PharmacokineticsKen-Ichi TakemaruAssistant ProfessorBST 7-182, 4-7976

takemaru@pharm.stonybrook.edu

Pharmacokinetics vs. Pharmacodynamics

Fig. 3-1. The relationship between dose and effect can be separated into pharmacokinetic (dose-concentration) and pharmacodynamic (concentration-effect) component (Basics & Clinical Pharmacology, Katzung, 2007)

Lithium vs. Activase (Alteplase)

Li+

MW = 7Mood-stabilizing drug (bipolar disorder)

Oral administration

Activase (Recombinant tissue plasminogen activator) MW = 59,050 (527 amino acids)

Blood clot-dissolving enzymeAdministered into the vascular compartment by IV injection

Pharmacokinetic Processes

• Pharmacokinetics is the study of drug movement throughout the body, which determines the concentration of a drug at its sites of action (‘what the body does to the drug’).

Figure 4-1. The four basic pharmacokinetic processes (Pharmacology for Nursing Care, Lehne, 2004).

Routes of Drug Administration

Table 3-1. Routes of drug administration (Principles of Pharmacology, Golan et al., 2005)

First-pass metabolism

Routes of Parenteral Drug Administration

Table 3-2. Routes of parenteral drug administration (Principles of Pharmacology, Golan et al., 2005)

1, Channels and pores: very few drugs with low molecular weights cross membrane via channels or pores.

2, Transport systems: important for certain drugs and also utilized for renal excretion.

3, Membrane diffusion: most drugs diffuse through the membrane. A drug must be lipid soluble, and polar molecules can not penetrate membranes.

Passage of Drugs across Membranes

Rate of diffusion = Partition coefficient x Surface Area x (C2-C1) Thicknessmembrane

C1: intracellular concentration of drugC2: extracellular concentration of drug

Figure 4-2. Structure of the cell membrane (Pharmacology for Nursing Care, Lehne, 2004)

In the absence of any other factors, a drug will enter a cell until the intracellular and extracellular concentrations of the drug are equivalent. Fick’s law of diffusion states the net drug flux across the membrane:

This equation applies to an ideal situation where there is an absence of complicating factors such as pH and charge gradients across the membrane.

Effect of pH on Drug Absorption

Weak acid Weak base

HA H+ + A- BH+ B + H+ Dissociation Curve: Aspirin

1 2 3 4 5 60

25

50

75

100

pH

Per

cen

t H

A

pK 3.5

Stomach Plasma

Henderson-Hasselbach equation

pH = pKa + log ________[A-][HA]

pH = pKb + log ________[B][BH+]

pH Trapping across Lipid Bilayers

Henderson-Hasselbach equation

pKa = pH + log ________[HA] [A-]

Fig. 3-2. pH trapping across lipid bilayers(Principles of Pharmacology, Golan et al., 2005)

In case of a weakly acidic drug with a pKa of 4 (stomach pH=1):

4 = 1 + log ________[HA] [A-]

3 = log ________[HA] [A-]

1,000 = ________[HA] [A-]

In stomach, the uncharged form of the drug exists 1,000 times the concentration of charged from; 99.9% of the drug is in the uncharged form. In contrast, in plasma, 99.9% of the drug is charged.

pH trapping is also applied to other situations such as breast milk (pH = 6 to 7) vs. plasma and urine (pH = 4.6 to 8) vs. plasma.

Morphine is a weak base with pK = 8.0. If the pH of breast milk is 6.8 and that of plasma is 7.4, calculate the distribution of morphine in plasma vs. breast milk.

From rearranging the Henderson-Hasselbach equation: [B]/[BH+] = 10 (pH-8.0)

PlasmapH = 7.4

Breast milkpH = 6.8

[Bp]/[BH+p] = 10 -0.6 = 0.251 [Bbm]/[BH+

bm] = 10 -1.2 = 0.0631

[B]/[BH+p] = 0.251 [B]/[BH+

bm] = 0.0631

[BH+p]/[B] = 1/0.251 = 3.98 [BH+

bm]/[B] = 1/0.0631 = 15.8

[BH+p]/[B] + 1 = 3.98 + 1 [BH+

bm]/[B] + 1 = 15.8 + 1

[BH+p] + [B]/[B] = 4.98 [BH+

bm] + [B]/[B] = 16.8

[Btotal p]/[B] = 4.98 [Btotal bm]/[B] = 16.8

[Btotal p] = 4.98 [B] [Btotal bm] = 16.8 [B]

The ratio [Btotal bm]/[Btotal p] = 16.8/4.98 = 3.37

Morphine is 3.37 times more concentrated in breast milk than in plasma.

Neutral form equilibrates across the plasma membrane

[Bp] = [Bbm] Let’s simplify this as [B]

Glomerular filtration

Passive tubular reabsorption

Active tubular secretion

Renal Elimination of Drugs and Metabolites

• Each kidney contains about one million functional filtering units called nephrons.• The kidneys filter about 45 gal (180 L) of blood each day.

Fig. 4-2. Renal excretion of drugs(Modern Pharmacology, Craig & Stitzel, 2004)

Nephron

Time Course of Drug Responses

Figure 4-14. Single-dose time course (oral) (Pharmacology for Nursing Care, Lehne, 2004)

Figure 4-15. Drug accumulation with repeated administration (Pharmacology for Nursing Care, Lehne, 2004)

(MEC)

Acetaminophen’s toxic conc. is 30 times greater than the MEC.Lithium’s toxic conc. is only 3 times greater than the MEC.

A drug with a half-life of 1 day.2 gm given once a day on days 1 through 9.

Peak conc.

Trough conc.

How do you predict time course of drug action?

What happens if you take 1 gm of the drug every 12 hours instead?

Single Compartment Model

Fig. 3-7, B. Schematic model of drug distribution and elimination (Principles of Pharmacology, Golan et al., 2005)

Fig. 8-6. Single-compartment pharmacokinetic model (Pharmacology, Rang et al., 2003)

A highly simplified model of a human being, which consists of a single well-stirred compartment (Vd) into which a quantity of drug Q introduced, and from which it can escape either by being metabolized or being excreted.

First-order kinetics

• First-Order Kinetics: a constant fraction of drug is handled per unit of time. Examples: 1. hepatic metabolism when enzymes are not saturated 2. glomerular filtration 3. radioactive decay

First-Order Kinetics

Fig. 8-7. Predicted behavior of single-compartment model following intravenous drug administration at time 0. (Pharmacology, Rang et al., 2003)

• The majority of drugs are eliminated by first-order kinetics.

• The rate of elimination is directly proportional to drug concentration.

Log10[X]

time

Slope = -k/2.3

First-Order Drug Elimination

The decrease in the plasma concentration of drug X over time can be written as:

-d[X]/dt = k[X] (k is the elimination rate constant)Rearrange: d[X]/[X] = -kdtIntegrate: ln[X] = -kt + CEvaluate the value of C (the constant of integration): when t = 0, [X] = [X]0, thus C = ln[X]0

Substitute: ln[X] = -kt + ln[X]0

ln[X] = ln [X]0 - kt or log [X] = log [X]0 - kt/2.3

ln([X]/[X]0) = -kt

[X] = [X]0 e-kt

[X] is the plasma drug concentration at any time t after administration([X]0 , initial plasma drug concentration; k, elimination constant).

Two-Compartment Model

Fig. 3-7, C. Schematic model of drug distribution and elimination (Principles of Pharmacology, Golan et al., 2005)

Fig. 8-6. Two-compartment pharmacokinetic model. The peripheral compartment represents the tissues while the central compartment represents the plasma. (Pharmacology, Rang et al., 2003)

Fig. 8-6. Kinetics of diazepam elimination in humans following a single oral dose. (Pharmacology, Rang et al., 2003)

The kinetics are biexponential, representing the process of transfer between plasma and tissues (fast phase) and elimination from the plasma (slow phase). Therefore, it still follows the first-order kinetics.

Fig. 8-12. Saturating kinetics of alcohol elimination in humans. The rate of disappearance of ethanol from the plasma is constant at about 4 mmol/l per hour irrespective of its plasma concentration because the rate of oxidation by the alcohol dehydrogenase reaches a maximum at low ethanol concentrations.

(Pharmacology, Rang et al., 2003)

Zero-Order Kinetics (Saturation Kinetics)• In a few cases, the time course of disappearance of drug from the plasma does not follow the exponential patterns but initially linear (i.e. drug is removed at a constant rate that is independent of plasma concentration).

• Zero-Order Kinetics: a constant amount of drug is handled per unit time. Examples: 1. absorption from a sustained release preparation 2. drug given by an infusion pump 3. hepatic metabolism of drug when enzymes are saturated

Bioavailability (F)Bioavailability (F)= fraction of a drug reaching the systemic circulation (quantity of drug reaching systemic circulation/quantity of drug administered)

For a drug administered orally, bioavailability is less than 100% for two main reasons; incomplete extent of absorption and first-pass elimination.

Hepatic extraction ratio (ER, first-pass elimination) = CLliver (L/h/70kg)

Q (hepatic blood flow; 90L/h/70kg)

F = extent of absorption (f) x (1-ER)

Fig. 3-3. Bioavailability following administration of a single dose of drug (Principles of Pharmacology, Golan et al., 2005)

Morphine is almost completely absorbed (f=1), so that loss in the gut is negligible. However, the hepatic extraction ratio for morphine is 0.67, so (1-ER) is 0.33. The bioavailability of morphine is therefore expected to be about 33%.

Represents the fluid volume required to contain the total amount of absorbed drug in the body at a uniform concentration equivalent to that in the plasma at a steady state.

Vd = Amount of drug in the body (Dose)

Concentration of drug in blood or plasma (Cp)

Volume of Distribution (Vd)

If 500 ug of digoxin were in the body of a 70-kg subject, a plasma concentration of approximately0.7 ng/ml would be observed. Vd = 500,000/0.7 = 714 L. Hence, digoxin distributes widelythroughout the body.

Ibuprofen typically exhibits a Vd of 10.8 L. Thus, this drug does not distribute widely to into tissues.

• Drugs (digoxin) that are highly distributed to extravascular compartments (fat and muscle) have a high Vd. • Drugs (ibuprofen) that are primarily retained within the vascular compartment have a relatively low Vd.

• Vd can be altered by liver or kidney disease.

Plasma Protein Binding

Fig. 3-5. Protein Binding and Drug Trapping(Principles of Pharmacology, Golan et al., 2005)

• Albumin is the most abundant plasma protein and is responsible for drug binding.

• Because a highly protein-bound drug tends to remain within the vasculature, such a drug often has a relatively low Vd.

• Co-administration of two or more drugs, each of which is highly bound to plasma protein, could results in higher-than-expected plasma concentration of the free form of either or both drugs.

• Phenytoin is approximately 90% bound to plasma proteins, leaving 10% of the concentration in the blood as free drug and available for pharmacological action and metabolism.

CL is the efficiency of irreversible elimination of drug from the body, and involves both metabolism and elimination of drug.

CL is defined as the volume of plasma that was completely cleared of the drug per unit of time (e.g., 100 mL/min).

CL = Rate of elimination / Concentration of drug in plasma (Cp)

CLSystemic = CLKidney + CLLiver + CLOther

Clearance (CL)

t1/2= the time required to change the amount of drug in the body by one-half during elimination.

Half-Life (t1/2) (1)

t1/2 = 4 hlog

scal

e

Need logarithmic increase in doseto get linear increase in duration of action. Doubling the dose increasesduration of action by one half-life.

MEC (minimum effective concentration)

[X] = [X]0 e-kt

[X]0/2 = [X]0 e-kt1/2

1/2 = e-kt1/2

2 = ekt1/2

ln 2 = kt1/2

t1/2 = 0.69/k

0.69 = kt1/2

Table 3-5. Factors affecting drug half-life(Principles of Pharmacology, Golan et al., 2005)

Half-Life (t1/2) (2)

Relationship between Clearance, Vd, and t1/2

Renal clearance describes the efficiency with which a drug or metabolite is excreted by the kidney.

c = drug concentration in plasma = gm/ml u = drug concentration in urine = gm/ml v = volume of urine = mlthen uv/t = gm/time = the rate of drug excretion into urine

From the first order elimination, the rate of drug excretion is -dX/dt = kX,and if elimination is occurring only by the kidney then

kX = uv/t, and since X = total amount of drug in body, X = c Vd,

Relating CL to t1/2 and Vd

Divide this by cuv/tc = ml/time = clearance = the volume of plasma that was completely cleared of the drug in the time interval.

CL = Vdk or t1/2 = 0.69 Vd/CL

Consequently, k c Vd = uv/t or k Vd = uv/tc = clearance or t1/2 = 0.69 Vd/clearance

• A rational dosage regimen is based on a target concentration that will produce the desired therapeutic effect.

• Absorption, distribution, metabolism, and excretion, each influences the design of an optimal dosing regimen for a drug.

• Therapeutic window is the safe opening between the minimum therapeutic concentration and the minimum toxic concentration of a drug

Dosage Regimen

Fig. 3-11. Therapeutic subtherapeutic, and toxic drug dosing (Principles of Pharmacology, Golan et al., 2005)

At steady state, the dosing rate (rate in) must equal the rate of elimination (rate out).

Maintenance Dosage

Dosing rate = Rate of elimination = CL x Target Concentration (TC)

If the drug is given by a route that has a bioavailability less than 100%, Dosing rate = CL x Target Concentration (TC) Bioavailability (F)

If intermittent doses are given, Maintenance dose = Dosing rate x Dosing interval

Maintenance Dosage: ExampleA target plasma theophylline concentration of 10 mg/L is desired to relive acute bronchial asthma in a patient. If the patient is a nonsmoker and otherwise normal except for asthma, clearance is 2.8 L/h/70 kg. Since the drug will be given via IV, F = 1.Dosing rate = CL x TC = 2.8 L/h/70 kg x 10 mg/L = 28 mg/h/70 kg

If the asthma attack is relieved, the clinician might want to maintain this plasma level using oral theophylline, which is given every 12 hours. Foral = 0.96.Maintenance dose = Dosing rate x Dosing interval = 28 mg/h x 12 hours = 350 mg F 0.96

If an 8-hour dosing interval was used, the ideal dose would be 233 mg; and if the drug was given once a day, the dose would be 700 mg.

Fig. 3-6. Relationship between frequency of dosing and maximum plasma concentrations when a steady state theophylline plasma level of 10 mg/L is desired. (Basics & Clinical Pharmacology, Katzung, 2004)

Initial (loading) dose to rapidly achieve a therapeutic concentration when the Vd is large

Loading dose = Vd x Target Concentration (TC)

Loading Dose

For example, warfarin has a Vd of 9.8 L (70 kg), when a TC is 1.01 mg/L, loading dose is calculated as

Loading dose = 9.8 L x 1.01 mg/L = 9.90 mg

• A loading dose may be desirable if the time required to attain steady state by the administration of drug at a constant rate (four elimination half-lives) is long relative to the temporal demands of the condition being treated such as arrhythmia.

• The use of a loading dose has significant disadvantages. Sensitive patients may be exposed to a toxic concentration of a drug. In addition, if the drug has a long half-life,it will take a long time for the concentration to fall.

Adjustment of Dosage with Compromised Clearance

Corrected dose = Average dose x Patient’s creatinine clearance (measure for glomerular filtration rate)

Drugs cleared by the renal route require adjustment of clearance in proportion to renal function (renal disease or reduced cardiac output).

Table 3-1. Pharmacokinetic and pharmacodynamic parameters for selected drugs (Basic & Clinical Pharmacology, Katzung, 2004)

The standard dose of a drug is based on trials in healthy volunteers with average ability to absorb, distribute, and eliminate the drug.

Problem 1

When 0.3 g of drug X is given by IV to a subject, the resulting drug concentration is 15 mg/L. Calculate the volume of distribution.

Amount of drug in the body (Dose)

Concentration of drug in plasma (Cp)Vd =

= 300 mg/15 mg/L = 20 L

An aminoglycoside has a normal elimination half-life of 90 minutes in young adults. However, in patients 70 to 90 years old, its elimination half-life is increased to 270 minutes. The normal dose of the aminoglycoside is 12 mg/kg per day. Assume that the volume of distribution per body weight is not changed by the patient’s age. What should be the daily dose for an 80-year-old patient?

t1/2 = 0.693 x Vd / CL

CL for older patients must be 3 times lower than that for young patients.

Dose = Cp x CL x time interval

In older patients, Dose/3 = Cp x CL/3 x time interval

Dose for older patients = 12 mg/kg per day / 3 = 4 mg/kg per day

Problem 2

Problem 3Mr. Walsh is taking 5 mg of warfarin every 24 hours for his chronic atrial fibrillation. The bioavailability of warfarin is 0.93, and its clearance is 0.192 L/h.1. Calculate the steady-state plasma concentrations of warfarin.2. Warfarin has a volume of distribution of 9.8 L. Calculate the loading dose.

1. The steady state is reached only when the amount of drug entering the system is equivalent to the amount being removed from it.

Css = Bioavailability x Dose Dosing interval x Clearance

= 1.01 mg/L= 0.93 x 5 mg24 h x 0.192 L/h

Bioavailability x Dose Dosing interval = Css x Clearance

2. Loading dose = 9.8 L x 1.01 mg/L = 9.90 mg

Problem 4A known abuser of narcotic analgesics is brought to the emergency department in adeep coma. His friends state that he took a large dose of IV morphine 6 hours earlier. A blood sample shows a morphine level of 0.25 mg/L. Assuming that the pharmacokinetics of morphine in this patient are similar to the population pharmacokinetics (volume of distribution 200 L, clearance 46.2 L/hr, half-life 3 hrs), approximately how much morphine did the patient inject?

[X] = [X]0 • e-kt

[X] = D/Vd • e- CL/Vdt

D = 200 mg

0.25 mg/L

Pla

sma

mor

ph

ine

con

c. (

log

scal

e)

Time (hr)0 63

0.5 mg/L

1 mg/LThe plasma drug concentration at any time t after administration, [X], is given by the following equation:[X] = [X]0 • e-kt ([X]0 , initial plasma drug concentration; k, elimination constant).k = CL/Vd

A compound is given as an intravenous bolus to a patient requiring a minimum plasma concentration of 40 mg/L for a therapeutic effect. Given the relevant characteristics of this compound below and assuming a one-compartment model, calculate how long the clinical effect will last following a one-gram dose. k = 0.1 hour -1; Vd = 8 L

Problem 5

Pla

sma

dru

g co

nc.

(lo

g sc

ale)

Time

40 mg/L

125 mg/L

0

Therapeutic window

[X] = [X]0 • e-kt

40 mg/L = 1000 mg / 8 L x e-0.1t

t = 11.4 hours

The plasma drug concentration at any time t after administration, [X], is given by the following equation:[X] = [X]0 • e-kt ([X]0 , initial plasma drug concentration; k, elimination constant).

C.A., an 8-year-old, 25-kg boy, is taking ethosuximide for treatment of absence seizures. Considering the relevant characteristics of the drug given below, calculate the maintenance dose required with a dosing interval of 12 hours.Vd = 0.69 L/kg; t1/2 = 30 hours; Peak target plasma concentration at steady state = 80 ug/mL; Bioavailability = 93%

Pla

sma

eth

osu

xim

ide

con

c. (

log

scal

e)

Days

Peak conc. 80 ug/mL

Trough conc.12 h

Problem 6

Ctrough = [X]0 • e-kt = 80 ug/mL x e -(0.693 / 30 h) x 12 h = 60.63 ug/mL

The amount of drug being eliminated during the dosing interval of 12 h = (Cpeak - Ctrough) x Vd x body weight (kg) = (80 ug/mL - 60.63 ug/mL) x 690 mL/kg x 25 kg = 334132.5 ug = 334.1325 mg

Maintenance dose = 334.1325 mg / Bioavailability = 334.1325 mg / 0.93 = 359.3 mg

[X] = [X]0 • e-kt ; k = 0.693 / t1/2