1. 重组蛋白
  2. Enzymes & Regulators
  3. Oxidoreductases (EC 1)
  4. MDO 蛋白, Mycobacterium sp. (His)

MDO 蛋白, Mycobacterium sp. (His)

目录号: HY-P702176
产品使用指南

MDO(甲醇脱氢酶)负责氧化甲醇产生甲醛。尽管体内电子受体仍未确定,但体外实验表明 N,N-二甲基-4-亚硝基苯胺 (NDMA) 可以作为合适的电子受体,并随后被还原为 4-(羟基氨基)-N,N-二甲基苯胺。MDO 蛋白, Mycobacterium sp. (His) 是重组的 MDO 蛋白,由 E. coli 表达,带有 N-6*His 标签。MDO 蛋白, Mycobacterium sp. (His) 全长 423 个氨基酸。

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描述

MDO (methanol dehydrogenase) is responsible for oxidizing methanol to produce formaldehyde. Although the in vivo electron acceptor remains unidentified, in vitro experiments indicate that N,N-dimethyl-4-nitrosaniline (NDMA) can serve as a suitable electron acceptor and is subsequently reduced to 4-(hydroxyamino) -N,N-Dimethylaniline. MDO Protein, Mycobacterium sp. (His) is the recombinant MDO protein, expressed by E. coli , with N-6*His labeled tag. The total length of MDO Protein, Mycobacterium sp. (His) is 423 a.a., .

研究背景

甲醇脱氢酶(MDO)是催化甲醇氧化产生甲醛的必需酶。虽然具体的体内电子受体仍不清楚,但体外研究已证明 N,N-二甲基-4-亚硝基苯胺 (NDMA) 具有此功能,在的过程。值得注意的是,MDO 在底物利用方面表现出多功能性,与乙醇和甲醛的活性与与甲醇的活性相当。此外,该酶以甲胺为底物时表现出弱活性。这种广泛的底物特异性表明 MDO 在多种代谢途径中的潜在作用,并强调其在适应不同醇和胺底物方面的适应性,强调其在细胞代谢和各种有机化合物氧化中的重要性(改编自所提供的通道)。

种属

Others

表达系统

E. coli

标签

N-6*His

蛋白编号

C5MRT8 (M1-Y423)

基因 ID

/

蛋白结构
N-term
6*His
MDO (M1-Y423)
Accession # C5MRT8
C-term
同用名
Methanol:N; N-dimethyl-4-nitrosoaniline oxidoreductase; MDO; Methanol dehydrogenase (nicotinoprotein); Methanol:NDMA oxidoreductase
纯度

Greater than 90% as determined by reducing SDS-PAGE.

内毒素含量

<1 EU/μg, determined by LAL method.

文件资料

MDO Protein, Mycobacterium sp. (His) 相关分类

Help & FAQs
  • Do most proteins show cross-species activity?

    Species cross-reactivity must be investigated individually for each product. Many human cytokines will produce a nice response in mouse cell lines, and many mouse proteins will show activity on human cells. Other proteins may have a lower specific activity when used in the opposite species.

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The specific activity calculator equation

Specific Activity (Unit/mg) = 106 ÷ Biological Activity (ED50)

Specific Activity (Unit/mg) = 106 ÷ Biological Activity (ED50)
Unit/mg = 106 ÷ ng/mL

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