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戊巴比妥钠诱导仓鼠发生的突变。

Sodium pentobarbital-induced mutations in the hamster.

作者信息

Ito T, Ingalls T H

出版信息

Arch Environ Health. 1981 Nov-Dec;36(6):316-20. doi: 10.1080/00039896.1981.10667644.

Abstract

When virgin Syrian hamsters aged 6 to 8 wk were mated during estrus and anesthetized with sodium pentobarbital (Nembutal) 4 to 5 hr before estimated ovulation, pregnancy wastage in the newly conceived litter was observed. This was manifested by polyspermy of moribund eggs and death of fertilized eggs, deficits of expected zygotes, and triploidy and tetraploidy in 3-day-old surviving embryos. Multiple factors determined the nature of this wastage, including dosage of Nembutal [sodium 5-ethyl-5-(1-methyl butal) barbiturate], route of its administration, degree of resulting respiratory depression, and ultimately a pH imbalance (below 7) in the mother. Polyspermy appeared to result from the penetration of degenerating ova by 10-20 sperm, triploidy to polar body retention, and tetraploidy to endoreduplication of chromosomes before cell cleavage; identification of 2N/4N mosaics supports this inference.

摘要

当6至8周龄的未交配叙利亚仓鼠在发情期交配,并在预计排卵前4至5小时用戊巴比妥钠(Nembutal)麻醉时,新受孕的窝仔出现妊娠丢失。这表现为濒死卵子的多精入卵和受精卵死亡、预期合子数量不足,以及3日龄存活胚胎中的三倍体和四倍体。多种因素决定了这种丢失的性质,包括戊巴比妥钠[5-乙基-5-(1-甲基丁基)巴比妥酸钠]的剂量、给药途径、由此产生的呼吸抑制程度,以及最终母体中的pH失衡(低于7)。多精入卵似乎是由于10至20个精子穿透退化的卵子所致,三倍体是由于极体保留,四倍体是由于细胞分裂前染色体的核内复制;2N/4N嵌合体的鉴定支持了这一推断。

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