A gene encoding for a polypeptide of 60 amino acids get mutated at `30^(th)` codon UAU becoming UAA The result would be

Options

A. A polypeptide of 29 amino acids
B. Two polypeptides one with 29 amino acids and second with 31 amino acids.
C. A polypeptide with 59 amino acids.
D. A polypeptide of 25 amino acids.

Answer

To solve the problem, we need to analyze the mutation of the gene encoding a polypeptide of 60 amino acids at the 30th codon. Here’s the step-by-step solution:

### Step 1: Understand the Original Codon
The original codon at the 30th position is UAU, which codes for the amino acid Tyrosine (Tyr).

### Step 2: Identify the Mutation
The mutation changes the 30th codon from UAU to UAA. UAA is one of the three stop codons (along with UAG and UGA) that signal the termination of protein synthesis.

### Step 3: Determine the Effect of the Mutation
Since UAA is a stop codon, the translation process will terminate at this point. This means that no further amino acids will be added to the growing polypeptide chain after the 29th amino acid.

### Step 4: Calculate the Length of the Resulting Polypeptide
The original polypeptide was supposed to be 60 amino acids long. However, due to the mutation at the 30th codon, the synthesis will stop after the 29th amino acid. Therefore, the resulting polypeptide will consist of 29 amino acids.

### Conclusion
The result of the mutation will be a polypeptide of 29 amino acids.

### Final Answer
The correct answer is: A polypeptide of 29 amino acids.

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