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Principles of Inheritance and Variation – Class 12 Biology notes, important questions and MCQs

NCERT Class 12 Biology, chapter 4. Notes, key terms, important questions with answers and MCQs.

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Chapter overview

Mendel's experiments on pea plants showed that traits pass through factors (genes) following laws of dominance, segregation and independent assortment. Chromosomes carry genes; sex determination and genetic disorders follow from them.

Mendelian inheritance

  • Monohybrid cross F₂: phenotype 3 : 1, genotype 1 : 2 : 1.
  • Dihybrid cross F₂: 9 : 3 : 3 : 1.
  • Incomplete dominance (snapdragon 1 : 2 : 1), co-dominance (ABO blood group), multiple alleles (Iᴬ, Iᴮ, i).

Chromosomes and disorders

  • Linkage reduces recombination; Morgan worked with Drosophila.
  • Sex determination: XX–XY in humans (father decides), XX–XO in grasshopper, ZW–ZZ in birds.
  • Disorders: haemophilia and colour blindness (X-linked recessive), sickle-cell anaemia (autosomal recessive), Down's syndrome (trisomy 21), Turner's (XO), Klinefelter's (XXY).

Key terms

Allele
alternative form of a gene
Test cross
cross of an organism with a homozygous recessive parent to find its genotype
Co-dominance
both alleles express fully in the heterozygote

Formulas

  • Monohybrid F₂: 3 : 1; dihybrid F₂: 9 : 3 : 3 : 1

Important questions with answers

What is a test cross and why is it used?2 marks

Crossing an organism showing the dominant trait with a homozygous recessive. It reveals whether the organism is homozygous (all offspring dominant) or heterozygous (1 : 1 ratio).

Explain sex determination in honey bees.3 marks

Honey bees show haplodiploidy. Fertilised eggs (2n = 32) develop into females (queen or workers); unfertilised eggs (n = 16) develop into males (drones) by parthenogenesis. Drones produce sperm by mitosis.

Explain Mendel's law of independent assortment with a dihybrid cross using a checkerboard, and state why linkage is an exception.5 marks

Cross RRYY (round yellow) × rryy (wrinkled green): F₁ is RrYy. F₁ makes four kinds of gametes in equal proportion: RY, Ry, rY, ry. A 4 × 4 Punnett square gives 16 combinations: 9 round yellow, 3 round green, 3 wrinkled yellow, 1 wrinkled green. Because the alleles of the two genes separate independently into gametes, new combinations appear. Linkage is an exception: genes close together on the same chromosome tend to be inherited together, so parental combinations appear more often than 9 : 3 : 3 : 1 predicts (Morgan's work on Drosophila).

Quick revision

  • Test cross ratio for heterozygote is 1 : 1.
  • Down's = trisomy 21.
  • Haemophilia is X-linked recessive.

Exam tips

  • For blood groups, IᴬIᴮ gives AB (co-dominance); ii gives O.
  • In pedigree charts, squares are males and circles females; shaded symbols are affected individuals.

Practice MCQs

  1. Mendel worked on

    1. Drosophila
    2. maize
    3. garden pea
    4. snapdragon
    Show answer

    Answer: (c) garden pea. Pisum sativum.

  2. F₂ phenotypic ratio of a monohybrid cross is

    1. 1 : 2 : 1
    2. 9 : 3 : 3 : 1
    3. 3 : 1
    4. 1 : 1
    Show answer

    Answer: (c) 3 : 1. Dominance.

  3. ABO blood group shows

    1. incomplete dominance
    2. linkage
    3. co-dominance and multiple alleles
    4. polygenic only
    Show answer

    Answer: (c) co-dominance and multiple alleles. Iᴬ, Iᴮ, i.

  4. Down's syndrome is caused by

    1. trisomy 21
    2. XO
    3. XXY
    4. deletion of 5
    Show answer

    Answer: (a) trisomy 21. Extra chromosome 21.

  5. A cross between Tt and tt gives the phenotypic ratio

    1. 3 : 1
    2. 1 : 2 : 1
    3. 1 : 1
    4. all tall
    Show answer

    Answer: (c) 1 : 1. Test cross.

  6. F₂ dihybrid ratio is

    1. 3 : 1
    2. 1 : 2 : 1
    3. 9 : 3 : 3 : 1
    4. 1 : 1 : 1 : 1
    Show answer

    Answer: (c) 9 : 3 : 3 : 1. Independent assortment.

  7. In snapdragon, F₂ of red × white shows red : pink : white as

    1. 3 : 0 : 1
    2. 9 : 6 : 1
    3. 1 : 2 : 1
    4. 2 : 1 : 1
    Show answer

    Answer: (c) 1 : 2 : 1. Incomplete dominance.

  8. A colour-blind man marries a normal (non-carrier) woman. Their daughters will be

    1. all colour blind
    2. half carriers
    3. all carriers
    4. all normal non-carriers
    Show answer

    Answer: (c) all carriers. Each daughter gets the father's Xᶜ.

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Written by Kundali IQ from standard, well-established facts to help with revision. This is not an official NCERT publication; always read your textbook too.