exam3 - UC Berkeley, CHEM C130/MCBC100A. FALL 2005....

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UNIVERSITY OF CALIFORNIA, BERKELEY CHEM C130/MCB C100A MIDTERM EXAMINATION #3. November 17, 2005 INSTRUCTOR: John Kuriyan THE TIME LIMIT FOR THIS EXAMINATION IS 1 HOUR AND 20 MINUTES SIGNATURE: _________________________________________________________________ Please SIGN your name ( in indelible ink ) on the line above. YOUR NAME: ________________________________________________________________ Please PRINT your name ( in indelible ink ) on the line above (& on the top right hand corner of every page). Also, please write all of your answers as legibly as possible . PLEASE CIRCLE THE NAME OF THE GSI FROM WHOM YOU WILL PICK UP YOUR GRADED MID-TERM EXAM: LUKE CHAO JOSEPH HICKEY AATHAVAN KARUNAKARAN ELIZABETH READ KYLE SIMONETTA This exam consists of 5 questions, each worth 20 points total, as indicated below, for a total of 100 points. This exam counts for 200 points out of the final score of 1000 for the course, and so your score on this exam will be multiplied by 2.0 when the final score is calculated. Question Part A Part B Part C Part D Your Total Maximum Score 1. (10) (5) (5) --------------- 20 2. (8) (7) ------------- --------------- 15 3. (10) (10) ------------- --------------- 20 4. (10) (10) (5) ---------------- 25 5. (10) (8) (2) ---------------- 20 TOTAL ---------- ---------- ---------- ----------- 100 UC Berkeley, CHEM C130/MCBC100A. FALL 2005. MID-TERM EXAM 3. YOUR NAME:_________________ ____________________ Page 1 of 13
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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 , 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
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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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This note was uploaded on 01/12/2010 for the course MCB 100A taught by Professor Kuryian during the Fall '09 term at University of California, Berkeley.

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exam3 - UC Berkeley, CHEM C130/MCBC100A. FALL 2005....

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