Question

How many minimum nucleotides would be required in the strand of DNA to encode a protein containing 301 amino acids?

• A. 900
• B. 903
• C. 906
• D. 897

Hint:

Formula: Total Nucleotides = (No. of Amino Acids $\times$ 3) + 3.
Always remember to add the "Stop Codon" ($+3$) unless the question specifically asks for "nucleotides coding for amino acids only."
The Start codon (AUG) is already included in the 301 amino acids because it codes for Methionine.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks for the minimum length of a DNA coding sequence required to produce a functional protein of a specific length. This involves understanding the relationship between the genetic code, transcription, and translation.
Key Formula or Approach:
1. Every amino acid is encoded by a triplet of nucleotides called a codon.
2. The total number of nucleotides for amino acids = (Number of amino acids) $\times$ 3.
3. A functional mRNA must also contain a stop codon (UAA, UAG, or UGA) to terminate translation, which does not code for an amino acid but occupies three nucleotides in the DNA template.

Step 2: Detailed Explanation:

• Coding for Amino Acids: To synthesize a protein containing 301 amino acids, the ribosome must read 301 sense codons. Since each codon consists of 3 nucleotides, the required length for the protein primary structure is: $301 \times 3 = 903$ nucleotides.

• Termination Signal: The "encoding" of a protein in a biological system is not complete without a stop signal. DNA contains a corresponding triplet that transcribes into a stop codon on the mRNA. This triplet is essential for the release of the polypeptide chain from the ribosome.

• Total Calculation: The minimum number of nucleotides required in the open reading frame (ORF) is the sum of the sense codons and the stop codon: $903 + 3 = 906$ nucleotides.

• Excluding non-coding regions: While real genes have promoters, 5' UTRs, and 3' UTRs, the term "encode a protein" in the context of "minimum nucleotides" typically refers to the coding sequence (CDS) from the start codon to the stop codon.

• Zygosity and Frame: Each amino acid is specifically positioned based on this triplet code. If even one nucleotide were missing (e.g., 905), a frameshift mutation would occur, and the resulting protein would not contain the 301 intended amino acids.

Step 3: Final Answer:
To encode 301 amino acids, 903 nucleotides are needed for the residues plus 3 nucleotides for the mandatory stop codon, totaling 906 nucleotides.