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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


Match List-I with List-II regarding Chargaff Rules.

List-I (Equality)List-II (Significance)
A. A = TI. Base pairing ratio
B. Purines = PyrimidinesII. Total bases balance
C. A+G = T+CIII. Constant diameter

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

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

    Remark: A equals T refers to the specific base pairing. Purines equal Pyrimidines ensures the total balance and constant diameter of the helix.

Match List-I with List-II regarding Transcription Factors.

List-I (Factor)List-II (Function in Prokaryotes)
A. SigmaI. Start signal recognition
B. RhoII. Termination signal recognition

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

  • A-I, B-II
  • A-II, B-I
  • A-I, B-I
  • A-II, B-II
    • Correct Option: A  [ A-I, B-II ]

    Remark: Sigma factor recognizes the promoter (start). Rho factor facilitates termination.

Match List-I with List-II regarding Types of Mutations.

List-I (Type)List-II (Effect on Frame)
A. Insertion of 1 baseI. Frame Shift
B. Deletion of 1 baseII. Frame Shift
C. Insertion of 3 basesIII. Non-Frame Shift (Amino acid insertion)

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

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

    Remark: Insertion or deletion of 1 base causes a frame shift. Insertion of 3 bases adds an amino acid but preserves the downstream reading frame.

Match List-I with List-II regarding Ribosomal Subunits (Prokaryotes).

List-I (Subunit)List-II (Role in Initiation)
A. Smaller SubunitI. Binds to Shine-Dalgarno sequence
B. Larger SubunitII. Joins later to form complete ribosome

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

  • A-I, B-II
  • A-II, B-I
  • A-I, B-I
  • A-II, B-II
    • Correct Option: A  [ A-I, B-II ]

    Remark: The smaller subunit binds to the mRNA first (Shine-Dalgarno). The larger subunit joins subsequently.

Match List-I with List-II regarding Lac Operon Status.

List-I (Condition)List-II (Result)
A. Lactose AbsentI. Repressor binds Operator (Switch Off)
B. Lactose PresentII. Repressor inactivated by Inducer (Switch On)

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

  • A-I, B-II
  • A-II, B-I
  • A-I, B-I
  • A-II, B-II
    • Correct Option: A  [ A-I, B-II ]

    Remark: No lactose means the Repressor is active and binds the Operator (Off). Presence of lactose inactivates the Repressor (On).

Match List-I with List-II regarding Human Genome Project Data.

List-I (Category)List-II (Fact)
A. Protein Coding GenesI. Less than 2 percent
B. Repetitive DNAII. Makes up large portion of genome
C. Total number of genesIII. Approximately 30,000

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

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

    Remark: Less than 2% of the genome codes for proteins. Repetitive DNA is abundant. Total gene count is approx 30,000.

Match List-I with List-II regarding VNTRs.

List-I (Feature)List-II (Description)
A. SizeI. 0.1 to 20 kb
B. CategoryII. Minisatellite
C. UsageIII. DNA Fingerprinting marker

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

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

    Remark: VNTR size ranges from 0.1 to 20 kb. It is a Minisatellite. It is used as a marker in DNA fingerprinting.

Match List-I with List-II regarding RNA Polarity.

List-I (End)List-II (Group)
A. 5-prime endI. Phosphate group
B. 3-prime endII. Hydroxyl (OH) group

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

  • A-I, B-II
  • A-II, B-I
  • A-I, B-I
  • A-II, B-II
    • Correct Option: A  [ A-I, B-II ]

    Remark: The 5-prime end has a free phosphate group. The 3-prime end has a free hydroxyl group.

Match List-I with List-II regarding DNA replication requirements.

List-I (Requirement)List-II (Source)
A. SubstratesI. dNTPs (Deoxyribonucleoside triphosphates)
B. EnergyII. dNTPs (High energy bonds)

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

  • A-I, B-II
  • A-II, B-I
  • A-I, B-I
  • A-II, B-II
    • Correct Option: A  [ A-I, B-II ]

    Remark: dNTPs serve dual purposes: they act as substrates for polymerization and provide energy for the reaction.

Match List-I with List-II regarding Transcription Unit Promoters.

List-I (Feature)List-II (Location)
A. UpstreamI. 5-prime end of coding strand
B. DownstreamII. 3-prime end of coding strand

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

  • A-I, B-II
  • A-II, B-I
  • A-I, B-I
  • A-II, B-II
    • Correct Option: A  [ A-I, B-II ]

    Remark: Upstream corresponds to the 5-prime end (Promoter). Downstream corresponds to the 3-prime end (Terminator).
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