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Unformatted text preview: 1 Binding of small to large molecules Receptor / ligand binding Binding of O 2 to hemoglobin Binding of carcinogens to DNA Binding of proteins to DNA Binding of ions to polymer fibers Examples: A + P k r k f PA At equilibrium, k f [A][P] = k r [PA] or ] P ][ A [ ] PA [ k k K r f 1. The large molecule has only one binding site. A: small molecule, P: polymer or large molecule Define ] P [ ] A [ bound (average number of binding sites on P that are occupied by A) For the reaction with only one binding site on P, ] PA [ ] P [ ] PA [ ] P [ ] A [ bound When = 0.5, [A] = 1/K. At equilibrium, ] A [ K ] A [ K ] P ][ A [ K ] P [ ] P ][ A [ K ] PA [ ] P [ ] PA [ 1 depends on T, P, and [A]. The curve of versus [A] is called binding isotherm. [A] T 2 2. The large molecule P has 3 identical binding sites and they are independent of each other. A + P k r k f PA (1) A + P k r k f PA (2) A + P k r k f PA (3) (A binds to the 1st site on P) (A binds to the 2nd site on P) (A binds to the nth site on P) [PA] t = [PA (1) ] + [PA (2) ] + [PA (3) ] = 3 [PA (1) ] (Since 3 sites are identical.) At equilibrium, ] P ][ A [ ] PA [ k k K ) 1 ( r f ] P ][ A [ ] PA [ k k K ) 2 ( r f ] P ][ A [ ] PA [ k k K ) 3 ( r f Thus, [PA] t = 3K[A][P] A + P k r1 k f1 PA (any site) In other words, for the reaction the macroscopic equilibrium association constant, K 1 , is K 3 ] P ][ A [ ] PA [ k k K t 1 r 1 f 1 (microscopic equilibrium association constant) Similarly, for the reaction PA(any site) + A k r2 k f2 PA 2 (any 2 sites) t t 2 2 r 2 f 2 ] PA ][ A [ ] PA [ k k K There are 3 ways of picking 2 sites on P, i.e., 12, 23, 13. Thus, there are 3 different PA 2 (i,j) in the solution. [PA 2 ] t = 3[PA 2 (1,2) ]. We have shown that [PA] t = 3 [PA (1) ]. Thus, K ] PA ][ A [ 3 ] PA [ 3 ] PA ][ A [ ] PA [ k k K ) 1 ( ) 2 , 1 ( 2 t t 2 2 r 2 f 2 3 PA 2 (any 2 sites) + A k r3 k f3 PA 3 (any 3 sites) For the reaction K 3 1 ] PA ][ A [ 3 ] PA [ ] PA ][ A [ ] PA [ k k K ) 2 , 1 ( 2 ) 3 , 2 , 1 ( 3 t 2 t 3 3 r 3 f 3 t 2 t 3 3 r 3 f 3 ] PA ][ A [ ] PA [ k k K There is only 1 way of picking 3 sites on P. Thus, [PA 3 ] t = [PA 2 (1,2,3) ]. We have shown that [PA 2 ] t = 3[PA 2 (1,2) ]. Thus, S S ) S ]( P [ ) S ]( P [ S ] P [ ] A [ bound 1 3 1 1 3 3 2 [A] bound = [PA] t + 2[PA 2 ] t + 3[PA 3 ] t t 2 t t ] S[PA 3 1 3 2S[PA] PA] [ S S 2 1 [PA] 2 t 2 S 1 3S[P] (S = K[A]) [P] = [P] + [PA] t + [PA 2 ] t + [PA 3 ] t t 2 t t ] S[PA 3...
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This note was uploaded on 01/17/2010 for the course BME 100 taught by Professor Yuan during the Fall '07 term at Duke.
 Fall '07
 YUAN

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