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478169867 page 1 2011 JH Steinbach - 2011 A brief introduction to the Hill equation and kinetic models. I. The concentration-effect relationship We often analyze a "dose-response" or "concentration-effect" relationship. This relationship describes how much effect (receptor activation, for example) is produced by various concentrations of a drug. Steady-state Steady-state conditions mean that the system is not changing with time. Of course, in biological studies a true (infinite) steady-state will not be present. Indeed, often it is the case that the system is far from steady-state, perhaps over a limited time it is "pseudo-steady-state." However, we will be considering steady-state responses. The basic parameter for a dose-response curve is the fraction of the maximal activation which is produced. The fraction can be expressed in two basic forms. Experimentally it is usually the response relative to the largest response which can be measured. Theoretically, however, it is possible to calculate the fraction of the maximal response which would be produced if all the receptors were fully activated. The difference between these two fractions will be clear later on. II. Kinetic models A kinetic model is a short-hand description of a particular idea we might have for how a receptor works. It shows "states" of the receptor (for example, a resting receptor with no ligands bound, or a receptor with 2 ligands bound and an open channel). The states are connected by arrows that show what can happen (a ligand can bind or unbind, a channel can open or close). The states show what we are allowing the receptor to do, and the arrows show how the states are connected. II.A. A single site occupancy model This is the simplest type of model: binding directly results in gating. The simplest form is to say that a receptor (R) has a single site to bind an activator (A), and when the site is occupied the receptor is active (R*). A+R κ1[A][R] ⇄ λ1[AR*] AR* The association rate constant is κ1 and the dissociation rate constant λ1. R AR* Rtot A AR* Atot κ1[A][R] λ1[AR*] κ1 λ1 K1 receptor with no drug bound receptor with drug bound (defined as active for this model) total receptor, Rtot = R + AR* free drug drug bound to receptor total drug, Atot = A + AR* association rate; units s-1 dissociation rate; units s-1 association rate constant; units M-2s-1 dissociation rate; units M-1s-1 dissociation constant, K1 = λ1/κ1 (units M) 478169867 page 2 2011 To calculate the steady-state concentration-response relationship for this model, we need to determine the fraction of the total receptors (R + AR*) that is active (AR*). At steady state, there is no change in AR* with time - the rate for entering AR* is the same as for leaving: λ1[AR*] = κ1[A][R] = κ1([Atot] - [AR*])([Rtot] - [AR*]) In most experimental conditions there is a vast excess of free activator (that is [A] >>> [AR*]) available for reaction, and so we assume that [A] is constant at Atot. (On some occasions this might not be true. However, we will use this assumption for the rest of this handout.) λ1[AR*] = κ1[A] ([Rtot] - [AR*]) [AR*]/[Rtot] = κ1[A]]/λ1 / {1+ κ1[A]/λ1} This is the Michaelis-Menten or Langmuir equation. The dissociation constant (K1 = λ1/κ1) is usually used so it looks like [AR*]/[Rtot] = [A]/K1 / {1+ [A]/K1} This is not a very useful equation for the receptors we study, but essential as the first step. II.B. A two site occupancy model In words, this model is: - there are 2 binding sites for a drug on each receptor - when a receptor has both sites occupied by drug the receptor is activated - when the receptor is activated we see a response. This is the simplest model for "cooperative" activation (a "two site occupancy model") in which an occupied receptor is active. The model looks like: A+R κ1[A][R] ⇄ λ1[AR] AR + A κ2[A][AR] ⇄ λ2[AR*A] AR*A The new rate constants are κ2 and λ2, giving dissociation constant K2.. The occupancies of the various states are: [AR] = [R][A]/K1 [AR*A] = [AR][A]/K2 = [R][A]/K1 x [A]/K2 so AR*A/Rtot = {[A]/K1 x [A]/K2 } / {1 + [A]/K1 + [A]/K2 + [A]/K1 x [A]/K2} = {[A]2/(K1 K2 )}/ {1 + [A]/(K1 + K2) + [A]2/(K1 K2 )} Numbers of binding site - A brief digression The number of bound agonists necessary to produce an activated receptor determines the steepness of the concentration-response curve. This is illustrated for a multi-site occupancy model in Fig. II-1. Only receptors with all sites occupied are active. 478169867 page 3 2011 N site occupancy model nA+R <-> ... <-> AnR* 1 n=1 n=2 n=3 n=5 n=50 0.1 AnR/Rtot 1.0 AnR/Rtot N site occupancy model nA+R <-> ... <-> AnR*n 0.5 0.01 n=1 n=2 n=3 n=5 n=50 0.001 0.0001 0.0 101 102 101 103 102 103 [drug] [drug] Fig. II-1. Increasing numbers of sites. In the left panel the fraction of receptors that are active is plotted against the drug concentration for a number of values for the number of binding site. Remember that only receptors with ALL sites occupied are active! The foot of the curve gets progressively more curved as n increases. On the log-log plot (right panel), it can be seen that the steepness of the curve approaches the number of sites, for small responses (the inset shows straight lines of increasing steepness). The curves differ less as the concentration gets higher and higher. II.C. Binding/gating models A wealth of experiments has shown that occupancy models only are accurate for competitive antagonists (drugs which act by binding to the binding site but having no other effect). An occupancy model is not able to explain the actions of agonists (drugs which bind to the receptor and then activate it). It is necessary to separate the binding step from the activation step. del Castillo and Katz proposed the first version of this type of model (del Castillo and Katz. Proc R Soc Lond B Biol Sci. 1957;146:36981.). In words, a binding/gating model might be: - there are 2 binding sites for a drug on each receptor - when a receptor has both sites occupied by drug then it can become activated - a doubly-occupied receptor activates and inactivates - when the receptor is activated we see a response. A+R κ1[A][R] ⇄ λ1[AR] AR + A κ2[A][AR] ⇄ λ2[AR*A] ARA β ⇄ α AR*A The new rate constants are the "activation" rate constant, β and "inactivation" rate constant, α. The equilibrium opening constant is W 2 = β/α (W 2 because two molecules of A are bound). [AR] = [R] [A]/K1 + [R] [A]/K2 [ARA] = [R] [A]/K1 x [A]/K2 [AR*A] = [R] [A]/K1 x [A]/K2 x W 2 so AR*A/Rtot = {[A]/K1x[A]/K2xW 2} / {1 + [A]/K1 + [A]/K2 + [A]/K1x[A]/K2 + [A]/K1x[A]/K2xW 2} 478169867 page 4 2011 II.D. Inhibition of responses II.D.1. Competitive inhibition - competition for the binding site A one-site occupancy model is readily extended to include a competitive antagonist (a drug that binds to the same site as A, but does not activate. [B]/M A+R+B RB [A]/K ⇄ AR* (Note that rate constants have been replaced by equilibrium constants, to simplify the presentation.) The steady state response is then described by the equation AR*/ Rtot = [A]/K1 / {1+ [A]/K + [B]/M } Two equations derived from this kinetic model are often used to derive an estimate for M. (Of course, it may not be proven that this simple model truly applies!) The Cheng-Prusoff equation This simple equation (Cheng Y, Prusoff WH (1973). Biochem Pharmacol 22 : 3099–108) has been used to estimate the dissociation constant for the antagonist, when K is known. This equation is derived from the IC50 - the concentration of B that reduces AR*/ Rtot to one-half the control binding in the presence of [A]. {AR*([B=IC50]/ Rtot } / {AR*([B=0]/ Rtot ) = 0.5 or {[A]/K / {1+ [A]/K + IC50/M }}/{[A]/K1 / {1+ [A]/K}} = 0.5 Simplify and rearrange the equation to get the "Cheng-Prusoff equation" M = IC50 /{1 + [A]/K} The Schild equation A similar relationship is the Schild equation, used in pharmacological studies of receptor activation (see Colquhoun D. Trends Pharmacol Sci. (2007). 28:608-14). Rather than using a single concentration of B, this relationship calculates a "dose ratio" at several [B]. The dose ratio is the ratio of the concentration of A required to give the control level of response when B is present. Put another way, how much more [A] does it take to overcome the presence of [B]? [A']/K / {1+ [A']/K +[B]/M } = [A0]/K / {1+ [A0]/K } Simplify and rearrange the equation to get the "Schild" equation dose ratio = [A'] / [A0] = 1+ [B]/M The Schild approach is usually implemented by a Schild plot, of log({dose ratio}-1) against log([B]), which has a slope of one and an intercept of -log(M) (log({dose ratio}-1) =0 or dose ratio = 2). It is interesting to note that the Schild approach can work with more complex kinetic models. For example, this is the result of a Schild analysis of the two-site binding/gating model with a competitive 478169867 page 5 2011 inhibitor: A+R+B A+BR+B BRB [B]/M [B]/M [A]/K ⇄ ⇄ A+AR BRA [A]/K ⇄ ARA W ⇄ AR*A (for simplicity, assume the binding sites are identical) AR*A/Rtot = ([A]/K)2 / {1 + 2[A]/K + ([A]/K)2 + ([A]/K)2 W + 2[B]/M + [B]/M x [A]/K} 2 site binding/gating model Competitive Schild plot 2 site binding/gating model Competitive inhibitor [B]=0 [B]=10 [B]=30 [B]=60 [B]=100 [B]=300 [B]=1000 1.0 Data Schild Regression 0.5 log(DR-1) Response/{max control) 1.0 0.5 0.0 -0.5 -1.0 0.0 100 101 102 103 104 1.0 1.5 2.0 2.5 3.0 log([B]) [A] Fig. II-2. Competitive inhibition. The left panel shows the response to various concentrations of A, in the presence of various concentrations of B. The curves were calculated with K = 100 units, M = 100 units and W = 20. An important point to note is that the EC50 concentration increases with increasing [B], but the maximal response is constant - this is a definition of competitive inhibition. The right hand panel shows a Schild plot, in which the target response was set to 40% of maximal (short dashed horizontal line in left panel) . The straight line shows the calculated Schild relationship. The estimate from the "data" is that M = 107 units, close to the 100 units used in the calculations. II.D.2. Noncompetitive inhibition - selective inhibition of the active state There are a number of drugs that selectively act on a particular state. One particular example is a drug that binds specifically to the active receptor (an "open channel blocker"). A+R+B [A]/K ⇄ A+AR [A]/K ⇄ ARA W ⇄ AR*A+B AR*AB AR*A/Rtot = ([A]/K)2 / {1 + 2[A]/K + ([A]/K)2 + ([A]/K)2 W + [B]/M x ([A]/K)2 W } [B]/M 478169867 page 6 2011 2 site binding/gating model Open channel block [B]=0 [B]=1 [B]=10 [B]=30 [B]=60 [B]=100 [B]=300 1.0 0.5 log(DR-1) 1.0 Response/{max control) 2 site binding/gating model Open channel Schild plot 0.5 0.0 Data Schild Regression Col 16 vs Col 17 Col 16 vs Col 17 -0.5 -1.0 0.0 100 101 102 [A] 103 104 1.0 1.5 2.0 2.5 3.0 log([B]) Fig. II-3 Selective active state block. The left panel shows the response to various concentrations of A, in the presence of various concentrations of B. The curves were calculated with K = 100 units, M = 100 units and W = 20. In contrast to a competitive inhibitor, there is little change in the EC50 concentration with increasing [B], but a significant decrease in the maximal response. The right hand panel shows a Schild plot, in which the target response was set to 10%, 20% or 40% of maximal (lines in left panel). The straight line shows the calculated Schild relationship. The estimate from the "data" is that M = 102 units, close to the 100 units used in the calculations when the target response is small (10% of control), but gets progressively worse at higher targets. At the highest values for [B], it is not possible for the response to reach the larger targets (see the left panel) and the values are not defined. 478169867 page 7 2011 III. The Hill equation The Hill equation (also known as the logistic equation) has the basic form: Y([A]) = ([A]/EC50)n / {1+ ([A]/EC50)n} The response, Y, is a function of the drug concentration, [A]. The parameters are the Hill coefficient (n) and the concentration of drug producing half of the maximal activation (EC50). Sometimes the equation includes a maximum observed response (Ymax): Y([A]) = Ymax ([A]/EC50)n / {1+ ([A]/EC50)n} or an offset (a non-zero minimal value) (Ymin): Y([A]) = Ymax ([A]/EC50)n / {1+ ([A]/EC50)n} + Ymin For simplicity (and for the Hill plot) the values are normalized to cover the available range of values (Ymax - Ymin): y([A]) = (Y-Ymin) / (Ymax - Ymin) = ([A]/EC50)n / {1+ ([A]/EC50)n} so the value of y runs from 0 to 1. The Hill equation also can be used to describe inhibition (that is, a value of Y which decreases with increasing X), simply by subtracting it from 1 Z([A]) = 1 – Y([A]) = 1 / {1+ ([A]/EC50)n} (note that is the same thing as letting n take a negative value). The Hill plot The “Hill plot” was introduced as a graphical method for analyzing data in terms of the Hill equation. Now that we can do non-linear fitting with computers, there is no reason for using a Hill plot (except to show that the data do not really conform to the Hill equation!). The major difficulties with the Hill plot are that it is extremely sensitive to errors in Ymax, and has a tendency to make no sense. Nonlinear fits of untransformed data are much more robust. The Hill plot is described here just to provide some historical context! To make a Hill plot, the data values are normalized to give a new variable, h. h = (y – 0) / (1 – y) Using the definition of y given above, h = ([A]/EC50)n The Hill plot is made on log-log coordinates: log(h) -vs.- log([A]) or {n(log([A]) – log(EC50))} -vs.- log([A]). 478169867 page 8 2011 The slope gives n and the value of X at which log(h) = 0 (or h=1 or y = 0.5) gives EC50. The Hill equation as a tool. The Hill equation has no relationship to any actual physical situation – for example it does not correspond to the predictions of any scheme for binding and gating. However, parameters estimated by fitting data with the Hill equation provide a commonly used summary of the data. The reason why the summary can be useful is that two critical features of the data are captured. The first is an estimate of the “midpoint” of the amount/effect relationship of y on [A] given by EC50. This gives a (rough) estimate of the affinity of the interaction between drug and site. The second is an estimate of the steepness of the amount/effect relationship of y on [A]: n. This gives an estimate of the minimal number of binding sites for A. Comparing the Hill equation to the kinetic models discussed: Two-site occupancy: ARA/Rtot = {[A]/K1 x [A]/K2 } / {1 + [A]/K1 + [A]/K2 + [A]/K1 x [A]/K2} Two-site binding/gating: AR*A/Rtot = {[A]/K1x[A]/K2xW} / {1 + [A]/K1 + [A]/K2 + [A]/K1x[A]/K2 + [A]/K1x[A]/K2xW} To simplify things, assume K1=K2=K (identical sites) then Two-site occupancy: ARA/Rtot = {([A]/K)2} / {(1 + 2[A]/K + ([A]/K)2)} Two-site binding/gating: AR*A/Rtot = {W([A]/K)2} / {(1 + 2[A]/K + ([A]/K)2)(1 + W)} The normalized Hill equation (with n=2): Y = ([A]/EC50)2 / {1+ ([A]/EC50)2} The denominator of the Hill equation has fewer terms. The equation for two-site binding/gating gets closer to the Hill equation when W >> 1. Two-site occupancy models only get closer if there is cooperative binding, so that the dissociation constant at the first site bound is much greater (lower affinity) than when the second drug molecule binds. These points will be illustrated in the next section. 478169867 page 9 2011 IV. Some simple models and plots. In the rest of this Introduction I will discuss some specific examples of simple models for receptor activation, and show how good (bad?) a job the Hill equation does in terms really describing the (calculated) data. Qualitatively, the amount of receptor activation is not necessarily directly related to the binding affinity for a drug to any particular "state" of the receptor. Therefore, the value of EC50 cannot directly be related to a molecular step (for example, binding of a drug to an inactive receptor). Similarly, the steepness of the relationship (n) is not directly related to the number of drug molecules which must bind to an effect to happen. However, it does provide an estimate of the smallest number of drugs which must bind before the response occurs. The best tool for getting a feeling of what your data might mean is puttering around with some predictions of various possible schemes. I have done some of this, in the attached plots. I calculated responses for 3 models which are simple but are applicable to the receptors we study. - two-site occupancy model - two-site binding model with only doubly- bound receptors able to be activated - two-site binding model with both singly- and doubly-bound receptors able to be activated In all of the plots, the points (connected by thin lines) are the predictions of the model and the thick lines are the predictions of the fit Hill equations. One objective of this handout is to indicate the value and shortcomings of using the Hill equation to understand your data. Another is to illustrate the value of plotting your data several ways to look at them. 478169867 page 10 2011 IV. A. Two-site occupancy model A+R κ1[A][R] ⇄ λ1[AR] AR + A κ2[A][AR] ⇄ λ2[AR*A] AR*A Assume: Initially sites 1 and 2 are identical. However, after binding to the first site the affinity of the second site changes by a factor F: K1 = λ1/κ1 K2 = λ2/κ2 = FK1 (Note that when F is smaller than 1, the second site has HIGHER affinity, and the converse.) K1 was given the arbitrary value of 100, and predictions were made over a range of “concentrations” around this midpoint. The basic equation for ACTIVE receptors is for the equilibrium fraction of receptors which are doublyliganded (ARA): AR*A / Rtot = {([A]/K1)2/F} / {1 + 2[A]/K1 + ([A]/(K1 × K1 × F)} (Eqn IV.A.1) (Notice that the factor of 2 in the denominator is there because there are initially 2 identical sites for binding.) The first plot is one often used: the ordinate gives the "response" (fraction of receptors as AR*A). Fig. IV.1 shows calculated concentration-effect curves for this model, with different values for F. Note that the value shifts the curves along the concentration axis. It is also interesting that the steepness of the curves changes, although the number of binding sites remains the same. 2 site occupancy model 2A+R <-> A+AR <-> A2R* F=1 F=0.001 F=0.03 F=30 F=100 1.0 A2R/Rtot Fig. IV.1. Two site occupancy. The first plot is one you are probably used to seeing: the ordinate gives the "response" (fraction of receptors as AR*A). Data for a number of assumed values for F are given. You might be surprised that the midpoint of the curve is NOT at 100 (=K1, dashed vertical line) when the two sites are equal. This is because of equation IV.A.1 - the value is 0.5 when [A]/K1 = 1+√2. The solid lines show fits of the Hill equation to the points. 0.5 0.0 10-2 10-1 100 101 102 103 104 105 106 [A] I fit the predictions with the Hill equation, using SIGMAPLOT, with the following results: F n EC50 0.001 1.92 3.28 0.03 1.67 21 1 1.19 254 30 100 1.02 1.006 6093 20090 For the linear plot the Hill equation (lines through the points) looks like a good description of the data. Fig. IV.2 shows When the data are plotted in log-log form. This plot emphasizes the shape at low levels of response. The slope of the lines approaches 2 for every plot, as would be expected because 2 A molecules must be bound for the receptor to be active. However the curvature of the plots clearly differs 478169867 page 11 2011 between the various values. 2 site occupancy model 2A+R <-> A+AR <-> A2R* Fig. IV-2. Two site occupancy. When F < 1, the sites are “positively cooperative” in binding, so the second site is occupied at a lower concentration than the first site. As F > 1, the sites are “negatively cooperative” in binding. This is formally equivalent to “non-identical, non-cooperative sites” (although other experiments can distinguish the cases). (For convenience, lines with slopes of 1 and 2 on the log-log plot are shown.) The Hill equation (solid lines) gives a rather poor description of the smaller responses. 1 A2R/Rtot 0.1 0.01 F=1 F=0.001 F=0.03 F=30 F=100 0.001 0.0001 10-2 10-1 100 101 102 103 104 105 106 [A] I also made a Hill plot of the data. As can be seen, the data clearly do not make straight lines on the plot. As F << 1, then the plot is straight with slope 2 over a greater range. In this case, essentially you look at titration of the second site being driven by binding to the first site. As F >> 1, the plot begins to look straight with a slope of 1. In this case, you look at the independent titration of the second site (since the first site is fully occupied at all of these concentrations). The plots indicate that the Hill coefficient is the slope of the tangent to the data at X = EC50 (when the Hill variable, h, is 1). 2 site occupancy model 2A+R <-> A+AR <-> A2R* Fig. IV-3. Two site occupancy Hill plot. One aspect to remember with a Hill plot is the actual range of experimentally accessible values; in physiological experiments it is often hard to measure to an accuracy of 0.1%, while a Hill value of 0.001 or 999 requires this accuracy. So, these plots are much less informative in the real world. 1000 100 Hill variable 10 1 F=1 F=0.001 F=0.03 F=30 F=100 0.1 0.01 0.001 10-2 10-1 100 101 102 103 104 105 106 [A] Occupancy of sites. A second way to look at the data for this situation is to ask, what fraction of the available binding sites are occupied? Fraction of sites occupied = {[A]/K1 + 2 ([A]/K1)2/F} / {2 + 2 [A]/K1 + 2 ([A]/K1)2/F} Fits of the Hill equation gave values: F n EC50 0.001 0.03 1.963 1.814 3.162 17.3 1 1.29 99 30 100 0.652 0.508 590 1254 478169867 page 12 Fig. IV-4. Two site occupancy model - binding curves. To clarify the plots, I left out the intermediate values of F. Two things are worth noting. First, when F = 0.001, the binding seems to be "high affinity." This happens because as soon as the first site is occupied, then so is the second (K1=100 and K2=0.1). The second is that when F=100 it is possible to see two steps in titration. The first ligand binds with K1=100, but the second has a dissociation constant of 10000, so the first site is essentially saturated before any binding occurs to the second. This "negative cooperativity" in binding also can be produced by a receptor that has two distinct and noninteracting sites, of different affinities. 2 site occupancy model 2A+R <-> A+AR <-> A2R* F=1 F=0.001 F=100 Sites occupied 1.0 0.5 0.0 0 0 1 10 100 [A] 1000 10000 2011 100000 1000000 478169867 page 13 2011 IV. B. Two-site binding/gating model This is the next simplest model. In this case, after binding has occurred the receptor can be either active or inactive. This is the simplest model which can explain the behavior of real receptors. Two site binding/gating model: A+R κ1[A][R] ⇄ λ1[AR] AR + A κ2[A][AR] ⇄ λ2[AR*A] ARA β ⇄ α AR*A Here we will assume that sites 1 and 2 are identical and do not interact (K1 = K2 = 100; F=1). However, after occupancy of both sites opening can occur with ratio W = β/α = (activation rate/inactivation rate). The basic equation is for the fraction of receptors with open channels (AR*A): (AR*A) / Rtot = {W[A]2/K2} / {1 + 2[A]/K + (1+W)[A] 2/ K2 } Eqn IV.B 2 site bind/gate model 2A+R <-> A+AR <-> A2R <-> A2R* W=1 W=0.01 W=10 W=100 W=1000 normalized response 1.0 0.5 0.0 10-2 10-1 100 101 102 103 104 105 106 Fig IV-5. Two-site binding/gating model. This plot is a typical plot of experimental data, in which the response is divided by the maximal response observed. A small value of W (poor activation) shifts the concentration-effect curve to the right, while a large value (strong activation) shifts it to the left. In terms of equation IV.B a very small value for W is negligible compared to 1 in the denominator, so in the limit as W->0 the equation approaches the simple occupancy model with F=1. Large W, however, dominates over 1 and so this becomes equivalent to a very large value for F. [A] Making W large is similar to making F small in case IV.A. Looking at the equations, you can see that large W and small F enter into the predictions in similar ways. The predicted shift in the EC50 values can be derived from Eqn IV.B. Define Q as the EC50 concentration of A. 0.5 = {WQ2/K2} / {1 + 2Q/K + (1+W)Q 2/ K2 } / {W/(1+W)} Arranging and simplifying this gives the quadratic: -(1+W) Q2/K2 + 2Q/K +1 = 0 The positive root for this is Q = K (1+√(2+W))/(1+W). When W=0 (no gating) then Q =K(1+√(2) (the same as for the occupancy model with F=1). When W gets very large, Q ~ K/(√(W) I fit the non-normalized data with the Hill equation. W 0.01 1 n 1.21 1.31 EC50 247 142 max 0.0098 0.5 10 1.53 42 0.9 100 1.81 11.1 0.99 1000 1.93 3.28 1 478169867 page 14 2011 The values for EC50 are pretty close to the predicted values from the quadratic. This plot is very deceptive. This is because it is normalized to the largest observed response. However, in terms of receptor activation, the maximal activation is very different (for these values from <0.1% to essentially 100%). Notice that this problem does not arise with an occupancy model, for which activity equals activity, and occupancy becomes 1.0 at high [A]. How big would the response be if all the receptors were actually activated? The next plot shows the predicted fraction of the total receptors which are actually activated for the different values of W. For small values of W only a small fraction of the receptors are activated, even when all the sites are occupied. When W=1, then half the receptors would be active, when W=0. 01 only 1% and so forth. This is a fundamental concern in studies of receptor activation: what is the activity of a single receptor when the response is maximal? Various approaches can be taken to answering this question, including recordings of single channel currents. 2 site bind/gate model 2A+R <-> A+AR <-> A2R <-> A2R* 1.0 A2R*/Rtot Fig. IV-6. Two-site binding/gating model. When the data are NOT normalized to the maximal observed response, the graph looks quite different. W=1 W=0.01 W=10 W=100 W=1000 0.5 0.0 10-2 10-1 100 101 102 103 104 105 106 [A] 2 site bind/gate model 2A+R <-> A+AR <-> A2R <-> A2R* Fig. IV-7. Two-site binding/gating model. The log-log plot shows that the slope of the relationship approaches 2 at low concentrations, as expected since two molecules of drug must bind. The curvature depends on how large W is compared to 1; large values give more rapid approach to the slope of 2. 100 A2R*/Rtot 10-1 10-2 W=1 W=0.01 W=10 W=100 W=1000 10-3 10-4 10-2 10-1 100 101 102 [A] 103 104 105 106 478169867 page 15 2011 IV.C. Gating with either one or two bound ligands. This is the last model considered here. What happens if a receptor can be activated when either 1 or 2 drug molecules are bound? There are now 2 active states: AR* and AR*A (and in the diagram the activation steps are now down). A+R κ1[A][R] ⇄ λ1[AR] κ2[A][AR] ⇄ λ2[AR*A] AR + A AR* y ARA α z AR*A β We assume that sites 1 and 2 are identical and not interacting (F=1). However, after occupancy of either one or both sites opening can occur with ratio W = β/α with two sites occupied and V = z/y with only one site. The basic equation is for the fraction of receptors with open channels ([AR*A] + [AR*]): ([AR*A] + [AR*]) / Rtot = {W x [A] 2/(K1K2) + V x 2[A]/K1]} / {1 + 2(1+V)[A]/K1 + (1+W) x [A] 2/(K1K2)} In each case W was held constant at 100 (strong gating with two ligands bound). The results of fitting with the Hill equation are: V n EC50 0 1.81 11 2 site bind/gate model either 1 (V) or 2 (W=100) open Response 0.5 0.0 10-3 10-2 10-1 100 101 [A] 102 1 1.63 9.9 10 1.11 4.3 100 1.0 0.49 Fig. VI-8. Gating with either 1 or 2 molecules bound. When V=0, no gating of monoliganded receptors occurs. As more and more gating occurs of monoliganded receptors, the curvature goes down and the EC50 shifts to lower concentrations. When gating is strong with only one bound molecule, then the curve is the same as a single site binding/gating model. V=0 V=0.1 V=1 V=10 V=100 1.0 0.1 1.79 11 103 104 478169867 page 16 2 site bind/gate model either 1 (V) or 2 (W=100) open Fig. IV-8. Gating with either 1 or 2 molecules bound. The log-log concentration-effect plot is quite informative. When V=W (=100) 99% of the receptors can be opened by a single bound drug, the same fraction as when 2 are bound. Accordingly, the curve is identical to one in which there is only one site, and at low drug concentrations the slope approaches 1. When V=0 there is no gating of monoliganded receptors, and the low-concentration slope is 2. At intermediate values of V, the slope at low concentrations approaches 1 and the maximal slope lies between 1 and 2. 1 Response 0.1 0.01 V=0 V=0.1 V=1 V=10 V=100 0.001 0.0001 10-3 10-2 10-1 100 101 102 103 2011 104 [A] END