A) \[3\text{ }\times \text{ }129\] base pairs
B) \[\left( 3\times 129 \right)+\left( 3\times 3 \right)=396\]base pairs
C) \[\left( 3\times 129 \right)+\left( 3\times 2 \right)=393\]base pairs
D) 4\[\times \]129 base pairs
Correct Answer: C
Solution :
[c] Three base pairs code for one amino acid and two more triplets are required for start and stop signals. \[\therefore \,\,\,\,(3\times 129)+(3\times 2)=393\]base pairs
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