Basic Pharmacokinetics - Chapter 3: Pharmacological Response

CHAPTER 3

Pharmacological Response

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 construct (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.

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Pharmacological Response

3.1 Pharmacological Response Drug must get into blood and blood is in contact with receptor.

3.1.1

One theory (A.J. Clarke) on the mechanism of action of drugs is the occupation theory. It suggests that the intensity of a pharmacological response (E) is proportional to the concentration of a reversible drug-receptor complex

THE HYPERBOLIC RESPONSE EQUATION A mathematical description of the occupation theory, assuming complete and instantaneous drug distribution, yields [ D ]E max E = --------------------KR + [D ]

(EQ 4-34)

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 complex and K R is the dissociation constant of the drug-receptor complex.

PKAnalyst Plot 1.0

0.8

0.6

E

If

0.4

E

is plotted against

[D]

a hyperbolic curve will result; the asymptote will be

E max .

0.2

0.0 0.0

0.8

1.6

2.4

3.2

4.0

D

a. If linear pharmacokinetics hold, the molar concentration of free drug at the 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. b. Because the mass of drug in the body is hyperbolic.

X = V ⋅ Cp ,

a plot of

E

against

X

will be

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., abscissa is logarithmic), has a sigmoid shape. If we plot E against Cp (of X , or D ) in this manner, the plot is virtually linear in the range E ⁄ E max = 0.2 → 0.8 ; and if this is the clinical range of responses, linear equations may be written. For example,

1.0

Response

0.8 0.6 0.4 0.2 0.0 10 -810-710 -610-510 -410-310-210-1

E = m ⋅ ln x + b

Conc.

where

m

(EQ 4-35)

is the slope

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Pharmacological Response

Plot of Response vs. Ln(C) is a straight line in the middle (if you squint), but only between 20% and 80% maximum response

3.1.2

This example equation shows that, in the clinical range, the intensity of a pharmacological response is proportional to the logarithm of the administered dose, providing response is measured at a consistent time after dosing. The proportionality constant (slope, m ) is a function of the affinity of the drug for the receptor. In fact, equation 4-35 yields a log-dose response plot. Note that doubling the dose does not double the response.

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.

Remember: Use only the data between 20% and 80% of maximum response for the straight part of both response vs. Ln(c) and response vs. t.

You should know what the various graphical relationships look like. Response vs. natural log of concentration is sigmoidal. (S shaped). We are interested in the middle almost straight part. The slope is dR ⁄ d ln c . Response vs. time is a straight line. The slope is

dR ⁄ dt .

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 dR- d---------ln c------- = ---------⋅ d ln c dt dt dR dR ⁄ dt ---------= -------------------d ln c d ln c ⁄ dt

NOTE: Only between 20% and 80% of maximum response!!!!!!

(EQ 4-36)

(EQ 4-37)

d---------ln cdR ⁄ dt (EQ 4-38) = ---------------------dt dR ⁄ d ln c You should be able to apply the equation y = mx + b to each of the above relationships. Given the slope (or having obtained the slope) and two of the three variables (y, x, b), you should be able to find the third.

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Pharmacological Response

3.2 Change in Response with Time 3.2.1

ONE-COMPARTMENT OPEN MODEL: INTRAVENOUS BOLUS INJECTION X = X0 e

– Kt

= De

– Kt

(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- E 0–b ----------= --------------- – Kt → E = E0 – Rt m m

(EQ 4-41)

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

Response follows plasma profile.

3.2.3

Because E is proportional to ln x at any time, a plot of E against t will be analogous 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 )

Duration of action is related to how long plasma concentration is above Minimum Effective Concentration.

Once equilibrium has been established, there is a minimum plasma concentration below which no pharmacological response is seen; this concentration is ( C p ) eff or MEC . For an intravenous bolus injection, the time to reach ( C p ) eff is t dur . ( C p )eff = ( C p ) 0 e

– Ktd ur

multiplying by the volume of distribution we obtain ln ( X eff ) = ln ( D ) – Kt dur

(EQ 4-42)

Rearranging,

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Pharmacological Response

t dur

D ln  -------- X eff = -------------------K

(EQ 4-43)

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 = ----------------------- – -----------------K K

(EQ 4-44)

Thus a plot of duration of action vs ln dose would result in a straight line with a slope of 1/K and an x intercept of

3.2.4

ln ( X ef f ) – ------------------- . K

PHARMACOKINETIC PARAMETERS FROM RESPONSE DATA

How can I get the elimination rate constant from pharmacological data? Use this “cookbook.”

The measurement of pharmacological effect provides a non-invasive means of obtaining the value of t 1 ⁄ 2 (but not V ).

Remember: Use only the data between 20% and 80% of maximum Response for both of these plots.

b. Find the slope

3.2.5

a. Obtain a log dose-response plot (Eq. 4-37). The response must always be measured at the same time after administering the dose. (m)

of this plot.

c. Obtain a response against time plot for a single dose (Eq. 4-36). d. Find the terminal slope e. Calculate

R K = ---- . m

f. Calculate

0.693 t 1 ⁄ 2 =  -------------  K 

( –R )

of this plot.

.

“DELAYED” RESPONSE

Two compartment model - biophase is in second compartment.

If a drug does not distribute instantaneously to all the body tissues (including the active site), the pharmacological response will not always parallel the drug concentrations 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 .

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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:

Parent compound (inactive) yields active daughter compound.

In the case of an inactive prodrug yielding an active metabolite, the response curves will mirror the active metabolite plasma profile (assuming the biophase is the plasma) and not the prodrug plasma profile.

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Pharmacological Response

3.3 Therapeutic Drug Monitoring Part of Pharmaceutical Care!

The pharmacokinetics of a drug determine the blood concentration achieved from 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. Pharmacological 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.

Need to keep plasma concentration in the therapeutic range to optimize therapy.

There exists a fundamental relationship between drug pharmacokinetics and pharmacologic response. The relationship between response and ln-concentration is sigmoidal. A threshold concentration of drug must be attained before any response 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 Concentration (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

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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

digoxin

0.8-2.0 ng ⁄ ml

gentamicin

2-10 µg ⁄ ml l

lidocaine

1-4 µg ⁄ ml

lithium

0.4-1.4 mEq ⁄ L

phenytoin

10-20 µg ⁄ ml

phenobarbitol

10-30 µg ⁄ ml

procainamide

4-8 µg ⁄ ml

quinidine

3-6 µg ⁄ ml

theophylline

10-20 µg ⁄ ml

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. Two components make up the process of therapeutic drug monitoring:

• Assays for determination of the drug concentration in plasma • Interpretation and application of the resulting concentration data to develop a safe and effective drug regimen.

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Pharmacological Response

The major potential advantages of therapeutic drug monitoring are the maximization 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.

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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 observation that for many drugs there is a relationship between plasma concentration and therapeutic response. There is a drug concentration below which the drug is ineffective, 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!

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Pharmacological Response

The process of therapeutic monitoring takes effort.

• • • • •

First the MD must order the blood assays. Second, someone (nurse, med tech, you) must take the blood. 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).

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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 1.0

1.

Plot Response vs Time on Cartesian (regular) Graph Paper.

13.

Use Response data between 20 and 80 percent of maximum (Pick the straight part) to do the linear regression on. (Rule of thumb: Connect first and last data point with a straight line. If all the points fall on one side of the line, its not straight!

Response

0.8

0.6

14.

dR Find the slope of the straight line, ------- , (eyeball the rise over the run or use linear regression as dT

0.4

required). Important: you must determine the best fit line through all of the points that you will 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 change in Y over the change in X.

0.2

0.0 10

10

10 Time

10

10

10

Response vs Ln(Concentration) Graph 1.

Turn semi-log paper on its side so that the numbers are on the top.

0

0.6 0.4 0.2 0.0

Response

Response

10010

10

0.8

10 10

1

10010

1.0

1010

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. 15.

Plot concentration on the x axis and response on the y.

16.

Find the slope of the line plotted this way by the rise over the run method. 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.

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Pharmacological Response

Rise = change in Response. Take the difference of the two responses coresponding to the concentrations picked. (R2-R1). 17.

R –R

rise 2 1 The slope of the line is m = -------- = ----------------run

0.693

Ln(Concentration) vs Time Graph (Pharmacokinetic Data) If you have concentration vs Time data: 1.

Plot Concentration vs time on semi-log paper (Y axis is concentration this time)

Concentration

10 100

1010

110 0

1

Time

18.

Find the slope as before, using semi log paper (Remeber the log is on the Y axis this time, so 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: 1.

Divide the slope of the Response vs Time graph by the slope of the Response vs ln(C) graph: dR ------dT dln(C) slope of r vs t ------------------------------------------ = --------------- = --------------- = m = – k dT slope of r vs ln(c) dR -------------dln(C)

Both methods should be equivalent. Additional problems are available in chapter 14, practice exams.

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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 Concentration BPM

Response vs time

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

Time (min)

ng C p  ------ ml

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.

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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

PROBLEM TABLE 4 - 3.

Minoxidil

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 d ln C

1. Graph and find ------------ (slope of (R)esponse vs. ln(C)oncentration graph).

dR dt

2. Graph and find ------- (slope of (R)esponse vs. (T)ime graph).

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 

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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.

PROBLEM TABLE 4 - 5.

Propranolol

Fall in Systolic BP (mmHg)

Cp

20

50

16

40

11

30

5

20

PROBLEM TABLE 4 - 6.

Propranolol

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 d ln C

1. Graph and find ------------ (slope of (R)esponse vs. ln(C)oncentration graph).

dR dt

2. Graph and find ------- (slope of (R)esponse vs. (T)ime graph).

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.

-------------  . 4. Calculate t 1 ⁄ 2 =  0.693 K 

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Pharmacological Response

3.5.1

ANSWERS:

OXPRANOLOL

1. Calculate the half life ( t 1 ⁄ 2 ) of oxpranolol from the pharmacological response table

Oxpranolol

Response (BPM)

22 20 18 16 14 12 10 1

10

3

2

10

10

Dose (mg)

TABLE 5.

X ln(Dose)

Dose

Y Response

X

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

ΣY = 79.5

ΣX = 98.25

ΣX = 22.03

2

2

X⋅Y

ΣXY = 359.82

2

( ΣX ) = 485.23

( Σ(x) ⋅ Σ(y )) – (n ⋅ Σ(x ⋅ y)) m = --------------------------------------------------------------------2 2 [Σ(x) ] – ( n ⋅ Σ (x ))

)

Σy y = ------ = 15.9 n

)

ΣX X = --------- = 4.41 n

Slope of the line from linear regression. Chapter 2.4.4

( 22.03 ⋅ 79.5 ) – ( 5 ⋅ 359.82 ) m = -------------------------------------------------------------------- = 7.93 485.32 – ( 5 ⋅ 98.25 )

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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

Response (BPM)

16 14 12 10 8 6 1.0

1.5

2.0

2.5

3.0

3.5

4.0

Time (hr)

R –R T1 – T 2

16 – 10 1.45 – 3.07

dR dt

1 2 The slope of this plot is m = ------------------ = --------------------------- = – 3.71 therefore, ------- = – 3.71 .

dR ------dt d ln c –3.71 –1 ----------= ----------- = – k = ------------- = – 0.4678hr dR dt 7.93 ----------d ln c

ln 2 0.693 t 1 ⁄ 2 = -------- = -------------------------= 1.48hr half life (89 min). –1 k 0.4678hr

2. Plot plasma concentration data on Cartesian graph paper directly as well as transforming Cp into ln C p .

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Pharmacological Response

Plasma concentration vs. Time

Oxpranolol

800 3

10

Concentration (ng/mL)

640

Concentration (ng/ml)

480

320

160

2

10

1

10

0

0 0

100

200

300

400

500

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.

Basic Pharmacokinetics

REV. 00.1.14

Copyright © 1996-2000 Michael C. Makoid All Rights Reserved

http://pharmacy.creighton.edu/pha443/pdf

3-19

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 (slope of (R)esponse vs. ln(C)oncentration graph). 1. Graph and find -----------mHg)

d ln C

R vs Ln(C) 80 60 40 20 0 10

10

10

Dose (mg)

dR -----------= 32.96 d ln C

dR dt

2. Graph and find ------- (slope of (R)esponse vs. (T)ime graph).

Basic Pharmacokinetics

REV. 00.1.14

Copyright © 1996-2000 Michael C. Makoid All Rights Reserved

http://pharmacy.creighton.edu/pha443/pdf

3-20

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 K = 0.028hr

–1

–------------0.91 = – ( 0.028 ) 32.96

-------------  . 4. Calculate t 1 ⁄ 2 =  0.693 K 

0.693 - t 1 ⁄ 2 =  ----------------------= 24.75hr – 1 0.028hr

Basic Pharmacokinetics

REV. 00.1.14

Copyright © 1996-2000 Michael C. Makoid All Rights Reserved

http://pharmacy.creighton.edu/pha443/pdf

3-21

Pharmacological Response

3.5.3

ANSWERS:

PROPRANOLOL

dR (slope of (R)esponse vs. ln(C)oncentration graph). 1. Graph and find -----------d ln C

20 15 10 5 0 10

10

dR -----------= 16.36 d ln C dR dt

2. Graph and find ------- (slope of (R)esponse vs. (T)ime graph).

25 20 15 10 5 0

1

2

3

4

5

6

dR ------- = – 2.93 dt

Basic Pharmacokinetics

REV. 00.1.14

Copyright © 1996-2000 Michael C. Makoid All Rights Reserved

http://pharmacy.creighton.edu/pha443/pdf

3-22

Pharmacological Response

dR ------dt 3. Find the ln(C)oncentration vs. (T)ime slope : ------------- : Note that your slope m = – K . dR -----------d ln C K = 0.179hr

–1

–------------2.93 = – 0.179 16.36

-------------  . 4. Calculate t 1 ⁄ 2 =  0.693 K 

0.693 - = 3.87hr t 1 ⁄ 2 =  ----------------------– 1 0.179hr

Basic Pharmacokinetics

REV. 00.1.14

Copyright © 1996-2000 Michael C. Makoid All Rights Reserved

http://pharmacy.creighton.edu/pha443/pdf

3-23