Calculated K iiv values (μM) for each of seven subjects for the inhibition of the P450 2C9 dependent 7-hydroxylation of (S)-warfarin by 100, 200, or 300 mg doses of fluconazole
Fluconazole Dose | |||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
0 mg
|
100 mg
|
200 mg
|
300 mg
| ||||||||||||||||||||||
|
Subject
|
CLf(c)a |
Flczb |
CLf(i) |
%(CLf(c) -
CLf(i))c | Kiiv
|
Flcz
|
CLf(i) |
%(CLf(c) - CLf(i))
| Kiiv
|
Flcz
|
CLf(i) |
%(CLf(c) - CLf(i))
| Kiiv
| ||||||||||||
| 1 | 106 | 10.4(0.8) | 62.4 | 41.1 | 14.8 | 27.1(3.4) | 46.1 | 56.5 | 20.8 | 27.8(3.3) | 35.4 | 66.6 | 13.9 | ||||||||||||
| 2 | 135 | 10.0(0.4) | 83.9 | 37.9 | 16.6 | 28.4(0.7) | 64.1 | 52.5 | 25.8 | 36.1(4.0) | 45.0 | 66.7 | 18.2 | ||||||||||||
| 3 | 94.3 | 7.2(1.7) | 84.8 | 10.1 | 64.3 | 17.3(3.1) | 48.1 | 49.0 | 18.0 | 30.4(4.5) | 30.4 | 67.8 | 14.5 | ||||||||||||
| 4 | 144 | 11.6(1.5) | 114 | 20.8 | 44.4 | 25.3(0.9) | 62.9 | 56.3 | 19.7 | 38.2(6.2) | 75.3 | 47.7 | 42.0 | ||||||||||||
| 5 | 133 | 5.5(1.6) | 86.7 | 34.8 | 10.2 | 16.0(0.7) | 74.9 | 43.7 | 20.5 | 24.9(3.3) | 40.2 | 69.8 | 10.8 | ||||||||||||
| 6 | 185 | 6.5(0.8) | 102 | 44.9 | 7.9 | 19.0(1.7) | 75.1 | 59.4 | 13.0 | 23.4(3.8) | 62.2 | 66.4 | 11.9 | ||||||||||||
| 7 | 65.0 | 10.0(0.9) | 55.3 | 14.9 | 56.7 | 20.8(1.1) | 31.5 | 51.5 | 19.6 | 26.0(4.4) | 22.8 | 64.9 | 14.0 | ||||||||||||
|
Mean
|
123(37)e
|
8.7[0.9]d
|
84.2(20.5)
|
29.2(13.7)
|
30.7(23.7)e
|
22.0[1.9]
|
57.1(17.9)
|
52.7(5.3)
|
19.6(3.8)e
|
29.5[2.2]
|
44.5(18.4)
|
64.3(7.4)
|
17.9(7.5)e
| ||||||||||||
↵ a All CL values are in ml · h−1 · kg−1 · 10−2. CLf(c) = control, CLf(i) = with fluconazole.
↵ b Steady state fluconazole plasma concentrations (μM) at 24-h intervals postdose.
↵ c % difference between CLf(c) and CLf(i) for each subject.
↵ d Numbers in parentheses are standard deviations while numbers in brackets are standard errors.
↵ e The means of the Kiiv values across the three fluconazole doses were not significantly different (p-value = 0.235).