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  #1  
Old 06-11-2011
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Drug Three Pharmacokinetic Questions

1.-What is the half-time of elimination for a drug that undergoes first-order elimination with a rate constant of 0.1 minute?

A. 10 minutes
B. 100 minutes
C. 0.1 minutes
D. 6.93 minutes
E. 693 minutes

2.-How many minutes are required for approximately 97% elimination of a drug undergoing first-order elimination with a half-time of 10 minutes?

A. 10
B. 30
C. 50
D. 70
E. 100

3.-A new drug is being tested in a clinical trials. The PK properties of the drug reveals that the half-life of the drug is 6 hours. If a continuous iv. infusion were started on a research subject, how long would it take to reach 75% of steady state?

A. 3 hours
B. 6 hours
C. 9 hours
D. 12 hours
E. 18 hours
F. 24 hours
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  #2  
Old 06-11-2011
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Default Ans.

Question 1. I choose D. 6.93 minutes, we have to remember that actually constant order for decay in first order kinetics is equal to 0.693, so solving for halflife would be t1/2=0.693/0.1.

Question 2. I choose C. 50 minutes. For 97% of elimination to occur, it is necessary for 5 halflives to pass, given that the half life of the drug is 10 minutes, you mujltiply it by 5.

Question 3. I choose D. 12 hours, since it is in a constant infusion, to get to 75% of steady state we need 2 t1/2 to pass by, since the halflife is 6 hours, we multiply by 2.

Great questions!!!
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Old 06-11-2011
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!-......i wonder from where u get the constant first order decay ???
2- c
3- 12 hrs
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Old 06-11-2011
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Default Constant order of decay

Quote:
Originally Posted by Al-Saoudi View Post
!-......i wonder from where u get the constant first order decay ???
2- c
3- 12 hrs
The constant order for decay in first order kinetics is the "0.7" value we find in the clearance equations: Cl=Vd x 0.7/halflife.
So, to solve the problem, we use the formula for elimination constant, which is Ke.
Ke=0.7(constant order for decay)/halflife, we clear the equation for halflife and we get the new equation: halflife=0.7/Ke.
The question states that Ke is 0.1, and 0.7 is really 0.693 rounded up, so
halflife=0.693/0.1
halflife=6.93 minutes
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Quote:
Originally Posted by NGaleas View Post
The constant order for decay in first order kinetics is the "0.7" value we find in the clearance equations: Cl=Vd x 0.7/halflife.
So, to solve the problem, we use the formula for elimination constant, which is Ke.
Ke=0.7(constant order for decay)/halflife, we clear the equation for halflife and we get the new equation: halflife=0.7/Ke.
The question states that Ke is 0.1, and 0.7 is really 0.693 rounded up, so
halflife=0.693/0.1
halflife=6.93 minutes
tnx alot
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Old 06-12-2011
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In case anyone is wondering that 0.693 comes from ln2 (use a calculator).
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Old 06-12-2011
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1. D
2. C
3. D

Good question, had to look up the relationship between elimination constant and t1/2 again. Now I won't forget it!

Thanks again bebix!
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Old 06-12-2011
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1.-What is the half-time of elimination for a drug that undergoes first-order elimination with a rate constant of 0.1 minute?
t1/2=0.693/Ke
t1/2=0.693/0.1
t1/2=6.93
D. 6.93 minutes

2.-How many minutes are required for approximately 97% elimination of a drug undergoing first-order elimination with a half-time of 10 minutes?
5half lives required to get appr. 97% so 10*5 =50

C. 50

3.-A new drug is being tested in a clinical trials. The PK properties of the drug reveals that the half-life of the drug is 6 hours. If a continuous iv. infusion were started on a research subject, how long would it take to reach 75% of steady state?
to get 75% of steady state we need 2 t1/2 as it's constant infu. so multiply the half life by 2
6*2=12
D. 12 hours
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Old 06-12-2011
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d, c, and d

good one!
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Old 06-13-2011
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1.D
2.c
3.D
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