Put the following steps in the correct order from 1-9

polypeptide chain is released from the P site
small ribosomal subunit binds to mRNA
initiation complex formed with addition of large ribosomal subunit
ribosomal subunits dissociate
codon recognition (non-initiating site)
translocation
transcription of mRNA from DNA
peptide bond formation
ribosome reads a stop codon

transcription of mRNA from DNA

small ribosomal subunit binds to mRNA

initiation complex formed with addition of large ribosomal subunit

translocation

codon recognition (non-initiating site)  

peptide bond formation

ribosome reads a stop codon

polypeptide chain is released from the P site

ribosomal subunits dissociate

Explanation:

The above describes the process of translation in the ribosome. After transcription of DNA to mRNA, the mRNA is taken to the ribosome to undergo translation, here the mRNA binds to the small ribosomal subuits and to other initiation factors; binding at the mRNA binding site on the small ribosomal subunit then the Large ribosomal subunits joins in.

Translation begins (codon recognition; initiating site) at the initiation codon AUG on the mRNA with the tRNA bringing its amino acid (methionine in eukaryotes and formyl methionine in prokaryotes) forming complementary base pair between its anticodon and mRNA's AUG start codon. Then translocation occurs with the ribosome moving one codon over on the mRNA thus moving the start codon tRNA from the A site to the P site, then codon recognition occurs (non-initiating site again) which includes incoming tRNA with an anticodon that is complementary to the codon exposed in the A site binds to the mRNA.

Then peptide bond formation occurs between the amino acid carried by the tRNA in the p site and the A site. When the ribosome reads a stop codon, the process stops and the polypeptide chain produced is released and the ribosomal subunits dissociates.

b is the correct answer

easy

explanation: