Role of Langmuir Equilibrium in Understanding Linked Equilibria: Saturatable Binding React, Lecture notes of Chemistry

Download Role of Langmuir Equilibrium in Understanding Linked Equilibria: Saturatable Binding React and more Lecture notes Chemistry in PDF only on Docsity! Linked equilibria: Langmuir -> the simples form of saturable binding Intro The remaining classes will be dedicated to understanding the behavior of molecular systems in which several equilibria are linked to one another. This linking of equilibria is at the core of biological signal processing. By combining binding equilibria and conformational changes living cells are able to shape their response to environmental stimuli. To me at least, this is where the transition between chemistry and biology occurs and the application of the ideas you will learn in this section to sensory systems and signaling pathways is a very hot topic of research at this moment. Just as in the beginning of this course when we talked about simple models to try to understand the principles that govern protein structure, we will again drastically simplify our models to capture the essence of the molecules behavior. It is important to realize that there are probably no real molecules that follow any of these models precisely and there will always be some sort of deviation form ideality. But before we get into these fancy linked equilibria, we will have to spend a day dealing with the default type of binding equilibrium. We get this type of binding behavior from a generic system with saturatable binding sites. A generic saturatable binding reaction. When we have a number of receptor molecules (or any other entities that can bind a ligand. The definition of what is the ligand and the receptor is completely arbitrary here) and we mix receptor and ligand together there can be two extreme scenarios. 1) Infinitely tight binding All ligands that are added are immediately bound, there is no free ligand until every single binding sites is occupied. 2) Infinitely weak binding The ligand is repelled by the receptor and we need to completely crowd the solution with ligand to force the ligand down the receptors throat. We will get saturation only at infinitely high concentrations of free ligand. Most ligand-receptor pairs fall somewhere in the middle. In these cases, what does a binding curve look like and what are the physical principles that shape it. Our ultimate goal is to find a simple model that predicts the fraction of receptors that have a ligand bound as a function of the ligand present. But before we go to such a model (which will turn out to be deceptively simple) lets think about the physics of this sort of binding reaction. What drives the ligand-receptor binding? Two general principles will determine the state of the binding equilibrium. 1) Potential energy of binding. The first one is the standard free energy of binding. This is the amount of energy released as the ligand binds to the receptor. (So and so much energy per hydrogen bond, so and so much per area of hydrophobic residue buried). This standard free energy also contains the price in conformational entropy we pay if we restrict the conformational freedom of the receptor or the ligand in the process of binding. This standard free binding energy is the same whenever a ligand of a given chemical type binds to a certain kind of receptor. In other words, it is purely dependent on the chemical properties of the ligand and the receptor but, by definition, independent of the concentration of the ligand – provided of course that the ligand does not show significant levels of self-association or possibly co- operative oligomerization. 2) Entropy of binding The second driving principle is the principle of maximum entropy. This entropy is different from the entropic contribution of changing the conformational freedom of the ligand or the receptor. To think about this entropy lets make a simple lattice model. We have a cube of 10x10x10=1000 sites. Lets say 990 of these sites are solvent and 10 of them are receptors. The entropy of the entire system then is the entropy of the receptor sites plus the entropy of the solvent sites. So how does the entropy of the whole system change when we move one ligand from a solution site to a receptor site in our lattice. † DStotal = Sreceptors - Ssolution DStotal = k lnWreceptors - k lnWsolution DStotal = k ln Wreceptors Wsolutions with W = N! n!(N - n)! DStotal = k ln Wreceptors Wsolutions DStotal = k ln Rectotal! Recoccu!(Rectotal - Recoccu)! Soltotal! Soloccu!(Soltotal - Soloccu)! Lets see how the entropy changes as we add a number of ligands to this system. The ligand can now either occupy solvent sites or receptor sites. We calculate the difference in entropy of adding all our ligands to receptor sites vs. adding all of them to solvent sites. What should be clear to you right away is that the multiplicity we generate by adding a ligand to the sites that have receptors will be much smaller than the multiplicity we would generate by adding the ligand to a solvent site. The reason is that there are many The overall shape of a Langmuir binding curve always looks exactly the same. The curve goes from 0 to 1 q. The curve starts out with its steepest slope and then gradually and continuous flattens until the saturation level (q=1 is reached). The only thing that happens if we change K is that we stretch or compress the X-axis. Does the curve reflect what we would have expected? At very low saturation levels of our receptor we only need to add very small amount of ligand to get a lot of it to bind to the receptor. I.e. the fraction of receptor that has a ligand bound grows rapidly as a function of the concentration of free ligand. As the saturation level (fraction bound) increases we have to keep adding more and more free ligand to get the same increase in the fraction of ligand-bound receptor. The reason for this is the increase in the entropic cost of binding the ligands to fewer and fewer open sites. How does the change of the Langmuir binding curve change as a function the binding constant? Below is a plot for three different binding constants. Which of the three curves comes from the weakest binder? Right, the light blue represents the weakest binder i.e. the ligand with the lowest standard free energy of binding. We need to increase the concentration of the free ligand a lot more, to achieve the same level of binding of this compound than we need for the other two compounds. Said a different way, the standard free energy of binding is lower and so we need to drive binding through mass action. To get K from an experiment: the easy way To ultimately get K we will measure q as a function of the free ligand concentration [X]. This is experimentally relatively easy. 1) We experimentally set the total concentration of our receptor 2) We follow some signal that is proportional to q 3) We can calculate [PX] from q and [Ptotal] and get the concentration of free ligand from [X]=[Xtotal]-[PX] Some times, if the binding constant is weak, we can even neglect the fraction of ligands bound to the receptor and simply use [Xtoal]=[X] and we get our plot of q vs. [X] Now how do we get K? The most obvious way is to determine the concentration of ligand at which we get the half maximal concentration of [PX]. In other words q = 1/2 In this case † 1 2 = q = K[X] 1+ K[X] This is only possible if K[X]=1 So: † K = 1 [X]q =1/ 2 A second way to get K An alternative way to get K is to look at the initial slope of the binding curve. † q = K[X] 1+ K[X] if K[X] <<1 q = K[X] 1+ K[X] = K[X] 1 = K[X] Our binding curve is q vs. [X] so the slope of this curve must be K. The best way to get K In practice this simple approach is often difficult to apply and may lead you into some traps. These problems are particularly severe in cases where the binding is weak and we need a very high concentration of X to achieve saturation. Such concentrations may not be achievable experimentally, because of aggregation or precipitation of the ligand. In these cases one would want to linearize the Langmuir equation and extract K from this linear plot. † q = K[X] 1+ K[X] invert 1 q = 1+ K[X] K[X] times[X] [X] q = 1+ K[X] K = 1 K + [X] so if we plot [X]/q versus [X] we should get a straight line and our intercept on the y-axis (i.e. [X]=0) is 1/K. This plot is superior to the simple method shown above for two reasons: 1) We get to check if the Langmuir model applies to our system. If our system follows the Langmuir model, we get a linear plot. 2) K is defined by all points in our measurement. So by fitting a least squares line to all our data points our random measurement errors are cancelled out and , equally important we get an estimate for the precision of our measurement. Some philosophy on Langmuirs and sensory systems We just have seen three completely different ways to get K from a Langmuir binding curve. Why are there so many different ways to get K? The reason is that each point on a Langmuir curve uniquely defines the whole curve. Two langmuirs with different binding constants never intersect – except, of course at [X]=0 and [X]=•. So, in principle, we could determine the binding constant of a Langmuir from any point on the binding curve and we could then calculate the shape of the entire curve.