- #1
havenly
- 9
- 0
- Homework Statement
- half life time
- Relevant Equations
- ๐1/2 =(๐๐2) ๐ฅ ๐๐๐๐ข๐๐ ๐๐ ๐๐๐ ๐ก๐๐๐๐ข๐ก๐๐๐/ clarence.
A 70 kg patient is treated with ciprofloxacin (2 x 500 mg / d) for an infection
pulmonary. Ciprofloxacin is an antibiotic eliminated primarily by the kidneys, its
apparent volume of distribution is 3 [1 / kg) and its clearance is 0.5 (Wh x kg).
Calculate the half-life of ofloxacin in this patient. Later, his situation
degrades: it produces edema and the volume of distribution of ofloxacin increases to 125 liters.
After calculating the new half-life of ofloxacin in this patient,
Choose
the correct statement among the following 5:
Please choose an answer:
A. The half-life is extended by 60 min. NS.
B. The half-life is shortened by 30 min. NS.
C. The half-life is lengthened by 4h10 min. NS.
D. The half-life time remains unchanged.
E. The half-life is extended by 100 min. NS.to calculate the first half-life I did:
T1/2 =(ln(2) x volume distribution)/ clearance
v=70Kg x 3L/KG =210L
clearance 0.5L/H KG x70KG=35
T1/2= (ln(2) x 210)/ 35 = 4.158h
to calculate the second half-life:
k= ln(2) / T1/2
K= ln2/4.158 =0.166
clearance= K. V
clearance= 0.166 X 125= 20.83
T1/2= (ln (2) x125)/ 20.83 =4.158h
I get that the half-life time remains the same; I don't see where my error is. you can help me?
pulmonary. Ciprofloxacin is an antibiotic eliminated primarily by the kidneys, its
apparent volume of distribution is 3 [1 / kg) and its clearance is 0.5 (Wh x kg).
Calculate the half-life of ofloxacin in this patient. Later, his situation
degrades: it produces edema and the volume of distribution of ofloxacin increases to 125 liters.
After calculating the new half-life of ofloxacin in this patient,
Choose
the correct statement among the following 5:
Please choose an answer:
A. The half-life is extended by 60 min. NS.
B. The half-life is shortened by 30 min. NS.
C. The half-life is lengthened by 4h10 min. NS.
D. The half-life time remains unchanged.
E. The half-life is extended by 100 min. NS.to calculate the first half-life I did:
T1/2 =(ln(2) x volume distribution)/ clearance
v=70Kg x 3L/KG =210L
clearance 0.5L/H KG x70KG=35
T1/2= (ln(2) x 210)/ 35 = 4.158h
to calculate the second half-life:
k= ln(2) / T1/2
K= ln2/4.158 =0.166
clearance= K. V
clearance= 0.166 X 125= 20.83
T1/2= (ln (2) x125)/ 20.83 =4.158h
I get that the half-life time remains the same; I don't see where my error is. you can help me?