HOW TO INTERPRET MASS SPECTROMETRY GRAPHS
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In order to analyze the
In order to analyze the characteristics of individual molecules, a mass spectrometer converts them to ions so that they can be moved about and manipulated by external electric and magnetic fields. If you want to learn more about how this is done, check out my video on MALDI-TOF and electrospray ionization, both videos linked in the end of this one. These molecular ions are energetically unstable and some will break up into smaller pieces. By analyzing these different sized pieces we can determine what molecule we are analyzing as well as what constituent parts it is made of.
The mass spectra graph is presented as a vertical bar graph, in which each bar represents an ion having a specific mass-to-charge ratio (m/z) and the length of the bar indicates the relative abundance of the ion (the most frequent being assigned as 100). Modern mass spectrometers can distinguish (resolve) ions differing by a single atomic mass unit (amu). The highest-mass ion is assumed to be the molecular compound in its entirety and any lower-mass ions are assumed to be fragments from that molecular ion. To decide where this fragmentation has occurred, one needs to consider the strength of the bonds inside the molecule.
The best way to make sense of all of this is with a few examples. Let us start nice and easy with carbon dioxide (CO2).
1. Before anything else, we calculate its total molecular mass.
1. C = 12, O = 16 → 12 + 16 x 2 = 44 amu
2. This will be the largest amu, displayed furthest to the right in the mass spectra graph
2. Second, in order to figure out how the ionization process might break up the compound, let us take a look at its chemical bonds
1. O-C-O → CO + O
2. This gives us molecular masses of 28 and 16
And when, we look at the graph, that is exactly right. In addition, we can infer that the complete carbon dioxide molecule is very stable since it has the highest relative abundance.
Okay, let us do one more example, this time using propane (C3H8).
1. Molecular mass
1. C = 12, H = 1 → 12 x 3 + 1 x 8 = 44 amu
2. Chemical bonds:
1. CH3-CH2-CH3 → CH3 + CH2-CH3
2. CH2-CH3 = 29 and CH3 = 15
But what about all these other lines… Well, these hydrogens can be knocked off as well. In addition, the bond between two of these carbons can be broken and then a hydrogen can be knocked as well. This is what we see in this graph here. However, this is why relative abundance is very important, since it shows us the most likely formations of these different ions.
As you can see, even with very simple compounds this quickly becomes very difficult to analyze manually. Now, imagine trying to analyze entire proteins or even just amino acids. However, a much better way to do it is to compare your results to an already prepared database or library. (Similarly to how you can recognize a face you already know insanely fast) This is how the technique can be utilized more efficiently, in order to accurately identify for example proteins. This however, requires a large enough data base to compare to, at least 10 or more of the reference spectra. Another way to ensure correct identification is to use so called peptide mass fingerprinting.
The mass spectra graph is presented as a vertical bar graph, in which each bar represents an ion having a specific mass-to-charge ratio (m/z) and the length of the bar indicates the relative abundance of the ion (the most frequent being assigned as 100). Modern mass spectrometers can distinguish (resolve) ions differing by a single atomic mass unit (amu). The highest-mass ion is assumed to be the molecular compound in its entirety and any lower-mass ions are assumed to be fragments from that molecular ion. To decide where this fragmentation has occurred, one needs to consider the strength of the bonds inside the molecule.
The best way to make sense of all of this is with a few examples. Let us start nice and easy with carbon dioxide (CO2).
1. Before anything else, we calculate its total molecular mass.
1. C = 12, O = 16 → 12 + 16 x 2 = 44 amu
2. This will be the largest amu, displayed furthest to the right in the mass spectra graph
2. Second, in order to figure out how the ionization process might break up the compound, let us take a look at its chemical bonds
1. O-C-O → CO + O
2. This gives us molecular masses of 28 and 16
And when, we look at the graph, that is exactly right. In addition, we can infer that the complete carbon dioxide molecule is very stable since it has the highest relative abundance.
Okay, let us do one more example, this time using propane (C3H8).
1. Molecular mass
1. C = 12, H = 1 → 12 x 3 + 1 x 8 = 44 amu
2. Chemical bonds:
1. CH3-CH2-CH3 → CH3 + CH2-CH3
2. CH2-CH3 = 29 and CH3 = 15
But what about all these other lines… Well, these hydrogens can be knocked off as well. In addition, the bond between two of these carbons can be broken and then a hydrogen can be knocked as well. This is what we see in this graph here. However, this is why relative abundance is very important, since it shows us the most likely formations of these different ions.
As you can see, even with very simple compounds this quickly becomes very difficult to analyze manually. Now, imagine trying to analyze entire proteins or even just amino acids. However, a much better way to do it is to compare your results to an already prepared database or library. (Similarly to how you can recognize a face you already know insanely fast) This is how the technique can be utilized more efficiently, in order to accurately identify for example proteins. This however, requires a large enough data base to compare to, at least 10 or more of the reference spectra. Another way to ensure correct identification is to use so called peptide mass fingerprinting.
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