2008 Exam 3 with Answers - BMB/ MICRB 460 Exam 3 Form A...

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Unformatted text preview: BMB/ MICRB 460 Exam 3 Form A April 10, 2008 Your name:________________________________________ 1. In this schematic of the events required to proceed from G0, through G1 and into S phase, I have left out several details after E2F is held inactive. Describe (or diagram) the molecular events leading to activation of E2F so that it can activate synthesis for S phase genes as specified in the last step. A. Quiescent cell + Growth Factor B. Signal transduction pathway (RTK, ras, raf, MAPK, etc) C. Activation of transcription factors by phosphorylation (TCF, SRF, etc) D. Transcription of early response genes E. Transcription of delayed response genes F. E2F is held inactive E2F stimulates synthesis of genes for S phase CyclinD/cdk levels rise Rb is phosphorylated E2F is released, and goes to nucleus to transcribe S phase genes positive feedback loop: newly made cyclin E phosphorylates Rb to release more E2F 2. Cyclohexamide blocks protein synthesis. Examine the schematic above and pick the true statement. a. Cyclohexamide would block the pathway between step A and B. a. Cyclohexamide would block the pathway between step B and C a. Cyclohexamide would block the pathway between step C and D. a. Cyclohexamide would block the pathway between step D and E. a. Cyclohexamide would block the pathway between step E and F. 3. A.Describe a method used by researchers to arrest yeast cells in a specific stage of the cell cycle. temperature-sensitive (conditional) mutation B. How can the cell cycle be reinitiated? lower temperature to permissive temp for growth Form A NAME:_____________________________________ 4. Cyclin levels oscillate during a normal cell cycle. The primary mechanism for decreasing cyclin levels is a. Inhibition of transcription a. Inhibition of protein translation a. Protein degradation a. Inhibitory protein phosphorylation a. Binding of CKIs (cyclin kinase inhibitors) such as p21 5. You can draw which of the following conclusions from the Lee and Nurses Fig. 2a? Fig. 2a: Southern blot of DNA prepared from a wild-type yeast strain (wt) or the cdc2Sp deletion strain containing a rescuing cDNA ( ). Lane 1, probed with yeast cdc2(2Sp); lane 2, wt probed with 2Sp; lane 3, probed with human CDC2 (2Hs); lane 4, wt probed with 2Hs. BamH I digestion generates a 7 kb fragment containing the cdc2Sp gene and a 2kb fragment containing the human cDNA. a. The human CDC2Hs gene is much smaller than the yeast cdc2 gene. a. The human and yeast cdc2 genes are quite different in nucleic acid sequence. a. The cells from lanes marked have their yeast cdc2 sequences deleted and contain the cDNA insert coding for the human CDC2 gene. a. Both A and B a. Both A and C a. Both B and C 6. Take a look again at Lee and Nurses Fig. 2a (above)? What would have been the authors conclusion if they had observed a band in lane 4....
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This note was uploaded on 10/19/2009 for the course B M B 460 taught by Professor Hanna-rose,w during the Spring '08 term at Pennsylvania State University, University Park.

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2008 Exam 3 with Answers - BMB/ MICRB 460 Exam 3 Form A...

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