缓激肽对豚鼠胃的舒张作用。

Relaxation of guinea-pig stomach by bradykinin.

作者信息

Downing O A, Morris J S

出版信息

Arch Int Pharmacodyn Ther. 1981 Feb;249(2):229-34.

Abstract

Intraluminal pressure changes were recorded from guinea-pig stomachs in vitro. Bradykinin (Bk) (0.4-9.6 nM) was generally found to produce a biphasic response (relaxation followed by contraction) in bathing solutions containing 1.1 mM calcium chloride. In bathing solutions containing 5.2 mM calcium chloride a relaxation without contraction was the most common observation. Relaxations to Bk were unaffected by selective concentrations of tetrodotoxin (5.6 microM), phentolamine (5.0-50.0 microM) and propranolol (5.0-50.0 microM). During virtually complete desensitization to Bk produced by 3 cumulative doses of 6.4 nM, relaxations to vagal stimulation were only reduced by 24.2% and those to adrenaline (2.0 microM) by 25.9%, indicating that Bk is unlikely to be the inhibitory transmitter released during vagal stimulation. Relaxations to ATP (10 microM) were reduced by 89.4% suggesting that Bk and ATP share a common step in the production of relaxation.

摘要

在体外记录豚鼠胃内的腔内压力变化。在含有1.1 mM氯化钙的浴液中，通常发现缓激肽（Bk）（0.4 - 9.6 nM）会产生双相反应（先松弛后收缩）。在含有5.2 mM氯化钙的浴液中，最常见的观察结果是只有松弛而无收缩。对Bk的松弛反应不受选择性浓度的河豚毒素（5.6 microM）、酚妥拉明（5.0 - 50.0 microM）和普萘洛尔（5.0 - 50.0 microM）的影响。在由3次累积剂量的6.4 nM Bk产生的几乎完全脱敏过程中，对迷走神经刺激的松弛反应仅降低了24.2%，对肾上腺素（2.0 microM）的松弛反应降低了25.9%，这表明Bk不太可能是迷走神经刺激期间释放的抑制性递质。对ATP（10 microM）的松弛反应降低了89.4%，这表明Bk和ATP在产生松弛反应过程中有一个共同步骤。