Answer11 - MCB 161 Study problems S. Harmer Regulation of...

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MCB 161 S. Harmer Study problems Regulation of Transcription I & II 7-29 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 charge 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. References: Benezra R, Davis RL, Lockshon D, Turner DL & Weintraub H (1990) The protein Id: a negative regulator of helix–loop–helix DNA binding proteins. Cell 61, 49–59. Li L, Zhou J, James G, Heller-Harrison R, Czech MP & Olson EN (1992) FGF inactivates myogenic helix–loop–helix proteins through phosphorylation of a conserved protein kinase C site in their DNA-binding domains. Cell 71, 1181–1194. 7-35 A. The shifted DNA molecules migrate to different positions on the polyacrylamide gel because the proteins encoded by the cDNAs are different sizes. Since the migration of protein–DNA complexes depends on the combined molecular mass, the shortest protein that binds (encoded by cDNA clone 2) retards the migration of the DNA fragment the least, and the longest protein (encoded by cDNA clone 4) retards migration the most. B. Since cDNA clones 2, 3, and 4 retard migration of the DNA, they must encode the binding domain for the DNA recognition sequence. The weak binding by the protein encoded by cDNA clone 2 suggests that it may be missing some portion that is essential for normal binding. The absence of binding by the protein encoded by cDNA clone 1 indicates that it does not encode the complete binding domain. Thus an essential portion of the binding domain must map between the 5 ‘ ends of cDNA clones 1 and 2. An additional portion of the binding domain may map between the 5 ‘ ends of clones 2 and 3, as suggested by the weak binding of the clone 2 protein relative to that of the clone 3 protein. These experiments do not define how far the binding domain extends in the 3 ‘ direction. The 3 ‘ end of the binding domain could be defined more precisely using a set of deletions that removed different extents of the 3 ‘ end of the gene, while leaving the 5 ‘ end intact. C. The presence of three distinct shifted bands in the mixture of cDNA clones 3 and 4 indicates that both forms of the protein must be binding to a single DNA molecule, thus retarding its migration to an intermediate extent. This observation suggests that the protein normally binds to its recognition sequence as a dimer. According to this idea, the new band
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Answer11 - MCB 161 Study problems S. Harmer Regulation of...

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