In an experiment employing the methods of the Ames test, two stra...

Question

In an experiment employing the methods of the Ames test, two strains of Salmonella are used. Strain A contains a base-substitution mutation, and Strain B contains a frameshift mutation. Four plates are prepared to test the mutagenicity of the compound ethyl methanesulfonate (EMS). Plate 1 is a control plate with Strain A and S9 extract but no EMS. Plate 2 is also a control plate and contains Strain B and S9 extract but no EMS. Plate 3 contains Strain A along with S9 extract and EMS, and Plate 4 contains Strain B, S9 extract, and EMS.

Characterize the expected distribution of colony growth on the four plates. Defend your growth prediction for each plate.

Accepted Answer

Hi, everybody. Let's look at our next question. Which of the following histidine concentrations is expected to have the isolated highest colony growth of mutants that are able to synthesize histidine in the Aimes test using ethyl methane sulfonate ems choice A no histidine B, low histamine, Concentra concentration, C intermediate histidine concentration or d high histidine concentration. Well, I know I found this one just a little bit tricky to think, think through because it's not the kind of question or the straightforward kind of question we often get asked about the aims test. So let's just kind of think through this situation. So we have our aims test using EMS which is a known mutagen. So we're expecting to have mutant colonies. So as we recall from our content video and our A MS test, we start with mutant colonies of salmonella that cannot synthesize histamine. Then we apply EMS and that's going to sort of reverse mutate a number of the colonies and generate colonies that can synthesize histamine. And our question is asking us in these different situations where we have different levels of histamine from no histamine at all to high histadine concentration. When will we have. And this one's this is kind of important the isolated highest colony growth of mutants that are able to synthesize histamine. So, under which situation will we see the most of these colonies that can synthesize histamine? So, not just bacterial colonies in general, but specifically bacteria that have been mutated by the EMS in a reverse mutation and are now able to synthesize histamine. Well, let's imagine then with each of our concentrations, when we have no histidine as in choice, a, only the reverse mutated colonies can grow in this situation where there's no histamine because our original colonies can't make it for themselves. And therefore, unless they spontaneously generate a mutation, which they can but won't happen very often that won't lead to many colonies. Um They won't grow at all. Whereas in a situation where we have any histidine on the plate, any of these situations in BC or D, low intermediate or high will have some of the original bacteria can grow all the way up to high histine concentration when presumably plenty of the original bacteria can grow because there's lots of histine to go around. So since we're asking about which situation will have the highest colony growth of those mutant, the reverse mutant colonies, we are going to look at choice. A no histamine because in that situation, since they are the only colonies can grow, they will have the highest level of growth. Whenever there's histamine present, they will face competition from the original bacteria and therefore wouldn't have as much growth as you would see when everything else is being screened out. So our answer choice here, which of the following con histadine concentrations is expected to have the isolated highest colony growth of mutants that are able to synthesize histadine. When we're doing an aims test with EMS is choice A no histamine. See you in the next video.

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