Question
1.A 70 kg volunteer is given an intravenous dose of an antibiotic, and serum concentration were determine at 2 hr and 5 hr after administration. The drug concentration were 1.2 and 0.3 g/ml respectively.
What's the biologic half life for this drug, assuming first order elimination kinetics?
2.a drug has an elimination half life of 6 hr and follow first order kinetics. If single 200 mg dose is given to an adult male patient (68kg) by iv bolus injection, what is percent of the dose is lost in hours?
What's the biologic half life for this drug, assuming first order elimination kinetics?
2.a drug has an elimination half life of 6 hr and follow first order kinetics. If single 200 mg dose is given to an adult male patient (68kg) by iv bolus injection, what is percent of the dose is lost in hours?
Answers
GPT-4o mini
To find the biological half-life and the percentage of a drug lost after a certain time period, we can apply concepts from pharmacokinetics.
### 1. Biological Half-Life Calculation
Given:
- \( C_1 = 1.2 \, \text{g/ml} \) at \( t_1 = 2 \, \text{hr} \)
- \( C_2 = 0.3 \, \text{g/ml} \) at \( t_2 = 5 \, \text{hr} \)
First, we need to calculate the elimination rate constant (\( k \)). The formula for the elimination rate constant from first-order kinetics is:
\[
k = \frac{\ln(C_1/C_2)}{t_2 - t_1}
\]
Substituting the values:
\[
k = \frac{\ln(1.2/0.3)}{5-2}
\]
Calculating \( \ln(1.2/0.3) \):
\[
\frac{1.2}{0.3} = 4 \implies \ln(4) \approx 1.386
\]
Now substituting into the equation for \( k \):
\[
k = \frac{1.386}{3} \approx 0.462 \, \text{hr}^{-1}
\]
Now, to find the biological half-life (\( t_{1/2} \)), we use the relationship:
\[
t_{1/2} = \frac{0.693}{k}
\]
Substituting the value of \( k \):
\[
t_{1/2} = \frac{0.693}{0.462} \approx 1.5 \, \text{hr}
\]
**Answer: The biological half-life of the drug is approximately 1.5 hours.**
---
### 2. Percent of the Dose Lost
Given:
- Half-life (\( t_{1/2} = 6 \, \text{hr} \))
- Dose administered = 200 mg
To find out how much of the drug is lost after a certain time period, we can use the half-life formula. The concentration of the drug remaining after time \( t \) is given by:
\[
C(t) = C_0 \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}}
\]
We will calculate for different time intervals to find out how much of the dose is lost.
Now, let's assume we want to calculate the amount lost after 6 hours:
\[
C(6) = 200 \left( \frac{1}{2} \right)^{\frac{6}{6}} = 200 \times \frac{1}{2} = 100 \, \text{mg}
\]
The amount lost:
\[
\text{Lost} = \text{Initial Dose} - C(6) = 200 \, \text{mg} - 100 \, \text{mg} = 100 \, \text{mg}
\]
Now, to find the percentage loss:
\[
\text{Percent Lost} = \frac{\text{Lost}}{\text{Initial Dose}} \times 100 = \frac{100}{200} \times 100 = 50\%
\]
**Answer: 50% of the dose is lost after 6 hours.**
### 1. Biological Half-Life Calculation
Given:
- \( C_1 = 1.2 \, \text{g/ml} \) at \( t_1 = 2 \, \text{hr} \)
- \( C_2 = 0.3 \, \text{g/ml} \) at \( t_2 = 5 \, \text{hr} \)
First, we need to calculate the elimination rate constant (\( k \)). The formula for the elimination rate constant from first-order kinetics is:
\[
k = \frac{\ln(C_1/C_2)}{t_2 - t_1}
\]
Substituting the values:
\[
k = \frac{\ln(1.2/0.3)}{5-2}
\]
Calculating \( \ln(1.2/0.3) \):
\[
\frac{1.2}{0.3} = 4 \implies \ln(4) \approx 1.386
\]
Now substituting into the equation for \( k \):
\[
k = \frac{1.386}{3} \approx 0.462 \, \text{hr}^{-1}
\]
Now, to find the biological half-life (\( t_{1/2} \)), we use the relationship:
\[
t_{1/2} = \frac{0.693}{k}
\]
Substituting the value of \( k \):
\[
t_{1/2} = \frac{0.693}{0.462} \approx 1.5 \, \text{hr}
\]
**Answer: The biological half-life of the drug is approximately 1.5 hours.**
---
### 2. Percent of the Dose Lost
Given:
- Half-life (\( t_{1/2} = 6 \, \text{hr} \))
- Dose administered = 200 mg
To find out how much of the drug is lost after a certain time period, we can use the half-life formula. The concentration of the drug remaining after time \( t \) is given by:
\[
C(t) = C_0 \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}}
\]
We will calculate for different time intervals to find out how much of the dose is lost.
Now, let's assume we want to calculate the amount lost after 6 hours:
\[
C(6) = 200 \left( \frac{1}{2} \right)^{\frac{6}{6}} = 200 \times \frac{1}{2} = 100 \, \text{mg}
\]
The amount lost:
\[
\text{Lost} = \text{Initial Dose} - C(6) = 200 \, \text{mg} - 100 \, \text{mg} = 100 \, \text{mg}
\]
Now, to find the percentage loss:
\[
\text{Percent Lost} = \frac{\text{Lost}}{\text{Initial Dose}} \times 100 = \frac{100}{200} \times 100 = 50\%
\]
**Answer: 50% of the dose is lost after 6 hours.**