1-Compartment Oral Dosing 400 mg of moxifloxacin is administered orally to Mr BB, a 68 yr old male...

1-Compartment Oral Dosing400 mg of moxifloxacin is administered orally to Mr BB, a 68 yr old male who weighs 75 kg. Blood samples were drawn following the dose and the plasma concentration determined. It isknown that about 20% of a moxifloxacin dose is excreted in theurine unchanged. A further 20% is excreted unchanged in the bileand the rest is metabolised to either M1 (sulpho) or M2 (acyl-glucuronide)

Questions to Consider• What model & equation best describes profile?• Calculate Cmax, Tmax, AUC, K, T½, V &Cl• Write an equation that will calculate

concentrations at anytime after the 1st dose.4. Calculate the absorption rate – ka.5. Comparison of Observed and Estimated [ ].


Exam Questions and Problem Sets will ask practical,

application questions

Recommend that you take this data and attempt

to reproduce the calculations on your own.

Graph Patient DataGraph Patient Data

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Using semi-log paper, or Excelgraph the data followingoral administration of

400 mg of moxifloxacin

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Graph Patient DataGraph Patient Data What model best describes this profile?

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Terminal elimination phase is log-linear…

1 Compartment Model with first order absorption (ka)and first order elimination (K)


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It is known that about 20% of a moxifloxacin dose is excreted in theurine unchanged. A further 20% is excreted unchanged in the bileand the rest is metabolised to either M1 (sulpho) or M2 (acyl-glucuronide)


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kH = kM1 + kM2

KNR = 20% excreted into bile as unchanged drug.


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K = ke + kH + kNR




What do you calculate first?Cmax, Tmax, AUC, K, T½, V or Cl




What do you calculate first?Cmax, Tmax, ? Does the maximum

concentration actually occur at 2 hours (Tmax)? Cmax and Tmax are

sampling time (data) dependant.

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1-Compartment Oral Dosing

Does the maximum concentration actually occur at 2 hours?Cmax and Tmax are sampling time (data) dependant and observed.

Could the true maximum concentration occur at 2.5 hr??




What do you calculate first?AUC, K and T½ can be easily calculated

from available data. K from slope of log [ ] vs. time or Excel. AUC, trapezoidal rule.


Since the concentrations between 4 and 24 hr appear log-linear, ANY pair of points in the terminal phase can be used

to calculate K and half-life, AFTER CONVERSION to LOG. In Excel this would likely result in K being calculated by

log-linear regression (SLOPE) of all points (4, 8, 12 and 24 hr).

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Using the concentrations at 4 and 24 hours calculate K(x1 = 4 hr and x2 = 24 hr)(y1 = log (C4) = 0.248)(y2 = log (C24) = -0.357)Slope = (y2 – y1) / (x2 – x1)

= (-0.357 – 0.248)/(24-4)= (- 0.605) / (20)= -0.0302

- K = slope x 2.303= - 0.0302 x 2.303

K = 0.06961 hr-1

T½ = 0.693 / K= 0.693 / 0.06961= 9.955 hours


Slope = (y2 – y1) / (x2 – x1)= (-0.357 – 0.248)/(24-4)= (- 0.605) / (20)= -0.0302

- K = slope x 2.303= - 0.0302 x 2.303

K = 0.06961 hr-1

T½ = 0.693 / K= 0.693 / 0.06961= 9.955 hours


Selection of the points from the terminal phase

for calculation of K.

Accurate calculation of K (T½) and all other parameters which

depend on K (AUC, Cl, etc) demand that the points used

to calculate K be found in the terminal phase

after absorption is complete.

When is absorption complete?

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If absorption occurs as a first order process

(constant percentage per unit time),

When does absorption stop?


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If we deal with ka as we would K,

Assume ka = 0.693, then T½ka = 1 hour.

This means that followinga 400 mg dose,

half of the dose (200 mg) will be absorbed in the first hour

And half of the remainingdose (100 mg) in the second.

If absorption occurs as a first order process

(constant percentage per unit time), When does

absorption stop?


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(mg/

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can be estimated as:

Time Amount (hr) (mg) 1 200 2 100 3 50 4 25 5 12.5 6 6.25

Peak

25 mg/hr

100 mg/hr

6.25 mg/hr


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Amount absorbed during each hour following the dose

can be estimated as:

Time Amount (hr) (mg) 1 200 2 100 3 50 4 25 5 12.5 6 6.25

Peak

so … when does absorption stop for this formulation?(i) At the Tmax – 2.6 hrs(ii) Prior to the peak(iii) ~ 1 hr after the peak(iv) Technically – never(v) It depends

25 mg/hr

100 mg/hr

6.25 mg/hr


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(mg/

L)Selection of the points

from the terminal phase for calculation of K.

In Excel®, selection of points in the terminal phase can be completed with statistics.

Evaluate the correlation coefficient (r) & the number of points.

A change in r value signals the end of absorption, with confirmation

by “visual inspection”.


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Selection of the points from the terminal phase

for calculation of K.Notice that as you select more points

to be used in the calculation of K, the K value begins to shrink, the T½ begins to increase and the correlation coefficient (r)

becomes smaller.As you use more points in the absorption phase

Best choice?

4 points? or

5 points?


K0.069 hr-1

T½10.04 hr

5 points?

r-0.9999


AUC(LP-)

AUCtrapezoid

AUC(0-24hr)

AUC(0-)


AUC Analysis following Oral Absorption

At time zero, even immediately following ingestion of the dose,

the plasma concentration is 0 mg/L.

Back extrapolation to t=0 is not appropriate,

and should not be done It will not aid in

the calculation of volume.



AUC should be calculated by trapezoidal rule

between all concentrations.

The sum of these partial areas will equal AUC 0-24hr or AUC 0-LP

AUC(0-24hr)



AUC from the last measured concentration should be calculated by the pharmacokinetic method ([ ]LP/K). = (0.44 mg/L) / 0.069 hr-1

= 6.377 mg*hr/L= AUC LP-

AUC(0-24hr)

26.623 mg*hr/L



Total AUC (0-) is the sum = 26.623 + 6.377 mg*hr/L= 33.000 mg*hr/L

AUC(0-24hr)

26.623 mg*hr/L

AUC(24hr-)

6.377 mg*hr/L


AUC(LP-)

AUCtrapezoid

AUC(0-24hr)

AUC(0-)



Time Plasma Conc (hr) (mg/L) 0.0 0.00 0.25 0.88 0.5 1.40 1 1.86 1.5 1.98 2 1.99 4 1.77 8 1.3412 1.0224 0.44


Now calculate Clearance.

Use the formula:ClT = Dose / AUC

Does this formula make any assumptions?



Clearance Cl = Dose / AUC= 400 mg / 33.00 mg*hr/L= 12.12 L/hr



Clearance Cl = Dose / AUC = K * V

Assumption?that the entire dose of 400 mg

was absorbed. If not what happens to V?



Clearance Cl = Dose / AUC = K * V

ClearanceFollowing oral absorption

Cl = F x Dose / AUCwhere F is the fraction absorbed.

So…how much was absorbed?

Calculation of ClearanceCalculation of Clearance

ClearanceCl = Dose / AUC

= 400 /33.00= 12.12 L/hr

Agrees with Estimate Using

Excel®=12.12 L/hr





Now calculate Volume.This can be done through the use of two equations:

Cl = Dose / AUC and Cl = K x VThis allows: K x V = Dose / AUCand so Volume = Dose /(K x AUC)

but again F? Volume = F x Dose /(K x AUC)

Calculation of VolumeCalculation of VolumeVolume

Volume = F x Dose /(K x AUC)

But because we do not know F

Volume/F = Dose /(K x AUC)V/F = 175.69 L

Notes:A volume of 175 L is

calculated when F = 1.

If F = 0.9 (90%) V = 158.12If F = 0.8 (80%) V = 140.55If F = 0.7 (70%) V = 122.98

If F = 0.6 …. … but what is F and how would this be

determined?

VolumeVolume = F x Dose /(K x AUC)

But because we do not know F we will assume a value of 1

(completely absorbed).

Volume/F = Dose /(K x AUC)=400 / (0.069 x 33.00)

V/F = 175.69 L

Calculation of VolumeCalculation of Volume VolumeVolume = F x Dose /(K x AUC)

But because we do not know F




If F = 0.9 (90%) V = 158.12If F = 0.8 (80%) V = 140.55If F = 0.7 (70%) V = 122.98

If F = 0.6 …. … but what is F and how would this be

determined?

Calculation of VolumeCalculation of Volume VolumeVolume = F x Dose /(K x AUC)or because we do not know F




If F = 0.9 (90%) V = 158.12

Monograph (CPS) and literature indicates that the fraction absorbed is approximately 90%.

Another way???

Calculation of VolumeCalculation of Volume VolumeVolume = F x Dose /(K x AUC)or because we do not know F


Notes:A volume of 175 L is close to that determined by back

extrapolation and calculation of a concentration of 2.3 mg/L at time zero.

This appears to have worked because (i) absorption was

relatively quick and (ii) elimination relatively

slow (T½=10 hr).

Concentration-time EquationConcentration-time Equation

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Equation must include F and ka in addition to

parameters found in the IV equation.

Profile for a 1 compartment model with first order input and first order elimination.


How do you calculate ka?

Determining the Absorption RateDetermining the Absorption Rate

How can we find the rate constant associated

with absorption - ka from the rising

concentration time data?


Terminal phasegenerally associated

with elimination of the drug.But what is going on here?Absorption? Elimination?

Absorption phasegenerally associated

with absorption of the drug, but what is going on here?Absorption? Elimination?


Terminal phasegenerally associated

with elimination of the drug.

Absorption &

Elimination

How can we look only at absorption?


Absorption & EliminationHow can we look only at absorption?

We need to be able to subtract the elimination rate,

to strip away one rate constant to reveal only the other.

This is called the method of residuals…sometimes called

feathering the curve or curve stripping.


Since the terminal phase is <pure>,

associated with only 1 rate constant we

can back extrapolate the terminal phase to find

extrapolated concentrations

at each of the times whereconcentrations were

observed during absorption.

Same method used with IV 2C models.


Estimation of Absorption RateUsing method

of residuals1. Back extrapolate

terminal phase using best r-value to guide selection.

2. Determine difference between obs. [ ] & back extrapolated [ ].

3. Plot differences

4. Slope estimates ka.

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Estimation of Absorption Rate Using method

of residuals1. Back extrapolate terminal phase

using best r-value to guide selection.

Recall we had a K value of 0.069 hr-1 based on the last 5 points or 0.0696 hr-1

using the 4 and 24 hr concentrations.

Using the selected K value and a point that is in the terminal phase (8, 12 or 24

hr), calculate the concentrations at 0, 0.25, 0.5, 1.0, 1.5 and 2 hr.





Recall we had a K value of 0.069 hr-1 based on the last 5 points or 0.0696 hr-1

using the 4 and 24 hr concentrations.

Using the selected K value and a point that is in the terminal phase (8, 12 or 24

hr), calculate the concentrations at 0, 0.25, 0.5, 1.0, 1.5 and 2 hr.

Completed using the K value of 0.069 hr-1 determined from the last 5 points generate

concentrations at every time point.





2. Determine the difference between the observed [ ] & back extrap. [ ].

Example: at 0.25 hr the back extrap conc is 2.27 mg/L and the observed concentration was 0.88 mg/L.The difference is: Diff. = 2.27 – 0.88

= 1.39 mg/L






3. Plot differences. Diff. must be plotted as log of concentrations.







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4. Slope estimates ka through conversion by –2.303. Slope = -1.027; ka = 2.365 hr-1



of residuals in Excel1. Back extrapolate


Using the K value determined from the

last 5 points generate concentrations at every time point.



of residuals in Excel1. Back extrapolate


2. Determine difference between obs [ ] & back extrapolated [ ].






3. Plot differences

4. Slope estimates ka.


Calculate expected concentrations to see how close the observed and expected concentrations are using our parameter estimates.Dose = 400 mg; T½ = 10.04; K = 0.06902 hr-1; ka = 2.365 hr-1 Volume = 175.69 L with F = 1.

ObservedTime Plasma Conc (hr) (mg/L) 0.0 0.00 0.25 0.88 0.5 1.40 1 1.86 1.5 1.98 2 1.99 4 1.77 8 1.3412 1.0224 0.44

Observed ExpectedTime Plasma Conc Plasma Conc (hr) (mg/L) (mg/L) 0.0 0.00 0.00 0.25 0.88 1.00 0.5 1.40 1.55 1 1.86 1.97 1.5 1.98 2.05 2 1.99 2.02 4 1.77 1.78 8 1.34 1.3512 1.02 1.0224 0.44 0.45


Agreement is good. Concentrations deviate by -0.11 to -0.15 mg/L in the first hour. This would indicate good estimation of K, with some error in ka. But volume and F are really still a combined term. What is the true F in this patient?

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Parameters to Parameters to Investigate FurtherInvestigate Further

F, Bioavailability and its calculationEffect of ka on profile

SR products.

Parameters to Parameters to Investigate FurtherInvestigate Further

F, Bioavailability and its calculationEffect of ka on profile

SR products.

1-Compartment Oral Dosing 400 mg of moxifloxacin is administered orally to Mr BB, a 68 yr old male...

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Transcript of 1-Compartment Oral Dosing 400 mg of moxifloxacin is administered orally to Mr BB, a 68 yr old male...