BIOLOGY WITH HAMZA · MDCAT REVISION
Inheritance
Name: ____________________ Date: ______________
Recall first. Check second. Return to the ideas you missed.
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Learning targets
- Connect segregation and independent assortment with meiosis.
- Construct mono- and dihybrid crosses systematically.
- Recognise when linkage, codominance or sex linkage changes simple expectations.
Core notes
- Segregation reflects separation of homologous chromosomes.
- Genotype and phenotype ratios answer different questions.
- Dihybrid probabilities can be multiplied only when events are independent.
- Dominance pattern, linkage and sex linkage can change expected ratios.
Read, cover and redraw
Identify alleles carried by each parent.
Separate alleles into valid gametes.
Combine one gamete from each parent.
Count genotype or phenotype outcomes.
Written recall
1. Why does Aa × Aa produce a 1:2:1 genotype ratio but a 3:1 phenotype ratio under complete dominance?
2. What evidence in offspring suggests that two genes are linked?
3. What is P(Aabb)?
4. Why not 9:3:3:1?
5. Must exactly ten of eighty be aabb?
Apply the idea
1. In AaBb × AaBb, the two loci assort independently and show complete dominance. What is the probability of an offspring recessive for both traits?
- 1/16
- 1/8
- 1/4
- 3/16
My answer: _____ Reason: ____________________________________
2. Assuming complete dominance, which parent genotype best explains the approximately equal offspring classes?
| Offspring phenotype | Number |
|---|---|
| A–B– | 52 |
| A–bb | 48 |
| aaB– | 49 |
| aabb | 51 |
- AABB
- AaBb
- AABb
- AaBB
My answer: _____ Reason: ____________________________________
3. A carrier mother and an unaffected father have children. For a recessive X-linked condition with full penetrance, what fraction of their sons is expected to be affected?
- None
- One quarter
- One half
- All
My answer: _____ Reason: ____________________________________
Answer key
Cover this section while attempting the questions.
Written recall
1. AA and Aa are different genotypes but share the dominant phenotype, so their probabilities combine into the phenotypic class.
2. Parental allele combinations occur more often than recombinant combinations, deviating from independent-assortment expectations.
3. P(Aa) × P(bb) = ½ × ½ = ¼.
4. One parent is bb, so the B-locus cross is not Bb × Bb.
5. No. Random samples vary around their expected counts.
Applied questions
1. A — For each Aa × Aa cross, the probability of the recessive genotype is 1/4. Independence allows multiplication: 1/4 × 1/4 = 1/16.
2. B — With independent assortment, AaBb makes four gamete types in equal proportions. Crossing it with aabb produces approximately a 1:1:1:1 phenotypic ratio.
3. C — Each son receives Y from his father and one of his mother’s X chromosomes. Half of her X chromosomes carry the allele, so half the sons are expected to be affected.