If a single bacteriophage infects one E. coli cell present on a lawn of bacteria and, upon lysis, yields 200 viable viruses, how many phages will exist in a single plaque if three more lytic cycles occur?

Everyone. Let's take a look at this question together, a single bacteriophage produces 50 viable viruses. Upon infecting a single equal eye that is present on a bacterial lawn, identify the total number of bacteriophages that one will observe on a single plaque. If in total three lyrics cycles occur. So we know that a single bacterial fage produces 50 viable viruses. So we have one bacterial fage equals 50 viable virus. It's And so we know that we have three little cycles occurring. So then we do 50 to the power of three Because there are those three lyrics cycles. And so we end up with 125,000 as the total number of bacteriophages that we will observe on a single plaque. Because in each lyric cycle we observe 50 viable viruses. And because we have three cycles, we multiply 50 by itself three times or 50 to the power of three, which gives us 125,000 as the total number of bacteriophages which is answer choice D The correct answer. I hope you found this video to be helpful. Thank you and goodbye.