主题：【求助】求助！The mean from the old (existing) solution is taken to be 100%一句该怎么理解

原文由 sunrain001 发表:
在欧盟《食品与饲料中农药残留分析方法确认与质量控制程序》标准品的核查一节中，有这样一段话The means from at least five replicate measurements for each of two solutions should not normally differ by more than ±5% . The mean from the old (existing) solution is taken to be 100%.请教后一句话 The mean from the old (existing) solution is taken to be 100%该怎么理解？
谢谢！

旧溶液的平均值视为100%。

谢谢楼上的回答，麻烦您可不可以再详细一点解说，将旧溶液的平均值视为100%是为了更方便与新配溶液作对比吗？该平均值指的是浓度还是什么？
附：原文整段如下：The means from at least five replicate measurements for each of two solutions should not normally differ by more than ±5% . The mean from the old (existing) solution is taken to be 100%. However, if the number of replicate determinations required  to distinguish a difference of  ±5% is unacceptably large for problematic analytes, the acceptable range may be increased to ±10%. The use of an internal standard may reduce the number of replicate injections required to achieve a ±5% difference. If a response of the old standard differs by more than ±5% (or ±10% in the case of problematic analytes) from the new, storage time or conditions must be adjusted as necessary on the basis of the results.

原文由 sunrain001 发表:
谢谢楼上的回答，麻烦您可不可以再详细一点解说，将旧溶液的平均值视为100%是为了更方便与新配溶液作对比吗？该平均值指的是浓度还是什么？
附：原文整段如下：The means from at least five replicate measurements for each of two solutions should not normally differ by more than ±5% . The mean from the old (existing) solution is taken to be 100%. However, if the number of replicate determinations required  to distinguish a difference of  ±5% is unacceptably large for problematic analytes, the acceptable range may be increased to ±10%. The use of an internal standard may reduce the number of replicate injections required to achieve a ±5% difference. If a response of the old standard differs by more than ±5% (or ±10% in the case of problematic analytes) from the new, storage time or conditions must be adjusted as necessary on the basis of the results.

我只是从翻译角度来讲的。这个existing substance在这里指什么还不太清楚，可不可以给全文？

另：
请将此帖转至英语翻译版，谢谢。
ruojun：不客气，谢谢帮忙翻译。

谢谢楼上各位关注！
这段话来自欧盟最新农药残留分析方法确认与质量控制程序（附件附上），上述一段话来自其中一段有关标准品的核查与替换，一共两段，如下：
Testing and replacement of standards
22.  Whenever any standard reaches its expiry date or is replaced, its purity should be checked. Existing stock and working solutions may be tested against newly prepared solutions by comparing the detector responses obtained from appropriate dilutions of individual standards or mixtures of standards. The purity of an old “pure” standard may be checked by preparing a new stock standard and comparing the detector responses obtained from freshly prepared dilu-
tions of old and new stock standards. Inexplicable differences in apparent concentration between old and new standards must be investigated.
23.  The means from at least five replicate measurements for each of
two solutions should not normally differ by more than ±5% . The mean from the old (existing) solution is taken to be 100%. However, if the number of replicate determinations required  to distinguish a difference of  ±5% is unacceptably large for problematic analytes, the acceptable range may be increased to ±10%. The use of an internal standard may reduce the number of replicate injections required to achieve a ±5% difference. If a response of the
old standard differs by more than ±5% (or ±10% in the case of problematic analytes) from the new, storage time or conditions must be adjusted as necessary on the basis of the results.
农药分析方法确认与质量控制程序

上文已经说得很清楚了，这个平均值是原标准液和新标准液相应的稀溶液检测器响应值的平均值。因为原标准液已过期或被取代，所以需要跟新制备的标准液进行比较，来确定原标准液的纯度。

当标准液已过期或被取代时，应当检测其纯度。方法就是重新制备一份标准液，将原标准液和新标准液分别制备新鲜配制的稀溶液，各重复进样至少5次，分别计算检测器响应值的平均值。将原标准液检测器响应值的平均值视为100％，则新标准液检测器响应值的平均值应该为95％-105％。（例如：原标准液在浓度为1mg/ml时检测器有响应，那新标准液就应该在浓度为0.95-1.05mg/ml时使检测器有响应。）在问题分析中，如果做到这一点所需要的重复进样次数大到难以接受的地步，则可以把范围扩大到±10％。使用内标可以减小重复进样的次数。如果原标准液和新标准液响应值均值的差异超过5％，则应当更改贮存时间和贮存条件。

这段话里有个矛盾的地方：从“The means from at least five replicate measurements for each of two solutions should not normally differ by more than ±5% . The mean from the old (existing) solution is taken to be 100%. ”来看应该是将原标准液检测器响应值的平均值视为100％，将新标准液跟它比较。但从“If a response of the old standard differs by more than ±5% (or ±10% in the case of problematic analytes) from the new, storage time or conditions must be adjusted as necessary on the basis of the results.”来看又应该是将新标准液视为100％，将原标准液跟它比较。我个人觉得应该是将新标准液视为100％，因为比较的目的是为了确定原标准液的纯度。不过我只是根据段落内容翻译的，具体是怎么做的还是应该请教具体做分析的人。

谢谢楼上详细解释，不胜感激！

楼上专家解释清楚明了，不胜感激，可有一处好像是专家粗心了：analyte是被分析物，所以应是分析不稳定的分析物时，可扩至±10%吧？

原文由 sunrain001 发表:
楼上专家解释清楚明了，不胜感激，可有一处好像是专家粗心了：analyte是被分析物，所以应是分析不稳定的分析物时，可扩至±10%吧？

不是粗心。这里的analyte的确是指被分析物，可是根据上文“Inexplicable differences in apparent concentration between old and new standards must be investigated. ”我觉得这里的problematic analyte不是指不稳定的被分析物，而是分析结果出现问题的被分析物，所以我意译为“问题分析”，但这只是从翻译角度来说的。