高效液相色谱法测定血清及尿中的速尿含量

本文通过简单的预处理,应用高效液相色谱法进行了血清及尿中的速尿含量测定。本法简便、快速、重现性好。采用了血清以乙腈除蛋白、尿稀释50倍后测定的预处理法。高效液相色谱条件为:反相柱,以含30%乙腈的5mM 己烷磺酸钠水溶液作流动相(pH3.5),流速2.5ml/min。使用荧光检测器,激发波长333nm、荧光检测波长415nm。血清及尿中速尿的回收率分别为93.5—101.7%及90.8%。重现性也得到良好的结果,用血清测定时,日内变异系数为3.48%,用尿测定时,日内变异系数为2.58%。

DETERMINATION OF FUROSEMIDE IN SERUM AND URINE BY HIGHPERFORMANCE LIQUID CHROMATOGRAPHY

A simple,rapid and reproducible method for the determination of furosem- ide in serum and urine by highperformance liquid chromatography has been de- veloped.Only 100 μl of serum is required.Serum proteins are precipitated with 100 μl of acetonitrile.After centrifugation,100 μl of the supernatant and the same volume of the counter ion solution(0.01 M 1-hexanesulfonic acid sodium salt,pH 2.8)are drawn into a microsyringe successively.Urine is diluted with water to 50 times.Fifty μl of the diluted urine and the same volume of the counter ion solution are drawn into the microsyringe successively.After mixing the two solutions in the microsyringe,the Whole mixture is directly injected into the chromatograph.The chromatograph used is Waters Associates consisted of a model 6000A pump,U6K injector. A Radial Pak μ Bondapak C_(18)(10 μm,10 cm × 8mm I.D.)cartridge is used as the analytical column.The flow rate of mobile phase,30% acetonitrile solution containing 5 mM 1-hexanesulfonic acid sodium salt(pH 3.5),is 2.5 ml/min.The eluate was monitored with Hitachi F 1000 fluorescence detector. The wavelength of excitation is 333 nm and that of emission is 415nm.The re- tention time for furosemide is 11 min.A calibration curve is linear in the range from 25 ng/ml to 10 μg/ml for the serum and from 1.25 μg/ml to 25 μg/ml for the urine.The limit of detection is 2.5 ng/ml of serum.The recoveries of ad- ded furosemide are 93.5-101.7% for the serum and 90.8% for the urine.For the serum,within-run and day-to-day reproducibility for 116.5 ng/ml,233 ng/ml, 466 ng/ml are 3,48%,3,40%,2.66%(n=10)and 2.070.%,1.44%,0.40%(n= 6)respectively.For the diluted urine,within-run for 500 ng/ml is 2.58% (n = 11).