HOW TO INTERPRET MASS SPECTROMETRY GRAPHS

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  • Опубліковано 15 чер 2022
  • 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.

КОМЕНТАРІ • 65

  • @biotechlucas4126
    @biotechlucas4126  2 роки тому +8

    What should I cover next?🤔

    • @confessmashaba1505
      @confessmashaba1505 2 роки тому +1

      how do we know the length of those peaks more especially on compounds??

    • @Confess_Tsaki
      @Confess_Tsaki 2 роки тому +1

      Let’s take they give you a compound and they tell you to draw mass spectrometry peaks from sketch

    • @biotechlucas4126
      @biotechlucas4126  2 роки тому +1

      @@Confess_Tsaki As far as I have understood, finding out the exact length of the peaks in this situation would be extremely difficult. You could calculate where the peaks are by looking at the bonds of the compound as well as how strong these bonds are. (For example the double bond in CO2 makes it strong, which is the reason why it does not fragment very often compared to propane) This information could help you figure out where the peaks are as well as approximating their length. But to get more exact than that I believe you need to do an actual mass spectrograph and check from that...
      That is everything I know on the topic, but please keep in mind that I am fairly new to it as well so there might be something obvious I am missing!

    • @emdm00011000
      @emdm00011000 Рік тому +1

      de novo peptide sequencing/ bottom up approach LC-MS-CID-MS/ CID: collision induced dissociation/ de novo peptide sequencing interpration algorithms & empirical rules/ de novo peptide sequencing machine learning programs

    • @biotechlucas4126
      @biotechlucas4126  Рік тому

      @@emdm00011000 Thank you so much for the suggestions! Will add it to my video topics list!

  • @jasmine.6021
    @jasmine.6021 18 днів тому +3

    why didn't this channel exist when I was doing my bachelor's 15 years ago!!!! very helpful

  • @adrianagoldova4305
    @adrianagoldova4305 Рік тому +7

    You are amazing! Thank you, brilliant work, very helpful!

    • @biotechlucas4126
      @biotechlucas4126  Рік тому +1

      That is so nice of you! Thank you so much! Happy I could help!😇👍

  • @fenanigans
    @fenanigans 5 місяців тому +3

    Thank you for making this video! I'm pretty sure you just saved me from failing my exam tomorrow!

  • @DoYouHaveAName1
    @DoYouHaveAName1 5 місяців тому +1

    Thank you for the effort you put into these videos :D

    • @biotechlucas4126
      @biotechlucas4126  5 місяців тому

      Thank you for showing your appreciation! Makes me genuinely happy!

  • @melatbrhane8577
    @melatbrhane8577 Рік тому +1

    thank you so much this was very helpful.

  • @paulmcc2006
    @paulmcc2006 6 місяців тому +2

    Nice explanation Lucas thanks!

  • @stephaniecantu6973
    @stephaniecantu6973 8 місяців тому +3

    Where can you access these databases to compare your spectra?

  • @zachkelly3868
    @zachkelly3868 9 місяців тому +1

    thank you so much!! very helpful!!!

  • @amarashmisewwandi5156
    @amarashmisewwandi5156 Рік тому +1

    Do u have any other mass spectroscopy explanation videos??...

  • @hannev.h.4956
    @hannev.h.4956 6 місяців тому +1

    I have a question: so on the right is the whole peptide who isn't separated into smaller ions? On left you have the peptide who is separeerde into different options of the peptide, so you get different ion formations. and these ions are extually amino acids?

  • @TurambaR_06
    @TurambaR_06 11 місяців тому

    thanks for the explanation!

  • @carenataocinqueporte
    @carenataocinqueporte Рік тому +1

    Excellent, thanks!

  • @koksalan75
    @koksalan75 12 днів тому

    Great video thanks. One question...if we look at m/z 191 for example, which is triterpanes as far as I remember, we see many peaks from C21 or so up to C37 or so..now, all those peaks from ~C21 to ~C37 are fragments that give off a m/z 191 fragmentation when going through the ionization, am I correct? But the main molecules would have a mass of ~296 (C21) up to 520(C37), also correct?

  • @alialfagih4805
    @alialfagih4805 24 дні тому

    Hello. Please how can I explain the rationale behind the buffer gas pressure difference between simulation (0.3 mTorr) and experimental conditions (approximately 5 Torr and 2.8 mTorr) in a linear ion trap

  • @eleclayton5890
    @eleclayton5890 8 місяців тому +2

    Are there any free databases of mass spec? I’m actually having a hard time right now in my PhD with a tough load of mass spec

  • @moodex2231
    @moodex2231 Рік тому +2

    thank you so much for your help 💜💜

  • @paulyaw
    @paulyaw 3 місяці тому

    Very nice. Thank you.

  • @hamiddelarue7413
    @hamiddelarue7413 Рік тому +1

    Thanks for explanation

  • @alokalliyanage7022
    @alokalliyanage7022 8 місяців тому +1

    Well explained👏

    • @biotechlucas4126
      @biotechlucas4126  8 місяців тому

      That's very kind of you and makes me happy to hear that!😇👍

  • @anagaytan7708
    @anagaytan7708 3 місяці тому +1

    Does the abundance have to be the same in the two graphs for them to be considered a match?

    • @biotechlucas4126
      @biotechlucas4126  3 місяці тому

      I must admit I'm not entirely sure but to my understanding they don't have to be exactly the same.

  • @jefferykill10
    @jefferykill10 Рік тому +1

    Thanks!

    • @biotechlucas4126
      @biotechlucas4126  Рік тому +1

      Thank you! You don't have to do that🙈♥️ But thank you so much! Let me know if I can help you out in some other way😇

  • @izaridho2019
    @izaridho2019 Рік тому +1

    Thank you. Finally understood, didnt have to pour over a book

  • @tenaciousgirl5201
    @tenaciousgirl5201 Рік тому +1

    thanks a lot

  • @aliceyong221
    @aliceyong221 11 місяців тому +1

    Thx!

  • @noraspeiser1865
    @noraspeiser1865 2 роки тому +2

    How do you actually determine the charge state of an ion in mass spectrometry. I mean, I know how the mass is determined. For example by TOF, quadruple etc. Is the charge of an ion determined by how far it is deflected by the magnetic field in the mass spectrometer?
    Hope my question is clear, thanks a lot

    • @biotechlucas4126
      @biotechlucas4126  2 роки тому +1

      To the best of my understanding, which to be fair is limited, the charge is always either plus or minus 1. This is a requirement for the spectrometry to work. How that is ensured however, I am not certain about... I apologize, I cannot be of more help!

    • @noraspeiser1865
      @noraspeiser1865 2 роки тому +1

      @@biotechlucas4126
      I found it out in the mean while :)
      It can be concluded based on the mass/charge ratio which you obtain in the spectrum. You have to look at the mass/charge ratios between individual peaks of the ion fragment/precursor and calculate back to the charge. So, if the difference between the m/z of 2 peaks is for example 0.3, the charge would be +3. If it is 0.5 the charge is +2 and if it’s 1 the charge of the fragment/precursor is also 1. It is calculated by taking the reciprocal of the difference in m/z.
      E.g., if the difference between to peaks is 0.3 then the charge is 1/0.3=3, if the difference is 0,5 then the charge is 1/0,5=2
      Thanks for your time and answer!

    • @biotechlucas4126
      @biotechlucas4126  2 роки тому +1

      @@noraspeiser1865 Thank you so much for sharing that answer! I am trying to learn and get better at this stuff too!

    • @noraspeiser1865
      @noraspeiser1865 2 роки тому +1

      @@biotechlucas4126 in MALDI ionization the charge is always plus 1, here we don't need to think about it. Only in ESI we need to consider the charge

    • @biotechlucas4126
      @biotechlucas4126  2 роки тому +1

      @@noraspeiser1865 Yeah, that was the impression I got as well when reading about MALDI in preparation for the video! But okay, that makes sense thank you so much once more, really appreciate you taking the time to share your knowledge with me!

  • @confessmashaba1505
    @confessmashaba1505 2 роки тому +1

    How do we know the length of the peaks for a compound let's take...

    • @biotechlucas4126
      @biotechlucas4126  2 роки тому +1

      If I understand your question correctly, the device should display the relative length of the peaks as well... These were just pictures I found online to use as general examples. Hopefully that helps!

  • @Esterified80
    @Esterified80 Місяць тому +2

    CO2 has double bonds between C and O

  • @koksalan75
    @koksalan75 4 місяці тому

    What if the fragment has +2 charge? Then the m/z ratio is divided by 2?

    • @biotechlucas4126
      @biotechlucas4126  4 місяці тому +1

      Correct!👍

    • @koksalan75
      @koksalan75 4 місяці тому

      @@biotechlucas4126 but then how will we know what the fragment is? Let's assume there is a peak at 50, this can come from directly 50 (+1 charge) or 100 devided by 2 (+2 charge). Then how does it work? I am kind of confused 😕

  • @nenescotillon1612
    @nenescotillon1612 Рік тому +1

    I love you.