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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).

Respuesta :

Answer:

9.378*10^8

Explanation:

The number of phages in each cycle [tex]= 175[/tex]

The number of host infected in each cycle [tex]= 175[/tex]

The number of phages released by each hosts in each turn  [tex]= 175[/tex]

There are total [tex]4[/tex] cycle .

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

[tex]= 175*175*175*175\\= 175^4\\= 9.378*10^8[/tex]

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