Pharmacological Response

CHAPTER 3 Pharmacological Response

Author: Michael Makoid and John CobbyReviewer: 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.

Basic Pharmacokinetics REV. 00.1.14 3-1Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf

Pharmacological Response

3.1 Pharmacological Response

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

One theory (A.J. Clarke) on the mechanism of action of drugs is the occupationtheory. It suggests that the intensity of a pharmacological response (E) is propor-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

(EQ 4-34)

where E is the intensity of the pharmacological response, is the maximum

attainable value of , is the molar concentration of free drug at the active com-plex and is the dissociation constant of the drug-receptor complex.

If is plotted against a hyperbolic curve will result; the asymptote will be.

a. If linear pharmacokinetics hold, the molar concentration of free drug at theactive site is proportional to the plasma concentration of the drug once equilibriumhas been established. Hence, a plot of against will also be hyperbolic.

b. Because the mass of drug in the body is , a plot of against will be

hyperbolic.

c. For a series of doses the value of at the same given time after dosing is propor-tional to the dose (D). Thus, a plot of against D will also be hyperbolic at a spe-cific time.

d. Any hyperbolic curve, if plotted on reverse semilogarithmic paper (i.e.,abscissa is logarithmic), has a sigmoid shape. If we plot against (of , or )in this manner, the plot is virtually linear in the range ; and if

this is the clinical range of responses, linear equations may be written. For exam-ple,

(EQ 4-35)

where is the slope

ED[ ]Emax

KR D[ ]+----------------------=

0.0 0.8 1.6 2.4 3.2 4.0

D

PKAnalyst Plot

E

0.0

0.2

0.4

0.6

0.8

1.0

Emax

E D[ ]KR

E D[ ]Emax

E Cp

X V Cp⋅= E X

X

E

E Cp X D

E Emax⁄ 0.2 0.8→=

E m xln b+⋅=

m

10-810-710-610-510-410-310-210-1

Conc.

0.0

0.2

0.4

0.6

0.8

1.0

Res

pons

e



o (or

tion-iables

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

This example equation shows that, in the clinical range, the intensity of a pharma-cological response is proportional to the logarithm of the administered dose, pro-viding response is measured at a consistent time after dosing. The proportionalityconstant (slope, ) 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 notdouble 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 con-centration and time relationship is studied in pharmacokinetics. The response andtime 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 mid-dle almost straight part. The slope is .

Response vs. time is a straight line. The slope is .

Natural log of concentration vs. time (drug given by IV bolus) is a straight line.The slope is .

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 twdata sets with which to get the other two).

(EQ 4-36)

(EQ 4-37)

(EQ 4-38)

NOTE: Only between 20% and 80% of maxi-mum response!!!!!!

You should be able to apply the equation to each of the above relaships. Given the slope (or having obtained the slope) and two of the three var(y, x, b), you should be able to find the third.

m

dR d cln⁄

dR dt⁄

d c dt⁄ln

dRdt------- dR

d cln----------- d cln

dt-----------⋅=

dRd cln----------- dR dt⁄

d cln dt⁄--------------------=

d clndt

----------- dR dt⁄dR d cln⁄----------------------=

y mx b+=



3.2 Change in Response with Time

3.2.1 ONE-COMPARTMENT OPEN MODEL: INTRAVENOUS BOLUS INJECTION

(EQ 4-39)

or

(EQ 4-40)

Substituting twice from eq. 4-35 once at time t and once at zero time

(EQ 4-41)

Hence a plot of the intensity of the pharmacological response at any time against time declines linearly. The slope is and the intercept is (the

initial intensity).

3.2.2 ONE-COMPARTMENT OPEN MODEL: ORAL ADMINISTRATION

Response follows plasma profile.

Because is proportional to at any time, a plot of against will be analo-gous to a plot of against . Hence will rise at first and then decline with time.When is large, the terminal slope will be .

3.2.3 DURATION OF EFFECTIVE PHARMACOLOGICAL RESPONSE

Duration of action is related to how long plasma concentration is above Minimum Effec-tive Concentration.

Once equilibrium has been established, there is a minimum plasma concentrationbelow which no pharmacological response is seen; this concentration is or

. For an intravenous bolus injection, the time to reach is .

multiplying by the volume of distribution we obtain

(EQ 4-42)

Rearranging,

X X0eKt–

DeKt–

= =

Ln X( ) Ln D( ) Kt–=

E b–m

------------E0 b–

m--------------- Kt–= E→ E0 Rt–=

E( )R– K– m⋅( )= E0

E xln E t

xln t E

t R–

tdur( )

Cp( )eff

MEC Cp( )eff

tdur

Cp( )eff Cp( )0eKtdur–

=

Xeff( )ln D( )ln Ktdur–=



ayed”.

allel

(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 :

(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 PHARMACOKINETIC PARAMETERS FROM RESPONSE DATA

How can I get the elimi-nation rate constant from pharmacological data? Use this “cook-book.”

The measurement of pharmacological effect provides a non-invasive means ofobtaining the value of (but not ).

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

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

b. Find the slope of this plot.

c. Obtain a response against time plot for a single dose (Eq. 4-36).

d. Find the terminal slope of this plot.

e. Calculate .

f. Calculate .

3.2.5 “DELAYED” RESPONSE

Two compartment model - biophase is in second compartment.

If a drug does not distribute instantaneously to all the body tissues (including theactive site), the pharmacological response will not always parallel the drug con-centrations in the plasma. In such a situation the response may parallel the mass ofdrug presumed to be in a second compartment , and hence seem “del

Eventually, however, once equilibrium is attained, the response will parplasma concentrations. In such a case, is proportional to .

tdur

DXeff

-------- ln

K--------------------=

tdurDose( )lnK

-----------------------Xeff( )ln

K-------------------–=

Xeff( )ln

K-------------------–

t1 2⁄ V

m( )

R–( )

KRm----=

t1 2⁄0.693

K-------------

=

X2( )

E X2ln



Thus a plot of against (or against ) will show a hysteresis loop with time,

most noticeably during an intravenous infusion.

3.2.6 RESPONSE OF ACTIVE METABOLITE:

Parent compound (inac-tive) yields active daughter compound.

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

E X1 E Cp



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phar-tion isponsecauseet the

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3.3 Therapeutic Drug Monitoring

Part of Pharmaceutical Care!

The pharmacokinetics of a drug determine the blood concentration achieved froma prescribed dosing regimen. During multiple drug dosing, the blood concentrationwill reflect the drug concentration at the receptor site; and it is the receptor siteconcentration that determines the intensity of the drug’s effect. Therefore, in to predict a patient’s response to a drug regimen, both the pharmacokineticpharmacological response characteristics of the drug must be understood. macological response is closely related to drug concentration at the site of aWe can measure plasma concentration and assume that the site of action is iequilibrium with the plasma since we usually do not measure drug concentratthe tissue or at the receptor site. This assumption is called “kinetic homogenand 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 andmacologic response. The relationship between response and ln-concentrasigmoidal. A threshold concentration of drug must be attained before any resis elicited at all. Therapy is achieved when the desired effect is attained bethe required concentration has been reached. That concentration would slower limit of utility of the drug, and is called the Minimum Effective Concenttion (MEC). Most drugs are not “clean”, that is exhibit only the desired theratic response. They may also exhibit undesired side effects, sometimes calledeffects at a higher, (hopefully a lot higher), concentration. At some concentrathese toxic side effects become become intolerable/and or dangerous patient.. That concentration, or one below it, would set the upper limit of utfor the drug and is called the Maximum Therapeutic Concentration or MinimToxic Concentration (MTC). Patient studies have generated upper (MTC)lower (MEC) plasma concentration ranges that are deemed safe and effectreating specific disease states. These concentrations are known as the “therrange” for the drug (Table 4-18).

When digoxin is administered at a fixed dosage to numerous subjects, the concentrations achieved vary greatly. Clinically, digoxin concentrations below

will elicit a subtherapeutic effect. Alternatively, when the digoxin conctration exceeds 2.0 side effects occur (nausea and vomiting, abdominalvisual disturbances). Drugs like digoxin possess a narrow therapeutic ibecause the concentrations that may produce toxic effects are close to

ng ml⁄ng ml⁄



ningrapeu-

required for therapeutic effects. The importance of considering both pharmacoki-netics and pharmacodynamics is clear.

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 thenmeasure the highest plasma drug concentration we would see that due to individualvariability, the resulting plasma drug concentrations differ. This variability can beattributed to factors influencing drug absorption, distribution, metabolism, andexcretion. Therefore, drug dosage regimens must take into account any diseasealtering state or physiological difference in the individual.

Therapeutic drug monitoring optimizes a patient’s drug therapy by determiplasma drug concentrations to ensure the rapid and safe drug level in the thetic 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.

TABLE 4-18 Average therapeutic drug concentration

DRUG RANGE

digoxin 0.8-2.0

gentamicin 2-10 l

lidocaine 1-4

lithium 0.4-1.4

phenytoin 10-20

phenobarbitol 10-30

procainamide 4-8

quinidine 3-6

theophylline 10-20

ng ml⁄

µg ml⁄

µg ml⁄

mEq L⁄

µg ml⁄

µg ml⁄

µg ml⁄

µg ml⁄

µg ml⁄



The major potential advantages of therapeutic drug monitoring are the maximiza-tion of therapeutic drug benefits and the minimization of toxic drug effects. Theformulation of drug therapy regimens by therapeutic drug monitoring involves aprocess for reaching dosage decisions.



rva- and inef-ug the

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3.3.1 THERAPEUTIC MONITORING: WHY DO WE CARE?

The usefulness of a drug’s concentration vs. time profile is based on the obsetion that for many drugs there is a relationship between plasma concentrationtherapeutic response. There is a drug concentration below which the drug is fective, the Minimum Effective Concentration (MEC), and above which the drhas untoward effects, the Minimum Toxic Concentration (MTC). That defines 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 drucentrations in these ranges. There is always the possibility that the range wdifferent in an individual patient.

For every pharmacokinetic parameter that we measure, there is a populationage and a range. This is normal and is called biological variation. People are ent. In addition to biological variation there is always error in the laboratory asthat we use to measure the parameters and error in the time we take the sEven with these errors, in many cases, the therapy is better when we attemonitor the patient’s plasma concentration to optimize therapy than if we dThis is called therapeutic monitoring. If done properly, the plasma concentraare rapidly attained and maintained within the therapeutic range throughoucourse of therapy. This is not to say all drugs should be monitored. Some have a such a wide therapeutic range or little to no toxic effects that the conctions 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, therapeuticmonitoring of key drugs can be economically beneficial to an institution. A recentstudy (DeStache 1990) showed a significant difference with regard to length ofstay in the hospital between the patients on gentamicin who were monitored (andtheir dosage regulated as a consequence) vs. those who were not. With DRGs thehospital was reimbursed a flat fee irrespective of the number of days the patientstayed in the hospital. If the number of days cost less than what the DRG paid, thehospital makes money. If the days cost more than the hospital loses money. Thisstudy showed that if all patients in the hospital who were on gentamicin were mon-itored, the hospital would save $4,000,000. That’s right FOUR MILLION per yI would say that would pay my salary, with a little left over, and that is only drug!



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. C

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699

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 pharmacokineticsto determine the optimum dosage regimen to maintain the patient’s plasmacentration within that range.

Selected References

1. Nagashima, R., O’Reilly, RA., and Levy, G, Kinetics of pharmacologic effects in man: the anticoagulant action of warfarinlin. Pharm. Therap, 10 22-35 (1969).

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 characts. J.Pharm.Sci, 61, 567-569 (1972).

4. Brunner, L., Imhof, P., and Jack, D. Relation between plasma concentrations and cardiovascular effects of oral oxpreno 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 pmide. Clin. Pharm. Therap., 20, 67-681 (1976).

6. Joubert, P., et al. Correlation between electrocardiographic changes, serum digoxin, and total body digoxin content. CliPharm. Therap., 20, 676-681 (1976).

7. Amery, A., et al. Relationship between blood level of atenolol and pharmacologic effect. Clin. Pharm. Therap., 21, 691-(1977).

Remember: We want the straight part!



3.4 Problems

What to do.---> We want to get pharmacokinetic data (elimination rate con-stant) from pharmacological response data (Response vs concontration andResponse vs time graphs) .

Response vs Time Graph

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 lin-ear 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!

14. Find the slope of the straight line, , (eyeball the rise over the run or use linear regression as

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.

Response vs Ln(Concentration) Graph

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

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.

0.0

0.2

0.4

0.6

0.8

1.0

10 10 10 10 10 10

Res

pons

e

Time

dRdT-------

10-810-710-610-510-410-310-210-1

Conc.

0.0

0.2

0.4

0.6

0.8

1.0

Res

pons

e

1

10

10

10

01

10

100

1

10 10 10

0

100

1 10

Concentration

Res

pons

e



Rise = change in Response.

Take the difference of the two responses coresponding to the concentrations picked. (R2-R1).

17. The slope of the line is

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)

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

the line is

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:

Both methods should be equivalent.

Additional problems are available in chapter 14, practice exams.

mriserun--------

R2 R1–

0.693------------------= =

1

10

10

10

0

Con

cent

rati

on

Time

1

10

100

mriserun-------- 0.693

t2 t1–-------------- 0.693

t– 12---

------------- k–= = = =

slope of r vs tslope of r vs ln(c)------------------------------------------

dRdT-------

dRdln(C)------------------------------ dln(C)

dT--------------- m k–= = = =



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) com-pared to placebo during physical exercise. The following approximate mean data is from 7 healthy volunteers: beats perminute (bpm) altered with time (t) after oral administration of three doses (D).

1. Calculate the half life of oxpranolol from the pharmacological response table.

2. Plot plasma concentration data on Cartesian graph paper directly as well as transforming into .

3. Plot plasma concentration data on semilog paper. Use linear regression to find the rate constant of elimination ofoxpranolol.

4. Calculate the half life obtained from the concentration data and compare it with the half life calculation based on thepharmacological response.

TABLE 4-19

Response vs Concentration

Response vs 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

Time (min)

30 699

60 622

120 413

150 292

240 152

360 60

480 24

Cpngml------

t1 2⁄( )

Cp Cpln



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.

From the preceding information, determine the following:

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

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

3. Find the ln(C)oncentration vs. (T)ime slope : : Note that your slope . If you are having problems

understanding this, refer to Sections 2.4.2 -2.4.4. K is the elimination rate constant.

4. Calculate .

PROBLEM TABLE 4 - 2. Minoxidil

Initial decrease in MAP Dose

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 Time

75 20

66 30

56 40

48 50

mmHg( ) mg( )

mmHg( ) hr( )

dRd Cln------------

dRdt-------

dRdt-------

dRd Cln------------------------- m K–=

t1 2⁄0.693

K-------------

=



Propranolol (Problem 4 - 4)

Citation?

Beta blockers can be considered first line drugs of choice in the treatment of hypertension in certain patients. The fol-lowing data was obtained regarding Propranolol used to treat hypertension in a group of patients.

From the preceding information, determine the following:



3. Find the ln(C)oncentration vs. (T)ime slope : : Note that your slope . If you are having problems

understanding this, refer to Sections 2.4.2 -2.4.4.

4. Calculate .

PROBLEM TABLE 4 - 5. Propranolol

Fall in Systolic BP (mmHg)

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

Cp

dRd Cln------------

dRdt-------

dRdt-------

dRd Cln------------------------- m K–=

t1 2⁄0.693

K-------------

=



3.5.1 ANSWERS: OXPRANOLOL

1. Calculate the half life of oxpranolol from the pharmacological response table

Slope of the line from linear regression. Chapter 2.4.4

TABLE 5.

Xln(Dose) Dose

YResponse

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

t1 2⁄( )

10 10 10

Dose (mg)

10

12

14

16

18

20

22

Oxpranolol

Res

pons

e (B

PM

)

1 2 3

X2

X Y⋅

ΣX 22.03= ΣY 79.5= ΣX2

98.25= ΣXY 359.82=

ΣX( )2485.23=

X Σ X

n--------- 4.41= =

)

y Σyn

------ 15.9= =

)

mΣ x( ) Σ y( )⋅( ) n Σ x y⋅( )⋅( )–

Σ x( )[ ]2n Σ x

2( )⋅( )–---------------------------------------------------------------------=

m22.03 79.5⋅( ) 5 359.82⋅( )–

485.32 5 98.25⋅( )–-------------------------------------------------------------------- 7.93= =



the slope is equal to the linear regression of the change in response vs. ln concentration.

The slope of this plot is therefore, .

half life (89 min).

2. Plot plasma concentration data on Cartesian graph paper directly as well as transforming into .

dRd cln----------- 7.93=

1.5 2.0 2.5 3.0 3.5 4.06

8

10

12

14

16

18

OXPRANOLOL

Time (hr) 1.0

Res

pon

se (

BP

M)

mR1 R2–

T1 T2–------------------ 16 10–

1.45 3.07–--------------------------- 3.71–= = =

dRdt------- 3.71–=

dRdt-------

dRd cln---------------------- d cln

dt----------- k–

3.71–7.93

------------- 0.4678hr1–

–= = = = t1 2⁄2ln

k-------- 0.693

0.4678hr1–

--------------------------- 1.48hr= = =

Cp Cpln



3. Plot plasma concentration data on semilog paper. Use linear regression to find the rate constant of elimination ofoxpranolol.

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 thepharmacological response.

compared to 1.48 hours (89 min) from the pharmacological response

method.

0 100 200 300 400 500

Time (min)

Plasma concentration vs. Time

Con

cent

ratio

n (n

g/m

l)

0

160

320

480

640

800

0 100 200 300 400 500

Time (min)

10

10

10

Oxpranolol

Con

cent

rati

on (

ng/m

L)

1

2

3

t1 2⁄0.693

0.00763------------------- 90min 1.5 hrs== =



3.5.2 ANSWERS: MINOXIDIL

NOTE: Answers will vary depending on whether linear regression is calculatedvia calculator or using the formula such as observed in Problem 1. Either methodcan be used. However, if you use the formula, you should be within 10% of thecalculated answer. A word of caution: if you choose to do linear regression viacalculator make sure you have valid data. This cannot be assured until you havegraphed all the data points given. Many a student has incorrectly calculatedparameters because he/she falsely assumes that all the points are valid. Blindlychoosing data points for linear regression will only lead to error. Every problemin this manual has been derived from actual journal articles and will therefore be“real” data. This real-world data is inexact.



10 10 10

0

20

40

60

80

mH

g) R vs Ln(C)

Dose (mg)

dRd Cln------------

dRd Cln------------ 32.96=

dRdt-------



3. Find the ln(C)oncentration vs. (T)ime slope : : Note that your slope .

4. Calculate .

20 25 30 35 40 45 50

Time (hr)

45

50

55

60

65

70

75

R vs T

dRdt------- 0.91–=

dRdt-------

dRd Cln------------------------- m K–=

K 0.028hr1–

=

0.91–32.96------------- 0.028( )–=

t1 2⁄0.693

K-------------

=

t1 2⁄0.693

0.028hr1–

------------------------ 24.75hr= =



3.5.3 ANSWERS: PROPRANOLOL



dRd Cln------------

10 10

0

5

10

15

20

dRd Cln------------ 16.36=

dRdt-------

0 1 2 3 4 5 6

5

10

15

20

25

dRdt------- 2.93–=



3. Find the ln(C)oncentration vs. (T)ime slope : : Note that your slope .

4. Calculate .

dRdt-------

dRd Cln------------------------- m K–=

K 0.179hr1–

=

2.93–16.36------------- 0.179–=

t1 2⁄0.693

K-------------

=

t1 2⁄0.693

0.179hr1–

------------------------ 3.87hr= =


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Pharmacological Response