F09 Exam 4 key-1

F09 Exam 4 key-1 - BSCI 110A Exam 4 December 2, 2009 Print...

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BSCI 110A Exam 4 December 2, 2009 Print Name _________________________________ Signature Pledge___________________________________ 1. Read all questions carefully. 2. You will have 50 minutes to complete this exam. 3. Do not start until directed. 4. Good luck!
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1. (6) In words or diagrams, explain the difference between a monocistronic transcript with introns and exons versus a polycistronic transcript. Eukaryotic protein coding genes are split into exons and introns which must be spliced to create a single, monocistronic open reading frame. Prokaryotic protein coding genes are often transcribed as a group of open reading frames on one polycistronic transcript. 2. Predict the effect of the following mutations on eukaryotic transcription and explain why: A. (2) Mutation of the TATA box to GAGA Decreased levels of transcription because TBP cannot bind the TATA box. B. (2) Deletion of 25 bases between the original start site (+1) and the TATA box The start of transcription will move up ~25 nucleotides. C. (2) Pol I transcription levels in the presence of a mutation that deletes TBP No transcription because Pol I uses TBP to initiate. 3. The 5’ ends of eukaryotic mRNAs are modified with a cap structure. A. (2) What is the cap? 7-methyl-GTP or 7-me-G or 7-methyl Guanosine B. (4) Describe two functions for the cap. RNA stability (protection from nucleases and degradation) and for translation initiation 4. (2) The start codon for translation is AUG meaning all proteins should begin with methionine. You purified a protein from the endoplasmic reticulum and found that its first amino acid was not methionine. How can this be? Proteins in the ER are targeted there by a signal peptide which gets cleaved off by signal peptidase
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This note was uploaded on 11/12/2010 for the course BSCI BSCI 110A taught by Professor Zwiebel during the Fall '09 term at Vanderbilt.

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F09 Exam 4 key-1 - BSCI 110A Exam 4 December 2, 2009 Print...

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