Labeling with stable
isotopes is the best method for calculating the flux of metabolites through
biological systems.

^{13}C is the isotope of choice for most small organic metabolites, and labeling with^{13}CO_{2}under physiological conditions gives us the most realistic picture of how the plant's metabolism works in its natural environment. This approach is known as whole plant kinetics . In order to calculate the degree of label incorporation into the pool of metabolites following a labeling experiment, it is first necessary to extract those metabolites from labeled plant tissue and separate them by gas or liquid chromatography. We can then collect the spectra one by one as they elute from the column and enter the mass spectrometer. The resulting mass spectra tell us the pool sizes of the different isotopologs of a given metabolite (i.e. related chemical species which differ only in that they bear one or more^{ 13}C isotopes compared to the all^{12}C reference molecule). Depending on how the molecule fragments under a typical ionization technique like electron impact, it may be impossible to say where in the molecule these^{13}C isotopes are located by mass spectrometry; we can only say how many isotopic labels are present. We can however determine the relative abundances of labeled species that have 1, 2, 3, or more isotopes compared to a reference compound (usually unlabeled). But this spectral data still doesn’t tell us what we really would like to know following a labeling experiment: what percentage of the total carbon atoms in a metabolite pool are represented by^{13}C instead of^{12}C?
The answer to this question is the

**% atom labeling**of a metabolite, and this information can then be used to calculate the flux of material through a pathway, the turnover time of a metabolite, and even elucidate the order of steps in a new pathway whose biochemistry is unknown. Each of these subjects is a world unto itself, so here I will just focus on how % atom labeling can be calculated from mass spectral data.
Let’s start with a visual
example using the plant volatile isoprene (molecular weight 68), an olefinic
five carbon molecule released by many plants into the atmosphere.

Depending on the ionization
technique, isoprene can produce really complex spectra because of its tendency
to rearrange, particularly the intact or molecular ion ([M]

^{+}) which tends to lose a H atom to form [M-H]^{+}. This makes it difficult to tell the difference between a [M-H]^{+ }ion^{ }that carries a single^{13}C label and an intact [M]^{+}ion with no labels since at unit resolution, they appear to have the same mass. For simplicity, let’s ignore its rearrangement for now and assume we can identify a single mass peak in the spectrum at 68 m/z that corresponds to the intact molecule, minus an electron. Using slightly more sophisticated ionization techniques such as chemical ionization or proton transfer, this is easily accomplished. Because the natural abundance of^{13}C is about 1.1%, we expect the natural [M+1]^{+}peak at m/z 69 to be about 5.5% of the intensity of the molecular ion (1.1% x 5 C atoms). There should also be an even smaller pool of this metabolite that carries two labels naturally, but its abundance will be much smaller, on the order of (1.1% x 5 C atoms)^{2}/200 or 0.15%. This is barely detectable for a small molecule like isoprene with only five carbons but can become more significant for larger ones. We need to be aware of the natural level of heavier isotopes in our sample for the following reason. In order to figure out the percentage of carbon centers labeled by isotopes in our experiment, we first need to subtract the naturally occurring^{13}C isotopes so we don’t count them as part of our experiment. This can be done by calculating the theoretical abundances of the M+1 and M+2 peaks as we have done above or determining these levels empirically by analyzing an unlabeled control sample.
Let’s imagine a very small sample
of isoprene consisting of only 16 molecules containing between 0 and 3

^{13}C labels (after subtracting out the naturally occurring isotopes). Let’s group them by isotopolog species as follows:
Since each molecule has 5
carbon atoms and there are 16 molecules, there are a total of 80 carbon atoms
in our sample. By adding up the

^{13}C atoms in the isotopologs containing 1 (*m/z*69), 2 (*m/z*70), or 3 (*m/z*71) labels, we can see there are a total of 10 labeled carbon centers. In other words, 10/80 = 12.5% of the C atoms are labeled by the heavier C isotope. Using this graphical solution, we can validate the method described below to see if we arrive at the same answer when all we have is the mass spectrum.
We can predict what the
molecular ion cluster might look like for this particular sample of molecules
if we have an instrument sensitive enough. Since there are 3 M+1 ions at

*m/z*69 for every 10 unlabeled ions (*m/z*68), two M+2s and a single M+3, we can normalize that all to 100 and expect a molecular ion cluster like this:
For comparison, we can look at the spectrum from an unlabeled sample which we used to subtract the naturally occurring 13C isotopes:

So far so good. In this
simple example, we can visually count the labeled atoms, see that 12.5% are
labeled, and predict the spectrum of the ion cluster. But what about a real
sample where we are dealing with much larger numbers of atoms where all we have
is the spectrum? How do we get to the % atom labeling when we can’t just count up the individual labels as in
the simplified example above?

The method I will describe
here is an approximation that assumes no rearrangement of the molecular ion and
can be performed with simple quadrupole instruments measuring spectra at unit
resolution. In other words, we do not know the exact mass of the fragments,
only their mass down to a single amu. Many techniques using high resolution spectrometers (time-of-flight, isotopic ratio mass spec for example) have been published that rely on more sophisticated approaches, but the advantage here is that a good approximation can be made with a common quadrupole instrument found in most laboratories. We need to draw up a table with the
abundance of the unlabeled molecular ion plus all the isotopomers we can
reliably detect. Here is an example of such as table:

After subtracting the natural isotopic abundance
(column B), we multiply the relative abundance of each isotopolog by the number
of

^{13}C atoms implied by the shift in mass (1 for the M+1 isotopolog, 2 for the M+2, etc.) (column C). Now we add up this column (column C total). This sum represents the total labeling contributed by each individual isotopomer. But in order to calculate the % atom labeling, we also need to know the total number of carbon centers available for labeling. For this, we need to know the number of carbon atoms in the molecule. For each isotopomer (and the unlabeled ion), we multiply its background subtracted abundance in column B by the number of carbon atoms in the molecule (column D). Now we add up this column as well, which represents the total number of carbon atoms which could bear an isotopic label. Note that these figures have all been normalized in the first step, so they don’t represent the real number of carbon atoms (only 80 in our example) but maintain the proportions of the different isotopologs. For the last step, we simply divide the total labeling by the total number of sites (column D total). In our simplified example, we can see that these steps give us the same % atom labeling we obtain by solving visually when the sample size is small enough to draw out. This means this method works for realistically large samples as well, such as we expect in a real labeling experiment, when we only have the spectrum and wish to calculate % atom labeling from isotopic peak abundances.
Now we can take a mass spectrum of a well-behaved molecular ion cluster (i.e. one that doesn't undergo confounding rearrangements), subtract the natural abundance of heavier isotopes, and calculate the percentage of all carbon centers that contain labels introduced during our labeling experiment. The next post in this series will discuss how we can use this kind of information to draw general conclusions about plant metabolic processes.

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