Limit of Detection (LOD)/Limit of Quantitation (LOQ)

Chromatography Forum: LC Archives: Limit of Detection (LOD)/Limit of Quantitation (LOQ)
Top of pagePrevious messageNext messageBottom of pageLink to this message  By mpaciolla on Thursday, May 9, 2002 - 06:26 am:

Would someone please share with me their views on how to determine LOD and LOQ. The ICH Q2 suggest at least four ways to do so: (1)Visual Examination, (2)S/N, (3)Standard Deviation of the response and slope, and(4)Recommended data. With (3), the standard deviation can be determined by standard deviation of the blank or based on the calibration curve. The latter requires residual standard deviation of a regression line or the deviation of y-intercepts of regression lines (plural). I'm not exactly sure how this is done. Is one way preferred over the others? Are the units for concentration and peak height critical? This can change the slope of the line. Any input would be greatly appreciated.


Top of pagePrevious messageNext messageBottom of pageLink to this message  By Marcelo B. on Thursday, May 9, 2002 - 07:53 am:

It would be interesting if you could provide more info regarding the application you're trying to carry out. LOD and LOQ could depend not only on the chromat. system (especially detector performance) but also on sample matrix, clean-up procedures, etc.
The simplest case would be the analysis of impurities in a bulk drug by simply dissolving the sample in the mobile phase. In this case one way to obtain LOD from a calibration curve
y = m.x + b
of the impurity within the expected range of conc. is to calculate the estimated std. deviation of the intercept Sb (see Miller and Miller, Statistics for Analytical Chemistry or any other book), to perform a t-test where the statistic is
t =absolute.value(b/Sb)
and to compare it to the value obtained from the table for the two-tailored test, 95% of confidence level and n-2 degree of freedom where n is the number of X,Y pairs used for the regression.
If the calculated statistic is lower than that obtained from the table, your won`t have experimental evidence for rejecting the hypothesis "intercept is not statistically different from zero". Thus you can get the LOD as the concentration required to obtain a peak response of Y(LOD)=3.Sb which will be
LOD = 3Sb/m, being m the slope of the calibration curve. All this calculation can be done by using Excel. Thus, one should verify this by preparing a sol. of such conc. and performing replicate injections. LOQ is similar but using Y(LOQ)=10.Sb

But there are many other situations, for instance bioanalytical method validation guidance (CDER-FDA) defines lower limit of quantitation (LLOQ) as the concentration that yield a peak having a response at least 5 times the response of any blank peak (due to the biological matrix, determined in at least 6 blank samples of diff. sources) and also LLOQ peak should be identifiable, discrete and reproducible with a RSD<= 20% and accuracy of 80-120%. So, its determination is absolutly experimental.
Hope this helps

Marcelo


Top of pagePrevious messageNext messageBottom of pageLink to this message  By mpaciolla on Thursday, May 9, 2002 - 08:37 am:

Marcelo,
Thanks for responding. We are simply determining the LOD/LOQ for drug degs and impurities by HPLC equipped with a UV. If we use the standard deviation of the blank, we get values that are below our experimental data. We haven't tried using the standard deviation of the y-intercept in the formula (10Sb/m) because it states y-interceptS of regression lineS - plural. Why is it plural? We make several injections at each concentration, therefore, are we suppose to treat the injections at each level separate in order to generate several regression lines?


Top of pagePrevious messageNext messageBottom of pageLink to this message  By Marcelo B. on Thursday, May 9, 2002 - 10:45 am:

I've been reviewing the ICH Q2B 6.3.2 and 7.3.2 and I understand from the text that it give you two alternative ways
1) to use the residual SD from a single calibration line
2) to use several intercepts values from different calibrations lines, That is, run different std preparations and calculate the intercept from each regression line and then use them to calculate the SD with the usual formula.

Both are different ways to estimatate the SD of the intercept, which also depend on the calibration model you intend to use -there was a great discussion regarding this two weeks ago.
I'm not sure if they are equivalent (I'd need to do some calculation first).
You should keep in mind that they are only estimates, you need to validate the estimated LOQ and LOD by injecting replicates samples of each estimated concentrations (see 6.4 for LOD and 7.4 for LOQ) and calculate RSD (LOD and LOQ) and also accuracy for the LOQ and verifying that meet previosuly established criteria in your validation protocol (for instance, RSD of LOQ<=5% and accuracy ±5% or whatever).
In my opinion, I'd prefer to run a single calibration experiment using 4-6 replicates (dilutions) of different conc. levels starting near an estimated LOQ covering the expected rage of analyte conc. (inject first 1 std. sol and calculate the conc. to obtain a peak response of .
In this way you can obtain more info from a single experiment (linearity, LOQ and LOD)as I pointed in my previous answer.

Regading the standard deviation of the blank:
What is a blank in your case? You'd need a drug neither with degr. nor impurities, I don't think this is practical. The other possibility is the solvent you are dissolving your samples, for me it does not make any sense to calculate LOD in such ways for drug substances. Blank refers to sample matrix, for instance a tablet placebo, free-drug plasma, etc.
I'd try with the 3SD and 10SD method.
I'm sure that someone else can give alternatives

Regards

Marcelo


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