A batch culture starts with 2 cells. Its first hour is a lag phase without multiplication; subsequently the cells grow exponentially with a one-hour doubling time. How many cells are present after 8 hours? Enter an integer.
The total time includes the lag. Active exponential growth therefore lasts $8-1=7$ hours, which is seven doublings. Cell number is $N=N_0 2^{t/t_d}=2(2^7)=\mathbf{256}$. Counting eight doublings would incorrectly include the nonmultiplying lag period.
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