In a study, bacterial culture of E. coli was infected with bacteriophage. How-ever, these cultures were protected from phage infection. Interestingly, it wasfound that the resistance is due to endonudeases present in E. coli, whichdeaved the phage DNA. How is the E. coli genomic DNA protected from theaction of these endonuclease enzymes?

A restriction endonuclease is a bacterial enzyme that cuts double stranded DNA by cleaving the…

Answered: In a study, bacterial culture of E.…

Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

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A pure culture of an unknown bacterium was streaked onto plates of a variety of media. You notice that the colony morphologyis strikingly different on plates of minimal media with glucose compared to that seen on trypticase soy agar plates. How can you explain these differences in colony morphology? Also, describe what happens when a nonsense mutation is introduced into the gene encoding transposase within a transposon and why is it more likely that insertions or deletions will be more detrimental to a cell than point mutations?
An ade+ arg+ cys+ his+ leu+ pro+ bacterial strain is knownto be lysogenic for a newly discovered phage, but the siteof the prophage is not known. The bacterial map isleucysarghisadeproThe lysogenic strain is used as a source of the phage, andthe phages are added to a bacterial strain of genotypeade- arg- cys- his- leu- pro-. After a short incubation,samples of these bacteria are plated on six differentmedia, with the supplementations indicated in thefollowing table. The table also shows whether colonieswere observed on the various media.PresenceMedium Ade Arg Cys His Leu Pro of colonies1 - + + + + + N2 + - + + + + N3 + + - + + + C4 + + + - + + N5 + + + + - + C6 + + + + + - NNutrient supplementation in medium(In this table, a plus sign indicates the presence of anutrient supplement, a minus sign indicates that asupplement is not present, N indicates no colonies, and Cindicates colonies present.)a. What genetic process is at work here?b. What is the approximate locus of the prophage?
The transfection reagent used to introduce plasmids to the HEK293 cells was a lipid reagent. Why is lipid utilized for this procedure?
Bacterial conjugation has medical consequences. For example, certain plasmids contain genes that code for toxins. The causative agent of a deadly form of food poisoning, E. coli 0157, synthesizes a toxin that causes massive bloody diarrhea and kidney failure. This toxin is now believed to have originated in Shigella, another bacterium that causes dysentery. Similarly, the growing problem of antibiotic resistance is partly attributable to the spread of antibioticresistant genes among bacterial populations. Antibiotic resistance develops because antibiotics are overused in medical practice and in livestock feeds. Suggest a mechanism by which this extensive use promotes antibiotic resistance.
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 E. coli, Base excision repair (BER) handles non-bulky lesions (e.g., deamination, oxidation) and nucleotide excision repair (NER) handles bulky lesions (e.g., Py dimers). Is this statement true or false? O True False
Using the plasmid map of pBCH2.0 provided above, predict how many DNA fragments would be formed if this plasmid was digested with restriction enzyme BamHI.
In a bacterial culture in which all cells are unable to synthesizeleucine (leu-), a potent mutagen is added, and the cells areallowed to undergo one round of replication. At that point, samplesare taken, a series of dilutions are made, and the cells areplated on either minimal medium or minimal medium containingleucine. The first culture condition (minimal medium) allowsthe growth of only leu+ cells, while the second culture condition(minimal medium with leucine added) allows growth of all cells.The results of the experiment are as follows: Culture Condition Dilution ColoniesMinimal medium 10-1 18Minimal medium + leucine 10-7 6What is the rate of mutation at the locus associated with leucinebiosynthesis?
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?
We transformed E coli cells with a plasmid modified to contain a 'virulence factor' which would allow growth on media containing the antibiotic kanamycin (Kan). The plasmid confers constitutive resistance to ampicillin (Amp) Assume you were given competent cells of known transformation efficiency (TE). Assume TE= 1×10[6] (note 10[6] means 10 to the power of 6). You want to have about 1000 colonies on the P-200 plate. How many nanograms of plasmid should you use in the transtormation reaction? О 1.5.00 O 2.0.05 О 3.50.00 О 4.200.00 О 5.20.00 О 6.0.50
A researcher has a Trp− auxotrophic strain of E. coliwith a mutation in a single gene. To identify thatmutant gene, she uses a genomic library made from a wild-type version of that same strain to find plasmids that rescue the mutant phenotype. The result issurprising. She recovers 10 plasmids that provide aTrp+ phenotype, but six of the plasmids contain geneX, while the other four contain gene Y. Our scientisthas encountered a phenomenon called multicopysuppression, related to the fact that plasmids are usually present in several copies per bacterium. Becausethe genes in the plasmids are present in more thantheir usual single copy in the bacterial chromosome,more than the usual amount of Protein X or Protein Yis being produced from the plasmids. Sometimes,overexpression of one protein can rescue the mutantphenotype caused by loss of a different protein.Suggest at least two ways that our scientist could determine which of the two genes, gene X or gene Y, actually corresponds to the…
In a process of production of a recombinant protein by E. coli cells, it was observed accumulation of acetate in the culture medium. In this situation, it can be said that: (a) certainly the process in question was being conducted in anaerobiosis (B).Acetate accumulation is advantageous for the process as the acetate formation reaction generates 1 molecule of ATP (c)Knowing that decreasing the temperature of the process causes a reduction in the rate of glycolysis, this could be a strategy to reduce the accumulation of acetate (d).the acetate formed can be re-assimilated by the cell if the glyoxylate pathway is activated at some point in the culture

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