Genetics – Dihybrid Crosses, Linked Genes & X-Linked Inheritance
Review of Mendelian Dihybrid Crosses
- Dihybrid = tracking two traits simultaneously instead of one.
- Each parent contributes 4 alleles total (2 per gene).
- Punnett‐square setup identical to single‐gene crosses, just doubled in dimensions (4×4 grid).
- Strategy for solving complex problems (e.g.
Labrador coat‐colour):
- Start with most recessive phenotype among the offspring.
- Because recessive = aa (homozygous), one recessive allele must have come from each parent.
- This instantly reveals half of each parent’s genotype.
- Build upwards: determine all possible parental genotypes that can generate every required puppy colour.
- After hypothesising parental genotypes, prove by a full dihybrid Punnett square and circle the genotypes that yield the demanded phenotypes.
- If three colours are required, ensure all three appear in predicted ratios.
Troubleshooting the Labrador‐Puppy Assignment
- Students report difficulty → instructor suggestion:
1) Start with recessive puppy, 2) remember offspring = half mom + half dad. - Quick tutorial offered; typical resolution time ≈ 2 minutes once approach understood.
Unexpected Post-Mendel Results & Introduction of Morgan Genetics
- Real organisms often deviated from Mendel’s law of independent assortment when genes sat on the same chromosome.
- Thomas Hunt Morgan (1910) used Drosophila melanogaster (fruit fly).
- Advantages:
- >100 eggs per mating.
- Life-cycle 10-15 days; \approx \frac{365}{15} \approx 24 generations / year.
- Cheap, tiny, minimal space/food.
- Clear sexual dimorphism → easy male/female separation.
- If genes truly assorted independently, a dihybrid test-cross would yield 1 : 1 : 1 : 1 phenotypic ratio (25 % each).
- Morgan’s actual ratios ~45 % : 45 % : 5 % : 5 % ⇒ evidence of linked genes.
Concept of Linked Genes & Crossing Over
- Genes close together on the same chromosome tend to be inherited together.
- Crossing‐over in prophase I of meiosis can separate linked genes.
- Probability of separation ∝ physical distance between loci.
- Analogy: magnets — closer magnets hold hands when strands cross; farther ones separate.
Chromosome Mapping
- Goal: create a linear map showing gene order & relative distances.
- Unit = centimorgan (cM) a.k.a. mapping unit (MU).
- 1\,\text{cM} = 1\% recombination frequency.
- Recombination frequency (RF) =
\text{RF} = \frac{\text{recombinant offspring}}{\text{total offspring}} \times 100\% - Steps to map (3+ genes):
- Convert each given % RF directly to MU.
- Locate the two genes with the largest separation → place at opposite ends.
- Position remaining genes between them using the remaining distances.
- Use subtraction to deduce missing intervals.
- Interpretation:
- Smaller distance ⇒ stronger linkage, lower recombinant %, higher chance of co-inheritance.
- Larger distance ⇒ weaker linkage, more recombination.
Simple Example (A, B, C)
- Data: A\leftrightarrow B = 22\%, A\leftrightarrow C = 10\%.
- Largest = 22 → draw A …..22 MU….. B.
- C is 10 MU from A → place C 10 MU to the right of A (12 MU left of B).
- Derived interval: C\leftrightarrow B = 22-10 = 12\,\text{MU}.
- Linkage hierarchy: A–C (10 MU) more tightly linked than C–B (12 MU).
Classroom Table Example (B, L, N, W)
- Table supplies pairwise MUs; sometimes one distance is missing & solved by subtraction.
- After mapping, typical exam questions:
- Draw the map (horizontal or vertical accepted).
- Indicate pair most likely & least likely to undergo recombination.
Non-numeric “eyeball” version
- Diagrams may show gene ticks without numbers; students infer closeness visually.
Sex-Linked (X-Linked) Inheritance — Core Rules
- Course simplification: all sex-linked traits are X-linked, none Y-linked.
- Male genotype: X^{?}Y → hemizygous (only one allele for X genes).
- Males can never be carriers; they either express or lack the trait.
- Female genotype: X^{?}X^{?} → may be homozygous or heterozygous (carrier).
- Transmission logic:
- Fathers pass their X only to daughters (sons get the Y).
- Mothers pass an X to both sons & daughters.
- Consequences:
- Affected father + unaffected mother ⇒
- All sons unaffected (received father’s Y).
- All daughters at least carriers (receive the single affected X).
- Recessive X-linked diseases appear predominantly in males; females require two mutant alleles.
Worked Problem: Duchenne (Pseudo-Hypertrophic) Muscular Dystrophy (MD)
- Statement: “seen only in boys born to apparently normal parents.”
- Diagnosis:
- Must be recessive (parents phenotype normal).
- Trait is X-linked.
- Legend: X^{D} = normal, X^{d} = MD.
- Parental genotypes producing an affected son:
• Father: X^{D}Y (normal).
• Mother: X^{D}X^{d} (carrier). - Punnett check (simplified row/column):
- Sons: X^{D}Y (normal), X^{d}Y (affected).
- Daughters: X^{D}X^{D} (normal), X^{D}X^{d} (carrier).
- Why boys only?
• Boys express disease with one mutant allele.
• Girls need two, which requires affected father and carrier/affected mother — improbable here because affected males die young and seldom reproduce.
Practical Tips & Ethical Notes
- Chromosomal mapping = relative, not absolute: think “Edmonton–Red Deer–Calgary” distances, not coordinate grid.
- Fruit-fly labs: ethical to minimise suffering; CO₂ used for sedation, not killing.
- Real-world relevance: linkage studies underpin modern genetic counseling & locate disease genes.
- Parents may feel guilt in X-linked disorders; important to communicate the involuntary nature of allele transmission.
Quick Reference Equations & Numbers
- Generations per year (fruit fly): 24\;\text{gen} = \frac{365\,\text{days}}{15\,\text{days · gen}^{-1}}.
- Recombination ↔ map units: 1\,\text{cM} = 1\% RF.
- Missing interval: d{AC} = d{AB} - d_{BC} (when A–B largest).
With these structured notes you can:
• Set-up and justify any dihybrid cross.
• Construct & interpret linkage maps.
• Solve recessive & dominant X-linked pedigree problems.
• Explain experimental choices (why fruit flies, why rapid cycles) and ethical considerations.