Exam2Key-06 - Genetics 511 Exam II 100 Points 1. (32...

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Genetics 511 Exam II Name: 100 Points 1. (32 points) BRIEFLY describe the functions of: TBP (4 points) TBP (TATA-box binding protein): bind TATA box, recognizes promoter and bends DNA to fit to the RNA polymerase to start transcription initiation. TAFs (6 points) -recruit TBP to TATA-less promoters -targets for different activators -[TAF250 is also involved in chromatin modifications (HAT, kinase activitities] TFIIB (3 points) linker (bridge) between TFIID and RNA polymerase II/TFIIF TFIIF (3 points) binds/recruits RNA polymerase II and interacts with non-template DNA TFIIH (6 points) -phosphorylates the C-terminal domain of the largest subunit of RNA polymerase II, which is required for transition from transcription initiation to elongation -causes full melting at the promoter by its helicase activity Mediator complex (4 points) -mediates interactions between transcriptional activators and basic transcription machinery (general transcription factor and RNA polymerase II or holoenzyme) -required for transcription activation by many transcription factors Nucleosome (6 points) -pack DNA to form chromatin -repress transcription -(regulate transcription depending on histone modifications) 2. (14 points): Transcription factor p53 can be ubiquitinated by ubiquitin ligase
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Mdm2, phosphorylated at S46 by HIPK2 and acetylated at several lysine residues (K370, 371, 372, 381 and 382) at the C-terminal. What are the functions of these three modifications (9 points)? How would you prove that phosphorylation at S46 is required for p53 acetylation at the C-terminal (5 points)?
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This note was uploaded on 03/08/2010 for the course GDCB 511 taught by Professor Yanhaiyin during the Spring '09 term at Iowa State.

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Exam2Key-06 - Genetics 511 Exam II 100 Points 1. (32...

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