Question:

Arrange the steps of translation initiation in prokaryotes:
A. Binding of 50S ribosomal subunit
B. Binding of initiator tRNA
C. Binding of mRNA to 30S subunit
D. Formation of complete 70S initiation complex

Show Hint

Always remember the assembly order: Small subunit first (with mRNA), then the starter tRNA, and finally the large subunit acts as the "lid" to close the complex.
Updated On: Mar 17, 2026
  • C $\rightarrow$ B $\rightarrow$ A $\rightarrow$ D
  • B $\rightarrow$ C $\rightarrow$ A $\rightarrow$ D
  • C $\rightarrow$ A $\rightarrow$ B $\rightarrow$ D
  • B $\rightarrow$ A $\rightarrow$ C $\rightarrow$ D
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is A

Solution and Explanation


Step 1: Understanding the Concept:
Translation initiation in prokaryotes involves the sequential assembly of the ribosomal subunits, mRNA, and initiator tRNA.
This highly ordered process is orchestrated by specific initiation factors (IFs).

Step 2: Detailed Explanation:
The sequential events unfold in the following order:
1. Binding of mRNA to 30S subunit (C): The small ribosomal subunit (30S), complexed with initiation factors, first binds to the mRNA.
This is facilitated by base-pairing between the Shine-Dalgarno sequence on the mRNA and the 16S rRNA.
2. Binding of initiator tRNA (B): Next, the initiator tRNA carrying formylmethionine (fMet-tRNA) binds to the start codon at the P site.
This completes the formation of the 30S initiation complex.
3. Binding of 50S ribosomal subunit (A): Once the 30S complex is stable, the large ribosomal subunit (50S) joins the assembly.
This step is accompanied by the hydrolysis of GTP and the release of initiation factors.
4. Formation of complete 70S initiation complex (D): The successful joining of the 50S subunit marks the end of the initiation phase.
This results in a fully assembled, translationally active 70S ribosome ready for the elongation phase.

Step 3: Final Answer:
The correct temporal sequence is C $\rightarrow$ B $\rightarrow$ A, which culminates in D.
Therefore, the correct option is (A) C $\rightarrow$ B $\rightarrow$ A $\rightarrow$ D.
Was this answer helpful?
0
0