In elongation, which step occurs first?

In elongation, which step occurs first?

• A. Binding of next charged tRNA (A site)
• B. Peptide bond formation (P->A)
• C. Translocation of ribosomes (tRNA a->P)
• D. Release of free tRNA

Answer: (A)

In elongation, the first thing the ribosome does in each cycle is bring in the next aminoacyl-tRNA to the ribosome’s A site, ensuring its anticodon pairs correctly with the mRNA codon. This fidelity step is essential because the peptide chain can only grow if the right amino acid is attached in the right place. Once the correct tRNA is in place, peptidyl transferase forms the peptide bond, transferring the growing polypeptide from the tRNA in the P site to the amino acid on the A-site tRNA. After the bond forms, the ribosome translocates, moving the tRNAs to the next positions (A to P, P to E) and allowing the now-deacylated tRNA to exit. Since the binding of the next charged tRNA must occur before any bond formation, it is the first step in the cycle.

In elongation, the first thing the ribosome does in each cycle is bring in the next aminoacyl-tRNA to the ribosome’s A site, ensuring its anticodon pairs correctly with the mRNA codon. This fidelity step is essential because the peptide chain can only grow if the right amino acid is attached in the right place. Once the correct tRNA is in place, peptidyl transferase forms the peptide bond, transferring the growing polypeptide from the tRNA in the P site to the amino acid on the A-site tRNA. After the bond forms, the ribosome translocates, moving the tRNAs to the next positions (A to P, P to E) and allowing the now-deacylated tRNA to exit. Since the binding of the next charged tRNA must occur before any bond formation, it is the first step in the cycle.