hw6_sol - 1. (10) Consider the reaction mechanism shown...

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1. (10) Consider the reaction mechanism shown below: Five out of the six rate constants in this scheme are known: k 1 = 1 s -1 , k -1 = 10 s -1 , k 2 = 2 s -1 , k- 2 = 20 s -1 , k 3 = 3 s -1 . Calculate k -3 . k 1 k 2 k 3 = k -1 k -2 k -3 , which gives k -3 = 0.03 s -1 (see Lecture notes) 2. Experiments show that folding of protein L is well described by the first order mechanism shown below: Here U labels the denatured (unfolded) state, F the native (folded) state. The folding rate at room temperature and P = 1 atm is k f = 60 s -1 , the unfolding rate is k f = 0.011 s -1 . A. (5) Calculate G 0 . 0 ln ln 21330 J/mol f u k GR T K R T k  ∆= = =   B. (10) It is hard to imagine that protein folding would be a single-step process without any intermediates. For example, α − helices or β -sheets may be formed before folding is complete. To see whether the existence of intermediates would affect the observed time dependence of the concentrations of the folded and the unfolded proteins, [U] and [F], consider the following reaction mechanism:
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where k 1 = 120.5 s -1 , k -1
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This note was uploaded on 09/06/2009 for the course CH 353M taught by Professor Lim during the Spring '08 term at University of Texas.

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hw6_sol - 1. (10) Consider the reaction mechanism shown...

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