BglII6. You have a circular plasmid available, with a single RE site for BglII (A^GATCT). You cleave theplasmid with BglII, and join it to both ends of a linear fragment of viral DNA which was obtained bydigesting the viral genome with the RE DpnII (^GATC), resulting in a larger plasmid.☑DpnIITextDpnIIOIf the resulting larger plasmid is incubated with BglII, will the enzyme be able to excise the fragmentof viral DNA? If no, why not? If yes, explain under which conditions. Motivate your answer by showingthe sequence of the plasmid DNA (in black) and the viral DNA (in blue) at both the join positions.Write out the sequences in dsDNA format, if necessary indicate any flanking nucleotides by N.

Excising a Viral DNA Fragment from a Plasmid Using BglII: Detailed Explanation with ReferencesThe…

Answered: BglII 6. You have a circular plasmid…

Biochemistry

6th Edition

ISBN:9781305577206

Author:Reginald H. Garrett, Charles M. Grisham

Publisher:Reginald H. Garrett, Charles M. Grisham

Chapter28: Dna Metabolism: Replication, Recombination, And Repair

Section: Chapter Questions

Problem 4P: Multiple Replication Forks in E. coli II On the basis of Figure 28.2, draw a simple diagram...

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Pstl. EcoRI Origin of replication (ori) Ampicillin Tetracycline resistance resistance (Amp) (Tet") pBR322 (4,361 bp) BamHI Pvull Sall Recombinant Plasmid DNA Bacterial cell contains... No plasmid DNA pBR322 (no insert) Recombinant plasmid 00,000. Host DNA Transformation of E. coli cells +AMP plate pBR322 Figure 7-5 +TET plate Based on the recombinant plasmid growth pattern (bottom row of blue table), which of the depicted plasmid's restriction sites was used to prepare this sample? Explain how you can tell.
gene. If the JM109 strain is transformed by the PBKSK plasmid, the strain will produce the B-galactosidase (from the lac gene) and will hydrolyze X-gal to produce the blue compound. Therefore, colonies that were transformed and contain the pBSKS wil you appear blue. IPTG & X-Gal & NO colonies Amp E. coli JM109 E. coli JM109 50 mM calcium chloride-15% glycerol lac lac lac IPTG & I Recovery X-Gal solution at -702C PBSKS White colonies E. coli JM109 E. coli JM109 ampR amp I amp lac lac Heat Shock Non-transformed 42°C E. coli JM109 E. coli JM109 amps amps lac lac IPTG & X-Gal lac I Recovery lac PBSKS BLUE colonies PBSKS ampRI (amp Transformed IPTG & X-Gal & BLUE colonies Amp Hypotheses: Circle the correct answer 1. If PBSKS is transformed into JM109 cells, colonies will be (able/not able) to grow in the presence of ampicillin. a. Why? _ 2. If PBSKS is transformed into JM109 cells, colonies in media with IPTG (will/will not) induce the production the B- galactosidase enzyme. a. Why?_ 3. If…
COMPLEMENTARY BASE PAIRING 1. A B. C. D. E. F OMPTON 2. A. Under each sequence of bases, write the sequence of bases complementary to each section of DNA in the virus 77174: ATG TGTCA AAGACGGT CCTGTTTTGTA CCGTCAGGATTGA 6 GTTTTCATGCCTCCAAATCTT 10 The DNA of each species has a different base composition. Find the base composition of each species, using what you know about complement Human DNA is approximately 20%C
I-Assume that a circular plasmid is 3200 base pairs in length and has restriction sites for HindIII restriction enzyme at the following locations: 400, 700, 1400, 2600. Give the expected sizes of the restriction fragments following complete digestion.
Multiple Replication Forks in E. coli II On the basis of Figure 28.2, draw a simple diagram illustrating replication of the circular E. coli chromosome (a) at an early stage, (b) when one-third completed, (c) when two-thirds completed, and (d) when almost finished, assuming the initiation of replication at oriC has occurred only once. Then, draw a diagram showing the E. coli chromosome in problem 3 where the E. coli cell is dividing every 20 minutes.
2AQ: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 bluecolony and one white colony. When you check the size of the inserts in each plasmid (blueand white), you are surprised to find that the plasmid from the blue colony contains a verysmall insert of approximately 60 bp, while the plasmid from the white colony does notappear to contain any insert at all. Explain these results.
PCHEM4321. An agarose gel electrophoresis pattern of the plasmid PSPM4321 digestion (restriction) is shown below. Draw a restriction map of a plasmid with the appropriate restriction sites based on the data given below. Hindlll Hindll BamHI +BamHI Figure 1: 1% agarose gel electrophoresis of pCHEM4321 40 24 16 12 12 8 4 4 + |
Genome for C. diphtheriae have about 2,500,000 nucleotides, 87% of them are coding. This ingle circular chromosome contains 2,389 genes from which 2,272 proteins are coded. It does not contain any plasmids. The genome contains Pathogenicity Islands (PAIs), which C. diphtheriae has 13. What is a PAI and what are their characteristics?
You first need to create a plasmid. What are the minimal components necessary to develop this plasmid? In addition to these components, please draw the plasmid. In this illustration, I am looking for the components, the direction of transcription (i.e., the direction the genes will be transcribed), and what should be transcribed last? Also, where would you specifically insert the P. falciparum gene in this plasmid, and why are you checking reading frames in your gene? Finally, please name your plasmid using the correct nomenclature too. You are excited you have this new plasmid; you want to transfect it into P. marinus and provide it as an oral vaccine to laboratory mice. However, even though your supervisor enjoys your enthusiastic attitude, they do not allow you to do this quite yet because all you have is this plasmid. Why doesn’t your supervisor allow you to use the laboratory mouse for this research regarding animal welfare guidelines? Your answer should include the 3 R’s of…
Kpnl 4. Plasmid Z has a size of 7 kb, and the map shows Kpnl (K) and Pstl (P) cut sites relative to each other. This plasmid was digested with three different restriction enzymes 2000 bp 3500 bp Kpnl (K), Pstl (P) and Bgll (B) either alone or in combination and the samples run on an agarose gel as shown below. Where does Bgll (B) cut this plasmid ? Does the plasmid have one recognition site or two for Bg|l? Describe the Bgll cut site in this plasmid relative to the Kpnl cut Plasmid Z -7 kb Pstl site. How many bases to the left or right of the Kpnl cut site would you observe the Bgll cut site. Explain briefly. 1500 bp Pstl Ladder Kpni Psti K/P Bgl K/B KPB 7000 bp 7000 bp 5600 bp 5500 bp 4900 bp 3500 bp 2000 bp 1500 bp 1500 bp 1500 bp 1400 bp 600 bp %3D
You have another circular plasmid. Complete and effective digestion of this plasmid with a restriction enzyme yields three bands: 4kb, 2kb, and 1 kb. In comparing the band intensity on an ethidium bromide-stained gel, you notice that the 4 kb and the 2 kb bands have the exact same brightness. The 1 kb band is exactly one fourth as bright as each of these. (Assume there is uniform staining with ethidium bromide throughout the gel.) How many times did the enzyme cut the plasmid? What is the size of the plasmid? Justify your answers to a and b above using a clearly labeled diagram showing the relative location of the cut-sites on the plasmid.
Help Please Supercoiled DNA migrates through an agorose gel faster than similarly sized linear DNA? True or False? Using the plasmid map, indicate where (bp region on plasmid) these enzymes cut on the plasmid. EcoR1 1. 1 Hindlll 2. 637 Bpu101 3. 14 BamHI 4. 256 To sequence a DNA plasmid, you need 1 µg of DNA. If your DNA concentration was 0.45 µg/µL. How many µl of your DNA 0.45 µg/µL is required to get 1 µg final? Answer in µl using 2 decimal places. Your Answer: Answer units >

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