DrVirgo
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What % of steady state?
94-97?
(whats the exact number? 95?)
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| robin082006
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In answering question we select 100% steady state.
As we do not bring calculator, I do by this way
For exact calucation (at 5 t1/2): = 50+25+12.5+6.25+3.125= 96.875%
At 4 t 1/2 = 50+25+12.5+6.25= 93.75%
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| DrVirgo
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This is how I remember...
Plasma conc. will reach:
50% of steady state in 1 half life
75% in 2 half lives
87.5 % in 3 half lives
90% in 3.3
94-97% in 4-5 half lives. (clinical steady state)
(thanks for your explanation... now I know the exact values for 4 and 5)! 
OK, now a question:
Drug X has a t 1/2 of 4 hours and is renally eliminated. After elimination fo one kidney, how long (in hours) would it take for Drug X to reach steady state?
A. 10
B. 20
C. 30
D. 40
E. 60
And explain... Thanks.
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| robin082006
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D
Because drug X is renally excreted, after elimination of one kidney, its haft life doubles--> T 1/2=8
hours
In answering real exam's question we select 5 T1/2 to reach steady sate--> 5*8=40
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| DrVirgo
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It takes 4-5 t 1/2 to reach steady state or 16 to 20 hours under normal conditions. If one kidney is removed, the elimination of Drug X is halved. Elimination and t 1/2 are inversely related. So the new t 1/2 for Drug X is 8 hours, and the new time to reach steady state is 32 to 40 hours.
This is the explanation from Q-Book, First Exam #31. it says the correct answer is B! Maybe a typo (wanted to confirm)... but since they said: "the new t 1/2 for Drug X is 8 hours, and the new time to reach steady state is 32 to 40 hours." I was wondering if the correct answer is closer to 30 or 40. I guess 40 is right... Thanks.
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