Case Study #5

IV and Oral Dosing of Theophylline

Answers provided by

Jeff Stark, Graduate Student

_{A patient is admitted to the clinic in status asthmaticus. Theophylline is administered i.v. and later on an oral dosing regimen is begun. Two oral dosage forms are considered. The first is a fast release tablet and the second is a slow release form. The slow release form may minimize the fluctuation in Cp and allow for less frequent dosing (improving patient compliance). Blood samples are drawn and Cp of theophylline is determined for the first dose of each of the dosage forms.}

_{Useful information:}

_{Therapeutic range is 10 –20 mg/L}

_{Half-life commonly 4 –5 hr}

_{Vd = 0.5 L/Kg}

_{Patient weighs 62.5 Kg}

_{Dose is 400 mg theophylline for each form}

_{1) IV bolus}

_{Cp=11.5 mg/L at 1hr and 6.7 mg/L at 6hr}

_{Calculate ke, t1/2, Vd, and Cl}

_{2) Fast release tablet}

_{Cp=9.35 mg/L at 3hr and 6.79 mg/L at 6hr}

_{Calculate the terminal slope. Does this represent ke or ka? If a complete Cp-t profile were available, outline how you could calculate the other rate constant.}

_{3) Slow release tablet}

_{Cp=5.99 mg/L at 6hr and 4.99 mg/L at 12 hr}

_{Calculate the terminal slope. What rate constant does this represent?}

_{Calculate the steady-state peak and trough levels for the IV and slow release dosing regimens. The dosing interval is 8hr and assume the bioavailability for the oral form is F=1. Compare the results.}

Answers

General: 400 mg dose for each dosage form

The predicted V_d is V_d = (0.5 L/kg)(62.5 kg) = 31.25 L

1) i.v. bolus

For an i.v. bolus dose, it is possible to determine k_e directly from the graph:

Then

This half life is determined from actual data is slightly above the population average. Before we can determine the clearance for this patient, we must find V_d. V_d is predicted to be 31.25 L. However, we should determine V_d for this patient using the y-intercept from the graph above (lnCp vs t).

and so

lnCp(0) is read from the graph and converted to Cp(0).

Here, Cp(0) = 12.81 mg/L

For this patient V_d = 31.22 L (surprisingly close to the predicted value). Clearance is then

Cl = k_e   V_d

= (0.108 hr^-1)(31.22 L) = 3.372 L/hr

2) Oral dose (fast release).

The equation for the Cp vs t curve after oral dosing is:

This is more complex since we must now consider absorption and the fraction of dose being absorbed.

Once drug enters the body, elimination begins. Thus, determining the elimination rate constant (k_e) and the absorption rate constant (k_a) is not that easy. However, in "normal" kinetics k_a>>k_e, i.e. the rate of absorption is much larger than the rate of elimination. If all of the drug is absorbed in, say, the 30 min time period just after administration, then at latter time points, only elimination is taking place. In this case, the slope of the terminal phase (start becoming familiar with this terminology) represents k_e.

m = -k_e if k_a>>k_e

For a fast release tablet, we might assume that k_a is very large. Remember, however, that k_a also depends on the ability of the drug to cross membranes and not just on the release rate from the tablet. We can use the data points given to calculate the terminal slope:

This value is very close to the k_e determined for the i.v. bolus dose. Thus, this must be an example of normal kinetics and k_a >> k_e for this dosage form and the terminal slope represents k_e. The absorption rate constant k_a may be determined by the method of residuals ("feather"). See below

3) Oral dose (slow release)

For a slow release tablet, k_a is predicted to be smaller than that for the fast release. How does this affect the terminal slope?

- Since drug cannot be eliminated until it is absorbed, the decline of the terminal slope may depend greatly on how fast absorption is taking place. If k_a << k_e, absorption is slower than the elimination rate. In such a case, the terminal slope reflects k_a rather than k_e and this is an example of "flip-flop" kinetics.

Using the data points given, the terminal slope is found to be 0.0304 hr^-1. We already know k_e = 0.108 hr^-1 from the i.v. data. So, the terminal slope for the slow release tablet reflects the absorption rate and k_a = 0.0304 hr^-1.

Regardless of whether the terminal slope represents k_a or k_e, the other rate constant, k_e or k_a, respectively, may be determined from the data. The method for this is termed the method of residuals or "feathering", which is a mathematical way of separating the two constants.

IV bolus: Steady state levels.

The general equation for plasma concentrations at steady state after multiple IV bolus doses is

where t is some time within the dosing interval, 0< t < tau . The maximum concentration occurs just after the dose at the state of the dosing interval. So, t = 0. This gives:

The minimum concentration occurs at the end of the dosing interval just before the next dose, t = tau.

Oral dose (slow release): Steady state levels

The general equation for concentrations following oral doses at steady state is:

where t is some time point within the dosing interval, < t < tau . Unlike the IV bolus case, the maximum concentration does not occur at t=0 but rather at t_max . The t_max at steady state is

=3.62 hours

It is now possible to calculate Cp_ss(max) following the oral doses.

The minimum concentration occurs at the end of the dosing interval (t = tau) just before the next dose.

=14.57 mg/L

The average concentration at steady state is not dependent on k_a in the oral dosing regimen. The only change for the oral dosing calculations is the inclusion of the bioavailability.

If F = 1, the average steady state levels for IV bolus and oral doses are the same. In this example,

Method of Residuals

Consider the Cp-time profile for a drug with "normal" kinetics. Since k_a >> k_e, we know that absorption takes place rapidly and the terminal phase reflects k_e, the elimination rate constant.

Although we can determine k_e directly from the graph, we must manipulate the data in some way in order to determine k_a. (One drug is absorbed, elimination begins. Thus, absorption and elimination occur simultaneously until all drug is absorbed. After this time only elimination occurs). The method of residuals is a way of separating k_a from k_e mathematically.

1) Extend the terminal slope back to y-axis and call this line Cp'

2) Determine values for Cp' (from this graph) at several time points. Then calculate values for (Cp' - Cp).

3) Plot (Cp' - C) on semi-log paper and draw a line through them. If not using semi-log paper, graph ln(C'p - cp) vs t.

The slope of this line reflects k_a (if k_a >> k_e)

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