Lecture 19. Tuesday, November 7. Ligand binding thermodynamics.

Lecture 19. Tuesday, November 7. Ligand binding thermodynamics.

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Restricted: For students enrolled in Chem130/MCB100A, UC Berkeley, Fall 2006 ONLY 1 John Kuriyan: University of California, Berkeley Chem C130/MCB 100A, Fall 2006, Lecture 19 Dissociation Constants/Association Constants In the last lecture we considered the equilibrium between a ligand, L, and its target, P (e.g., a protein). The equilibrium constant for the binding reaction is: K A = [ PL ] [ P ][ L ] association constant The equilibrium constant for the dissociation reaction is: K D = [ P ][ L ] [ PL ] dissociation constant and a D K K 1 = The standard free energy change upon binding is given by: G ° = -Rt ln K A In biochemistry we usually use the dissociation constant rather than the association constant, because it is readily related to the saturation.
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Restricted: For students enrolled in Chem130/MCB100A, UC Berkeley, Fall 2006 ONLY 2 For every binding interaction that is studied we develop one or more binding assays. E.g. radio active assay This allows us to plot a binding isotherm:
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Restricted: For students enrolled in Chem130/MCB100A, UC Berkeley, Fall 2006 ONLY 3 This is usually replotted in terms of the fractional saturation, ƒ :
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Restricted: For students enrolled in Chem130/MCB100A, UC Berkeley, Fall 2006 ONLY 4 The dissociation constant, K D , is equal to the ligand concentration at half saturation. In this plot, [ L ] is the free ligand concentration [ ] TOTAL = [ ] + [ P.L ] where [ ] is the concentration of the protein ligand complex. In most applications we assume that [ ] >> [ ] and so: [ ] TOTAL [ ] [ ] TOTAL is usually what we know, because we add it to the reaction. It s easy to see why K D is the half saturating ligand concentration: f = [ P . L ] [ P ] + [ P . L ] = conc. of bound protein conc. of total protein Because K D = [ P ][ L ] [ P . L ] [ P . L ] = [ P ][ L ] K D f = [ P ] [ L ] K D [ P ] + [ P ] [ L ] K D
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Restricted: For students enrolled in Chem130/MCB100A, UC Berkeley, Fall 2006 ONLY 5 f = [ L ] K D 1 + [ L ] K D Clearly, if [L] = K D , then ƒ = 1 2 The binding isotherm characterized by a simple equilibrium such as this is called hyperbolic. ƒ is a pure number, so [ L ] K D must be dimensionless. This must mean that K D has units of concentration. But, any equilibrium constant must itself be dimensionless. Δ G = + RT In K D energy units = energy units × pure number The discrepancy arises because: In practise, because the standard state concentrations are all unity we ignore them and treat the dissociation constant as it it has units of concentration.
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Lecture 19. Tuesday, November 7. Ligand binding thermodynamics.

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