Home › Biology Class 12 › Molecular Basis of Inheritance

Molecular Basis of Inheritance

Biology · Class 12pp. 79–1072 sets × 25 questions

Key facts and formulas for Molecular Basis of Inheritance, 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.

Start the 50-question practice

Key facts for this chapter

IdeaRule or fact
Base pairingA=T (2 H-bonds), G≡C (3 H-bonds); Chargaff A=T, G=C only in double-stranded DNA
DNA helix numbersPitch 3.4 nm, about 10 bp per turn, 0.34 nm per bp; strands antiparallel
NucleosomeOctamer (H2A, H2B, H3, H4) × 2 wrapped by about 200 bp DNA; H1 on linker DNA; histones are basic
Genetic material experimentsGriffith transformation; Avery et al: DNase stops it; Hershey-Chase: 32P labels DNA, 35S labels protein
ReplicationSemiconservative (Meselson-Stahl, Taylor); polymerase 5'→3'; lagging strand in Okazaki fragments joined by ligase; primase makes RNA primer
RNA polymerases (eukaryotes)Pol I: 28S, 18S, 5.8S rRNA; Pol II: hnRNA; Pol III: tRNA, 5S rRNA, snRNA
hnRNA processingSplicing of introns, 5' methyl guanosine cap, 3' poly-A tail of 200 to 300 A
Genetic codeTriplet, 64 codons, 61 sense and 3 stop (UAA, UAG, UGA); AUG start; degenerate, universal
MutationsPoint: silent, missense, nonsense; insertion or deletion of one or two bases causes frameshift
lac operoni repressor, z beta-galactosidase, y permease, a transacetylase; lactose is the inducer; polycistronic mRNA
Genome and fingerprintingAbout 3 × 10⁹ bp, coding under 2%; chromosome 1 has 2968 genes, Y has 231; VNTR is a mini-satellite probe

Three sample questions

Q1. In a double-stranded DNA sample, adenine makes 30 per cent of the bases. The percentages of thymine, guanine and cytosine are, respectively,
  1. 20, 30, 20
  2. 30, 20, 20
  3. 30, 30, 10
  4. 20, 25, 25
Hint and solution

Hint: In double-stranded DNA, A pairs with T and G pairs with C. Find T first, then share what is left between G and C.

  1. A = T = 30% — A pairs with T by two hydrogen bonds, so thymine is also 30 per cent.
  2. A + T = 30 + 30 = 60% — Together A and T make 60 per cent of the bases.
  3. G + C = 100 − 60 = 40% — The rest, 40 per cent, is G and C.
  4. G = C = 40 ÷ 2 = 20% — G pairs with C, so they are equal. Thymine 30, guanine 20, cytosine 20. Option B.
Answer: B. A pairs with T, so T = 30. A + T = 60, leaving 40 for G + C, and G = C = 20.
Q2. Which of the following statements about the DNA double helix is correct?
  1. The pitch is 0.34 nm and one turn has about 10 base pairs
  2. The two strands run parallel, both 5' to 3'
  3. G and C are joined by two hydrogen bonds
  4. The pitch is 3.4 nm and one turn has about 10 base pairs
Hint and solution

Hint: Recall the numbers: pitch is ten times the gap between base pairs. Also remember the strand direction and which pair has three bonds.

  1. Pitch 3.4 nm, 10 bp per turn — The helix repeats after 3.4 nm, and one repeat holds about 10 base pairs.
  2. 3.4 nm ÷ 10 bp = 0.34 nm per bp — So the gap between neighbouring pairs is 0.34 nm. This removes the first option, which gives 0.34 nm as pitch.
  3. Strands antiparallel; G≡C 3 bonds — The strands run in opposite directions, and G-C has three bonds. That removes the second and third options.
  4. Only pitch 3.4 nm is correct — The correct statement is pitch 3.4 nm with about 10 base pairs per turn. Option D.
Answer: D. Pitch 3.4 nm, about 10 bp per turn, so 0.34 nm between pairs. The strands are antiparallel. G-C has three hydrogen bonds, A-T two.
Q3. The total DNA in a diploid human cell is 6.6 × 10⁹ bp and the spacing between base pairs is 0.34 nm. The approximate length of this DNA is
  1. 2.2 metres
  2. 0.22 metres
  3. 22 metres
  4. 2.2 centimetres
Hint and solution

Hint: Multiply the number of base pairs by the gap per pair, then convert nanometres to metres (1 nm = 10⁻⁹ m).

  1. Length = bp × 0.34 nm — Total length is the number of base pairs times the spacing of one pair.
  2. 6.6 × 10⁹ × 0.34 = 2.244 × 10⁹ nm — Multiply 6.6 by 0.34 to get 2.244, keeping the 10⁹.
  3. 1 nm = 10⁻⁹ m, so 2.244 × 10⁹ × 10⁻⁹ = 2.244 m — The powers of ten cancel, leaving 2.244 metres.
  4. About 2.2 m — The DNA of one cell is about 2.2 metres long. Option A.
Answer: A. 6.6 × 10⁹ × 0.34 × 10⁻⁹ m = 2.244 m.

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.

← Principles of Inheritance and VariationEvolution →
Start practice