Practicing Success

Target Exam

CUET

Subject

Biology

Chapter

Molecular Basis of Inheritance

Question:

Match List - I with List - II.

List - I List - II
(A) Initiation factor (I) Tailing
(B) Introns (II) Rho(p)
(C) Termination factor (III) Sigma (o)
(D) Adenylate residue (IV) Splicing

Choose the correct answer from the options given below:

Options:

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

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

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

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

Correct Answer:

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

Explanation:

The correct answer is Option (4) - (A) - (III), (B) - (IV), (C) - (II), (D) - (I)

List - I List - II
(A) Initiation factor (III) Sigma (σ)
(B) Introns (IV) Splicing
(C) Termination factor (II) Rho(ρ)
(D) Adenylate residue (I) Tailing

The RNA polymerase is only capable of catalysing the process of elongation. It associates transiently with initiation-factor (σ) and termination-factor (ρ) to initiate and terminate the transcription, respectively. Association with these factors alter the specificity of the RNA polymerase to either initiate or terminate .

The second complexity is that the primary transcripts contain both the exons and the introns and are non-functional. Hence, it is subjected to a process called splicing where the introns are removed and exons are joined in a defined order. hnRNA undergoes additional processing called as capping and tailing.

In tailing, adenylate residues (200-300) are added at 3'-end in a template independent manner.