Compound | ACh Response | N | 3 | 5 | 17 | IC50 | Hilln | P versus ACN | P versus DMSO |
---|---|---|---|---|---|---|---|---|---|
mean ± S.D. | μM | ||||||||
ACN | 0.27 ± 0.09 | 11 | αOH | α | βCN | 1.64 | 1.75 | 1-165 | |
B164 | 0.28 ± 0.10 | 3 | αOH | β | βCN | (1.55) | NS | 1-160 | |
B372 | 0.33 ± 0.10 | 10 | βOH | α | βCN | (1.83) | NS | 1-165 | |
B260 | 0.46 ± 0.07 | 3 | βOH | β | βCN | (2.69) | 1-150 | 1-160 | |
PROG | 0.50 ± 0.13 | 6 | O | x | βCH3CO | 3.03 | 1.22 | 1-160 | 1-165 |
ent-ECN | 0.54 ± 0.18 | 5 | αCN | β | αOH | (3.37) | 1-150 | 1-160 | |
ECN | 0.56 ± 0.17 | 5 | βCN | α | βOH | (3.58) | 1-150 | 1-160 | |
AND | 0.57 ± 0.10 | 3 | αOH | α | βOH | (3.61) | 1-150 | 1-160 | |
ent-ACN | 0.61 ± 0.18 | 8 | βOH | β | αCN | 4.64 | 1.39 | 1-160 | 1-160 |
B163 | 0.66 ± 0.20 | 5 | αOH | β | αCN | (4.76) | 1-150 | 1-160 | |
βEST | 0.69 ± 0.04 | 4 | OH | x | βOH | (5.30) | 1-165 | 1-165 | |
DHEAS | 0.76 ± 0.12 | 4 | βSO4 | x | O | (6.96) | 1-160 | 1-150 | |
αEST | 0.79 ± 0.12 | 15 | OH | x | αOH | 11.7 | 1.15 | 1-165 | 1-160 |
3α5αP | 0.84 ± 0.08 | 4 | αOH | α | βCH3CO | 12.3 | 1.55 | 1-165 | 1-160 |
DMSO | 0.96 ± 0.08 | 7 | 1-165 | ||||||
Bath | 1.03 ± 0.16 | 19 | 1-165 | NS |
The table is organized with the compound producing the greatest block (ACN) at the top and that producing the least (bath solution) at the bottom. The structures of the drugs used are shown in Fig. 6. The first column gives the name of the compound. The second column shows the mean residual response to 100 μM acetylcholine after 30-s preexposure to the steroid (all at 3 μM concentration) for N cells. The next three columns give the substituent and orientation at the 3, 5, and 17 positions, respectively. The next two columns show the measured IC50 values and Hill coefficient for inhibition or the calculated IC50 value (in parentheses). The calculated IC50 value was estimated from the mean residual current with 3 μM compound (ρ) and the mean value for the Hill coefficient for all the blocking curves determined (1.4), using the equation IC50 = 1.4 th root {(1 − ρ)/ρ} × 3 μM.