Short on time? This page is the recap. For explanations, figures and worked examples, use the full study guide.
Recap
- Genotype = the alleles an individual has (AA, Aa, aa). Phenotype = the observable trait. Homozygous = two identical alleles; heterozygous = two different.
- A dominant allele shows its phenotype in the heterozygote. Often it codes for a working protein, and one copy is enough.
- Law of segregation: each gamete gets one allele of each gene, because homologous chromosomes separate in meiosis I.
- Law of independent assortment: genes on different chromosomes are sorted independently, because each homologous pair lines up at metaphase I independently of the others. Linked genes are the exception.
- Fertilization joins two haploid gametes, restores the diploid number and creates new allele combinations.
- Pedigrees: squares are males, circles females, filled symbols affected.
Key relationships
| Cross | Genotypic ratio | Phenotypic ratio |
|---|---|---|
| Aa × Aa | 1 AA : 2 Aa : 1 aa | 3 : 1 |
| Aa × aa (test cross) | 1 Aa : 1 aa | 1 : 1 |
| AA × aa | all Aa | all dominant |
| AaBb × AaBb | 9 genotypes | 9 : 3 : 3 : 1 |
| AaBb × aabb (test cross) | 1 : 1 : 1 : 1 | 1 : 1 : 1 : 1 |
| Rule or equation | Use |
|---|---|
| P(A and B) = P(A) × P(B) | independent events: separate genes, separate children |
| P(A or B) = P(A) + P(B) | mutually exclusive routes to one outcome, e.g. P(Aa) = 1/4 + 1/4 |
| χ² = Σ (o − e)² ÷ e | o = observed count, e = expected count |
| df = categories − 1 | compare χ² with the critical value at p = 0.05 |
| Pedigree clue | Points to |
|---|---|
| two unaffected parents with an affected child | recessive |
| two affected parents with an unaffected child | dominant |
| recessive trait in a daughter whose father is unaffected | autosomal, not X-linked |
Assumptions behind the numbers
- Mendel’s ratios assume complete dominance, unlinked genes, and that every genotype survives equally well.
- Expected ratios are probabilities. Small samples scatter around them.
- The chi-square test needs counts in categories, not percentages or continuous measurements.
Mistakes to avoid
- Giving a genotypic ratio when the question asks for a phenotypic one, or the reverse.
- Saying P(carrier) = 1/2 for an unaffected child of two carriers; it is 2/3.
- Using number of offspring − 1 as degrees of freedom.
- Writing “the null hypothesis is proved” or “accepted”; say “fail to reject”.
- Writing a null hypothesis that does not mention the expected ratio or the variables.
- Applying independent assortment to genes on the same chromosome.
Quick self-check
- AaBb × aabb: probability of an aabb offspring? (1/2 × 1/2 = 1/4)
- A monohybrid F₂ of 400 has 290 dominant and 110 recessive. χ² against 3 : 1? (Expected 300 and 100; χ² = 100/300 + 100/100 = 1.33; df = 1; 1.33 < 3.84, so fail to reject.)
- Why is a test cross made with a homozygous recessive? (It gives only recessive alleles, so each offspring’s phenotype reveals the allele from the unknown parent.)
Next: practice questions.