Genetics: From Genes to Genomes, 5th edition
5th Edition
ISBN: 9780073525310
Author: Leland H. Hartwell, Michael L. Goldberg, Janice A. Fischer, Leroy Hood, Charles F. Aquadro
Publisher: McGraw-Hill Education
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Chapter 15, Problem 9P
Bacteriophage λ, after infecting a cell, can integrate into the chromosome of the cell if the repressor protein, cI, binds to and shuts down phage transcription immediately. (A strain containing a bacteriophage DNA integrated into the chromosome is called a lysogen.) The alternative fate is the production of many more viruses and lysis of the cell. In a mating, a donor strain that is a lysogen was crossed with a lysogenic recipient cell, and no phages were produced. However, when the lysogen donor strain transferred its DNA to a nonlysogenic recipient cell, the recipient cell burst, releasing a new generation of phages. Why did the mating with a nonlysogenic cell result in phage growth and release, but the infection of a lysogenic recipient did not?
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Chapter 15 Solutions
Genetics: From Genes to Genomes, 5th edition
Ch. 15 - For each of the terms in the left column, choose...Ch. 15 - The following statement occurs early in this...Ch. 15 - One of the main lessons of this chapter is that...Ch. 15 - All mutations that abolish function of the Rho...Ch. 15 - The promoter of an operon is the site to which RNA...Ch. 15 - You are studying an operon containing three genes...Ch. 15 - Prob. 7PCh. 15 - You have isolated two different mutants reg1 and...Ch. 15 - Bacteriophage , after infecting a cell, can...Ch. 15 - Mutants were isolated in which the constitutive...
Ch. 15 - For each of the E. coli strains containing the lac...Ch. 15 - For each of the growth conditions listed, what...Ch. 15 - For each of the following mutant E. coli strains,...Ch. 15 - Maltose utilization in E. coli requires the...Ch. 15 - Seven E. coli mutants were isolated. The activity...Ch. 15 - Prob. 16PCh. 15 - Six strains of E.coli mutants 16 that had one of...Ch. 15 - a. The original constitutive operator mutations in...Ch. 15 - Prob. 19PCh. 15 - Figure 15.16 on p. 525 shows that in the lac...Ch. 15 - Why is the trp attenuation mechanism unique to...Ch. 15 - a. How many ribosomes are required at a minimum...Ch. 15 - The following is a sequence of the leader region...Ch. 15 - For each of the E. coli strains that follow,...Ch. 15 - One mechanism by which antisense RNAs act as...Ch. 15 - For each element in the list that follows,...Ch. 15 - Great variation exists in the mechanisms by which...Ch. 15 - Many genes whose expression is turned on by DNA...Ch. 15 - Prob. 29PCh. 15 - The E.coli MalT protein is a positive regulator of...Ch. 15 - Prob. 31PCh. 15 - To find genes that are turned on or off in...Ch. 15 - Prob. 33PCh. 15 - Prob. 34PCh. 15 - Prob. 35PCh. 15 - The researchers who investigated bioluminescence...Ch. 15 - Prob. 37PCh. 15 - Quorum sensing controls the expression of...
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- In conjugation, when a bacterial F factor is transferred: the donor cell turns into a recipient cell the donor cell becomes F- chromosomal DNA is transferred in the mating the recipient cell becomes F- the recipient cell becomes F+arrow_forwardThe linear dsDNA genome of λ binds on the LamB receptor of E. Coli and conducts a normal lysogenic cycle. Exposure to stress will cause the excision of λ prophage from the E. Coli genome. The excised λ genome is then replicated, packaged, and released from the cell as mature λ phage particles and ready to infect other bacterial cells. Among λ phage particles,the transducing phage mediates a specific type of recombination. Understand this scenario and answer the following questions. 1. What are the basic requirements for the insertion of λ into the E. Coli genome? 2. What special features are found in the λ insertion site? 3. What type of recombination occurs with λ insertion in the E. Coli genome? 4. How you will differentiate λ transducing phage from normal λ phage? 5. What exclusive mechanism λ phage utilizes for recombination?arrow_forwardWhen various strains of λ phage are seeded on a lawn of E. coli, they can form clear or turbid plaques. (b) For mutant λ phages that can only form clear plaques, give two different types of mutation in the phage that can explain the clear plaque phenotype.arrow_forward
- Consider three genes in E. coli: thr+, ara+, and leu+ (which give the cell the ability to synthesize threonine, arabinose, and leucine, respectively). All three of these genes are close together on the E. coli chromosome. Phages are grown in a thr+ ara+ leu+ strain of bacteria (the donor strain). The phage lysate is collected and used to infect a strain of bacteria that is thr− ara− leu −. The recipient bacteria are then tested on selective medium lacking leucine. Bacteria that grow and form colonies on this medium (leu+ transductants) are then replica-plated on medium lacking threonine and on medium lacking arabinose to see which are thr+ and which are ara+. Another group of the recipient bacteria are tested on medium lackingthreonine. Bacteria that grow and form colonies on this medium (thr+ transductants) are then replica-plated on medium lacking leucine and onto medium lacking arabinose to see which are ara+ and which are leu+. Results from these experiments are as follows:…arrow_forwardA donor strain of bacteria with alleles a* b* c* is infected with phages to map the donor chromosome using generalized transduction. The phage lysate from the bacterial cells is collected and used to infect a second strain of bacteria that are a b c. Bacteria with the a* allele are selected, and the percentage of cells with cotransduced b* and c* alleles are recorded. Selected Cells with cotransduced Donor Recipient allele allele (%) a* b* a b c a* 25 b* c* a* 3 c* Is gene b or gene c closer to gene a? Explain your reasoning.arrow_forwardBelow is a diagram of the general structure of the bacteriophagel chromosome. Speculate on the mechanism by which it forms aclosed ring upon infection of the host cell. 5'GGGCGGCGACCT:double@stranded region-3' 3'- double@stranded region:CCCGCCGCTGGA5'arrow_forward
- A donor strain of bacteria with alleles a+ b+ c+ is infected with phages to map the donor chromosome using generalized transduction. The phage lysate from the bacterial cells is collected and used to infect a second strain of bacteria that are a− b− c−. Bacteria with the a+ allele are selected, and the percentage of cells with cotransduced b+ and c+ alleles are recorded. Donor Recipient Selected allele Cells with cotransduced allele (%) a+ b+ c+ a− b− c− a+ 25 b+ a+ 3 c+ Is gene b or gene c closer to gene a? Explain your reasoning.arrow_forwardIn 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?arrow_forwardThe figure below shows a partial chromosome map of an E. coli Hfr strain. Each mark = 10 minutes between conjugation transfer time. If transfer of genes begins at “*” relative to the origin of transfer, what is one of the predicted results from this map? It would take less than 30 minutes to transfer all of the genes that are shown. gal and azi will rarely be transferred together. gal and ton will rarely be transferred together. Ten minutes after transfer of ton, lac will be transferred. This strain will produce very few gal recombinants.arrow_forward
- 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?arrow_forwardNine rII− mutants of bacteriophage T4 were used inpairwise infections of E. coli K(λ) hosts. Six of themutations in these phages are point mutations; theother three are deletions. The ability of the doubly infected cells to produce progeny phages in large numbers is scored in the following chart.1 2 3 4 5 6 7 8 91 − − + + − − − + +2 − + + − − − + +3 − − + − + − −4 − + − + − −5 − − − + +6 − − − −7 − + +8 − −9 −The same nine mutants were then used in pairwise infections of E. coli B hosts. The production of progenyphages that can subsequently lyse E. coli K(λ) hosts isnow scored. In the table, 0 means the progeny do notproduce any plaques on E. coli K(λ) cells; − meansthat only a very few progeny phages produce plaques;and + means that many progeny produce plaques(more than 10 times as many as in the − cases).1 2 3 4 5 6 7 8 91 − + + + + − − + +2 − + + + + − + +3 0 − + 0 + + −4 − + − + + +5 − + − + +6 0 0 − +7 0 + +8 − +9 −a. Which of the mutants are the three deletions? Whatcriteria did…arrow_forwardA donor strain of bacteria with genotype leu+ gal− pro+ is infected with phages. The phage lysate from the bacterial cells is collected and used to infect a second strain of bacteria that are leu− gal+ pro−. The second strain is selected for leu+, and the following cotransduction data are obtained: Donor Recipient Selected allele Cells with cotransduced allele (%) leu+ gal− pro+ leu− gal+ pro− leu+ 47 pro+ leu+ 26 gal− Which genes are closest, leu and gal or leu and pro?arrow_forward
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