Chapter 2 — Classical Genetics
Key Concepts Quick Recap
| Term | Definition |
|---|
| Gene | Unit of inheritance; segment of DNA coding for a specific trait |
| Allele | Alternative forms of a gene at the same locus |
| Homozygous | Both alleles identical (AA or aa) |
| Heterozygous | Alleles different (Aa) |
| Dominant | Allele that expresses in heterozygous condition |
| Recessive | Allele that expresses only in homozygous condition |
| Genotype | Genetic constitution (e.g. Aa, TT) |
| Phenotype | Observable trait (e.g. tall, round seeds) |
| Monohybrid cross | Cross involving one pair of contrasting characters |
| Dihybrid cross | Cross involving two pairs of contrasting characters |
| Test cross | Cross of unknown genotype with homozygous recessive |
| Incomplete dominance | F1 intermediate phenotype |
| Codominance | Both alleles express equally in F1 |
| Multiple alleles | More than two alleles for one gene in a population |
| Linkage | Genes on same chromosome tend to be inherited together |
I. Choose the Best Answer (MCQs)
1. The ratio of phenotypes in a monohybrid cross (F2) is: (a) 1:2:1 (b) 3:1 (c) 9:3:3:1 (d) 1:1
Answer(b) 3:1 Monohybrid cross (Aa × Aa): F2 genotype ratio = 1 AA : 2 Aa : 1 aa → phenotype ratio = 3 dominant : 1 recessive.
2. Law of Independent Assortment holds when genes are located on: (a) Same chromosome (b) Sex chromosomes only (c) Different (non-homologous) chromosomes (d) Mitochondrial DNA
Answer(c) Different (non-homologous) chromosomes Law of Independent Assortment states that genes on non-homologous chromosomes segregate independently during meiosis. Genes on the same chromosome show linkage and do not assort independently.
3. In ABO blood group system, blood group AB is an example of: (a) Incomplete dominance (b) Codominance (c) Epistasis (d) Pleiotropy
Answer(b) Codominance In AB blood group, both I^A and I^B alleles are expressed equally — both antigens A and B are present on the RBC surface. Neither allele is dominant over the other.
4. In a dihybrid cross, the F2 phenotypic ratio is: (a) 3:1 (b) 1:2:1 (c) 9:3:3:1 (d) 1:1:1:1
Answer(c) 9:3:3:1 Dihybrid cross (AaBb × AaBb): F2 = 9 A_B_ : 3 A_bb : 3 aaB_ : 1 aabb.
5. The cross between F1 hybrid and homozygous recessive parent is called: (a) Back cross (b) Test cross (c) Reciprocal cross (d) Dihybrid cross
Answer(b) Test cross Test cross (Mendel's backcross): Aa × aa → 1 Aa : 1 aa (1:1 ratio). Used to determine if unknown phenotype is homozygous dominant (AA) or heterozygous (Aa). If all offspring show dominant phenotype → AA; if 1:1 ratio → Aa.
6. Snapdragon (Antirrhinum) shows pink flower in F1 when red × white are crossed. This is an example of: (a) Codominance (b) Epistasis (c) Incomplete dominance (d) Overdominance
Answer(c) Incomplete dominance Neither red (R^R) nor white (R^W) allele is completely dominant. F1 (R^R R^W) = pink (intermediate). F2: 1 red : 2 pink : 1 white (1:2:1 phenotypic ratio).
7. Genes that are present on the same chromosome are called: (a) Allelomorphs (b) Linked genes (c) Epistatic genes (d) Pleiotropic genes
Answer(b) Linked genes Linked genes are on the same chromosome and tend to be inherited together. Linkage reduces recombination frequency. Morgan studied linkage in Drosophila.
II. Short Answer Questions
Q1. State Mendel's Laws of Inheritance.
Answer
Law 1 — Law of Dominance: When two homozygous parents differing in one character are crossed, the F1 offspring shows only the dominant character. The recessive character is hidden (but not lost).
Law 2 — Law of Segregation (Law of Purity of Gametes): During gamete formation, the two alleles of a gene separate (segregate) so each gamete receives only one allele. The two alleles reunite at fertilisation. This is Mendel's most important law — applies to all sexually reproducing organisms.
Law 3 — Law of Independent Assortment: When two pairs of contrasting characters are crossed, the alleles of each pair segregate independently of alleles of the other pair, producing all possible combinations in gametes. Applies only when genes are on different chromosomes.
Q2. A red-flowered plant (RR) is crossed with a white-flowered plant (rr). Show the F1 and F2 generations using a Punnett square.
Answer
Parental cross: RR × rr → F1: all Rr (red, since R is dominant)
F1 × F1: Rr × Rr
F2 genotype ratio: 1 RR : 2 Rr : 1 rr F2 phenotype ratio: 3 Red : 1 White
Q3. Distinguish between incomplete dominance and codominance.
Answer
| Feature | Incomplete Dominance | Codominance |
|---|
| F1 phenotype | Intermediate (neither parent) | Both parental phenotypes expressed simultaneously |
| F2 phenotype ratio | 1:2:1 | 1:2:1 |
| Example | Snapdragon (red × white = pink) | ABO blood group (I^A I^B = AB) |
| Allele interaction | Neither fully dominates | Both express equally |
In incomplete dominance the phenotype is a blend; in codominance both phenotypes coexist (e.g. both A and B antigens on RBCs in AB blood group).
Q4. Explain multiple allelism with the ABO blood group example.
Answer
Multiple allelism occurs when more than two alleles exist for a single gene locus in a population (though any individual carries only two).
ABO blood group: The I gene has three alleles: I^A, I^B, i.
- I^A and I^B are codominant; both are dominant over i.
| Genotype | Blood Group | Antigen on RBC | Antibody in Plasma |
|---|
| I^A I^A or I^A i | A | A | Anti-B |
| I^B I^B or I^B i | B | B | Anti-A |
| I^A I^B | AB | A and B | None |
| ii | O | None | Anti-A and Anti-B |
III. Worked Examples
Example 1 — Dihybrid cross
In pea: Round (R) dominant over wrinkled (r); Yellow (Y) dominant over Green (y).
Cross: RRYY × rryy F1: all RrYy (Round Yellow) F1 × F1: RrYy × RrYy
Gametes of each: RY, Ry, rY, ry (4 types)
F2 phenotypic ratio: 9 Round Yellow : 3 Round Green : 3 Wrinkled Yellow : 1 Wrinkled Green
9 R_Y_ : 3 R_yy : 3 rrY_ : 1 rryy
Example 2 — Test cross
A tall plant (genotype unknown) × dwarf plant (tt):
- If all offspring are tall → parent was TT
- If offspring ratio is 1 tall : 1 dwarf → parent was Tt
TT × tt → all Tt (tall) → parent = TT
Tt × tt → 1 Tt : 1 tt → parent = Tt
Common Mistakes
- Forgetting that Law of Segregation applies to all organisms; Law of Independent Assortment only to genes on different chromosomes.
- Confusing 3:1 (monohybrid F2 phenotype) with 1:2:1 (monohybrid F2 genotype or incomplete dominance phenotype).
- Writing test cross as F1 × F1 — it is F1 × homozygous recessive.
- Saying codominance = incomplete dominance — they are different.
Exam Checklist
- Mendel's three laws stated correctly with one example each.
- Monohybrid F2: phenotype 3:1, genotype 1:2:1.
- Dihybrid F2: 9:3:3:1.
- Test cross = unknown × homozygous recessive → 1:1 ratio.
- ABO blood group table: genotype + phenotype + antigen + antibody.
