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 12EQ

Let’s suppose a new strain of P1 phage has been identified that packages larger pieces of the E. coli chromosome. This P1 strain packages pieces of the E. coli chromosome that are 5 minutes long. If two genes are 0.7 minute apart along the E. coli chromosome, what would be the cotransduction frequency using a normal strain of P1 and using the new strain of P1 that packages larger pieces? What would be the experimental advantage of using this new P1 strain?

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In E. coli, the gene bioD+ encodes an enzyme involved in biotin synthesis, and galK+ encodes an enzyme involved in galactose utilization. An E. coli strain that contained wild-type versions of both genes was infected with P1 phage, and then a P1 lysate was obtained. This lysate was used totransduce (infect) a strain that was bioD− and galK−. The cellswere plated on a medium containing galactose as the sole carbonsource for growth to select for transduction of the galK+ gene.This medium also was supplemented with biotin. The resultingcolonies were then restreaked on a medium that lacked biotin tosee if the bioD+ gene had been cotransduced. The following resultswere obtained:What topic in genetics does this question address?
In E. coli, the gene bioD+ encodes an enzyme involved in biotin synthesis, and galK+ encodes an enzyme involved in galactose utilization. An E. coli strain that contained wild-type versions of both genes was infected with P1 phage, and then a P1 lysate was obtained. This lysate was used totransduce (infect) a strain that was bioD− and galK−. The cellswere plated on a medium containing galactose as the sole carbonsource for growth to select for transduction of the galK+ gene.This medium also was supplemented with biotin. The resultingcolonies were then restreaked on a medium that lacked biotin tosee if the bioD+ gene had been cotransduced. The following resultswere obtained:What information do you know based onthe question and your understanding of the topic?
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.

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Genetics: Analysis and Principles

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