7. Superhelical Density Bacteriophage A infects E. coli by integrating its DNA into the bacterial chromosome. The suc- cess of this recombination depends on the topology of the E. coli DNA. When the superhelical density (ơ) of the E. coli DNA is greater than -0.045, the probability of integration is <20%; when o is less than -0.06, the probability is –70%. | Plasmid DNA isolated from an E. coli culture is found to have a length of 13,800 bp and an Lk of 1,222. Calculate o for this DNA and predict the likelihood that bacteriophage A will be able to infect this culture.

7. Superhelical Density Bacteriophage A infects E. coli by integrating its DNA into the bacterial chromosome. The suc- cess of this recombination depends on the topology of the E. coli DNA. When the superhelical density (ơ) of the E. coli DNA is greater than -0.045, the probability of integration is <20%; when o is less than -0.06, the probability is –70%. | Plasmid DNA isolated from an E. coli culture is found to have a length of 13,800 bp and an Lk of 1,222. Calculate o for this DNA and predict the likelihood that bacteriophage A will be able to infect this culture.

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

14th Edition

ISBN:9781305073951

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

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

Chapter8: Dna Structure And Function

Section: Chapter Questions

Problem 3DAA: HersheyChase Experiments The graph shown in FIGURE 8.5 is reproduced from an original 1952...

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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…
HersheyChase Experiments The graph shown in FIGURE 8.5 is reproduced from an original 1952 publication by Hershey and Chase. Bacteriophage were labeled with radioactive tracers and allowed 10 infect bacteria. The virusbacteria mixtures were then whirled in a blender to dislodge any viral components attached to the exterior of the bacteria. Afterward, radioactivity from the tracers was measured. FIGURE 8.5 Detail of Alfred Hershey and Martha Chases 1952 publication describing their experiments with bacteriophage. Infected bacteria refers to the percentage of bacteria that survived the blender. How did the researchers know that the radioisotopes in the fluid came from outside of the bacterial cells and not from bacteria that had been broken apart by whirling in the blender?
HersheyChase Experiments The graph shown in FIGURE 8.5 is reproduced from an original 1952 publication by Hershey and Chase. Bacteriophage were labeled with radioactive tracers and allowed 10 infect bacteria. The virusbacteria mixtures were then whirled in a blender to dislodge any viral components attached to the exterior of the bacteria. Afterward, radioactivity from the tracers was measured. FIGURE 8.5 Detail of Alfred Hershey and Martha Chases 1952 publication describing their experiments with bacteriophage. Infected bacteria refers to the percentage of bacteria that survived the blender. After 4 minutes in the blender, what percentage of each isotope was extracellular?
HersheyChase Experiments The graph shown in FIGURE 8.5 is reproduced from an original 1952 publication by Hershey and Chase. Bacteriophage were labeled with radioactive tracers and allowed 10 infect bacteria. The virusbacteria mixtures were then whirled in a blender to dislodge any viral components attached to the exterior of the bacteria. Afterward, radioactivity from the tracers was measured. FIGURE 8.5 Detail of Alfred Hershey and Martha Chases 1952 publication describing their experiments with bacteriophage. Infected bacteria refers to the percentage of bacteria that survived the blender. The extracellular concentration of which isotope increased the most with blending?
HersheyChase Experiments The graph shown in FIGURE 8.5 is reproduced from an original 1952 publication by Hershey and Chase. Bacteriophage were labeled with radioactive tracers and allowed 10 infect bacteria. The virusbacteria mixtures were then whirled in a blender to dislodge any viral components attached to the exterior of the bacteria. Afterward, radioactivity from the tracers was measured. FIGURE 8.5 Detail of Alfred Hershey and Martha Chases 1952 publication describing their experiments with bacteriophage. Infected bacteria refers to the percentage of bacteria that survived the blender. Do these results imply that viruses inject DNA or protein into bacteria? Why or why not?
HersheyChase Experiments The graph shown in FIGURE 8.5 is reproduced from an original 1952 publication by Hershey and Chase. Bacteriophage were labeled with radioactive tracers and allowed 10 infect bacteria. The virusbacteria mixtures were then whirled in a blender to dislodge any viral components attached to the exterior of the bacteria. Afterward, radioactivity from the tracers was measured. FIGURE 8.5 Detail of Alfred Hershey and Martha Chases 1952 publication describing their experiments with bacteriophage. Infected bacteria refers to the percentage of bacteria that survived the blender. Before blending what percentage of each isotope. 35S and 32P, was extracellular (outside the bacteria)?
Please answer this asap. Thanks, You have discovered a new plasmid RK21 in a unique bacterial community. As a first step towardunderstanding this plasmid, you digest the plasmid with three restriction enzymes: SspI, XhoI andSmaI. You run the digested plasmid DNA on an agarose gel, along with an uncut sample of theRK21 plasmid DNA as a control.Unfortunately you forget to load a DNA ladder, and obtain the following results. Assumecomplete digestion of all samples or all the digests worked completely
P2. Calculate the frictional coefficient of a molecule of DNA of 20 base pairs in water at 20C; assume that the hydrodynamic behavior of the DNA itself can be approximated to be rod-like; the viscosity of the solvent is 0.01 g cm-1 s -1
C. Which of the following statements are true about double-stranded DNA? Write TRUE or FALSE 1. A + G =C + T 4. C = G 2. A+T=C+ G 5. A = G 3. A+ C = T+G
The transformation results below were obtained with 10 ul of intact plasmid DNA at nine concentrations. The following numbers of colonies are obtained when 100 ul of transformed cells are plated on selective medium: Fill in the following table: Concentration # colonies DNA mass of Fraction of Mass Transformation PGREEN (Concentration x volume OR X spread = x 10ul plasmid solution) PGREEN in cell Cell efficiency Y÷ A suspension suspension spread = 100 ul - total vol cell susp. (Colonies - Mass spread) C x Z = A See (510 ul) HINT: this calculation is constant Given= X Given=Y С. Z. 0.00001 ug/ul | 4 0.00005 ug/ul 12 0.0001 ug/ul 0.0005 ug/ul 32 125 0.001 ug/ul 442 0.005 µg/ul 0.01 ug/ul 0.05 ug/ul 0.1 ug/ul 542 507 475 516 0.5 ug/ul 505
For each of the following ( A & B ) provide the method of transfer and a brief explanation as to why the method would not take place under the conditions described . 1. Which method of DNA transfer between bacteria would not take place if the donor and recipient were separated by a filter with a pore size of 0.45 um or another physical barrier 2. Which method of transfer would be blocked by the presence of high concentrations of DNAase ( enzymes capable of degrading DNA ) ?
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