有關Ceruloplasmin之問題!!真的很不懂! 20點

Question

以下為本PAPER之實驗步驟：Fresh blood was obtained from a slaughterhouse and immediately fractionated (through step 4) as in [13]. Blue fractions from both DEAE-cellulose columns, in each case eluted by 0.1 M phosphate [Na2HPO4-KH2PO4] buffer (pH 7.0) were pooled and dialyzed vs 0.006 M phosphate buffer (pH 6.8). The solution was then applied to a 2.5 × 30 cm column of hydroxylapatite equilibrated with 0.006 M phosphate buffer (pH 6.8) and eluted with a 0.06-0.5 M phosphate buffer gradient (pH 6.8). Fractions with A610/A280 > 0.024 from the form of the enzyme that eluted second (vide infra) were pooled and concentrated to ~3 ml by pressurized ultrafiltration. This concentrated solution was applied to a 2.5 × 40 cm Sephadex G-150 column equilibrated with 0.03 M phosphate buffer (pH 7.0) and eluted with the same buffer. Fractions with A610/A280 > 0.035 were pooled and concentrated to 4 ml, then rechromatographed on the G-150 column to give pure ceruloplasmin with A610/A280 = 0.046. We have also found that the scheme for purification of human ceruloplasmin in [ 14] can be used to purify bovine ceruloplasmin as well. Both of these methods permit bovine ceruloplasmin and amine oxidase to be purified from the same batch of plasma. Using the N,N\'-dimethyl-p-phenylenediamine assay in [ 15], the average oxidase activity at 22℃ of two preparations was (2)= 9.2/μmol ceruloplasmin (assuming e610 = 1 0 4 M-1 cm-1 [ 1-3 ]).

(1)= 0.5 for visible CD samples, which were diluted 10-fold for use in the UV. Spectral band width was maintained at 4 nm in the visible region; UV CD spectra were recorded with a constant slit width of 1 mm (2 nm spectral band width at 250 nm).

EPR measurements were made at 77 K using a Varian E-9 spectrometer. Ceruloplasmin was dialyzed extensively against Chelex-treated phosphate buffer (pH 7.0) prior to EPR or atomic absorption measurements.

Cu(EDTA)2- (20% excess Na2EDTA) in distilled, deionized water served as the standard for double integration. The copper content of ceruloplasmin was determined with a Perkin-Elmer 305A atomic absorption spectrometer.

(1)之圖片為http://photo.pchome.com.tw/s07/l/a/lamb_mmmm/book15/p114965265067.jpg
(2)之圖片為http://photo.pchome.com.tw/s07/l/a/lamb_mmmm/book15/p114965265124.jpg

我想問的是...像A610/A280 > 0.024代表的是什麼意思? 實際操作的話要怎麼取得?
...而且他為何要取0.024，而非其他數質???

更正：
(1)之圖片為http://photo.pchome.com.tw/lamb_mmmm/114965265067
(2)之圖片為http://photo.pchome.com.tw/lamb_mmmm/114965265124

Answer 1

A: absorbance, 吸光度以UV/visible spectrometer 分光光度計測量A610 ^1cm http://photo.pchome.com.tw/lamb_mmmm/114965265067 是指在 610 nm 波長下 sample 在光線穿透距離為1cm 的cuvette 中所得之吸光度 Absorbanceceruloplasmin 在 610 nm 有吸收 此為與其他蛋白質不同特殊的吸收A280 是指在 280 nm 波長下 sample 所得之吸光度 Absorbance 一般蛋白質在280nm會有吸收 (tryptophan 的吸收)A610/A280 這比值是 ceruloplasmin 與蛋白質的比值在粗蛋白分離時 因為有其他蛋白質存在 而這些不是ceruloplasmin的蛋白質量比ceruloplasmin多 所以A280值會比A610大因為ceruloplasmin 本身也是蛋白質 所以也會在280 nm吸收 因此為求要ceruloplasmin量高的吸收折中表示 所以要求A610/A280 > 0.024 以示ceruloplasmin的純度與含量