# Quiz 2 Practical Kinetics II ( Lab 5 & 7)

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Questions: 5 | Attempts: 182

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Group of MUST Pharmacy
Pharmacokinetics
Practical Quiz 2

• 1.

### A women 50 kg given IV two compartment 2 mg/kg and conc. of elemination is 5 mg/l the conc. of distribution when volume of central compartment equal 10 mg/l is

• A.

3 mg/l

• B.

5 mg/l

• C.

0 mg/l

• D.

20 mg/l

B. 5 mg/l
Explanation
The concentration of distribution when the volume of the central compartment is equal to 10 mg/l is 5 mg/l. This means that the drug is evenly distributed throughout the central compartment, resulting in a concentration of 5 mg/l.

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• 2.

### By using Sigma Minus method,A Single IV dose of new Antibiotic was given to 50 kg women at dose level 10 mg/kg. Urine and Blood samples were removed periodically and examined for parent drug. the data is tabulated Excretion rate constant is:

• A.

1.2/hr

• B.

0.9/hr

• C.

1.046/hr

• D.

0.9/min

C. 1.046/hr
Explanation
The excretion rate constant is a measure of how quickly the drug is eliminated from the body. In this case, the rate constant is given as 1.046/hr. This means that, on average, 1.046 units of the drug are eliminated from the body every hour. A higher rate constant indicates a faster elimination of the drug, while a lower rate constant indicates a slower elimination. Therefore, the given answer of 1.046/hr suggests that the drug is being eliminated at a moderate rate from the body of the patient.

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• 3.

### A  Single IV dose of new Antibiotic was given to 50 kg women at dose level 20 mg/kg. Urine and Blood samples were removed periodically and examined for parent drug found elemination rate constant equal 0.679 hr-1 at 10 hours the rate of drug accumulation in the urine when total amount of drug excreted in urine is 0.9 g

• A.

0.687 mg/l

• B.

6.87 mg/l

• C.

0.687 mg/hr

• D.

6.87 X 10-4 mg/hr

C. 0.687 mg/hr
Explanation
The rate of drug accumulation in the urine can be calculated by multiplying the elimination rate constant (0.679 hr-1) by the amount of drug excreted in urine (0.9 g). This will give the rate of drug excretion per hour. Therefore, the correct answer is 0.687 mg/hr.

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• 4.

• A.

0,25 hr-1

• B.

0.24 hr-1

• C.

0.52 hr-1

• D.

0.14 hr-1

B. 0.24 hr-1
• 5.

### A drug follows two compartment open model was given IV at plasma drug  concentration 15 mg/l and after 3 hrs if found plasma concentration 10 mg/l when conc. of distribution equal 2.5 mg/l and β equal 0.109 hr-1 ,so slope of distribution equal

• A.

0.43 hr-1

• B.

0.31 hr-1

• C.

0.12 hr-1

• D.

0.25 hr-1

B. 0.31 hr-1
Explanation
In a two-compartment open model, the distribution phase represents the movement of the drug from the central compartment (plasma) to the peripheral compartment (tissues). The elimination phase represents the removal of the drug from the body.

The given information states that after 3 hours, the plasma concentration of the drug decreased from 15 mg/l to 10 mg/l. The concentration of distribution at this time is given as 2.5 mg/l and the elimination constant (β) is given as 0.109 hr-1.

To calculate the slope of the distribution phase, we can use the equation:

Slope of distribution = (Concentration at time 0 - Concentration at time 3 hours) / (Time 3 hours - Time 0)

= (15 mg/l - 2.5 mg/l) / (3 hours - 0 hours)

= 12.5 mg/l / 3 hours

= 4.17 mg/l/hr

Therefore, the slope of the distribution phase is approximately 4.17 mg/l/hr, which is equivalent to 0.31 hr-1.

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• Current Version
• Mar 22, 2023
Quiz Edited by
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• May 01, 2012
Quiz Created by
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