Genetics: Analysis and Principles
Genetics: Analysis and Principles
6th Edition
ISBN: 9781259616020
Author: Robert J. Brooker Professor Dr.
Publisher: McGraw-Hill Education
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Chapter 7, Problem 10EQ
Summary Introduction

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An experiment must be designed that could show that the recombinant bacterium has been transduced or transformed.

Introduction:

Bacteria can transmit its genetic matter naturally by various mechanisms, one of which is transduction and another one is transformation. Transduction is a technique by which the bacterial genetic material moves one bacterial cell to another via bacteriophages. The transformation is a technique, which takes foreign DNA (deoxyribonucleic acid) from its surrounding environment.

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In Hershey-Chase experiment, bacteriophages protein coats were tagged with radioactive isotope S-32. These phages were used to infect E. coli cells and the cells were further centrifuged to form pellets. Why was the radioactivity level of S-32 found greater outside the cells compared to the E. coli cell pellets? Explain briefly. If the experiment is repeated in the same manner but this time the phage protein coats are labelled with isotope X and the phage DNA with isotope Y, which isotope’s radioactivity will be found in greater amounts in the E. coli cell pellets after centrifugation? Explain briefly.
In order to determine the genetic material of a T2 phage, Alfred Hershey and Martha Chase conducted experiments using T2 phages that infected bacteria. In one treatment, they grew phages with radioactive sulfur. In another treatment, they grew phages with radioactive phosphorous. They allowed both types of phages to infect bacterial cells. After infection, they found that only bacteria infected with phages grown with radioactive phosphorous showed any radioactivity. Why did they use radioactive sulfur and phosphorous for this experiment? * O Sulfur is part of the DNA molecule but not part of a protein molecule. Sulfur and phosphorous are some of the most reactive molecules and are easily traced. Sulfur and phosphorous are able to survive the centrifuge, a crucial component of the experiment. O Phosphorous is part of the DNA molecule but not part of a protein molecule.
A researcher is studying the rII locus of phage T4. FourrII− strains are obtained: A, B, C, and D. In the first experiment, E. coli strain K(λ) is coinfected with two rII− strains simultaneously and the results are recorded.    Infection with A and B phage = lysis occurs  Infection with A and C phage = lysis occurs  Infection with B and C phage = no lysis occurs  Infection with B and D phage = no lysis occurs  Infection with C and D phage = no lysis occurs    In a second experiment, coinfections are performed first in E. coli strain B, then the progeny phage are used to infect E. coli strain K(λ).    Progeny of A and B phage = plaques form  Progeny of B and C phage = plaques form  Progeny of C and D phage = plaques form  Progeny of B and D phage = no plaques from Which conclusions are consistent with these data? Why?  A) Strains A and B carry mutations in the same gene.      B) Strains B and D both carry the same mutation. C) Strains B, C, and D carry mutations in the same gene. D)…

Chapter 7 Solutions

Genetics: Analysis and Principles

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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