ND STONY BROOK UNIVERSITY LOOSELEAF GENETICS: FROM GENES TO GENOMES
6th Edition
ISBN: 9781260406092
Author: HARTWELL, Leland, HOOD, Leroy, Goldberg, Michael
Publisher: Mcgraw-hill Education/stony Brook University
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Chapter 16, Problem 10P
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.
a. Why did the mating with a nonlysogenic recipient result in phage growth and release, but the infection of a lysogenic recipient did not? |
b. Explain how this phenomenon relates to the PaJaMo experiment in Fig. 16.6. |
c. Explain how this phenomenon relates to hybrid dysgenesis, described in Problem 29 of Chapter 13. |
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E. coli cells are simultaneously infected with two strains of phage λ. One strain has a mutant host range, is temperature sensitive, and produces clear plaques (genotype h st c); another strain carries the wildtype alleles (genotype h+ st+ c+). Progeny phages are collected from the lysed cells and are plated on bacteria. The following numbers of different progeny phages are obtained: Progeny phage genotype Number of plaques h+ c+ st+ 321 h c st 338 h+ c st 26 h c+ st+ 30 h+ c st+ 106 h c+ st 110 h+ c+ st 5 h c st+ 6 a. Determine the order of the three genes on the phage chromosome. b. Determine the map distances between the genes. c. Determine the coefficient of coincidence and the interference
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Chapter 16 Solutions
ND STONY BROOK UNIVERSITY LOOSELEAF GENETICS: FROM GENES TO GENOMES
Ch. 16 - For each of the terms in the left column, choose...Ch. 16 - The following statement occurs early in this...Ch. 16 - One of the main lessons of this chapter is that...Ch. 16 - All mutations that abolish function of the Rho...Ch. 16 - The figure at the beginning of this chapter shows...Ch. 16 - The promoter of an operon is the site to which RNA...Ch. 16 - You are studying an operon containing three genes...Ch. 16 - You have isolated a protein that binds to DNA in...Ch. 16 - You have isolated two different mutants reg1 and...Ch. 16 - Bacteriophage , after infecting a cell, can...
Ch. 16 - Mutants were isolated in which the constitutive...Ch. 16 - Suppose you have six strains of E. coli. One is...Ch. 16 - The previous problem raises some interesting...Ch. 16 - For each of the E. coli strains containing the lac...Ch. 16 - For each of the following growth conditions, what...Ch. 16 - For each of the following mutant E. coli strains,...Ch. 16 - Maltose utilization in E. coli requires the...Ch. 16 - Seven E. coli mutants were isolated. The activity...Ch. 16 - Cells containing missense mutations in the crp...Ch. 16 - Six strains of E.coli mutants 16 that had one of...Ch. 16 - a. The original constitutive operator mutations in...Ch. 16 - In an effort to determine the location of an...Ch. 16 - Prob. 23PCh. 16 - The footprinting experiment described in Fig....Ch. 16 - Why is the trp attenuation mechanism unique to...Ch. 16 - a. How many ribosomes are required at a minimum...Ch. 16 - The following is a sequence of the leader region...Ch. 16 - For each of the E. coli strains that follow,...Ch. 16 - Prob. 29PCh. 16 - For each element in the list that follows,...Ch. 16 - Among the structurally simplest riboswitches are...Ch. 16 - Great variation exists in the mechanisms by which...Ch. 16 - Many genes whose expression is turned on by DNA...Ch. 16 - In 2005, Frederick Blattner and his colleagues...Ch. 16 - The E.coli MalT protein is a positive regulator of...Ch. 16 - Prob. 36PCh. 16 - Prob. 37PCh. 16 - Prob. 38PCh. 16 - Prob. 39PCh. 16 - Prob. 40PCh. 16 - Prob. 41PCh. 16 - The researchers who investigated bioluminescence...Ch. 16 - Prob. 43PCh. 16 - Quorum sensing controls the expression of...Ch. 16 - Scientists are currently screening a chemical...
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- Bacterial cells were coinfected with two types of bacteriophage lambda: One carried the c+ allele and theother the c allele. After the cells lysed, progeny bacteriophage were collected. When a single such progeny bacteriophage was used to infect a new bacterialcell, it was observed in rare cases that some of theresulting phage progeny were c+ and others were c.Explain this result.arrow_forwardDetermine the map distances between the genes.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
- The 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_forwardConsider 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_forward
- The 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_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. 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 information do you know based onthe question and your understanding of the topic?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 topic in genetics does this question address?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_forwardDetermine the coefficient of coincidence and the interferencearrow_forward
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