人 CD57 最初称为 HNK-1，是一种糖蛋白，发现于 15- 20% 的 PBL包括 60% 的 NK 细胞和一部分 T 细胞 (1)。正在研究表达高水平 CD8 的 CD57 阳性 PBL 的免疫调节作用 (2)。
同种型：小鼠 IgM
免疫原：人类 PBL
特异性：抗体 NK-1 识别约 110 kd 的 CD57 分子。
参考资料：
1.白细胞分型 V（S.F. Schlossman 等人编）牛津大学出版社，牛津 (1995) p. 1412-1414。
2。 E.C.Y. Wang, et al, (1995) J Immunol 155: 5046-5056.
USD $235.00In cart: 0Catalog #:209-040Form:FITCSize:120 testsAlternate Name:HNK-1, Leu-7Clone:NK-1Applications:FC, FXApplication KeyFCFlow CytometryWBWestern blotEIAEnzyme ImmunoassayIHC免疫组化ICC免疫细胞化学FPS固定石蜡切片PS石蜡切片IP免疫沉淀FX固定细胞或组织所示应用已在内部测试或报告。产品可能适用于未指明的应用。
1类别：抗人抗体下载 sp209_040.pdfPDF 文件30 KBO其他形式抗-CD57 / NK-1（生物素）抗-CD57 / NK-1（Pur/PF）抗-CD57 / NK -1（纯/西澳）主页产品新产品抗人抗体抗小鼠抗体细胞因子相关免疫检查点同型对照祖细胞/干细胞重组蛋白二步试剂TNF 超家族信息热门话题美国订单/价目表国际订单程序MSDSanti-CD57生物素化人类 CD57 最初称为 HNK-1，是一种糖蛋白，发现于 15- 20% 的 PBL 包括60% 的 NK 细胞和一部分 T 细胞 (1)。正在研究表达高水平 CD8 的 CD57 阳性 PBL 的免疫调节作用 (2)。
同种型：小鼠 IgM
免疫原：人类 PBL
特异性：抗体 NK-1 识别约 110 kd 的 CD57 分子。
参考资料：
1.白细胞分型 V（S.F. Schlossman 等人编）牛津大学出版社，牛津 (1995) p. 1412-1414。
2。 E.C.Y. Wang, et al, (1995) J Immunol 155: 5046-5056。
235.00 美元在购物车中：0Catalog #:209-030Form:BiotinSize:100 µgAlternate Name:HNK-1, Leu-7Clone:NK-1Applications:FC, FXApplication KeyFC流式细胞仪WBWestern blotEIA酶免疫分析IHC免疫组织化学ICCI免疫细胞化学FPS固定石蜡切片PS石蜡切片IP免疫沉淀FX固定细胞或组织所示应用已在内部测试或报告。产品可能适用于未指明的应用。
1类别：抗人抗体下载sp209_030-1.pdfPDF 文件30 KBO其他形式抗-CD57 / NK-1 (FITC) 抗-CD57 / NK-1 (Pur/PF) 抗-CD57 / NK-1（普尔/西澳）主页产品新产品抗人抗体抗小鼠抗体细胞因子相关免疫检查点同型对照祖细胞/干细胞重组蛋白二步试剂TNF 超家族信息热门话题美国订单/价目表国际订单程序MSDS抗-CD57纯化（含抗生素）人 CD57 最初称为 HNK-1，是一种糖蛋白，发现于15-20百分比的 PBL 包括 60% 的 NK 细胞和一部分 T 细胞 (1)。正在研究表达高水平 CD8 的 CD57 阳性 PBL 的免疫调节作用 (2)。
同种型：小鼠 IgM
免疫原：人类 PBL
特异性：抗体 NK-1 识别约 110 kd 的 CD57 分子。
参考资料：
1.白细胞分型 V（S.F. Schlossman 等人编）牛津大学出版社，牛津 (1995) p. 1412-1414。
2。 E.C.Y. Wang, et al, (1995) J Immunol 155: 5046-5056。
200.00 美元在购物车中：0Catalog #:209-020Form:Pur/WASize:100 µgAlternate Name:HNK-1, Leu-7克隆：NK-1应用：FC、FX应用关键FC流式细胞术WB蛋白质印迹EIA酶免疫测定IHC免疫组织化学ICCI免疫细胞化学FPS固定石蜡切片PS石蜡切片IP免疫沉淀FX固定细胞或组织所示应用已在内部测试或报告。产品可能适用于未指明的应用。
1类别：抗人抗体下载 sp209_020.pdfPDF 文件28 KBO其他形式抗-CD57 / NK-1（生物素）抗-CD57 / NK-1（FITC）抗-CD57 / NK-1 (纯/全氟)主页产品新产品抗人抗体抗小鼠抗体细胞因子相关免疫检查点同型对照祖细胞/干细胞重组蛋白二步试剂TNF 超家族信息热门话题美国订单/价目表国际订单程序MSDS抗-CD56纯化（不含防腐剂）人 CD56 是来自 Ig 超家族的粘附分子，它是仅限于 NK免疫系统中的细胞。据信，NK 细胞形成了抵御肿瘤细胞和感染细菌和病毒的细胞的第一道防线。抗体 ERIC-1 识别 CD56 分子的干区表位 (4)。
同种型：小鼠 IgG1 kappa
免疫原：人视网膜母细胞瘤肿瘤组织，膜部分 (1)
特异性：抗体 ERIC-1 识别 NK 细胞上发现的 CD56 分子。
参考文献：
1) S.P. Bourne, et al, (1991) J Neuro- Oncol 10：111-119。
2) T.L. Whiteside 和 R.B. Herberman，（1994）临床和诊断实验室免疫学 1：125-133。
3) H. Spits, et al, (1995) Blood 85: 2654-2670。
4) R. Gerardy-Schahn, M Eckhardt, (1994) Int J Cancer:补充剂 8：38-42。
200.00 美元购物车中：0产品目录 #：208-820Form：Pur/PFSize：100 µg Alternate Name：NCAMClone：ERIC-1Applications：EIA、FC、IHC（丙酮）、WBApplication KeyFCFlow CytometryWBWestern blotEIAEnzyme免疫分析IHC免疫组织化学ICCI免疫细胞化学FPS固定石蜡切片PS石蜡切片IP免疫沉淀FX固定细胞或组织所示应用已在内部测试或报告。产品可能适用于未指明的应用。
1类别：抗人抗体祖细胞/干细胞下载sp208_820.pdfPDF 文件21 KBO其他形式抗-CD56 / ERIC-1（生物素）抗-CD56 / ERIC-1（Pur/WA）A simple dilution is one in which a unit volume of a liquid material of interest is combined with an appropriate volume of a solvent liquid to achieve the desired concentration. The dilution factor is the total number of unit volumes in which your material will be dissolved. The diluted material must then be thoroughly mixed to achieve the true dilution. For example, a 1:5 dilution (verbalize as \"1 to 5\" dilution) entails combining 1 unit volume of solute (the material to be diluted) + 4 unit volumes of the solvent medium (hence, 1 + 4 = 5 = dilution factor). The dilution factor is frequently expressed using exponents: 1:5 would be 5e-1; 1:100 would be 10e-2, and so on. Example 1: Frozen orange juice concentrate is usually diluted with 4 additional cans of cold water (the dilution solvent) giving a dilution factor of 5, i.e., the orange concentrate represents one unit volume to which you have added 4 more cans (same unit volumes) of water. So the orange concentrate is now distributed through 5 unit volumes. This would be called a 1:5 dilution, and the OJ is now 1/5 as concentrated as it was originally. So, in a simple dilution, add one less unit volume of solvent than the desired dilution factor value. Example 2: Suppose you must prepare 400 ml of a disinfectant that requires 1:8 dilution from a concentrated stock solution with water. Divide the volume needed by the dilution factor (400 ml / 8 = 50 ml) to determine the unit volume. The dilution is then done as 50 ml concentrated disinfectant + 350 ml water. Top of page A serial dilution is simply a series of simple dilutions which amplifies the dilution factor quickly beginning with a small initial quantity of material (i.e., bacterial culture, a chemical, orange juice, etc.). The source of dilution material for each step comes from the diluted material of the previous. In a serial dilution the total dilution factor at any point is the product of the individual dilution factors in each step up to it. Example: In a typical microbiology exercise the students perform a three step 1:100 serial dilution of a bacterial culture (see figure below) in the process of quantifying the number of viable bacteria in a culture (see figure below). Each step in this example uses a 1 ml total volume. The initial step combines 1 unit volume of bacterial culture (10 ul) with 99 unit volumes of broth (990 ul) = 1:100 dilution. In the second step, one unit volume of the 1:100 dilution is combined with 99 unit volumes of broth now yielding a total dilution of 1:100x100 = 1:10,000 dilution. Repeated again (the third step) the total dilution would be 1:100x10,000 = 1:1,000,000 total dilution. The concentration of bacteria is now one million times less than in the original sample. Top of page Very often you will need to make a specific volume of known concentration from stock solutions, or perhaps due to limited availability of liquid materials (some chemicals are very expensive and are only sold and used in small quantities, e.g., micrograms), or to limit the amount of chemical waste. The formula below is a quick approach to calculating such dilutions where: Example: Suppose you have 3 ml of a stock solution of 100 mg/ml ampicillin (= C1) and you want to make 200 ul (= V2)of solution having 25 mg/ ml (= C2). You need to know what volume (V1) of the stock to use as part of the 200 ul total volume needed. V1 = the volume of stock youll start with. This is your unknown.C1 = 100 mg/ ml in the stock solutionV2 = total volume needed at the new concentration = 200 ul = 0.2 mlC2 = the new concentration = 25 mg/ ml By algebraic rearrangement: V1 = (V2 x C2) / C1 V1 = (0.2 ml x 25 mg/ml) / 100 mg/ml and after cancelling the units, V1 = 0.05 ml, or 50 ul So, you would take 0.05 ml = 50 ul of stock solution and dilute it with 150 ul of solvent to get the 200 ul of 25 mg/ ml solution needed (remember that the amount of solvent used is based upon the final volume needed, so you have to subtract the starting volume form the final to calculate it.) Top of page 4. Moles and Molar solutions (unit = M = moles/L) Sometimes it may be more efficient to use molarity when calculating concentrations. A mole is defined as one gram molecular weight of an element or compound, and comprised of exactly 6.023 x 10^23 atoms or molecules (this is called Avagadros number). The mole is therefore a unit expressing the amount of a chemical. The mass (g) of one mole of an element is called its molecular weight (MW). When working with compounds, the mass of one mole of the compound is called the formula weight (FW). The distinction between MW and FW is not always simple, however, and the terms are routinely used interchangeably in practice. Formula (or molecular) weight is always given as part of the information on the label of a chemical bottle. The number of moles in an arbitrary mass of a dry reagent can be calculated as: Molarity is the unit used to describe the number of moles of a chemical or compounds in one liter (L) of solution and is thus a unit of concentration. By this definition, a 1.0 Molar (1.0 M) solution is equivalent to one formula weight (FW = g/mole) of a compound dissolved in 1 liter (1.0 L) of solvent (usually water). Example 1: To prepare a liter of a simple molar solution from a dry reagent Multiply the formula weight (or MW) by the desired molarity to determine how many grams of reagent to use: Example 2: To prepare a specific volume of a specific molar solution from a dry reagent A chemical has a FW of 180 g/mole and you need 25 ml (0.025 L) of 0.15 M (M = moles/L) solution. How many grams of the chemical must be dissolved in 25 ml water to make this solution? #grams/desired volume (L) = desired molarity (mole/L) * FW (g/mole) by algrebraic rearrangement, #grams = desired volume (L) * desired molarity (mole/L) * FW (g/mole) #grams = 0.025 L * 0.15 mole/L * 180 g/mole after cancelling the units, #grams = 0.675 g For more on molarity, plus molality and normality: EnvironmentalChemistry.com More examples of worked problems: About.com: Chemistry Top of page 5. Percent Solutions ( = parts per hundred or grams/100 ml) Many reagents are mixed as percent concentrations as weight per volume for dry reagent OR volume per volume for solutions. When working with a dry reagent it is mixed as dry mass (g) per volume and can be simply calculated as theconcentration (expressed as a proportion or ratio) x volume needed = mass of reagent to use. Example 1: If you want to make 200 ml of 3NaCl you would dissolve 0.03 g/ml x 200 ml = 6.0 g NaCl in 200 ml water. When using liquid reagents the percent concentration is based upon volume per volume, and is similarly calculated asconcentration x volume needed = volume of reagent to use. Example 2: If you want to make 2 L of 70 actone you would mix 0.70 ml/ml x 2000 ml = 1400 ml acetone with 600 ml water. To convert fromsolution to molarity, multiply thesolution by 10 to express the percent solution grams/L, then divide by the formula weight. Example 1: Convert a 6.5solution of a chemical with FW = 325.6 to molarity, To convert from molarity to percent solution, multiply the molarity by the FW and divide by 10: Example 2: Convert a 0.0045 M solution of a chemical having FW 178.7 to percent solution: Top of page 6. Concentrated stock solutions - using \"X\" units Stock solutions of stable compounds are routinely maintained in labs as more concentrated solutions that can be diluted to working strength when used in typical applications. The usual working concentration is denoted as 1x. A solution 20 times more concentrated would be denoted as 20x and would require a 1:20 dilution to restore the typical working concentration. Example: A 1x solution of a compound has a molar concentration of 0.05 M for its typical use in a lab procedure. A 20x stock would be prepared at a concentration of 20*0.05 M = 1.0 M. A 30X stock would be 30*0.05 M = 1.5 M. 7. Normality (N): Conversion to Molarity Normality = n*M where n = number of protons (H+) in a molecule of the acid. Example: In the formula for concentrated sulfuric (36 N H2SO4), there are two protons, so, its molarity= N/2. So, 36N H2SO4 = 36/2 = 18 M.1. Simple Dilution (Dilution Factor Method based on ratios)
2. Serial Dilution
![](\"dxy_expriment/201812/serial_dilution_fig.jpg\")
3. Making fixed volumes of specific concentrations from liquid reagents:
