Interference Using PDA and fluorescence detectors in series

[HW Mueller]

Danko, you are getting closer.

[danko]

[Uwe Neue]

I don't see that we have gotten to an answer yet. Gregory's citation applies to an scanning UV/Vis detector, and not to a PDA.

The way I see it is that the reference wavelength at 400 is just smack in the middle of the fluorescence emission spectrum. Also, the quadratic (or parabolic) response curve speaks for the argument that I had presented.

If possible, I would set the PDA reference wavelength above the fluorescence spectrum, at 500 nm or higher. If the curved UV response disappears, or at least diminishes, my argument about the wrong choice of the reference wavelength of the PDA was correct.

I hope that Gregory is still listening in and has not taken off into cyberspace.

[lmh]

Sorry, I'm feeling very stupid at the moment.

The absorbance value measured by the machine is log(I/I0). It is curving upwards, positive squared term in the quadratic fit?

Just to check I'm understanding: The point about the log10(I/Io) thing is this: If only 10% of light comes out of one flow cell, the reading is 1. If we add a 2nd flow cell in the light-path, then 10% of incident light comes out that one too, 10% of 10% is 1%, and the absorbance reading is 2.

So the upward curve means that if we add a 2nd flow-cell in the light beam, the 2nd one absorbs a greater proportion of incident light than the first??? How??

You're telling me that either (1) the percentage of light removed by the solution depends on the amount of light entering it; the less light enters, the more of it is absorbed. If so, I don't understand the physics. Or (2) there's a positive interaction between fluorescent molecules so that if we add a second one in the vicinity of the first, together they absorb more than they could absorb were they both in the light-beam but unaware of each other's existance. Please could someone explain how?

And let's view the extreme. This upward curve means there comes a point (probably at not too high a concentration) where it's completely impossible to get any light through the solution at all. This really doesn't fit with either Lambert-Beer absorbance or fluorescence. Absorbance is always a proportion of total light, never everything. And fluorescence is merely re-emitting of absorbed light at a different wavelength, and in all directions.

I can see how fluorescence might make an absorbance value come out a trace too low (if you're measuring absorbance at the emission wavelength), but I can't see how fluorescence can make absorbance come out too high.

Please help, someone, in nice simple terms... Thanks!

[HW Mueller]

Imh, all you need to know is A = abc and Aq = F,

A is the absorbance (the machine does the I/Io for you, it puts out A so for an explanation here, forget this ratio)
a is the specific absorbance
b is the path length (of the cell)
c is the concentration of the absorbing substance
q is the quantum yield
F is the fluorescence intensity

The only difficult thing here is to get your units straight. Practically, it is important to stay away from situations which are far from ideal. The formulars given here represent the ideal.

[Uwe Neue]

To lmh:
The adsorbance measured by the instrument is A = log (I0/I). It is a small positive value for small concentrations.

Most UV detectors for HPLC are single-beam detectors, with a single flow cell. So in general, there is no reference beam, and the signal I0 is not an actually measured signal, but the signal that the detector "remembers" when it set itself to 0. If you do this, then there is generally not a problem, until you run out of light.

However, Gregory has observed that there is curvature in his PDA response curve. He was not entirely clear on this, but I believe that he is getting an upward curvature, i.e. an overproportionally higher response at higher concentrations. With the model above, this is can not be explained.

He also says that the analyte is fluorescing. OK, this means that some light of a different wavelength is created inside the cell. Now we need to remember that he is using a PDA, which means that after the flow cell, the light is divided up into different wavelength. The fluorescence maximum is typically 100 nm higher then the absorbance maximum (he is measuring at 305 nm). So at the absorbance wavelength of the PDA, one should not see any of the light generated by the fluorescence, and everything should still be OK.

However, there is a problem. Modern PDA's use a reference wavelength out of the entire spectrum to reduce noise caused by lamp fluctuations and other events. If this reference wavelength is in the range where the analyte fluoresces, this creates a problem. The PDA thinks that the lamp is brighter (= higher I0) therefore it concludes (using the definition of the absorbance noted at the beginning) that the absorbance is higher then it really is, which in turn causes an upward curvature of the response curve.

Note that the reference wavelength was set at 400 nm, which could very well still be within the fluorescence spectrum. My recommendation is to either switch the reference wavelength off (if this is possible), or to go to a higher reference wavelength beyond the fluorescence spectrum.

I hope that this was clear enough.

[lmh]

Uwe Neue,

Thanks for the very clear explanation; your clarity helps me to understand the argument, and provides the opportunity to test it!

I'm still unhappy about this effect causing a measurable upward curve. When I tried a quick excel model of it, with an assumption that fluorescence light climbed linearly with concentration, reaching 5% of the strength of the true reference beam by a concentration corresponding to an absorbance of 1.0, the deviation from linearity was immeasurably small, and in any case, more downward than upward (negative x-squared term). The slope of the straight-ish line is, of course, steeper upwards than it ought to be.

If I allow the fluorescent light to reach 50% of the strength of the true reference beam by an absorbance of 1.0 (a highly unlikely situation), the line is only just noticeably curvy, and still heading downwards, not upwards.

I wish I were a mathematician!

[Uwe Neue]

Very good point, lmh!!!

However, you will get an upward curvature, if there is quenching in the fluorescence spectrum. So we really need the inputs from Danko and HW as well, unless there are other non-linearites in the system that are caused by the use of a wide wavelength window.

[HW Mueller]

In a thinking experiment, using Uwe´s suggestion as I understood it, there should be an increase in the absorbance signal. First my understanding of Uwe´s case: Your absorbance detector is good enough to cleanly distinguish between the absorbance wavelength and the wavelength of the fluorescence. It measures absorbance without interference by fluorescence. But: Your detector is also monitoring the light of a wavelength different from the absorbing wavelength, the reference. This it does for compensating variations in lamp output, etc. Now one might have the special cleverness (sarcastic) to set this reference on the fluorescence. When increasing concentrations of analyte go through the detector the fluorescence will go up. The detector measures more light at the reference and will think that the light source is putting out more light (or whatever). But, since the light source didn´t change, the absorbance is where it should be. The machine thus must think the absorbance went up (the reference gets more light so corrects the Io up, the ratio I/Io goes down, Io/I goes up, and absorbance is up).
So, either my reasoning is hung up somewhwere or Imh´s model is different from what I described here..

[lmh]

My model was this:

Set up a column in Excel containing increasing values of "concentration". These are proportional to absorbance.

Given a constant, true Io (say = 100), calculate (from the absorbance) the true I

Now calculate a fluorescence to increase linearly with concentration

Add this fluorescence to the reference beam Io (=100) to give a gradually increasing reference beam.

Calculate the measured absorbance as log(Io/I) using the true I measurements and the inflated-by-fluorescence measured Io values.

Now plot the measured absorbance against concentration. The line is higher than the true absorbance line, setting off from the origin with a greater slope. However, it curves back downwards, very slightly, for any value of fluorescence that I tried.

There's no guarantee my model is right: I am not an expert on hplc detectors, absorbance, physics, or maths! (and I know nothing about fluorescence)

[Peter Apps]

It seems to me that any model to explain the deviations from linearity will depend on whether the deviations are positive or negative. Looking back through the thread the only person who knows this is Gregory, and he aint telling !!!

Peter

[HW Mueller]

Imh, so you got a positive deviation throughout! Tough birth.
Gregory, we will suppress our chuckles . . . . what did you really do?

[lmh]

Sorry all, I think I'm getting hung-up on two sentences from early in this thread:

(1) Uwe Neue:
"Is your response curving upward (positive squared term) or downward (negative squared term)?"

(2) Gregory: (in reply)
"The absorbance response curve for the fluorescent molecule is concave shaped and fits fairly well with a quadratic function."

It's difficult to express the difference between a slope and a rate of change of slope. I wasn't thinking about whether the slope is correct; if you have a calibration curve with known standards, who cares? We're aiming to quantify an analyte, not measure its extinction coefficient. On the other hand, if there is a significant deviation from linearity, we want to understand why. And there is no way that curve should be curving upwards. Unfortunately "concave" is still ambiguous!

[Uwe Neue]

It IS possible for the calibration curve to curve upward. If you send me an e-mail, I'll send you my spreadsheet with the model.