cdetermination of LOQ from Linearity

[ydna6969]

Hi ,,

Just trying out this way of calculating LOQ.

Concentration range from 240 - 740 µg/ml (area mAU 29540893 - 88365800)

Intercept = 290228
Slope = 120312
Standard error = 348366

So would LOQ = (348366x10)/120312 = 28.955 ug/ml

Though Y=MX+C at this level is Y=120312*28.955+290228 = 3773861 and peak would clearly be above S/N 10:1

Any thoughts much appreciated

[Fernando]

Hi ydna6969

I think your LOQ is very high, almost 30 µg/ml, I would try the range for linearity 7 - 14 - 28 µg/ml or more points (near the origin).
The second thing you have to do is run a blank and measure de noise (N), I inject it many times and take an average. Then calculate the LOQ= S/N. With S: signal height and N: noise, if it is equal or higher than 10 you are OK.

Best regards

Fernando

[lmh]

You're really interested in the vertical error (error in the y-component rather than error in the slope) at the actual limit of quantification - how does this error compare to the acceptable error? The measured SE is an estimate. Since your calibration curve actually starts at a concentration ten times your estimate, the estimate is a considerable extrapolation from where the data were collected. In these circumstances it's probably best to go back and measure a new set of calibration points at a lower concentration range, around the estimate you've got. Unfortunately LODs and LOQs can be an iterative process where the first go just tells you that the method is a lot better than you expected!

Personally I don't much trust statistics unless they "look sensible" anyway. With LODs for example, I do want to see a peak that looks peakish to me (and that I can persuade the instrument software to integrate), at the expected LOD, before I really believe that it's a LOD and not just a number. You may feel similarly...

[Peter Apps]

Two things strike me as odd.

Your calibration range is very short - the highest standard is only about three times the concentration of the lowest. This makes it very difficult to determine if the response is anywhere near linear over the range that you need to extrapolate to. How many points did you manage to squeeze into that range ?

The SE is larger than the intercept - which implies poor repeatability (points scattered either side of the line). What is the repeatability at a single calibration level ?

Peter

[ydna6969]

Hi ,
I agree,

this is a bad example to use (50-120 %)

Repeatability is fine ,

My usual approach to LOQ and LOD is 0.04-0.05% of nominal and check S/N ratio.
This extrapolation doesn't seem right to me I must be missing something though
Other suitable data below (10-120%) approx. 10-120ug/ml

Slope = 24892
Intercept = 346
Se = 2516

LOQ = (10*2516)/24892=1.01 ug/ml..

Not correct as 10ug.ml (10%) = 250489 area and I can get Loq to 0.05ug/ml (approx. 1250 area)

Or I am using the incorrect SE from ANOVA ?

X variable SE = 25
Intercept SE = 1703

[ydna6969]

Further calcs ,

I maybe wrong but don't you use SE of X variable

but in this case I get 0.01ug/ml .. no LOQ s/n >10 at that level at all !

[HPLCaddict]

LOQ = (10*2516)/24892=1.01 ug/ml..
Not correct as 10ug.ml (10%) = 250489 area and I can get Loq to 0.05ug/ml (approx. 1250 area)

Do I understand correctly that you are trying to predict s/n values from calculated peak areas?

[ydna6969]

The area is what the Loq OF SN>10 just for information .

The regression line is conc v area.

This is where I am getting abit lost ,,, how can you 'predict'loq levels of a regrsssion conc v area ?

[Peter Apps]

The S:N >10 criterion for LOQ does not take into account the variation in peak size caused by sample prep and analysis - consequently it always gives optimistically low values, and is popular with the authors of papers as a result.

The extrapolation of the calibration, and the estimate of the standard error or standard deviation of the intercept does take analytical variability into account, and gives higher and much more realistic estimates for LOQ and LOD. As mentioned in previous posts it works well only if you have calibration points close to the LOQ, a good range of calibration, and a close fit to whatever line you extrapolate.

Peter

[lmh]

question of which error you're looking at: I usually plot calibration curves with the response on the y-axis and concentration or amount on the x-axis, but some software does it the other way round, or offers the choice.
When fitting the calibration curve, it's important that the least-squares regression is done in the same direction as the likely errors. You are choosing the best fit that has the least error, so it wouldn't be logical to make the curve as close as it can be to the points sideways when the error you're trying to minimise is up and down. For this reason, I think the standard error has to be measured initially in the direction of the response (which I put on the y-axis).

The potential confusion is that actually you don't want to know the standard error in the response. You want to know the standard error in the final measurement, which is amount or concentration, and therefore the x-axis in my calibration curves. Fortunately, the uncertainty in concentration can be calculated from the uncertainty in response by using exactly the method you'd use to find the actual concentration in a sample from the actual response: dividing by the slope of a linear calibration curve. This, then, is what you do when finding the limit of quantification. The extra constant is just there so that you get the error-window that someone once decided was acceptable, but theoretically you could use any multiplier you wanted, dependent on what level of precision you want.