Zero Order Absorption
*
Types of chemical administration or exposure that provide (or approximate) zeroorder chemical input include:
 Intravenous (IV) infusion
 Prolonged breathing of chemical vapors
 Prolonged skin exposure to constant chemical concentration
 Implantable controlled/sustained release system
*
*
Model Equations
D = k0 * T
k0 = D/T
*
During Chemical Exposure (t≤T)
 dA/dt = (rate in) – (rate out)
= (absorption rate) – (elimination rate)
dA/dt = k0 – (k*A)
dA/dt + (k*A) = k0
*
A = c1 + (c2 * ekt)
 The unknown constant terms are then evaluated by employing the initial condition (A=0 at t=0) and the differential equation to get:
c1 = c2 = k0/k
or
A = (k0/k)(1ekt)
*

 As before, it is more useful to write the final equation in terms of plasma concentration, which can be done through the transformation C = A/V, giving:
C = [k0/V*k] [1ekt] for t≤T
*
After Chemical Exposure Stopped (t>T)
 After chemical exposure has stopped, the differential mass balance can be written:
dA/dt = k*A
A = AT at t = T
*
A = AT * ek(tT) = (k0/k)(1 ekT)ek(tT)
C = CT * ek(tT) = (k0/V*k)(1 ekT)ek(tT)
CT = (k0/V*k)(1 ekT)
*
*
31.bin
SteadyState Concentration (CSS)
 Css occurs when the rate of chemical elimination exactly equals the rate of chemical input
 Chemical Elimination Rate = Chemical Input Rate
CL * Css = k0
V * k * Css = k0
Css = k0/V*k
Note: that the equations for C at t≤T can be rewritten in terms of Css:
C = (k0/V*k)(1ekt) = Css(1ekt) for t≤T
*
 Concentration drops exponentially immediately after exposure stops, so after infusion (t>T):
C = CT * ek(tT)
 taking the natural logarithm of both sides gives:
lnC = lnCT k(tT)
or
lnC = {lnCT + kT} – kt
*
*
Effect of Model Parameters on Plasma Concentrations
Dosing Rate (k0): Concentration proportional to k0 at all times
*
Effect of Model Parameters on Plasma Concentrations
Distribution Volume (V): Concentration proportional to 1/V at all times
*
Effect of Model Parameters on Plasma Concentrations
Elimination Rate Constant (k) and Halflife (t1/2): Affect approach to plateau and rate of drop after stopping input, k and t1/2 have opposite effects
*

 The approach to steady state (plateau) conditions is dependent on halflife in a manner similar to elimination:
at t1/2 → 50% of way to equilibrium → C=0.5Css
at 2t1/2 → 75% of way to equilibrium → C=0.75Css
at 5t1/2 → ~97% of way to equilibrium → C~0.97Css
at 7t1/2 → ~99% of way to equilibrium → C~0.99Css
at 10t1/2 → ~99.9% of way to equilibrium → C~0.999Css
*
*
Determination of Chemical Input Rate (k0)

 IV Infusion
 Known dose (D) infused over known time period (T) → k0 = D/T
 k0 = D/T → 100mg/2hr = 50 mg/hr
 Skin Exposure Absorption
 k0 = Absorption Rate = Pskin*Sskin [Cskin – Cblood]
*
Determination of Chemical Input Rate (k0)

 Lung Exposure Absorption

 Low Solubility Gas (Kblood/air <<1, perfusion limited)
 K0 = Absorption Rate = Qlungblood [Cblood out – Cblood in] ~ Qlung blood [Kblood/air*Cair inCvein]
 High Solubility Gas (Kblood/air >>1, ventilation limited)
 K0 = Absorption Rate = Qlung air [Cair in – Cair out] ~ Qlungair *Cair in
*
Example: Estimate the toluene absorption rate for rats exposed to 100ppm toluene via lung absorption

 kblood/air = 18; MW = 92.15; Qlung air = 3.75 L/hr; Qlung blood = 3.75 L/hr

 remember: C(mg/L) = C(ppm)*MW/24450
 Cair in = 100ppm * [92.15/24450] = 0.377 mg/L
 kblood/air = 18, therefore kblood/air >> 1 → High solubility gas
 k0 = Qlung air * Cair in ~ 3.75 L/hr * 0.377mg/L = 1.4mg/hr
*

 Rats are exposed to 100 ppm toluene for 4 hrs, after which the exposure to toluene stops. k0 = 1.4 mg/hr (from previous slide); T = 4hrs; V = 1.0L; t1/2,elim = 0.65 hr; k=ln2/t1/2,elim = 0.693/0.65hr = 1.07 hr1
 Estimate C for t<T: calculate C at 2hr after exposure (t = 2hr)
 Estimate C at t=T: calculate CT at 4hr after exposure starts (t = T = 4hr)
 Estimate theoretical Css: Calculate Css if exposure lasted for a very long time (e.g., T>10 t1/2,elim)
 Estimate C for t>T: calculate C at 6hr after exposure starts (t=6hr), same as C at 2hr after exposure ends (tT = 6hr – 4hr = 2hr)
 Estimate t for given C if t<T: calculate t when C first rises to 0.5mg/L
 Estimate tT for given C if t>T: calculate tT when C falls back down to 0.5mg/L

 Rats are exposed to 100 ppm toluene for 4 hrs, after which the exposure to toluene stops. k0 = 1.4 mg/hr (from previous slide); T = 4hrs; V = 1.0L; t1/2,elim = 0.65 hr; k=ln2/t1/2,elim = 0.693/0.65hr = 1.07 hr1

 Estimate C at t=T: calculate CT at 4hr after exposure starts (t = T = 4hr)
CT = [k0/Vk][1ekt]
CT = [1.4mg/hr/(1.0L * 1.07hr1)][1e1.07hr1*4hr]
CT = 1.31mg/L * 10.0138
CT = 1.29 mg/L

 Rats are exposed to 100 ppm toluene for 4 hrs, after which the exposure to toluene stops. k0 = 1.4 mg/hr (from previous slide); T = 4hrs; V = 1.0L; t1/2,elim = 0.65 hr; k=ln2/t1/2,elim = 0.693/0.65hr = 1.07 hr1

 Estimate C for t<T: calculate C at 2hr after exposure (t = 2hr)
C = [k0/Vk][1ekt]
C = [1.4mg/hr/(1.0L * 1.07hr1)][1e1.07hr1*2hr]
C = 1.31mg/L * 10.1177
C = 1.16mg/L

 Rats are exposed to 100 ppm toluene for 4 hrs, after which the exposure to toluene stops. k0 = 1.4 mg/hr (from previous slide); T = 4hrs; V = 1.0L; t1/2,elim = 0.65 hr; k=ln2/t1/2,elim = 0.693/0.65hr = 1.07 hr1

 Estimate C at t=T: calculate CT at 4hr after exposure starts (t = T = 4hr)
CT = [k0/Vk][1ekt]
CT = [1.4mg/hr/(1.0L * 1.07hr1)][1e1.07hr1*4hr]
CT = 1.31mg/L * 10.0138
CT = 1.29 mg/L

 Rats are exposed to 100 ppm toluene for 4 hrs, after which the exposure to toluene stops. k0 = 1.4 mg/hr (from previous slide); T = 4hrs; V = 1.0L; t1/2,elim = 0.65 hr; k=ln2/t1/2,elim = 0.693/0.65hr = 1.07 hr1

 Estimate theoretical Css: Calculate Css if exposure lasted for a very long time (e.g., T>10 t1/2,elim)
Css = [k0/Vk]
Css = [1.4mg/hr/(1.0L * 1.07hr1)]
Css = 1.31mg/L

 Rats are exposed to 100 ppm toluene for 4 hrs, after which the exposure to toluene stops. k0 = 1.4 mg/hr (from previous slide); T = 4hrs; V = 1.0L; t1/2,elim = 0.65 hr; k=ln2/t1/2,elim = 0.693/0.65hr = 1.07 hr1

 Estimate C for t>T: calculate C at 6hr after exposure starts (t=6hr), [same as C at 2hr after exposure ends (tT = 6hr – 4hr = 2hr)]
C = CT * ek(tT)
C = 1.29mg/L * e1.07hr1(2hr)
C = 1.29mg/L * 0.1177
C = 0.152 mg/L
*

 Rats are exposed to 100 ppm toluene for 4 hrs, after which the exposure to toluene stops. k0 = 1.4 mg/hr (from previous slide); T = 4hrs; V = 1.0L; t1/2,elim = 0.65 hr; k=ln2/t1/2,elim = 0.693/0.65hr = 1.07 hr1

 Estimate t for given C if t<T: calculate t when C first rises to 0.5mg/L
C = [k0/Vk][1ekt]
0.5mg/L = 1.31mg/L * e1.07hr1(t)
(0.5mg/L)/(1.31.mg/L) = 0.3817 = 1 – e1.07hr1(t)
e1.07hr1(t) = 1 – 0.3817 = 0.6183
ln e1.07hr1(t) = ln0.6813
1.07hr1 * t = 0.4807 → 0.4807/1.07hr1 = 0.45 hr
or 0.45 hr after start of exposure
*

 Rats are exposed to 100 ppm toluene for 4 hrs, after which the exposure to toluene stops. k0 = 1.4 mg/hr (from previous slide); T = 4hrs; V = 1.0L; t1/2,elim = 0.65 hr; k=ln2/t1/2,elim = 0.693/0.65hr = 1.07 hr1

 Estimate C at t=T: calculate CT at 4hr after exposure starts (t = T = 4hr)
CT = [k0/Vk][1ekt]
CT = [1.4mg/hr/(1.0L * 1.07hr1)][1e1.07hr1*4hr]
CT = 1.31mg/L * 10.0138
CT = 1.29 mg/L

 Rats are exposed to 100 ppm toluene for 4 hrs, after which the exposure to toluene stops. k0 = 1.4 mg/hr (from previous slide); T = 4hrs; V = 1.0L; t1/2,elim = 0.65 hr; k=ln2/t1/2,elim = 0.693/0.65hr = 1.07 hr1

 Estimate tT for given C if t>T: calculate tT when C falls back down to 0.5mg/L
C = CT * ek(tT)
0.5 mg/L = (1.29 mg/L) * e(1.07/hr)(tT)
(0.5mg/L)/((1.29 mg/L) = 0.3876 = e(1.07/hr)(tT)
ln0.3876 = lne(1.07/hr)(tT)
1.07hr1 * (tT) = 0.9478
tT = (0.9478/1.07hr1) = 0.86hr or 0.86hr after exposure ends
*
Analysis of plasma concentration data (C vs t)
 Parameters from PostExposure Measurements (t≥T):
 for t>T: C = CT * e–k(tT)
 taking the natural logarithm of both sides gives:
lnC = ln CT – k(tT)
 Plot postexposure data in the form:
y = lnC vs x = tT
 To yield: slope = m = k and y = intercept = lnCT
*
*

 CT = [k0/Vk][1ekT]

 V = [k0/CT*k ][1ekT]
 CL = V * k
*
*
Example: Estimate toluene k, t1/2, CT, V and CL from rat concentrations following a 4hr exposure to 100ppm toluene (from earlier graph, data provided below). Previous example gave k0 ~ 1.4mg/hr
Time, t (hr)  Concentration, C (mg/L) 
0.00  0.000 
0.25  0.278 
0.50  0.537 
1.00  0.830 
2.00  1.114 
3.00  1.243 
4.00  1.277 
4.08  1.201 
4.50  0.722 
5.00  0.392 
5.50  0.261 
6.00  0.153 
*
How do we go about solving this problem?
 First, calculate (tT) and lnC for time points after exposure stopped (t≥4hrs)
Time, t (hr)  t – T (hr)  C (mg/L)  lnC 
4.00  0  1.277  0.245 
4.08  0.08  1.201  0.183 
4.50  0.50  0.722  0.326 
5.00  1.00  0.392  0.936 
5.50  1.50  0.261  1.343 
6.00  2.00  0.153  1.877 
 Next, plot lnC vs (tT)
 Determine useable points (points falling on linear portion of line)
*
 Perform linear regression analysis
m = 1.07hr1 → k = m = (1.07hr1) → k = 1.07 hr1
t1/2 = 0.693/k → 0.693/1.07hr1 → t1/2 = 0.65hr
b = 0.228 → CT = eb = e0.228 → CT = 1.26mg/L
V = [k0/(CT*k)][1ekT] = [(1.4mg/hr)/(1.26mg/L * 1.07hr1)][1 – e1.07hr1 * 4hr] → V = 1.02L
CL = V*k = 1.02L * 1.07hr1 = 1.10 L/hr (intercept method)
*
Parameters from TwoPost Exposure Plasma Samples (monitoring after accidental exposure)
Ct1 at t1, Ct2 at t2 with t1 and t2 ≥ T
Slope = (ln Ct1 – ln Ct2) / (t1 – t2)
= [ln(Ct1 / Ct2)]/(t1 – t2)
k = slope = [ln(Ct1 / Ct2)]/(t2 – t1)
Ct1 = [k0/V*k] [1ekT] ek(t1T)
V = [k0/(Ct1*k)][1ekT]ek(t1T)
Example: Use 4ht and 6hr measurements from toluene example.
Ct1 = 1.277 mg/L at t1 = 4.0hr
Ct2 = 0.153 mg/L at t2 = 6.0hr
k = [ln(1.277mg/L / 0.153mg/L)]/(6.0 – 4.0 hr) = 1.06 hr1
t1/2 = 0.693/1.06hr1 = 0.65hr
V = [(1.4mg/hr)/(1.277mg/L * 1.07hr1)][1e1.07hr1*4hr)][e 1.06hr1(4hr – 4hr)]
= 1.01 L
CL = V*k = 1.01 L * 1.06hr1 = 1.08 L/hr
*
Parameters from measurements during exposure period (t≤T)
 In order to perform this analysis, it is necessary to first estimate Css
 Analysis only works for data from long chemical exposure period that clearly reaches a steadystate plateau
 Css estimate is more accurate when 2 – 3 plateau points can be averaged
 For a chemical fitting a onecompartment model, concentrations during the exposure period are given by:
C = Css * (1ekt)
 This equation can be rearranged as follows:
C = Css – Css*e–kt → Css – C = Css * e–kt
 Then taking the natural logarithm of each side gives:
ln(Css – C) = lnCss – kt
 Which is now in a linear form with
Slope = m = k
Intercept = b = lnCss
*
ln(CSS – C)
t
lnCSS
slope = k
 then if one performs a linear regression with
y = ln(Css – C) and x = t
 the values of k and V can be determined by
k = m
Css = k0/(V*k) → V = k0/(Css * k)
 one should always plot the data to determine if the onecompartment model is appropriate and to identify unusable points
*
ln(CSS – C)
t
Late points that
jump around are
too close to C
ss
for
use in the analysis.
If early points do not
fall on line, may need
different model.
Example: Estimate toluene k, t1/2, V and CL from rat concentrations during 4hr exposure to 100ppm toluene (as previously estimated, k0 ~ 1.4 mg/hr)
Estimate the Css
Calculate ln(CssC) for each measurement
Plot ln(CssC) vs t
Determine useable points
Perform linear regression to calculate necessary variables
Time, t (hr)  Concentration, C (mg/L) 
0.00  0.000 
0.25  0.278 
0.50  0.537 
1.00  0.830 
2.00  1.114 
3.00  1.243 
4.00  1.277 
Example: Estimate toluene k, t1/2, V and CL from rat concentrations during 4hr exposure to 100ppm toluene (as previously estimated, k0 ~ 1.4 mg/hr)
Estimate the Css
Since concentration is still rising between last two timepoints can only use the last timepoint to estimate Css
Css ~ 1.277 mg/L
Time, t (hr)  Concentration, C (mg/L) 
0.00  0.000 
0.25  0.278 
0.50  0.537 
1.00  0.830 
2.00  1.114 
3.00  1.243 
4.00  1.277 
*
Example: Estimate toluene k, t1/2, V and CL from rat concentrations during 4hr exposure to 100ppm toluene (as previously estimated, k0 ~ 1.4 mg/hr)
Estimate the Css: Css~ 1.277 mg/L
Calculate ln(CssC) for each measurement
Time, t (hr)  Concentration, C (mg/L)  ln(Css – C) 
0.00  0.000  0.245 
0.25  0.278  0.001 
0.50  0.537  0.301 
1.00  0.830  0.805 
2.00  1.114  1.814 
3.00  1.243  3.381 
4.00  1.277 
Example: Estimate toluene k, t1/2, V and CL from rat concentrations during 4hr exposure to 100ppm toluene (as previously estimated, k0 ~ 1.4 mg/hr)
Estimate the Css
Calculate ln(CssC) for each measurement
Plot ln(CssC) vs t
Determine useable points
*all points are useable
*can use linear regression to describe data
*
Example: Estimate toluene k, t1/2, V and CL from rat concentrations during 4hr exposure to 100ppm toluene (as previously estimated, k0 ~ 1.4 mg/hr)
Perform linear regression to calculate necessary variables
m = 1.18hr1 → k = m = (1.18hr1) → k = 1.18hr1
t1/2 = 0.693/k → 0.693/1.18hr1 →t1/2 = 0.59hr
b = 0.315 → Css,intercept = eb = e0.315 → Css = 1.37 mg/L
V = k0/(Css * k) = (1.4 mg/hr)/[1.28mg/L * 1.18hr1) → V = 0.93L
CL = V*k = 0.93L * 1.18hr1 → CL = 1.10L/hr
AUC Calculations from C vs t Data

 Calculate AUC(t1 →tf) using the trapezoidal rule for consecutive points
 AUC(t1 →tf) = ½[Ct1 + Ct2][t1t2]
 Use C0 = 0 at t=0, even if C0 is not actually measured
 If CT is not measured, estimate it from postexposure regression
 Evaluate the AUC of the tail region just as for bolus IV injection
 AUCtail = AUC(tf →∞) = Ctf/k
*
AUC Calculations from C vs t Data
 As a shortcut method, a reasonable estimate of AUC during postexposure period is given by
 AUC(T →∞) ~ CT/k
 There is unfortunately no general shortcut method for the exposure period (t<T)
 If Css can be estimated, then total AUC ~ Css * T
*
Evaluation of V and CL from AUC (AUC Method)

 AUC = D/(V*k) = D/CL (always true for linear kinetics)

 Remember: D = k0 * T yields:
V = [k0 * T]/[AUC * k]
CL = [k0 * T]/AUC
Example: Determine AUC, V, and CL (using AUC Method) for 100ppm rat toluene exposure data examined earlier (use k = 1.07hr1 and k0 = 1.4mg/hr as determined previously)
 Make table showing t1, t2, Ct1, Ct2 (include t=0 and t=∞)
 Calculate AUC for each interval and for tail region (after tf)
 Sum AUC values to get running sum of AUC and total AUC
 Calculate V and CL by the AUC Method (compare to intercept method used previously)
Example: Determine AUC, V, and CL (using AUC Method) for 100ppm rat toluene exposure data examined earlier (use k = 1.07hr1 and k0 = 1.4mg/hr as determined previously)
 Make table showing t1, t2, Ct1, Ct2 (include t=0 and t=∞)
t1 (hr)  t2 (hr)  Ct1 (mg/L)  Ct2 (mg/L) 
0.00  0.25  0.000  0.278 
0.25  0.50  0.278  0.537 
0.50  1.00  0.537  0.830 
1.00  2.00  0.830  1.114 
2.00  3.00  1.114  1.243 
3.00  4.00  1.243  1.277 
4.00  4.08  1.277  1.201 
4.08  4.50  1.201  0.722 
4.50  5.00  0.722  0.392 
5.00  5.50  0.392  0.261 
5.50  6.00  0.261  0.153 
6.00  ∞  0.153 
Example: Determine AUC, V, and CL (using AUC Method) for 100ppm rat toluene exposure data examined earlier (use k = 1.07hr1 and k0 = 1.4mg/hr as determined previously)

 Make table showing t1, t2, Ct1, Ct2 (include t=0 and t=∞)
 Calculate AUC for each interval and for tail region (after tf)
AUC(0→0.25hr) = ½[0 + 0.278mg/L][0.25 – 0 hr] = 0.035 mg hr/L
AUCtail = (0.153mg/L)/(1.07hr1) = 0.143 mg hr/L
3. Sum AUC values to get running sum of AUC and Total AUC
t1 (hr)  t2 (hr)  Ct1 (mg/L)  Ct2 (mg/L)  AUC(t1→t2) (mg hr/L)  AUC run SUM (mg hr/L) 
0.00  0.25  0.000  0.278  0.035  0.035 
0.25  0.50  0.278  0.537  0.102  0.137 
0.50  1.00  0.537  0.830  0.342  0.478 
1.00  2.00  0.830  1.114  0.972  1.450 
2.00  3.00  1.114  1.243  1.178  2.629 
3.00  4.00  1.243  1.277  1.260  3.889 
4.00  4.08  1.277  1.201  0.099  3.988 
4.08  4.50  1.201  0.722  0.404  4.392 
4.50  5.00  0.722  0.392  0.278  4.670 
5.00  5.50  0.392  0.261  0.163  4.834 
5.50  6.00  0.261  0.153  0.104  4.937 
6.00  ∞  0.153  0.143  5.080 
As a comparison, using the shortcut method for the postexposure period gives:
AUC(T→∞) ~ CT/k = (1.277mg/L)/(1.07hr1) = 1.193 mg hr/L
AUC = AUC(0→T) + AUC(T→∞)
= 3.889 mg hr/L + 1.193 mg hr/L
= 5.082 mg hr/L
4. Calculate V and CL by the AUC Method (compare with intercept method)
V = [k0 * T] / [AUC * k]
= [1.4mg/hr * 4hr] / [5.080mg hr/L * 1.07hr1]
= 1.03 L (compared to 1.02 L)
CL = [k0 * T] / AUC
= [1.4 mg/hr * 4hr] / 5.080 mg hr/L
= 1.10 L (compared to 1.10 L)
*
Chemical
Absorption
Rate
Time
0
Instantaneous
Absorption
ZeroOrder
Absorption
T
C
k
V
k
0
Short Exposure
T
C
Time
Prolonged Exposure
T
C
Time
C
T
C
T
Short Infusion
T
lnC
Time
Prolonged Infusion
T
lnC
Time
lnC
T
lnC
T
slope = k
slope = k
T
C
Time
Large
k
0
Small
k
0
T
lnC
Time
Large
k
0
Small
k
0
Slope not
affected
by k
0
T
C
Time
Small
V
Large
V
T
lnC
Time
Small
V
Large
V
Slope not
affected
by V
T
C
Time
Small t
1/2
Large k
Large t
1/2
Small k
T
lnC
Time
Slope, C
�
,
and approach
to plateau all
affected
Large t
1/2
Small k
Small t
1/2
Large k
0.0
1.0
2.0
3.0
4.0
5.0
6.0
0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
Toluene Concentration in Plasma (mg/L)
Time (hr)
100 ppm Toluene
200 ppm Toluene
0.1
1.0
10.0
0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
Toluene Concentration in Plasma (mg/L)
Time (hr)
100 ppm
Toluene
200 ppm
Toluene
lnC
Time after Chemical Input Stopped (t – T)
b = lnC
T
m = k
Late points may jump
around near limit of
drug assay, not reliable
=> do not use
lnC
t – T
Early high points suggest need
for more than one compartment
=> we will consider later
Useable
Points
2.0
1.5
1.0
0.5
0.0
0.5
0.0
0.5
1.0
1.5
2.0
2.5
y = 0.22844 + 1.0674x R= 0.99808
lnC
Time after Exposure Ended, t – T (hr)
C
t
If possible, average multiple
plateau points to estimate C
ss
in order to improve accuracy.
l
n
(
C
SS
–
C)
t
l
nC
SS
s
lope
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k
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(
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SS
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3.5
3.0
2.5
2.0
1.5
1.0
0.5
0.0
0.5
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
y = 0.31522 + 1.1777x R= 0.99534
ln(C
ss
C)
Time (hr)