exam-11-key - Bacterial Genetics GMS6038 Final Exam Fall...

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Bacterial Genetics GMS6038 Final Exam Fall 2011 You should be able to complete the exam in 2 hours, but there is extra time allowed if necessary. This is a closed book, closed note exam. All backpacks and notebooks must be against the wall, not at your workstation. Follow the rules of the exam center. You may use the rest room one person at a time. Any cheating will result in a 0 for the exam and failure of the course. THIS IS NOT AN ESSAY EXAM. Thoughtful answers will be short. The points allotted for each part of each question mainly correspond to the number of facts/concepts are being looked for in the answer. For example, a 1 or 2 point question should be able to be answered in a single (compound) sentence. To complete the exam, simply type your text beneath each question. You may use the printed exam form to organize your thoughts. It will be turned in when you are done.
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Shown below is pGULIG-13. Note the following genetic elements: At the top of the map is the origin of replication of the R6K plasmid. Moving clockwise: From 1:00-3:00 the cat gene expressed from the dnaK promoter. From 3:00-5:00 is the pir gene expressed by the Lambda phage pL promoter. At the bottom is the Lambda CI gene expressed by an unnamed promoter that this positively regulated by the V. harveyi LuxR protein. From 7:00-9:00 is the bla gene expressed from the wild-type lac promoter. From 10:00-11:00 is the aph gene driven by the wild-type araBAD promoter. There are no other promoters on this plasmid other than those indicated on the map. Assume that all genes have appropriate translation initiation and termination sequences. The bars crossing the circle represent typical factor-independent terminators that will prevent transcriptional read through between these different elements. The host E. coli K12 strain has the normal chromosomal genes except that it has no F plasmid and no Lambda phage . For purposes of this exam, this E. coli has the complete set of genes for the Vibrio harveyi quorum sensing system in the chromosome, including autoinducer production, sensing, and signal transduction.
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The following questions pertain to pGULIG-13. 1. A. (4 points) If the E. coli culture is growing at a moderately low cell density, explain what is governing the plasmid copy number. The copy number is regulated by the OriR6K origin of replication. This requires the Pir protein to be expressed from the pir gene. The pir gene is expressed by the Lambda left promoter (pL), which will be expressed unless it is repressed by the Lambda CI protein. The CI gene is expressed from a LuxR positively regulated promoter. Therefore, the question is if the LuxR protein will be expressed for activation of CI expression. LuxR is the quorum sensing master regulator, so it will only be expressed at high cell density. At high cell density it will be expressed and will activate CI, which will repress pir, which will prevent plasmid replication (lower copy number). However, the question is about low cell density, so LuxR will not be expressed, CI will not be expressed, and pir will be
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This note was uploaded on 12/13/2011 for the course GMS 6038 taught by Professor Gulig during the Fall '11 term at University of Florida.

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exam-11-key - Bacterial Genetics GMS6038 Final Exam Fall...

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