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Principles of Inheritance and Variation

Biology · Class 12pp. 51–762 sets × 25 questions

Key facts and formulas for Principles of Inheritance and Variation, three worked sample questions, and two full practice sets. Every question has a hint, a reason for every option and a step-by-step solution, with a calculator or animation where one helps. English and Hinglish.

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Key facts for this chapter

IdeaRule or fact
Law of SegregationAlleles separate at gamete formation, so each gamete is pure and carries one allele.
Test cross and back crossBack cross is F1 × either parent. Test cross is with the homozygous recessive, and 1 : 1 means a heterozygote.
Dihybrid F29 : 3 : 3 : 1 with 9 genotypes and 4 phenotypes. Multiply the chances of independent genes.
Gamete and class countsGametes = 2 to the power of heterozygous pairs. n genes in F2 give 3ⁿ genotypes and 2ⁿ phenotypes.
Gene interaction ratiosComplementary 9 : 7, recessive epistasis 9 : 3 : 4, dominant epistasis 12 : 3 : 1, duplicate dominant 15 : 1.
ABO blood groupsAlleles IA, IB and i. IA and IB are co-dominant. Group O is ii. Bombay hh looks like O.
Linkage and mappingRecombination is at most 50%. The largest value marks the end genes and distances add up.
Sex determinationHumans XX–XY, birds ZZ–ZW (egg decides), grasshopper XO, honey bee haplodiploid, Drosophila X : A ratio.
Sex-linked inheritanceX-linked recessive shows criss-cross inheritance. Y-linked passes father to all sons. Daughter of colour blind father is a carrier.
Chromosomal disordersDown is trisomy 21, Edwards 18, Patau 13, Klinefelter XXY, Turner X0. Barr bodies = X chromosomes − 1.

Three sample questions

Q1. Which of the following statements about Mendel's work are correct?
(a) He worked on garden pea from 1856 to 1863.
(b) He started with 14 true-breeding varieties.
(c) He was the first to apply statistics to a biological problem.
(d) He worked with very few plants.
  1. (a), (b) and (d) only
  2. (a), (b) and (c) only
  3. (b), (c) and (d) only
  4. (a) and (d) only
Hint and solution

Hint: Mendel is famous for using large numbers of plants so that his ratios were reliable. Check statement (d) first.

  1. (a) 1856–1863 pea: true — Mendel worked on garden pea for about seven years, from 1856 to 1863. So (a) is correct.
  2. (b) 14 true-breeding lines: true — He began with 14 true-breeding pea varieties, which gave the seven contrasting pairs. So (b) is correct.
  3. (c) first statistics: true; (d) few plants: false — He was the first to apply statistical analysis to a biological problem. He used very large numbers of plants, so (d) is wrong.
  4. Correct set = (a), (b), (c) — The statements that hold are (a), (b) and (c). That is Option B.
Answer: B. Mendel worked 1856 to 1863, used 14 true-breeding varieties and was first to use statistics. He used very large numbers of plants, so (d) is wrong.
Q2. Which of the following was NOT one of the seven contrasting characters studied by Mendel in pea?
  1. Flower position
  2. Seed shape
  3. Pod colour
  4. Pod length
Hint and solution

Hint: Recall the seven pairs: stem height, flower colour, flower position, pod shape, pod colour, seed shape, seed colour. Which option is missing from this list?

  1. 7 pairs: height, flower colour/position, pod shape/colour, seed shape/colour — Write the seven characters Mendel used. Each had two contrasting forms.
  2. Flower position, seed shape, pod colour are all in the list — Options A, B and C each match one of the seven characters.
  3. Pod length is not in the list — Only the shape and colour of the pod were studied, not its length.
  4. Answer: pod length — The character that was not studied is pod length. Option D.
Answer: D. The seven: stem height, flower colour, flower position, pod shape, pod colour, seed shape, seed colour. Pod length is not one of them.
Q3. A tall F2 plant from a Tt × Tt cross is chosen at random and self-pollinated. What fraction of its progeny is expected to be dwarf, on average?
  1. 1/6
  2. 1/4
  3. 1/3
  4. 1/2
Hint and solution

Hint: A tall F2 plant is TT or Tt. Find the chance it is Tt first, then the chance a selfed Tt gives dwarf.

  1. Tall F2 = 1 TT : 2 Tt — Among the three tall boxes of a 3 : 1 ratio, one is TT and two are Tt. So the chance of Tt is 2/3.
  2. Tt × Tt → 1/4 tt — A selfed Tt plant gives 1/4 dwarf (tt) offspring. A TT plant gives no dwarf.
  3. P(dwarf) = 2/3 × 1/4 = 2/12 — Multiply the two chances because both must happen.
  4. = 1/6 — Simplify 2/12 to 1/6. Option A.
Answer: A. Two-thirds of tall F2 plants are Tt, and a selfed Tt gives ¼ dwarf. So 2/3 × 1/4 = 1/6.

How to use this chapter

Read the NCERT pages first, then attempt Set 1 without the hints. Note every wrong answer in a mistake list, read the reason for each option, and retry Set 2 two days later. A topic is solid when you can say why each wrong option is wrong.

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