CN108358956B - 荧光探针EuШ-dtpa-bis(adenine)及其在检测尿液中乳清酸中的应用 - Google Patents
荧光探针EuШ-dtpa-bis(adenine)及其在检测尿液中乳清酸中的应用 Download PDFInfo
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Abstract
本发明公开了一种荧光探针EuШ‑dtpa‑bis(adenine)及其在检测尿液中乳清酸中的应用。取二乙三胺五乙酸,乙酸酐,吡啶在65℃下搅拌回流24h。冷却,减压抽滤,洗涤,干燥。将得到的二乙三胺五乙酸二酐与三乙胺,二甲基甲酰胺(DMF),腺嘌呤,于100℃搅拌回流24h。冷却,旋蒸,洗涤,干燥。得到的二乙三胺五乙酸‑双(腺嘌呤)与Eu(NO3)3·6H2O于60℃加热搅拌2h,得到目标产物。将EuШ‑dtpa‑bis(adenine)作为探针结合荧光方法检测乳清酸。本发明方法简单新颖,成本低,效率高,且可应用在实际尿样当中。
Description
技术领域
本发明属于分析化学领域,尤其涉及一种新型荧光探针的合成和在真实尿液中检测乳清酸中的应用。
背景技术
乳清酸,也称为维生素B13,在医疗领域有着广泛的应用。然而,在新出生的婴儿中,一种常染色体隐性遗传病-乳清酸尿症,偶尔出现。该病的患者由于两种重要的酶,乳清酸磷酸核糖转移酶和乳清酸核苷酸脱羧酶有缺陷而几乎不能合成嘧啶类核苷酸,从而使乳清酸不能转变为尿苷酸,导致乳清酸大量出现在血液和尿液中。健康人尿液中乳清酸含量约为 1~4mg/24h,而乳清酸尿症患者尿液中乳清酸含量高达400~1400mg/24h。所以患者在出生数月内就表现出明显症状,比如低色素巨成红细胞性贫血以及身体发育和智力发育障碍。所以医学上急需发展一种方法用来鉴别和定量分析乳清酸。
通常,检测乳清酸的常规方法有串联质谱法、毛细管区带电泳法、差分脉冲极谱法、气相色谱-质谱法和高效液相色谱法。这些方法通常被认为是快速且敏感的。但是这些方法通常前处理过程复杂,信号不稳定并且需要昂贵的仪器设备。对于一些偏远地区或欠发达地区,使用上述方法检测乳清酸是困难的。因此,迫切需要开发一种高选择性,高敏感性,方便且低成本的乳清酸检测方法。
发明内容
本发明的目的之一是设计合成一种可用于有效检测尿液中乳清酸的新型荧光探针EuIII- dtpa-bis(adenine)。
本发明的目的之二是提供一种操作简单,成本低,敏感快速,且选择性好的检测乳清酸的方法。
本发明采用的技术方案是:本发明提供的荧光探针EuШ-dtpa-bis(adenine),其制备方法包括如下步骤:
1)将二乙三胺五乙酸、乙酸酐和吡啶混合均匀,在65℃下搅拌回流24h,冷却至室温,减压抽滤,依次用乙酸酐和无水乙醚洗涤,60℃下干燥,得二乙三胺五乙酸二酐(dtpaa)。优选的,按摩尔比,二乙三胺五乙酸:乙酸酐:吡啶=1:4:6。
2)将二乙三胺五乙酸二酐、三乙胺、无水DMF和腺嘌呤混合均匀,在100℃下搅拌回流24h,冷却至室温,旋转蒸发,依次用乙腈和无水乙醚洗涤,减压抽滤,于50℃干燥,得二乙三胺五乙酸-双(腺嘌呤)。优选的,按摩尔比,二乙三胺五乙酸二酐:三乙胺:腺嘌呤=1:3:2。
3)将二乙三胺五乙酸-双(腺嘌呤)和Eu(NO3)3·6H2O分别加去离子水溶解,然后混合,于60℃搅拌加热2h,冷却,得EuШ-dtpa-bis(adenine)。优选的,按摩尔比,二乙三胺五乙酸-双(腺嘌呤):Eu(NO3)3·6H2O=1:1。
上述的荧光探针EuШ-dtpa-bis(adenine)在定性和定量检测尿液中乳清酸中的应用。
定性检测尿液中乳清酸的方法:取尿液,加入上述新型荧光探针EuШ-dtpa-bis(adenine) 的水溶液,充分混合,在280nm下进行荧光检测,观察荧光光谱的变化。可以发现,随着尿液中乳清酸含量增加,探针的荧光强度明显降低。
定量检测尿液中乳清酸的方法:取50uL浓度是5.0×10-3mol/L的上述新型荧光探针 EuШ-dtpa-bis(adenine)的水溶液于10mL比色管中,用尿液定容至5mL。在280nm下进行荧光检测。
本发明的有益效果是:
1.本发明针对被检测物乳清酸的结构特点,利用碱基互补配对原则对dtpa进行修饰,设计合成了一种新型的荧光探针。
2.通过本发明的方法,该探针可以对乳清酸进行灵敏且特异性检测。与其它检测乳清酸的方法相比,具有简单,快速,成本低,选择性好,不受外界电磁场影响等优点。
附图说明
图1是荧光探针EuШ-dtpa-bis(adenine)的合成路线图。
图2a是dtpa的傅里叶变换红外光谱(FT-IR)图。
图2b是腺嘌呤(adenine)的傅里叶变换红外光谱(FT-IR)图。
图2c是dtpa-bis(adenine)的傅里叶变换红外光谱(FT-IR)图。
图3是dtpa-bis(adenine),EuШ-dtpa-bis(adenine)和EuШ-dtpa-bis(adenine)+乳清酸(OA)的紫外吸收光谱图。
图4a是荧光探针对乳清酸(OA)检测的荧光光谱图。
图4b是荧光探针对乳清酸(OA)检测的荧光光谱对比柱状图。
图5是荧光探针对乳清酸(OA)分别与不同物质混合的干扰荧光光谱对比图。
图6是荧光探针对实际尿样(Ur)中乳清酸检测的荧光光谱对比图。
具体实施方式
实施例1新型荧光探针EuШ-dtpa-bis(adenine)
(一)制备方法
1、二乙三胺五乙酸二酐(dtpaa)的制备
称取7.8670g(0.02mol)二乙三胺五乙酸(dtpa),乙酸酐16.0mL(0.08mol),吡啶10.0 mL(0.12mol)置于三颈圆底烧瓶中,在65℃下缓慢搅拌加热,冷凝回流24h。停止加热和搅拌,冷却至室温后将产物减压抽滤,依次用乙酸酐和无水乙醚分别洗涤三次(3×10mL),并减压抽滤,将产物于干燥箱中60℃干燥,即得二乙三胺五乙酸二酐(dtpaa)。
2、二乙三胺五乙酸-双(腺嘌呤)(dtpa-bis(adenine))的制备
取二乙三胺五乙酸二酐(dtpaa)1.9635g(5.5mmol),2.334mL的三乙胺(16.5mmol),无水DMF(50mL),腺嘌呤1.4864g(11mmol),于三颈圆底烧瓶中。恒温100℃条件下,快速搅拌,冷凝回流24h。反应完全后静置,冷却到室温后,旋转蒸发除去溶剂,得乳白色固体物质,减压抽滤,依次用乙腈和无水乙醚分别洗涤三次(3×10mL)。于50℃条件下干燥,即得二乙三胺五乙酸-双(腺嘌呤)(dtpa-bis(adenine))。
3、荧光探针EuШ-dtpa-bis(adenine)的制备
称取0.7838g二乙三胺五乙酸-双(腺嘌呤)(dtpa-bis(adenine))于三颈圆底烧瓶中,加30 mL去离子水溶解。同时称取0.5576g的Eu(NO3)3·6H2O(1.25mmol)置于烧杯中,加30mL 去离子水溶解,然后转移至上述圆底烧瓶中,于60℃搅拌加热2h。将反应后的溶液冷却至室温,然后转移至250mL容量瓶中,将上述圆底烧瓶用去离子水洗涤三次,洗涤液全部转移至容量瓶中。最后用去离子水定容至刻度。得到EuШ-dtpa-bis(adenine)储备液,此时浓度为5.0×10-3mol/L。合成过程如图1所示。
(二)检测
(1).Dtpa、腺嘌呤、dtpa-bis(adenine)(dtpa-BA)的FT-IR图如图2a,2b,2c所示。对比图2a 和图2c可以发现,C=O的吸收峰分别位于1752cm-1和1631cm-1处。与图2a相比,图2c中C=O 的吸收峰蓝移121cm-1。对比图2b和图2c可以发现,N-H的伸缩振动峰分别出现在3294cm-1和3392cm-1处。与图2b相比,图2c中N-H的伸缩振动峰红移98cm-1。这两种特征峰的移动说明酰胺键的形成,即荧光探针被成功合成。
(2).Dtpa-bis(adenine)(dtpa-BA),EuШ-dtpa-bis(adenine)(EuШ-dtpa-BA)和EuШ-dtpa- bis(adenine)+乳清酸(EuШ-dtpa-BA+OA)的紫外吸收光谱图如图3所示。从图3可以看出,dtpa- bis(adenine)在262nm处有一个吸收峰,而EuШ-dtpa-bis(adenine)在262nm处依然有一个吸收峰,且吸光度只微弱降低。这说明,Eu3+的加入几乎不会改变dtpa-bis(adenine)的吸光度。然而当乳清酸加入到EuШ-dtpa-bis(adenine)溶液中后,EuШ-dtpa-bis(adenine)的吸光度明显降低。可以预测EuШ-dtpa-bis(adenine)的荧光强度在加入乳清酸之后会发生很大改变。
实施例2荧光探针EuШ-dtpa-bis(adenine)在检测乳清酸中的应用
(一)荧光探针对乳清酸检测的荧光光谱
实验条件:取一定量的乳清酸用去离子水配置成浓度为5.0×10-3mol/L的溶液,作为乳清酸储备液。
取三支10mL比色管,分别加入150uL乳清酸储备液,50uL实施例1制备的荧光探针储备液,然后用去离子水将三支比色管定容至5mL。此时溶液中探针的浓度为5.0×10-5mol/L,乳清酸溶液的浓度为1.5×10-4mol/L。在280nm波长光的激发下观察荧光光谱的变化。
结果如图4a,4b所示。在280nm波长光的激发下,荧光探针在320nm处发射出强荧光,而乳清酸在320nm处几乎不发出荧光。当把乳清酸加入到探针溶液中后,探针的荧光被明显猝灭。图4b能更直观的比较320nm处荧光强度的差别。
(二)不同共存物质与乳清酸混合对荧光探针EuШ-dtpa-bis(adenine)检测影响
实验条件:取5支比色管,分别加入1.5mL浓度为5.0×10-4mol/L的尿酸(UA),组氨酸 (His),抗坏血酸(AA),马尿酸(Hipa),肌酸酐(Cre)溶液,再分别加50uL荧光探针储备液及 150uL乳清酸储备液,然后用去离子水定容至5mL。此时溶液中探针的浓度为5.0×10-5mol/L,乳清酸及其他共存物质溶液的浓度为1.5×10-4mol/L。在280nm波长光的激发下观察荧光光谱的变化。
结果如图5所示。从图5中可以看出,探针溶液在320nm处发出强荧光,当乳清酸加入到探针溶液中后,探针的荧光明显被猝灭。然而当尿酸,组氨酸,抗坏血酸,马尿酸,肌酸酐等共存物质分别加入探针和乳清酸的混合溶液中后,混合溶液的荧光几乎不发生改变。这说明,尿液中其他与乳清酸共存的物质不会干扰探针对乳清酸的检测。从表1中可以更加清楚的观察荧光强度的变化。
表1
(三)荧光探针EuШ-dtpa-bis(adenine)在实际尿样中检测乳清酸
实验条件:取9支10mL比色管,第1支为空白尿样,第2支加入150uL乳清酸储备液,第3支加入50uL探针储备液,其他每支分别加入50uL探针储备液,10uL,25uL,50uL,100uL,150uL,250uL的乳清酸储备液,然后用尿样定容至5mL。在280nm波长光的激发下观察荧光光谱的变化。
结果如图6所示。在280nm波长光的激发下,纯尿样在380nm附近发射相对较强的荧光。当尿液中加入乳清酸,尿液荧光强度明显降低。乳清酸本身不是荧光物质,因此,当人尿液的荧光强度明显降低时,乳清酸可能在尿液中大量产生。但是这种方法可能不能准确地鉴别尿液中是否产生大量的乳清酸。然而,当探针溶液被添加到尿液中时,在320nm附近发出强荧光。由于探针溶液发射峰的位置远离纯尿液的发射峰的位置,因此基本上可以认为探针在320nm处的荧光不会受到纯尿液的干扰。当在含有探针的尿液中加入乳清酸时,探针的荧光明显被猝灭。此外,随着乳清酸浓度的增加,探针在320nm处的荧光强度逐渐降低。因此,可以推测,所提出的荧光探针可以用作在实际尿样中检测乳清酸。从表2中可以更加清楚的观察荧光强度的变化。
表2
Claims (5)
1.定性检测尿液中乳清酸的方法,其特征在于,方法如下:取尿液,加入荧光探针EuШ-dtpa-bis(adenine)的水溶液,混合均匀,于280 nm波长光激发下进行荧光检测;所述荧光探针EuШ-dtpa-bis(adenine)为铕-二乙三胺五乙酸-双(腺嘌呤)。
2.如权利要求1所述的方法,其特征在于,所述荧光探针EuШ-dtpa-bis(adenine)的制备方法包括如下步骤:
1)将二乙三胺五乙酸、乙酸酐和吡啶混合均匀,在65 ℃下搅拌回流24 h,冷却至室温,减压抽滤,依次用乙酸酐和无水乙醚洗涤,60 ℃下干燥,得二乙三胺五乙酸二酐dtpaa;
2)将二乙三胺五乙酸二酐、三乙胺、无水DMF和腺嘌呤混合均匀,在100 ℃下搅拌回流24 h,冷却至室温,旋转蒸发,依次用乙腈和无水乙醚洗涤,减压抽滤,于50 ℃干燥,得二乙三胺五乙酸-双(腺嘌呤)dtpa-bis(adenine);
3)将二乙三胺五乙酸-双(腺嘌呤)和Eu(NO3)3·6H2O 分别加去离子水溶解,然后混合,于60 ℃搅拌加热2 h,冷却,得EuШ-dtpa-bis(adenine)。
3.如权利要求2所述的方法,其特征在于,步骤1)中,按摩尔比,二乙三胺五乙酸 :乙酸酐 :吡啶 = 1:4:6。
4.如权利要求2所述的方法,其特征在于,步骤2)中,按摩尔比,二乙三胺五乙酸二酐 :三乙胺 :腺嘌呤 = 1:3:2。
5.如权利要求1或2所述的方法,其特征在于,步骤3)中,按摩尔比,二乙三胺五乙酸-双(腺嘌呤):Eu(NO3)3·6H2O = 1:1。
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