推荐回答(1个)
欲知封闭水域会否出现藻类疯长,如蓝藻爆发等现象,应监测水域中的藻类数量以及水质,由于对藻类等浮游植物采用计数的方法测定误差较大,耗时费力,对检测人员的工作经验要求相对较高,一般可测定水中的叶绿素a含量代替藻类测定。当水中的叶绿素a含量突然增高,而且水中含有大量氮、磷等营养物质,加上阳光照射强烈,气候炎热等因素,该水域极有可能发生藻类疯长,可通知各有关部门尽早采取应对措施。
因为藻类是一类含叶绿素的、光合自养的、无胚的原植体植物,在浮游藻类里叶绿素a的含量大约占有机物比重的1~2%,是估算藻类生物量的较好指标。可预先测定藻类计数和叶绿素含量的相关关系,以叶绿素a的含量来推算藻类的数量,即通过测定水中的叶绿素来快速了解藻类的大致数量。
测定叶绿素a的仪器和方法有许多种,分光光度法测定叶绿素a是一简便易行的测定方法,水样经离心或过滤浓缩、研磨、丙酮提取后,定容,取上清液分别测量750nm、645nm、663nm、652nm等几个波长下的吸光度值,根据经验公式可分别计算出叶绿素a、叶绿素b和总叶绿素的含量。
分光光度法测定叶绿素a,与测定其他物质稍有不同,如:测磷只需测定单一波长的吸光度值,再以该吸光度值代入由标准溶液测得的校准曲线计算含磷量。而叶绿素a无法使用校准曲线,需用几个波长下的吸光度值,根据经验公式来分别计算出各项指标的含量。因为叶绿体色素由叶绿素a、叶绿素b等物质组成,试液是多组分的混合溶液,在试液中分离这几类物质的难度较大,且无必要。叶绿素a在645nm和663nm 处均有吸收,在645nm处吸光系数较小,为16.75,在663nm 处较大,为82.04;叶绿素b 在645nm和663nm 处亦都有吸收,但在645nm处吸光系数较大,为45.60,在663nm 处较小,为9.27。由此可知:叶绿素a的吸收峰值出现在663nm 处,该吸收曲线延伸到645nm处,在此波长处的吸收系数不如在663 nm 处大,因此在计算公式中求算叶绿素a的含量时,需扣除叶绿素b在663nm和645nm 处的吸光度值,再进行计算。
标准分析方法要求,叶绿体色素提取液不可浑浊,在710nm或750nm波长下测量吸光度,其值应小于叶绿素a吸收峰的吸光度值的5%,否则应重新过滤。假定样品在663nm处的吸光度值为0.03,则在750nm处的吸光度值不得大于0.0015,对试液的清澈程度要求很高,测量710nm或750nm的目的是避免悬浮物质的干扰,一般测量水中的浑浊度所采用的波长为680nm,为避免在680nm处仍有叶绿素a产生的吸收值,故将测量浑浊度的波长选在710nm以上。在计算公式中,凡参与计算的各吸光度值都应减去710nm处的吸光度值,以扣除悬浮物质的干扰。
采用分光光度法测定叶绿素含量,对测量仪器分光光度计的波长精确度要求较高。如果波长与原吸收峰波长相差1nm,则叶绿素a的测定误差为2%,叶绿素b为19%,使用前必须对分光光度计的波长进行校正。校正方法除按仪器说明书外,还应以纯的叶绿素a和b来校正。
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