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# exam3 - UC Berkeley CHEM C130/MCBC100A FALL 2005 MID-TERM...

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QUESTION 1 . (20 points total) Question 1A ( 10 points ) Consider a protein consisting of 100 residues. Assume the following conditions for the process of protein folding in a vacuum (i.e., there are no solvent molecules) : - in the unfolded state each residue in the protein has 4 possible conformations, each of which is allowed (i.e., we neglect problems arising from the chain running into itself). - in the folded state the protein has only one unique conformation - in the unfolded state there are no interactions between residues (except for the covalent bonded structure) What is the entropy change upon folding in vacuum? Specify the units that you are using, and show all the details of the calculation. ANSWER: The number of possible conformations in the unfolded state = 4 100 Assume each conformation has equal energy. Then probability of each individual conformation: p i = 1 4 100 S = Nk B p i conformations ln p i = N k B 1 4 100 ln 1 4 100 i where N= total number of molecules in sample. The sum is over 4 100 identical terms, so: S = N k B 4 100 ln 1 4 100 The entropy per molecule, S N , is: S N = k B ln 1 4 100 The entropy per mole is S = R 100 ( ) ln4 =1.152kJmol -1 K 1 Entropy of folded molecule = 0 S = -1.152 kJ mol -1 K -1 UC Berkeley, CHEM C130/MCBC100A. FALL 2005. MID-TERM EXAM 3. YOUR NAME:_________________ ____________________ Page 2 of 13
Question 1B (5 points) In the folded state each residue makes one favorable interaction with the rest of the protein. The strength of each interaction is the same for all residues. For the situation described in 1A, what is the minimal value of the energy of interaction of each residue in the protein in order for the protein to fold spontaneously in a vacuum? Calculate the value of the energy at 300K, and give your answer in units of kJ mol -1 .

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