All material is copyrighted and the property of mehlmanmedical.
Copyright © 2020 mehlmanmedical.
Privacy Policy and Terms and Conditions
Bioavailability (F)
- The amount of an administered drug that reaches the circulation unaltered.
- 100% for IV administration.
- Oral bioavailability always <100% (due to gut excretion and first-pass metabolism).
- First-pass metabolism –> The drug will be absorbed by gut lymphatics, where it travels to the liver and is metabolized before reaching the circulation.
Loading dose (LD)
A 46-year-old patient is receiving an unknown drug intravenously. It is known that the drug is distributed into total body water. What dose of the drug is needed to obtain an initial plasma level of 4 mg/L in a patient weighing 70 kg?
a) 42 mg
b) 168 mg
c) 200 mg
d) 210 mg
e) 280 mg
(The answer is below in blue)
In order to answer this question, we need to determine the loading dose of the drug.
The loading dose is the initial higher dose of the drug given to achieve a certain plasma concentration earlier before maintaining it with a lower, maintenance dose.
It is calculated from the following equation: LD = (Cp x Vd) / F, where Cp is the plasma level, Vd is the volume of distribution, and F is the bioavailability of the drug.
The loading dose is proportional to the volume of distribution because the larger the volume of distribution the more drug needs to be given to achieve the same plasma concentration.
The bioavailability of the drug is the opposite. More drug needs to be given if the bioavailability is lower (i.e., drugs with low bioavailability less readily enter systemic circulation so more drug needs to be given).
For drugs administered intravenously, F = 1. Since the drug is distributed in total body water, Vd is about 60% of total body weight, or 42 kg (0.6 x 70 kg). 42 kg of water is 42 L of water.
Therefore, Vd = 42 L.
LD = 4 mg/L x 42 L / 1 = 168 mg.
The correct answer is b.
Plasma drug concentration
A 20-year-old male hospitalized for a severe infection is receiving an IV antibiotic that distributes in total body water. He was initially given 1,800 mg of the drug. His body weight is 75 kg. Which of the following represents the plasma concentration of the antibiotic in this patient:
a) 14 mg/L
b) 24 mg/L
c) 40 mg/L
d) 45 mg/L
e) 75 mg/L
(The answer is below in blue)
The answer to this question can be calculated from the equation for the loading dose of the drug.
The loading dose is the initial higher dose of the drug given to achieve a certain plasma concentration earlier before reducing the dosage to a maintenance dose.
The loading dose is calculated as: LD = ( Cp x Vd ) / F, where Cp is the plasma level, Vd is the volume of distribution, and F is the bioavailability of the drug.
The greater the volume of distribution of the drug, the more drug needs to be given to achieve the same plasma concentration, whereas more drug needs to be given if the bioavailability is lower.
For drugs administered intravenously, F = 1 as all the drug enters systemic circulation.
Since the drug is distributed in total body water, Vd is about 60% of total body weight, or 0.6 x 75 kg = 45 kg. 45 kg of water is 45 L of water. Loading dose is 1,800 mg.
1,800 mg = Cp x 45 L / 1
Cp x 45 L = 1,800 mg
Cp = 1,800 mg / 45 L = 40 mg/L.
The correct answer is c.
Maintenance dose
A drug is administered to a 75-year-old patient whose renal function has decreased by 50%. Which of the following best describes the change that would be expected in this patient:
a) t1/2 would be shorter
b) Vd would decrease
c) Loading dose would increase
d) Clearance of the drug would increase
e) Maintenance dose would be lower
(The answer is below in blue)
The patient in question has decreased renal function, meaning his renal clearance is lower.
The maintenance dose of the drug is the dose of drug given to the patient to maintain a certain plasma concentration.
The equation for the maintenance dose is: MD = Cl x Cp x td / F, where Cl is the clearance, Cp is the steady state concentration, td is the dosing interval and F is the bioavailability.
It can be concluded from this equation that the maintenance dose would be lower if the renal clearance was lower.
It also makes sense intuitively as a decrease in renal clearance of the drug indicates that more drug remains in the body for extended periods of time so lower maintenance doses would be needed to maintain a certain plasma concentration. Vd would not change. t1/2 would increase due to decreased clearance.
Note that the loading dose equation, in comparison, does not take into account the renal clearance.
The correct answer is e.
Steady state concentration
A 50-year-old patient is started on an IV infusion of a drug at the rate of 300 mg/h. The clearance of the drug is 50 L/h. Which of the following represents the steady state concentration of the drug in question:
a) 250 mg/L
b) 75.6 mg/L
c) 60 mg/L
d) 6 mg/L
e) 3.6 mg/L
(The answer is below in blue)
The maintenance dose of the drug is the dose of drug given to the patient to maintain a certain plasma concentration.
It can be calculated as: MD = Cl x Cp x td / F, where Cl is the clearance, Cp is the steady state concentration, td is the dosing interval and F is the bioavailability.
Since the drug is given as an IV infusion, there is no need to consider the dosing interval or bioavailability so it can be simplified as:
MD = Cl x Cp
Cp = MD / Cl = 300 mg/h / 50 L/h = 6 mg/L
The correct answer is d.
Volume of distribution / Clearance
A clinical study is done to determine the characteristics of a novel drug that is distributed in total body water. Which of the following represents the total amount of the drug in a body of a 35-year-old participant whose body mass is 80 kg if the plasma concentration of the drug is 10 mg/L:
a) 480 mL
b) 480 mg
c) 640 mL
d) 640 mg
e) 800 mg
(The answer is below in blue)
Once a drug enters the body, it distributes within body fluid compartments. Not all drugs distribute equally (i.e., some may remain in plasma only, others may distribute in interstitial fluid, while some others may distribute in total body water).
The volume of distribution is the ratio of the total amount of drug in the body and the concentration of the drug (remember, for any solution c = amount / V; where V = amount / c).
Therefore, Vd = amount of drug in the body / Cp –> amount of drug in the body = Vd x Cp.
Since water represents approximately 60% of body weight, it can be calculated that this patient has 48 kg of water in his body (0.6 x 80 kg), or 48 L of water.
Therefore, Vd is 48 L.
Cp = 10 mg/L
Amount of drug in the body = 48 L x 10 mg/L = 480 mg.
The correct answer is b.
A 25-year-old patient is being treated with a drug that distributes in the total body water. Currently, the amount of drug in his body is 600 mg and the plasma concentration of drug is 10 mg/L. Which of the following represents this patient’s body weight:
a) 60 kg
b) 66 kg
c) 80 kg
d) 88 kg
e) 100 kg
(The answer is below in blue)
Individual drugs distribute differently between the body’s fluid compartments. The volume of distribution represents to what degree the drug distributes within the body and can be calculated as the ratio of the total amount of drug in the body and the concentration of the drug (remember, for any solution c = amount / V => V = amount / c).
Therefore, the volume of distribution of a drug can be calculated from the following formula:
Vd = amount of drug in the body / Cp
In this patient, Vd = 600 mg / 10 mg/L = 60 L.
Since the drug is distributed in the total body water, the total volume of water in this patient is 60 L. The weight of 60 L of water is 60 kg. Since water represents approximately 60% of body weight, it can be calculated that his body weight is 60 kg / 0.6 = 100 kg.
The correct answer is e.
A novel drug is being studied in a clinical trial. It is determined that the half-life of a drug is 4 h and that the Vd of a drug in one of the participants is 30 L. Which of the following represents clearance of the drug in this patient:
a) 0.18 L/h
b) 3.15 L/h
c) 5.25 L/h
d) 7.50 L/h
e) 10.71 L/h
(The answer is below in blue)
First-order elimination of the drug means that a constant fraction of the drug is eliminated per unit of time. Therefore, we can determine the half-life of the drug, which is the time frame in which the concentration of the drug in the body halves. The equation used to calculate the half-life of the drug is:
t1/2 = 0.7 x Vd / Cl
It makes sense that clearance is inversely proportional to the half-life (i.e., greater clearance means more drug is removed from the body, making the half-life shorter).
The clearance of the drug can be calculated from the equation for the half-life:
t1/2 = 0.7 x Vd / Cl
Cl = 0.7 x Vd / t1/2
Cl = 0.7 x 30 L / 4 h = 5.25 L/h
The correct answer is c.
Half-life
A 40-year-old patient with a body mass of 70 kg is treated with the drug that is distributed in total body water. The drug has clearance of 10 L/h. Which of the following best represents the half-life of the drug in question:
a) 1 h
b) 2 h
c) 3 h
d) 4 h
e) 5 h
(The answer is below in blue)
The half-life is calculated for the drugs that follow first-order kinetics, meaning that a constant fraction of the drug is eliminated per unit of time. It represents the time frame in which the concentration of the drug in the body halves.
The equation used to calculate the half-life of the drug is:
t1/2 = 0.7 x Vd / Cl
Since the drug is distributed in total body water, Vd is the total volume of water in the body. Since water accounts for approximately 60% of body weight, the weight of water in this patient’s body is approximately 42 kg (0.6 x 70 kg). 42 kg have the volume of 42 L. Cl is 10 L/h.
Therefore, t1/2 = 0.7 x 42 L / 10 L/h = 2.94 h, or roughly 3 hours.
The correct answer is c.
Rate of elimination
A 35-year-old patient is participating in a clinical trial involving a novel drug that distributes in total body water. The plasma concentration of the drug, its clearance and half-life are measured. Which of the following represents the correct way of calculating the rate of elimination of the drug from the body:
a) clearance x plasma concentration
b) clearance / plasma concentration
c) clearance x volume of distribution / half-life
d) clearance x half-life
e) clearance x half-life / volume of distribution
(The answer is below in blue)
The rate of elimination can be calculated from the following equation:
Cl = rate of elimination / Cp
Rate of elimination = Cl x Cp; where Cl is clearance and Cp is plasma concentration.
It makes sense that the rate of elimination of the drug is proportional to the clearance and the concentration of the drug. If there is more drug in the plasma and more plasma gets cleared, more drug will be eliminated.
The correct answer is a.
A 40-year-old patient is treated with a novel drug that distributes in total body water. The body mass of the patient is 80 kg. It is determined that the half-life of the drug is 5 hours. The plasma drug concentration is 10 mg/L. Which of the following represents the rate of elimination of the drug:
a) 6.72 L/h
b) 67.2 mg/h
c) 48 L/h
d) 160 mg/h
e) 40 mg/h
(The answer is below in blue)
The rate of elimination can be calculated from the following equation:
Cl = rate of elimination / Cp
Rate of elimination = Cl x Cp; where Cl is clearance and Cp is plasma concentration.
The rate of elimination of the drug is proportional to the clearance and the concentration of the drug (the more drug in the plasma and the higher the clearance of such plasma – the more drug gets eliminated).
First-order elimination of the drug means that a constant fraction of the drug is eliminated per unit of time, making it possible to determine the half-life of the drug, which is the time frame in which the concentration of the drug in the body reduces by 50%.
The equation used to calculate the half-life of the drug is:
t1/2 = 0.7 x Vd / Cl, where t1/2 is the half-life and Vd volume of distribution.
Clearance of the drug can be calculated from that equation:
t1/2 = 0.7 x Vd / Cl
Cl = 0.7 x Vd / t1/2
The drug is distributed in total body water, which accounts for 60% of the body weight, or 48 kg (0.8 x 80 kg). 48 kg of water have the volume of 48 L. Therefore, Vd = 48 L.
Cl = 0.7 x 48 L / 5 h = 6.72 L/h
Rate of elimination = 6.72 L/h x 10 mg/L = 67.2 mg/h
The correct answer is b.
