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Microbiology · Bacterial growth: generation time calculation
Sample itemEscherichia coli has a generation time of 20 minutes in log phase. If 2 x 10^3 cells in log phase are inoculated into broth and lag is ignored, what is the approximate count after 3 hours?
Original practice item, not a recalled exam question. Nothing here is recorded.
The answer and the reasoning
C. 1.0 x 10^6
Three hours is 180 minutes, so there are 180 / 20 = 9 generations. Growth is exponential: N = N0 x 2^n = 2 x 10^3 x 2^9 = 2000 x 512 = 1,024,000, or about 1.0 x 10^6. Using 2^3 or multiplying by 180 are common errors.
Option by option
A. 3.6 x 10^5
This treats growth as linear (2000 x 180), but bacteria multiply geometrically.
B. 1.6 x 10^4
This uses 2^3 (three hours treated as three generations), which ignores the 9 doublings.
C. 1.0 x 10^6 (correct)
Correct. 2000 x 512 is about 1.0 x 10^6.
D. 5.1 x 10^5
This uses 1000 x 512 and forgets the starting inoculum of 2000 cells.
Source note: Original NextPrep sample item — pending expert review. This item is original practice material awaiting independent expert review, so confirm important facts in a standard textbook.
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