midterm2_key

# midterm2_key - 10.37 Exam 2 25 April 2007 100 points...

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10.37 Exam 2 25 April, 2007 100 points Problem 1: 35 points A protein and ligand bind reversibly with K d = 10 nM . The association rate constant k = 2 x 10 4 M 1 s -1 . The two species are mixed at an initial protein concentration of 3 nM on and an initial ligand concentration of 0.2 nM. a) At equilibrium, what fraction of the ligand will be complexed with protein? (15 points) b) At what time will the fraction of ligand in complex reach 95% of the equilibrium value? (20 points) Justify any assumptions you make to simplify equations. P + L ←→ C k off [ P ] eq [ L ] eq K d = = k on [ C ] eq Using a batch reactor mole balance and looking at the reaction stoichiometry, it is easy to see that given the initial conditions any unit of complex formed takes away a unit of protein and ligand: [ P ] + [ C ] = [ P ] 0 [ L ] + [ C ] = [ L ] 0 Using these in the equilibrium equation we can get a quadratic equation in [C] eq . ( [ P ] 0 [ C ] eq )( [ L ] 0 [ C ] eq ) K d = [ C ] eq 0 = [ C ] eq 2 −( [ L ] 0 + [ P ] 0 + K d ) [ C ] eq + [ P ] 0 [ L ] 0 = 0 [ C ] = ( [ L ] 0 + [ P ] 0 + K d ( [ L ] 0 + [ P ] 0 + K d ) 2 4[ P ] 0 [ L ] 0 = 0 eq 2 Reject the positive root, it is too large (larger than the initial amount of ligand and protein). [ C ] eq = 0.0456 nM [ C ] eq = 0.228 = 23% [ L ] 0 At this point, it is interesting to look at different approximations to the expression. Good Approximation: [P] 0 >> [C] This leads to ( [ P ] 0 )( [ L ] 0 [ C ] eq ) K d [ C ] eq Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].

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[ C ] = [ L ] 0 [ P ] 0 0.462 nM eq K d + [ P ] 0 [ C ] eq = 0.231 = 23% [ L ] 0 Bad Approximation: [L] 0 >> [C] ( [ P ] 0 [ C ] eq )( [ L ] 0 ) K d [ C ] eq [ C ] eq = [ L ] 0 [ P ] 0 = 0.0588 K d + [ L ] 0 [ C ] eq = 0.294 = 29% [ L ] 0 The error in [C] eq of the bad approximation is about 30% of the true answer, whereas the good approximation is only off by about 1%. By noticing that the “good” approximation is a good approximation, the dynamic equation becomes easier to solve. (As an aside, an even better approximation would be just to neglect the second order term that is O([C] eq 2 ).) Start with the full dynamic equation: d [ C ] = k on [ L ][ P ] k off [ C ] = k on ( [ L ] 0 [ C ] ) ( [ P ] 0 [ C ] ) k off [ C ] dt Make an appropriate approximation: ( [ P ] 0 [ C ] ) [ P ] 0 d [ C ] k on [ L ] 0 [ P ] 0 k on [ P ] 0 [ C ] k off [ C ] dt Rearrange and solve using the integrating factor: d [ C ] + [ C ] ( k on [ P ] 0 +
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midterm2_key - 10.37 Exam 2 25 April 2007 100 points...

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