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Molecular Basis of Inheritance MCQs

Class XII Biology NCERT Based NEET Practice

📘 Concept Based 📝 Exam Level 🤖 AI Explanations
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668 questions in this chapter
Question 631 of 668
📘 CLASS XII
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
A equals T refers to the specific base pairing. Purines equal Pyrimidines ensures the total balance and constant diameter of the helix.
Question 632 of 668
📘 CLASS XII
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
Sigma factor recognizes the promoter (start). Rho factor facilitates termination.
Question 633 of 668
📘 CLASS XII
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)
Insertion or deletion of 1 base causes a frame shift. Insertion of 3 bases adds an amino acid but preserves the downstream reading frame.
Question 634 of 668
📘 CLASS XII
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
The smaller subunit binds to the mRNA first (Shine-Dalgarno). The larger subunit joins subsequently.
Question 635 of 668
📘 CLASS XII
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)
No lactose means the Repressor is active and binds the Operator (Off). Presence of lactose inactivates the Repressor (On).
Question 636 of 668
📘 CLASS XII
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
Less than 2% of the genome codes for proteins. Repetitive DNA is abundant. Total gene count is approx 30,000.
Question 637 of 668
📘 CLASS XII
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
VNTR size ranges from 0.1 to 20 kb. It is a Minisatellite. It is used as a marker in DNA fingerprinting.
Question 638 of 668
📘 CLASS XII
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
The 5-prime end has a free phosphate group. The 3-prime end has a free hydroxyl group.
Question 639 of 668
📘 CLASS XII
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)
dNTPs serve dual purposes: they act as substrates for polymerization and provide energy for the reaction.
Question 640 of 668
📘 CLASS XII
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
Upstream corresponds to the 5-prime end (Promoter). Downstream corresponds to the 3-prime end (Terminator).