Pharmacokinetics – multi-compartment models – method of residuals, intercepts and slopes

This is an example of using the method of residuals to calculate the values for intercepts A and B and slopes a, b, k, k12, and k21.

Data

 

Time (hr) and Plasma concentration (Cp; ug/mL)

Time (hr)	Cp (ug/mL)
0	70
0.25	53.8
0.5	43.3
0.75	35
1.0	29.1
1.50	21.2
2	17
2.5	14.3
3	12.6
4	10.5
5	9
6	8
7	7

 

 

A drug can be delivered via intravenous injection to an average sized person (i.e., 70 kg). Blood samples can be withdrawn from 0-7 hr (as is shown from the data above). An assay such as an ELISA or HPLC can be conducted to quantify the levels of drug within the blood. The objective here is to use the method of residuals (aka “peeling” and “feathering”) to calculate the values of A and B and slopes a, b, k, k12, and k21. You can readily do this via MatLab using cftool as well, but we will be using a different method to obtain the bioexponential fit, called the method of residuals here.

 

Example of a curve that has 2 slopes that would fit well with Cp(t) = A*exp(-k1*t)+B*exp(-k2*t):

http://www.bishopkingdom.com/aichatbox/id/568/

 

Note that the above figure is plotted on a semi-logarithmic scale (y-axis is logarithmic and x-axis is linear). Because there are 2 slopes, this means that the drug is distributed in two compartments. The first phase is the distribution phase and the second phase is the elimination phase.The distribution phase is generally faster than the elimination phase, but this is not necessarily always the case. a is the slope of the distribution phase and it is larger than b, which is the slope of the elimination phase. The half-life for each of the phases can be calculated as t1/2 = 0.693/a and 0.693/b. You can then extrapolate what the y-intercept is for each of these phases. The y-intercept for the distribution phase, with slope a, is A. The y-intercept for the distribution phase, with slope b, is B. In this sense: Cp(t) = A*exp(-a*t)+B*exp(-b*t).

 

The residual plasma concentration can be calculated from each of the rows given in the table above. The residual plasma concentration is Cp-Cp’. Cp’ can be calculated from the rows. Note that if you have x number of rows of data, you will then have x-1 rows of data for the residual plasma concentrations. The Cp-Cp’ is then plotted against time which will represent the distribution a and b phases, respectively, which are the slopes (slope of phase a is a and slope of phase b is b). Note that these latter 2 lines (Cp-Cp’ versus time) will be linear. The intercept of these lines are A and B as well.

Now that we know a, b, and A, and B, we can then calculate what k (overall elimination constant), and k12 (phase a), and K21 (phase b).

k = ab(A+B)/(Ab+Ba)

k12 = AB(b-a)^2/((A+B)(Ab+Ba))

k21 = Ab+Ba/(A+B)

 

Now, let’s apply this to the data above. In order to do so, you know that the later time points will not involve Ae^(-at) because if you drew a linear fit of the earlier time points which is also the distribution phase, then the line quickly drops to zero. For example, with the data above:

 

With this in mind, we can estimate what b and B are, such that we know what Cp(t)=Be^(-bt) is. For example, the dotted gray line appears to hit the y-axis at t=0 close to 17, so let’s use 17 for B.

Then we can plug in Cp(t=any value we know)=17e^(-bt) and solve for b. For example:

Cp(t=7hr)=7 ug/mL = 17ug/mL*e^(-b*7 hr), thus [ln(7 ug/mL  /   17 ug/mL]/-7 =b =0.12676 hr^-1.

Now we know that the elimination phase is: 17ug/mL*e^(-0.12676*t). The fitted line equation for the elimination phase is -1.57x+17.

We can now actually do the method of residuals with our table now.Note that the residual Note, however, that in the Residual amount column we shift the row down by one so that the time points are off set. Some books will refer to this as Cp-Cp’ but I prefer thinking about it in this fashion.

 

Time (hr) and Plasma concentration (Cp; ug/mL) with Residual Information

Time (hr)	Cp (ug/mL)	Elimination phase: linear regression: Cp_elimination=-1.57t+17	Residual amount: Cp-Cp_elimination
0	70	17
0.25	53.8	16.6	53
0.5	43.3	16.2	37.2
0.75	35	15.8	27.1
1.0	29.1	15.4	19.2
1.50	21.2	14.6	13.7
2	17	13.9	6.6
2.5	14.3	13.1	3.1
3	12.6	12.3	1.2
4	10.5	10.7
5	9	9.2
6	8	7.6
7	7	6.0

 

The table does not have to be completed for the elimination phase because we calculated the distribution phase with already making a good assumption that the distribution phase is not affecting the later time points.

Now we have a new linear fitted line for the elimination phase using the table above:

 

As you may see, the elimination phase’s y-intercept at time = 0 hr is A which is close to 62. Similar to how we did before we can then choose a time > 0 within the elimination phase (i.e., 1 hr) to calculate a:

Cp(t=1 hr)=62 ug/mL*e^(-a*1 hr)

ln(1/62)/-1=a = 4.13 hr^-1

 

 

We now know that for this data:

Cp(t)=62 *e^(-4.13t)+17e^(-0.13*t)

After you know these values, you can plug in any time, t, and know what the estimated Cp value is.