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- 详细信息
- 文献和实验
- 技术资料
- 免疫原:
Recombinant protein encompassing a sequence within the center region of human RAB26. The exact sequence is proprietary.
- 亚型:
IgG
- 形态:
Liquid
- 保存条件:
Store as concentrated solution. Centrifuge briefly prior to opening vial. For short-term storage (1-2 weeks), store at 4ºC. For long-term storage, aliquot and store at -20ºC or below. Avoid multiple freeze-thaw cycles.
- 克隆性:
Polyclonal
- 标记物:
Unconjugated
- 适应物种:
Human, Rat
- 保质期:
12 months from the shipping date of the product.
- 抗原来源:
Human
- 目录编号:
GTX118872
- 级别:
Primary Antibodies
- 库存:
Available
- 供应商:
GeneTex
- 宿主:
Rabbit
- 应用范围:
WB, ICC/IF, IHC-P
- 浓度:
1 mg/ml (Please refer to the vial label for the specific concentration.)
- 靶点:
RAB26
- 抗体英文名:
RAB26 antibody [N2C3-2]
- 抗体名:
RAB26 抗体 [N2C3-2]
- 规格:
100 μl/25 μl
| 规格: | 100 μl | 产品价格: | ¥4000.0 |
|---|---|---|---|
| 规格: | 25 μl | 产品价格: | ¥1700.0 |
RAB26 antibody [N2C3-2] detects RAB26 protein by immunofluorescent analysis.
Sample: DIV9 rat E18 primary hippocampal neuron cells were fixed in 4% PFA at RT for 15 min.
Green: RAB26 stained by RAB26 antibody [N2C3-2] (GTX118872) diluted at 1:500.
Red: beta Tubulin 3/ Tuj1, stained by beta Tubulin 3/ Tuj1 antibody [GT1338] (GTX631831) diluted at 1:500.
Blue: Fluoroshield with DAPI (GTX30920).
Sample (30 ug of whole cell lysate)
A: U87-MG
12% SDS PAGE
GTX118872 diluted at 1:1000
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文献和实验Chang YC et al., Mol Cancer 2017 (PMID:28784136)
Antibody Staining of Dissected Gonads
Antibody Staining of Dissected Gonads - From R. Francis - -Adapted by Min-Ho Lee- Antibody staining (in Tubes) 1. Using a pasteur pipette (drawn in flame and cut open), transfer worm carcasses with attached gonads
RUVKUN Antibody Staining Protocol revised 10/29/90
RUVKUN Antibody Staining Protocol revised 10/29/90 Fixation: 1. C. elegans N2 grown in liquid or plates. Harvest worms. Wash for 2 hours at room temp. in 1X M9 or PBS. 2. Fix in 5 ml of FRESH THAT DAY 1% paraformaldehyde:
i 的概率,任何时刻一个变量被选取是随机的)的二项分布。这一二项分布可以很好地近似为参数泊松分布,而且计算起来也更容易。 7. 见注 2。对于这种情况,λ 可以计算如下。不同于等概率情况,λi 在这里是不同的。首先,为了计算 λ,我们需要计算所有的 n = 2110 个参数 λi,这个工作量很大,但是这里包含了很多重复,这一特征可以用来简化计算。而且,让我们确定一些由五元组(n1,n2,n3,n4,n5)确定的十肽,由密码子的数量可知,它们的概率在所描述的十肽中分别为 1/64,2/64,3/64
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