If a single bacteriophage infects one

e. coli cell present on a lawn of bacteria and, upon lysis, yields 325 viable viruses, how many phages will exist in a single plaque if 5 more lytic cycles occur?

Question

Harris

Biology

28 December, 09:20

If a single bacteriophage infects one

e. coli cell present on a lawn of bacteria and, upon lysis, yields 325 viable viruses, how many phages will exist in a single plaque if 5 more lytic cycles occur?

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Answers (1)

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Areli Mosley · Answer 1

The answer is 3,625,908,203,125. This calculated by (325) ^5.

This is exponential growth where the population of the virus doubles in every lytic cycle and decimates the E. coli population. In some hours the lawn of E. coli cells will be completely mowed down. The other life cycle of bacteriophages, other than lytic, is lysogenic cycle that does not lyse the bacteria but only integrates its DNA in the genome and is replicated when the bacteria divides.

If a single bacteriophage infects onee. coli cell present on a lawn of bacteria and, upon lysis, yields 325 viable viruses, how many phages will exist in a single plaque if 5 more lytic cycles occur?

If a single bacteriophage infects one

e. coli cell present on a lawn of bacteria and, upon lysis, yields 325 viable viruses, how many phages will exist in a single plaque if 5 more lytic cycles occur?