Skip to main content

My Courses

AI Tools Channels Home

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

Previous problem

Klug 12th Edition

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

Problem 18

Chapter 6, Problem 18

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.

Verified Solution

Video duration:

1m

This video solution was recommended by our tutors as helpful for the problem above.

650

views

Was this helpful?

Video transcript

Hello, everyone. Here we have a question asking what is the difference between the R two Mutant pages and the Wild type pages? So the difference between the R two Mutant pages and the Wild type pages is whether or not they can Leis E Coli B or E Coli K 12. So the R two Mutant and the Wild type can both wise E coli B. So both E Coli B but only the wild type can lies the E Coli K Because the R two mutant cannot. So our answer here is D The mutant type can only lie sequelae B but could not lice equal IK 12. Thank you for watching. Bye.

Previous problem Next problem

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?

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

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?

Textbook Question

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 -

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

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?