Multiple IV Dosing Pharmacokinetic Research, Bioequivalence Studies, Drug – Drug Interactions...

Multiple IV DosingMultiple IV DosingPharmacokinetic

Research,Bioequivalence

Studies,Drug – Drug Interactions

Generally all evaluated

through single dose studies.

But patients receive and most drug therapy is given in a multiple dose regimen

Multiple doses generally result in accumulation of drug,… but how much, how fast ?

Problem:A patient with renal dysfunction (CrCl~ 20 mL/min) is given tobramycin IV. The half-life of tobramycin in this patient is 8 hours. Although tobramycin is normallyinfused over at least 30 minutes, we will administer 100 mg of tobramycin to our patient by IV bolus every 8 hours. Calculate the amount of tobramycin in the body after each dose for the first 10 doses.

Dose Time Amount Amount Increase(n) (hr) in the Eliminated in Body

Body During Store (mg) Dose Int (mg)

1. Zero 0.000.01 100.008.00

Tobramycin, 100 mg administered by IV bolus every 8 hours. Half-life of of tobramycin in this patient is 8 hrs. Calculate the amount in the body after each of the first 10 doses.



1. Zero 0.000.01 100.008.00

2. 8.0116.0

3. 16.0124.00




1. Zero 0.000.01 100.008.00 50.00 50.00 50.00

2. 8.01 150.0016.0 75.00 75.00 25.00

3. 16.01 175.0024.00 87.50 87.50 12.50

Increaseis difference

between dose and what was eliminated



Body During Store (mg) Dose Int. (mg)

4. 24.01 187.5032.00 93.75 93.75 6.25

9. 64.01 199.6172.0 99.80 99.80 0.20

nth. 200.00100.00 100.00 zero



Body During Store (mg) Dose Int. (mg)

nth. 200.00100.00 100.00 zero

nth + 1 200.00100.00 100.00 zero

Eventually, at some point accumulation stopsThis is called steady-state (ss)


Dose Amount Amount Increase(n) in the Body Eliminated in Body

End of Interval During Store(mg) Dose Int. (mg)

0 01 50 50 502 75 75 253 87.5 87.5 12.54 93.75 93.75 6.259 99.80 99.8 0.20nth. 100.00 100.00 = Dose 0.00

At steady state the amount eliminated during an interval isequal to the dose, or with oral dosing the amount of the dose absorbed.

IN = OUT




0 01 50 50 502 75 75 253 87.5 87.5 12.54 93.75 93.75 6.259 99.80 99.8 0.20nth. 100.00 100.00 = Dose 0.00

At steady state the amount eliminated during an interval isequal to the dose, or with oral dosing the amount of the dose absorbed.

IN = OUT

How did SS occur?How did SS occur?

During each During each dosing interval dosing interval (8 hour period) (8 hour period) the body appearsthe body appears

to eliminate to eliminate (metabolise, clear)(metabolise, clear)

more drug…more drug…

How is this How is this possible?possible?




0 01 50 50 502 75 75 253 87.5 87.5 12.54 93.75 93.75 6.259 99.80 99.8 0.20nth. 100.00 100.00 = Dose 0.00

When did SS occur?When did SS occur?Increase in body stores continued with each dose, in fact with each dose Increase in body stores continued with each dose, in fact with each dose

body stores increased by half as much as the previous dose…… why?why?




0 01 50 50 502 75 75 253 87.5 87.5 12.54 93.75 93.75 6.259 99.80 99.8 0.20nth. 100.00 100.00 = Dose 0.00

When did SS occur?When did SS occur?Steady State is considered to have been achieved when conc. (amount) Steady State is considered to have been achieved when conc. (amount)

are within 10%. For this patient this occurs between the 3are within 10%. For this patient this occurs between the 3rdrd and 4 and 4thth dose. dose.

Dose Time Amount (n) (hr) in the Amount in the Body

Body Following a First and SINGLE Dose (mg)

1. Zero 0.000.01 100.008.00 50.00 16.0 25.0024.0 12.5032.0 6.2540.0 3.1248.0 1.5654.0 0.78

Can we develop an equation Can we develop an equation which will predict these amounts / concentrations ?which will predict these amounts / concentrations ?

Dose Time Amount (n) (hr) in the Equation for Amount in the Body

Body after the First Dose (mg)

1. Zero 0.000.01 100.00 Amount = Dose8.00 50.00 Amount = Dose1 * e-Kt = Dose1 * e-Kτ


τ (tau) represents dosing interval

Dose Time Amount(n) (hr) in the Equation for Amount in the Body

Body after the Second Dose (mg)

1. 8.00 50.00 Amount = Dose1 * e-Kt = Dose1 * e-Kτ

2. 8.01 150.00 Amount = Dose2 + Dose1 * e-Kτ

= 8hr single dose +0 hr single dose = 50 + 100 = 150 mg =

What is the concentration immediately following What is the concentration immediately following The second dose (8.01 hrs)?The second dose (8.01 hrs)?

Time Am’t in (hr) the Body (mg)0.01 100.008.00 50.0016.0 25.0024.0 12.5032.0 6.253 = 2 + 1


Body after the Second Dose (mg)

1. 8.00 50.00 Amount = Dose1 * e-Kt = Dose1 * e-Kτ


16.0 75.00 Amount = [Dose2 + Dose1 * e-Kτ] * e-Kτ

= Dose2 e-Kτ + Dose1 * e-2Kτ

= 8hr single dose +16hr single dose = 50.0 + 25.0 = 75 mg =


3 = 2 + 1

Time Am’t in (hr) the Body (mg)0.01 100.008.00 50.0016.0 25.0024.0 12.50


Body after the Third Dose (mg)

2. 16.0 75.00 Amount = Dose2 e-Kτ + Dose1 * e-2Kτ

3. 16.01 175.00 Amount = Dose3 +Dose2 e-Kτ + Dose1 * e-2Kτ

= Dose (1 + e-Kτ + e-2Kτ) = 0 hr + 8hr + 16 hr = 100 + 50 + 25 = 175 mg =


6 = 1 + 2 + 3

Time Am’t in (hr) the Body (mg)0.01 100.008.00 50.0016.0 25.0024.0 12.5032.0 6.25


Body following the Third Dose (mg)


= Dose (1 + e-Kτ + e-2Kτ)24.0 87.50 Amount = Dose3 e-Kτ +Dose2 e-2Kτ +Dose1e-3Kτ

= Dose (e-Kτ + e-2Kτ +e-3Kτ) = 8 hr + 16 hr + 24 hr = 50 + 25 = 12.5 = 87.5 mg =


6 = 1 + 2 + 3



Body Second & Third Doses (mg)


16.0 75.00 Amount = [Dose2 + Dose1 * e-Kτ] * e-Kτ

= Dose2 e-Kτ + Dose1 * e-2Kτ


= Dose (1 + e-Kτ + e-2Kτ)

24.0 87.50 Amount = Dose3 e-Kτ +Dose2 e-2Kτ +Dose1e-3Kτ

= Dose (e-Kτ + e-2Kτ +e-3Kτ)



Body Third Dose (mg)


= Dose (1 + e-Kτ + e-2Kτ)24.0 87.50 Amount = Dose3 e-Kτ +Dose2 e-2Kτ +Dose1e-3Kτ



This is a geometric progressionfor both the initial concentration

following each dose (Cmax) and the end-of interval concentration (Cmin).

But what does this show us?


Body (mg)

3. 24.0 87.50 Amount = Dose3 e-Kτ +Dose2 e-2Kτ +Dose1e-3Kτ



But what does this show us?First, that the concentration in a

multiple dose profile can be obtained by adding the conc.from various times following

a single dose.For example: [ ]8hr after dose 3 = [3]8hr + [2]16hr + [1]24hr


Body (mg)

3. 24.0 87.50 Amount = Dose3 e-Kτ +Dose2 e-2Kτ +Dose1e-3Kτ



For example: [ ]8hr after dose 3 = [3]8hr + [2]16hr + [1]24hr = 50 + 25 + 12.5 = 87.5 mg



Second that we have a Geometric ProgressionSolution to the Geometric Progression

These equations will calculate the concentration for the Cmax and Cmin of any dose (n) in a multiple dosing regimen.

Cmax after doses where SS has not been achieved – blue arrows

SinceCminn = Cmaxn e-Kτ

Can we develop an equation which will Can we develop an equation which will predict concentrations at any time?predict concentrations at any time?

Can we calculate the concentration at any time

after any dose?

Since

Cminn = Cmaxn e-Kτ

ThenCt = Cmaxn e-Kt

and where n = 6 to concentrations corresponding to the red arrows above.


Therefore the more general equation which will calculate

concentrations at any time t afterany dose (1 through ) is:

and when n is large the term

e-nKτ approaches zero and then:

Predicts for concentrations at steady state.

Ct = e -0.086*4

Ct = 100*[(1 - 0.1250)/(1-0.50)]*0.7

= 100 * [(0.875)/(.50)] * 0.707 = 100 * 1.75 * 0.707 = 175 * 0.707 = 123.7 mg/L


Using our equation:

1000 (1- e-3*0.0866* 8)

10 (1- e-0.0866* 8)

What is 175 ..??

Example: Calculate the concentration 4 hrs after dose 3 in a patient.

Dose = 1000 mg, V=10 L; τ = 8 hr T½ = 8 hr (K = 0.0866hr-1)

With Multiple Doses Drugs Accumulate.With Multiple Doses Drugs Accumulate.Can we predict the degree of accumulation quickly, Can we predict the degree of accumulation quickly, without an equation?without an equation?

Dose Amount Amount in the Body in the BodyStart of Interval End of Interval

(n) (mg) (mg)

1 100 50 2 150 75 3 175 87.5 4 187.5 93.75 9 193.75 99.80nth. 200.00 100.00

We have seen that in a patient given 1000 mg Q8H (τ)that when the T½ is 8 hr

(K = 0.0866hr-1) and V = 10L, the degree of

accumulation is a factor of 2.

Dose Q2H Q4H Q6H Q8H Q12H Q24H

1st. 10.00 10.00 10.00 10.00 10.00 10.00 2nd. 3rd.@ 24.@ 48@ 96

Can we predict how much accumulation WILL occur?Can we predict how much accumulation WILL occur?

Consider giving 100 mg of tobramycin IV, at different dosing intervals, when half-life is 8 hours and Volume is 10 L.

Conditions : Dose =100; V = 10 L half-life = 8 hr

CMAX

Using MD 1C-IV Bolus Excel Sheet determine degree of accumulation for Q2, Q4, Q8, Q12, Q16 and Q24 hr regimens.

Conditions : Dose = 200; V = 20 L half-life = 8 hr


Accumulation depends on the rate (frequency) at which we give the drug (dosing interval).



CMAX


1st 10.00 10.00 10.00 10.00 10.00 10.00 2nd 18.41 17.07 15.95 15.00 13.54 3rd 25.48 22.07 19.48 17.50 14.79 @ 24 56.25 31.13 22.84 18.75 14.79 11.25@ 48 61.17 33.77 24.44 19.85 15.39 11.41@ 96 62.83 34.13 24.66 20.00 15.47 11.43


1 10.00 10.00 10.00 10.00 10.00 10.002. 18.41 17.07 15.95 15.00 13.54 3. 25.48 22.07 19.48 17.50 14.79 @ 24 56.25 31.13 22.84 18.75 14.79 11.25@ 48 61.17 33.77 24.44 19.85 15.39 11.41@ 96 62.83 34.13 24.66 20.00 15.47 11.43


Notice that at Q8H,

degree of accumulation is a factor of 2

Accumulation depends on the rate (frequency) at which we give the drug (dosing interval).


Initial concentration is : Dose/ Vd = 100/ 10 = 10 mg/L


1st. 10.00 10.00 10.00 10.00 10.00 10.00 2nd. 3rd.@ 24.@ 48@ 96

Can we predict how much accumulation WILL occur?Can we predict how much accumulation WILL occur?Repeat the process, but this time use a half-life of 6 hours

(patient with a different CrCl). Administer 100 mg of tobramycin IV, at different dosing intervals, when half-life is 6 hours and Volume is 10 L.


CMAX

Using MD 1C-IV Bolus Excel Sheet determine degree of accumulation for Q2, Q4, Q8, Q12, Q16 and Q24 hr regimens.

Conditions : Dose = 200; V = 20 L half-life = 6 hr


1st. 10.00 10.00 10.00 10.00 10.00 10.00 2nd. 17.94 16.30 15.00 13.97 12.50 10.63 3rd. 24.24 20.37 17.50 15.54 13.13 10.66@ 24. 46.08 25.96 19.38 16.17 13.13 10.66@ 48 48.36 26.96 19.96 16.56 13.32 10.66@ 96 48.48 27.03 20.00 16.58 13.33 10.67




When is the degree of accumulation of 2 observed now?


1st. 10.00 10.00 10.00 10.00 10.00 10.00 2nd. 17.94 16.30 15.00 13.97 12.50 10.63 3rd. 24.24 20.37 17.50 15.54 13.13 10.66@ 24. 46.08 25.96 19.38 16.17 13.13 10.66@ 48 48.36 26.96 19.96 16.56 13.32 10.66@ 96 48.48 27.03 20.00 16.58 13.33 10.67




CMAX

2 x


1 10.00 10.00 10.00 10.00 10.00 10.002. 18.41 17.07 15.95 15.00 13.54 3. 25.48 22.07 19.48 17.50 14.79 @ 24 56.25 31.13 22.84 18.75 14.79 11.25@ 48 61.17 33.77 24.44 19.85 15.39 11.41@ 96 62.83 34.13 24.66 20.00 15.47 11.43


Accumulation depends on the rate at which we give the drug (dosing interval τ) and the rate at which the drug is eliminated (K)

When the Dosing Interval and half-life are equal, accumulation is 2xWhen the Dosing Interval and half-life are equal, accumulation is 2x

Half-life = 8 hours


1 10.00 10.00 10.00 10.00 10.00 10.002. 18.41 17.07 15.95 15.00 13.54 3. 25.48 22.07 19.48 17.50 14.79 @ 24 56.25 31.13 22.84 18.75 14.79 11.25@ 48 61.17 33.77 24.44 19.85 15.39 11.41@ 96 62.83 34.13 24.66 20.00 15.47 11.43



When Dosing interval is greater than the half-life, more drug is eliminated during a dosing interval

and LESS accumulation occursHalf-life = 8 hours


1 10.00 10.00 10.00 10.00 10.00 10.002. 18.41 17.07 15.95 15.00 13.54 3. 25.48 22.07 19.48 17.50 14.79 @ 24 56.25 31.13 22.84 18.75 14.79 11.25@ 48 61.17 33.77 24.44 19.85 15.39 11.41@ 96 62.83 34.13 24.66 20.00 15.47 11.43



When Dosing Interval is shorter than the half-life, less drug is eliminated during a dosing interval (because it is short)

and MORE accumulation occursHalf-life = 8 hours


1 10.00 10.00 10.00 10.00 10.00 10.002. 18.41 17.07 15.95 15.00 13.54 3. 25.48 22.07 19.48 17.50 14.79 @ 24 56.25 31.13 22.84 18.75 14.79 11.25@ 48 61.17 33.77 24.44 19.85 15.39 11.41@ 96 62.83 34.13 24.66 20.00 15.47 11.43

The ratio of the Peak Concentration at Steady State The ratio of the Peak Concentration at Steady State and the Peak Concentration after the first doseand the Peak Concentration after the first dose

is a direct measure of the degree of accumulation.is a direct measure of the degree of accumulation.

Cmax Cmax ssss

----------- = ----------- = MMaximum aximum AAccumulation ccumulation FFactoractor

Cmax Cmax 11

MAF = 6.28 3.41 2.47 2.00 1.55 1.14



Cmax Cmax ssss 11

---------------- = MAF = -----------= MAF = -----------Cmax Cmax 11 1 - e1 - e

-K-K

MAF rules of thumb if = T½… then MAF = 2

if > T½… then MAF < 2 if < T½… then MAF > 2

CmaxCmaxssss = Cmax = Cmax11 x MAF x MAF

CminCminssss = Cmin = Cmin11 x MAF x MAF



Cmax Cmax ssss 11

---------------- = MAF = -----------= MAF = -----------Cmax Cmax 11 1 - e1 - e

-K-K

MAF considers only … the dosing interval

K … the half-life

Building the Multiple Dose EquationBuilding the Multiple Dose Equation

11

MAF = ----------- … MAF = ----------- … predicts accumulationpredicts accumulation

1 - e1 - e -K-K

Dose

VCt = ---------- e

-Kt 1x -------------

1 – e -K

First Dose Conc. Accumulation Steady State Equation

and the Steady State equation predicts ONLYONLY steady state concentrations

We need the other piece to calculate concentrations following ANYANY dose


Dose

VC = ---------- e -Kt

Provides for all concentrations from first dose until steady state.

1- e -nK

x -------------

1 – e -K

In this equation what happens to the term

1- e -nk when n (number of doses) is large?


And the term

1- e -nK

-------------

1 – e -K

Can convert the single dose IV equation to a multiple dose equation

Dose Amount Amount Increase Cumulative(n) in the Body Eliminated in Body Increase

End of Interval During Storein Body Store(mg) Dose Int. (mg) (mg)

0 01 50 50 50 502 75 75 25 753 87.5 87.5 12.5 87.54 93.75 93.75 6.25 93.259 99.80 99.8 0.20nth. 100.00 100.00 = Dose 0.00

How long does it take to get to steady state?When did SS occur?When did SS occur?

Steady State is considered to have been achieved when the Steady State is considered to have been achieved when the concentration or amount in the body is within 10% of the totalconcentration or amount in the body is within 10% of the totalamount of drug in the body. For this patient total was 100mg. amount of drug in the body. For this patient total was 100mg.

90% achievedBetween 3rd

& 4thdose

How long does it take to get to steady state?

T½ = 8 hrT½ = 8 hrDose Time Q2H 1. 0 10.00 2. 2 18.41 4. 6 31.43 10. 22 40.6411. 24 56.2512. 26 57.31 3.2T½49. 96 62.83

When did SS occur?When did SS occur?Steady State is considered to have been achieved when the Steady State is considered to have been achieved when the

Concentration is within 10% of the true steady state concentration.Concentration is within 10% of the true steady state concentration.Compare patients with TCompare patients with T½ of 6 and 8 hrs for time to reach SS.½ of 6 and 8 hrs for time to reach SS.

90% of 96 hr [ ] = 43.63 56.55

T½ = 6 hrT½ = 6 hrDose Time Q2H 1. 0 10.00 2. 2 17.94 4. 6 29.21 10. 14 40.8411. 16 42.4212. 18 43.67 3T½13. 20 44.6649. 96 48.48



Concentration is within 10% of the true steady state concentration.Concentration is within 10% of the true steady state concentration.90% of true steady state will occur at ~ 3.3 half-lives.90% of true steady state will occur at ~ 3.3 half-lives.

Number Cumulative Of Increase T½ Body Store

(mg) 1 50 2 75 3 87.5 4 93.25 5 96.875 6 98.4375SS 100

Sometimes, SS is not considered to have been achieved until concentrations exceed

95% of the true steady state concentrations.This occurs between 4 and 5 half-lives.

This gives rise to the statement that steady state is achieved in 3-5 half-lives.

Each T½ reduces the gap between current concentrations and SS by half.

Dosing interval (τ) does not affect time to SS





(mg) 1 50 2 75 3 87.5 4 93.25 5 96.875 6 98.4375SS 100

When is 90% of true (eventual)Steady State Achieved?

e-Kt determines proportion lost





(mg) 1 50 2 75 3 87.5 4 93.25 5 96.875 6 98.4375SS 100


e-Kt determines proportion loste-K# determines proportion lost for a set number (#) of half-lives

Example:If K = 0.693 and # = 2 T½

= e (-0.693 x 2)

= 0.25





(mg) 1 50 2 75 3 87.5 4 93.25 5 96.875 6 98.4375SS 100


e-K# determines proportion lost … then 1 - e-K# will determine

proportion of steady-state achieved.Example:

If K = 0.693 and # = 2 T½= 1 -e (-0.693 x 2)

= 0.75Or expressed as a %75% of SS after 2 T½





(mg) 1 50 2 75 3 87.5 4 93.25 5 96.875 6 98.4375SS 100


e-K# determines proportion lost

and … 1 - e (-0.693 x #)

determines proportion of steady-state achieved.

and … 100 x (1 - e (-0.693 x #) )determines percent of steady-state achieved,

where # is the number of half-lives.% SS = 100 x (1 - e (-0.693 x #) )





(mg) 1 50 2 75 3 87.5 4 93.25 5 96.875 6 98.4375SS 100


% SS = 100 x (1 - e (-0.693 x #) )

Therefore, 90% of true SS is achieved…90 = 100 x (1 - e (-0.693 x #) )

0.9 = 1 - e (-0.693 x #)

e (-0.693 x #) = 0.1ln(e (-0.693 x #) = ln(0.1)-0.693 x # = -2.30259

# = 3.322





(mg) 1 50 2 75 3 87.5 4 93.25 5 96.875 6 98.4375SS 100


After 3.322 half-lives… frequently stated as 3.3 half lives

Or in hours: Time to SS (hr) = 3.322 x T½

If the T½ for your drug is 10 hoursit will take (10 * 3.322) hours

… or 33.22 hrs to achieve 90% of SS.





(mg) 1 50 2 75 3 87.5 4 93.25 5 96.875 6 98.4375SS 100


After how many half-lives (#) is 95% of Steady State Achieved?

ln(e (-0.693 x #) = ln(0.05)-0.693 x # = -2.9957

# = 4.322

(95% of SS achieved after 4.322 half-lives)or in hours: Time to SS (hr) = 4.322 x T½





(mg) 1 50 2 75 3 87.5 4 93.25 5 96.875 6 98.4375SS 100

When is 90% of SS Achieved?

Shown slightly differently, Recall that:the time to eliminate 50% of body stores is

T½ = 0.693/KAnd 0.693147 is ln(0.5) (negative sign omitted)

Therefore, the time to reduce the amountof drug in the body to 10% would be

Time in hours = ln(0.1)/K = 2.302/K or since K = 0.693/T½, then

Time (hr) = (ln(0.1)*T½) / 0.693 or = (3.322)*(T½)

In practice, some multiple dose regimens are initiated with

a loading dose. Why do we use loading doses?

To achieve steady state faster..?Or to achieve a therapeutic concentration sooner?

Is a loading dose more useful for a drug with a longer or shorter half-life?

Multiple IV Doses & Multiple IV Doses & Concentrations at Steady StateConcentrations at Steady State

Therapeutic Drug Monitoring

It is known that for many drugs, optimal effect is achieved with a certain minimum concentration. It might also be known that toxicity is more likely to occur with concentrations that exceed a certain concentration. Therefore, it becomes the intent of a multiple dose regimen to have peak and trough concentrations fall within this therapeutic range.

Half-life is 8 hours, drug is administered every 8 hours.Dose is 400 mg. VD = 40 L.

What dose will achieve a SS peak concentration of 30 mg/L?




What dose will achieve a steady state concentration of 30 mg/L?

Dose of 400 mg produces a peak of 400 mg/ 40 L = 10 mg/LSince half-life and Dosing interval are equal … MAF = ???And the dosing regimen of 400 mg q8h will produce SS [ ] of ? Cpmax ss = Cp1 * MAF

= 10 mg/L * 2= 20 mg/L




What dose will achieve a steady state concentration of 30 mg/L?

A dosing regimen of 400 mg q8h will produce SS [ ] of ? Cpmax ss = Cp1 * MAF

= 10 mg/L * 2= 20 mg/L

To acheive a SS peak of 30 mg/L we will need to administer 600 mg every 8 hours in this patient.




If 600 mg is administered IV every 8 hours, what will the steady-state trough (Cmin) concentrations be?


Given Cmaxss = 30 mg/L

Since dosing interval and half-life are equal … half the concentration is eliminated during a dosing interval.

Cpmaxss = 30 mg/L * 0.5

= 15 mg/LAt SS [ ] fluctuate between 15 mg/L and 30 mg/L

Therapeutic Drug Monitoring … tougher Question

Following the first 80 mg dose of an aminoglycoside administeredby IV bolus*, the concentrations at 2 & 6 hours were observed to be 4 and 2 mg/L. At steady state the desired peak is 8 mg/L andthe trough is 1 mg/L. What dosing regimen (dose and interval) will yield these desired concentrations?


What do you need to know?What predicts degree of accumulationWhat is the Half-life?What should the dosing interval be?

* Aminoglycosides normally administered by IV infusion over 30-60 min



MAF is based on τ and T½Time [mg/L] 2 4 6 2T½ = ??? K = 0.1733 hr-1


Following the first 80 mg dose of an aminoglycosisde administeredby IV bolus, the concentrations at 2 & 6 hours were observed to be4 and 2 mg/L. At steady state the desired peak is 8 mg/L and the trough is 1 mg/L. What dosing regimen (dose and interval) will yield these desired concentrations?



MAF is based on τ and T½Dosing interval based on Peak 8 mg/LTrough 1 mg/L3 T½’s (8,4,2,1) = 12 hr.

MAF = 1.143




Peak would be the time zero concentration … What was the concentration at time zero following the first dose?

K= 0.17328 hr-1 C2 = 4; C0 = C2e(-Kt)

= 4 e(-0.17328 x 2)= 5.66 mg/L




What would the current dose (80 mg) yield for a Steady State Peak? CSS-peak = 5.66 mg/L x MAF

= 5.66 x 1.1428= 6.46 mg/L




What dose is necessary to produce a peak of 8 mg/L (simple ratio)? Dose = (8.0 / 6.46) x (80 mg)

= 1.237 x 80= 98.99 mg

This dose is unreasonable and so should be rounded up to 100 mg.




Best Regimen is 100 mg administered every 12 hours. Calculate the SS peak. (use either complete equation or peak1 x MAF

CmaxSS = (dose/volume) x MAF

= (100 mg / 14.14L) x 1.143= 8.08 mg/L




Best Regimen is 100 mg administered every 12 hours.

Calculate the SS Trough (use CmaxSS x e(-Kτ)

CmaxSS = 8.08 e(-Kτ)

= 8.08 e (-0.1733 x 12)

= 1.01 mg/L



Calculation of Parameters at SS – IV dosingCalculation of Parameters at SS – IV dosing

Dose Dose Dose Dose 1 2 6 15

Time [ ] Time [ ] Time [ ] Time [ ] (hr) (mg/L) hr) (mg/L) (hr) (mg/L) (hr) (mg/L)

0 10.00 8 15.00 40 19.69 112 20.00 2 8.40 10 12.62 42 16.56 114 16.82 4 7.07 12 10.61 44 13.92 116 14.15 6 5.95 14 8.92 46 11.71 118 11.90 8 5.00 16 7.50 48 9.85 120 10.00

Evaluate the following concentrations determined in an patient following the 1st, 2nd, 6th and 15th 100mg doses of a drug given Q8H.

Steady State?? First Dose Half way 96.9% ~ True SS 1 T½ 5 T½ 14 T½

Calculation of Parameters at SS – IV dosingCalculation of Parameters at SS – IV dosing



0 10.00 8 15.00 40 19.69 112 20.00 2 8.40 10 12.62 42 16.56 114 16.82 4 7.07 12 10.61 44 13.92 116 14.15 6 5.95 14 8.92 46 11.71 118 11.90 8 5.00 16 7.50 48 9.85 120 10.00


Does the half-life change?Does the volume change?Does clearance change?

Does AUC change?

Calculation of Parameters at SSCalculation of Parameters at SS– IV dosing– IV dosing



0 10.00 8 15.00 40 19.69 112 20.00 2 8.40 10 12.62 42 16.56 114 16.82 4 7.07 12 10.61 44 13.92 116 14.15 6 5.95 14 8.92 46 11.71 118 11.90 8 5.00 16 7.50 48 9.85 120 10.00



Does AUC change?

Protein binding.

Calculation of Parameters at SS Calculation of Parameters at SS – IV dosing– IV dosing



0 10.00 8 15.00 40 19.69 112 20.00 2 8.40 10 12.62 42 16.56 114 16.82 4 7.07 12 10.61 44 13.92 116 14.15 6 5.95 14 8.92 46 11.71 118 11.90 8 5.00 16 7.50 48 9.85 120 10.00



Does AUC change?Dose / AUC or K*V




0 10.00 8 15.00 40 19.69 112 20.00 2 8.40 10 12.62 42 16.56 114 16.82 4 7.07 12 10.61 44 13.92 116 14.15 6 5.95 14 8.92 46 11.71 118 11.90 8 5.00 16 7.50 48 9.85 120 10.00



Does AUC change? Let’s calculate AUC




0 10.00 8 15.00 40 19.69 112 20.00 2 8.40 10 12.62 42 16.56 114 16.82 4 7.07 12 10.61 44 13.92 116 14.15 6 5.95 14 8.92 46 11.71 118 11.90 8 5.00 16 7.50 48 9.85 120 10.00


AUC (mg*hr/L)AUC(0-t) 57.74 86.61 113.63 115.47AUC(0-) 115.47 173.21 227.37 230.95

Does this make sense?

In = Out @ SS



1st Dose 2nd Dose 6th Dose 15th DoseHalf way 96.9% ~ True SS

1 T½ 5 T½ 14 T½AUC (mg*hr/L)AUC(0-t) 57.74 86.61 113.63 115.47AUC(0-) 115.47 173.21 227.37 230.95

Elimination of 100 mg following the first dose produced an AUC(0-) of 115.47 mg*hr/L.

At SS, a dose (100 mg) is eliminated during a dosing interval, in this case 8 hours. At steady state we are eliminating a dose.

Therefore, the AUC(0-τ) at SS should be equivalent to the

AUC(0-) following the first dose.


Bioequivalence

1st Dose 2nd Dose 6th Dose 15th DoseHalf way 96.9% ~ True SS

1 T½ 5 T½ 14 T½AUC (mg*hr/L)AUC(0-t) 57.74 86.61 113.63 115.47AUC(0-) 115.47 173.21 227.37 230.95

Just as AUC(0-) following the 1st dose can be use to estimate bioequivalence so can AUC(0-τ), as long as you are at steady state.

Steady state will require at least 3 T½ of dosing, and the greater the time allowed to achieve steady, the smaller the difference

between AUC1(0-) and AUCSS

(0-τ).


No change in other parameters

We have seen that:1. [ ] following single doses can be Summed to create MD profile

(Additivity of [ ])2. At Steady State In (Dose) = Out (elimination)3. MAF, predicts the degree of accumulation (K and τ)4. (1/1-e(-Kτ)), converts a single dose equation to multiple dose

5. (1-e(-nKτ))/(1-e(-Kτ)) allows Calculation of [ ] at any time6. Time to Reach steady state determine by K (Cl/V) -3.3 T½7. AUC(0-τ) over a dosing interval @ SS = AUC(0-) 1st dose.

8. Just as AUC(0-) 1st dose can be use to estimate bioequiavalence so can AUC(0-τ), as long as you are at steady state (min3 T½ dosing) .

9. Using MAF we can design an IV dosing regimen that will achieve desired peak and trough concentrations at SS.

10. At Steady state all other parameters remain unchanged (ClR,ClH, proportion metabolised.)

Multiple Dosing Summary … Multiple Dosing Summary … so far so far – IV dosing– IV dosing

Multiple IV Dosing Pharmacokinetic Research, Bioequivalence Studies, Drug – Drug Interactions...

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