Problem 3. Competition experiment.0.5 nM radioligand L* is bound to 1 ml samples of the same membrane samples used in the above experiments in the presence or absence of varying amounts of either compound A or compound B. After establishing equilibrium the following counts were bound to the membranes (after subtracting nonspecific binding):Concentration A (nM)CPM bound with AConcentration B (nM)CPM bound with B0730010730010.05721000.05663650.1712210.1608350.3679080.3456260.5648900.5365011584011243343417153104295324455663610208581034763085883011971002808100363300961300121a). Which compound A or B is more a more potent displacer of the radioligand? Justify your answer why you think so.b). Plot the fractional binding on a semilog plot and estimate the IC50s for both A and B.c). Plot the results on a logit-log plot (pseudo-Hill plot) and determine the IC50s for both compounds A and B. Use these values for the next step. How close are the computed values to your estimate in part b?d). Use the Cheng-Prusoff equation and the value of Kdfor ligand L determined by the Scatchard plot in question 2 to calculate the Ki values for compounds A and B.Problem Set PHSC5100 Concepts in Pharmaceutical Science Fall 2015Dr. Ralph Loring each question is worth 4 points for a total of 20. Revised 9/24/15Remember to show your work. This must be your own work and not someone else’s or a computer program (although it is perfectly OK to use a computer to plot your data and find slopes and intercepts). If you use Excel or some other spreadsheet explain what you are doing on a separate page. Remember we can give partial credit if you show you understand the concepts but made a math mistake while if you only give a wrong answer without any explanation you will get no credit for that problem. Please show any equations that you use. This is due October 22nd and represents 20% of your total grade. Note that this is ideal data and several values (but not all) will come out as whole numbers (or close to it due to rounding errors). Therefore a good policy would be to carry all calculations out to 4-5 place accuracy but report your final values at 2-3 place accuracy. Problem 1. Rate of dissociation Rate of association and Kd(4 points total)1 ml samples of tissue membranes were incubated overnight in a nearly saturating amount (100 nM) of radioligand L*. At time 0 1000 x [L*] of unlabeled ligand (100mM) is added to each sample and then bound ligand is separated from unbound (and non-specific binding is subtracted by using an excess competing ligand before adding the radioligand in separate samples) at 30 minute intervals to give the following counts of specifically bound ligand as a function of time: min Specific CPM bound 0 662550 30 538158 60 437120 90 355052 120 288392 150 234247 180 190268 210 154545 a. Calculate K-1 (unit = min-1) if the specific activity of the ligand is 150 Ci/mMole 1 Ci = 2.22 x 1012 disintegrations/min and the efficiency of the counter is 0.5 counts/min per disintegration/min. Calculate the number of CPM (Counts per Minute) expected for 1 pmole of bound ligand. Assume that [LR]eq = binding at time 0. Show your plot of the data and how you did your calculations. b. Why does the concentration of radioligand not matter in this experiment as long as almost all the binding sites are occupied at time 0 and that the nonspecific binding is not so high as to cause large errors in determining specific binding?c. From k-1 calculate the T½ of L* unbinding from the receptor. T½ = -ln(2)/-k-1 for dissociation. d. What technical problems might you have (if any) if your binding assay requires a 20 minute centrifugation step to separate bound from unbound? Calculate or estimate how much of a problem it would be.Problem 1 Part 2. At time zero 1 ml samples of the same tissue membranes used in problem 1 are mixed with 10 nM of radioligand L* and then the counts bound to the membranes are determined at 2 minute intervals. After subtracting nonspecific binding (determined in other samples) the following data are obtained: Time (min) Specific CPM bound 0 0 2 155836 4 272312 6 359369 8 424437 10 473071 12 509421 14 536590 16 556897 18 572075 20 583419 e. Use 10 nM [LR]eq = 3.70563828 pMole (derived from a Scatchard plot) and the K-1 value from problem 1a to calculate K 1 (unit = L/mole-1 min-1) and then from both values calculate Kd. Show your graph and your work. The values for the radioligand the volume and the counting efficiency are the same as above and throughout this entire problem set questions 1-3. Hint: remember the difference between moles and molar.f. Calculate the T½ of the ligand binding to the receptor at 10 nM ligand. T½ of association = ln(2)/Kong. What would the graph look like for [L*]=1 nM? Plot the data and calculate the T½of binding under these conditions. h. If you were setting up an equilibrium binding assay at 1 nM radioligand would 1 hour of binding be sufficient? Explain your answer. Also give an estimate of how long you would expect to wait to have 87.5% of binding at equilibrium (remember T1/2 = time to reach 50% of binding at equilibrium so 87.5% = 3 x T1/2 or Percent binding =100(1 – 1/(2x)) where x is the number of elapsed T1/2s as long as x>0). How long would it take to reach approximately 99% of the equilibrium binding? Problem 2. Equilibrium binding Scatchard plot Hill plot Kd and Bmax. The following data are from an experiment measuring equilibrium binding of various concentrations of the same radioligand L to the same 1 ml samples of membranes used in problem 1. Nonspecific binding was determined in the presence of 10 mM unlabeled ligand. Total Ligand Conc. (nM) Total cpm bound Non- specific cpm bound 0.1 15595 833 0.3 46551 2498 0.5 77164 4163 1 151938 8325 3 410233 24975 5 565970 41625 10 700239 83250 a. Plot specific binding (Total bound – nonspecific bound) against [L*]. Make initial estimates of Kd and Bmax from this graph (2A). Plot the specific binding on a semilog plot (2B) to see if that helps. b. Construct a Scatchard plot (2C) of the data and derive new estimates of Kd and Bmax. For these calculations consider that free [L*] = counts added/sample – specific cpm bound. c. Does the Kd from this Scatchard differ from that obtained in plot 2A and if so why? How does this Kd compare with that obtained measuring the rates of association and dissociation in Problem 1?d. Construct a Hill plot (Log [B/(Bmax-B) vs. log[L]). What can you conclude from the slope and intercept of this plot? SHOW YOUR WORK for all your calculations. Problem 3. Competition experiment.0.5 nM radioligand L* is bound to 1 ml samples of the same membrane samples used in the above experiments in the presence or absence of varying amounts of either compound A or compound B. After establishing equilibrium the following counts were bound to the membranes (after subtracting nonspecific binding): Concentration A (nM) CPM bound with A Concentration B (nM) CPM bound with B 0 73001 0 73001 0.05 72100 0.05 66365 0.1 71221 0.1 60835 0.3 67908 0.3 45626 0.5 64890 0.5 36501 1 58401 1 24334 3 41715 3 10429 5 32445 5 6636 10 20858 10 3476 30 8588 30 1197 100 2808 100 363 300 961 300 121 a). Which compound A or B is more a more potent displacer of the radioligand? Justify your answer why you think so. b). Plot the fractional binding on a semilog plot and estimate the IC50s for both A and B.c). Plot the results on a logit-log plot (pseudo-Hill plot) and determine the IC50s for both compounds A and B. Use these values for the next step. How close are the computed values to your estimate in part b?d). Use the Cheng-Prusoff equation and the value of Kdfor ligand L determined by the Scatchard plot in question 2 to calculate the Ki values for compounds A and B. Problem 4. Analysis of antagonist potency in tissue (Schild analysis). Below are the results of stimulating an isolated rat ileum (part of the small intestine) with an agonist. The agonist’s molar concentrations are presented in the leftmost (first) column (Example 3E-07 = 3×10-7 M = 300 nM). The normalized control responses to the agonist are presented in the second column as a percentage of the maximum muscle contraction. These measurements are repeated in the presence of an antagonist at varying concentrations. The molar concentration of the antagonist is placed at the top of each subsequent column going to the right and the responses of the tissue are shown below. For Antag =0 For Antag =2×10-9M For Antag = 2×10-8M For Antag= 2×10-7 M Agonist (M) 0.003 98.67 88.22 0.001 99.56 96.12 0.0005 99.13 92.52 0.0003 98.55 88.13 42.83 0.0001 99.60 95.79 19.98 0.00005 99.21 91.91 11.10 0.00003 98.68 87.21 42.61 6.97 0.00001 96.15 19.84 2.44 0.000005 92.59 11.01 0.000003 88.24 40.54 6.91 0.000001 18.52 2.42 5E-07 10.20 3.00E-07 42.86 6.38 1.00E-07 20.00 2.22 5.00E-08 11.11 3.00E-08 6.98 1.00E-08 2.44 The experiment involves taking smooth muscle in a bath connected to a force transducer. Full activation with agonist gives a 10 gm contraction with maximum agonist and no antagonist. Varying doses of agonist are given in the presence or absence of 3 doses of antagonist. Some data is missing. The data is recorded as % of maximum response:Questions:a) Is the antagonist a competitive or noncompetitive antagonist? Justify your answer by showing a plot. b) Define pA2 in your own words. c) Calculate pA2 of the antagonist. (Hint: To calculate pA2 you need to compare equivalent responses. What plot can you use to linearize the responses so you can compare equivalent responses? Show your linearized results and how you used them to calculate pA2) Convert pA2 to a concentration. d) Why do you need to compare equivalent responses? Why is this analysis useful? Give two examples
