To code the 50 aminoacids in a polypeptide chain, what will be the minimum number of nucleotides in its cistron?

Options

A. 50
B. 153
C. 306
D. 300

Answer

To determine the minimum number of nucleotides required in a cistron to code for a polypeptide chain of 50 amino acids, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Codons**: Each amino acid in a polypeptide is coded by a sequence of three nucleotides, known as a codon.

2. **Calculating Codons for Amino Acids**: To code for 50 amino acids, we need to consider that we also require a stop codon. Therefore, the total number of codons needed is:
\[ \text{Total Codons} = \text{Number of Amino Acids} + \text{Stop Codon} = 50 + 1 = 51 \text{ codons} \]

3. **Calculating Nucleotides**: Since each codon consists of 3 nucleotides, the total number of nucleotides required is:
\[ \text{Total Nucleotides} = \text{Total Codons} \times 3 = 51 \times 3 = 153 \text{ nucleotides} \]

4. **Considering Cistron Structure**: A cistron refers to a segment of DNA that can code for a protein and is typically double-stranded. Therefore, we must account for both strands of the DNA. Thus, we multiply the number of nucleotides by 2:
\[ \text{Minimum Nucleotides in Cistron} = 153 \times 2 = 306 \text{ nucleotides} \]

5. **Final Answer**: The minimum number of nucleotides in the cistron required to code for a polypeptide chain of 50 amino acids is **306**.