# Podgorski - Pharmacokinetics I Flashcards

1
Q

Rate equation

A

Rate = Constant * [Drug]^n where n is the kinetic order.

2
Q

Drug kinetic order definition:

A

Dependence of rate of a process on the exponent of the drug concentration

3
Q

Zero order vs. first order

A

If n=0, [Drug]^0 = 1. The process is not dependent on the drug concentration. The process
proceeds at a constant rate. For drug elimination, a constant amount is lost per unit time.

If n=1, [Drug]^1 = [Drug] and the rate of the process is directly dependent on the drug
concentration. For drug elimination, a constant percent or fraction is lost per unit time.

4
Q

Processes the follow first order kinetics

A

hepatic metabolism under ordinary circumstances -therapeutic plasma concentrations below Km

renal excretion under ordinary circumstances -therapeutic plasma concentrations below Km

5
Q

Examples of zero-order kinetics

A

Alcohol and aspirin overdose (overwhelms elimination process)

Both are first order at lower concentrations

Phenytoin (Dilantin)

• 100% bioavailable by oral and IV dosing
• liver at almost maximum metabolic capacity when Dilantin is in the therapeutic range
6
Q

Blood alcohol level considered to be the threshold for a fatal level

A

Blood alcohol levels of 500 mg/dl are considered to be the threshold for a fatal level.

7
Q

Graphic test for zero order

A

Straight-line behavior when concentration data are plotted on a normal (nonlogarithmic) Y axis.

8
Q

Graphic test for first order

A

Straight-line behavior when concentration data are plotted on a logarithmic Y axis.

First order behavior for alcohol is seen only when blood levels are below 10 mg/dl.

Between 10 and 100 mg/dl is a transition region between zero and first order.

9
Q

rate of drug input =

A

rate of drug input = f(D/T)

10
Q

Equation for first order disappearance

A

Amount (time=t) = Amount(start) * e^-kt

Amount at t = starting amount * e^-kt

11
Q

Significance of t1/2

A

Time required for amount to decrease to 1/2 of the starting amount.

Each drug has its own characteristic t1/2 value that is determined by the associated elimination process(es)

t1/2 value for a given drug assumes a normal patient with normal liver and kidney function.

t1/2 determines the rate at which blood concentration rises during a constant infusion and falls after administration is stopped.

12
Q

Factors altering t1/2 value

A

Changes in volume of distribution – due to age, obesity, etc.
Induction/downregulation of metabolizing enzymes
Drug accumulation in pathologic fluids
Cardiac failure
Kidney failure
Liver failure

13
Q

Ke=

A

ke = (0.7)/t1/2

14
Q

Significance of rate constant: Ke

A

If t1/2 = 2 day
k = (0.693)/2 day = 0.347/day

This indicates that about 35% of the starting amount will be eliminated in 1 day.

15
Q

Rate of Drug Output =

A

Rate of Drug Output = X ke

16
Q

How are t1/2 and ke related?

A

t1/2 and ke are inversely related

17
Q

Objective of therapeutic dosing

A

To maintain highest plasma drug concentration below its toxic concentration and at
the same time the lowest concentration above the minimally effective level

To obtain steady state plasma concentrations within the therapeutic window.

18
Q

Equation for total body amount

X=

A

X = C Vd

19
Q

Input = output

F(D/T) =

A

= Css Vd Ke

= Css Vd 0.7/(t1/2)

= Css CL

20
Q

A

21
Q

Rule of thumb to get to steady state

A

Rule of thumb -
4 x t1/2 to get to steady-state.

Usually split into two half doses

22
Q

Dose maintenance =

A

Dose maintenance = Clearance x Csteady state

23
Q

Clearance =

A

Clearance = Vd Ke

24
Q

If a drug is cleared 80% by the liver and 20% by the kidney and the regular
daily dose is 200 mg/day:
1. How much (in mg/day) is cleared by kidney and liver of a healthy patient
2. What should be the daily dose for the patient with 50% kidney function?

A

A1: Liver 160 mg/ml; kidney 40 mg/ml

A2: 180mg/ml: 160 mg/ml (liver) + 20 mg/ml (kidney)

25
Q

Area under curve: 2 formulae
CL=
f=

A

CL=Dose/AUCiv

f=AUCoral/AUCiv

26
Q

Vd =

INPUT RATE = OUTPUT RATE

f(D/T) =

f(D/T) =

Clearance =

f(D/T) =

Dose maintenance =

A

INPUT RATE = OUTPUT RATE

f(D/T) = (Css)(Vd)(ke)

f(D/T) = (Css)(Vd)(0.7/t1/2)

Clearance = (Vd)(ke)

f(D/T) = (Css)(CL)

Dose maintenance = Clearance x Csteady state

27
Q

Q1. Drug V is administered at 4 x 1 mg tablets every 4 hours. The t1/2 for drug V is 1 day. How long will it take to reach the steady state in the patient?

A

A1. Four days. By 4 x t1/2, we have reached 93.75% of the steady state (recall that the t1/2 for the approach to steady state is the same as the t1/2 for elimination). Thus, we use 4 x t1/2 as the time to steady state as a rule of thumb.

28
Q

Q2, Another patient with increased drug sensitivity requires only 2 x 1 mg tablets of drug V every 4 hours. How long will this patient require to reach steady state?

A

A2. Four days. The kinetics of approach to steady state is not dependent on input or maintenance dosage. The second patient’s steady state level will be one-half of the first’s, however.

29
Q

Q3. You decide that 4 days is too long to wait in your patient. You wish to immediately achieve the steady state concentration in your patient (from Q1). How much do you give?

A

A3. : You want to give your patient, at one time, the
amount of drug he will have in his body at steady
above, we know that the maintenance input rate is 4 mg/4 hr or 24 mg/day.

```Equations to use –
INPUT RATE = OUTPUT RATE
f(D/T) = (Css)(Vd)(ke)
ke = 0.7/t1/2
Substituting
24 mg/day = (Css)(Vd)(0.7/1 day)
(Css)(Vd) = 34.3 mg
(Css)(Vs) is the amount in the body at steady state, so the Loading Dose is 34.3 mg.```
30
Q

Q5. Data for metoprolol (Lopressor®): Oral availability, 38%; Clearance, 63 L/hr/70 kg; Vd, 290 L/70 kg; t1/2, 3.2 hr; Target concentration, 25 ng/mL. Calculate a maintenance dosing rate.

A

A5. f(D/T) = (Css)(CL)
0.38(D/T) = (25 ng/mL)(63 L/hr/7 kg)
(1000 mL/L)(1 μg/1000 ng)
(D/T) = (4145 μg/hr/70 kg)(1 mg/
1000 μg)
For a 70 kg person, (D/T) = 4.1 mg/hr or 100
mg/day. So a reasonable prescription would be 50
mg tablet 2 times a day (take with the morning and
evening meal).

31
Q

Q5. Patient NR has been hospitalized with a cardiac arrhythmia that has been treated satisfactorily using an i.v. infusion rate of 20 mg/hr of Drug X (t1/2 = 5 hr). Kidney problems unrelated to the cardiac problem suddenly change the t1/2 to 10 hr. If the patient continues to receive 20 mg/hr, what will happen to the steady state blood level?

A
```A5. It will double.
f(D/T) = (Css)(Vd)(0.7/t1/2)
Css = (f(D/T)(t1/2))/(Vd)(0.7)
If t1/2 increases by 2x, then Css must increase by
2x since everything else is constant.```
32
Q

Q6. If the therapeutic steady state level of Drug X in the previous question is 100 μg/ml, and toxicity becomes evident at a level of 150 μg/ml, how long from the time of the change in renal function will it take for the signs of toxicity to be manifested?

A

A6. 10 hr (one half-life)
The steady state level will rise from 100 to 200
μg/ml as shown above. The approach to the new
steady state level is controlled by the new t1/2
value. In one half-life, we go half of the distance
from the starting point of 100 toward the new
steady state level of 200. So, in one half-life, the
level is 150 μg/ml where the signs of toxicity
become evident.

33
Q

Q7. Continuing from Q6, what is the level after two

half-lives?

A

A7. In the second half-life, we go half of the
remaining distance from the new starting value of
150 toward 200. Thus the level after 2 x t1/2 is 175
μg/ml.

34
Q

Q8. In the patient above, suppose that, by the time that the treating physician figured out that the t1/2 had doubled, the plasma level was at 200 μg/ml. How do we treat this patient to get the plasma level back to 100 μg/ml and then maintain the correct steady state level?

A

A8. First, stop administering the drug. If we wait one half-life of 10 hours, first order kinetics indicate that the plasma level will drop 50% from the 200 μg/ml level to 100 μg/ml which is the first goal. To maintain this level, we must decrease the input rate from the original 20 mg/hr to 10 mg/hr to compensate for the doubling of the half-life in this patient (assuming the kidney problem is still present).