Toxicokinetic CalculationsToxicokinetic Calculations

Extent of distributionExtent of distribution The parameter that reflects the extent of The parameter that reflects the extent of

distribution is the apparent volume of distribution is the apparent volume of distribution, Vdistribution, Vdd, where:, where:

VVdd = Dose/C = Dose/Cp,0p,0

Where dose = total amount of drug in the body Where dose = total amount of drug in the body , while C, while Cp,0p,0 is the concentration of drug in is the concentration of drug in

plasma at 0 hrs after injection. plasma at 0 hrs after injection.

Example:Example: After an IV bolus dose of 500 mg, the following data After an IV bolus dose of 500 mg, the following data

was collected: Find the elimination rate and the was collected: Find the elimination rate and the apparent volume of distribution.apparent volume of distribution.

Solution:Solution: First, we should be familiar with the first order First, we should be familiar with the first order

kinetics where:kinetics where:

Time (hr)12346810

Cp, mg/L725133201494

dCdCpp/dt = -K/dt = -Kel. el. CCpp

dCdCpp/C/Cpp = -K = -Kelel.dt.dt Integrationof the above equation gives:Integrationof the above equation gives: CCpp = C = Cp0p0 e e-Kel.t -Kel.t , or, or

lnClnCpp = lnC = lnCp0p0 - K - Kelel.t.t

Between time tBetween time t11 and t and t22, we have, we have

ln Cln Cp1p1 – lnC – lnCp2p2 = k = kelel (t (t22 – t – t11)) It follows that:It follows that: KKelel = (lnC = (lnCp1p1 – lnC – lnCp2p2)/(t)/(t22 – t – t11)) Plotting lnCPlotting lnCp p versus time should yield a straight versus time should yield a straight

line with a slope equals kline with a slope equals kelel

After plotting the curve, extrapolation should After plotting the curve, extrapolation should yield Cyield Cp0p0..

Finally the apparent volume can be calculated Finally the apparent volume can be calculated from the relation:from the relation:

VVdd = Dose/C = Dose/Cp,0p,0

KKelel = (lnC = (lnCp1 p1 - lnC- lnCp2p2)/(t)/(t22 – t – t11))

KKelel = (ln87.1 – ln4.17)/(10 – 0) = (ln87.1 – ln4.17)/(10 – 0)

KKelel = 0.304/hr = 0.304/hr

VVdd = Dose/C = Dose/Cp,0p,0

VVdd = 500/87.1 = 5.74 L = 500/87.1 = 5.74 L

Half life of eliminationHalf life of elimination From first order kinetics we have:From first order kinetics we have: lnClnCpp = lnC = lnCp0p0 – k – keleltt lnClnCp p - lnC- lnCp0p0 = -k = -keleltt lnClnCpp/C/Cp0p0 = -k = -keleltt The half life of elimination is defined as the time The half life of elimination is defined as the time

required for the concentration to decrease to one half. required for the concentration to decrease to one half. This means CThis means Cp0p0 = 2C = 2Cpp

Substituting in the last equation above gives:Substituting in the last equation above gives: ln ½ = -kln ½ = -keleltt1/21/2 Or: tOr: t1/21/2 = 0.693/k = 0.693/kelel

The steps to take are: The steps to take are:

1.1. Draw a line through the points (this tends to Draw a line through the points (this tends to average the data) average the data)

2.2. Pick any CPick any Cpp and t and t11 on the line on the line

3.3. Determine CDetermine Cpp/2 and t/2 and t22 using the line using the line

4.4. Calculate tCalculate t1/21/2 as (t as (t22 - t - t11) )

And finally calculate kAnd finally calculate kelel = 0.693/t = 0.693/t1/21/2

Cp/2 Cp/2 in in 1 half-life i.e. 50.0 % lost 50.0 %1 half-life i.e. 50.0 % lost 50.0 %Cp/4 Cp/4 in in 2 half-lives i.e. 25.0 % lost 75.0 %2 half-lives i.e. 25.0 % lost 75.0 %Cp/8 Cp/8 in in 3 half-lives i.e. 12.5 % lost 87.5 %3 half-lives i.e. 12.5 % lost 87.5 %Cp/16 Cp/16 in in 4 half-lives i.e. 6.25 % lost 93.75 %4 half-lives i.e. 6.25 % lost 93.75 %Cp/32 Cp/32 in in 5 half-lives i.e. 3.125 % lost 96.875 %5 half-lives i.e. 3.125 % lost 96.875 %Cp/64 Cp/64 in in 6 half-lives i.e. 1.563 % lost 98.438 %6 half-lives i.e. 1.563 % lost 98.438 %Cp/128 Cp/128 in in 7 half-lives i.e. 0.781 % lost 99.219 % 7 half-lives i.e. 0.781 % lost 99.219 %

Thus over 95 % is lost or eliminated after 5 half-livesThus over 95 % is lost or eliminated after 5 half-lives

ExampleExample If the rate of elimination of a drug is 0.3/hr, If the rate of elimination of a drug is 0.3/hr,

find the half life of elimination.find the half life of elimination. tt1/21/2 = 0.693/k = 0.693/kelel

tt1/21/2 = 0.693/0.3 = 0.693/0.3

tt1/21/2 = 2.31 hr = 2.31 hr

ClearanceClearance

At t = 0, e-kel*t = 1 and at t = ∞, e-kel*t = 0 , Therefore:

Or, V = Dose/(AUC * kel)

In the same mannar,In the same mannar,

From these equations and since From these equations and since CL=DoseCL=Doseiviv/AUC/AUCiviv , ,

and since AUCand since AUC0-0-Dose/(Dose/(VVdd*K*Kelel))

It Turns out that CL = VIt Turns out that CL = Vdd*K*Kelel

AUC by Trapezoidal ApproachAUC by Trapezoidal Approach

The area of a trapezoid is calculated as:The area of a trapezoid is calculated as:

A = ½ sum of the two parallel sides * heightA = ½ sum of the two parallel sides * height

However, look at the AUC where we have However, look at the AUC where we have plasma concentrations usually taken within plasma concentrations usually taken within some interval, say from 1-10 hrs:some interval, say from 1-10 hrs:

It is evident that the overall AUC value should It is evident that the overall AUC value should involve the trapezoids from 0-1 and from 10-involve the trapezoids from 0-1 and from 10-infinity:infinity:

Example Calculation of AUCExample Calculation of AUC

A dose of A dose of 250 mg250 mg was administered to healthy was administered to healthy volunteer. Seven blood samples were collected volunteer. Seven blood samples were collected at 0.5, 1, 2, 4, 6, 8, 10 hours. Plasma was at 0.5, 1, 2, 4, 6, 8, 10 hours. Plasma was separated from each blood sample and separated from each blood sample and analyzed for drug concentration. The collected analyzed for drug concentration. The collected data are shown in the table below. Use these data are shown in the table below. Use these data with the trapezoidal rule shown in the data with the trapezoidal rule shown in the related equations to calculate each AUC related equations to calculate each AUC segment including the last segment. segment including the last segment.

Time (hr) Cp (mg/L) Δ (AUC mg.hr/L) AUC (mg.hr/L)

 0 

     

 0.5 

5.42    

 1 

4.61    

 2 

3.28    

 4 

1.28    

 6 

0.65    

 8 

0.32    

 10 

0.14    

 ∞ 

     

SolutionSolution::

1.1. First we plot the data with ln CFirst we plot the data with ln Cpp against time against time

2.2. Extrapolate to y axis and find CExtrapolate to y axis and find Cppoo

3.3. From the slope find KFrom the slope find Kelel

4.4. Start calculating Start calculating Δ(AUC mg.hr/L) segments using segments using the equationthe equation

For the first segment and then go on for the other ones For the first segment and then go on for the other ones using the same equationusing the same equation

4. Find4. Find

5. Find AUC(0-∞) 5. Find AUC(0-∞)

The following table summarizes the results:The following table summarizes the results: Dose = 250 mg Dose = 250 mg CCpopo = 6.65 mg/L = 6.65 mg/L

kkelel = 0.386 hr-1 = 0.386 hr-1 AUC(0-10 hr) = 17.39 mg.hr/L AUC(0-10 hr) = 17.39 mg.hr/L AUC(0-∞) = 17.75 mg.hr/L AUC(0-∞) = 17.75 mg.hr/L

Time (hr)Cp (mg/L)Δ (AUC mg.hr/L)AUC (mg.hr/L)

0 6.65   SegmentSummation 

0.5 5.42 3.02 3.02

14.612.51 5.52

23.283.95 9.47

41.284.56 14.03

60.651.93 15.96

80.320.97 16.93

100.140.46 17.39

∞ 0 0.3617.75

For the first segment we have: {(6.65+5.42)/2}*(0.5 – 0) = 3.02For the second segment we have: {(5.42+4.61)/2}*(1 – 0.5) = 2.51

CL = rate of elimination/plasma concentrationCL = rate of elimination/plasma concentration Rate of elimination = change of amount per Rate of elimination = change of amount per

time = d(amount)/dttime = d(amount)/dt Therefore CL = {d(amount)/dt}/CTherefore CL = {d(amount)/dt}/Cpp

Taking in consideration that:Taking in consideration that:

ExampleExample

What IV bolus dose is required to achieve a What IV bolus dose is required to achieve a plasma concentration of 2.4 µg/ml (2.4 mg/L) plasma concentration of 2.4 µg/ml (2.4 mg/L) at 6 hours after the dose is administered. The at 6 hours after the dose is administered. The elimination rate constant, kelimination rate constant, kelel is 0.17 hr is 0.17 hr-1-1) and ) and

the apparent volume of distribution, V, is 25 L the apparent volume of distribution, V, is 25 L

ExampleExample

If CIf Cp p afterafter 2 hours is 4.5 mg/liter and C2 hours is 4.5 mg/liter and Cp p afterafter 6 6

hours is 3.7 mg/liter, after a 400 mg IV bolus hours is 3.7 mg/liter, after a 400 mg IV bolus dose what are the values of kdose what are the values of kelel and V. and V.

mg/L

ExampleExample

What is the concentration of a drug 0, 2 and 4 What is the concentration of a drug 0, 2 and 4 hours after a dose of 500 mg. Known hours after a dose of 500 mg. Known pharmacokinetic parameters are apparent pharmacokinetic parameters are apparent volume of distribution, Vvolume of distribution, Vdd is 30 liter and the is 30 liter and the

elimination rate constant, kelimination rate constant, kelel is 0.2 hr is 0.2 hr-1-1

Make PredictionsMake Predictions

Once we have a model and parameter values we can use this Once we have a model and parameter values we can use this information to make predictions. For example we can information to make predictions. For example we can determine the dose required to achieve a certain drug determine the dose required to achieve a certain drug

concentrationconcentration..

Finding a dose necessary to achieve a Finding a dose necessary to achieve a certain Ccertain Cpp

Plasma drug concentration after multiple IV Plasma drug concentration after multiple IV dosesdoses

The anticipated plasma concentration is meant The anticipated plasma concentration is meant to be before the steady state is reached. The to be before the steady state is reached. The equation used for such a calculation is:equation used for such a calculation is:

The steady state is reached when the number of The steady state is reached when the number of doses exceeds 5 half lives, but is surely doses exceeds 5 half lives, but is surely attainable when n = ∞attainable when n = ∞

Example calculation of plasma drug Example calculation of plasma drug concentration after multiple IV dosesconcentration after multiple IV doses

CCpp = {100/14}{[(1- e = {100/14}{[(1- e-12*0.23*4-12*0.23*4)/(1-e)/(1-e-0.23*12-0.23*12)]e)]e-0.23*3-0.23*3))

CCpp = 3.8 mg/L = 3.8 mg/L

Steady state from first principlesSteady state from first principles

At steady state the rate of drug administration is At steady state the rate of drug administration is equal to the rate of drug elimination. equal to the rate of drug elimination. Mathematically the rate of drug administration Mathematically the rate of drug administration can be stated in terms of the dose (can be stated in terms of the dose (DD) and ) and dosing interval (dosing interval (). It is always important to ). It is always important to include the salt factor (include the salt factor (SS) and the ) and the bioavailability (bioavailability (FF). The rate of drug ). The rate of drug elimination will be the clearance of the plasma elimination will be the clearance of the plasma concentration at steady state:concentration at steady state:

For IV RouteFor IV Route

t = to

t’ = t

Css = Cp0 = Cpt/e-kel*t

However, now t = t’ – t , since to = t

Rearranging gives:

For Non-IV RoutesFor Non-IV Routes

From multiple doses to steady stateFrom multiple doses to steady state

We have the equation for multiple doses We have the equation for multiple doses where:where:

When n is infinity then the value eWhen n is infinity then the value e -nKel-nKel = 0, the = 0, the equation then becomes:equation then becomes:

Therefore, at steady stateTherefore, at steady state

The plasma concentration (The plasma concentration (CCpp) at any time () at any time (tt) )

within a dosing interval (within a dosing interval () at steady state is ) at steady state is represented by the equation:represented by the equation:

Example:Example: Calculate the concentration of drug in plasma Calculate the concentration of drug in plasma

2 hrs after the last dose of a series of doses 2 hrs after the last dose of a series of doses (6hrs interval and 100 mg each) that brought (6hrs interval and 100 mg each) that brought the patient to a steady state. Kthe patient to a steady state. Kel el = 0.3/hr and= 0.3/hr and

VVdd =5.6 L. =5.6 L.

Substitution gives:Substitution gives: CCp p = {100 (e= {100 (e-0.3*2-0.3*2) / 5.6(1 – e) / 5.6(1 – e-0.3*6-0.3*6)})}

= 54.88/4.67 = 11.74 mg/L= 54.88/4.67 = 11.74 mg/L

CCppmaxmax and C and Cpp

minmin

At steady state we have:At steady state we have:

Now, the maximum plasma concentration for Now, the maximum plasma concentration for each dose administration occurs at t = 0, while each dose administration occurs at t = 0, while the minimum plasma concentration at steady the minimum plasma concentration at steady state occurs at t = state occurs at t = , back to equation of C , back to equation of Cpt pt at at

steady state steady state

Applying the conditions for t = 0 and t = Applying the conditions for t = 0 and t = and and taking in consideration that etaking in consideration that e-Kelt-Kelt = 1 when t = = 1 when t = 0, and e0, and e-Kelt-Kelt = e = e-Kel-Kel when t = when t = , we get: , we get:

Immediately after many doses, after t = 0, we have:Immediately after many doses, after t = 0, we have:

Immediately before many doses, after t = Immediately before many doses, after t = , we have: , we have:

Therefore,Therefore,

An example may be helpful: tAn example may be helpful: t1/21/2 = 4 hr; IV dose 100 = 4 hr; IV dose 100

mg every 6 hours; V = 10 liter mg every 6 hours; V = 10 liter

Dose/Vd = Cpo1 = 100/10 = 10 mg/L

Where Where Cpo1 is the plasma concentration at zero time

after the first dose

kkelel = 0.693/4 = 0.17 hr = 0.693/4 = 0.17 hr-1-1 R = eR = e-kel -kel ** = e = e-0.17 x 6-0.17 x 6 = 0.35 = 0.35 therefore therefore

Therefore the plasma concentration will fluctuate Therefore the plasma concentration will fluctuate between 15.5 and 5.4 mg/liter during each dosing between 15.5 and 5.4 mg/liter during each dosing interval when the plateau is reached. interval when the plateau is reached.