Homework 1 (PHA 5127)

Transcription

1 Homework 1 (PHA 5127) 1. The elimination rate constant of a drug is 1 hr -1 : k e =1 hr -1 A. Half-life: t 1/2 = ln(2)/k e = 0.693/1 hr -1 = hr B. C 1 =5ng/ml First-order elimination: k e = (ln(c 1 )-ln(c 2 ))/(t 2 -t 1 ) ln(c 1 )-ln(c 2 )=k e (t 2 -t 1 ) ln(c 1 /C 2 )=k e (t 2 -t 1 ) C 1 /C 2 =e ke (t2-t1) C 2 =C 1 /e ke (t2-t1) =(5ng/ml)/(e 1 (2-1) )=(5ng/ml)/e 1 =(5ng/ml)/2.72=1.84ng/ml C. Zero-order elimination: k e =C 0 /2t 1/2 C 2 = C 0 -k e t 2 We have to know the initial plasma concentration C The drug is eliminated faster in patient A, than in patient B, thus, t 1/2 <T 1/2 or k e >K e. This can be due to a different volume of distribution or different clearance knowing the relationship Cl=k e V d. In other words, the elimination rate can be different because or the clearance or the volume of distribution, or both are different. A. Assuming that the same doses were given, if the initial plasma concentrations are the same in both patients (C 01 =C 02 ), the volumes of distributions have to be the same (V d1 =V d2 ), but the clearances have to be different. B. Assuming again that the same doses were given, if the initial concentrations are different (C 01 C 02 ), the volumes of distributions have to be different (V d1 V d2 ). Patient B has a larger volume of distribution, thus, D:\pha5127_Dose_Opt_I\Homeworks\Homework1\Fall-00\ans-hmwk1.doc 1

2 the elimination of the drug will take more time. Besides, its clearance can be different as well. 3. Time (hr) Cp (ng/ml) Plotting the data: This log of plasma drug concentration versus time profile shows a biphasic curve, which can be described by a two-compartment model. Many drugs are D:\pha5127_Dose_Opt_I\Homeworks\Homework1\Fall-00\ans-hmwk1.doc 2

3 characterized by a two-compartment model, which consists of a central compartment and a peripheral compartment. The central compartment includes the organs with high blood flow, while the peripheral compartment includes the organs with less blood flow. In case of a two-compartment model the drug distribution is not as fast as in case of a one-compartment model when we assume, that the body behaves as a big tank. In case of a two-compartment model the distribution takes some time. This distribution is described by the initial portion of the plot (in our case from 0 to 7 hrs) and is called the distribution phase. For many drugs the distribution phase is very short, and so the central compartment can be ignored in which case a one-compartment model can adequately represent the plasma concentrations observed. In our case the distribution phase is not short at all, so we can not neglect the distribution phase. You were not introduced to this type of model called a two-compartment model, so obviously you questioned how should you calculate the slope that varies, depending on which portion of the curve is examined. In a two-compartment model the slope of the initial portion of the curve is related to the rate of distribution, and only the slope of the terminal portion of the curve describes the elimination rate. Thus, to calculate the half-life of the drug the slope of the terminal portion of the curve has to be calculated. Because you were not told this we accept the value of the rate constant as long as you calculated as the slope of the curve. Some of the students calculated the slope of the line traced over all the points, some calculated from just two different time points. We accept all these answers till then we see you know what you did, but please note that in case of a two-compartment model to determine the elimination rate constant the slope of the terminal portion of the curve has to be calculated. To answer the questions in problem #3 we consider just the data points from 7 hours in order to have a monophasic curve, a one-compartment model: Time (hr) Cp (ng/ml) D:\pha5127_Dose_Opt_I\Homeworks\Homework1\Fall-00\ans-hmwk1.doc 3

4 A. The elimination rate constant: K e =(ln(c 7 )-ln(c 30 ))/(t 30 -t 7 )=(ln(59)-ln(10))/(30-7)=ln(59/10)/23=0.077 hr -1 B. The half-life: t 1/2 = ln(2)/k e = 0.693/0.077 hr -1 = 9 hr C. The initial plasma concentration: Determined by extrapolation: C o =97.38 ng/ml D. The volume of distribution: V d =D/C o = 5 mg / ng/ml = 5000 µg / µg/l= L E. AUC by trapezoidal rule: AUC(0-7hr)=(C o +C 7 )(t 7-0)/2=( )*7/2 µg/l/hr= 547 µg/l/hr D:\pha5127_Dose_Opt_I\Homeworks\Homework1\Fall-00\ans-hmwk1.doc 4

5 AUC(7-10hr)=(C 7 +C 10 )(t 10 -t 7 )/2= (59+45)*(10-7)/2= 156 µg/l/hr AUC(10-14hr)=(C 10 +C 14 )(t 14 -t 10 )/2= (45+33)*(14-10)/2=156 µg/l/hr Calculated in the same way: AUC(14-17hr)=90 µg/l/hr AUC(17-20hr)=70.5 µg/l/hr AUC(20-25hr)=87.5 µg/l/hr AUC(25-30hr)=62.5 µg/l/hr AUC(30 hr - )=C 30 /k e = 10/0.077 µg/l/hr = 130 µg/l/hr AUC total = 1300 µg/l/hr 4. Zero-order elimination: C-C 0 =-k t C=C 0 -k t C=C 0 /2 at t=t 1/2 C 0 /2=C 0 -k t 1/2 t 1/2 =(C 0 -C 0 /2)/k t 1/2 =C 0 /2k Half-life depends on concentration. First-order elimination: C=C 0 e -k t C=C 0 /2 at t=t 1/2 C 0 /2=C 0 e -k t1/2 1/2=e -k t1/2 ln2=k t 1/2 t 1/2 =ln2/k Half-life does not depend on concentration. The equations did not have to be deduced I just did so to show from where they actually come from. D:\pha5127_Dose_Opt_I\Homeworks\Homework1\Fall-00\ans-hmwk1.doc 5