Ten E. coli cells with ${}^{15}\mathrm{N}$-dsDNA are incubated in medium containing ${}^{14}\mathrm{N}$ nucleotide. After $60$ minutes, how many E. coli cells will have DNA totally free from ${}^{15}\mathrm{N}$?

E. coli has a doubling time of 20 minutes. Thus, in 60 minutes, there would be 3 cycles of doubling. That means, after 60 minutes, there would be 80 E. coli cells, if we start from 10 initial cells.

In the first cycle, all 20 cells would be hybrid DNA containing ${\mathrm{N}}^{15}/{\mathrm{N}}^{14} \mathrm{state}$. The second cycle would yield some 40 cells and out of those, 20 cells would be totally free from ${\mathrm{N}}^{15}$ DNA. In the last cycle, there would be 80 cells created and 60 would have ${\mathrm{N}}^{14}/{\mathrm{N}}^{14}$ state and only 20 cells would have ${\mathrm{N}}^{15}/{\mathrm{N}}^{14} \mathrm{state}$.