Question

If a single bacteriophage infects one E. coli cell present on a lawn of bacteria and, upon lysis, yields 175 viable viruses, how many phages will exist in a single plaque if 3 more lytic cycles occur? Enter your answer to three significant digits (for example: 1.11*10^3).

Answer 1

9.378*10^8

Step-by-step explanation:

The number of phages in each cycle
`= 175`

The number of host infected in each cycle
`= 175`

The number of phages released by each hosts in each turn
`= 175`

There are total
`4` cycle .

So the total number of phages that exist in a single plaque if 3 more lytic cycles occur

`= 175*175*175*175\\= 175^4\\= 9.378*10^8`