24小时热门版块排行榜

>论坛更新日志 (2804)
>虫友互识 (370)
>导师招生 (132)
>文献求助 (112)
>硕博家园 (74)
>休闲灌水 (73)
>考博 (27)
>论文投稿 (23)
>博后之家 (22)
>基金申请 (19)
>公派出国 (18)
>教师之家 (17)
>考研 (17)
>标准与专利 (15)
>找工作 (15)
>论文道贺祈福 (14)

返回列表

lifuchengol

应助: (幼儿园)
在线:
虫号: 1074609

[交流] 【求助】求助：关于仪器检测限相关问题？已有1人参与

请问一个关于检测限（LOD）和定量限（LOQ）的问题
我的一篇文章投到analytical chimica acta之后，编辑一直问一个关于LOD和 LOQ的问题。我定义了检测限为3倍信噪比，定量限为10倍信噪比。我做的时候是标准样品倍比稀释，浓度分别为40, 20, 10, 5, 2.5μg/kg，然后用HPLC检测，得到的LOD5μg/kg,为LOQ10μg/kg。但是编辑认为我既然定义了检测限为3被信噪比，定量限为10 被倍信噪比，那么定量限应该为检测限的3.3倍，可是我的浓度本来就是倍比稀释的啊，信噪比大于10的最低浓度为10μg/kg，信噪比大于3的最低浓度为5μg/kg。
我不知道怎么解释他们才能听懂，还是我应该改变检测限和定量限的定义呢？着急，谢谢好心人！
如下是编辑的comments
In your response to my comments you have written "We defined LOD as the lowest drug concentration in the samples can be measured with a signal to the background signal (S/N) ratio of 3/1(S/N), and LOD as the lowest drug concentration in food samples can be measured with a signal to the background signal (S/N) ratio of 10/1(S/N). After the detection, we found that the LOD was 5 g/kg and the LOQ was 10 g/kg." If 5 µg/kg gives a signal ise ratio of 3/1, 10 µg/kg cannot give a signal ise ratio of 10/1 (it would give just a signal ise ratio of 6/1) !!! Either your value for the LOD or your value for the LOQ is wrong.

回复此楼