Zero Order Absorption

Types of chemical administration or exposure that provide (or approximate) zero-order 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 * e-kt)

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

A = (k0/k)(1-e-kt)

- 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] [1-e-kt] 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 * e-k(t-T) = (k0/k)(1- e-kT)e-k(t-T)

C = CT * e-k(t-T) = (k0/V*k)(1- e-kT)e-k(t-T)

CT = (k0/V*k)(1- e-kT)

31.bin

Steady-State 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)(1-e-kt) = Css(1-e-kt) for t≤T

Concentration drops exponentially immediately after exposure stops, so after infusion (t>T):

C = CT * e-k(t-T)

taking the natural logarithm of both sides gives:

lnC = lnCT -k(t-T)

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 Half-life (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 half-life 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 in-Cvein]

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 hr-1

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 (t-T = 6hr – 4hr = 2hr)
Estimate t for given C if t<T: calculate t when C first rises to 0.5mg/L
Estimate t-T for given C if t>T: calculate t-T 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 hr-1

- Estimate C at t=T: calculate CT at 4hr after exposure starts (t = T = 4hr)

CT = [k0/Vk][1-e-kt]

CT = [1.4mg/hr/(1.0L * 1.07hr-1)][1-e-1.07hr-1*4hr]

CT = 1.31mg/L * 1-0.0138

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 hr-1

- Estimate C for t<T: calculate C at 2hr after exposure (t = 2hr)

C = [k0/Vk][1-e-kt]

C = [1.4mg/hr/(1.0L * 1.07hr-1)][1-e-1.07hr-1*2hr]

C = 1.31mg/L * 1-0.1177

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 hr-1

- Estimate C at t=T: calculate CT at 4hr after exposure starts (t = T = 4hr)

CT = [k0/Vk][1-e-kt]

CT = [1.4mg/hr/(1.0L * 1.07hr-1)][1-e-1.07hr-1*4hr]

CT = 1.31mg/L * 1-0.0138

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 hr-1

- 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 * 1.07hr-1)]

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 hr-1

- Estimate C for t>T: calculate C at 6hr after exposure starts (t=6hr), [same as C at 2hr after exposure ends (t-T = 6hr – 4hr = 2hr)]

C = CT * e-k(t-T)

C = 1.29mg/L * e-1.07hr-1(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 hr-1

- Estimate t for given C if t<T: calculate t when C first rises to 0.5mg/L

C = [k0/Vk][1-e-kt]

0.5mg/L = 1.31mg/L * e-1.07hr-1(t)

(0.5mg/L)/(1.31.mg/L) = 0.3817 = 1 – e-1.07hr-1(t)

e-1.07hr-1(t) = 1 – 0.3817 = 0.6183

ln e-1.07hr-1(t) = ln0.6813

-1.07hr-1 * t = -0.4807 → -0.4807/-1.07hr-1 = 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 hr-1

- Estimate C at t=T: calculate CT at 4hr after exposure starts (t = T = 4hr)

CT = [k0/Vk][1-e-kt]

CT = [1.4mg/hr/(1.0L * 1.07hr-1)][1-e-1.07hr-1*4hr]

CT = 1.31mg/L * 1-0.0138

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 hr-1

- Estimate t-T for given C if t>T: calculate t-T when C falls back down to 0.5mg/L

C = CT * e-k(t-T)

0.5 mg/L = (1.29 mg/L) * e-(1.07/hr)(t-T)

(0.5mg/L)/((1.29 mg/L) = 0.3876 = e-(1.07/hr)(t-T)

ln0.3876 = lne-(1.07/hr)(t-T)

-1.07hr-1 * (t-T) = -0.9478

t-T = (-0.9478/-1.07hr-1) = 0.86hr or 0.86hr after exposure ends

Analysis of plasma concentration data (C vs t)

Parameters from Post-Exposure Measurements (t≥T):
for t>T: C = CT * e–k(t-T)
taking the natural logarithm of both sides gives:

lnC = ln CT – k(t-T)

Plot post-exposure data in the form:

y = lnC vs x = t-T

To yield: slope = m = -k and y = intercept = lnCT

- CT = [k0/Vk][1-e-kT]

- V = [k0/CT*k ][1-e-kT]

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 (t-T) 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 (t-T)
Determine useable points (points falling on linear portion of line)

Perform linear regression analysis

m = -1.07hr-1 → k = -m = -(-1.07hr-1) → k = 1.07 hr-1

t1/2 = 0.693/k → 0.693/1.07hr-1 → t1/2 = 0.65hr

b = 0.228 → CT = eb = e0.228 → CT = 1.26mg/L

V = [k0/(CT*k)][1-e-kT] = [(1.4mg/hr)/(1.26mg/L * 1.07hr-1)][1 – e-1.07hr-1 * 4hr] → V = 1.02L

CL = V*k = 1.02L * 1.07hr-1 = 1.10 L/hr (intercept method)

Parameters from Two-Post 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] [1-e-kT] e-k(t1-T)

V = [k0/(Ct1*k)][1-e-kT]e-k(t1-T)

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 hr-1

t1/2 = 0.693/1.06hr-1 = 0.65hr

V = [(1.4mg/hr)/(1.277mg/L * 1.07hr-1)][1-e-1.07hr-1*4hr)][e -1.06hr-1(4hr – 4hr)]

= 1.01 L

CL = V*k = 1.01 L * 1.06hr-1 = 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 steady-state plateau
Css estimate is more accurate when 2 – 3 plateau points can be averaged

For a chemical fitting a one-compartment model, concentrations during the exposure period are given by:

C = Css * (1-e-kt)

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)

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 one-compartment model is appropriate and to identify unusable points

ln(CSS – C)

Late points that

jump around are

too close to C

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(Css-C) for each measurement

Plot ln(Css-C) 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(Css-C) 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(Css-C) for each measurement

Plot ln(Css-C) 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 = -1.18hr-1 → k = -m = -(-1.18hr-1) → k = 1.18hr-1

t1/2 = 0.693/k → 0.693/1.18hr-1 →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 * 1.18hr-1) → V = 0.93L

CL = V*k = 0.93L * 1.18hr-1 → 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][t1-t2]
- Use C0 = 0 at t=0, even if C0 is not actually measured
- If CT is not measured, estimate it from post-exposure 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 short-cut method, a reasonable estimate of AUC during post-exposure period is given by
AUC(T →∞) ~ CT/k
There is unfortunately no general short-cut 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 = 1.07hr-1 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 = 1.07hr-1 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 = 1.07hr-1 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)/(1.07hr-1) = 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 short-cut method for the post-exposure period gives:

AUC(T→∞) ~ CT/k = (1.277mg/L)/(1.07hr-1) = 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 * 1.07hr-1]

= 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

Instantaneous

Absorption

Zero-Order

Absorption

Short Exposure

Time

Prolonged Exposure

Time

Short Infusion

lnC

Time

Prolonged Infusion

lnC

Time

lnC

slope = -k

Time

Large

Small

lnC

Time

Large

Small

Slope not

affected

by k

Time

Small

Large

lnC

Time

Small

Large

Slope not

affected

by V

Time

Small t

1/2

Large k

Large t

1/2

Small k

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

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)

If possible, average multiple

plateau points to estimate C

in order to improve accuracy.

(

–

lope

–

(

–

ate

poi

s t

d a

e to

ana

f ea

poi

s d

o no

ll o

eed

ffe

ode

-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

-C)

Time (hr)

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