pharm final dr j

Intravenous Bolus Administration(Monitoring Drug in the Urine)

Non invasive technique. Perhaps more convenient to collect samples and

sample size is generally not a problem; however, one can not be selective about the sampling

time. Urinary data allows direct measurement of

bioavailability, both absolute and relative, without the need of fitting the data to a mathematical model.

SCHEME

X Xu (amount of drug in urine, mg)

Amount of drug inthe blood

XK, (hr-1)

Xu (mg)

Setup

Dose, mgor mg.kg-1

where,

X = mass (amount) of drug in the body (mg, mcg, etc.);

Xu = mass (amount) of drug in the urine (mg, mcg, etc.); and

Ku = excretion/elimination rate constant (hr-1, min-1, etc.)

Differential Equation forthe above setup

dXu

----- = Ku.X (Eq.3.20)

dtUpon integration of Eq.3.20, we get

(Xu)t = X0 (1 - e- Ku.t) (Eq.3.21)

where,

(Xu) = cumulative mass or amount (mg) of drug excreted into urine at time, t.

X0 = Administered dose (mg, mcg, etc.).

Ku = Excretion/elimination rate constant (hr-1)

Equation (Eq.3.21) clearly suggests that the cumulative mass of drug excreted into urine increases asymptotically with time.

Cumulative amountof drug in the urine

Xu, (mg)

Time (hr.)

Cumulative amount of drug in the urine following the administration of a intravenous

bolus dose

t = 0 t = t

R.L. Plot

t = ∞

(Xu)∞

We know from equation (Eq 3.21) that

(Xu)t = X0 (1 - e- Kut) (Eq.3.21)

when t = ; e- Kut = 0

therefore,

(Xu) = X0 (1 - 0) (Eq.3.22)

or

(Xu)= X0 = Administered dose

Please note that this is applicable only when the drug is totally removed in an unchanged form, as is assumed in this situation.

Cumulative amountof drug in the urine

Xu, (mg)

Time (hr.)

(Xu), amountEliminated in urine

at t = ∞

Cumulative amount of drug in urinefollowing the administration of a drug

R.L. Plot

t = 0 t = ∞

At anytime, t, the amount of drug in body (X)t + mass of drug in urine (Xu)t = Dose

Amount of drug inthe blood or urine

X, (mg)

Time (hr)

Drug in the urine

Drug in the blood or body

(X)t + (Xu)t = Dose

t = 0 t = t

At any time, t,(X)t + (Xu)t + (Xmu)t = Dose (X0)

Amount of drug inthe blood or urine

(mg)

Time (hr)

Drug in the urinein an unchanged

form

Drug in the blood or body

Drug in the urineas a metabolite

t = 0 t = t

Use of Urinary Excretion Data

There are two methods that permit us to compute some of the pharmacokinetic parameters of a drug from the urinary excretion data.

A.R.E. (Amount of drug Remain to be Excreted) or 'sigma-minus method'

Rate of Excretion method

The A.R.E. Method or Sigma Minus Method Theory: We know from earlier discussion

(Equations 3.21 and 3.22) that

(Xu)t = X0 (1 - e- Kut) (Eq.3.21)

or

(Xu)t = X0 - X0 e- Kut

We also know that when t = ,

(Xu) = X0 = Dose (Eq.3.22)

Therefore, subtracting Eq.3.21 from Eq.3.22 yields the following:

(Xu) = X0

_

(Xu)t = - X0 ± X0e-Kut

_________________

(Xu) - (Xu)t = X0e- Kut (Eq.3.23)

where, (Xu) - (Xu)t = Amount of drug

remain to be excreted or the amount of drug in the blood/body or X.

R.L. Plot

(Xu)∞- (Xu)t = X (amount of drug

remain to be excreted)

(mg)

Time, hr.

(Xu) - (Xu)t = X0e– Kut

(Eq.3.23)

Plot of the First Order Process

t = 0 t = t

S.L. Plot

(Xu)∞- (Xu)t = X(amount of drugremained to be

excreted)(mg)

Time, hr.

Ku

Slope = - ------- 2.303

Ku.tlog [(Xu)∞ - (Xu)t] = log (Xu)∞ – ------- (Eq.3.23)

2.303

Intercept = (Xu)∞= (X0)

Plot of the First Order Process

Ku = K

t = 0 t = t

Obtain t1/2 and Ku or K by using the approach described in previous lectures.

Please note that one can not obtain the apparent volume of distribution from urinary excretion

data. Also, note that on the S.L. plot shown earlier,

Intercept = (Xu)= (X0) only when there is absence of metabolite (s)

Limitations of the A.R.E Method Must collect urine samples until such time that,

practically, no more drug appears in the urine (i.e. t = 7 t1/2).

Must not lose any urine samples or urine from any samples in order to determine, (Xu) (need

to know the exact volume of urine at each time interval).

Time consuming method for drugs with long elimination half-life (t1/2).

Cumulative build up of error.

When the administered dose of a drug is not completely removed in an unchanged form

(i.e. K is Ku and (Xu) is (X0) or dose

Xu (amount of unchanged drug in the urine or amount

excreted)

X

Xmu (amount of metabolite in the urine)

Example of parallel process or reaction

Ku

Km

where,

X = mass or amount (mg) of drug in the body,

Xu = mass or amount (mg) of unchanged drug in the urine,

Xmu = mass or amount (mg) of metabolite in the urine, and,

Ku and Km are the excretion and metabolite rate constants (hr-1), respectively (K = Ku + Km )

dXu

----- = Ku.X

dt

Using the La Place transformation technique,

the equation can be integrated into

Ku.X0

(Xu)t = ------- (1 - e- Kt)

Kwhere,

X0 = the administered dose (mg)

K = the elimination rate constant (hr-1)

when t = infinity (); (Xu) = (Xu) and

e-Kt ----> 0; therefore, equation reduces to

Ku.X0

(Xu) = -------

K

Ku. X0

substitute for ---------

K

in to earlier equation for (Xu)

(Xu)t = (Xu) (1 - e- Kt)

(Xu)t = (Xu) - (Xu) e- Kt

rearrangement of the equation yields

(Xu) - (Xu)t = (Xu) e- Kt

taking the logarithmic form of the equation yields

K. tlog [(Xu) - (Xu)t] = log (Xu) - -------

2.303

Intercept Slope

R.L. Plot

(Xu)∞- (Xu)t

(amount of drug remain to be

excreted)(mg)

Time, hr.

(Xu) - (Xu)t = (Xu)∞e– Kt

Plot of the First Order Process

t = 0 t = t

S.L Plot

(Xu)∞- (Xu)t

(amount of drugremain to be

excreted)(mg)

Time, hr.

KSlope = - -------

2.303

K.tlog [(Xu)∞ - (Xu)t] = log (Xu)∞ – ------

2.303

Intercept = (Xu)∞

Plot of the First Order Process

t = 0 t = t

Example of PharmacokineticAnalysis of Urinary Excretion Data

ExampleIntravenous bolus dose of 80.0 mg was administered. Drug is eliminated entirely by urinary excretion as unchanged drug following one compartment model and first order eliminationAssumptionsOne compartment open model. Entire dose is eliminated in an unchanged form (i.e., excretion rate constant and elimination rate constant are congruent) a by the first order process and passive diffusion (80 mg I.V. Bolus dose)Follow Table 3; page # 35 of your handout

Time interval of urine

collection (hr)

Volume of Urine

collected (mL)

Drug concen. in Urine

(mg/mL)

Mass (amount)of drug in urine

Xu (mg)

0 - 1 200 0.200 40.0

1 - 2 50 0.400 20.0

2 - 3 50 0.200 10.0

3 - 4 100 0.050 5.00

4 - 5 25 0.100 2.50

5 - 6 125 0.010 1.25

6 - 12 250 0.005 1.25

Cumulativeamount of

drug excretedXu (mg)

Time (t), hr

Amount remainto be excreted

ARE (mg)[(Xu)∞ - (Xu)t

Averagetime (t)

(hr)

(ΔXu/Δt)t

(mg/hr)

40.00 1.00 40.00 0.5 40.00

60.00 2.00 20.00 1.5 20.00

70.00 3.00 10.00 2.5 10.00

75.00 4.00 5.00 3.5 5.00

77.50 5.00 2.50 4.5 2.50

78.75 6.00 1.25 5.5 1.25

80.0* 12.0 0.00 9.0 0.21

Next two figures (page # 37 and 38 of your handout) represent the graphs of cumulative amount of drug eliminated against time (page 37) and amount of drug remaining to be excreted against time (page 38; ARE; S.L. plot) for the data presented in Table on previous slide (Table 3)

R. L. Plot

(Xu)∞

S. L. Plot for A.R.E Method

(Xu)∞

Rate of Excretion MethodTheory:We know from earlier differential

equation (Eq.3.20)

dXu

------ = KuX (Eq.3.20)

dt

We also know that

X = X0 e- Kut (Eq.1.18)

Therefore, substituting for X in Eq.3.20 from Eq.1.18, yields the following:

dXu

----- = Ku.X0 e- Kut (Eq.3.24)

dtIn practice equation 3.24 becomes

(dXu)t --

-------- = Ku.X0 e- Ku.t (Eq.3.25)

dt

Where,

(dXu/dt)t = average rate of excretion (mg/hr or mcg/hr, etc.)

-

t = average time between urine collection (hr, min.)

Ku = excretion rate constant (hr-1, min-1)

X0 = Dose (mg)

Both equations (Eqs 3.24 and 3.25) suggest that the rate of excretion of a drug declines mono-exponentially with time.

R.L Plot

(dXu/dt)t

(mg/hr)

Time (t), hr.

(dXu)t ---------- = Ku.X0 e- Kut

dt(Eq.3.25)

Plot of the Rate of ExcretionAgainst the Average Time (t)

t = 0 t = t

S. L Plot

(dXu/dt)t mg/hr

Time, hr.

Ku

Slope = - ------ 2.303

Ku.tlog (dXu/dt)t = log Ku.X0 – ------- (Eq.3.25)

2.303

Intercept = Ku.X0

Plot of the Rate of Excretion Against the Average Time

t = 0 t = t

Ku = K

Computation of Rate of ExcretionUsing the data presented in Table 3 (column 5 and 6), the rate of excretions were calculated and reported in column 9. The rate of excretion reported in column 9 are plotted against average time (column 8) values

Cumulativeamount of

drug excretedXu (mg)

Time (t), hr

Amount remainto be excreted

ARE (mg)[(Xu)∞ - (Xu)t

Averagetime (t)

(hr)

(ΔXu/Δt)t

(mg/hr)

40.00 1.00 40.00 0.5 40.00

60.00 2.00 20.00 1.5 20.00

70.00 3.00 10.00 2.5 10.00

75.00 4.00 5.00 3.5 5.00

77.50 5.00 2.50 4.5 2.50

78.75 6.00 1.25 5.5 1.25

80.0* 12.0 0.00 9.0 0.21

A plot (S. L. plot) on page 42 of your handout represents the data for the rate of excretion (dxu/dt) against average time (t) presented in Table 3 (page 35).

Using this graph or ARE method plot on S. L. paper, one can obtain some of the pharmacokinetic parameters of a drug

Find t1/2 and Ku or K from the S. L. plot of --

Rate of excretion vs. average time (t)

S. L. Plot of Rate of Excretion vs. Average

Time

Intercept = initial rateof excretion = Ku .X0

Intercept------------ = K or Ku

(X0)

Elimination Half life (t1/2)t1/2 = 1 hr

(from the plot of rate of excretion vs. time)

Elimination/Excretion Rate Constant.

Ku = 0.693/t1/2 = 0.693/1 hr = 0.693 hr-1

Intercept = Ku.X0 (mg/hr)

= initial rate of excretion/elimination

From the graph, intercept = 56 mg/hr

Intercept

------------ = X0 (administered dose)

Ku

56 mg.hr-1

------------- = X0 (administered dose)0.693 hr-1

80.8 mg = X0 = Dose(Overestimate is due to the averaging technique)

Intercept

-------------- = Ku or K

X0 (dose)

56 mg/hr

-------------- = 0.70 hr-1 = Ku

80 mg

Intercept = Ku.X0

Intercept = 0.693 hr-1 x 80 mg

Intercept = 55.44 mg/hr

Rate of Excretion Method

(When K Ku or (Xu)∞ X0) When administered dose of a drug is not totally removed in an unchanged form (i.e., K is Ku and/or (Xu) is X0

dXu

----- = Ku.X dt

where,

dXu/dt = the rate (mg/hr) of excretion X = the mass or amount (mg) of drug in the

body

Ku = the excretion rate constant (hr-1)

However, according to earlier equation

X = X0.e– Kt

(When drug is monitored in the blood) substitute for the term X in the rate equation

dXu

----- = Ku.X0 e- Kt

dtIn the logarithmic form, the equation becomes

dXu K. t

log ------ = log Ku.X0 - -------dt 2.303

S.L. Plot

(dXu/dt)t mg/hr

Average time, hr.

KSlope = - -------

2.303

K.tlog (dXu/dt)t = log Ku.X0 – ------

2.303

Intercept = Ku.X0

Plot of Rate of ExcretionAgainst Average Time (t)

General Comments onRate of Excretion Method

1. Method tends to give overestimate of the intercept value.

2. The over estimation can be minimized by having more frequent urine samples (not always

practical and easy).The table (next slide) indicates that the more frequent urine samples we have, the less error involved in estimating pharmacokinetic parameters.

The table indicates that the more frequent urine samples we have, the smaller is the error involved in estimating

pharmacokinetic parameters

# of Half-lives % Over estimate

3.00 190.0

2.00 80.00

1.00 20.00

0.50 6.00

0.25 0.03

Clearance Concepts What is clearance (Cl)? Why is this parameter clinically very

important? Relationship between the apparent volume of

drug distribution (V) and the elimination rate constant (K).

Relationship between the rate of elimination and/or rate of excretion and the plasma or serum concentration.

What are the units for clearance? The types of clearances. Different methods of determining the different

types of clearance.

What is an extraction ratio? How is it related to clearance?

Creatinine clearance (Cl)cr and is significance in clinical practice.

Methods of determination of the creatinine clearance (Cl)cr.

Inulin clearance and its use in clinical practice. Relationship between creatinine clearance and

the systemic clearance.

Use of creatinine clearance in evaluating organ responsible for the elimination of drugs and

use of this information in adjusting the dose of a drug.

Use of an equation in adjusting the dose of a drug.

Use of nomograms in adjusting dose of a drugTables of few drugs with recommended doses

in renally impaired patients.

IntroductionThis parameter, perhaps, has the greatest

potential for clinical applications. Furthermore, this is the most useful parameter available for the evaluation of the elimination mechanism and eliminating organs (kidney and liver). The utility of clearance measurement lies in its intrinsic model independence.

Drugs are eliminated from the body by metabolism and excretion. The liver is the major site of drug metabolism; however, other tissues also contain drug metabolizing enzymes and, therefore, contribute to the biotransformation of selected drugs. Kidneys, on the other hand, are involved in the elimination of virtually every drug or drug metabolite.

Some drugs, such as gentamicin and cephalexin, are eliminated from the body almost solely by renal excretion. Many other drugs are eliminated in part by the kidneys.

And, even when the drug elimination from the body involves biotransformation, the corresponding drug metabolites are usually cleared by the kidneys.

Therefore, kidneys play an important role in removal of unchanged drug and/or the metabolites from the body. Some drugs are excreted in the bile and may be eliminated in the feces.

The clearance concept was developed by renal physiologists in early 1930 as an empiric measure of kidney function.

The pharmacokinetic basis of the term was defined at about the same time with the recognition that the concept could be more generally applied to other organs and elimination pathways.

The next couple of lectures will describe some aspects of the current understanding and applications of clearance with emphasis placed upon the renal excretion of drugs.

Clearance (Definitions) The most general definition of clearance (Cl) is

that 'it is a proportionality constant that describes the relationship between a substance's rate of elimination (amount per unit time) at a time and its corresponding concentration in an appropriate body fluid or concentration at that time.’

The clearance can also be defined as "the hypothetical volume of blood (plasma or serum) or other biological fluids from which the drug is totally and irreversibly removed per unit time."

The larger the hypothetical value, the more efficient is the eliminating organ (kidney and/or liver).

The limiting factor is the volume of the blood that is presented to the eliminating organ.

For kidneys, the upper limit of blood flow is 19 ml.min-1.kg-1

For liver, the upper limit of blood flow is about 1.5 Liter.min-1.

Systemic (Cls) or Total Body Clearance (TBC) This is the sum of all individual organ

clearances that contribute to the overall elimination of drugs.

However, the organ clearance that can be routinely determined independently, in human, is the renal clearance because this is the only organ for which we can easily determine an elimination rate. Removal process is elimination (excretion and metabolism).

Renal clearance, (Clr)The drug or the fraction thereof that is removed

from the blood (plasma/serum) by the process of renal excretion.Metabolic clearance, (Clm)

The drug or fraction thereof that is removed from the blood (plasma/serum) by the process of metabolism.Hepatic clearance, (ClH)

The drug or fraction thereof that is removed from the blood (plasma/serum) by the process of metabolism.The organ responsible for metabolism is liver.

It can be shown that the total body clearance (ClTBC) or systemic clearance (Cl)s of a drug is the summation of all the organ clearances.

Hence, (Cl)s is often partitioned into renal (Cl)r and non-renal (Cl)nr clearance.

(Cl)s = (Cl)nr + (Cl)r

Although there may be many sites of drug elimination besides kidney and liver, these two organs are quantitatively the most important and, therefore, have been most thoroughly studied.

Non-renal clearance (Cl)nr is often considered to be equivalent to hepatic clearance (Cl)H.

The unit for all clearances is ml.min-1 or lit.hr-1

ml.min-1.kg-1 or lit.hr-1.kg-1

ml.min-1.1.73 m2 (i.e. the normal body surface area)

Clearance Concepts (Rate and Concentration) Just like parameter apparent volume of

distribution (V or Vd) is necessary to relate the plasma or serum concentration (Cp or Cs) to the mass or amount (X) of the drug in the body at a time, there is also a need to have a parameter that relates the plasma or serum concentration (Cp or Cs) to the rate of drug excretion or elimination (dXu/dt) at a time. The term clearance (Cl) is that proportionality constant.

Rate of excretion =Clearance x Plasma or Serum concentration

(dXu/dt)t = Cl x (Cp)t or (Cs)t

(Eq.3.26)

Rearranging equation 3.26 yields the

following

(dXu/dt)t mcg/hr

Cl = ----------- = ---------- = mL.hr-1 (Cp)t mcg/mL

(Eq.3.27)

For example, if the rate of elimination and the average plasma concentration at a time are 1 mg/hr and 1 mg/liter, respectively; then

1 mg.hr-1

Cl = ----------- = 1 lit.hr-1

1 mg.lit-1

For a given rate of excretion, the type of clearance depends upon the site of drug measurement (blood, plasma, serum). Generally, when a first order process and passive diffusion are applicable; as the concentration of drug in the body (serum, plasma) increases, so does its rate of elimination; clearance, however, remains independent of the dose administered.

Organ Clearance

Eliminatingorgan

Qml/hr

Qml/hr

CA Cv

E. R.

Consider the following situation in the body following the administration of a drug and a well perfused organ (kidney or liver) capable of eliminating the drug

Where,

Q = blood flow rate (ml/min) through an eliminating organ,

CA = the drug concentration (mcg/ml) in the arterial blood entering the organ, and

Cv = the drug concentration (mcg/ml) in the venous blood leaving an organ

If the organ eliminates or metabolizes some or all of the drug entering an organ then

CV < CA

CA x Q = mcg.ml-1 x ml.min-1

the rate (mcg/min) at which drug enters the organ

Cv x Q = mcg.ml-1 x ml.min-1

the rate (mcg/min) at which drug leaves the organ

Based on steady state and mass balance considerations, the instantaneous rate of drug elimination from the organ is equal to the difference between the rate at which drug enters an organ and the rate at which it leaves an organ.

This is equal to the product of the blood flow rate (Q) and the arterial-venous concentration difference (CA - CV).

Clorgan = Rate of elimination

= (CA.Q - CV.Q)

Clorgan = Rate of elimination

= Q (CA- CV)

(Eq.3.28)

If we obtain the ratio of rate of elimination to the rate at which drug enters an organ, we would give rise to a dimensionless term called Extraction ratio, (E.R).

Rate of elimination Q (CA - CV)

E.R = ------------------------ = ---------------

Rate in Q (CA)

(CA – CV)

E.R = ------------ (Eq.3.29)

CA

The Extraction ratio (E.R) quantifies the efficiency of an organ with respect to drug elimination.

If an organ is incapable of eliminating the drug, CV will be equal to CA, and the extraction ratio = 0.

On the other hand, if the organ is so efficient in metabolizing or eliminating the drug that CV 0, then the extraction ratio approaches unity (E.R will have a value from 0 to 1.0).

We can also look at the extraction ratio (E.R) as an index of how efficiently the organ clears a drug from the blood flowing through an organ.

For example, an extraction ratio of 0.8 indicates that 80% of the blood flowing through the organ will be completely cleared of drug.

Following this line of reasoning, we can define the organ clearance of a drug as the product of the extraction ratio (E.R.) and the flow rate (Q).

Organ clearance =

Blood flow rate (Q) x E.R

Cl = Q x E.R (Eq.3.30)

It is possible to estimate clearance by direct determination of the parameters in equations. However, the practical difficulty involved in applying this approach usually precludes its use.

First, an accurate estimation of organ flow rate (Q) is difficult to obtain. Moreover, the total flow rate may not necessarily be constant over the study period.

Also, measuring the concentration of drug in arteries (CA) and veins (CV) is not very easy experimentally, particularly, in human.

Estimation of Systemic ClearanceTotal Clearance

From earlier definition, we know that

Rate of elimination

(ClS) = -------------------------

(Cp)

or

(dxu/dt)t

(ClS) = ----------

(Cp)t

This relationship makes it relatively easy to determine the renal clearance (Clr) of any drug that is excreted, to some measurable extent, in unchanged form in the urine.

1. Determine the elimination and/or excretion rate of drug by methods discussed in previous lectures.

2. Determine the plasma concentration (Cp) at a point of urine collection interval.

Integrating the right hand side of the equation from t = 0 to t = , we obtain the following:

(dxu/dt)t . dt

0

Cl = ------------------

(Cp).dt

0

For intravenous bolus administration

Dose

(Cl)s = ---------- (Eq.3.31)

(AUC)0

When a single organ is solely responsible for the complete elimination of all of a xenobiotic, then the total clearance can be measured by application of either eq.3.31 under steady state condition.

Calculating Renal Clearance (Clr) andMetabolic Clearance (Clm)

Renal clearance (Clr)

1. (Cl)r = Ku.V (Eq. 3.32)

Where,Ku is the excretion rate constant (hr-1)

V is the apparent volume of distribution (ml, lit, lit.kg-1, etc.)

or

(Xu)

2. (Cl)r = ---------- (Eq.3.33)

(AUC)0

Where, (Xu) is the amount of unchanged drug excreted in urine at t = and,

(AUC)0 is the area under the plasma

concentration time curve from t = 0 to t =

or

(% excreted) x (Dose)

3. (Cl)r = ---------------------------

(AUC)0 (Eq.3.34)

Metabolic Clearance (Clm)

1. (Cl)m = Km.V (Eq. 3.35)Where,

Km is metabolite rate constant (hr-1)V is the apparent volume of distribution

Or

(Xmu)

2. (Cl)m = Km.V = ----------

(AUC)0 (Eq. 3.36)

Where,

(Xmu) is the amount (mg) or mass of metabolite in urine at time t =

or

(Cl)m = Km.V

(% of metabolite in urine) x (Dose)

=---------------------------------------------

(AUC)0

(Eq. 3.37)

Please note that when the drug is removed completely in an unchanged form [i.e., (Xu) = (X0)] then renal clearance (Cl)r is equal to systemic clearance (Cl)s. Analogously, if drug is completely eliminated as a metabolite [i.e., (Xmu) = Dose (X0)] then metabolic clearance (Cl)m is equal to systemic clearance (Cl)s.

(Systemic clearance)

(Cl)s = (Cl)r + (Cl)m (Eq. 3.38)

or

(Cl)m = (Cl)s - (Cl)r

or

(Cl)r = (Cl)s - (Cl)m

(Systemic clearance)

(Cl)s = (Cl)r + (Cl)m (Eq. 3.38)

or

(K x V) = (Ku x V) + (Km x V)or

(K x V) = V (Ku + Km)

K x V

-------- = (Ku + Km)

V

Determination of the Area Under the Plasma Concentration versus Time Curve (AUC)0

(Application of Trapezoidal Rule)

Intravenous Bolus Administration

One Compartment Model

R. L. plot

Concentrationmcg/ml

Time (hr.)t = 0 t = t* t = ∞

Use of Trapezoidal Rule for the determination of the areaunder the plasma concentration time curve (AUC)0

∞

The last observed Concentration value

Plasma Concentration Versus Time Plot

Cp.dt = (AUC)0 = (AUC)0

t* + (AUC)t*

0

(Eq. 3.39)

(AUC)0t* can be determined by the application of

trapezoidal rule and

(AUC)t*can be obtained by using an equation

R. L. plot

Concentrationmcg/ml

Time (hr.)t = 0 t = t* t = ∞

Use of Trapezoidal Rule for the determination of the areaunder the plasma concentration time curve (AUC)0

∞

Last observed concentration

Plasma Concentration Versus Time Plot

Trapezoid

How to Use Trapezoidal Rule

t1 (Cp)0 + (Cp)1

Cp.dt = (AUC)0t1 = --------------- x (t1 - t0)

t0 2

(AUC)0t1 = Average (Cp) x dt

(AUC)0t1 = mcg.mL-1 x hr = mcg.mL-1.hr

t2 (Cp)1 + (Cp)2

Cp.dt = (AUC)t1t2 = -------------- x (t2 – t1)

t1 2

(AUC)1t2 = Average (Cp) x dt

(AUC)1t2 = mcg.mL-1 x hr = mcg.mL-1.hr

Follow this procedure and determine (AUC) for each trapezoid until the last observed plasma concentration value

(AUC)0t* = Sum of all the trapezoid value

Determination of (AUC)t*

This may be obtained by using the following equation

Cp*

(AUC)t* = Cp.dt = ----- (Eq.3.40)

t* K

Where,

Cp* = the last observed plasma concentration (mcg.ml-1)

K = the elimination rate constant (hr-1)

Therefore,

Cp.dt = (AUC)0 = (AUC)0

t* + (AUC)t*

0

EXAMPLE(Problem Set #1; question #4)

t1 (Cp)0 + (Cp)1

Cp.dt = (AUC)0t1 = ---------------- x (t1 - t0)

t0 2

12.0 + 11.6

= -------------- x (0.25 - 0.0)

2

= 11.8 mcg/mL x 0.25 hr

(AUC)0t1 = 2.95 mcg.m-1.hr

Average of two concentration values

t2 (Cp)1 + (Cp)2

Cp.dt = (AUC)t1t2 =--------------- x (t2 - t1)

t1 211.6 + 8.4

= ------------- x (0.5 - 0.25) 2

= 10.0 mcg/mL x 0.25 hr

(AUC)t1t2 = 2.50 mcg.mL-1.hr

t3 (Cp)2 + (Cp)3

Cp.dt = (AUC)t2t3 = -------------- x (t3 - t2)

t2 28.4 + 7.2

= ----------- x (0.75 - 0.50) 2 = 7.8 mcg/mL x 0.25 hr

(AUC)t2t3 = 1.95 mcg.mL-1.hr

Follow this procedure until the last observed serum concentration (i.e. Cs = 0.09 mcg/mL).Determine the cumulative (AUC) by adding the individual (AUC) values up to the last observed concentration (i.e. Cs = 0.09 mcg/mL at 8 hrs)

In this example,

8hr Cp.dt = (AUC)t0

t8 = 19.177 mcg.mL-1.hr0

Cp* (mcg/mL) Cp.dt = (AUC)8

= -----------------8 K (hr-1)

0.09 mcg/mL = ----------------

0.577 hr-1

(AUC)8 = 0.156 mcg.mL-1.hr

Cp.dt = (AUC)0 = (AUC)0

8 + (AUC)t8

0

(AUC)0 = 19.177 + 0.156

(AUC)0 = 19.333 mcg.mL-1.hr

Alternately, for drugs that are administered as an intravenous bolus,

Dose

(Cl)s = ---------- (Eq. 3.31)

(AUC)0

Dose Dose

(AUC)0 = ------- = ------

(Cl)s V.K

(Cp)0

(AUC)0 = ------ (Eq. 3.41)

K

Elimination Mechanism The appearance of drug in the urine is the net

result of filtration, secretion, and re-absorption processes.Rate of excretion =

Rate of filtration + Rate of secretion- Rate of re-absorption

If a drug is only filtered and all the filtered drug is excreted into the urine thenThe rate of excretion = Rate of filtration

Renal clearance (Cl)r = fu x G.F.R.Where,fu = fraction of unbound drug andG.F.R = glomerular filtration rate.

Creatinine and inulin (an exogenous polysaccharide) are not bound to plasma proteins nor secreted into urine. Therefore, renal clearance for each of these substances is a clear measure of G.F.R. and, therefore, kidney function.

Creatinine Clearance (Cl)cr

This is a renal clearance (Cl)r applied to endogenous creatinine.

It is used to monitor renal function, and this is a valuable parameter for calculating dosage regimens in elderly patients or those suffering from renal dysfunction. The normal creatinine clearance (Cl)cr values are as follows:

For male: 120 ± 20 ml/min

For female: 108 ± 20 ml/min

Methods of Measuring Renal Function Creatinine is an end product of muscle

metabolism, and appears to be eliminated from the body by the kidneys.

The normal range of serum creatinine concentration is between 1 and 2 mg/100 mL (1-2 mg%).

Renal excretion of endogenous creatinine is largely dependent on glomerular filtration and closely approximates the glomerular filtration rate as measured by inulin, both in healthy individuals and individuals with impaired renal function.

Since creatinine production bears a direct relation to the muscle mass of an individual, creatinine clearance measurements, where possible, are normalized to a body surface area of 1.72 m2 in order to obtain a more comparative measurement for different individuals.

Although inulin clearance is generally accepted as the most accurate method for the estimation of glomerular filtration rate, which is approximately 125 ml/min in healthy individuals, its practical utility for evaluating renal function is limited.

Creatinine is formed from muscle metabolism in the body and circulates in the plasma of individuals with normal renal function at a concentration of approximately 1 mg %.

Creatinine is cleared via kidneys by filtration to yield the creatinine clearance of about 130 ml/min. This value depends partially on body size, degree of activity, muscle mass, and age.

Renal function can be measured in several ways.

The most common method involves determining circulatory levels and excretion of creatinine or creatinine clearance (Cl)cr.

Direct Measurement

The mass of endogenous creatinine excreted into the urine, collected over a given time interval (Δt), is determined.

The mean serum creatinine concentration (Cs)cr over that interval is calculated from sample determinations; this represents the concentration halfway through the interval. In practice, Δt = 24 hr and, as (Cs)cr is relatively constant, the serum sample is taken at any convenient time.

(ΔXu/Δt)t

(Cl)cr = ----------- (Eq.3.42)

(Cs)cr If the rate of excretion of creatinine is 1.3 mg/min and serum creatinine concentration is 0.01 mg/ml then the creatinine clearance is 130 ml/min.Please note that one must obtain both creatinine excretion rate and average serum creatinine concentration to accurately measure creatinine clearance.

For Adults (Cockcroft-Gault Equation)

Weight (kg) x (140 - age)

(Cl)cr = -------------------------------

72 x (Cs)cr (mg%) (Eq.3.45)

The age is expressed in years; body weight in Kg; and serum creatinine concentration in mg/100 mL. The same equation is used for male and female patients, but the value determined should be reduced by 15% for female patients

For female : 0.85 x above value

The use of serum creatinine to determine renal function has been reviewed in considerable

detail by Lott et al. Normal serum creatinine concentrations vary

from 0.6 to 1.0 mg per 100 ml (mg/dL) in women and 0.8 to 1.3 mg/dL in men

Significance of Creatinine Clearance Normal creatinine clearance value indicates

that kidney is functioning normally. In some disease states or pathological conditions

or in elderly population, the creatinine clearance is likely to alter. This, consequently, results in lower values for creatinine clearance.

If conditions described in #2 exist then we must consider adjusting the dose of a drug if that drug is eliminated by kidney.

Failure to do so will result in much higher blood levels (may be toxic levels) for that drug.

Alternative to adjusting the dose of a drug is to decrease the frequency at which the normal

dose is administered e.g., if a normal dose is 250 mg Q.I.D or 1 tablet Q.I.D, change this to 250 mg T.I.D or B.I.D, etc.

Lower creatinine clearance value will affect other so called 'constant' parameters.

They include the elimination or excretion rate constant (K or Ku), elimination half life (t1/2)

and, possibly, the apparent volume of distribution of a drug.

This, in turn, will influence the value of any other pharmacokinetic parameter whose value depends on K, Ku, and t1/2 and (For one compartment model).

These parameters include Cp at any time t, area under the plasma concentration time curve, (AUC)0

, and clearance (Cl)s.Will the plasma concentration change when time = 0 (i.e. Cp)0?

(Cp) = (Cp)0 e-Kt

Will the rate of elimination change when t = 0 (i.e. dxu/dt at time = 0)?

(dXu/dt) = K(X0) e-Kt

Will the rate of elimination change when t = t (i.e. dxu/dt at time = t)?

For Intravenous Bolus(One Compartment)

(Cp) = (Cp)0.e- Kt and (Cp)0 = X0/V

t1/2 = 0.693/K and (Cl)s = K.V

(Dose)

(AUC)0 = --------

(Cl)s

How will the plot of AUC against (Cl)s look like?

For I.V. Bolus (One Compartment)(A drug with identical doses)

The next slide is a S.L. plot of plasma concentration against time following the administration of the same drug and dose in three subjects with different degree of renal impairment

Administration of an identical dose of a drug,as an IV bolus, to three different subjects

with renal insufficiency

(Cp) = (Cp)0.e - Kt

Administration of an identical dose of a drug by anextravascular route to three subjects with renal insufficiency

Minocycline

Cefazolin

(t1/2)hr.

Comments on Table # 5 and 6(pages 66 and 67 of the handout)

Acyclovir (ZoviraxR)

Treatment of initial and recurrent mucosal and cutaneous HSV-1 and HSV-2 infections in immuno-copmpromised adults and children and for severe initial clinical episodes of genital herpes.

About 62 to 91% of the dose, following intravenous administration, is excreted renally in changed form.

Creatinine Clearance(ml.min-1.1.73 m2)

Eliminationhalf-life (t1/2)

(hr)

Total Body Clearance

(ml.min-1.1.73 m2)

> 80 2.5 327

50 to 80 3.0 248

50 to 15 3.5 190

0 (Anuric) 19.5 29

Using the data provided in the table, plot Elimination half life against creatinine clearance Creatinine clearance against systemic clearance Elimination half life against systemic clearance

Creatinine Clearance

(ml.min-1.1.73 m2)

Dose(mg.kg-1)

Dosing interval(τ), hours

> 50 5.0 8.0

25 to 50 5.0 12.0

10 to 25 5.0 24.0

0 to 10 2.5 24.0

Ceftazidime (FortazR, TazicefR, TazidimeR)

Cephalosporin related antibiotic that is used for lower respiratory tract infection, skin infections, urinary tract infections, etc. It is excreted, almost exclusively, by glomerular filtration.

Recommended MaintenanceDoses in Renal Insufficiency

Creatinine Clearance (ml.min-1)

Recommended Unit Dose

Frequency of Dosing (τ)

50 to 31 1 gram Every 12 hours

30 to 16 1 gram Every 24 hours

15 to 6 500 mg Every 24 hours

< 5 500 mg Every 48 hours

Note unitfor (Cl)cr

The nomogram for netilmicin, an aminoglycoside antibiotic

Calculation of Adjusted Daily Dose

Adjusted daily dose =

Patient’s creatinine clearance

Normal daily dose x -----------------------------------------

Normal creatinine clearance

Some Profiles to Ponder

(AUC)0∞

ng.ml-1.hr

Dose, mg or mg.kg-1

Subject A

Subject B

Subject C

Which subject is a normal subject?

(AUC)0∞ = Dose/VK = Dose/Cl

For a drug totally removed by the kidneys, systemic clearance against creatinine clearance

For a drug removed by the kidneys, elimination rate constant of a drug against creatinine clearance

For a drug removed by the liver, elimination half life against the creatinine clearance

Plot of ratio of rate of excretion/rate of elimination at a time against administered dose

Plot ARE at a time against dose administered Plot ARE against number of elimination half

lives Plot of (AUC)0

∞ against Dose administered

Plot of (AUC)0∞ against Systemic clearance

Rate of elimination against average time (R.L and S.L plots)

Rate of elimination at a time against dose administered

Rate of elimination against # of elimination half lives