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 question regarding the mutation of a gene encoding a polypeptide of 60 amino acids, we need to analyze the effect of the mutation on the polypeptide chain.

### Step-by-Step Solution:

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

2. **Identify the Mutated Codon**:
- The mutated codon is UAA. UAA is one of the three stop codons (the others being UAG and UGA) that signal the termination of protein synthesis.

3. **Effect of the Mutation**:
- Since UAA is a stop codon, the translation process will terminate at this point. This means that the polypeptide synthesis will stop prematurely.

4. **Determine 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 translation will stop before reaching the 60th amino acid.
- The polypeptide will consist of the first 29 amino acids (from the 1st to the 29th codon) and will not include the 30th amino acid or any subsequent amino acids.

5. **Conclusion**:
- The resulting polypeptide will be 29 amino acids long.

### Final Answer:
The result of the mutation will be a polypeptide of 29 amino acids.