A host cell has successfully taken up a recombinant PUC18 plasmid but th plasmid bears mutated origin of replication. Which of the following will happen to that host cell following amplification process on a media containing ampicillin and X-gal? * It fails to form a colony on the media It survives in the media It would not be able to digest X-gal II III O I and Il only O I and III only O Il and III only O I, Il and I

A host cell has successfully taken up a recombinant PUC18 plasmid but th plasmid bears mutated origin of replication. Which of the following will happen to that host cell following amplification process on a media containing ampicillin and X-gal? * It fails to form a colony on the media It survives in the media It would not be able to digest X-gal II III O I and Il only O I and III only O Il and III only O I, Il and I

Biology: The Unity and Diversity of Life (MindTap Course List)

15th Edition

ISBN:9781337408332

Author:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr

Publisher:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr

Chapter15: Studying And Manipulating Genomes

Section: Chapter Questions

Problem 2SQ

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With regard to conjugation, a key difference between F+ and Hfrcells is that an Hfr cella. is unable to conjugate.b. transfers a plasmid to the recipient cell.c. transfers a portion of the bacterial chromosome to the recipient cell.d. becomes an F− cell after conjugation.
You are cloning the genome of a new DNA virus into pUC18.You plate out your transformants on ampicillin plates containing X-gal and pick one blue colony and one white colony.When you check the size of the inserts in each plasmid (blueand white), you are surprised to fi nd that the plasmid fromthe blue colony contains a very small insert of approximately60 bp, while the plasmid from the white colony does notappear to contain any insert at all. Explain these results.
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= 1x10[7] (note 10[7] means 10 to the power of 7). You want to have about 1000 colonies on the P-200 plate. How many nanograms of plasmid should you use in the transformation reaction? Select only one answer. 1. 0.05 2. 20.00 3. 0.50 4. 50.00 5. 200.00 6. 5.00
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
Acridine orange is a chemical that inhibits the replication ofF- factor DNA but does not affect the replication of chromosomalDNA, even if the chromosomal DNA contains an Hfr. Let’s supposethat you have an E. coli strain that is unable to metabolizelactose and has an F factor that carries a streptomycin-resistantgene. You also have an F− strain of E. coli that is sensitive to streptomycinand has the genes that allow the bacterium to metabolize lactose. This second strain can grow on a lactose-containing medium.How would you generate an Hfr strain that is resistant tostreptomycin and can metabolize lactose? (Hint: F factors occasionally integrate into the chromosome to become Hfr strains, and occasionally Hfr strains excise their DNA from the chromosome to become F+ strains that carry an F′ factor.)
A strain with the ability to conjugate is mixed with a recipient strain. Recipients are noted to become recombinant but F-. This indicates a. unsuccessful completion of conjugation b. the donor strain has a defective plasmid c. the donor strain has a plasmid integrated in the chromosome
You are analyzing the intracellular DNA intermediates formed during the replication of the single-stranded (ss) DNA phage M13. In a replication mutant, you find accumulation of DNA molecules with the following properties:Results of a 0.7% agarose gel electrophoresis run at pH 8.0 – gel is running top tobottom, with the origin at the top of the diagram. The smudges indicate several bandswith slightly different migration rates. open circle M13closed circle M13 Left lane — markers (wild-type open circle M13 and closed circle M13)2nd lane — DNA extracted from mutant M13 phage cells3rd lane — DNA extracted from mutant M13 phage cells denatured4th lane — DNA extracted from normal M13 phage cellsRight lane — DNA extracted from normal M13 phage cells denaturedProfiles from a 5 – 20% sucrose gradient centrifuging of the mutant M13 phage DNA The arrow on the neutral (pH 7.4) gradient shows the position of the covalently closed circular DNA. On the alkaline (pH 12.5) gradient, note two peaks…
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A linear DNA fragment and a plasmid has three restriction sites for EcoRIhow many fragments will be produced from linear DNA and plasmid respectively.
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