NEET MCQs Practice on Molecular Basis of Inheritance Class-XII Biology, Free unlimited practice on Molecular Basis of Inheritance Class-XII Biology Mcqs

Match List-I with List-II regarding RNA Polymerase Subunits.

List-I (Factor)	List-II (Role)
A. Sigma Factor	I. Elongation
B. Rho Factor	II. Initiation
C. Core Enzyme	III. Termination

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

A-II, B-III, C-I
A-I, B-II, C-III
A-II, B-I, C-III
A-III, B-II, C-I

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

Remark: Sigma factor initiates transcription. Rho factor terminates it. The Core enzyme carries out elongation.

Match List-I with List-II regarding DNA vs RNA.

List-I (Feature)	List-II (Molecule)
A. 2-prime OH group	I. DNA
B. Thymine	II. RNA
C. Double Stranded (usually)	III. DNA
D. Catalytic activity (Ribozyme)	IV. RNA

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

A-II, B-I, C-III, D-IV
A-I, B-II, C-III, D-IV
A-II, B-III, C-I, D-IV
A-III, B-I, C-IV, D-II

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

Remark: RNA has a 2-prime OH group making it reactive. DNA contains Thymine. DNA is typically double-stranded. RNA can act as a catalyst (Ribozyme).

Match List-I with List-II regarding HGP Goals.

List-I (Goal)	List-II (Action)
A. Sequencing	I. Address ELLS issues
B. Storage	II. Determine order of 3 billion bases
C. ELSI	III. Store information in databases

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

A-II, B-III, C-I
A-I, B-II, C-III
A-II, B-I, C-III
A-III, B-II, C-I

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

Remark: Sequencing determines the base order. Storage involves databases (Bioinformatics). ELSI addresses Ethical, Legal, and Social Issues.

Match List-I with List-II regarding Histone Amino Acids.

List-I (Amino Acid)	List-II (Property)
A. Lysine	I. Basic, Positive charge
B. Arginine	II. Basic, Positive charge
C. Glutamic Acid	III. Acidic, Negative charge

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

A-I, B-II, C-III
A-III, B-I, C-II
A-II, B-III, C-I
A-I, B-III, C-II

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

Remark: Lysine and Arginine are basic and positive, abundant in histones. Glutamic acid is acidic and negative.

Match List-I with List-II regarding tRNA structure.

List-I (Feature)	List-II (Shape/Part)
A. 2D Structure	I. Inverted L
B. 3D Structure	II. Clover-leaf
C. Acceptor Arm	III. Binds Amino Acid
D. Anticodon Loop	IV. Reads mRNA codon

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

A-II, B-I, C-III, D-IV
A-I, B-II, C-III, D-IV
A-II, B-III, C-I, D-IV
A-II, B-I, C-IV, D-III

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

Remark: The 2D structure is a clover-leaf. The 3D structure is an inverted L. The Acceptor Arm binds the amino acid. The Anticodon loop reads the mRNA.

Match List-I with List-II regarding DNA Replication Directions.

List-I (Process)	List-II (Direction)
A. Synthesis of New Strand	I. 5-prime to 3-prime
B. Reading of Template Strand	II. 3-prime to 5-prime
C. Bond formation	III. 3-prime to 5-prime (relative to new strand)

[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: New strands are always synthesized 5-prime to 3-prime. The template is read 3-prime to 5-prime. Phosphodiester bond formation involves attacking the 3-prime end, directionally extending 5-to-3, but chemically linking 3-to-5.

Match List-I with List-II regarding Central Dogma.

List-I (Process)	List-II (Flow)
A. Replication	I. DNA to DNA
B. Transcription	II. DNA to RNA
C. Translation	III. RNA to Protein
D. Reverse Transcription	IV. RNA to DNA

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

A-I, B-II, C-III, D-IV
A-II, B-I, C-III, D-IV
A-I, B-III, C-II, D-IV
A-IV, B-III, C-II, D-I

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

Remark: Replication: DNA to DNA. Transcription: DNA to RNA. Translation: RNA to Protein. Reverse Transcription: RNA to DNA (Retroviruses).

Match List-I with List-II regarding Early Scientists.

List-I (Scientist)	List-II (Discovery)
A. Friedrich Miescher	I. Nuclein
B. Altmann	II. Nucleic Acid (Renaming)
C. Watson & Crick	III. Double Helix

[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: Miescher discovered Nuclein. Altmann renamed it Nucleic Acid. Watson and Crick proposed the Double Helix.

Match List-I with List-II regarding Genetic Code Specifics.

List-I (Number)	List-II (Item)
A. 64	I. Total Codons
B. 61	II. Sense Codons (coding for amino acids)
C. 3	III. Stop Codons
D. 20	IV. Standard Amino Acids

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

A-I, B-II, C-III, D-IV
A-II, B-I, C-III, D-IV
A-I, B-III, C-II, D-IV
A-IV, B-II, C-III, D-I

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

Remark: There are 64 total codons. 61 code for amino acids (Sense). 3 are Stop codons. There are 20 standard amino acids.

Match List-I with List-II regarding S-strain vs R-strain.

List-I (Strain)	List-II (Feature)
A. S-strain	I. Smooth, Virulent, Capsule present
B. R-strain	II. Rough, Non-virulent, No capsule

[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: S-strain is Smooth, Virulent, and Capsulated. R-strain is Rough, Non-virulent, and Non-capsulated.

Molecular Basis of Inheritance: Class-XII

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