time plasma conc (hr) (mg/l) 119.5 214.4 5 5.8 8 2.4 12 0.7 18 below limit of detection 1...

Time Plasma Conc (hr) (mg/L)

1 19.5 2 14.4 5 5.8 8 2.4 12 0.7 18 Below limit of detection

1 Compartment IV Analysis& Renal Elimination

500 mg of tobramycin is administered by iv bolus 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 is known that tobramycin is excreted primarily unchanged in the urine. The following plasma concentrations were observed.

1. What model best describes this profile?2. Calculate AUC, K, Vd, T½ and Cl.3. Write an equation that will calculate

concentrations at anytime after the 1st dose.4. How much is in the body at time zero?5. How much is in the body at 12 hours?6. Where did the tobramycin go?7. How much is in the urine at 12 hrs?8. What is the renal clearance of tobramycin?9. Is tobra a high or low extraction drug?



1 Compartment IV Analysis500 mg of tobramycin is administered by iv bolus 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 is known that tobramycin is excreted primarily unchanged in the urine. The following plasma concentrations were observed.



Graph Patient DataGraph Patient Data

0 2 4 6 8 10 12 14

Using semi-log paper,graph the data following

Iv bolus of 500 mg oftobramycin.

100

60

30

10

6

4

2

1



Graph Patient Data in Excel®Graph Patient Data in Excel®Using semi-log paper,

graph the data followingIV bolus of 500 mg of

tobramycin.



What model best describes this profile?

0.1

1.0

10.0

100.0

0 4 8 12Hours

Conc

entra

tion

(mg/

L)


0 2 4 6 8 10 12 14

100

60

30

10

6

4

2

1

What model best describes this profile?

… a 1 compartment model with iv bolus input




1. What model best describes this profile?2. Calculate AUC, K, Vd, T½ and Cl.3. Write an equation the will calculate


What do you calculate first?ClT½

AUCVdK






What do you calculate first?Cl Dose/AUC or K x Vd

T½ - 0.693 KAUC –/ K

Vd - [ ]0 by back extrapolation of KK …?


0 2 4 6 8 10 12 14

100

60

30

10

6

4

2

1

Slope = -K/2.303

Slope based on log of the concentrations

Time Conc Log (hr) (mg/L) Conc 1 19.5 1.29 2 14.4 1.16 5 5.8 0.76 8 2.4 0.38 12 0.7 -0.15 18 BLOQ

Slope = (y2 – y1) / (x2 – x1)=


0 2 4 6 8 10 12 14

100

60

30

10

6

4

2

1

Slope = -K/2.303



Slope = (y2 – y1) / (x2 – x1)= (-0.15 – 1.29)/(12-1)= (- 1.44) / (11)= -0.1309

- K = slope x 2.303


0 2 4 6 8 10 12 14

100

60

30

10

6

4

2

1

Slope = -K/2.303



Slope = (y2 – y1) / (x2 – x1)= (-0.15 – 1.29)/(12-1)= (- 1.44) / (11)= -0.1309

- K = slope x 2.303= - 0.1309 x 2.303

K = 0.3015 hr-1


0 2 4 6 8 10 12 14

100

60

30

10

6

4

2

1

Slope = -K/2.303



K = 0.3015 hr-1

T½ = 0.693 / K= 0.693 / 0.3015= 2.298 hr

1 Compartment IV Analysis

K

T½






Now calculate Volume.Since it was an IV dose,

and appears to be a 1 compartment model,we will assume instantaneous distribution

and so we will back-extrapolate to t=0to determine the initial concentration.

Calculation of VolumeCalculation of Volume

0 2 4 6 8 10 12 14

100

60

30

10

6

4

2

1

Back extrapolation can be completed a number of ways

1. GraphicallyIntercept


1 19.5 2 14.4

0 2 4 6 8 10 12 14

100

60

30

10

6

4

2

1


1. Graphically2. By equation

[ ]t = C0 e-Kt

or [ ]0 = C1 e+Kt

=

Intercept



1 19.5 2 14.4

0 2 4 6 8 10 12 14

100

60

30

10

6

4

2

1



[ ]t = C0 e-Kt

or [ ]0 = C1 e+Kt [ ]0 = 19.5 e(0.3015 x 1)

= 26.36 mg/LVolume = Dose / [ ]0

=

Intercept



1 19.5 2 14.4

0 2 4 6 8 10 12 14

100

60

30

10

6

4

2

1



[ ]t = C0 e-Kt

or [ ]0 = C1 e+Kt [ ]0 = 19.5 e(0.3015 x 1)

= 26.36 mg/LVolume = Dose / [ ]0

= 500 / 26.36 = 18.97 L

Intercept


Calculation of VolumeCalculation of Volume Back extrapolation can be completed a number of ways


3. By ExcelUsing Intercept function

=10^(INTERCEPT(C5:C9,A5:A9))






Now calculate AUC.Since it was an IV dose, we must realise

that the data set starting at 1 hour is missing concentration data from 0 to 1 hour.

Therefore, AUC by trapezoidal rule must use a time zero concentration of 26.38 mg/L.

Calculation of AUCCalculation of AUC Trapezoidal RuleAUC =((C0+C1)/2)*(t1-t0)

The value of 9.75 for the AUC is based on a no concentration

or a concentration of 0.00 at time zero.

Was the Concentration zero? Or should we determine

the time zero concentration

Trapezoidal Rule (Excel)AUC = ((B4+B5)/2)*(A5-A4)The value of 9.75 in this cell

is based on a no concentration or a concentration of 0.00

at time zero in cell B4.

Calculation of AUCCalculation of AUC Trapezoidal RuleIf the initial concentration of

26.38 is used in the calculation of AUC from time 0 to 1 hr, a

more appropriate AUC of 22.94 mg*hr/L is estimated.AUC0-12h is 88.69 mg*hr/L.

= 88.69

You must also calculate AUC

from 12 hr to . This can onlybe done using

the PCK method:[ ]LP/K.

Calculation of AUCCalculation of AUC AUC from 12hr to Recall that K was calculated

as 0.3015 hr-1 using concentrations at 1 hr and 12 hrs.Using this K value, AUC12- is:

[ ]LP/K. = 0.7/0.3015 = 2.32 mg*hr/L.

= 88.69

The sum of the AUC (0-12hr)

and AUC (12hr-).

is:AUC (0-) = 88.69 + 2.32

= 91.01 mg*hr/L.

Calculation of AUCCalculation of AUC Other Estimates of AUC (0-)

Several AUC (0-) estimates are possible based on the

estimate of K and the method used.

1. Kinetic method Using the determined back-extrapolated t=0

concentration of 26 .38 mg/L AUC0- can also be estimated as:

[ ]t=0/K. = 26.38/0.3015 = 87.496 mg*hr/L.

Calculation of AUCCalculation of AUC Other Estimates of AUC (0-)

Several AUC (0-) estimates are possible based on the

estimate of K and the method used.

2. Estimates of K In the Excel sheet using the

last 2 … or last 3 …or last 4…K values can be estimated

ranging from of 0.30809 hr-1

0.30245 hr-1

0.30185hr-1

and 0.301806 hr-1

(these are effectively the same~ differ by less than 0.007 hr-1)

These variations will produce small differences in AUC

Calculation of AUCCalculation of AUC Other Estimates of AUCExcel sheet shows some of the possible estimates

of K and using 0.0318 hr-1 and calculates AUC12h-

As 2.32 mg*hr/L.

Also shown are

Calculation of AUCCalculation of AUCOther Estimates of AUCExcel sheet shows some of the possible estimates

of K and using 0.0318 hr-1 and calculates AUC12h-

As 2.32 mg*hr/L.

Also shown are two estimates of AUC0-:

1. Based on kinetic method from the t=0

determined concentration and

2. Based on sum of Trap Rule AUC

& kinetic method from LP

Calculation of AUCCalculation of AUC Other Estimates of AUC

AUC0- is 91.01 mg*hr/L calculated by trapezoidal rule.

But if we use the PCK method from time zero (26.38 mg/L)

we would calculate AUC0- as 87.41 mg*hr/L.

Which is more correct?






Now calculate Clearance.

This can be completed by at least 2 methods:

1. Dose / AUC2. K x Vd

Calculation of ClearanceCalculation of Clearance Estimates of ClearanceAlthough a variety of estimates

of K have been calculated all are very close (~0.3018 hr-1)Using the volume calculated from the t=0 concentration determined by back-extrap using this K value (Excel Sheet)

Cl = K x Vd = 0.3018 x 18.95 L

= 5.720 L/hr

Calculation of ClearanceCalculation of Clearance Estimates of ClearanceAlthough a variety of estimates

of K have been calculated all are very close (~0.3018 hr-1)

Cl = 5.720 L/hr (K x V)

When we use AUC calculatedby the PCK method, clearanceestimate is identical to estimate

determined by K*Vd.

Clearance is 5.72 L/hr and

AUC0- of 87.41 mg*hr/L.is the more correct estimate.

[ ]t = C0 e-Kt

Where K = 0.3018 hr-1

and C0 can always be calculated from Dose/V

Ct = (Dose / V) e-Kt

and for our patient K = 0.3018 hr-1

V = 18.95 L

Equation describing profileEquation describing profileProfile is a 1 compartment

model with bolus input.






At time zero, immediately following the dose, there should be

500 mg in the body.The entire dose is delivered into the body.


and for our patientK = 0.3018 hr-1

Dose = 500 mgV = 18.95 L

So the concentration at 12 hr would be:

Amount in the body at 12 hoursAmount in the body at 12 hours

C12hr =



Dose = 500 mgV = 18.95 L

So the concentration at 12 hr would be:


C12hr = (500 / 18.95) e(-0.3018 x 12)

= 28.38 e(-0.0267)

= 0.7054 mg/L

and the concentration reported at 12 hours was 0.7 mg/L

Amount in the body at 12 hoursAmount in the body at 12 hoursC12hr = (500 / 18.95) e(-0.3018 x 12)

= 28.38 e(-0.0267)

= 0.7054 mg/L

and the concentration reported at 12 hours was 0.7 mg/L

Since the volume is 18.95 L, the amount in the body at 12 hours is:

Amount = [ ] x V= 0.7 mg/L x 18.95 L= 13.265 mg

Only 13.265 mg of the initial 500 mg dose remains.

We could have also calculated the amount in the body directly

from the equation:Instead of


We would haveAmtt =


We could have also calculated the amount in the body directly

from the equation:Instead of


We would haveAmtt = Dose e-Kt


Dose = 500 mgSo the amount in the body

at 12 hr would be:


Amt12hr= (500) e(-0.3018 x 12)

= 500 e(-0.0267)

= 13.36 mg

Where did the tobramycin go?Where did the tobramycin go?500 mg of tobramycin is administered by iv bolus 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 is known that tobramycin is excreted primarily unchanged in the urine. The following plasma concentrations were observed.

Where did the tobramycin go?Where did the tobramycin go?

Amount = [ ] x V= 0.7 mg/L x 18.95 L= 13.26 mg

Only 13.3 mg of the initial 500 mg dose remains.


Amt12hr= (500) e(-0.3018 x 12)

= 500 e(-0.0267)

= 13.36 mg

Does this mean that 486.7 mg is in the urine in the first 12 hours following the dose?

Calculation of ClearanceCalculation of Clearance Since renal elimination appearsto be the only way tobramycin

is cleared from the body,

Total Body Clearance should equal renal clearance…?

Clearance is 5.7 L/hr

At constant concentration, clearancecan be estimated from blood flow

and the difference betweenarterial and venous concentrations.

ClH = Q [ (Ca – Cv) / Ca ]

But in our situation the concentration changing.In the first hour concentration started at

26.38 mg/L and declined to 19.51 mg/L and we collected 130.26 mg of tobramycin in the urine.

In the second hour concentration started at 19.51 mg/L and declined to 14.43 mg/L

and we collected 96.33 mg of tobramycin in the urine.

EquationsConc = Dose / VV = Dose/Conc

Cl = Q x ERER = Cl / Q

Renal Clearance of tobramycinRenal Clearance of tobramycin

As the concentration declines we recover less in the urine and because urinary excretion is

the only method of elimination, drug excretion into the urine is tied to elimination from the body.

Rate of appearance of drug in the urine is proportional to the amount in the bodyTherefore:

Where X is the amount of drug in the body at time t and Xu is the amount of drug

eliminated in the urine to time t.




dXu

dt--- = ke X

Since X, the amount drug in the bodyat time t changes according to

X = X0 e (-Kt)

or as we have previous writtenAmtt = Dose e (-Kt)

Then …

and a semi-log plot of excretion rate of drug in the urine vs. time

should be linear and have a slope of –K/2.303.

This is the same slope as for a plasma concentration curve.




dXu

dt--- = ke X0 e(-Kt)

Evaluation of excretion rate

vs. time.Note:

Time scale expandedConc. & Amount/timeDifference produces

“Amount Lost”both total and duringeach 1 hour interval.

An excretion rateper hour(mg/hr)

is then calculated.


Plots of excretion rate (mg/hr) vs. time and concentration vs. time are parallel. Slope of the log of excretion rate

vs. time is a function of the overall elimination rate (K), NOT the urinary excretion rate constant – ke.


The urinary excretion rates that were calculated are not instantaneous rates,

Rather, they are average ratesdetermined over an interval.

The average excretionrate closely approximates the

instantaneous rate seen at the mid-point of the urine collection interval.

Therefore, urinary excretion rates should be plotted at the midpoints

of the urine collection period.




dXu

dt--- = ke X0 e(-Kt)

Xu

t--- = ke X0 e(-Kt)

Just as the plasma concentration-time profile can be characterized by both an

elimination rate constant (K) and a clearance term (ClT or TBC) , so can

urinary excretion data: renal excretion rate constant (ke) and a renal clearance term (ClR).

We have seen that the rate at which drug is cleared into the urine is proportional to the plasma concentration at the mid point

of the urine collection interval.




(dXu / dt ) CMID

ClR =

We have seen that the rate at which drug is cleared into the urine is proportional to the plasma concentration at the mid point

of the urine collection interval.

In practice renal clearance is estimated by dividing the urinary excretion rate by the

plasma concentration at the mid point of the urinary collection interval.

where XU is the amount of drug appearing in the urine




(dXu / dt ) CMID

ClR =

(Xu / t ) CMID

ClR =

In practice renal clearance is estimated by dividing the urinary excretion rate by the

plasma concentration at the mid point of the urinary collection interval.

where XU is the amount of drug appearing in the urine. Equations

Conc = Dose / VV = Dose/Conc



(Xu / t ) CMID

ClR =

Rearrangement yields:(Xu / t ) = ClR x CMID

which indicates that ClR will be the slope of a straight-line relationship between

the amount recovered in the urine during an interval and the

midpoint plasma concentration.


The excretion rate(mg/hr) has been

plotted against the plasma

concentration at the midpoint of

the interval.


The slope is renal clearance (ClR) and you can see that it equivalent

to total clearance (ClT) calculated previously.


Example: Blood Urine Urine Urine Amount ExcretionSample Collection Volume Conc. Excreted Rate Time Interval (hr) Start-Stop (mL) (mg/mL) (mg) (mg/hr) 1.0 0 – 2 hr 120 1.89 226.6 113.6 4.0 2 – 6 hr 250 0.77 192.5 9.0 6 – 12 hr 400 0.17

In practice, renal clearance is calculated from a series of urine collections with blood samples

taken at the mid-point of the interval.

Urinary Excretion Rate is calculated from the urine volume,

the drug concentration in the urine and period over which the urine is collected.


Example: Blood Urine Urine Urine Amount ExcretionSample Collection Volume Conc. Excreted Rate Time Interval (hr) Start-Stop (mL) (mg/mL) (mg) (mg/hr) 1.0 0 – 2 hr 120 1.89 226.6 113.6 4.0 2 – 6 hr 250 0.77 192.5 48.13 9.0 6 – 12 hr 400 0.17 68.0 11.33

In practice, renal clearance is calculated from a series of urine collections with blood samples

taken at the mid-point of the interval.

Urinary Excretion Rate is calculated from the urine volume,

the drug concentration in the urine and period over which the urine is collected.


Calculated Renal

Clearance5.723 L/hr

Another method for calculating Renal Clearance

Since dXu / dt is equivalent to keX, Substitution in the equation:

Results in

Where X/C = V, and so…

ClR = keV




(dXu / dt ) CMID

ClR =

(ke X ) CMID

ClR =

Equation SummaryConc = Dose / VV = Dose/Conc


(Xu / t ) = ClR x CMID

ClR = keV

Renal Clearance of tobramycinRenal Clearance of tobramycinRenal blood flow = 25% CO

1.5 L/min.Hct = 0.5 in our patientTherefore, renal plasma flow isQplasma = 1.5 L/min x 0.5

= 0.75 L/min= 45 L/hr.

Using plasma because renal clearance was based

on plasma concentrations and the volume of distribution

was based on plasma concentrations




ClR = keV



= 0.75 L/min= 45 L/hr.

Tobramycin renal clearance in our patient was 5.72 L/hr

ER = ClR / Qplasma

=




ClR = keV



= 0.75 L/min= 45 L/hr.

Tobramycin renal clearance in our patient was 5.72 L/hr

ER = ClR / Qplasma

= 5.72 / 45= 0.127

Low Extraction … compareTheophylline (liver) 5%Creatinine (renal) 14.6%Ciprofloxacin (renal) 71%

time plasma conc (hr) (mg/l) 119.5 214.4 5 5.8 8 2.4 12 0.7 18 below limit of detection 1...

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