Q.57 In an actively growing population from a single bacterium, 1,048,576 cells are
present after 20th generation. How many cells were there in 5th generation?
Step-by-Step Solution
Bacteria double per generation via binary fission, creating geometric progression: 1 → 2 → 4 → 8 → … → 2^n.
Generation 0: 1 cell
After 20 generations: 2^{20} = 1,048,576 cells (verifies input).
Cells at generation g: 2^g.
For 5th generation: 2^5 = 32.
Alternatively, divide total by doublings from generation 5 to 20: 1,048,576 ÷ 2^{15} = 32.
Core Formula and Verification
The equation N_n = 2^n applies, with N_0 = 1. Here, 2^{20} = 1,048,576 confirms the scenario. For the 5th generation, N_5 = 2^5 = 32 cells.
| Generation | Cells (2^n) |
|---|---|
| 0 | 1 |
| 5 | 32 |
| 10 | 1,024 |
| 20 | 1,048,576 |
Detailed Calculation Methods
- Direct method: Compute 2^5 = 32.
- Ratio method: 1,048,576 ÷ 2^{15} = 32 (15 doublings from gen 5 to 20).
- Logarithmic check: n = \log_2(1,048,576) = 20, so gen 5 is 2^5.
No options provided, but common MCQ traps include:
- 1,024 (2^{10}, wrong generation).
- 512 (2^9, off-by-one error).
- 16 (2^4, undercount).
Correct: 32.
CSIR NET Application
This tests exponential growth in microbiology for exams like CSIR NET Life Sciences. Practice verifies 2^{10} = 1,024, 2^{20} = 1,048,576. Real bacteria like E. coli double every 20-30 minutes under ideal conditions.
