Calculate impurity content using linearity

[tubish81]

Hii,
We calculate impurities using regression equation. We don’t Just use 100% standard solution. We prepare LOQ, 100% and 120% solutions in release test and calculate using y=mx+b
Authority asked “Why do you substract intercept value from impurities area? Please show scientific source.”
Example:

Impurity content % = (A-b)/m x 100/5 x 100/L

A: Impurity area from test solution
b: intercept value from standard solutions linearity
m: slope from standard solutions linearity
100/5: dilution factor
L: Label claim

How can we response the question?

Thank you

[sbashkyrtsev]

Could you elaborate on what you're trying to achieve? What's the end goal?

Do I get it correctly that:
1. You have another source of pure Impurity and you make a solution with it at LOQ, 100% and 120% relative to the concentration that you expect to find? And in your y=mx + b, y is concentration (%) of the impurity in the injection?
2. You take your test vial with L% impurity and dilute it another 20x?

[tom jupille]

Did "Authority" fail math class?

Once you agree that your calibration line is "A = mx + b", then solving for "x" (concentration) gives "x = (A - b)/m", which is exactly what your formula says. Your "scientific source" could be a basic algebra textbook.

I would be much more concerned about inclusion of the LOQ point.

[DR]

If you're feeling cheeky, reply with "As far as I know, not all lines pass through the origin. If you have a scientific source indicating otherwise, I'd be happy to consider it."

[James_Ball]

If you're feeling cheeky, reply with "As far as I know, not all lines pass through the origin. If you have a scientific source indicating otherwise, I'd be happy to consider it."

Good answer.

Authority=Lawyer=Little knowledge of linear algebra

But Tom's answer is the correct one. Simple linear curve fit math, when solving the equation you get -b. I think you would only not have -b if you are forcing through the origin, which usually leads to bias high results for lower concentrations.

[tubish81]

Thank you for your good answers:)

It is a finished product. For example, we will calculate the unknown impurity in molecule A. We draw linearity using the working standard of molecule A. We also calculate the unknown impurity with this regression equation. I know it's more accurate to calculate with a 100% standard, but it's a study before me and we're expected to explain it. I answered the question before, and they persistently asked it again. There are many scientific articles that calculate in this way. I don't know if anyone who asks this question will accept these articles:)

[DR]

Tom's answer plus a "it is very commonly done this way in this industry" should suffice.

[lmh]

The problem that is probably bothering them is that if your line passes at any distance significantly above the origin, then you could have small peaks that are visible to the naked eye but don't show up at all because they give negative values. If, for example, your linearity were so poor that the intercept were 10% of the size of the pure compound peak, then you could have individual impurities of up to 10% without them showing up.

10% is obviously ridiculous. You're probably being very professional about this, but somewhere out there is a lab who are doing the same thing, with an intercept that's 1%, or 2%, and who are busy selling products as 98% pure when they have 15 impurities at around the 1% mark by peak-area, and one impurity around 4%.

The final answer is that your method is good provided the line passes close to the origin. There is an LC-GC article somewhere that touches on the situation, but in reverse. The question was "when is it acceptable to force my calibration curve through the origin?" and the answer was "when it passes within a statistically insignificant distance of the origin anyway".

In your situation, the question is how far down the curve can I go before the fact that I did/didn't force to the origin makes a statistically-significant difference to the result? So the way that I'd answer the question is to stress that you have already tested the LOQ. You therefore have a method that is valid down to the LOQ. Your method may fail on impurities that are a long way below the LOQ, but by definition these are not of any practical relevance because in choosing your method, you ensured that it had an adequate LOQ to pick up all relevant impurities. And you are routinely including the LOQ as part of your three-point calibration curve, so your method is valid within the region LOQ < result < 120%, the range of the calibration curve. And that's in ICH Q1(R1).