6. For a 1:1 ligand-receptor binding interaction that is modeled by X + Y XY :(A)Show how the fraction of ligand-receptor binding (fo) is represented by the[X][X]+ K,equation of f,(i.e., derive this expression for this reaction beginningegwith the relationship for K = f([X].[X],[XY]) under equilibrium conditions).(B)shown to the right for thisreaction, what is theestimated value of Keg? Giveyour answer in units of M.For the data plot0.90.80.70.6문 0.50.40.30.20.10.0000 0.0050 0.0100 0.0150 0.0200 0.0250 0.0300 0.0350 0.0400[X) (mM)

Lets consider the equilibrium reaction; X + Y ⇌KdKa XY Here 'X' is the ligand, 'Y' is the receptor and 'XY' is the receptor-ligand complex. Ka and Kd are equilibrium constants. 'Ka' is the equilibrium association constant and 'Kd' is the equilibrium dissociation constant. Ka=[XY][X][Y] ------(eq 1) and Kd=[X][Y][XY]i.e. Kd=1KaThe fraction of ligand-receptor binding , fb is the ratio of receptors bound to the ligand to the total number of receptors . fb=concentration of ligand receptor complexconcentration of total receptorsfb =[XY][XY]+[Y] --------(eq 2) (eq 1) can be written as ; [XY]=Ka[X][Y]substituting this value of [XY] to (eq 2) fb=Ka[X][Y]Ka[X][Y] + [Y](cancel out [Y] from numerator and denominator) =Ka[X]Ka[X]+1(cancel out Ka from numerator and denominator) =[X][X]+1Ka(we know Kd=1Ka)So ; fb=[X][X]+KdSince the reaction is at equilibrium; concentrations of all species are at equilibrium concentration; i.e. [X]=[X]eq and we already know Kd is an equilibrium constant.So the equation at equilibrium is represented as ; fb=([X][X]+Kd)eqThere are 2 equilibrium constants (Keq) concerning the reversible binding of a ligand to its receptor; Ka and Kd. Kd has the unit M while Ka has the unit M-1. Here we are asked to find the Keq with the unit M-1 so we have to find Ka. Kd is the concentration of ligand when half of the receptors are bound with ligands. i.e. Kd is the value of [X] when fb=0.5 In the graph, plot a line to the curve from the point 0.5 in Y axis parallel to X axis. From the point where this line intercepts the curve, draw a line perpendicular to the X axis. The point of interception of this line at X axis gives the value of Kd. As shown above, by plotting we get; Kd= 0.00250 mM = 0.00250 x 10-3 M Ka=1Kd =10.00250×10-3M =400000M-1 So the value of Keq in M-1 (i.e. Ka) is 400000M-1