Concepts of Genetics (12th Edition)
Concepts of Genetics (12th Edition)
12th Edition
ISBN: 9780134604718
Author: William S. Klug, Michael R. Cummings, Charlotte A. Spencer, Michael A. Palladino, Darrell Killian
Publisher: PEARSON
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Chapter 6, Problem 20PDQ

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.

  1. (a) What is the recombination frequency between the two mutants?

Chapter 6, Problem 20PDQ, Using mutants 2 and 3 from Problem 19, following mixed infection on E. coli B, progeny viruses were

  1. (b) 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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Consider the following experiment. First, large populations of two mutant strains of Escherichia coli are mixed, each requiring a different, single amino acid. After plating them onto a minimal medium, 45 colonies grew. Which of the following may explain this result? A) The colonies may be due to back mutation (reversion). B) The colonies may be due to recombination. C) Either A or B is possible. D) Neither A nor B is possible.
A fluctuation test was carried out to determine the rate of mutation to Azetidine resistance (a toxic proline analog) in S. typhimurium. Twenty tubes of rich medium were each inoculated with a few wild-type cells and the cultures were grown to 10° cells / ml. A 0.1 ml sample of each culture was then plated on each plate (a total of 20 plates) to detect AztR mutants. The results are shown in the following table. Calculate the mutation rate of S. typhimurium to AztR. Culture # # AztR mutants Culture # # AztR mutants 11 12 3 4. 13 14 15 4. 30 303 97 69 14 16 17 18 19 20 10 19
Four E. coli strains of genotype atb¯ are labeled 1, 2, 3, and 4. Four strains of genotype a¯b* are labeled 5, 6, 7, and 8. The two genotypes are mixed in all possible combinations and, after incubation, are plated to determine the frequency of a*b+ recombinants. The following results are obtained, where M=many recombinants, L=low number of recombinants, and 0=no recombinants: 1 3 4 M M 6. M M L M 8 L L Based on these results, assign a sex type to each strain. Answer Bank 2 F- Hfr F+ 3 4

Chapter 6 Solutions

Concepts of Genetics (12th Edition)

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genetic recombination strategies of bacteria CONJUGATION, TRANSDUCTION AND TRANSFORMATION; Author: Scientist Cindy;https://www.youtube.com/watch?v=_Va8FZJEl9A;License: Standard youtube license