Question

1. An antibiotic has a variable half-life from 2 to 6 h in the general population. A patient was given an IV infusion at a rate of 15 mg/h. Blood samples were taken at 8 and 48 h. Plasma concentrations were 5.5 and 6.5 mg/L respectively. Estimate the half-life of the drug in this patient.

Answer 1

Solution

The equation to be used =

log(Css -  Cp ) / Css = - kt / 2.3 ; k = -2.3/t X  log (Css -  Cp ) / Css

Where Cp is the plasma drug concentration taken at time t; Css is the approximate steady-state plasma drug concentration in the patient.

Because the second plasma sample was taken at 24 hours, or 24/6 = 4 half-lives after infusion, the plasma drug concentration in this sample is approaching 95% of the true plasma steady-state drug concentration assuming the extreme case of t 1/2 = 6 hours. By substitution into Equation ,
   log ( 6.5 -  5.5) =k(8)
6.5 2.3

k=0.234 hr-1   t 1/2 = 0.693 / 0.231 = 2.96 hr

The elimination half-life calculated in this manner is not as accurate as the calculation of t 1/2 using multiple plasma drug concentration time points after a single IV bolus dose or after stopping the IV infusion.

However, this method may be sufficient in clinical practice. As the second blood sample is taken closer to the time for steady state, the accuracy of this method improves.

At the 30th hour, for example, the plasma concentration would be 99% of the true steady-state value (corresponding to 30/6 or 5 elimination half-lives), and less error would result in applying Equation .

When Equation was used as in the example above to calculate the drug t 1/2 of the patient, the second plasma drug concentration was assumed to be the theoretical . As demonstrated below, when t 1/2 and the corresponding values are substituted,

log(Css - 5.5 ) / Css = - 0.231 X 8 / 2.3

log(Css - 5.5 ) / Css = 0.157

Css = 6.5 mg/L

Answer 2

To estimate the half-life of the drug in this patient, we can use the formula for exponential decay:

C(t) = C(0) * e^(-kt)

Where: C(t) is the concentration at time t, C(0) is the initial concentration, k is the rate constant, and t is the time.

Given: C(0) = 5.5 mg/L (at 8 hours), C(t) = 6.5 mg/L (at 48 hours), and The infusion rate is 15 mg/h.

We can rearrange the equation to solve for the rate constant (k):

k = (-1/t) * ln(C(t)/C(0))

Now we can calculate the rate constant:

k = (-1/40) * ln(6.5/5.5) ≈ -0.0172 h^(-1)

The half-life (t1/2) can be calculated using the following formula:

t1/2 = ln(2) / k

Now we can calculate the half-life:

t1/2 = ln(2) / (-0.0172) ≈ 40.3 hours

Therefore, the estimated half-life of the drug in this patient is approximately 40.3 hours.