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

E. coli divides every $20$ mins. In $60$ mins, there are $3$ generations ($n=3$).Total cells $=10\times 2^3=80$ cells.In semi-conservative replication, the original ${}^{15}\text{N}$ strands never disappear; they are just partitioned.Gen 0: $10$ cells (all ${}^{15}\text{N}{/}^{15}\text{N}$).Gen 1 ($20$ min): $20$ cells (all ${}^{15}\text{N}{/}^{14}\text{N}$ hybrid).Gen 2 ($40$ min): $40$ cells ($20$ hybrid, $20$ pure ${}^{14}\text{N}$).Gen 3 ($60$ min): Each hybrid gives $1$ hybrid + $1$ light. Each light gives $2$ light.From $20$ hybrids $\to$ $20$ hybrid + $20$ light.From $20$ lights $\to$ $40$ light.Total light (free from ${}^{15}\text{N}$) $=20+40=60$ cells.