Question

Rod shaped E. coli is 2 µm long and has diameter of 0.8 µm. The average density of E. coli is 1.1 x 10³ g/L. The mass of single E. coli cell in pg is ........... (decimal digits up to 1 place)

Hint:

Use the formula for the volume of a cylinder and convert units to calculate the mass of a single cell.

Answer 1

Correct Answer: 1

Step 1: Calculate the volume of the E. coli cell.
The shape of the E. coli cell is a cylinder, so the volume \( V \) of a cylinder is given by: \[ V = \pi r^2 h, \] where \( r \) is the radius and \( h \) is the height. The radius is half the diameter: \[ r = \frac{0.8}{2} = 0.4 \, \mu\text{m}. \] The height is given as 2 µm. Thus, the volume of the E. coli cell is: \[ V = \pi \times (0.4)^2 \times 2 = 1.0 \, \mu\text{m}^3. \]

Step 2: Convert volume to cubic centimeters.
Since 1 µm = \( 10^{-6} \) meters, \( 1 \, \mu\text{m}^3 = 10^{-18} \, \text{cm}^3 \), so: \[ 1.0 \, \mu\text{m}^3 = 1.0 \times 10^{-18} \, \text{cm}^3. \]

Step 3: Calculate the mass of the E. coli cell.
The density of E. coli is given as \( 1.1 \times 10^3 \, \text{g/L} \), or \( 1.1 \times 10^3 \, \text{g/m}^3 \). The mass is: \[ \text{Mass} = \text{Density} \times \text{Volume} = 1.1 \times 10^3 \times 1.0 \times 10^{-18} = 1.1 \times 10^{-15} \, \text{g}. \]

Step 4: Convert the mass to picograms.
1 gram = \( 10^{12} \) picograms, so: \[ 1.1 \times 10^{-15} \, \text{g} = 1.1 \times 10^{-3} \, \text{pg} = 2.8 \, \text{pg}. \]

Step 5: Conclusion.
The mass of a single E. coli cell is 2.8 pg.