Question

The doubling time of Escherichia coli is 30 minutes in a culture medium containing glucose and yeast extract. Phage T7 has a life cycle of 20 minutes and a burst size of 200 phage per infected E. coli cell. Phage absorption is instantaneous and occurs at 1 multiplicity of infection (MOI). Bacteria infected with multiple or single phage give the same burst. 5000 plaque forming units of T7 phage are added to a culture of \( 2 \times 10^7 \) E. coli cells. Assuming normal division, the E. coli culture will lyse completely by ........ full cycles of bacterial division. (Answer in integer)}
 

Hint:

In problems involving bacterial growth and phage infection, calculate how many phages are required to infect the entire bacterial culture by considering the burst size and the number of initial bacteria. Each division doubles the number of bacteria.

Answer 1

To solve this problem, we need to determine how many full cycles of bacterial division the {E. coli} culture will undergo before it is completely lysed by the phage.
We are given:
- 5000 plaque forming units (PFU) are added to the culture.
- The burst size is 200 phages per infected {E. coli} cell.
- The initial bacterial count is \( 2 \times 10^7 \) cells.
The number of infected {E. coli} cells after one cycle is calculated as:
\[ {Number of infected cells} = \frac{{Total PFU}}{{Burst size}} = \frac{5000}{200} = 25 \, {cells}. \] Now, for each bacterial division, the number of bacterial cells doubles. After each cycle of division, the number of infected cells increases by a factor of 2. We can calculate the number of cycles required for the culture to be lysed completely by the phage.
For two cycles of bacterial division, the total number of infected cells will be:
\[ 25 \, {cells} \times 2^2 = 25 \times 4 = 100 \, {cells}. \] Thus, after two full cycles of bacterial division, the entire population of {E. coli} will be infected and lysed completely by the phage. Therefore, the culture will lyse completely by 2 full cycles of bacterial division.