左旋千金藤立定对溴隐亭诱导的哺乳期大鼠促乳素水平低下的拮抗作用  

Antagonism of l-stepholidine against bromocriptineinhibition on prolactin level in lactational rats

作　　者：尤春来[1] 韩兆丰[1] 屈玉书 王义明[1] 陈丽娟 金国章 

机构地区：[1]辽宁中医学院药理教研室 [2]中国科学院上海药物研究所

出　　处：《中国药理学报》1996年第4期382-384,共3页Acta Pharmacologica Sinica

摘　　要：目的:研究左旋千金藤立定(SPD)对溴隐亭(Bro)诱导的促乳素(PRL)水平低下的对抗作用。方法:哺乳期母鼠sc Bro 0.5mg·kg^(-1)·d^(-1),PRL显著降低,乳腺组织发育不良,而且仔鼠体重增长缓慢。用放免法测定母鼠PRL,检查乳腺发育状况,评价SPD的对抗作用。结果:SPD30及100mg·kg^(-1)·d^(-1)ip,能够显著对抗Bro诱导的母鼠PRL降低,分娩后d15PRL为11±4及23±6μg·L^(-1)(生理盐水为7±2),而且乳腺组织发育正常,仔鼠在出生后d11—15内迅速生长发育。结论:SPD能阻断大鼠脑垂体前叶的D_2受体,是一个D_2受体的拮抗剂。AIM: TO study the antagonism of /-stepholidine (SPD) against bromocriptine (Bro)-inhibition on prolactin (PRL) level. METHODS: Bro (0.5 mg ·kg-1 ·d-1, sc) reduced the PRL and caused a dysplasia of mammary gland in lactational rats. The weight growing of newborn rats was retarded. The PRL of the lactational rats was assessed by immunoradiometric assay (IRMA); the weight of newborn rats and development of mammary glands in lactational rats were also examined. Antagonism of SPD was evaluated. RESULTS: SPD (30 & 100 mg·kg-1 · d-1, ip) obviously antagonized the Bro that induced lowering the PRL level in lactational rats, the PRL was 11 ± 4 & 23 ±6 μg · L-1 (NS 7 ± 2) respectively on d 15 of postpartum and the development of mammarygland in lactational rats was normal. The newborn rats grew rapidly in 11 - 15 d. CONCLUSION: SPD possessed an antagonism with Bro inhibition on D2 receptors located in the pituitary gland, and was an antagonist of dopamine D2 receptors.

关 键 词：左旋千金藤立定 溴隐亭 促乳素 多巴胺D2 受体 

分 类 号：R965.1[医药卫生—药理学] R966[医药卫生—药学]