Monday, November 05, 2007

Integration for Idiots, Part III: Adjacent Peaks

(The Perils of Two) Adjacent Peaks

Series by Ali and TBV


In Part I, we learned the effects of altering the left and right integration limits, and in Part II, we learned about background and background subtraction, both involving a single peak.

Here in Part III, we'll see some of the surprising results when you have more than one peak and they get close enough together.

Figure 11 shows the addition of another peak with identical o/oo value to our original peak )-27). They are positioned so that their tails just overlap. Although not perceptable to the naked eye, this has raised the left hand tail of our original peak slightly. This minor co-elution has resulted in a measured value of -27.4. The true value for this peak is -27.


Figure 12 shows a more significant co-elution. This has caused a measured value of -29. The true value for this peak is -27.


Figure 13 shows a significant co-elution. This has resulted in a measured value of -32.8. The true value for this peak is -27.


In Part IV, we'll see what can happen when there are three peaks (and you only think there are two).


Reminder: you can try scenarios yourself with our spreadsheet. (It's hours of Fun!)



9 comments:

Larry said...

TBV and Ali -

NOW I'm lost.

First, why is this problem different than the problem in Part I? Isn't Part III another example of superimposing one peak over another? (Only in Part I there was a lot of superimposition and in Part III there's only a little bit of superimposition)

Second, how come the numbers you're showing increase from Figure 11 to 12 to 13, but the shaded area under our peak decreases from Figure 11 to 12 to 13?

(I KNEW this would happen. I didn't pay close enough attention in Parts I and II, and now I've fallen behind and I'm going to have to hire a tutor.)

tbv@trustbut.com said...

Larry, as I touched on in answer to your comment in Part II, yes, mathematically they are similar, though here we are dealing with 4 different peaks, two each C12 and C13, with offsets.

Still it seems tutorially useful to show what happens.

You second point shows why it's useful, as you've hit on one of the counter-intuitive lessons of the example.

The reported results are not related to the total size of the shaded area, but of the ratios.

You'd be quite right to wonder if the sloped background subtraction in the right peak is really doing the correct thing, but that is the way it would be typically computed automatically. You see, if you knew what the correct values for the peaks were, you could probably adjust the integration markers and the background subtraction to get the right result.

And if I had some ham, I could have ham and eggs. If I had some eggs.

Once peaks run into each other and there are unknowns, you don't really know what the truth is.

You are approaching the understanding of a point WM-A made in one slide that is taking us 20 to work towards. But we are not there yet.

TBV

Larry said...

TBV, so there are TWO curves in each curve! OK. Actually, if I stare at the drawings long enough, I get double-vision and momentarily I can see what you're talking about.

Then the shaded area you're showing in the Part III examples is the area of the C12 curve, which is why the values go up when the area goes down?

As long as this is the idiot's class ... why are the values expressed as a negative number? If C13 is a positive amount of carbon 13 and C12 is a positive amount of C12, shouldn't C13/C12 be a positive number?

Another question you'll hopefully get to: I understand that it's hard to measure the area of these curves ... but do these difficulties introduce an equal margin of error (plus or minus) or do they tend to skew the measured results in one direction? I'm sure you know why I'm asking.

Larry said...

Y'know, that's a pretty good question I just asked.

It would appear from the examples you've set up that, if you have two pairs of IRMS peaks that are close together, and you needed to measure the C13/C12 value of the second set of peaks, you would tend to guess high more often than you'd guess low. And vice versa.

Ali said...

What you're looking at is the C12 plot, obtained from a sensor which measures the intensity for this isotope(also known as the m44). Another separate sensor picks up the C13 intensity (also known as m45). The software calculates the area between two limits, after removing the background selected (for both curves) and then works out the ratio of the C13 area to the C12 area. This is then converted to a o/oo value by 1000*((C13/C12) - PDB))/PDB, where PDB is a reference standard. Due to the slight time difference between the two plots, you have to be careful with your integration limits and background removal so that you capture the full area of both when performing the integration. The eventual o/oo value is very sensitive to the C13/C12 ratio.

tbv@trustbut.com said...

if you have two pairs of IRMS peaks that are close together, and you needed to measure the C13/C12 value of the second set of peaks, you would tend to guess high more often than you'd guess low. And vice versa.

If you know it's only two peaks, that's true, I think. But if there are more buried peaks, things go kerblooey. We'll see some simple cases of this in Part IV.

TBV

Larry said...

Ali -

I'm lost again.

I don't follow your 1:02 AM post. There are SEPARATE sensors for picking up the C12 and the C13 ions? Then what exactly was the problem in Part I of this course? I thought the Part I problem was that the C12 curve overlapped the C13 curve and you couldn't easily distinguish the two curves. But if the two curves are each determined by separate sensors, then you have the data for each curve. Why can't you refer to this data to determine the left and right integration limits?

I can see where having separate sensors does not solve all our problems when we're dealing with multiple sets of C13-C12 curves.

I don't know what you mean when you say that the software calculates the area between two limits. Is that the same as the area of the peak?

What is a 0/00 value? (thought you couldn't divide by zero)

I asked why the numbers shown for each pair of curves is always a negative number. From the formula, the number could be negative if the PDB reference standard is greater than C13/C12. But if that's the case, then the absolute value of the number shown for a curve should get smaller as C13 gets bigger, and I thought the opposite was the case.

And while I'm asking questions ... I thought that there would be a lot less C13 than C12. Wouldn't the C13 curve be a lot smaller than the C12 curve?

tbv@trustbut.com said...

Larry, pull up the spread sheet.

There are three graphs, the 44, the 45, and the 44/45.

The displays we see in the LDP are always the 44s. Sometimes we get the 44/45.

If you have pure peaks, that is indeed all you need. Note the predicate: If you have pure peaks

Yeah, area of the peak.

"o/oo" is the symbol the relevant geeks use for the ratio. Maybe because one has the extra neutron?

Why it's always discussed negatively has always confused me too, but I never chose to figure it out.

Look closely at the graph scales in the spreadsheet. The m=44 goes from zero to 8E-11, while the m=45 goes from 0 to 8E-9. Big difference in scale.

TBV

Ali said...

Larry,

Good question and one which I couldn't figure out myself for a long time ... 'why not just set up your integration limits independantly on each plot ... problem solved'

First of all o/oo is shorthand for parts per 1000. I don't know who thought it up, but as parts per 100 (percent) is written as o/o, parts per thousand is indicated by adding an extra 'o', giving o/oo.

Back to the C13 time difference. First off, given the very small signals (pico amps), in the real world, the signal to noise ratio is worse than that of the C12 plot (nano amps), so it may not be that easy to always determine exactly where the C13 peak starts and stops. That's one reason. The other is to do with the differences between the C12 and C13 plots. For example, if you have two C12 peaks of equal magnitudes, one with a o/oo value of -10 and one with a o/oo value of -50, their corresponding C13 peaks will have different magnitudes. The -50 C13 peak will be smaller than the -10 peak ( because it has less C13). If those two peaks were co-eluting, even modestly, their relatively different magnitudes could shift the trough between them away from the bigger peak, toward the smaller peak. This could make the already smaller peak look even smaller than it actually was. That's another reason and that's why the preferred way is to look at the 45/44 (C13/C12) instantaneous ratio plot to see where the C13 starts and the C12 stops. Great if you have well separated peaks. A problem if you don't. That's when things get difficult.

o/oo is usually negative because it represents the ratio of C13/C12, relative to the standard PDB, which has a higher C13/C12 than many of the substances we're concerned with in these discussions. So when you calculate your substance o/oo as 1000*((C13/C12)-PDB)/PDB, if your substance C13/C12 ratio is less than the PDB C13/C12 ratio, you're going to get a negative number and it will be expressed in parts per 1000.