Q.65 A rod shaped bacterium has a length of 2 µm, diameter of 1 µm and density the same as that
of water. If proteins constitute 15% of the cell mass and the average protein has a mass of
50 kDa, the number of proteins in the cell is _________
(1 Da = 1.6 x 10-24g)
Number of Proteins in a Rod-Shaped Bacterium Cell: Calculation and Explanation
Rod-shaped bacteria, modeled as cylinders, allow precise volume and mass calculations for estimating protein content. Given specific dimensions and assumptions, the number of average 50 kDa proteins is approximately 2,945,000.
Problem Details
- Cell shape: Cylinder
- Length: 2 µm
- Diameter: 1 µm (radius = 0.5 µm)
- Density = water = 1 g/cm³
- Proteins = 15% of total mass
- Average protein mass = 50 kDa
- Conversion: 1 Da = 1.6 × 10-24 g
Step-by-Step Calculation
Step 1 — Calculate Cell Volume
Cylinder formula: V = πr²h
r = 0.5 µm, h = 2 µm
V = π(0.5)²(2) ≈ 1.57 µm³ = 1.57 × 10-12 cm³
Step 2 — Cell Mass
Mass = volume × density = 1.57 × 10-12 g
Step 3 — Protein Mass
Protein = 15% of cell mass
= 0.15 × 1.57 × 10-12 g = 2.36 × 10-13 g
Step 4 — Mass of One Protein
50,000 Da × 1.6 × 10-24 g/Da = 8 × 10-20 g
Step 5 — Number of Proteins
Total proteins = (2.36 × 10-13) / (8 × 10-20) ≈ 2,945,243
Rounded → 2,945,000 proteins
Answer Verification
The estimated value matches typical bacterial protein concentrations (~2–4 million proteins per cell), validating the calculated result for a ~1.57 µm³ bacterium.
Final Answer
A rod-shaped bacterium of size 2 µm × 1 µm contains approximately 2.95 × 106 proteins.
