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产碱杆菌属菌株O-1中苯磺酸、4-甲苯磺酸和邻氨基苯甲酸降解的初始步骤

Initial steps in the degradation of benzene sulfonic acid, 4-toluene sulfonic acids, and orthanilic acid in Alcaligenes sp. strain O-1.

作者信息

Thurnheer T, Zürrer D, Höglinger O, Leisinger T, Cook A M

机构信息

Microbiology Institute, Swiss Federal Institute of Technology, ETH-Zentrum, CH-8092 Zürich.

出版信息

Biodegradation. 1990;1(1):55-64. doi: 10.1007/BF00117051.

Abstract

Alcaligenes sp. strain O-1 grew with benzene sulfonate (BS) as sole carbon source for growth with either NH4+ or NH4+ plus orthanilate (2-aminobenzene sulfonate, OS) as the source(s) of nitrogen. The intracellular desulfonative enzyme did not degrade 3- or 4-aminobenzene sulfonates in the medium, although the enzyme in cell extracts degraded these compounds. We deduce the presence of a selective permeability barrier to sulfonates and conclude that the first step in sulfonate metabolism is transport into the cell. Cell-free desulfonation of BS in standard reaction mixtures required 2 mol of O2 per mol. One mol of O2 was required for a catechol 2,3-dioxygenase. When meta ring cleavage was inhibited with 3-chlorocatechol in desalted extracts, about 1 mol each of O2 and of NAD(P)H per mol of BS were required for the reaction, and SO3(2-) and catechol were recovered in high yield. Catechol was shown to be formed by dioxygenation in an experiment involving 18O2. 4-Toluene sulfonate was subject to NAD(P)H-dependent dioxygenation to yield SO3(2-) and 4-methylcatechol, which was subject to meta cleavage. OS also required 2 mol of O2 per mol and NAD(P)H for degradation, and SO3(2-) and NH4+ were recovered quantitatively. Inhibition of ring cleavage with 3-chlorocatechol reduced the oxygen requirement to 1 mol per mol of OS SO3(2-) (1 mol) and an unidentified organic intermediate, but no NH4+, were observed.

摘要

产碱杆菌属菌株O-1能够以苯磺酸盐(BS)作为唯一碳源生长,以NH₄⁺或NH₄⁺加邻氨基苯磺酸盐(2-氨基苯磺酸盐,OS)作为氮源。细胞内的脱磺酶不会降解培养基中的3-或4-氨基苯磺酸盐,尽管细胞提取物中的酶能够降解这些化合物。我们推断存在对磺酸盐的选择性渗透屏障,并得出结论:磺酸盐代谢的第一步是转运进入细胞。在标准反应混合物中,BS的无细胞脱磺反应每摩尔需要2摩尔O₂。儿茶酚2,3-双加氧酶每摩尔需要1摩尔O₂。当在脱盐提取物中用3-氯儿茶酚抑制间位环裂解时,每摩尔BS的反应大约需要1摩尔O₂和1摩尔NAD(P)H,并且亚硫酸盐(SO₃²⁻)和儿茶酚的回收率很高。在涉及¹⁸O₂的实验中表明儿茶酚是通过双加氧作用形成的。4-甲苯磺酸盐发生依赖于NAD(P)H的双加氧反应生成SO₃²⁻和4-甲基儿茶酚,后者发生间位裂解。OS的降解每摩尔也需要2摩尔O₂和NAD(P)H,并且SO₃²⁻和NH₄⁺能够定量回收。用3-氯儿茶酚抑制环裂解可将每摩尔OS的需氧量降低至1摩尔,观察到了SO₃²⁻(1摩尔)和一种未鉴定的有机中间体,但未观察到NH₄⁺。

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