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Tutorial Exercise D Angela Arsova 1. Make a literature in PubMed or ChEMBL and see if you can find the determined EC50/Ki/Kd-values for fenoterol, propranolol, xamoterol and acebutolol, remember to note the method by which the EC50/Ki/Kd values have been determined. Jozwiak et al. 2011 Fenoterol EC50 1-600 nM Acebutolol EC50 1μM Propranolol IC50 1-50 nM ChEMBL search Type in compound name and click at ‘compounds’ Click the ChEMBL ID Scroll down and find a pie chart with the bioactivity summary (e.g click at EC50) 2. Discuss any discrepancies between your calculated EC50 values and the Ki/Kd values in the literature. Litterature: Fenoterol EC50 1-600 nM Acebutolol EC50 1μM Propranolol IC50 1-50 nM EC50 and IC50 EC50: The EC50 is defined quite simply as the concentration of agonist that provokes a response halfway between the baseline (Bottom) and maximum response (Top). It is impossible to define the EC50 until you first define the baseline and maximum response. IC50: In many experiments, you vary the concentration of an inhibitor. With more inhibitor, the response decreases, so the dose-response curve goes downhill. With such experiments, the midpoint is often called the IC50 ("I" for inhibition) rather than the EC50 ("E" for effective). This is purely a difference in which abbreviation is used, with no fundamental difference. Kd and Ki Kd: Equillibration dissociation constant for labeled compound ([R]*[L])/ [RL] = Kd The Kd has a meaning that is easy to understand. Set [Ligand] equal to Kd in the equation above. The Kd terms cancel out, and you will see that [Receptor]/ [Ligand×Receptor]=1, so [Receptor] equals [Ligand×Receptor]. Since all the receptors are either free or bound to ligand, this means that half the receptors are free and half are bound to ligand. In other words, when the concentration of ligand equals the Kd, half the receptors will be occupied at equilibrium. If the receptors have a high affinity for the ligand, the Kd will be low, as it will take a low concentration of ligand to bind half the receptors. Ki: Equillibration dissociation constant for the unlabeled compound 3. Explain why the concentration response curves for fenoterol, propranolol, xamoterol and acebutolol looks different when IBMX is present. Is there any difference in the max response and EC50 values for each drug when IBMX is present? Concentration-response curves (AUC) 26 Fenoterol Fenoterol + IBMX Propranolol Propranolol + IBMX Xamoterol Xamoterol + IBMX Acebutolol Acebutolol + IBMX AUC 24 22 20 18 -10 -5 log(M) Fenoterol EC50 Fenoterol 7.533e-010 Fenoterol + IBMX 5.048e-011 Propranolol ~ 2.215e-006 Xamoterol 0.0001086 Xamoterol + IBMX 1.490e-007 Acebutolol 7.518e-006 Propranolol + IBMX 1.166e-008 Acebutolol + IBMX 1.748e-007 Concentration-response curves (12 min) 1.0 Fenoterol Fenoterol + IBMX Propranolol Propranolol + IBMX Xamoterol AUC 0.9 0.8 Xamoterol + IBMX 0.7 Acebutolol Acebutolol + IBMX 0.6 -10 -5 log(M) Fenoterol EC50 Fenoterol 7.067e-010 Fenoterol + IBMX 4.264e-011 Xamoterol ~ 3.686e+009 Propranolol ~ 5.872e-007 Xamoterol + IBMX 1.364e-007 Acebutolol 1.699e-006 Propranolol + IBMX 6.076e-008 Acebutolol + IBMX 2.033e-007 4. Explain what happens to the fenoterol curves when you add increasing concentrations of propranolol. Which type of ligand is propranolol? Concentration-response curves (AUC) 28 Fenoterol + buffer Fenoterol + propranolol (-9) Fenoterol + propranolol (-8.5) Fenoterol + propranolol (-8) Fenoterol + propranolol (-7.5) Fenoterol + propranolol (-7) Forskolin AUC 26 24 22 20 18 -13 -12 -11 -10 -9 -8 -7 -6 -5 -4 log(M) drug EC50 Fenoterol + buffer 1.043e-009 Fenoterol + propranolol (-8) 6.215e-008 Fenoterol + propranolol (-9) 1.923e-009 Fenoterol + propranolol (-7.5) 2.058e-007 Fenoterol + propranolol (-8.5) 5.859e-009 Fenoterol + propranolol (-7) 8.950e-007 Forskolin 1.801e-006 Concentration-response curves (12 min) 1.0 Fenoterol + buffer Fenoterol + propranolol (-9) Fenoterol + propranolol (-8.5) Fenoterol + propranolol (-8) Fenoterol + propranolol (-7.5) Fenoterol + propranolol (-7) Forskolin 12 min 0.9 0.8 0.7 0.6 -13 -12 -11 -10 -9 -8 -7 -6 -5 -4 log(M) drug EC50 Fenoterol + buffer 1.004e-009 Fenoterol + propranolol (-8) 6.953e-008 Fenoterol + propranolol (-9) 2.152e-009 Fenoterol + propranolol (-7.5) 2.280e-007 Fenoterol + propranolol (-8.5) 6.228e-009 Fenoterol + propranolol (-7) 1.022e-006 Forskolin 1.923e-006 5. What does the Schild plot tell you about propranolol? Schild plot Log ((A'/A)-1) 3 AUC 12 min 2 1 0 -1 -10 -9 -8 -7 -6 log(M) propranolol AUC Best-fit values Slope Y-intercept when X=0.0 X-intercept when Y=0.0 1/slope 95% Confidence Intervals Slope Y-intercept when X=0.0 X-intercept when Y=0.0 12 min 1.376 ± 0.08988 11.61 ± 0.7219 -8.439 0.7266 1.383 ± 0.09489 11.73 ± 0.7621 -8.481 0.7229 1.090 to 1.662 9.317 to 13.91 -8.636 to -8.281 1.081 to 1.685 9.308 to 14.16 -8.698 to -8.312 GraphPad Prism Help File SchildSlope quantifies how well the shifts correspond to the prediction of competitive interaction. If the competitor is competitive, the SchildSlope will equal 1.0. You should consider constraining SchildSlope to a constant value of 1.0. antagonist term, [B], is now raised to the power S, where S denotes the Schild slope factor. If the shift to the right is greater than predicted by competitive interactions, S will be greater than 1. If the rightward shift is less than predicted by competitive interaction, then S will be less than 1. The slope of a Schild plot should equal 1 if all of the assumptions underlying the method of analysis are fulfilled. A slope which is significantly greater than 1 may indicate positive cooperativity in the binding of the antagonist, depletion of a potent antagonist from the medium by receptor binding or non-specific binding (e.g. to glassware or partitioning into lipid), or lack of antagonist equilibrium. 6. Are you familiar with other types of cAMP assays? Which and what are the pros/cons compared to the real-time cAMP biosensor (EPAC)? FlashPlate® cAMP dynamic 2 assay radiolabeled cAMP and a AB against cAMP AB competion based assay based on FRET EPAC vs other cAMP assays Pros • Real-time • Cheap! Cons • Time-delay (due to handeling) • Occupy the machine in at least 30 min 7. If you were to characterize a drug from a pharmaceutical company targeting a GPCR signaling through the cAMP assay how would you then do it? Can you use the EPAC cells for that? Real-time cAMP assay HEK293 Excitatio n Wild type human β2-AR Donor EPAC149 cAMP biosensor FRET Acceptor Emission 2013.09.17 Slide no 22 Real-time cAMP assay cAMP cAMP cAMP cAMP cAMP cAMP cAMP cAMP cAMP cAMP cAMP ( d o n o r /a c c e p to r ) cAMP cAMP c A M P re s p o n s e 0 .9 0 .8 0 .7 0 .6 0 .5 0 30 60 90 120 150 T im e ( m in ) FRET Acceptor signal 2013.09.17 Donor signal cAMP = FRET = donor/acceptor Slide no 23 Receptor reserve/spare receptors • Full agonists need to activate only a small fraction of the receptors in the tissue to elicit a response (the rest of the receptors are ‘spare’) • Spare receptors make tissues more sensitive to an agonist Relationship between EC50 and Kd • Full agonist: Kd agonist concentration is not an approximation of EC50 agonist concentration because maximal response of the full agonist does not correlate with 100% receptor occupancy • For example the full agonist maybe needs to bind only 20% of the receptors to elicit a maximal response and this does not correlate with 100% binding of the receptors from which you can determine the Kd Relationship between EC50 and Kd • Partial agonist: Kd agonist concentration is an approximation of EC50 agonist concentration because the concentration producing maximal response approximates the saturation binding concentration of the agonist • The partial agonist will need to bind 100% of the receptors to reach a maximal response