1. Receptors - Quantitation Quantitation of Receptor Occupation and Response for School of Biomedical Engineering, Drexel University Prof. Philip Lazarovici, Ph.D., M.Sc., B.Sc. Department of Pharmacology and Experimental Therapeutics School of Pharmacy, Faculty of Medicine, Hebrew University, Jerusalem, Israel

2. Receptors - Quantitation Equations The binding equation:R+L+I RL + RI (eq. 1)  The binding equation: R+L+I RL + RI (eq. 1) R – concentration of free (unoccupied) binding sites L – concentration of the (radioactive) ligand I – concentration of unlabelled drug RL, RI – concentrations of the receptor bound with each ligand Total/added ligand concentration:L T = L + RL (eq. 2)  Total/added ligand concentration: L T = L + RL (eq. 2) RL – the specific binding (not the total binding!)

3. Receptors - Quantitation Types of Experiments to Demonstrate the BINDING EQUATION L T = L + RL a)Saturation experiments a)Saturation experiments in which L T is increased and RL determined at equilibrium (I=0). b) Kinetic experiments b) Kinetic experiments in which RL is determined as a function of time with L T being held constant. c) Inhibition experiments c) Inhibition experiments in which RL is determined as the concentration of an unlabelled drug is increased. L T is held constant.

4. Equilibrium dissociation constant: K D = (R*L) / RL (eq. 4) Total receptor concentration: R T = R + RL (eq. 5) therefore:K D * RL = R T * L – RL * L RL = (R T * L) / (K D + L) therefore:K D * RL = R T * L – RL * L RL = (R T * L) / (K D + L)  mathematically equivalent to Michaelis-Menten equation and Langmuir adsorption isotherm)  When L=K D, then RL = ½ R T  If RL is small compared to L T, then L is taken as being equal to L T. a) Saturation Experiments – Basic Relationships Saturation Experiments Saturation experiments Saturation experiments – R (receptor concentration) is held constant and RL is determined at equilibrium as a function of L: R + L RL R + L RL (eq. 3) Fig 1 (eq. 6)

5. Saturation Experiments Rosenthal (Scatchard) Analysis RL = (R T * L) / (K D + L)  Linear description of the non-linear equation: RL = (R T * L) / (K D + L).  When the molecular weight and concentration of the macromolecule (ligand) is known (and small) – the Scatchard plot is used.  The Rosenthal plot is used when the above parameters are not known.  The Rosenthal plotRL/L = R/K D = - RL/K D + R T /K D  The Rosenthal plot is the equation RL/L = R/K D = - RL/K D + R T /K D, but with different symbols: RL = B (bound), L = F (free), R T = B max (maximal number of binding sites): B/F = - (1/K D )*B + B max /K D (eq. 7) (eq. 7)

6. Saturation Experiments Non-linear Rosenthal Plots  Specific case (and not the only one) – existence of two classes of non-interacting binding sites – described by part B.  The high affinity is not apparent in figure A ( B vs F) but is very apparent in figure B – Rosenthal plot (B/F vs B). Generated data PlotsA B C

7. Saturation Experiments K D 1, B max 1, K D 2, B max 2  Methods to determine the 4 constants: K D 1, B max 1, K D 2, B max 2 : a)Drawing a line through the top and portions of the graph (fig C). b)“Trial and error” method – estimation the value of the four constants and then calculate the expected experimental points and observe the “goodness” of fit - B = B 1 + B 2 = [(B max 1 * F) / (K D 1 + F)] + [(B max 2 * F) / (K D 2 + F)] (eq. 8) c) Determining the limiting slopes for each end of the curve. d) Computer programs, e.g. NLIN (a Statistical Analysis System – SAS).

8. Saturation Experiments Hill Plots  Hill’s equation  Hill’s equation used to describe binding of oxygen to hemoglobin (1910): log [B / (B max – B)] = n log F – log K’ D B = [(B max * F n ) / (K D + F n ).  The latter is derived from the equation: B = [(B max * F n ) / (K D + F n ).  n H - Hill coefficient log [B / (Bmax – B)]log F n H  n H - Hill coefficient – the slope of the plot log [B / (Bmax – B)] vs log F (if the slop is not linear n H is the slope at B=0.5B max.  n H n H  n H can never exceed the number of sites (for highly cooperative sites n H is the number of sites); it can be lower than the number of sites (when the sites are not strongly cooperative). EXAMPLE: n H of 1.8 could imply of two sites with moderately strong cooperativity, but it could also represent six sites with weak positive cooperativity.  n H  n H lower than 0.8 (value considered different from unity) – negative cooperativity (also other interpretations are valid) or artifacts!

9. Kinetic Experiments b) Kinetic Experiments b) Kinetic Experiments – Basic Relationships Kinetic experiments Kinetic experiments – binding is determined as a function of time: R + L RL R + L RL (eq. 8) d(RL) / dt = k +1 *R*L – k -1 *RL  The net rate at which RL is being formed is equal to the rate of formation of RL minus the rate of dissociation of RL: d(RL) / dt = k +1 *R*L – k -1 *RL (eq. 9)  At equilibrium d(RL) / dt equals zero and therefore: k -1 / k +1 = R*L / RL k -1 / k +1 = R*L / RL (eq. 10)  Combining the latter with the equation K D = R*L / RL : K D = k -1 / k +1  These kind of experiments are also useful to determine the time when apparent equilibrium has been reached, so that saturation and inhibition experiments may be properly carried out. k +1 k -1

10.  k -1 – association rate constant –  k -1 – association rate constant – can be determined by two ways: a)Second order treatment of kinetic data – only the initial rate of binding is considered (the bound concentration is assumed to be low, and therefore k -1 *RL is zero). d(RL) / dt = k +1 *R*L – k -1 *RL dB / dt = k +1 *F*R F = L T – B R = B max – B  Since we can write the equation d(RL) / dt = k +1 *R*L – k -1 *RL also as dB / dt = k +1 *F*R, and since F = L T – B (eq. 2) and R = B max – B (eq. 5) therefore: dB / dt = k +1 (L T – B) (B max - B) dB / dt = k +1 (L T – B) (B max - B)  Integrating between the limits of B = 0 at R = 0 and B = B at t = t : k +1 = 1/[t(L T – B max )] * ln {[B max (L T – B)] / [L T (B max – B)]} (eq.11 k +1 = 1/[t(L T – B max )] * ln {[B max (L T – B)] / [L T (B max – B)]} (eq.11) k +1  Therefore k +1 can be calculated when B was determined (B max can be determined in independent saturation experiments). Kinetic Experiments Association Rate Constant

11.  k +1 – association rate constant –  k +1 – association rate constant – can be determined by two methods: a)Second order treatment of kinetic data - already discussed. d(RL) / dt = k +1 *R*L – k -1 *RLk’ +1 = k +1 * L T b)Pseudo first-order method – includes the reverse reaction (k -1 RL = 0), but assumes that the concentration of the radioactive ligand is constant (L T = F). Therefore, from the equations: d(RL) / dt = k +1 *R*L – k -1 *RL and k’ +1 = k +1 * L T we can conclude that: dB / dT =k’ +1 *R – k -1 *B dB / dT = k’ +1 *R – k -1 *B B e  At equilibrium, the bound concentration (B) is related to the equilibrium concentration – B e by: ln [B e /(B e – B)] = (k’ +1 + k’ -1 )t = k ob * t (eq. 12) ln [B e /(B e – B)] = (k’ +1 + k’ -1 )t = k ob * t (eq. 12) Kinetic Experiments Association Rate Constant

12. ln [B e /(B e – B)] = (k’ +1 + k’ -1 )t = k ob * t ln [B e /(B e – B)] = (k’ +1 + k’ -1 )t = k ob * t (eq. 12) Kinetic Experiments Association Rate Constant ln [B e /(B e – B)]tk ob.  A plot of ln [B e /(B e – B)] vs t will have a slope of k ob.  Knowing k -1 (from prior experiments) we can calculate k +1 : k ob - k -1 = k’ +1 = k +1 * L T k ob - k -1 = k’ +1 = k +1 * L T  Another method is do several experiments at different radioligand concentrations and determine k ob for each L T. Then a plot of k ob vs L T will have a slope of k +1. The intercept will be k -1.

13. Kinetic Experiments Dissociation Rate Constant  K -1 – dissociation rate constant –  K -1 – dissociation rate constant – can be determined by two methods: a)Pseudo first-order method- already discussed. d(RL) / dt = k +1 *R*L – k -1 *RL b) In conditions where the rebinding of the radioligand is prevented, meaning k +1 *R*L = 0. (using eq. 9: d(RL) / dt = k +1 *R*L – k -1 *RL )  Two modes of obtaining the above state: 1. 50 fold or grater dilution of the reaction mixture. 2. An excess of non-radioactive ligand is added, so all the free binding sites become occupied by the unlabelled ligand, and R=0 (excess means 100 times the concentration which will occupy 50% of the sites). dB / dt = – k -1 *B (eq. 13)  Eq. 9 can be written as: dB / dt = – k -1 *B (eq. 13)

14. Kinetic Experiments Dissociation Rate Constant  dB / dt = – k -1 *B  dB / dt = – k -1 *B (eq. 13) ln (B / B 0 ) = – k -1 * t  And experimentally: the binding reaction is allowed to proceed for some arbitrary time. Then, at t=0, rebinding of the radioligand is prevented and the amount bound is determined at various times thereafter. If B=B 0 at t=0, the integration of equation 13 gives: ln (B / B 0 ) = – k -1 * t (eq. 14).  The plot of ln (B / B 0 ) vs t has a slope of –k -1.  Or: k -1 = 0.693 / t 1/2, where t 1/2 is the time at which B = 0.5 B 0. Time (min)

15. Inhibition Experiments c) Inhibition Experiments c) Inhibition Experiments – Basic Relationships Inhibition experiments Inhibition experiments – R T and L T are kept constant, while the concentration of the unlabelled ligand, I, is being varied. R+L+I RL + RI  By increasing I, RI will increase, thus reducing RL due to a decrease in R ( R+L+I RL + RI ).  Therefore we can derive the equation (homework problems): B = [B max * F] / [F + K D (1+I/K I )] (eq.15)  Inhibition constantK I = (R * I) / RI  Inhibition constant - K I = (R * I) / RI potency  Inhibition studies – allow to determine the potency of any drug for the receptor (at the binding site of the radioligand).

16. Inhibition Experiments Theoretical Example The THEORETICAL Data: I : 0 2 5 10 20 40 80 200 B : 100 91 80 67 50 33 20 9  IC 50 = 20.  As concentration of I is increased over 100 fold range, from 1/10 the IC 50 to 10 times the IC 50, the amount of binding decreases from 91% to 9%.  The IC 50 is not equal to the K I (IC 50 = 20, K I = 10). IC 50 - the concentration which inhibits 50% of the binding or the concentration of I which gives ½ the binding in the absence of I.

17. Inhibition Experiments The Relationship between K I and IC 50 B = [B max * F] / [F + K D (1+I/K I )]  B = [B max * F] / [F + K D (1+I/K I )] (eq.15)  Therefore we can conclude that:  [B max * F] / [F + K D (1+IC 50 /K I )] = ½ * (B max * F)/(F + K D )  2F + 2K D = F + K D (1 + IC 50 / K I )  F + K D = K D * IC 50 / K I  K I = (K D * IC 50 ) / (F + K D )  K I = IC 50 / (1 + F/K D ) (eq. 16)  The equation is used to calculate K I from experimentally determined IC 50. the unlabelled ligand and the radioligand are the sameK D = K I  Special case of inhibition experiment: the unlabelled ligand and the radioligand are the same. If, in addition, K D = K I, then the experiment is analogous to a saturation experiment (in which the specific activity of the radioligand is lowered at each higher concentration of the unlabelled ligand).

18. Inhibition Experiments Determination of IC 50 the standard dose-response semi log plot  The simplest method: the standard dose-response semi log plot: the log of the concentration (abscissa) is plotted against the percent bound (ordinate).

19. Inhibition Experiments The Logit Transformation  The logit transformation  The logit transformation (Rodbart and Frazier, 1975): logit (P) = ln [P/(100-P)] logit (P) = ln [P/(100-P)] (eq. 17) P – P – the percent bound  IC 50 – the concentration when logit=0 (P=50). Can be determined graphically or by linear regression analysis. Log I

20. Inhibition Experiments Hill Plot  When the data is expressed as percent bound – the plot is log [P/(100-P)] vs log I and the slope is n H (the Hill coefficient).  The log-logit plot and the Hill plot are identical, except that the natural logarithm (ln) is used in the former and the common logarithm (log) is used in the latter.  Hill’s equation  Hill’s equation used to describe binding of oxygen to hemoglobin (1910): log [B / (B max – B)] = n log F – log K’ D  The logit transformation  The logit transformation (Rodbart and Frazier, 1975): logit (P) = ln [P/(100-P)] logit (P) = ln [P/(100-P)] (eq. 17) P – P – the percent bound REMINDER BOX

21. Inhibition Experiments Multiple Classes of Binding Sites  In certain types of receptor binding studies, particularly those concerned with receptor subtypes, biphasic inhibition curves are observed.  Possible explanation – the unlabelled ligand has different affinities for two classes of binding sites.  This data can be analyzed by the Eadie-Hofstee plot – mathematically equivalent to the Rosenthal and Scatchard plots.

22. Inhibition Experiments Rosenthal (Scatchard) and Hill Plots 3 H – clonidine  The table contains data obtained in a saturation binding experiment using 3 H – clonidine as the radioligand and membranes prepared from rat salivary gland as the receptors. Fig. 1 Fig. 2

23. Inhibition Experiments Association and Dissociation Rate Constants 3 H-DHA  Kinetic experiments in the guinea pig cerebral cortex: column A – time, column B – specific binding. The radioligand – 3 H-DHA at a concentration of 0.36 nM. B e = 720 cpm. Amount of non-specific binding – 900 cpm, and did not change as a function of time.

24. Inhibition Experiments Study Questions and Problems (1) 1. Given the following experimental data, determine K D, B max and n H. Specific activity = 12 Ci/mmol, volume = 1 ml, efficiency = 38%.

25. Inhibition Experiments Study Questions and Problems (2) 2. Plot the following inhibition data an a semi-log plot and estimate non- specific binding. Then determine the IC 50 using a logit-log plot.

26. Inhibition Experiments Study Questions and Problems (3) 3. List the assumptions involving in analyzing saturation data by the Rosenthal (Scatchard) method to obtain a K D and B max. 4. Calculate the maximum specific activity for a radioactive compound labelled with a single atom of tritium. t 1/2 = 12.3 years; Ci = 2.22*10 12 dpm; Avogadro’s number = 6.023*10 23. 5. Given a system with competing equilibria, R + L RL B = [B max * F] / [F + K D (1+I/K I )] derive the equation B = [B max * F] / [F + K D (1+I/K I )].