Study+Questions+Lecture+13+_+14+transcription+regulation+section+KEY

Study+Questions+Lecture+13+_+14+transcription+regulation+section+KEY

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Study Questions – Lecture 13/14 Transcriptional regulation *Question 1 (MBOG 7-27) Many gene regulatory proteins form dimers of identical or slightly different subunits on the DNA . Suggest two advantages of dimerization. There are at least two advantages of dimerization. First, the binding affinity of a dimer can be higher because the number of potential contacts with DNA is double that possible with a monomer. Second, combinatorial pairing of different subunits can increase the number of DNA- binding specificities that are available to a cell. *Question 2 (MBOG 7-29) The differentiation of muscle cells of the developing embryo is controlled by myogenin, a HLH gene regulatory protein that functions as a heterodimer with another member of the MyoD Family of HLH proteins . The activity of myogenin must be carefully controlled lest it trigger premature expression of the muscle program of cell differentiation. The myogenein gene is turned on in advance of the time when it is needed, but myogenin is prevented from functioning by its tight binding to Id, an HLH protein that lacks a DNA binding domain, and by phosphorylation of its DNA-binding domain Explain how dimerization with Id and phosphorylation of the DNA binding domain might act to keep myogenin nonfunctional. The phosphate interferes with the function of the DNA-binding domain by adding a negative charge and by creating steric problems. Typically, DNAbinding domains are positively charged,which helps them bind to the negatively charged DNA. Addition of a negative chargen would increase charge repulsion between the DNA and the protein, interfering with its function. If the phosphate were added to the binding surface itself, it would likely interfere directly with the interaction of myogenin with the DNA. A heterodimer formed between myogenin and a truncated HLH protein lacking a DNA-binding domain would be unable to bind to DNA tightly because it would make only half of the necessary contacts.
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Question 3 . Shown below is the sequence of a newly identified transcription factor called “DNA121”. The single letter amino acid code for the underlined amino acids in the sequence below is as follows: Q = Glutamine; L = Leucine; R = arginine; K = lysine) Genes encoding altered DNA121 proteins with specific deletions were produced. The genes were evaluated for their ability to activate transcription of a target gene in transient assays. The protein products were also produced by in vitro translation and were tested for DNA binding in gel shift assays. The resulting data are given in the table below. a) Write the letter corresponding to the most likely function of each region of the protein next to the number designating the domain (one letter per blank). Domain 1
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Study+Questions+Lecture+13+_+14+transcription+regulation+section+KEY

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