PHARMACODYNAMICS II. The total number of receptors, [R T ] = [R] + [AR] + [BR] (A = agonist, B = antagonist, R = receptors) = T. Antagonist present

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1 Pharmacology Semester 1 page 1 of 5 PHARMACODYNAMICS II Antagonists Are structurally similar to the binding site of a receptor and thus show affinity towards the receptor. However, they have zero intrinsic efficacy (you don t see any effect it is silent). They compete with the agents for the receptor binding site. They can bind reversibly, and thus display competitive inhibition They can bind irreversibly (non-surmountable) via covalent bonds. The agonist cannot displace the antagonist from the binding site. Antagonists are important for receptor classification. This is because, theoretically, an antagonist is designed to work at a specific type of receptor, whereas an agonist may have multiple effects on a wide range of receptors in different tissues. Competitive antagonism We have 2 drugs acting at the one receptor, both drugs wanting to occupy the same binding site. Agonist: [A] + [R] [AR] Antagonist: [B] + [R] [BR] The total number of receptors, [R T ] = [R] + [AR] + [BR] (A = agonist, B = antagonist, R = receptors) The Gladdum equation can be used to describe the occupancy curve of an agonist in the presence of an antagonist. [AR] = [R ] T [B ] [A'] + K A 1 + K B Where [A'] = new K A of the agonist (the concentration at which 50% of receptors are occupied by an agonist in the presence of an antagonist. K B = concentration of antagonist at which 50% of the receptors are occupied by antagonist. [A' ] No antagonist present, B=0 K A A' Antagonist present Concentration of agonist All the antagonist (B) has done is change the location of the curve to the right (there has been no change in shape and no change in efficacy. Essentially, what B has done is dilute the concentration of A, so that you need more agonist to get the same response. A very important thing to note: The occupancy curve is not the same as the response curve IN THEORY. However, since their shapes are the same, we tend to work more with response curves because measuring the response is much easier than measuring the occupancy of a drug. Hence, instead of working with K A (which is the concentration of agonist required to occupy 50% of the receptors), we tend to work more with EC50 (the concentration of agonist required to get 50% of a response). Remember that K A and EC50 are not always the same. Also, K A is very hard to measure (because occupancy is hard to measure). We tend to measure EC50 more easily by just reading off the response curve. Why do we want to measure K B? K B is specific for a type of receptor, and hence it is a good way to characterise a particular receptor. We can find the K B very easily by plotting a Schild plot.

2 Pharmacology Semester 1 page 2 of 5 The schild plot The Schild plot can only be used when we are working with competitive antagonist. The following are the criteria for a competitive antagonist: In the presence of an antagonist, the response curve must be shifted to the right The efficacy of the antagonist must not be changed (the maximum response is still possible) The shape of the curve must be identical If all these are correct, the slope of the Schild plot should be equal to 1 To plot a Schild plot, we take the following steps (an example is given): Step 1: Plot the response curves of an agonist in the presence of varying concentrations of antagonist. Curve A is the control and has no antagonist Curve B has a concentration of antagonist of 0.3 nm Curve C has a concentration of antagonist of 1.0 nm Curve D has a concentration of antagonist of 3.0 nm Response A B C D Concentration of agonist Step 2: The next step is to find the EC50 of each of the curves by reading off the graph: EC50 of curve A = 0.3 µm EC50 of curve B = 1.0 µm EC50 of curve C = 3.0 µm EC50 of curve D = 10.0 µm Step 3: Find the concentration ratio of each of the curves (except curve A) The concentration ratio is a measure of how much the curve has shifted to the right. A concentration ratio of 2 means that there is a two fold shift of the normal curve (curve with no antagonist present) to the right. A CR of 5 means a five fold shift of the curve to the right. The concentration ration can be calculated by: EC50 of the shifted curve CR= EC50 of the normalcurve In our example, EC50 of the normal curve (curve A) is 0.3 CR of curve B = 1.0/0.3 = 3.3 CR of curve C = 10

3 Pharmacology Semester 1 page 3 of 5 CR of curve D = 33 Step 4: The equation of the Schild plot is: We plot log(cr 1) vs log[antagonist] In our example: log(cr 1) = log[antagonist] - logk B Curve [Antagonist] log[antagonist] CR CR 1 Log(CR 1) nm A 0 B C D Log(CR 1) logk B Log[Antagonist] When log(cr 1) = 0, i.e. when CR = 2, then log[antagonist] = logk B. This means that the intersection of the line with the x axis is equal to logk B K B can thus be defines as the concentration of antagonist required to shift the dose-response curve 2 fold to the right (i.e. we want to increase the EC50 of the antagonist by 2 times). In this example, K B turns out to be 0.31nM A clinical example Propranalol if a β adrenoreceptor antagonist. It has a K B of M This means that we need a concentration of M of propranalol to occupy ½ of the β receptors. This will lead to a 2 fold shift in the normal curve of agonist (lets say isoprenalol) acting on the β receptor. We measure the response as being the heart rate. Under normal conditions (with no propranalol) isoprenalol can give a heart rate of 50 beats per minute at a given concentration of 5mM (the EC50). In the presence of M propranalol, the EC50 of the curve has been shifted 2 fold to the right, i.e. the new EC50 is 10mM. Heart rate M of propranalol 50 25

4 Pharmacology Semester 1 page 4 of Concentration of isoprenalol We can interpret the outcome of using the antagonist in 2 ways: At the same concentration of isoprenalol (5mM) we can get 50 heart beats without an antagonist or 25 heart beats in the presence of antagonist. To maintain the heart rate at 50 beats, we need to increase the concentration of isoprenalol (or if this were a natural β agonist, work much harder) Irreversible antagonists Irreversible antagonists bind irreversibly to the receptor. If the tissue has spare receptors, a small concentration of irreversible antagonist will bind to some receptors, but the remaining number of receptors is sufficient to maintain a maximal response. As we gradually increase the concentration of antagonist, we find that more and more of the receptors are being irreversibly bound, resulting in fewer free receptors for the agonist. As a result, the maximal response starts to diminish, until we add enough antagonists to fill up all the receptors, leaving none left for the agonist to bid to. This would result in total inhibition of the agonist, with 0 response. Response As you can see, the curves are quite different to a competitive (reversible) antagonist The curves are not the same shape The maximal response is not maintained We cannot construct a schild plot for this antagonist! Partial agonists A partial agonist has a lower maximal response than a full agonist (it has a lower intrinsic efficacy). Let me give an analogy: Normal curve, no partial agonist present Increase in initial response due to partial agonist helping low concentrations of full agonist. Competitive inhibition of the full agonist by the partial agonist because the partial agonist is occupying sites that the full agonist wants to have Increasing concentration of full agonist

5 Pharmacology Semester 1 page 5 of 5 Say we have 10 punching bags. If we have one strong adult (full agonist) present, he punches one bag and gets a response of say 5 If we have 1 adult and 1 kid (partial agonist), the adult can give a response of 5 but the kid hits the bag and gets a response of 2. The total repines is 7. If we have 1 adult and 3 kids, the total initial response is 11. If we have 1 adult and 9 kids, the total initial response is 23. See how the presence of a partial agonist at low concentrations of full agonist will raise the initial response. However, say we have 10 kids and 1 adult. The 10 kids will probably all be using the bags and so the maximum response is 20. The 1 adult does not get a go. If we increase the number of adults, they will be able to kick the kids off and use the bags themselves, but the kids will be trying to push in and play. The effectiveness of the adults is reduced because they have to waste time telling the kids to get lost. With heaps of adults, say 50 and 10 kids, the kids will be too scared to push in and the adults can go about their business punching bags to their maximum capabilities. When we have a partial agonist in conjunction with a full agonist, initially, we get an increased response due to the additive effects of partial agonist with small concentrations of full agonist. However, the partial agonist is also acting as a competitive antagonist because it occupies the same binding site as the full agonist but has lower efficacy. Every receptor that it occupies prevents the full agonist from binding and producing a better response. That is why we see a shift in the curve to the right. The partial agonist is preventing the full agonist from working to its full potential by binding to sites. With enough concentration of agonist, the partial agonist will be overwhelmed and the full agonist can operate to give a maximum response. Efficacy is the property of a tissue. A drug can have a better efficacy on one tissue than another due to: Differences in receptor number Differences in transducer coupling