Question

Which of the following represents the correct sequence of events in bacterial translational initiation?

a. 30S binds Shine-dalgarno --> N-formylMet-tRNA binds AUG --> 5QS binds 2nd aminoacyl-tRNA binds in the P site
b. 50S binds Shine-dalgarno --> N-formylMet-tRNA binds AUG --> 30S binds 2nd aminoacyl-tRNA binds in the A site
c. 30S binds Shine-dalgarno --> N-formylMet-tRNA binds AUG --> 50S binds 2nd aminoacyl-tRNA binds in the A site
d. 30S binds Shine-dalgarno --> 50S binds N-formylMet-tRNA binds AUG --> 2nd aminoacyl-tRNA binds in the A site

Answer 1

c) 30S binds Shine-dalgarno → N-formylMet-tRNA binds AUG → 50S binds → 2nd aminoacyl-tRNA binds in the A site.

Step-by-step explanation:

The first step in formation of 70S initiation complex in bacterial translational initiation includes binding of 30S rRNA subunit to Shine-dalgarno sequence present on mRNA.

Soon after that N-formylMet-tRNA binds AUG (start codon) which is present on mRNA.

There are two sites in 30S rRNA which are P site and A site. The first tRNA mentioned above i.e. N-formylMet-tRNA is adjusted in P site. Then after the larger rRNA subunit which is known as 50S joins the complex.

Once the 70S initiation complex has formed, a 2nd aminoacyl-tRNA enters the complex and gets adjusted in A site located in 30S rRNA.