Question

What would happen if a gene encoding a polypeptide of 50 amino acids has UAC at 25${}^t$${}^h$ position mutated to UAA?

• A. A polypeptide of 25 amino acids will be formed.
• B. A polypeptide of 24 amino acids will be formed.
• C. A polypeptide of 49 amino acids will be formed.
• D. Two plypeptides of 24 and 25 amino acids will be formed.

Answer 1

Answer: (B)

A polypeptide of 24 amino acids will be formed.

The codons UAA is a stop codon and cause termination of protein synthesis. When this stop codon will occupy the ribosomal A site, it causes hydrolysis of the terminal peptidyltRNA bond to release the polypeptide and the last tRNA and finally dissociation of the 70S ribosome into its 30S and 50S subunits.

The whole process results in premature termination of polypeptide at 25th codon. Since, 24 codons are translated properly, the polypeptide will have 24 amino acids, not 49 or 25 amino acids. "AUG" is the most common start codon. The start codon specify the first codon of mRNA transcript that is translated by a ribosome. It codes for methionine in eukaryotes and a modified Met in prokaryotes. Translation initiation requires presence of initiation codon and promoter sequences (pribnow box, TATA box, etc) which are not present in middle of any gene. Hence, after premature termination at 25th codon, the gene can not be translated at its 26th codon; two polypeptides of 24 and 25 amino acids can not be formed. Thus, correct answer is B.