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THERMODYNA MIC CONCEPTS For any diagramed equilibrium, you should be able to identify favorable and unfavorable thermodynatin contributions that arisc from structural changes that occur. BREIFLY DESCRIBE (one sentence) 1) ONE STRUCTURAL change that contributes FAVORABLY and 2) ONE STRUCTURAL change that contributes UNFAVORABLY Underncath cach EXPLAIN how the structural change contributes BP SPECIFIC and includo whether and how AH or -TAS is involved ASSUME THE MOST GENERAL CASE (don't make up some unusual feature just to provide an explanation) Assume the temperature is 20°C Ext Protein unfolding FAVORABLE UNFAVORABLE Protein chain goes from having single Protein chain trades interactions conformation to having many conformations. with itself to interacting with water EXPLAIN EXPLAIN Incr easing the number of available Intraprotein interactions are usually Conformational states is entropically favorable enthalpically favorable than peptide- water interactions. FOLDING CONCEPTS Using understandable sentences or phrases (dangling key words will be given NO CREDIT), A. What is the most important element or property of a protein sequence that determines the 3t D path of the polypeptide for particular fold? B. Briefly DESCRIBE how,in molecular terms this (the answer from A)specifies the fold C. Briefly DESCRIBE two different types of interactions that stabilize single compacted structure from the family of structures specified by (the answer from A). EXPLAIN how they do this for full credit. PROTEIN-DNA INTERACTIONS-Binding A stress- -regulated transcription factor DUB1 (1 10 residues, MW~12,000 Da) regulates certain promoters. Oxidation increases DUB1 DNA affinity, which represses the promoters by blocking access to them by RNA polymerase, while heat shock (a shift in temperature from 25°C to 42C°) causes DUB1 to dissociate from the promoter allowing transcription to initiate. Using gel filtration chromatography, the molecular masses of the following species were measured 25°C 42°C DUBI in deoxygenated buffer 24,000 12,000 25 bp DNA site from promoter 16,000 16,000 DUB1 in deoxygenated buffer+1 DNA 40,000 12,000 16,000 DUB1 in air-saturated buffer 24,000 48,000 12,000 24,000 DUB1 peroxide treated 48,000 24,000 DUB1 peroxide treated DNA site 80,000 24,000 16,000 DUB1. peroxide treated DNA site BME 40,000 12,000 16,000 +BME thiol containing reducing agent A fragment of the natural promoter has DNA sequence shown below. The relative intensity of hydoxy- radical cleavage of each strand in the presence of DUB1 is given above the sequence. 5' COGACTGC 3' 1. How many DUB1 proteins bind to this region of the promoter? 2. What is the DNA recognition sequence of DUB1 monomer? 3. What is interesting about the sequence of the footprinted region? A small amount of the above DNA duplex was titrated increasing amounts of deoxygenated DUBI solution in bandshift (EMSA) experiment to give the following gel. The DUB1 concentrations in micromolar (10* M) are given below. as are the relative intensities of the 0.23 0.38 0.46 0.69 0.79 0.80 bands. The arrow indicates the 4a direction of electrophoresis. 4. In the spaces on the right of the gel, indicate the 4b bands corresponding to free 1.0 0.77 0.62 0.54 0.31 0.21 0.20 . DNA ("free") and protein- 2 5 10 15 20 25 [DUB1],uM DNA complex ("bound"). 5. What is the approximate Kn? uM 6. At 25°C, how many kcal/mol binding energy does this correspond to? 7. Considering only the data above, CIRCLE the DNA structure below that most likely appears in the DUB1/25 bp DNA crystal structure. n X Interestingly, in vitro, oxidation does not change DUB1 affinity for the foot printed DNA fragment shown above, but greatly enhances binding to the entire promoter. Deletion of a short 20bp fragment 525 bp upstream of the foot- printed region "fragment B") above reduced bindingto the levels exhibited by the non- Oxidized protein used in the EMSA experiment. The sequence of the fragment is TCAACTAGCACGTTCTAGGA 8. UNDERLINI the bases that would be protected by DUBI in footprinting experiment. 9. There are two reasons that DUB1 does not bind this fragment as well as the fragment on the previous page (Ko for fragment B 10-4M) CIRCLE the two bases that are the reason for the lower affinity. 10. Using one sentence each, explain why the bases cause this affinity change. 11. What is the number of DUB1 monomers bound to the entire promoter? 12. Given the Kp for the dcoxygenated DUB1 to the smaller promoter fragment what is the lowest approximate Kp (highest affinity) you could expect for oxidized DUB1 binding the entire promoter?- 10 M -10°M -10-1°M 13. In reality, the affinity will be somewhat lower than this. Give ONE REASON WHY this is the case (Hint: it has to do with the DNA) 14. What happens to DUB1 stability upon binding to DNA? INCREASES/DECREASES 15. What happens to DUB1 stability upon oxidation? INCREASES/DECREASES Right are AFM A B C topographical images of DUB1 the presence of promoter DNA. . zz 16. Which image corresponds to the following situations? (A, B or C) i. At 2 25 C with peroxide ii. At 25°C, no oxygen iii. .At42°C

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A. A protein sequence or primary structure consists of a long chain of amino acids joined with a strong peptide (covalent) bond. Each one of them has different physiochemical properties i.e. some are polar or non-polar; positively or negatively charged and forms elements such as alpha helix/beta sheets. Three-dimensional structure (conformations) is determined by the physiochemical properties of amino acids, the order of amino acids and types of bonds (covalent or non-covalent). The particular fold consisting of polypeptide chains has three types of weak non-covalent bonds i.e. hydrogen bonds, ionic bonds, van der wall interactions which go parallel with each other and also with covalent bond and determines the stability of protein...

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