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NCERT Exercises and Solutions: Principles of Inheritance and Variation

🎓 Class 12 Biology CBSE Theory Ch 4 – Principles of Inheritance and Variation ⏱ ~8 min
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આ MCQ મોડ્યુલ આના પર આધારિત છે: NCERT Exercises and Solutions: Principles of Inheritance and Variation

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NCERT Exercises and Solutions: Principles of Inheritance and Variation

5.13 Chapter Summary — Key Concepts at a Glance

Mendel's Three Laws

Dominance, Segregation, Independent Assortment — discovered through pea experiments (1856–63).

Monohybrid Cross

F2 phenotype = 3:1, genotype = 1:2:1; test cross gives 1:1.

Dihybrid Cross

F2 phenotype = 9:3:3:1; gametes RY, Ry, rY, ry equal probability.

Incomplete Dominance

Heterozygote = intermediate phenotype (snapdragon pink); 1:2:1 ratio.

Codominance

Both alleles fully expressed; ABO blood AB phenotype.

Multiple Alleles

≥3 alleles in population; ABO has IA, IB, i.

Pleiotropy

One gene → multiple effects (PKU, sickle cell).

Chromosomal Theory

Sutton & Boveri (1902) — genes on chromosomes.

Linkage & Recombination

Morgan's Drosophila work; map distance in cM.

Sex Determination

XY (humans), XO (grasshopper), ZW (birds).

Mendelian Disorders

Sickle cell, thalassemia, haemophilia, PKU.

Chromosomal Disorders

Down (+21), Klinefelter (XXY), Turner (XO).

5.14 Key Terms

TermMeaning
AlleleAlternative form of a gene at a given locus
GenotypeGenetic makeup (e.g., TT, Tt, tt)
PhenotypeObservable trait (e.g., tall, dwarf)
HomozygousTwo identical alleles (TT or tt)
HeterozygousTwo different alleles (Tt)
DominantAllele expressed in heterozygote
RecessiveAllele masked in heterozygote
Test crossCross with homozygous recessive
LinkageTendency of nearby genes to be inherited together
RecombinationNew allele combinations from crossing-over
PleiotropyOne gene → many effects
TrisomyExtra copy of one chromosome (e.g., +21)
MonosomyMissing copy of one chromosome (e.g., XO)
Non-disjunctionFailure of chromosomes to separate in meiosis

🧬 Interactive: Cross Type Identifier

Enter the F2 ratio you observed and identify the inheritance pattern:

5.15 NCERT Exercises — With Worked Solutions

Exercise 1 (NCERT)

Mention the advantages of selecting pea plant for experiment by Mendel.

Advantages of pea plant for Mendel's experiments:
  1. Several pure-breeding varieties with sharply contrasting traits.
  2. Self-pollinating — pure lines can be maintained naturally.
  3. Cross-pollination is possible by hand emasculation — controlled crosses are easy.
  4. Short life cycle — multiple generations in a single year.
  5. Many seeds per pod — large samples for statistical analysis.
  6. Each character has only two distinct, contrasting forms (no intermediates).
  7. Pea plants are hardy and easy to grow in temperate climates.

Exercise 2 (NCERT)

Differentiate between (a) Dominance and Recessiveness (b) Homozygous and Heterozygous (c) Monohybrid and Dihybrid.

(a) Dominance vs Recessiveness:
  • Dominant allele expresses its phenotype in heterozygous condition (Tt looks like TT).
  • Recessive allele is masked in heterozygote, expressed only when homozygous (tt).
(b) Homozygous vs Heterozygous:
  • Homozygous — two identical alleles (TT or tt). Pure-breeding.
  • Heterozygous — two different alleles (Tt). Hybrid, not pure-breeding.
(c) Monohybrid vs Dihybrid:
  • Monohybrid — cross involving ONE pair of contrasting traits. F2 phenotype 3:1.
  • Dihybrid — cross involving TWO pairs of contrasting traits. F2 phenotype 9:3:3:1.

Exercise 3 (NCERT)

Define and design a test-cross.

Definition: A test cross is a cross between an individual showing the dominant phenotype (genotype unknown — could be homozygous TT or heterozygous Tt) and an individual that is homozygous recessive (tt). The phenotypic ratio of offspring reveals the unknown genotype.

Design: Suppose we want to know if a tall pea is TT or Tt.
Cross: Tall (TT or Tt) × Dwarf (tt)
Outcomes:
  • If parent was TT → all offspring Tt (100% tall).
  • If parent was Tt → offspring 1 Tt (tall) : 1 tt (dwarf) = 1:1 ratio.
The 1:1 ratio reveals the heterozygous parent.

Exercise 4 (NCERT)

Using a Punnett square, work out the distribution of phenotypic features in the F1 generation after a cross between a homozygous female and a heterozygous male for a single locus.

Let the locus have alleles A (dominant) and a (recessive). Two interpretations are common:

Case 1: Homozygous dominant female × Heterozygous male (AA × Aa)
Female gametes: A. Male gametes: A, a.
F1: ½ AA + ½ Aa → 100% dominant phenotype; genotypes 1:1 (AA : Aa).

Case 2: Homozygous recessive female × Heterozygous male (aa × Aa)
Female gametes: a. Male gametes: A, a.
F1: ½ Aa (dominant) + ½ aa (recessive) → 1 dominant : 1 recessive.

The phenotypic distribution depends on which homozygous form (AA or aa) the female has. Always interpret the question with both cases.

Exercise 5 (NCERT)

Briefly mention the contribution of T.H. Morgan in genetics.

T.H. Morgan's contributions:
  1. Used Drosophila melanogaster (fruit fly) to test Mendel's laws — short life cycle, easy lab breeding, only 4 chromosome pairs.
  2. Discovered linkage — when two genes are on the same chromosome, they are inherited together more often than expected by independent assortment.
  3. Discovered recombination via crossing-over between linked genes.
  4. Showed white-eye gene in Drosophila is X-linked — proving genes are physically located on chromosomes (confirming Sutton & Boveri's chromosomal theory).
  5. Set the stage for genetic mapping — his student Sturtevant constructed the first chromosome map in 1913.
  6. Won the Nobel Prize in Physiology or Medicine in 1933 for these discoveries.

Exercise 6 (NCERT)

What is pedigree analysis? Suggest how such an analysis can be useful.

Definition: Pedigree analysis is the study of inheritance of a trait or genetic disorder in several generations of a family, displayed as a tree-like diagram (pedigree chart). Squares represent males, circles represent females; affected individuals are filled.

Uses:
  • Trace inheritance pattern — autosomal dominant, autosomal recessive, X-linked, etc.
  • Identify carriers of recessive disorders.
  • Estimate risk for future children.
  • Useful for genetic counselling of couples planning pregnancy.
  • Detect novel mutations.
  • Critical when controlled crosses (like Mendel did) are unethical in humans.

Exercise 7 (NCERT)

How is sex determined in humans?

Mechanism — XY system:
  • Females have XX — homogametic (eggs all carry X).
  • Males have XY — heterogametic (sperms carry either X or Y, in 50:50 ratio).
Outcome:
  • X (egg) + X (sperm) → XX → female.
  • X (egg) + Y (sperm) → XY → male.
Key point: The sex of the offspring is determined by the FATHER's sperm — specifically whether it carries an X or Y chromosome. The Y chromosome contains the SRY gene that triggers male development.

Exercise 8 (NCERT)

A child has blood group O. If the father has blood group A and mother blood group B, work out the genotypes of the parents and the possible genotypes of the other offspring.

Step 1: Child has blood group O → genotype ii. Each parent must have contributed an i allele.
Step 2: Father is A → must be IAi (heterozygous, since he passed an i to the O child).
Step 3: Mother is B → must be IBi (heterozygous, similarly).

Cross: IAi × IBi
Possible offspring (Punnett):
GenotypePhenotypeProbability
IAIBAB¼
IAiA¼
IBiB¼
iiO¼
All four blood groups are equally possible from this cross — a famous example of multiple-allele inheritance with codominance.

Exercise 9 (NCERT)

Explain the following terms with example: (a) Co-dominance (b) Incomplete dominance.

(a) Co-dominance: Both alleles in a heterozygote are fully and equally expressed — both phenotypes appear distinctly together.
Example: Human ABO blood group. A heterozygous person with genotype IAIB has BOTH antigen A AND antigen B on the surface of red blood cells — blood group AB. Neither A nor B masks the other.

(b) Incomplete dominance: Neither allele is fully dominant; the heterozygote shows an intermediate phenotype between the two homozygous parents.
Example: Snapdragon (Antirrhinum) flower colour. Cross of Red (RR) × White (rr) gives Pink (Rr) F1. F2 shows 1 Red : 2 Pink : 1 White — phenotype ratio = genotype ratio.

Key difference: In incomplete dominance, the heterozygote phenotype is INTERMEDIATE; in codominance, BOTH parental phenotypes appear together (no blending).

Exercise 10 (NCERT)

What is point mutation? Give one example.

Definition: A point mutation is a change in a single base pair (nucleotide) of DNA. It may be a substitution (one base replaced by another), insertion, or deletion of one base.

Example: Sickle cell anaemia
  • The β-globin gene undergoes a substitution mutation: GAG → GTG at codon 6.
  • This changes the amino acid from glutamic acid → valine.
  • The mutated haemoglobin (HbS) polymerises under low oxygen, causing red blood cells to take a sickle shape.
  • Affected (HbS HbS) individuals suffer severe anaemia, organ damage, and reduced life span.
A single base change in a gene of ~3 billion nucleotides — the smallest possible mutation — but with profound consequences.

Exercise 11 (NCERT)

Who had proposed the chromosomal theory of inheritance?

The chromosomal theory of inheritance was proposed independently in 1902 by:
  • Walter Sutton (American geneticist)
  • Theodor Boveri (German biologist)
They observed that the behaviour of chromosomes during meiosis closely paralleled the behaviour of Mendel's "factors" (genes) — pairing, segregation, and independent assortment. Thomas Hunt Morgan later experimentally confirmed the theory through Drosophila experiments demonstrating linkage and X-linked inheritance.

Exercise 12 (NCERT)

Mention any two autosomal genetic disorders with their symptoms.

1. Sickle Cell Anaemia (autosomal recessive):
  • Cause: Glu→Val substitution at position 6 of β-globin chain.
  • Symptoms: Chronic anaemia, fatigue, episodes of severe pain (sickling crises), organ damage, jaundice, swelling of hands/feet, reduced life span.
2. Thalassemia (autosomal recessive):
  • Cause: Mutations reducing α- or β-globin synthesis.
  • Symptoms: Severe anaemia from infancy, failure to thrive, enlarged liver and spleen, bone deformities, requires regular blood transfusions.
3. Phenylketonuria (PKU) — autosomal recessive (alternative):
  • Cause: Defective enzyme converting phenylalanine to tyrosine.
  • Symptoms: Mental retardation, seizures, light skin/hair, characteristic odour. Treatable with phenylalanine-restricted diet.
📐 Activity 5.5 — Final Synthesis

Setup: Two heterozygous tall pea plants with violet flowers (TtVv) are crossed. Both genes assort independently.

Predict: (a) What gametes does each parent produce? (b) Phenotypic ratio of F2? (c) What fraction of F2 are tall AND have violet flowers?

(a) Each TtVv parent produces 4 gamete types in equal frequency: TV, Tv, tV, tv (1/4 each).

(b) Phenotypic ratio of F2 = 9 Tall-Violet : 3 Tall-white : 3 dwarf-Violet : 1 dwarf-white (the classic 9:3:3:1).

(c) Tall AND violet = 9/16 ≈ 56.25%. Out of 1600 plants, expect 900 tall-violet.

Reasoning: P(tall) × P(violet) = 3/4 × 3/4 = 9/16 — multiplied because the two genes are independent.

🎯 Competency-Based Questions — Chapter Review

Q1. The principle of "purity of gametes" is also called:L1 Remember

  • (a) Law of Dominance
  • (b) Law of Segregation
  • (c) Law of Independent Assortment
  • (d) Law of Pleiotropy
Answer: (b). Each gamete carries only one allele of a gene — they are "pure" — never contaminated. This is Mendel's Law of Segregation.

Q2. True/False: Sickle cell anaemia is caused by chromosomal non-disjunction. L2 Understand

FALSE. Sickle cell anaemia is caused by a single point mutation (substitution) in the β-globin gene — chromosome number is normal (46). Non-disjunction causes chromosomal disorders like Down syndrome (47) or Turner syndrome (45).

Q3. A man with blood group AB marries a woman with blood group O. What blood groups are NOT possible in their children? L3 Apply

Father: IAIB; Mother: ii.
Father's gametes: IA or IB. Mother's gametes: only i.
Possible offspring: IAi (A) or IBi (B).
NOT possible: AB and O. AB requires both IA and IB (mother has neither); O requires ii (father has no i).

Q4. Analyse: Why are Mendel's laws sometimes called "laws of inheritance" rather than just "rules"? L4 Analyse

Reason:
  • Mendel's findings were universal — they apply to all sexually reproducing diploid organisms (plants, animals, fungi, humans).
  • They make predictive statements that can be tested by experiment with reproducible ratios.
  • They have a physical basis in chromosome behaviour (later confirmed by Sutton, Boveri, Morgan).
  • The Law of Segregation is exception-free.
The Law of Independent Assortment, however, has known exceptions (linkage), so some textbooks now call it a "tendency" rather than absolute law.

Q5. HOT (Create): A 35-year-old couple is concerned about Down syndrome risk in their next child. Design a counselling conversation including risk assessment, prenatal options, and ethical considerations. L6 Create

Counselling structure:
  1. Risk explanation: Risk of Down syndrome rises with maternal age — at 30 it's ~1/940, at 35 ~1/365, at 40 ~1/100. The cause is age-related non-disjunction in maternal meiosis.
  2. Screening options:
    • First trimester: blood test (β-hCG, PAPP-A) + ultrasound (nuchal translucency).
    • Second trimester: triple/quad screen.
    • Non-invasive prenatal testing (NIPT) — cell-free fetal DNA from maternal blood (>99% accuracy).
  3. Diagnostic options:
    • Chorionic villus sampling (CVS) at 10–13 weeks.
    • Amniocentesis at 15–18 weeks.
    • Both give definitive karyotype but carry small miscarriage risk (~0.5%).
  4. Ethical considerations: Respect couple's values; discuss continuation vs termination if positive; emphasise that Down syndrome individuals have rich and meaningful lives; explore support resources.
  5. Decision support: Provide unbiased information, no judgement, ample time to discuss.
Genetic counselling is a profession — counsellors are trained to provide non-directive support to families facing such decisions.

🧠 Assertion–Reason Questions — Chapter Review

A: The 9:3:3:1 ratio in dihybrid cross is a manifestation of independent assortment.

R: The two genes are located on different chromosomes.

Answer: (A). Both true; R explains A. When genes are on different chromosomes (or far apart on the same chromosome), they assort independently in meiosis, leading to the classic 9:3:3:1 F2 ratio.

A: Chromosomal disorders cannot be inherited from parents.

R: They arise from non-disjunction during meiosis in parents.

Answer: (D). A is FALSE — chromosomal disorders CAN be inherited (especially translocation Down syndrome where one parent is a balanced translocation carrier). R is TRUE — most chromosomal disorders arise from non-disjunction in parental meiosis. So they ARE inherited from parents in the sense that non-disjunction in parental gametogenesis produces them.

A: Mendel's laws hold for human inheritance just as for pea plants.

R: Genes follow the same chromosomal behaviour during meiosis in all sexually reproducing diploid organisms.

Answer: (A). Both true; R explains A. The universality of Mendelian inheritance reflects the universality of meiosis as a fundamental cellular mechanism — in pea, fly, mouse, and human alike.

Frequently Asked Questions - NCERT Exercises and Solutions: Principles of Inheritance and Variation

What are the most-asked NCERT exercise questions in Chapter Principles of Inheritance and Variation?
NCERT Class 12 Biology Chapter on Principles of Inheritance and Variation exercises cover definitions, mechanisms, labelled diagrams, comparison tables, and application-based questions. The MyAiSchool solution set provides full step-by-step solutions for every NCERT question, aligned with the CBSE board exam pattern. Students should master scientific terminology, mechanism flowcharts, and concept comparison to score full marks.
How should students approach diagram-based questions in Principles of Inheritance and Variation?
For diagram-based questions in NCERT Class 12 Biology Chapter on Principles of Inheritance and Variation: (1) draw clean, proportional, large diagrams with sharp pencil lines, (2) label all parts horizontally on the right side using a ruler, (3) use scientific terminology (Latin/Greek names where applicable), (4) write a 1-2 line description if asked. The MyAiSchool solutions provide editable reference diagrams aligned with NCERT textbook figures.
What types of CBSE board questions come from Principles of Inheritance and Variation?
CBSE Class 12 Biology board questions from Principles of Inheritance and Variation typically include: (1) 1-mark MCQs on definitions and processes, (2) 2-mark short-answer mechanism/comparison questions, (3) 3-mark labelled-diagram questions, (4) 5-mark long-answer essays combining mechanism + diagram + significance. The MyAiSchool exercise set tags each question by mark weight and Bloom level (L1-L6).
How do I compare two biological processes in 5-mark questions?
For 5-mark comparison questions in NCERT Class 12 Biology on Principles of Inheritance and Variation: (1) use a two-column table with feature headings down the left side, (2) compare on at least 5-6 features (definition, location, mechanism, control, outcome, significance), (3) include one labelled diagram if relevant, (4) end with biological significance. The MyAiSchool solutions follow this CBSE-aligned tabular format consistently for full marks.
What are common mistakes in Principles of Inheritance and Variation exercises?
Common mistakes in NCERT Class 12 Biology Principles of Inheritance and Variation include: (1) confusing similar terms (mitosis vs meiosis, syngamy vs triple fusion, etc.), (2) skipping intermediate steps in mechanisms, (3) missing examples or species names, (4) writing essays when tables are expected, (5) forgetting biological significance. The MyAiSchool solutions highlight these traps with red flags so students avoid losing marks.
How does the MyAiSchool solution differ from other NCERT solution sets?
MyAiSchool Class 12 Biology Principles of Inheritance and Variation solutions use NEP 2024-aligned pedagogy with Bloom Taxonomy tagged questions (L1 Remember to L6 Create), step-by-step working with biological reasoning, fully labelled SVG diagrams, comparison tables, interactive simulations, and Competency-Based Questions (CBQs) for board exam practice. Each solution is verified against NCERT and CBSE marking schemes.
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