Review of Pharmacology - 9E (2015)

= 40 L × 4 mg/L
= 160 mg
Clearance plays no role in the determination of loading dose. It is given to confuse you.
General Pharmacology
101. Ans. (c) 3.2 days (See below)
We want to decrease the plasma concentration of digoxin from 4 ng/ml to 1 ng/ml. It will take two half lives. Thus time
required will be 2 × t 1/2
i.e. 2 × 1.6 = 3.2 days.
102. Ans. (d) Constant intravenous infusion (Ref: Katzung, 10/e p40, 43, 44)
• When a drug is administered less frequently, it produces marked variation in the plasma concentrations. In
this question the drug has a half life of 6 hours. If we repeat the dose at 6 hourly intervals, there will be 100%
variation in the plasma concentration.
• More frequent dosing will minimize the variation between maximum and minimum plasma concentrations.
As the margin of safety of this drug (given in the question) is very low (maximum tolerable concentration is
only 1.5 times the effective concentration), constant i.v. infusion is the best route.
103. Ans. (d) Approximately 32 hours (See below)
• For a drug following first order kinetics, rise in plasma concentration as well as fall in plasma concentration is similar.
When the steady state is attained and the drug administration is stopped, it will be eliminated from the body. 50%
will be eliminated in one half life, 75% in 2 t 1/2
, 87.5% (50 + 25 + 12.5%) in 3 t 1/2
and 93.75% (50 + 25 + 12.5 + 6.25%) in
four half lives.
• When constant i.v. infusion is administered, plasma concentration increases in the same manner. In one half life, it is
50% of the steady state and to reach 93.75% of steady state, 4 half lives will be required.
• As half life of this drug is 8 hours, approximately 32 hours (4 × 8) will be taken.
104. Ans. (b) 4 mg/L (Ref: Katzung, 10/e p40, 43, 44)
• Half life of this drug is 2 hours and its plasma concentration is 3 mg/L after 4 hours (9AM to 1 PM).
• This means, after 2 half lives ( 4 hours) plasma concentration is 3 mg/L. We know, by constant i.v. infusion, plasma
concentration attained is 75% of the steady state in 2 half lives. So, if 3 mg/L is 75% of steady state, it will amount to
4 mg/L.
105. Ans. (c) 40 hr (Ref: Katzung, 10/e p38, 39)
V
t d
1 =× 0.693
2 CL
106. Ans. (b) 30 mg/hr (Ref: Goodman and Gilman 12/e p36-37)
In this question 80 percent of drug is eliminated by renal route and 20 percent by non-renal routes (10 percent by hepatic
metabolism and 10 percent by biliary secretion).
This patient has 50 percent renal function (60 ml/min of GFR instead of 120 ml/min). Thus, the drug that can be eliminated
in this person is 20 percent (Non-renal route) + 40 percent (Renal route; 50 percent of 80 percent)
= 60 percent
Thus, the dose rate should be 60 percent of the original.
i.e. 50 mg/hr × 60 percent = 30 mg/hr.
107. Ans. (c) Increasing rate of elimination as plasma concentration increases. (Ref: Katzung 10/e p38)
108. Ans. (b) 93% (Ref: KDT 6/e p32)
• Drugs elimination after different t ½
:
General Pharmacology
Half Life
Elimination
First 50%
Second 75% (50 + 25)
Third 87.5% (50 + 25 + 12.5)
Fourth 93.75% (50 + 25 + 12.5 + 6.25)
109. Ans. (b) Barbiturates (Ref: KDT 6/e p31)
110. Ans. (b) Phenytoin (Ref: KDT 6/e p31)
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