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

Answer:

9.378*10^8

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

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