Question

The approximate length of the open reading frame coding for a 55 kDa bacterial protein will be

• A. 5000 nucleotides
• B. 5500 nucleotides
• C. 1500 nucleotides
• D. 3000 nucleotides

Hint:

• 1 amino acid $\approx$ 110 Da
• 1 amino acid = 3 nucleotides
• Quick formula: \(\frac{\text{Protein (Da)}}{110} \times 3\)

Answer 1

The Correct Option is C

Concept:
• Average molecular weight of one amino acid $\approx$ 110 Da
• 1 amino acid is coded by 3 nucleotides

Step 1: Calculating number of amino acids.
Given protein size = 55 kDa = 55,000 Da \[ \text{Number of amino acids} = \frac{55000}{110} = 500 \]

Step 2: Calculating length of ORF.
Each amino acid requires 3 nucleotides: \[ \text{Total nucleotides} = 500 \times 3 = 1500 \]

Step 3: Selecting the correct answer.
Thus, the approximate ORF length is 1500 nucleotides.