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Anti-EYA2 Polyclonal Antibody (HC008014)

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概述
货号HC008014
品牌abinScience
描述
Anti-EYA2 Polyclonal Antibody (HC008014) is a rabbit polyclonal antibody in ELISA, IHC, WB. Suitable for Human, Bovine, Mouse, and Chicken.
Highlights
  • ●Affinity Purified — Minimal background and high purity for reliable results.
  • ●Multi-Application — Validated across multiple applications.
  • ●Multi-Species — Cross-reactive for translational research.
种属反应性Human, Mouse
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 EYA2 (Glu266-Leu538).
内毒素水平Please contact with the lab for this information.
纯化方式Purified by antigen affinity column.
Accession号O00167, Q58DB6, O08575, Q9YHA0
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
稳定性和存储Use a manual defrost freezer and avoid repeated freeze thaw cycles. Store at 2 to 8°C for frequent use. Store at -20 to -80°C for twelve months from the date of receipt.
背景

Protein phosphatase EYA2 is a ~59 kDa protein. Functions both as protein phosphatase and as transcriptional coactivator for SIX1, and probably also for SIX2, SIX4 and SIX5. Tyrosine phosphatase that dephosphorylates 'Tyr-142' of histone H2AX (H2AXY142ph) and promotes efficient DNA repair via the recruitment of DNA repair complexes containing MDC1. 'Tyr-142' phosphorylation of histone H2AX plays a central role in DNA repair and acts as a mark that distinguishes between apoptotic and repair responses to genotoxic stress. Its function as histone phosphatase may contribute to its function in transcription regulation during organogenesis. Plays an important role in hypaxial muscle development together with SIX1 and DACH2; in this it is functionally redundant with EYA1.

1. Fougerousse, F. et al. (2002) Journal of muscle research and cell motility 23, 255-64. PMID: 12500905
2. Patrick, AN. et al. (2013) Nature structural & molecular biology 20, 447-53. PMID: 23435380
3. Krishnan, N. et al. (2009) The Journal of biological chemistry 284, 16066-16070. PMID: 19351884
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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