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(PHA 5127)
Answers for Homework 1
- Give the equation that may be used to calculate the half-life when (1) two time points and the corresponding concentrations are known, or (2) clearance and volume of distribution are known, or (3) the elimination rate constant is known.
(i)
ke = -m
1 2
2 1 1 2
(ln 1 ln 2 ) (ln ln ) t t
C C
t t
C C
t1/2 = (ln ln )
ln 2 ( ) ln 2 2 1
1 2 C C
t t ke โ
(ii) Cl = ke โ
Vd
d
e V
Cl k =
Cl
V
k
t d e
ln 2 ln^2 1 / 2
(iii) k e
t ln 2 1 / 2 =
- Define volume of distribution and clearance.
Vd - relates the amount of drug in the body at any given time point to the
concentration observed in the plasma. The volume the body would have to be in order to
have the same concentration in all of the body (1-compartment) as that in the plasma.
Controlled by plasma and tissue protein binding.
Cl - the volume (of body fluid) cleared of the drug in a given time period. Units: e.g.
L/hr. All elimination routes are contained in the total clearance.
- A drug is given as an i.v. bolus injection. The rate of change in the plasma concentration is described by -dC/dt = K * C. Show mathematically that the half-life does not depend on the concentration.
k C C
dC = โ โ
(1st^ order elimination)
This may be rearranged to give:
kdt C
dC =โ
] ]
t t
C ln C (^) C (^) 0 =โ tk 0
0
ln ln 0 0
C โ C =โ tk + t โ
k
tk C
C
0
ln
e^ tk C
C โ
0
C = C โ
e โ^ tk 0 By definition, half-life is the time required for the concentration to drop to 1/2 the initial value. So, at what time point is C = C 0 /2?
C (^) C e t 1 / 2 k 0
0 2
= โ
^ โ
= e โ t^1 /^2 โ
^ k 2
Taking the natural log (ln) of each side gives
= โ t (^) 1 / 2 โ
k 2
ln
{ โ =โ t^ โ
k
1 / 2 0
ln 1 ln 2
k
t ln 2 1 / 2 =^ ; independent of concentration
- For first-order kinetics, if the initial plasma concentration is 11.5 mg/L and the half โ life is 3.2 hours, what is the plasma concentration going to be two and half hours after injection?
t1/2 = 3.2 hr Cp t = Cp โ
e โ^ kt () 0
Cp ( t ) ( 11. 5 mg / L ) e (^0.^22 hr^ )(^2.^5 hr ) 6. 63 mg / L
1 = โ
= โ โ 1 1 / 2
ln 2 0. (^693) โ = = = hr t hr
ke
- What kind of information does AUC reveal?
AUC - how much drug in the body over time
- A patient was given a bolus i.v. injection of 1000mg of a drug. The volume of distribution is 30L. When the plasma concentration (Cp) was assayed, the following data was obtained: Time t[min] Cp[mg/L] 5 32. 10 32. 15 31. 25 30. 50 27. 100 22. 150 17. 200 14. 250 11. 300 9. 355 7. 400 6. 500 4. 600 2. 800 1. 1000 0.
AUC(0-5) 174.5 mg/L/hr AUC(5-1000) 7794.375 mg/L/hr AUC(1000-) 125.2172 mg/L/hr AUC(total) 8094.092 mg/L/hr
