If a single bacteriophage infects one E. coli cell present on a l...

Question

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?

Dilution Factor Assay Results
(a) 10⁴ All bacteria lysed
(b) 10⁵ 14 plaques
(c) 10⁶ 0 plaques

Accepted Answer

Hello everyone and welcome to today's video. So which of the following is counted in plaque assay. Well remember from previous videos that like I say is going to involve whole cells and it's going to involve involve viruses being incubated together. And this is going to allow the virus to attach and enter the host cell to infected. And then eventually in the life cycle of the virus license is going to be observed. Which is going to release many viruses into the plate. And then we're going to be measuring the license in this plate. So let's go over now or answer choices so that we may solve this problem. Answer choice A states the number of viruses that can be eliminated by an antibiotic here. We never mentioned the presence of an antibiotic. So we're going to cancel this out. Then we have the rate of growth of viruses in relation to the nutrients available. Well this is not what we're really measuring. Remember that we're measuring the license. So we're going to cancel this out as well. And Tracy says the sensitivity of the virus to a temporary change in the explanation that we have here. We never mentioned temperature changes or measuring this. So we're going to cancel this out and then we're left with answer choice. The number of viruses that light state their host cells. Well, according to our explanation, this is exactly what we're measuring. So we're going to highlight answer choice C as the final answer to our question. I really hope this video helped you and I have to see you on the next one

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