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

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

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.