1-C: Renal and Hepatic Elimination

1-C: Renal and Hepatic Elimination400 mg of moxifloxacin 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 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)

The intent of this slide pack is to evaluate Renal Clearance of a drug … How much moxi is in the urine ?But first, although we have already analyzedplasma data, as a review calculate volume, AUC,

K, half-life and clearance. Then:1. How much unchanged moxi is in the urine

at 12 hrs? ..at 36 hrs? Calculate K and ke. from urine data.

2. What is the ClR of moxifloxacin?

1-C: Renal and Hepatic Elimination400 mg of moxifloxacin 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 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)

1-Compartment IV Dosing

Analysis of Concentration –Time Data

This data set provided is identical in many ways to the tobramycin data set

analysed in Slide Pak 07.

While the solution to volume, AUC, K, half-life and clearance

is shown briefly in the following slides, you should complete the calculation of these parameter values independently

… for practice.

Class time will not be spent reviewing the method of calculation of

the initial concentration, AUC, K, half-life or clearance.

You need to calculate the initial concentration, AUC, K, half-life and clearance.

You must FIRST determine the log of the concentration and then from any pair of points (or using all data in Excel)

determine the slope. K is be determined from inspection of the data, graphical methods or by calculation of the slope.

Estimation of K. K is the slope of the line (t=2 to 12 hr) K = -2.303[log(C2) – log(C1)] / (t2 - t1)

= -2.303[log(1.16) – log(2.32)] / (12 – 2)= -2.303[0.064 – 0.365] / (10)= -2.303[ - 0.301]/10= 0.0693hr-1 T½ = 10.0 hr. Excel = 0.694 hr-1

Using K, we can then estimate the initial concentration by extrapolation. Then calculate the volume & AUC by trapezoidal rule

between each time point (or the pharmacokinetic method). A variety of estimates of K are possible given the concentrations.

The Excel sheet (posted) provides a range of answers. Those calculated by hand should be very similar.

Volume150.12 L

K0.0694 hr-1

1-Compart.IV Dosing

0.0694 hr-1

T½9.98 hr

AUC38.38 mg*hr/L

Cl 10.42L/hr

Volume150.12 L

400 mg of moxifloxacin 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 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)

Parameter Estimates K = 0.0694 hr-1

T½ = 9.98 hrAUCI =38.38 mg*hr/L

V = 150.12 L

1-C: Renal and Hepatic Elimination

0 4 8 12 16 20 24 28 32 36Hours

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

Terminal elimination phase is log-linear…

1 Compartment Model

with first order elimination (K)

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)

0 4 8 12 16 20 24 28 32 36Hours

kH = kM1 + kM2

KNR = 20% excreted into bile as unchanged drug.

0 4 8 12 16 20 24 28 32 36Hours

K = ke + kH + kNR

1-C: Renal and Hepatic Elimination400 mg of moxifloxacin 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 isknown that about 20% of a moxifloxacin dose is excreted in theurine unchanged.

Plasma Urine Amount Urinary

Time Conc Collection Urine Urine Recovered Excretion (hr) (mg/L) Period Volume conc in Urine Rate

(hr) (mL) (mg/L) (mg) (mg/hr) 0 no sample 2 2.32 0 – 4 290 0.060 17.40 6 1.76 4 - 8 295 0.045 13.28 12 1.16 8 - 16 490 0.036 17.64 24 0.50 16 - 32 1060 0.015 15.90 36 0.22 32 - 40 555 0.006 3.330

Urine collection following the 400-mg IV dose yields:

(hr) (mL) (mg/L) (mg) (mg/hr) 0 no sample 2 2.32 0 – 4 290 0.060 17.40 17.40 mg / 4 hr 6 1.76 4 - 8 295 0.045 13.28 12 1.16 8 - 16 490 0.036 17.64 24 0.50 16 - 32 1060 0.015 15.90 36 0.22 32 - 40 555 0.006 3.330

Calculate the Urinary excretion Rate (mg/hr):Amount recovered in urine / duration of urine collection.

(hr) (mL) (mg/L) (mg) (mg/hr) 0 no sample 2 2.32 0 – 4 290 0.060 17.40 4.350 6 1.76 4 - 8 295 0.045 13.28 3.319 12 1.16 8 - 16 490 0.036 17.64 2.205 24 0.50 16 - 32 1060 0.015 15.90 0.994 36 0.22 32 - 40 555 0.006 3.330 0.416

Calculate the Urinary excretion Rate (mg/hr):Amount recovered in urine / duration of urine collection.

When evaluating Aminoglycoside Renal Clearance (following IV dose with no metabolism and complete renal excretion)

we observed:ClR = ClT and therefore, ClNR = 0 L/hr

It would have also been apparent that the amount excreted in the urine after infinite time (Ae0-)

would equal the dose.

… that an Excretion Rate vs. Time plot had a slope proportional to K

…that an Excretion rate vs. mid point plasma concentration had a slope equal to Renal clearance (ClR).

But now, moxi has other routes of elimination…?

Renal Clearance of moxifloxacinRenal Clearance of moxifloxacin

Calculations to complete1. Plot Excretion rate vs. Time.2. Calculate K3. Plot Excretion rate vs.

mid-point plasma conc.4. Calculate Renal clearance5. Calculate ke6. Determine Ae0-

7. Establish some principles of Elimination

1-C: Renal and Hepatic EliminationPlot Excretion Rate vs. Time.

Plot at the mid point of the collection period as this is the time which most accurately reflects the rate

Plot Excretion Rate vs. Time1-C: Renal and Hepatic Elimination

Slope = -K/2.303

Urine DataSlope = -K/2.303

= - 0.030 K = -0.030 * -2.303

= 0.0687 hr-1

T½ = 10.084 hr

Change in the Rate at whichdrug appears in the urine

( Ex. Rate) is proportional to concentration in serum

and slope is –K/2.303.

Calculations to complete1. Plot Excretion rate vs. Time.2. Calculate K3. Plot Excretion rate vs.

1-C: Renal and Hepatic EliminationPlot Excretion Rate vs. Mid-point Plasma concentration.

Plot the excretion rate vs. the concentration at the mid point of the collection period

Plot Excretion Rate vs. Mid-point Plasma concentration.

Slope = ClR (L/hr)

Renal Clearance & Urinary Excretion

Slope = ClR (L/hr)= 1.864 L/hr

Renal Clearance & Urinary Excretion.

Slope = ClR (L/hr) CLR = 1.864 L/hrPlasma Data; ClT = 10.42 L/hr

What does this tell us?

ClR = 1.864 L/hr ClT = 10.42 L/hr

ClR / ClT = 0.1789

If 17.89% of ClT is determined by ClR

what is ClNR?

what is ClNR?ClT = ClR + ClNR

ClNR = 10.4 – 1.8ClNR = 8.55 L/hr

Calculations to complete• Plot Excretion rate vs. Time.• Calculate K• Plot Excretion rate vs.

ClT = ClR + ClNR

and K = ke + knr

and K and ClT are related by the

proportionality constant -Volume

what is ke?

ClT = ClR + ClNR

and K = ke + knr

and K and ClT are related by the proportionality constant - Volume

what is ke?

ClT = ClR + ClNR

and K = ke + knr

and K and ClT are related by the proportionality constant - Volume

what is ke?

ClR ke ClT K----- = -----

ClT = 10.42 L/hr and ClR = 1.864 L/hrK = 0.0687 hr-1 and ke = 0.01242 hr-1

ClR ke ClT K----- = -----

However, now there is at least one other source of elimination (kNR which is likely made up of kH (or kM1 and kM2) and kB

and since …

If ke is determining how much drug appears in the urine then …

However, now there is at least one other source of elimination (kNR which is likely made up of kH (or kM1 and kM2) and kB

and since …

If ke is determining how much drug appears in the urine then …

Ae0- ke

DoseIV K

and all will have a ratio of ~ 0.1789 in this patient.

ClR ke Ae0- ClT K DoseIV

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

ClR ke ClT K----- = -----

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

Over the first 40 hours, 67.55 mg of moxi is collected in the urine.Based proportion of drug being eliminated into the urine (ke/K)

the total amount expected in the urine would be 400 mg x 0.1789 = 71.56 mg.

Would 4.01 mg be expected to appear in the urine after 40 hr?(40 hr )

67.55 mg

At 40 hours, 0.16 mg/L of moxi remains in plasma.The estimated volume is 150.12 L.

Therefore the amount remaining in the body at 40 hours is:0.16 mg/L x 150.12 L =

67.55 mg

At 40 hours 0.16 mg/L of moxi remains in plasma.The estimated volume is 150.12 L.

Therefore the amount remaining in the body at 36 hours is:0.16 mg/L x 150.12 L = 24.6 mg

The amount that should appear in the urine (40 hr ) would be:0.1789 x 24.6 =

67.55 mg

At 40 hours 0.16 mg/L of moxi remains in plasma.The estimated volume is 150.12 L.

Therefore the amount remaining in the body at 40 hours is:0.16 mg/L x 150.12 L = 24.6 mg

The amount that should appear in the urine (40 hr ) would be:0.1789 x 24.6 = 4.29 mg

67.55 mg

Principals of EliminationPrincipals of Elimination Equations

K = ke + kH + kNR

TBC = ClT = ClR + ClH + ClNR

ClH = ClM1 + ClM2

kH = kM1 + kM2

ClR = keV

----- = ----- = ----- = ---------ClR ke ke Ae0- ClT K k10 DoseIV

Change in the Rate at whichdrug appears in the urine

( Ex. Rate) is proportional to concentration in serum

and slope is –K/2.303.Slope of log Excretion Rate vs.

Time is proportional to K (-2.303)

the amount excreted in the urine after infinite time (Ae0-)

equals ke/K

…that an Excretion rate vs. mid point plasma concentration

has a slope equal to Renal clearance (ClR).

Summary of Urinary EliminationSummary of Urinary Elimination

equals ke/K

Change in the Excretion Rate at which drug appears in the urine [Slope (Log) Ex. Rate) vs.Time]

is parallel the slope of the Concentration in serum vs. Time

Slope is –K/2.303.

This is true regardless of the proportion of the dose excreted

into the urine ... as long as Ae0- >0

equals ke/K

…that an Excretion rate vs. mid point plasma concentration

has a slope equal to Renal clearance (ClR).

K is the overall rate constant which determines the half-life.

ke determines the proportion of the dose excreted into the urine.

Ae0- ke

DoseIV K -------- = -----

Second use of Excretion Rate:

If Excretion Rate is plotted (not Log Excretion Rate)

vs. mid point plasma concentration

…slope is equal to Renal clearance (ClR).

Minimum requirement is a single blood sample drawn at the mid

point of the urine collection interval.

What do you need to calculate K?

(i) At least 2 blood (plasma) concentrations drawn

at known times.

(ii) At least 2 urine collections of known intervals and time

since the dose.

Plot Log Excretion Rate vs.

What do you need to calculate ke?

(i) A complete urine collection and a known dose

Ae0- ke

DoseIV K

(ii) Through relationships with other variables

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

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

ClR ke Ae0-

ClT K DoseIV

What do you need to calculate ClR?

Minimum requirement is a single blood sample drawn at the mid

point of the urine collection interval.

If Urinary Excretion Rate is plotted

vs. mid point plasma concentration

…slope is equal to Renal clearance (ClR).

Renal Excretion is based on:

Filtration at the glomerulus. This is a function of a number of competing pressures.

The hydrostatic pressure of blood is the overall driving force filtering blood. Working against this pressure is the osmotic pressure of the

blood to hold on to the fluid and the hydrostatic pressure of the capsule. The result of these

completing pressures is a net ~10 mmHg pushing fluid through the capsule and

into the proximal tubule.

In addition to filtration, there is active secretion, in the proximal renal tubule and reabsorption

most likely in the distal tubules.

Renal Function & Drug Clearance ClR

Remaining Slides in this file

are for interest only.Creatinine clearance will

be covered in the 2nd term.

Filtration at the glomerulus

Approximately 125 mL of protein free filtrate is removed from the blood every minute as it passes through the glomeruli of the kidneys. However, only drugs that have a low molecular weight and are not protein bound can pass through the glomeruli [since protein is not filtered, drug bound to protein is not filtered]. Since most drugs are small molecules, only a high degree of protein binding will reduce the amount of drug found in the protein free filtrate.

Active secretion

Active secretion, In addition to glomerular filtration, some drugs may be actively secreted into the proximal renal tubules. Since secretion is an active process, the secretion rate does not appear to be influenced by protein binding and is therefore, generally felt to be “nonrestrictive”.

ReabsorptionReabsorption is a passive process &may occur at any point in the Tubule but is more likely occur in the distal tubules as the urine and the concentration of the drugs within it, becomes more concentrated. Reabsorption can occur for any drug but it is more likely to occur for unionized, non-polar drugs. As a result, the reabsorption of weakly basic or acidic drugs may be dependant on urine pH, which will determine the degree to which the drug exists in the unionized state.

Renal function is often estimated.

There are at least three compounds that have been widely used to estimate kidney function: Inulin (MW ~ 5000);

total eliminated by GFR. PAHA [para amino hippuric acid]

filtered and secreted.Creatinine,

endogenous, filtered and secreted.

Creatinine, easy & most popular.Compoun used to estimate renal function.It is an endogenous end product of muscle metabolism.

Calculation of CrCl.

To properly estimate creatinine clearance a 24-hour collection of urine is completed and the creatinine measured. The amount of creatinine excreted is determined based on the urine volume.

Calculation of CrCl

Creatinine Clearance (CrCl) is then determined based on a single serum creatinine measurement, assuming that the measured serum creatinine is the average concentration of creatinine or that the serum creatinine concentration is constant over the 24-hr collection period.

Calculation of CrCl…Example

Consider a patient with a serum creatinine 1.1 mg/dL and following a 24 hr urine collection is found to have a creatinine concentration of 1.4 mg/mL and 1250 mL of urine.

CrCl (mL/mIn = (100 x U x V ) / SCr x 1440

Where U is the concentration of creatinine in the urine in mg/dL, V is the urine volume in mL and 100 is the conversion from dL,and SCr is the creatinine concentration in plasma, in mg/dL.1440 is the conversion to units of mL/min from mL/24 hrs

Calculation of CrCl…Example

Consider a patient with a serum creatinine 1.1 mg/dL and following a 24 hr urine collection is found to have a creatinine concentration of 1.4 mg/mL and 1250 mL of urine.

CrCl (mL/mI) = (100 x U x V ) / SCr x 1440= (100 x 1.4 x 1250) / (1.1 x 1440)= 175000 / 1584= 110.47 mL/min= 110 mL/min

Although creatinine is easy & popular, as endogenous endproduct of muscle metabolism, creatinine production is directly related to muscle mass. Therefore, it is necessary to correlate serum creatinine with urinary creatinine clearance in males and females of various ages.

There are many nomograms that describe a relationship between serum creatinine and renal function. One is the Cockcroft & Gault equation.Cockcroft DW, Gault MH. Prediction of Creatinine clearance from Serum Creatinine. Nephron 1976; 16: 31-41.

Prediction of Creatinine Clearance from Serum CreatinineDONALD W COCKCROFT & M HENRY GAULT

Nephron 1976; 16: 31-41.

Derived Equation (x 0.85 for females)

When SCr is reported in mg/100 mL CrCl is calculated in mL/min.

When Serum Creatinine is reported in μmol/L the equation is:

and CrCl is calculated in mL/sec and must be multiplied by 60 to yield mL/min.

1-C: Renal and Hepatic Elimination

initial concentration

plasma concentration

urine data

renal clearance

volume auc

moxifloxacin dose

analyzedplasma data

unchanged moxi

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