait (B) is also 3/4.
2
The two genes are independent, so multiply: 3/4 × 3/4 = 9/16, the 9 in 9 : 3 : 3 : 1.
🎯 Exam Points
①Law of segregation: the offspring of Aa × Aa have genotypes 1 : 2 : 1 and phenotypes 3 : 1
②Law of independent assortment: for two genes on different chromosomes, AaBb × AaBb gives 9 : 3 : 3 : 1
③Test cross: cross a dominant individual of unknown genotype with a recessive purebred (aa) and read the genotype from the offspring ratio
④Linkage: genes on the same chromosome are inherited together, so independent assortment does not hold
⑤If the offspring show a 3 : 1 phenotype ratio, both parents are Aa (heterozygous)
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