Mendelian Genetics & Probability
- Apply the laws of segregation and independent assortment to predict offspring
- Use Punnett squares and probability rules to solve mono- and dihybrid crosses
- Explain how a test cross reveals the genotype of a dominant-phenotype individual
Alleles, genotype, and the law of segregation
A gene can exist as different versions called alleles. A diploid organism carries two alleles per gene — its genotype — which may be homozygous (two of the same, like AA or aa) or heterozygous (two different, Aa). The phenotype is the visible trait. A dominant allele (written uppercase) masks a recessive one (lowercase) in a heterozygote. Mendel’s law of segregation states that the two alleles for a gene separate during meiosis, so each gamete carries only one allele — the rule that makes Punnett squares work.
The monohybrid cross
Cross two heterozygotes, Aa × Aa. Each parent’s alleles segregate, so each produces ½ A and ½ a gametes. Filling the Punnett square gives genotypes in a 1 AA : 2 Aa : 1 aa ratio. Because A is dominant, AA and Aa both show the dominant phenotype, so the phenotype ratio is 3 dominant : 1 recessive — the signature 3:1 of a heterozygous cross. Notice that ¼ of the offspring are homozygous recessive and express the hidden trait.
Independent assortment and the probability rules
Mendel’s law of independent assortment says that alleles of different genes are sorted into gametes independently (they assort separately in metaphase I). This lets you treat each gene as its own monohybrid cross and combine them with two probability rules. The rule of multiplication ("AND"): the chance of two independent events both happening is the product of their individual chances. The rule of addition ("OR"): the chance of either of two mutually exclusive events is the sum of their chances. These turn even complicated crosses into quick arithmetic.
In pea plants, round seeds (R) are dominant to wrinkled (r) and yellow (Y) is dominant to green (y). Cross two dihybrids: RrYy × RrYy. Predict the phenotype ratio of the offspring.
- 1.Treat each gene separately. For seed shape, Rr × Rr gives ¾ round (R_) and ¼ wrinkled (rr). For color, Yy × Yy gives ¾ yellow (Y_) and ¼ green (yy).
- 2.Combine phenotypes with the multiplication rule. Round yellow: ¾ × ¾ = 9⁄16. Round green: ¾ × ¼ = 3⁄16. Wrinkled yellow: ¼ × ¾ = 3⁄16. Wrinkled green: ¼ × ¼ = 1⁄16.
- 3.Express the four fractions over the common denominator 16 as a ratio: 9 : 3 : 3 : 1.
The 9:3:3:1 phenotype ratio is the fingerprint of a dihybrid cross between two heterozygotes (AaBb × AaBb). If a problem gives you that ratio, both parents were heterozygous for both genes.
Two heterozygous pea plants are crossed: Aa × Aa. What fraction of the offspring are expected to show the recessive phenotype?
A test cross for the dihybrid above: what fraction of the offspring of RrYy × RrYy are homozygous recessive for both genes (rryy)?
- 1.To be rryy, the offspring must be rr AND yy.
- 2.From Rr × Rr, the chance of rr is ¼. From Yy × Yy, the chance of yy is ¼.
- 3.Apply the multiplication rule for both happening together: ¼ × ¼ = 1⁄16.
A plant with purple flowers could be homozygous (PP) or heterozygous (Pp), since purple (P) is dominant to white (p). A breeder crosses it with a white plant (pp) and gets offspring that are roughly 1 purple : 1 white. What was the purple parent’s genotype?
For "what fraction of offspring…" questions, do not draw a 16-box dihybrid grid. Solve each gene as a small monohybrid cross (¾ or ¼) and multiply. It is faster and far less error-prone under time pressure.
Answer the 2 checkpoints as you read.
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