Pharmacological Response

Pharmacological Response CHAPTER 3 Author: Michael Makoid and John Cobby Reviewer: Phillip Vuchetich OBJECTIVES After completing this chapter, the student will be able to: 1. Given patient data of the following types, the student will be able to properly con- struct (III) a graph and compute (III) the slope using linear regression: response (R) vs. concentration (C), response (R) vs. time (T), concentration (C) vs. time (T) 2. Given any two of the above data sets, the student will be able to compute (III) the slope of the third by linear regression. 3. Give response vs. time and response versus concentration data, the student will be able to compute (III) the terminal (elimination) rate constant and half life of the drug. 3-1 Basic Pharmacokinetics REV. 00.1.14 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf

Pharmacological Response 3.1 Pharmacological Response One theory (A.J. Clarke) on the mechanism of action of drugs is the occupation Drug must get into blood and blood is in contact theory. It suggests that the intensity of a pharmacological response (E) is propor- with receptor. tional to the concentration of a reversible drug-receptor complex 3.1.1 THE HYPERBOLIC RESPONSE EQUATION A mathematical description of the occupation theory, assuming complete and instantaneous drug distribution, yields [ D ]E max E = --------------------- - (EQ 4-34) KR + [D ] where E is the intensity of the pharmacological response, Emax is the maximum attainable value of E , [ D ] is the molar concentration of free drug at the active com- PKAnalyst Plot 1.0 plex and K R is the dissociation constant of the drug-receptor complex. 0.8 0.6 [D] If is plotted against a hyperbolic curve will result; the asymptote will be E E 0.4 E max . 0.2 0.0 0.0 0.8 1.6 2.4 3.2 4.0 a. If linear pharmacokinetics hold, the molar concentration of free drug at the D active site is proportional to the plasma concentration of the drug once equilibrium has been established. Hence, a plot of E against Cp will also be hyperbolic. X = V ⋅ Cp , b. Because the mass of drug in the body is a plot of against will be E X hyperbolic. c. For a series of doses the value of X at the same given time after dosing is proportional to the dose (D). Thus, a plot of E against D will also be hyperbolic at a specific time. d. Any hyperbolic curve, if plotted on reverse semilogarithmic paper (i.e., 1.0 abscissa is logarithmic), has a sigmoid shape. If we plot E against Cp (of X , or D ) 0.8 0.6 in this manner, the plot is virtually linear in the range E ⁄ E max = 0.2 → 0.8 ; and if Response 0.4 this is the clinical range of responses, linear equations may be written. For exam- 0.2 ple, 0.0 10 -810-710 -610-510 -410-310-210-1 E = m ⋅ ln x + b Conc. (EQ 4-35) where is the slope m 3-2 Basic Pharmacokinetics REV. 00.1.14 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf

Pharmacological Response This example equation shows that, in the clinical range, the intensity of a pharma- Plot of Response vs. Ln(C) is a straight line in cological response is proportional to the logarithm of the administered dose, pro- the middle (if you viding response is measured at a consistent time after dosing. The proportionality squint), but only constant (slope, m ) is a function of the affinity of the drug for the receptor. In fact, between 20% and 80% equation 4-35 yields a log-dose response plot. Note that doubling the dose does not maximum response double the response. 3.1.2 INTERRELATIONSHIPS BETWEEN CONCENTRATION, TIME AND RESPONSE Pharmacological Response (R), Concentration (C), and Time (t) are interrelated. The response and concentration relationship is studied in pharmacology. The concentration and time relationship is studied in pharmacokinetics. The response and time relationship is applied in therapeutics. You should know what the various graphical relationships look like. Response vs. Remember: Use only the data between 20% natural log of concentration is sigmoidal. (S shaped). We are interested in the mid- and 80% of maximum dle almost straight part. The slope is dR ⁄ d ln c . response for the straight part of both response dR ⁄ dt . Response vs. time is a straight line. The slope is vs. Ln(c) and response vs. t. Natural log of concentration vs. time (drug given by IV bolus) is a straight line. The slope is d ln c ⁄ dt . You should be able to obtain the slope of each of these relationships from data sets. You should be able to obtain the third slope’s relationship given the other two (or data sets with which to get the other two). dR ------ = ---------- ⋅ d ln c dR- ---------- (EQ 4-36) - - d ln c dt dt dR ⁄ dt- dR - ---------- = ------------------- (EQ 4-37) d ln c ⁄ dt d ln c dR ⁄ dt - d ln c ---------- = --------------------- - (EQ 4-38) dR ⁄ d ln c dt You should be able to apply the equation y = mx + b to each of the above relation- NOTE: Only between 20% and 80% of maxi- ships. Given the slope (or having obtained the slope) and two of the three variables mum response!!!!!! (y, x, b), you should be able to find the third. 3-3 Basic Pharmacokinetics REV. 00.1.14 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf

Pharmacological Response 3.2 Change in Response with Time 3.2.1 ONE-COMPARTMENT OPEN MODEL: INTRAVENOUS BOLUS INJECTION – Kt – Kt X = X0 e = De (EQ 4-39) or Ln ( X ) = Ln ( D ) – Kt (EQ 4-40) Substituting twice from eq. 4-35 once at time t and once at zero time E – b E0 – b ----------- = -------------- – Kt → E = E0 – Rt - - (EQ 4-41) m m Hence a plot of the intensity of the pharmacological response at any time ( E ) against time declines linearly. The slope is –R = ( –K ⋅ m ) and the intercept is E0 (the initial intensity). 3.2.2 ONE-COMPARTMENT OPEN MODEL: ORAL ADMINISTRATION Because E is proportional to ln x at any time, a plot of E against t will be analo- Response follows plasma profile. gous to a plot of ln x against t . Hence E will rise at first and then decline with time. When t is large, the terminal slope will be –R . DURATION OF EFFECTIVE PHARMACOLOGICAL RESPONSE ( t dur ) 3.2.3 Once equilibrium has been established, there is a minimum plasma concentration Duration of action is related to how long below which no pharmacological response is seen; this concentration is ( C p ) eff or plasma concentration is MEC . For an intravenous bolus injection, the time to reach ( C p ) eff is t dur . above Minimum Effec- tive Concentration. – Ktd ur ( C p )eff = ( C p ) 0 e multiplying by the volume of distribution we obtain ln ( X eff ) = ln ( D ) – Kt dur (EQ 4-42) Rearranging, 3-4 Basic Pharmacokinetics REV. 00.1.14 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf

Pharmacological Response ln  -------- D  X eff = ------------------- - t dur (EQ 4-43) K The duration of effective pharmacological response is proportional to the (natural) logarithm of the dose. A second rearangement of equation 4-42 results in : ln ( Dose ) ln ( Xeff ) t dur = ----------------------- – ------------------ - (EQ 4-44) K K Thus a plot of duration of action vs ln dose would result in a straight line with a ln ( X ef f ) slope of 1/K and an x intercept of – ------------------- . K 3.2.4 PHARMACOKINETIC PARAMETERS FROM RESPONSE DATA The measurement of pharmacological effect provides a non-invasive means of How can I get the elimination rate constant obtaining the value of t 1 ⁄ 2 (but not V ). from pharmacological data? Use this “cook- a. Obtain a log dose-response plot (Eq. 4-37). The response must always be mea- book.” sured at the same time after administering the dose. (m) b. Find the slope of this plot. Remember: Use only the data between 20% and 80% of maximum c. Obtain a response against time plot for a single dose (Eq. 4-36). Response for both of these plots. ( –R ) d. Find the terminal slope of this plot. R e. Calculate K = --- . - m t 1 ⁄ 2 =  ------------  0.693 f. Calculate . - K 3.2.5 “DELAYED” RESPONSE If a drug does not distribute instantaneously to all the body tissues (including the Two compartment model - biophase is in active site), the pharmacological response will not always parallel the drug con- second compartment. centrations in the plasma. In such a situation the response may parallel the mass of drug presumed to be in a second compartment ( X 2 ) , and hence seem “delayed”. Eventually, however, once equilibrium is attained, the response will parallel plasma concentrations. In such a case, E is proportional to ln X 2 . 3-5 Basic Pharmacokinetics REV. 00.1.14 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf

Pharmacological Response Thus a plot of E against X1 (or E against Cp ) will show a hysteresis loop with time, most noticeably during an intravenous infusion. 3.2.6 RESPONSE OF ACTIVE METABOLITE: In the case of an inactive prodrug yielding an active metabolite, the response Parent compound (inactive) yields active curves will mirror the active metabolite plasma profile (assuming the biophase is daughter compound. the plasma) and not the prodrug plasma profile. 3-6 Basic Pharmacokinetics REV. 00.1.14 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf

Pharmacological Response 3.3 Therapeutic Drug Monitoring The pharmacokinetics of a drug determine the blood concentration achieved from Part of Pharmaceutical Care! a prescribed dosing regimen. During multiple drug dosing, the blood concentration will reflect the drug concentration at the receptor site; and it is the receptor site concentration that determines the intensity of the drug’s effect. Therefore, in order to predict a patient’s response to a drug regimen, both the pharmacokinetics and pharmacological response characteristics of the drug must be understood. Phar- macological response is closely related to drug concentration at the site of action. We can measure plasma concentration and assume that the site of action is in rapid equilibrium with the plasma since we usually do not measure drug concentration in the tissue or at the receptor site. This assumption is called “kinetic homogeneity” and is the basis for clinical pharmacokinetics. There exists a fundamental relationship between drug pharmacokinetics and phar- Need to keep plasma concentration in the macologic response. The relationship between response and ln-concentration is therapeutic range to sigmoidal. A threshold concentration of drug must be attained before any response optimize therapy. is elicited at all. Therapy is achieved when the desired effect is attained because the required concentration has been reached. That concentration would set the lower limit of utility of the drug, and is called the Minimum Effective Concentra- tion (MEC). Most drugs are not “clean”, that is exhibit only the desired therapeutic response. They may also exhibit undesired side effects, sometimes called toxic effects at a higher, (hopefully a lot higher), concentration. At some concentration, these toxic side effects become become intolerable/and or dangerous to the patient.. That concentration, or one below it, would set the upper limit of utility for the drug and is called the Maximum Therapeutic Concentration or Minimum Toxic Concentration (MTC). Patient studies have generated upper (MTC) and lower (MEC) plasma concentration ranges that are deemed safe and effective in treating specific disease states. These concentrations are known as the “therapeutic range” for the drug (Table 4-18). When digoxin is administered at a fixed dosage to numerous subjects, the blood concentrations achieved vary greatly. Clinically, digoxin concentrations below 0.8 ng ⁄ ml will elicit a subtherapeutic effect. Alternatively, when the digoxin concentration exceeds 2.0 ng ⁄ ml side effects occur (nausea and vomiting, abdominal pain, visual disturbances). Drugs like digoxin possess a narrow therapeutic index because the concentrations that may produce toxic effects are close to those 3-7 Basic Pharmacokinetics REV. 00.1.14 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf

Pharmacological Response required for therapeutic effects. The importance of considering both pharmacokinetics and pharmacodynamics is clear. TABLE 4-18 Average therapeutic drug concentration DRUG RANGE 0.8-2.0 ng ⁄ ml digoxin 2-10 µg ⁄ ml l gentamicin 1-4 µg ⁄ ml lidocaine 0.4-1.4 mEq ⁄ L lithium 10-20 µg ⁄ ml phenytoin 10-30 µg ⁄ ml phenobarbitol 4-8 µg ⁄ ml procainamide 3-6 µg ⁄ ml quinidine 10-20 µg ⁄ ml theophylline Note that drug concentrations may be expressed by a variety of units. Pharmacokinetic factors that cause variability in plasma drug concentration are: • drug-drug interactions • patient disease state • physiological states such as age, weight, sex • drug absorption variation • differences in the ability of a patient to metabolize and eliminate the drug If we were to give an identical dose of drug to a large group of patients and then measure the highest plasma drug concentration we would see that due to individual variability, the resulting plasma drug concentrations differ. This variability can be attributed to factors influencing drug absorption, distribution, metabolism, and excretion. Therefore, drug dosage regimens must take into account any disease altering state or physiological difference in the individual. Therapeutic drug monitoring optimizes a patient’s drug therapy by determining plasma drug concentrations to ensure the rapid and safe drug level in the therapeutic range. • Assays for determination of the drug concentration in plasma Two components make up the process of • Interpretation and application of the resulting concentration data to develop a safe and effective therapeutic drug drug regimen. monitoring: 3-8 Basic Pharmacokinetics REV. 00.1.14 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf

Pharmacological Response The major potential advantages of therapeutic drug monitoring are the maximiza- tion of therapeutic drug benefits and the minimization of toxic drug effects. The formulation of drug therapy regimens by therapeutic drug monitoring involves a process for reaching dosage decisions. 3-9 Basic Pharmacokinetics REV. 00.1.14 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf

Pharmacological Response 3.3.1 THERAPEUTIC MONITORING: WHY DO WE CARE? The usefulness of a drug’s concentration vs. time profile is based on the observa- tion that for many drugs there is a relationship between plasma concentration and therapeutic response. There is a drug concentration below which the drug is inef- fective, the Minimum Effective Concentration (MEC), and above which the drug has untoward effects, the Minimum Toxic Concentration (MTC). That defines the range in which we must attempt to keep the drug concentration (Therapeutic Range). The data in Table 4-18 are population averages. Most people respond to drug concentrations in these ranges. There is always the possibility that the range will be different in an individual patient. For every pharmacokinetic parameter that we measure, there is a population average and a range. This is normal and is called biological variation. People are different. In addition to biological variation there is always error in the laboratory assays that we use to measure the parameters and error in the time we take the sample. Even with these errors, in many cases, the therapy is better when we attempt to monitor the patient’s plasma concentration to optimize therapy than if we don’t. This is called therapeutic monitoring. If done properly, the plasma concentrations are rapidly attained and maintained within the therapeutic range throughout the course of therapy. This is not to say all drugs should be monitored. Some drugs have a such a wide therapeutic range or little to no toxic effects that the concentrations matter very little. Therapeutic monitoring is useful when: • a correlation exists between response and concentration, • the drug has a narrow therapeutic range, • the pharmacological response is not easily assessed, and • there is a wide inter-subject range in plasma concentrations for a given dose. In this era of DRGs, where reimbursement is no longer tied to cost, therapeutic monitoring of key drugs can be economically beneficial to an institution. A recent study (DeStache 1990) showed a significant difference with regard to length of stay in the hospital between the patients on gentamicin who were monitored (and their dosage regulated as a consequence) vs. those who were not. With DRGs the hospital was reimbursed a flat fee irrespective of the number of days the patient stayed in the hospital. If the number of days cost less than what the DRG paid, the hospital makes money. If the days cost more than the hospital loses money. This study showed that if all patients in the hospital who were on gentamicin were monitored, the hospital would save $4,000,000. That’s right FOUR MILLION per year. I would say that would pay my salary, with a little left over, and that is only one drug! 3-10 Basic Pharmacokinetics REV. 00.1.14 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf

Pharmacological Response • The process of First the MD must order the blood assays. therapeutic monitoring • Second, someone (nurse, med tech, you) must take the blood. takes effort. • Someone (lab tech, you) must assay the drug concentration in the blood. • You must interpret the data. • You must communicate your interpretation and your recommendations for dosage regimen change to the MD. This will allow for informed dosage decisions. • You must follow through to ensure proper changes have been made. • You must continue the process throughout therapy. Therapeutic drug monitoring, in many cases, will be part of your practice. It can be very rewarding. Thus, if we have determined the therapeutic range, we could use pharmacokinetics to determine the optimum dosage regimen to maintain the patient’s plasma concentration within that range. Selected References 1. Nagashima, R., O’Reilly, RA., and Levy, G, Kinetics of pharmacologic effects in man: the anticoagulant action of warfarin. Clin. Pharm. Therap, 10 22-35 (1969). Remember: We want the straight part! 2. Wagner, J.G, Relations between drug concentration and response. J. Mond. Pharm., 4, 279-310 (1971). 3. Gibaldi M. and Levy, G. Dose-dependent decline of pharmacologic effects of drugs with linear pharmacokinetics characteristics. J.Pharm.Sci, 61, 567-569 (1972). 4. Brunner, L., Imhof, P., and Jack, D. Relation between plasma concentrations and cardiovascular effects of oral oxprenolol in man. Europ. J. Clin. Pharmacol., 8, 3-9 (1975). 5. Galeazzi, R.L., Benet, L.Z., and Sheiner, L.B. Relationship between the pharmacokinetics and pharmacodynamics of procainamide. Clin. Pharm. Therap., 20, 67-681 (1976). 6. Joubert, P., et al. Correlation between electrocardiographic changes, serum digoxin, and total body digoxin content. Clin. Pharm. Therap., 20, 676-681 (1976). 7. Amery, A., et al. Relationship between blood level of atenolol and pharmacologic effect. Clin. Pharm. Therap., 21, 691-699 (1977). 3-11 Basic Pharmacokinetics REV. 00.1.14 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf

Pharmacological Response 3.4 Problems What to do.---> We want to get pharmacokinetic data (elimination rate constant) from pharmacological response data (Response vs concontration and Response vs time graphs) . Response vs Time Graph Plot Response vs Time on Cartesian (regular) Graph Paper. 1. Use Response data between 20 and 80 percent of maximum (Pick the straight part) to do the lin- 13. 1.0 ear regression on. (Rule of thumb: Connect first and last data point with a straight line. If all 0.8 the points fall on one side of the line, its not straight! Response 0.6 dR Find the slope of the straight line, ------ , (eyeball the rise over the run or use linear regression as - 14. dT 0.4 required). Important: you must determine the best fit line through all of the points that you will 0.2 use. Eyeball method: Get the line as close to the points as possible placing as many points above the line as below the line. Take two points on the line (not data points) to calculate the 0.0 10 10 10 10 10 10 Time change in Y over the change in X. Response vs Ln(Concentration) Graph Turn semi-log paper on its side so that the numbers are on the top. 1. 100 100 10 10 1 10 1.0 10 10 0.8 0 0.6 Response 0.4 0.2 Response 0.0 10 10 10-810-710-610-510 -410-310 -210-1 Conc. 1 Concentration 1 10 0 1 What we are attempting to do is get the logarithm part of the paper on the x axis and have the numbers get bigger as you go from left to right. Plot concentration on the x axis and response on the y. 15. Find the slope of the line plotted this way by the rise over the run method. 16. Run is change in ln(C). If you take any two concentrations such that C2 = 2*C1 then the run is (ln(C2) - ln(C1)). Using rules of logs, when two logs are subtracted, the numbers are devided, thus: = ln(C2/C1). If C2 = 2*C1 then ln(C2/C1) = ln(2) = 0.693. 3-12 Basic Pharmacokinetics REV. 00.1.14 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf

Pharmacological Response Rise = change in Response. Take the difference of the two responses coresponding to the concentrations picked. (R2-R1). R –R rise 2 1 The slope of the line is m = ------- = ----------------- - - 17. run 0.693 Ln(Concentration) vs Time Graph (Pharmacokinetic Data) If you have concentration vs Time data: Plot Concentration vs time on semi-log paper (Y axis is concentration this time) 1. 100 10 Concentration 10 10 110 0 1 Time Find the slope as before, using semi log paper (Remeber the log is on the Y axis this time, so 18. you find two concentrations such that c2 = 2*c1 and put it in the rise this time. Thus the slop of rise 0.693 0.693 the line is m = ------- = ------------- = ------------ = – k - - - run t2 – t1 –t1 -- - 2 If you have pharmacological response data: Divide the slope of the Response vs Time graph by the slope of the Response vs ln(C) graph: 1. dR ------ - slope of r vs t dT dln(C) ----------------------------------------- = --------------- = --------------- = m = – k - slope of r vs ln(c) dR - dT -------------- dln(C) Both methods should be equivalent. Additional problems are available in chapter 14, practice exams. 3-13 Basic Pharmacokinetics REV. 00.1.14 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf

Pharmacological Response 3.5 Oxpranolol Brunner et al, Europ. J. Clin. Pharmacol., 8, 3-9 (1975). In humans, the pharmacological response to oxpranolol (a beta blocker) is a decrease in beats per minute (bpm) compared to placebo during physical exercise. The following approximate mean data is from 7 healthy volunteers: beats per minute (bpm) altered with time (t) after oral administration of three doses (D). TABLE 4-19 Response vs Response vs Concentration time BPM Dose (mg) BPM Time (hr) 10 40 17.6 1 13.5 60 13.9 2 16 80 10.2 3 19 120 6.6 4 21 160 TABLE 4-20 Oxpranolol plasma concentration following 160 mg IV dose C p  -----  ng -  ml Time (min) 30 699 60 622 120 413 150 292 240 152 360 60 480 24 1. Calculate the half life ( t 1 ⁄ 2 ) of oxpranolol from the pharmacological response table. 2. Plot plasma concentration data on Cartesian graph paper directly as well as transforming Cp into ln C p . 3. Plot plasma concentration data on semilog paper. Use linear regression to find the rate constant of elimination of oxpranolol. 4. Calculate the half life obtained from the concentration data and compare it with the half life calculation based on the pharmacological response. 3-14 Basic Pharmacokinetics REV. 00.1.14 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf

Pharmacological Response Minoxidil (Problem 4 - 1) Shen et al. Clin. Pcol. Ther 17:593-8 (1975) Minoxidil is a potent antihypertensive which lowers the mean arterial blood pressure (MAP) in certain patients. PROBLEM TABLE 4 - 2. Minoxidil Initial decrease in MAP ( mmHg ) Dose ( mg ) 17 2.5 40 5.0 53 7.5 63 10.0 76 15 Minoxidil PROBLEM TABLE 4 - 3. 25 mg I.V. Bolus yielded: Decrease in MAP ( mmHg ) Time ( hr ) 75 20 66 30 56 40 48 50 From the preceding information, determine the following: dR 1. Graph and find ------------ (slope of (R)esponse vs. ln(C)oncentration graph). d ln C dR 2. Graph and find ------ (slope of (R)esponse vs. (T)ime graph). - dt dR ------ - dt - 3. Find the ln(C)oncentration vs. (T)ime slope : ------------ : Note that your slope m = – K . If you are having problems dR ------------ d ln C understanding this, refer to Sections 2.4.2 -2.4.4. K is the elimination rate constant. 4. Calculate t 1 ⁄ 2 =  0.693  . ------------ - K 3-15 Basic Pharmacokinetics REV. 00.1.14 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf

Pharmacological Response Propranolol (Problem 4 - 4) Citation? Beta blockers can be considered first line drugs of choice in the treatment of hypertension in certain patients. The following data was obtained regarding Propranolol used to treat hypertension in a group of patients. Propranolol PROBLEM TABLE 4 - 5. Cp Fall in Systolic BP (mmHg) 20 50 16 40 11 30 5 20 Propranolol PROBLEM TABLE 4 - 6. I.V. Bolus dose of Propranolol Fall in Systolic BP (mmHg) Time (hr) 24 1 20 2 19 3 9 6 From the preceding information, determine the following: dR 1. Graph and find ------------ (slope of (R)esponse vs. ln(C)oncentration graph). d ln C dR 2. Graph and find ------ (slope of (R)esponse vs. (T)ime graph). - dt dR ------ - dt - ------------ : Note that your slope m = – K . If you are having problems 3. Find the ln(C)oncentration vs. (T)ime slope : dR ------------ d ln C understanding this, refer to Sections 2.4.2 -2.4.4. 4. Calculate t 1 ⁄ 2 =  0.693  . ------------ - K 3-16 Basic Pharmacokinetics REV. 00.1.14 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf

Pharmacological Response 3.5.1 ANSWERS: OXPRANOLOL 1. Calculate the half life ( t 1 ⁄ 2 ) of oxpranolol from the pharmacological response table Oxpranolol 22 20 Response (BPM) 18 16 14 12 10 3 1 2 10 10 10 Dose (mg) TABLE 5. X Y 2 X⋅Y X ln(Dose) Dose Response 3.689 40 10 13.61 36.89 4.094 60 13.5 16.76 55.27 4.382 80 16 19.20 70.11 4.787 120 19 22.92 90.96 5.075 160 21 25.75 106.58 ΣX = 22.03 ΣY = 79.5 ΣXY = 359.82 2 ΣX = 98.25 2 ( ΣX ) = 485.23 ΣX Σy ) ) X = -------- = 4.41 - y = ----- = 15.9 - n n ( Σ(x) ⋅ Σ(y )) – (n ⋅ Σ(x ⋅ y)) Slope of the line from linear regression. Chapter 2.4.4 m = -------------------------------------------------------------------- - 2 2 [Σ(x) ] – ( n ⋅ Σ (x )) ( 22.03 ⋅ 79.5 ) – ( 5 ⋅ 359.82 ) m = ------------------------------------------------------------------- = 7.93 - 485.32 – ( 5 ⋅ 98.25 ) 3-17 Basic Pharmacokinetics REV. 00.1.14 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf

Pharmacological Response dR - ---------- = 7.93 the slope is equal to the linear regression of the change in response vs. ln concentration. d ln c OXPRANOLOL 18 16 Response (BPM) 14 12 10 8 6 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Time (hr) R –R 16 – 10 dR 1 2 The slope of this plot is m = ----------------- = -------------------------- = – 3.71 therefore, ------ = – 3.71 . - - - T1 – T 2 1.45 – 3.07 dt dR ------ - ----------- = d ln c = – k = –3.71 = – 0.4678hr –1 dt ln 2 0.693 - ---------- - ------------ - t 1 ⁄ 2 = ------- = -------------------------- = 1.48hr half life (89 min). - –1 dR - dt 7.93 k 0.4678hr ---------- d ln c 2. Plot plasma concentration data on Cartesian graph paper directly as well as transforming Cp into ln C p . 3-18 Basic Pharmacokinetics REV. 00.1.14 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf

Pharmacological Response Plasma concentration vs. Time Oxpranolol 800 3 10 640 Concentration (ng/mL) 480 2 10 320 Concentration (ng/ml) 160 1 10 0 100 200 300 400 500 0 0 100 200 300 400 500 Time (min) Time (min) 3. Plot plasma concentration data on semilog paper. Use linear regression to find the rate constant of elimination of oxpranolol. Using linear regression, as described above, the elimination rate constant is approximately 0.007797 min-1 * (60 min/hr) = 0.4678 hr-1 4. Calculate the half life obtained from the concentration data and compare it with the half life calculation based on the pharmacological response. 0.693 - t 1 ⁄ 2 = ------------------ = 90min = 1.5 hrs compared to 1.48 hours (89 min) from the pharmacological response 0.00763 method. 3-19 Basic Pharmacokinetics REV. 00.1.14 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf

Pharmacological Response 3.5.2 ANSWERS: MINOXIDIL NOTE: Answers will vary depending on whether linear regression is calculated via calculator or using the formula such as observed in Problem 1. Either method can be used. However, if you use the formula, you should be within 10% of the calculated answer. A word of caution: if you choose to do linear regression via calculator make sure you have valid data. This cannot be assured until you have graphed all the data points given. Many a student has incorrectly calculated parameters because he/she falsely assumes that all the points are valid. Blindly choosing data points for linear regression will only lead to error. Every problem in this manual has been derived from actual journal articles and will therefore be “real” data. This real-world data is inexact. dR 1. Graph and find ------------ (slope of (R)esponse vs. ln(C)oncentration graph). d ln C mHg) R vs Ln(C) 80 60 40 20 0 10 10 10 Dose (mg) dR ------------ = 32.96 d ln C dR 2. Graph and find ------ (slope of (R)esponse vs. (T)ime graph). - dt 3-20 Basic Pharmacokinetics REV. 00.1.14 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf

Pharmacological Response R vs T 75 70 65 60 55 50 45 20 25 30 35 40 45 50 Time (hr) dR ------ = – 0.91 - dt dR ------ - dt - 3. Find the ln(C)oncentration vs. (T)ime slope : ------------ : Note that your slope m = – K . dR ------------ d ln C –1 K = 0.028hr – 0.91 ------------- = – ( 0.028 ) 32.96 4. Calculate t 1 ⁄ 2 =  0.693  . ------------ - K t 1 ⁄ 2 =  -----------------------  = 24.75hr 0.693 -  – 1 0.028hr 3-21 Basic Pharmacokinetics REV. 00.1.14 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf

Pharmacological Response 3.5.3 ANSWERS: PROPRANOLOL dR 1. Graph and find ------------ (slope of (R)esponse vs. ln(C)oncentration graph). d ln C 20 15 10 5 0 10 10 dR ------------ = 16.36 d ln C dR 2. Graph and find ------ (slope of (R)esponse vs. (T)ime graph). - dt 25 20 15 10 5 0 1 2 3 4 5 6 dR ------ = – 2.93 - dt 3-22 Basic Pharmacokinetics REV. 00.1.14 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf

Pharmacological Response dR ------ - dt - ------------ : Note that your slope m = – K . 3. Find the ln(C)oncentration vs. (T)ime slope : dR ------------ d ln C –1 K = 0.179hr – 2.93 ------------- = – 0.179 16.36 4. Calculate t 1 ⁄ 2 =  0.693  . ------------ - K t 1 ⁄ 2 =  -----------------------  = 3.87hr 0.693 -  – 1 0.179hr 3-23 Basic Pharmacokinetics REV. 00.1.14 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf

Pharmacological Response

by Siddharth Nath

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