Genetics – B.Pharm Admission Study Notes
(IIT‑JEE / Competitive Exam Focus)
Why Genetics matters for Pharmacy?
• Understanding drug–gene interactions (pharmacogenetics).
• Designing targeted therapies (gene therapy, monoclonal antibodies).
• Interpreting clinical trials that involve genetic markers.
1. Classical (Mendelian) Genetics
| Law | Statement | Key Points | Example |
|---|---|---|---|
| Law of Segregation | Each individual carries two alleles; they segregate during gamete formation. | Alleles are independent. | Flower colour in pea plants: R (red) & r (white). |
| Law of Independent Assortment | Genes for different traits assort independently (except for linkage). | Applies to genes on different chromosomes or far apart on the same chromosome. | Plant height (T/t) & flower colour (R/r). |
| Law of Dominance | One allele can mask the expression of another. | Dominant allele expressed unless homozygous recessive. | R > r. |
| Law of Recessive Inheritance | Recessive phenotype appears only in homozygous state. | Requires two recessive alleles. | rr → white flowers. |
1.1 Monohybrid Cross
| Genotype | Gametes | Offspring Probabilities |
|---|---|---|
Aa |
A, a |
1/4 AA, 1/2 Aa, 1/4 aa |
| Phenotypic ratio: 3:1 (dominant : recessive) |
1.2 Dihybrid Cross
- Traits:
A/aandB/b - Gametes:
AB,Ab,aB,ab - Punnett square: 16 cells
- Phenotypic ratio: 9:3:3:1
- 9 →
ABphenotype - 3 →
AboraBphenotypes (single dominance) - 1 →
abphenotype
- 9 →
1.3 Linkage & Recombination
| Concept | Key Idea | Formula |
|---|---|---|
| Linkage | Genes on the same chromosome are inherited together. | |
| Recombination frequency (r) | r = (crossovers / total meioses) × 100 |
Recombination frequency is a measure of map distance (in centiMorgans, cM). 1 cM ≈ 1% recombination. |
| Map distance | Distance = (recombination frequency / 100) × 100 cM |
Example: 5% → 5 cM |
| Interference | I = 1 - (observed double crossovers / expected) |
I ranges 0 (no interference) to 1 (complete interference). |
Tricky Point
- Independent Assortment vs Linkage: If two loci are linked, the 9:3:3:1 ratio collapses to a 10:6:4 (or 9:3:3:1 + recombination fraction).
- Crossing‑over can create new combinations, but only if the two loci are separated by at least 1–2 cM.
1.4 Sex‑Linked Traits
| Gene | Chromosome | Inheritance Pattern |
|---|---|---|
| X‑linked recessive | X | Male hemizygous (X^rY) shows trait; female heterozygous (X^rX) is carrier. |
| X‑linked dominant | X | Both sexes express, but carriers (males) transmit to all daughters, none to sons. |
Example: Red‑eye in rabbits
X^rY(red eye) → male;X^rX^w→ female carrier.- Offspring phenotypes follow classic sex‑linked ratios.
1.5 Multiple Alleles, Codominance & Incomplete Dominance
| Term | Definition | Example |
|---|---|---|
| Multiple alleles | More than two alleles at a locus. | Human blood group: IA, IB, i. |
| Codominance | Both alleles expressed simultaneously. | IAIB → AB blood type. |
| Incomplete dominance | Heterozygote phenotype is intermediate. | Rr → pink flowers. |
1.6 Epistasis
| Type | Interaction | Phenotypic Ratio (typical) |
|---|---|---|
| Complementary | Two genes required for phenotype. | 9:7 |
| Duplicate | Either gene can produce phenotype. | 15:1 |
| Dominant | One gene masks the other. | 12:3:1 |
| Recessive | One gene’s recessive allele masks the other. | 9:3:3:1 but with rr suppressed |
Derivation (Complementary)
- Genes A & B:
A(AA/Aa) &B(BB/Bb) are required for normal phenotype. - Offspring genotype (assuming heterozygous parents
AaBb×AaBb) gives 9/16 normal, 7/16 mutant.
2. Modern Genetics
2.1 DNA & Gene Structure
| Feature | Description |
|---|---|
| Base pairing | A↔T, G↔C |
| Chargaff’s rule | A = T, `G = C |