|
Title: A mutation in E. coli results in the loss of both restriction endonucleases and modification ... Post by: jibbyjam1 on Feb 20, 2018 A mutation in E. coli results in the loss of both restriction endonucleases and modification enzymes. Would you expect any difference in the frequency of gene transfer via transduction FROM Salmonella INTO this E. coli strain?
A. No-since the Salmonella strain is normal, the rate of production of transducing virus particles would still be the same, resulting in the same frequency of gene transfer. B. Yes-the loss of the restriction endonucleases would leave the recipient E. coli unable to break down invading' viral DNA from the transducing phage. This would lead to higher rates of successful transduction. C. Yes-the loss of the modification enzymes would leave the recipient E. coli unable to tag its own DNA as self, leaving the viral DNA untagged and recognizable as foreign, and targeted for destruction. This would lead to higher rates of successful transduction. D. No-transduction efficiency isn't affected by either restriction endonucleases or modification enzymes, so there'd be no effect on the overall rate. E. Yes-the loss of the restriction endonucleases would leave the recipient E. coli unable to break down invading viral DNA from the transducing phage, AND the loss of the modification enzymes would leave the recipient E. coli unable to tag its own DNA as self, leaving the viral DNA untagged and recognizable as foreign, and targeted for destruction. Together, thesewould lead to higher rates of successful transduction. The clustered, regularly interspaced short palindromic repeats (CRISPR) system in bacterial cells has been called the immune system of bacteria. CRISPR protect bacteria from a repeat infection from the same phage because bacterial cells A. recognize proteins on the surface of the phage and secrete enzymes that digest the phage. B. recognize proteins on the surface of the phage and secrete proteins that block the binding of the phage. C. integrate fragments from the phage DNA in their own chromosomes and target for destruction any DNA that contains the same fragments in the future. D. modify the attachment sites for the phages so that new infections cannot take place. Most temperate phages integrate into the host chromosome, whereas some replicate as plasmids. Which kind of relationship do you think would be more likely to maintain the phage in the host cell, and why? A. Plasmids-they're smaller, so they'd be easier to replicate by the host cell. B. Integrated-the host cell would be less likely to view this DNA as foreign on subsequent rounds of replication, and would retain it more easily. C. Plasmids-these structures often carry other genes that may give the host cell a selective advantage over cells that don't have them. D. Integration-because plasmids are frequently lost during cell division, which could leave a daughter cell without the virus genome. You add an unknown phage to a mixture of F+ and F- cells of E. coli and plate out the bacteria. The bacterial colonies that grow are all F-. How can you explain this phenomenon? A. The phage bound to a receptor on the sex pilus, and therefore only infected the F+ cells (leaving the F- cells alone). B. The phage bound to a receptor on the F- cells, leaving only them alive and allowing the F+ cells to die off. C. The phage integrated (lysogenized) the F- cells, giving them a selective advantage over the F+ cells. D. F+ cells are uniquely susceptible to phage attack for unknown reasons. Electron microscopy is useful for counting viruses and distinguishing between infective and non-infective virions. Indicate whether this statement is true or false. Title: A mutation in E. coli results in the loss of both restriction endonucleases and modification ... Post by: Arab12006 on Feb 20, 2018 Content hidden
Title: A mutation in E. coli results in the loss of both restriction endonucleases and modification ... Post by: jibbyjam1 on Feb 20, 2018 This calls for a celebration :raised_hands:
|