Final Answers - 2009

# Final Answers - 2009 - Final Exam MCB C100A / Chem C130...

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Your Name_____________________________ Your signature__________________________ Your GSI ʼ s name________________________ Final Exam MCB C100A / Chem C130 Fall 2009 This exam has a total of 19 pages including cover Point distribution: Problem part A part B part C part D total 1 24 12 - - 36 2 14 12 10 - 36 3 18 6 6 - 30 4 8 8 7 - 25 5 6 7 6 6 25 6 8 9 8 - 25 7 10 15 - - 25 8 7 6 12 - 25 9 8 8 7 - 25 10 6 6 7 6 25 Final Exam MCB C100A / Chem C130 Fall 2009 page 1 of 19

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John Kuriyan - 1 - Q. JK1 (25 points total) Part A. (16 points) Shown below is the binding isotherm for a protein binding to a ligand. The graph shows the behavior of log(f/1-f), where f is the fractional saturation, as a function of log [L], where [L] is the concentration of the ligand. The protein is allosteric, and switches from a “T” state when the ligand concentration is low to an “R” state when the ligand concentration is high. (i) (2 points) Based on the graph, what is the value of the dissociation constant, K D1 , of the T state? Sketch on the graph how you estimate the answer. (ii) (2 points) Based on the graph, what is the value of the dissociation constant, K D2 , of the R state? Sketch on the graph how you estimate the answer. (iii) (2 points) Does the protein show positive or negative cooperativity? Explain.
John Kuriyan - 2 - QJK1, Part A, continued (iv) (4 points) Based on the graph, estimate the value of the Hill coefFcient, n H , of the protein. Explain how you work out the answer by writing down the deFnition of the Hill coefFcient and show how you used the graph to estimate the Hill coefFcient. (v) (2) Based on the graph, what is the most reasonable answer for the number of subunits (i.e., ligand binding sites) in the protein? Circle the best answer and explain your choice. (a) 1 (b) 2 (c) 3 (d) 4 (e) 5 (f) 6 (vi) (4 points) Bisphosphoglycerate is an allosteric modulator of human hemoglobin. When the pH of a solution is lowered the afFnity of BPG for hemoglobin increases. Provide two different reasons for why this is the case. (2 points each)

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John Kuriyan - 3 - QJK1 Part B. (9 points total) (i) (3 points) A drug molecule binds to its intended protein target with a dissociation constant, K D , of 10 -12 M. The drug is delivered at concentration of 1 nanomolar. What is the fractional saturation of the intended target? (ii) (3 points) At the same concentration of the drug, an unintended target is also bound by the drug, and it is found that the unintended target has a fractional saturation of 10%. What is the dissociation constant of the drug for the unintended target? (iii) (3 points) Medicinal chemists decided to modify the chemical structure of the drug so that the fractional saturation for the unintended target would be reduced to 2% when the modiFed drug is given at 1 nanomolar concentration. By how much would the binding free energy have to be reduced to achieve this effect? That is, calculate the value of !! G 0 = ! G 2 0 " ! G 1 0 where ! G 1 0 and ! G 2 0 are the standard binding free energies of the original and modiFed drugs, respectively.
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## Final Answers - 2009 - Final Exam MCB C100A / Chem C130...

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