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由5-脂氧合酶-白三烯途径从5,8,11-二十碳三烯酸衍生而来的产物。

Products derived from 5,8,11-eicosatrienoic acid by the 5-lipoxygenase-leukotriene pathway.

作者信息

Jakschik B A, Morrison A R, Sprecher H

出版信息

J Biol Chem. 1983 Nov 10;258(21):12797-800.

PMID:6313677
Abstract

Analysis of products derived from 5,8,11-eicosatrienoic acid via the 5-lipoxygenase-leukotriene pathway showed that this fatty acid is readily converted to leukotriene (LT)A3. When 10,000 X g supernatant from rat basophilic leukemia cell homogenates was incubated with 30 microM fatty acid, 5,8,11-eicosatrienoic acid produced 6.2 +/- 1.1 nmol of LTA3 and arachidonic acid 15.5 +/- 1.9 nmol of LTA4 (n = 4). However, only insignificant amounts of LTB3 were formed (0.15 +/- 0.04 nmol of LTB3 and 4.2 +/- 0.4 nmol of LTB4, n = 4). These data indicate that the LTA-hydrolase requires not only the three double bonds of the triene but also the double bond at C-14 to efficiently convert LTA to LTB. These findings have significant implications for essential fatty acid deficiency.

摘要

对通过5-脂氧合酶-白三烯途径从5,8,11-二十碳三烯酸衍生的产物进行分析表明,这种脂肪酸很容易转化为白三烯(LT)A3。当将大鼠嗜碱性白血病细胞匀浆的10,000×g上清液与30μM脂肪酸一起孵育时,5,8,11-二十碳三烯酸产生了6.2±1.1 nmol的LTA3,而花生四烯酸产生了15.5±1.9 nmol的LTA4(n = 4)。然而,仅形成了少量的LTB3(0.15±0.04 nmol的LTB3和4.2±0.4 nmol的LTB4,n = 4)。这些数据表明,LTA水解酶不仅需要三烯的三个双键,还需要C-14位的双键才能有效地将LTA转化为LTB。这些发现对必需脂肪酸缺乏具有重要意义。

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