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Anti-Human ENPP7 Polyclonal Antibody (HD356014)

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概述
货号HD356014
品牌abinScience
描述
Anti-Human ENPP7 Polyclonal Antibody (HD356014) is a rabbit polyclonal antibody detecting ENPP7. Suitable for Human.
种属反应性Human
应用ELISA, IHC, WB
宿主Rabbit
克隆类型Polyclonal
同种型IgG
免疫原E. coli - derived recombinant Human ENPP7 (Asn31-Leu413).
靶标Alk-SMase, Alkaline sphingomyelin phosphodiesterase, E-NPP 7, EC:3.1.4.12, ENPP7, Ectonucleotide pyrophosphatase/phosphodiesterase family member 7, Intestinal alkaline sphingomyelinase, NPP-7
内毒素水平Please contact with the lab for this information.
纯化方式Purified by antigen affinity column.
Accession号Q6UWV6
状态Liquid
保存溶液 0.01M PBS, pH 7.4, 50% Glycerol, 0.05% Proclin 300.

Please refer to the specific buffer information in the hardcopy of datasheet or the lot-specific COA.

产品使用信息
应用方法 稀释比例
ELISA 1:5000-1:20000
IHC 1:50-1:500
WB 1:500-1:2000
背景

Ectonucleotide pyrophosphatase/phosphodiesterase family member 7 (ENPP7) is a ~51 kDa protein. Choline-specific phosphodiesterase that hydrolyzes sphingomyelin releasing the ceramide and phosphocholine and therefore is involved in sphingomyelin digestion, ceramide formation, and fatty acid (FA) absorption in the gastrointestinal tract. Also has phospholipase C activity and can also cleave phosphocholine from palmitoyl lyso-phosphatidylcholine and platelet-activating factor (PAF) leading to its inactivation. Does not have nucleotide pyrophosphatase activity. May promote cholesterol absorption by affecting the levels of sphingomyelin derived from either diet or endogenous sources, in the intestinal lumen.

1. Duan, RD. et al. (2003) Journal of lipid research 44, 1241-50. PMID: 12671034
2. Duan, RD. et al. (2003) The Journal of biological chemistry 278, 38528-36. PMID: 12885774
3. Wu, J. et al. (2004) American journal of physiology. Gastrointestinal and liver physiology 287, G967-73. PMID: 15205117
4. Wu, J. et al. (2006) The Biochemical journal 394, 299-308. PMID: 16255717
5. Gorelik, A. et al. (2017) The Journal of biological chemistry 292, 7087-7094. PMID: 28292932
NoteFor research use only
图片
参考文献
公式
质量 (g) = 浓度 (mol/L) × 体积 (L) × 分子量 (g/mol)
填写 质量、浓度、体积中的任意 2 项 + 分子量,自动计算未知值。
质量
=
浓度
×
体积
分子量 *
g/mol
公式
C₁ × V₁ = C₂ × V₂
填写 4 项中的任意 3 项,自动计算未知值。
母液
C₁ (起始浓度)
×
V₁ (起始体积)
=
工作液
C₂ (终浓度)
×
V₂ (终体积)

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