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Molecular Basis of Inheritance: Class-XII


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MCQs on Molecular Basis of Inheritance: Class-XII for NEET Practice


Degeneracy of the genetic code means _______

[Molecular-Basis-of-Inheritance] [class-xii ]

  • One codon codes for multiple amino acids
  • Multiple codons code for one amino acid
  • Code is overlapping
  • Code is universal
  • Correct Option: B  [ Multiple codons code for one amino acid ]

    Remark: Degeneracy = redundancy.

_______ is not a stop codon.

[Molecular-Basis-of-Inheritance] [class-xii ]

  • UAA
  • UAG
  • UGA
  • UGG
  • Correct Option: D  [ UGG ]

    Remark: UGG is not a stop codon (Trp).

Match List-I with List-II.

List-I (Scientists)List-II (Discovery)
A. Friedrich MiescherI. Double Helix Model
B. AltmannII. X-ray Diffraction Data
C. Watson and CrickIII. Nuclein
D. Wilkins and FranklinIV. Nucleic Acid (Acidic nature)

[Molecular-Basis-of-Inheritance] [class-xii ]

  • A-III, B-IV, C-I, D-II
  • A-III, B-I, C-IV, D-II
  • A-IV, B-III, C-I, D-II
  • A-II, B-IV, C-I, D-III
  • Correct Option: A  [ A-III, B-IV, C-I, D-II ]

    Remark: Friedrich Miescher (1869) was the first to identify DNA as an acidic substance inside the nucleus and named it "Nuclein". Later, Altmann renamed it "Nucleic Acid" due to its acidic properties. Maurice Wilkins and Rosalind Franklin produced the critical X-ray diffraction data of DNA. Based on this data, James Watson and Francis Crick proposed the famous Double Helix model of DNA structure in 1953.

Match List-I with List-II.

List-I (DNA Structure)List-II (Characteristics)
A. Phosphodiester bondI. Stability of Helix
B. N-Glycosidic bondII. Backbone of DNA
C. Hydrogen bondIII. Links Base to Sugar
D. Stacking interactionIV. Base Pairing

[Molecular-Basis-of-Inheritance] [class-xii ]

  • A-II, B-III, C-IV, D-I
  • A-III, B-II, C-IV, D-I
  • A-II, B-III, C-I, D-IV
  • A-IV, B-III, C-II, D-I
  • Correct Option: A  [ A-II, B-III, C-IV, D-I ]

    Remark: The Phosphodiester bond connects nucleotides to form the sugar-phosphate backbone of the DNA strands. The N-Glycosidic bond links the nitrogenous base to the pentose sugar. Hydrogen bonds form between complementary bases (A-T, G-C) holding the two strands together. Stacking interactions between the stacked base pairs provide additional stability to the helical structure.

Match List-I with List-II.

List-I (Nucleosome)List-II (Component/Role)
A. Histone OctamerI. Seals DNA entry/exit (Linker)
B. H1 HistoneII. Core of Nucleosome (8 molecules)
C. DNAIII. Basic Amino Acids
D. Lysine & ArginineIV. Negatively charged polymer

[Molecular-Basis-of-Inheritance] [class-xii ]

  • A-II, B-I, C-IV, D-III
  • A-II, B-IV, C-I, D-III
  • A-III, B-I, C-IV, D-II
  • A-I, B-II, C-IV, D-III
  • Correct Option: A  [ A-II, B-I, C-IV, D-III ]

    Remark: The Histone Octamer forms the core of the nucleosome, consisting of two molecules each of H2A, H2B, H3, and H4. The H1 Histone acts as a linker or plug, sealing the DNA as it enters and exits the nucleosome. DNA is negatively charged due to its phosphate groups, allowing it to wrap around the positively charged histones. Lysine and Arginine are the abundant basic amino acids that give histones their positive charge.

Match List-I with List-II.

List-I (Griffith Experiment)List-II (Outcome)
A. S-strain injectedI. Mice Live
B. R-strain injectedII. Mice Die
C. Heat-killed S-strainIII. Mice Die (Transformation)
D. Heat-killed S + Live RIV. Mice Live

[Molecular-Basis-of-Inheritance] [class-xii ]

  • A-II, B-I, C-IV, D-III
  • A-II, B-IV, C-I, D-III
  • A-I, B-II, C-IV, D-III
  • A-III, B-I, C-IV, D-II
  • Correct Option: A  [ A-II, B-I, C-IV, D-III ]

    Remark: Injecting the virulent S-strain (Smooth) kills the mice. Injecting the non-virulent R-strain (Rough) allows the mice to live. Heat-killed S-strain alone is non-virulent, so mice live. However, mixing Heat-killed S-strain with Live R-strain causes the mice to die because the R-strain is transformed into the virulent S-strain by the DNA from the heat-killed bacteria.

Match List-I with List-II.

List-I (Hershey-Chase Exp.)List-II (Detail)
A. Radioactive Sulfur (S-35)I. Separates viral coats from bacteria
B. Radioactive Phosphorus (P-32)II. Labels Viral Protein Coat
C. BlendingIII. Labels Viral DNA
D. CentrifugationIV. Separates based on density (supernatant/pellet)

[Molecular-Basis-of-Inheritance] [class-xii ]

  • A-II, B-III, C-I, D-IV
  • A-III, B-II, C-I, D-IV
  • A-II, B-III, C-IV, D-I
  • A-I, B-IV, C-II, D-III
  • Correct Option: A  [ A-II, B-III, C-I, D-IV ]

    Remark: Hershey and Chase used Radioactive Sulfur (S-35) to label the protein coat (capsid) because amino acids contain sulfur. Radioactive Phosphorus (P-32) was used to label DNA because DNA contains phosphate. Blending agitates the mixture to shear off empty viral coats. Centrifugation spins the mixture to separate the heavy bacterial pellet (containing DNA) from the lighter viral supernatant.

Match List-I with List-II.

List-I (Replication Enzymes)List-II (Function)
A. DNA HelicaseI. Synthesizes RNA primer
B. TopoisomeraseII. Joins DNA fragments
C. PrimaseIII. Unwinds DNA helix
D. DNA LigaseIV. Relieves supercoiling/tension

[Molecular-Basis-of-Inheritance] [class-xii ]

  • A-III, B-IV, C-I, D-II
  • A-III, B-I, C-IV, D-II
  • A-IV, B-III, C-I, D-II
  • A-II, B-IV, C-I, D-III
  • Correct Option: A  [ A-III, B-IV, C-I, D-II ]

    Remark: DNA Helicase unzips or unwinds the DNA double helix. Topoisomerase (like DNA Gyrase) relieves the tension or supercoiling that builds up ahead of the replication fork. Primase synthesizes short RNA primers needed to start replication. DNA Ligase joins the discontinuous Okazaki fragments on the lagging strand.

Match List-I with List-II.

List-I (Replication Components)List-II (Characteristics)
A. Leading StrandI. Synthesized discontinuously
B. Lagging StrandII. Prevents re-annealing of strands
C. Okazaki FragmentsIII. Synthesized continuously 5-3
D. SSB ProteinsIV. Small DNA segments on Lagging strand

[Molecular-Basis-of-Inheritance] [class-xii ]

  • A-III, B-I, C-IV, D-II
  • A-I, B-III, C-IV, D-II
  • A-III, B-II, C-IV, D-I
  • A-IV, B-I, C-II, D-III
  • Correct Option: A  [ A-III, B-I, C-IV, D-II ]

    Remark: The Leading Strand is synthesized continuously in the 5 to 3 direction towards the replication fork. The Lagging Strand is synthesized discontinuously away from the fork. Okazaki Fragments are the short DNA segments produced on the lagging strand. Single-Strand Binding (SSB) proteins stabilize the separated strands and prevent them from snapping back together.

Match List-I with List-II.

List-I (Transcription Unit)List-II (Description)
A. PromoterI. Located towards 3-prime end of coding strand
B. TerminatorII. Has polarity 3-prime to 5-prime
C. Template StrandIII. Located towards 5-prime end of coding strand
D. Coding StrandIV. Sequence same as RNA (except T/U)

[Molecular-Basis-of-Inheritance] [class-xii ]

  • A-III, B-I, C-II, D-IV
  • A-I, B-III, C-II, D-IV
  • A-III, B-I, C-IV, D-II
  • A-II, B-IV, C-I, D-III
  • Correct Option: A  [ A-III, B-I, C-II, D-IV ]

    Remark: The Promoter is located upstream at the 5-prime end of the coding strand. The Terminator is located downstream at the 3-prime end of the coding strand. The Template Strand has 3-prime to 5-prime polarity and guides synthesis. The Coding Strand has the same sequence as the RNA (except Thymine is replaced by Uracil) and runs 5-prime to 3-prime.