Additional Equilibria linked to Langmuirs Review from last class Last class we talked about the “default” type of saturable binding equilibrium: Langmuir style binding. Here are the features of the Langmuir curve again: 1) The function always started with θ (the fraction of receptors in the bound form) at zero and monotonically rose until we reached a value of 1 (saturation). 2) The Langmuir curve was always steepest at very low concentrations of free ligand and became continuously flatter as we went to higher and higher concentrations of free ligand and degrees of saturation of the receptor. 3) If the binding constant is large, (strong binding) the receptor gobbles up most of the added ligand very quickly so the concentration of free ligand stays low until most of the receptors has a ligand bound. The half-saturation point (1/K) is already reached at relatively low concentrations of ligand. If the binding constant is small (weak binding) the curve is relatively flat and approaches saturation only at very high levels of ligand concentration. We also need a rather high concentration of free ligand before we reach the half-saturation point (1/K). And here is the formula for a simple Langmuir: ! "=[PX ][Ptotal]=K[X]1+ K[X ] Lets see again how this formula represents the characteristics of the binding process. 1) If K[X] is very small relative to 1 ! "=K[X ]1+ K[X ]K [X ]#0$ # $ $ $ "=K[X ]1= K[X ] as a result the slope of the Langmuir at very low levels of [X] is K. 2) If K[X] is very large relative to 1 ! "=K[X ]1+ K[X ]K [X ]#$% # % % % "=K[X ]K[X ]= 1! 3) If K[X]=1 ! "=K[X ]1+ K[X ]=11+ 1= 0.5[X ] = 1/K 4) Also the K we are using is a Kbind which is equivalent to 1/Kdisc Linking Langmuirs to other types of equilibria Today we will look at what happens if we link a simple saturable binding system to another equilibrium.Here is the first example of an idealized zinkfinger protein. These proteins are rather small and they cannot form a hydrophobic core large enough to stabilize them. Instead they derive much of their stability from binding to a zinc ion. But if the protein is not folded there is no binding site for the zinc ion. So we are looking at a sequential reaction in which we have two linked equilibria. In this linked equilibrium we have two equilibrium constant: - L is the dimension less equilibrium constant of unfolding this definition will make our calculations easier later on. - K is the equilibrium constant of binding and has the dimension of M-1. The part of the equilibrium Folded+X -> FoldedX is the same as in our Langmuir. Lets see what effect the linking of the equilibrium of folding has on our binding equilibrium. I.e. how does our plot of θ as a function of [X] change? Here is how we setup the equations ! K =[FX ][F ][X ]andL =[U][F ] Now we express θ(x) in terms of [FX],[U],[F].! "(x) =[FX]TotalPr otein=[FX][F] + [U] + [FX]=[FX][F][F][F]+[U][F]+[FX][F]substituting for K and L usin g[FX][F]= K # [X] and[U][F]= L"(x) =K[X ]1+ L + K[X] This sort of almost looks like a Langmuir. Lets see if we can make it into a proper Langmuir. ! "(x) =K[X ]1+ L + K[X]divide both top and bottom by 1+ [L]"(x) =K[X ]1+ L1+ L1+ L+K[X ]1+ L=K1+ L# [X]1+K1+ L# [X]defining Kapp=K1+ Lwe get"(x) =[FX]Pr oteintotal=Kapp# [X]1+ Kapp# [X] So the result is that we are getting a true Langmuir behavior and if we did not know that there was this added equilibrium of unfolding, we would have no idea that this is going on. Lets look at the results:Overall look of the curve The result we are getting is true Langmuir behavior. Its not just sort of like a Langmuir, it is!!!! a Langmuir. If we did not know that there was this added equilibrium of unfolding and we had no expectation for the numerical value of K, our experiment would give us no indication that something more complicated is going on. This is an important lesson in physical biochemistry. Distinctly different models can lead to qualitatively identical behavior. And if we have free parameters that we can fit, we can often get very good fits between our data and the mathematical description of our model. So just be aware that even though your data fits your model, there may be another model that results in the same mathematical equation and therefore fits your data just as well. To distinguish between a simple saturation of one species and the more complicated situation of the linked equilibrium, we would have to perform additional experiments, like monitoring for the presence of unfolded protein. L very small (No unfolding) In the extreme case of the protein never unfolding (this is equivalent to our original Langmuir) L is very small and the denominator equals 1 so that K=Kapp Another conclusion is, that we cannot increase our Kapp by hooking up an equilibrium to the folded form of the protein. The reason is that the denominator never becomes smaller than 1. So K>Kapp L =1 Kapp = K/2L very large (protein mostly unfolded) If L is very large relative to 1, Kapp = K/L Our apparent binding constant becomes rather small in other words our point of half maximal binding (1/Kapp) is achieved only at much higher concentrations of ligand. One way to think about it is that the addition of the unfolding equilibrium depletes the concentration of [F] available for binding of a ligand, As a result we have to provide a large driving force in the form of a high [X] to pull the Unfolded protein into the complex via folded form. So we need a lot more free ligand until we get to the state where half of our total protein is in the bound form, so our binding constant 1/[X]1/2 is smaller than in the absence of the unfolding equilibrium. Another example Here we have a receptor R that binds a ligand. In ligand-bound form and only in the ligand bound form, the receptor can then equilibrate between an inactive complex and an active form of the complex. Ligand bound in the active form is only released via first going to the inactive form. This is a linear binding scheme! R*X cannot release the ligand What we want to determine is the fraction of active molecules as a function of ligand concentration and of the two equilibrium constants. So the name of the game is ending up with an equation that contains only [X], K and L You guys should be able to do this now. What is the fraction of active molecules f*(X)?! f * ([X]) =[R * X][Rtotal]=[R * X][R] + [RX] + [R * X]we also haveK