Section 15 — Board Exam-Pattern Questions (CBSE Class 12 Biology · Chapter 4)
Chapter 4 (Principles of Inheritance and Variation) is one of the highest-yield chapters in the CBSE Class 12 Biology Board exam. Its parent unit (Genetics and Evolution) carries about 20 marks, and a substantial share typically comes from this chapter — spread across MCQs, 2/3-mark short answers, and frequently a 5-mark question on Mendelian crosses, sex-linked inheritance, or genetic disorders.
This section curates 24 high-yield Board questions in CBSE answer-writing format and mark-weighting:
| Marks | Number of Qs | Typical paper allocation |
|---|---|---|
| 1 mark | 7 questions (Q1–Q7) | 2–3 / paper |
| 2 marks | 6 questions (Q8–Q13) | 2 / paper |
| 3 marks | 6 questions (Q14–Q19) | 1–2 / paper |
| 5 marks | 5 questions (Q20–Q24) | 1 / paper (very high probability) |
| Total | 24 questions | ~24 marks worth |
How to use: Cover the answer, attempt yourself, then check your response against the model answer.
No quiz at the end: Pair with Sections 14 (Solved Examples) and 16 (NEET-Pattern Practice Questions).
Mini Memory Capsule — What CBSE Always Asks
Most CBSE questions on Chapter 4 fall into 5 recurring themes:
1. Mendel's laws (2- or 3-mark) — state and explain Law of Dominance, Segregation, Independent Assortment. Marks for clear definitions + examples.
2. Punnett-square problems (3- or 5-mark) — monohybrid (TT×tt), dihybrid (RRYY×rryy), test cross. Often includes incomplete dominance or codominance.
3. ABO blood groups (3- or 5-mark) — multiple alleles + codominance. Cross-based problems (e.g., AB × O parents).
4. Sex-linked inheritance (3- or 5-mark) — haemophilia / colour blindness pedigree. Often asks "why more affected in males?"
5. Genetic disorders (3- or 5-mark) — sickle-cell, thalassemia, PKU, Down/Turner/Klinefelter. Cause + mode of inheritance + symptoms.
1-Mark Questions (Q1–Q7)
Q1. [1 mark] Name the FIRST law of Mendel.
Answer: Law of Dominance.
Q2. [1 mark] What is a test cross?
Answer: Crossing an individual showing the dominant phenotype (unknown genotype) with a homozygous recessive (tt) to find out whether it is TT or Tt.
Q3. [1 mark] Define pleiotropy.
Answer: A single gene affecting several phenotypic traits at once — e.g. sickle-cell anaemia.
Q4. [1 mark · Assertion-Reason]
Assertion (A): Haemophilia is more common in males than females. Reason (R): Haemophilia is an X-linked recessive disorder.
Options: (a) Both A and R true and R explains A. (b) Both true but R doesn't explain A. (c) A true, R false. (d) A false, R true.
Answer: (a) — both are true and R correctly explains A. A male has only one X, so a single defective allele makes him affected.
Q5. [1 mark] What is the genotype of a person with blood group AB?
Answer: I^A I^B — both alleles are expressed (codominance), so both A and B antigens appear on the RBCs.
Q6. [1 mark] Name the chromosomal abnormality responsible for Down syndrome.
Answer: Trisomy 21 — an extra copy of chromosome 21 (47 chromosomes in all).
Q7. [1 mark] Why are Mendelian disorders typically rare?
Answer: They arise from single-gene mutations. Most familiar ones (sickle-cell anaemia, thalassemia, PKU, cystic fibrosis) are recessive, so the affected condition needs the defective allele from both parents — which is uncommon. (Dominant Mendelian disorders, e.g. myotonic dystrophy, also exist but the mutant alleles themselves stay rare.)
2-Mark Questions (Q8–Q13)
Q8. [2 marks] Differentiate between incomplete dominance and codominance with one example each.
Answer:
| Feature | Incomplete Dominance | Codominance |
|---|---|---|
| Heterozygote | Intermediate / blended phenotype | Both alleles expressed together |
| Example | Snapdragon: Rr is pink | ABO: I^A I^B gives AB (both antigens) |
Q9. [2 marks] A couple has two daughters. Will the third child necessarily be a son? Justify scientifically.
Answer: No. Each child's sex is decided independently by whether an X- or Y-bearing sperm fertilises the egg. The father contributes X or Y with equal (50%) probability, and earlier children have no effect. So the third child is again 50% likely to be a son.
Q10. [2 marks] Mention any TWO Mendelian disorders and indicate their mode of inheritance.
Answer: Any two, e.g. sickle-cell anaemia (autosomal recessive) and haemophilia (X-linked recessive). Thalassemia and PKU are also autosomal recessive; colour blindness is X-linked recessive.
Q11. [2 marks] Why is haemophilia more common in MEN than in WOMEN?
Answer: Haemophilia is X-linked recessive. A man (XY) has only one X, so a single defective allele (X^h Y) makes him affected — the Y carries no clotting-factor gene to compensate. A woman (XX) must inherit the defective allele on both X's (X^h X^h), which is far rarer.
Q12. [2 marks] What is non-disjunction? What disorders does it cause?
Answer: Non-disjunction is the failure of homologous chromosomes (Meiosis I) or sister chromatids (Meiosis II) to separate, giving gametes with an extra (n+1) or missing (n−1) chromosome. It causes Down syndrome (trisomy 21), Klinefelter (47, XXY) and Turner (45, X0) syndromes.
Q13. [2 marks] A father with blood group AB marries a woman with blood group O. State the possible blood groups of their children.
Answer: Father AB is I^A I^B; mother O is ii. The mother gives only i, so children are I^A i (A) or I^B i (B) — group A or B only, never AB or O.
3-Mark Questions (Q14–Q19)
Q14. [3 marks] Explain the Law of Independent Assortment with a suitable example and Punnett square.
Answer:
Statement: Alleles of different genes assort independently of one another during gamete formation (true for genes on different chromosomes, which separate independently in Meiosis I).
Example — dihybrid cross: RRYY (round-yellow) × rryy (wrinkled-green) gives F1 RrYy, all round-yellow. F1 makes four gamete types (RY, Ry, rY, ry) in equal amounts, and F1 × F1 gives an F2 ratio of 9 round-yellow : 3 round-green : 3 wrinkled-yellow : 1 wrinkled-green (9:3:3:1).
Q15. [3 marks] Explain the chromosomal theory of inheritance. Who proposed it?
Answer: The theory states that genes (Mendel's factors) lie on chromosomes, which are the physical carriers of heredity passed from parents to offspring. It was proposed independently by Walter Sutton and Theodor Boveri (1902–1903).
Sutton noted that factors and chromosomes behave in parallel: both occur in pairs, both separate during gamete formation, both assort independently, and both reunite at fertilisation. Thomas Hunt Morgan later confirmed the theory through his Drosophila work on sex-linkage and linkage.
Q16. [3 marks] Describe sickle-cell anaemia with respect to its (i) cause, (ii) mode of inheritance, (iii) symptoms.
Answer:
(i) Cause: A point mutation in the beta-globin gene at codon 6 (GAG → GTG), changing glutamic acid to valine (Glu → Val).
(ii) Inheritance: Autosomal recessive — HbS HbS is affected, HbA HbS is a carrier (usually symptom-free), HbA HbA is normal.
(iii) Symptoms: Anaemia and fatigue, painful vaso-occlusive crises, organ damage (spleen, kidney, lungs), jaundice and a higher risk of infection.
Q17. [3 marks] Define linkage and recombination. How are they related to genetic mapping?
Answer:
Linkage: the tendency of genes lying close together on the same chromosome to be inherited together, departing from independent assortment.
Recombination: the formation of new (non-parental) allele combinations through crossing over in Prophase I.
Mapping: crossing over partly breaks linkage. Recombination frequency = (recombinants / total) × 100, and it is proportional to the distance between genes (1 cM = 1% recombination). Sturtevant used these frequencies to build the first linkage map of the Drosophila X chromosome.
Q18. [3 marks] Describe the inheritance of ABO blood groups in humans, mentioning all 6 genotypes and 4 phenotypes.
Answer: ABO has three alleles (multiple alleles): I^A, I^B and i. I^A and I^B are codominant with each other and both are dominant over i.
| Genotype | Phenotype |
|---|---|
| I^A I^A, I^A i | A |
| I^B I^B, I^B i | B |
| I^A I^B | AB (both antigens — codominance) |
| ii | O |
O is the universal donor (no A/B antigens); AB is the universal recipient (no anti-A/anti-B antibodies).
Q19. [3 marks] Differentiate between Klinefelter syndrome and Turner syndrome.
Answer:
| Feature | Klinefelter | Turner |
|---|---|---|
| Karyotype | 47, XXY | 45, X0 |
| Sex | Male | Female |
| Cause | Extra X | Missing X |
| Features | Tall, gynecomastia, small sterile testes | Short stature, sterile, no menstruation |
5-Mark Questions (Q20–Q24)
Q20. [5 marks] A homozygous tall pea plant (TT) with red flowers (RR) is crossed with a homozygous dwarf plant (tt) with white flowers (rr). (T = tall, t = dwarf; R = red, r = white; both T and R are dominant.) What are:
(a) The F1 genotype and phenotype? (b) The F2 phenotypic ratio? (c) The expected number of "tall-white" individuals out of 320 F2? (d) Show the F2 Punnett square.
Answer:
(a) F1 is all TtRr — tall, red (both dominant).
(b) F1 × F1 gives four gamete types (TR, Tr, tR, tr) and an F2 ratio of 9 tall-red : 3 tall-white : 3 dwarf-red : 1 dwarf-white = 9:3:3:1.
(c) Tall-white = 3/16 × 320 = 60 individuals.
(d) F2 Punnett square:
| TR | Tr | tR | tr | |
|---|---|---|---|---|
| TR | TTRR | TTRr | TtRR | TtRr |
| Tr | TTRr | TTrr | TtRr | Ttrr |
| tR | TtRR | TtRr | ttRR | ttRr |
| tr | TtRr | Ttrr | ttRr | ttrr |
Counts: 9 tall-red, 3 tall-white, 3 dwarf-red, 1 dwarf-white.
Q21. [5 marks] Describe the principles of pedigree analysis. Discuss the criteria for identifying (a) autosomal dominant, (b) autosomal recessive, and (c) X-linked recessive inheritance.
Answer:
A pedigree charts a trait across generations of a family. Standard symbols: square = male, circle = female, filled = affected, half-filled = carrier, horizontal line = mating, vertical line = offspring; it is used to read the inheritance pattern and for genetic counselling.
(a) Autosomal dominant: appears in every generation (no skipping); about half the children of an affected parent are affected; both sexes equally affected; father-to-son transmission possible. (e.g. Huntington's disease.)
(b) Autosomal recessive: can skip generations; affected children may have unaffected carrier parents; both sexes equally affected; commoner in consanguineous families. (e.g. sickle-cell anaemia, PKU.)
(c) X-linked recessive: males affected far more often; passes from carrier mothers to affected sons; father-to-son transmission is impossible (sons get the Y); daughters of affected fathers are carriers. (e.g. haemophilia, colour blindness.)
Q22. [5 marks] Explain the chromosomal theory of inheritance and Morgan's experimental verification using Drosophila. Why is Drosophila considered an ideal organism for genetic studies?
Answer:
Theory: proposed independently by Walter Sutton (1902) and Theodor Boveri (1903) — genes lie on chromosomes, the physical carriers of heredity. Factors and chromosomes behave alike: both occur in pairs, separate during gamete formation, reunite at fertilisation, and assort independently.
Morgan's verification (Drosophila, 1910s):
- Sex-linkage: a white-eyed male crossed to red-eyed females gave white eyes only in males in F2 — the gene is on the X.
- Linkage: some genes were inherited together rather than independently — they lie on the same chromosome.
- Recombination/mapping: crossing over breaks linkage, and recombination frequency reflects gene distance; Sturtevant built the first gene maps.
Why Drosophila is ideal: short life cycle (~2 weeks), many offspring, easy lab culture, small genome (2n = 8), many visible mutations, and easily distinguished sexes. Morgan won the Nobel Prize in 1933.
Q23. [5 marks] Discuss the inheritance of ABO blood groups. Explain why a child of blood group O cannot have parents both with blood group AB.
Answer:
The ABO gene (chromosome 9) has three alleles — I^A, I^B and i. I^A and I^B are codominant, and both are dominant over i.
| Genotype | Phenotype | Antigens | Antibodies |
|---|---|---|---|
| I^A I^A, I^A i | A | A | anti-B |
| I^B I^B, I^B i | B | B | anti-A |
| I^A I^B | AB | A + B | none |
| ii | O | none | anti-A and anti-B |
Why an O child can't come from two AB parents: an O child must be ii, i.e. it needs an i allele from each parent. But AB parents are I^A I^B and carry no i allele. So AB × AB gives:
| I^A | I^B | |
|---|---|---|
| I^A | I^A I^A (A) | I^A I^B (AB) |
| I^B | I^A I^B (AB) | I^B I^B (B) |
The offspring are 1 A : 2 AB : 1 B — never O. This is why an O child rules out two AB parents in paternity testing.
Q24. [5 marks] Describe mutations and chromosomal aberrations. Mention various types with one example each. Why are mutations important?
Answer:
Mutations are heritable changes in the DNA sequence.
Point mutations:
- Substitution — one base replaced: silent (same amino acid), missense (different amino acid, e.g. sickle-cell, GAG → GTG, Glu → Val at codon 6), or nonsense (a premature stop codon).
- Insertion/deletion (indel) — adds or removes bases and shifts the reading frame (frameshift), usually destroying protein function.
Chromosomal aberrations:
- Numerical — aneuploidy: trisomy 21 (Down), 47,XXY (Klinefelter), 45,X0 (Turner).
- Numerical — polyploidy: extra full sets, common in plants (e.g. triploid banana).
- Structural: deletion (e.g. Cri-du-chat), duplication, inversion, translocation (e.g. chronic myeloid leukaemia, chr 9–22).
Causes: replication errors and mutagens — physical (UV, X-rays), chemical (EMS, nitrous acid), and biological (transposons).
Importance: mutations are the source of genetic variation and the raw material for natural selection and evolution. Most are harmful or neutral, but rare beneficial ones drive adaptation and the diversity of life.
End of Section 15
You have now worked through 24 high-yield CBSE Board questions spanning Mendel's laws, deviations, sex-linked inheritance, mutations, and genetic disorders. If you can answer the five 5-markers (Q20–Q24) from memory with their Punnett squares, you have effectively secured the chapter's contribution to your Board paper.
Final tip: On Board day, draw Punnett squares wherever the cross is asked. Examiners give marks for visible structure. Also tabulate comparison-type questions (Klinefelter vs Turner, incomplete vs codominance, etc.).