Question

For E. coli, the dry cell weight of one billion cells is approximately:

• A. 250 mg
• B. 150 mg
• C. 350 mg
• D. 450 mg

Hint:

Always remember: Dry cell weight of bacteria is typically $20% - 25%$ of the wet cell weight. One billion ($10^9$) E. coli cells contain approximately $0.25 \text{ mg}$ (or $250\ \mu\text{g}$) of dry organic mass.

Answer 1

The Correct Option is A

Step 1: Concept
Determining bacterial biomass requires understanding the correlation between cell counts, wet cell weight, and dry cell weight.

Step 2: Standard Physical Values of an E. coli Cell
- The wet weight of a single E. coli cell is approximately $1 \text{ pg} = 10^{-12} \text{ g}$.
- The water content of an actively dividing bacterial cell is about $75%$.
- Therefore, the dry cell weight (DCW) is roughly $25%$ of its wet weight:
$$\text{Dry weight of 1 cell} = 0.25 \times 10^{-12} \text{ g} = 0.25 \text{ pg}$$

Step 3: Calculation for One Billion ($10^9$) Cells
Using the standard textbook definition:
$$\text{Dry weight of } 10^9 \text{ cells} = 10^9 \times \left(0.25 \times 10^{-12} \text{ g}\right) = 0.25 \times 10^{-3} \text{ g} = 0.25 \text{ mg} = 250\ \mu\text{g}$$

Step 4: Resolving Paper Mismatch
The official question paper options are listed in milligrams (mg) instead of micrograms ($\mu$g). The numerical value corresponding to $0.25 \text{ mg}$ (or $250\ \mu\text{g}$) is represented by the digits 250 in Option (A). Due to a standard units typographical error in the paper (mg instead of $\mu$g), Option (A) is the intended correct answer. Final Answer: (A)