Chapter 2

Background Mathematical Material

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Graphing Data on Linear and Semi-Log Graph Paper

A very important part of pharmacokinetic analysis is the ability to graph data and interpret the resulting graphs. The graph can provide a useful picture of the data and provide an insight into the underlying mathematical model. Graphs should be drawn carefully. Straight lines should be straight --- use a ruler!!

Pharmacokinetic data involves models with rate processes. The most simple of these would be a single first order decline described with a single exponential term. We may use an equation such as:

Cp = 40 • exp(- 0.23 • t)

Equation 2.4.1 Concentration (Cp) versus time

to describe drug concentrations after an IV bolus dose. These data are drawn as three graphs on this page. The first graph is a linear (or Cartesian) plot of the data versus time. Notice the smooth decline in concentration with time.

Linear plot of Cp <i>versus</i> time

Figure 2.4.1 Linear Plot of Cp versus Time

Click on the figure to view the interactive graph

The second graph is a plot of the natural log (ln) of the concentration values versus time. Now we get a straight line graph. You might try 'rearranging' the equation above to verify that a straight line is to be expected.

Linear plot of Cp <i>versus</i> time

Figure 2.4.2 Linear Plot of ln(Cp) versus Time

Click on the figure to view the interactive graph

The third graph is on different paper. This is semi-log graph paper. On this graph paper the scale on the y-axis is proportional to the log of the number not the number itself, like the scale on the slide rule on the previous page. Notice that the distance from 1 to 2 is the same as the distance from 2 to 4 or from 4 to 8. Again, we have a straight line but without the need to calculate the natural log of each number.

Linear plot of Cp <i>versus</i> time

Figure 2.4.3 Semi-log Plot of Cp versus Time

Click on the figure to view the interactive graph


For many calculations involving rate constants you will be determining the slope of these lines. The second and third graphs (Figure 2.4.2 and 2.4.3) are a lot easier to use in terms of calculating slopes than Figure 2.4.1, since they each represent a straight line. We will revisit the equation for this line later. Remember semi-log graph paper has a normal x- axis scaling but the y- axis scaling is proportional to the log of the number not the number itself. It saves you from taking the log of each number before you plot it. Note: Once you take the log of a number you loose the units.

Example use of semi-log graph paper: Plot the data, draw a line "through the data", and calculate the slope of the line.

Table 2.4.1 Example Cp versus Time Data
Time (hr) 1 2 4 8 12
Cp (mg/L) 20 15 6.8 3.2 1.3

Figure 2.4.4 Semi-log Plot of Cp versus Time

A best-fit line is drawn through the data points and extended to the extremes of the graph paper (for better accuracy). Values for Cp1 and Cp2 are read from the y-axis and t1 and t2 values are read from the x-axis. These values can be used to estimate a value for the slope of the line and also the rate constant for the drug elimination. From the Y-axis the first point is 0, 22.8 and from the X-axis the second point is 12.8, 1 (in the format x-value, y-value). Always try to use the extremes of the line for the best accuracy.

From Figure 2.4.4 Value
Cp1 22.8
Cp2 1.0
t1 0
t2 12.8

The slope can be calculated using the equation:

Equation for semi-log slope using log

Equation 2.4.2 Slope of the line on a semi-log plot

This equation uses logarithms with base 10 for the calculation of slope.

Thus the slope of the line in Figure 2.4.4 is: [log(1.0) - log(22.8)]/(12.8 - 0) = (0.000 - 1.358)/12.8 = - 0.106 hr-1

In the study of pharmacokinetics first order rate processes and rate constants are common. When the rate constant, k, is calculated from the slope drawn on a semi-log plot, k is found to equal to -slope • 2.303 [Note ln(10) = 2.303]. Thus, the rate constant is calculated as -slope x 2.303. The calculation of k is easier with calculators (etc.) if you use ln (logarithm with base 'e') in the calculation. At the same time you can change the sign by calculating the numerator as ln(Cp1) - ln(Cp2). Thus k can be calculated using the equation:

Equation for k from slope using ln

Equation 2.4.3 Equation for estimating First Order Rate Constants

Thus k = [ln(22.8) - ln(1.0)]/(12.8 - 0) = [3.127 - 0.000]/12.8 = 3.127/12.8 = 0.244 hr-1

We can perform this calculation using a different approach. Remember when I was talking about using logarithms to do multiplication and division. We can use that approach to re-arrange Equation 2.4.3. Therefore instead of subtracting the two logarithms we can divide one by the other and take the log of the quotient.

Equation 2.4.4 Equation for estimating k

Thus k = ln(22.8/1.0)/(12.8 - 0) = ln(22.8)/12.8 = 3.127/12.8 = 0.244 hr-1

Although I prefer to use the approach in Equation 2.4.3 the advantage of the approach in Equation 2.4.3 is that you don't have to deal with a double negative when Cp2 is less than one and you only need to take one ln value. For example, if Cp1 and Cp2 were 12.5 and 0.13 at 0 and 12 hours, respectively. Using Equation 2.4.2 gives k = [ln(12.5) - ln(0.13)]/12 = [2.526 - -2.040]/12 = [2.526 + 2.040]/12 = 4.566/12 = 0.381 hr-1. Using Equation 2.4.3 gives k = ln(12.5/0.13)/12 = ln(96.15)/12 = 4.566/12 = 0.381 hr-1. Note, each approach gives the same answer.


Drawing a best-fit line through the Data

Drawing a line through the data doesn't mean through just two data points but through all the data points. Be especially careful about picking two adjacent data points. Sometimes the first and last point can work but the last point, the lowest concentration data point will probably be inaccurate. The best approach is to put the line through all the data. There should be points above and below the line. Maybe an equal number above and below. Drawing this 'best-fit' line is like averaging the data. That is why it is important to use points from the (extremes of the) line to calculate the slope. Any individual data point may have some assay or other error associated with it. It you look at adjacent pairs of data on a graph you will notice that a line drawn through these pairs of data can have quite different slopes. Drawing the 'best-fit' line averages all the data to give the best value for the slope and intercept. Don't ignore that line after you have drawn it.

Click the arrow to play the movie


Practice and Assignments

Two exercises asking you to draw straight lines through data on both linear and on semi-log graph paper.

Linear Graph
Semi-log Graph

Try them out if you need practice graphing data and calculating intercept and slope in these graphs.


Graph Paper Resources

Graph Paper in PDF Format. If you have Adobe Reader or Preview (in Mac OS X) installed clicking on the links below will provide the graph paper indicated.

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