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Ch. 6 - Genetic Analysis and Mapping in Bacteria and Bacteriophages

Chapter 6, Problem 19

If further testing of the mutations in Problem 18 yielded the following results, what would you conclude about mutant 5? Mutants Results 2, 5 - 3, 5 - 4, 5 -

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Hi everyone. Let's look at our next question. It says the development of resistance among E. Coli in the absence of stimulus such as bacteriophage T. one is called. And we have choices of spontaneous mutation, induced mutation, immediate mutation and continuous mutation. Well our question asked us about the development of resistance in the absence of stimulus. So when you have no stimulus to the to a mutation to develop a mutation, The term we're looking for is choice a spontaneous mutation. So spontaneous mutation is something that can just arise in nature without needing a mutation. Um or the presence of something like the bacteria HT1 to cause that mutation. So let's look through our other answer choices. And induced mutation is one in which you need the presence of a mutation to make that happen. So in this case we're talking about something that arises without anything around. So that's why a spontaneous mutation happens naturally without something that would spur on mutations. So that's why Choice B is not our answer here. Choice C says immediate mutation. Their immediate would mean like right away. But that's not we're not talking about timing. We're talking about whether they arise naturally. So that's why C. Isn't correct choice D. Is continuous mutation in this sense? I think the question to be asking about continuous variation when you have a variation in a trait um along the spectrum. So between two extremes. So something like um height in humans versus blood type. So you're talking about mutations that cause variation um by degrees. But again that's not describe what we're talking about here here, we're talking about the development of resistance which would be, you know, either yes or no the bacteria, the resistant or not. So Choice D. Is not correct. So again, the development of resistance among sequelae in the absence of stimulus such as bacteria of HT1 is called choice a spontaneous mutation. See you in the next video.
Related Practice
Textbook 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
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Textbook Question
In recombination studies of the rII locus in phage T4, what is the significance of the value determined by calculating phage growth in the K12 versus the B strains of E. coli following simultaneous infection in E. coli B? Which value is always greater?
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Textbook Question
In an analysis of rII mutants, complementation testing yielded the following results: Mutants Results (+/- lysis) 1, 2 + 1, 3 + 1, 4 - 1, 5 - Predict the results of testing 2 and 3, 2 and 4, and 3 and 4 together.
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Textbook Question
Using mutants 2 and 3 from Problem 19, following mixed infection on E. coli B, progeny viruses were plated in a series of dilutions on both E. coli B and K12 with the following results. What is the recombination frequency between the two mutants? Strain Plated Dilution Plaques E. coli B 10⁻⁵ 2 E. coli K12 10⁻¹ 5
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Textbook Question
Using mutants 2 and 3 from Problem 19, following mixed infection on E. coli B, progeny viruses were plated in a series of dilutions on both E. coli B and K12 with the following results. Another mutation, 6, was tested in relation to mutations 1 through 5 from Problems 18–20. In initial testing, mutant 6 complemented mutants 2 and 3. In recombination testing with 1, 4, and 5, mutant 6 yielded recombinants with 1 and 5, but not with 4. What can you conclude about mutation 6?
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Textbook Question
During the analysis of seven rII mutations in phage T4, mutants 1, 2, and 6 were in cistron A, while mutants 3, 4, and 5 were in cistron B. Of these, mutant 4 was a deletion overlapping mutant 5. The remainder were point mutations. Nothing was known about mutant 7. Predict the results of complementation (+ or -) between 1 and 2; 1 and 3; 2 and 4; and 4 and 5.
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