Short on time? This page is the recap. For explanations, figures and worked examples, use the full study guide.
Recap
- Detecting a deviation: compare observed counts with the Mendelian prediction using chi-square. If χ² exceeds the critical value, reject the Mendelian null hypothesis and look for a biological cause.
- Incomplete dominance: heterozygote is a blend (red × white → pink). Codominance: heterozygote shows both phenotypes fully (blood group AB, IᴬIᴮ).
- Linked genes are on the same chromosome and tend to be inherited together. Crossing over in prophase I makes recombinants. Recombination frequency gives map distance; adding distances orders genes.
- Sex-linked genes are on X or Y. XY males have one X, so one recessive X-linked allele shows. Fathers pass X to every daughter and Y to every son. Y-linked traits go father to all sons.
- Other systems: birds are ZW (female) and ZZ (male); bees are haplodiploid (haploid males from unfertilized eggs).
- Pleiotropy: one gene, several effects (sickle-cell allele: anaemia, organ damage, malaria resistance in heterozygotes). The effects never separate.
- Non-nuclear inheritance: mitochondria and chloroplasts are randomly shared out at division and usually come from the egg or ovule, so their traits are typically maternal.
Key relationships
| Pattern | Data clue |
|---|---|
| Incomplete dominance or codominance | heterozygote × heterozygote gives 1 : 2 : 1 phenotypes |
| Linked genes (test cross) | two large and two small classes instead of 1 : 1 : 1 : 1 |
| X-linked recessive | more affected males; never father to son; can skip through carrier daughters |
| Y-linked | only males; every son of an affected father |
| Pleiotropy | several traits always inherited together |
| Mitochondrial or chloroplast | reciprocal crosses differ; offspring match the mother |
| Equation | Meaning |
|---|---|
| Recombination frequency (%) = recombinants ÷ total × 100 | 1% = 1 map unit |
| χ² = Σ (o − e)² ÷ e | df = categories − 1 |
Assumptions behind the numbers
- Map distances from one cross are reliable for genes fairly close together; recombination frequency never exceeds about 50%.
- The heterozygous parent in a mapping test cross must have a known arrangement of alleles (from known parents) to tell parental from recombinant classes.
- Maternal inheritance of organelles is “typical”, not universal.
Mistakes to avoid
- Calling a pink heterozygote codominant: a blend is incomplete dominance.
- Dividing recombinants by parental types instead of by the total.
- Labelling the large classes as recombinants.
- Saying an affected father can pass an X-linked trait to his son.
- Assuming the female is always the sex with two identical sex chromosomes (birds are the reverse).
- Explaining traits that always go together with linkage when no recombinants ever appear (think pleiotropy).
Quick self-check
- 45 recombinants among 300 test-cross offspring: map distance? (45 ÷ 300 × 100 = 15 map units)
- Pink × pink (incomplete dominance): proportion pink? (1/2)
- Why are Z-linked recessive traits more common in female birds? (Females are ZW, so one recessive allele on their single Z is expressed.)
Next: practice questions.