Mass Spectrum - Online Tutor, Practice Problems & Exam Prep

Mass spectrometry analyzes protein structures by generating mass spectra, which plot relative abundance against m/z ratios. Key fragments, known as b ions and y ions, arise from peptide bond cleavage. B ions contain the N-terminal residue and are read left to right, while y ions contain the C-terminal residue and are read right to left. This directional analysis reveals the primary structure of proteins, with y ions typically being more stable and prominent in spectra. Understanding these concepts is crucial for interpreting mass spectrometry data effectively.

Mass Spectrum

So now that we've reviewed the basics of mass spectrometry, in this video, we're going to talk about a mass spectrum. A mass spectrum is literally just a plot of the mass spectrometry data or the mass spec data, and it plots the relative abundance on the y-axis and the m/z ratio on the x-axis. If we take a look at our example below on the left-hand side, we see a mass spectrum, where on the y-axis, we have the relative abundance, and on the x-axis, we have the m/z ratio. Throughout our mass spectrum, we see a bunch of different peaks. From our previous lesson, we know that most of these peaks correspond to smaller peptide fragments that result from cleavage of our original peptide at only one peptide bond. Later in our course, we will discuss more details about how a peptide actually fragments during mass spectrometry. But before we get there, I want you to know that a mass spectrum can be analyzed to reveal the primary protein structure. Recall that the primary protein structure is the amino acid composition, the number and types of amino acids, as well as the amino acid sequence, or the order of amino acids from the N-terminal end to the C-terminal end of a protein. Amino acids are identified in a mass spectrum by looking at the m/z ratio differences between appropriate sets of peaks.

Let's examine our example below. Notice that above our mass spectrum, we have a pentapeptide displayed, or a peptide with 5 amino acid residues represented by five different squares. We know that in general, proteins are considered from their N-terminal ends to their C-terminal ends. Most of the peptide molecules in mass spectrometry are fragmented at only one peptide bond. If our peptide does not fragment, all of the peptide bonds remain intact, and all amino acid residues stay intact as a single unit. This single unit will have its largest possible mass, and the largest m/z ratios, appearing furthest to the right on our spectrum. This peak in light blue represents our unfragmented pentapeptide. The reason this peak is blue is because the leading N-terminal residue is also blue. While some peptide molecules can go unfragmented, most will be fragmented at only one peptide bond. Imagine this peptide bond being fragmented; the residue on the end is cleaved off, and this fragment is detected by the mass spectrum. This fragment will have the next largest m/z ratio, corresponding to a highlighted peak.

The only difference between this yellow peak and the original blue peak of the unfragmented peptide is the mass of this one residue. Taking the mass difference, Δmass between these two peaks, which is 99.08, gives us the mass of this residue that was cleaved off. By analyzing all of these molecular weights, we find that a mass of 99.08 corresponds to a valine amino acid residue. This confirms that the blue residue is valine. Now, imagine another peptide bond being cleaved; the mass difference comes to 137.18, matching with asparagine (137.18 is the molecular mass). We continue this process for other cleavages, obtaining different amino acids based on the mass differences until all residues are identified, from the N-terminal end to the C-terminal end, as demonstrated by our analysis from right to left in the spectrum.

One of the actual limitations of mass spectrometry is that it struggles to differentiate leucine from isoleucine because they are isomers, having the same chemical atoms in different arrangements, thus sharing the same exact mass. This inability to differentiate is a limitation of mass spectrometry. This concludes our initial introduction of a mass spectrum; in our next lesson video, we'll further explore how peptides fragment during mass spectrometry. Before that, let's get some practice, so I'll see you guys in our practice video.

Use the mass spectrum below to determine the sequence of the peptide.

Hey, guys. So here's a pretty straightforward practice problem that says, use the mass spectrum below to determine the sequence of the peptide. And notice that we have all of these blanks here to fill in our sequence, and we have the N-terminal on the left and the C-terminal on the right. Also, notice that below, we have a mass spectrum where we have the relative abundance on the y-axis and the m/z ratio or the mass on the x-axis. Now, really the key to this problem is just remembering that when we analyze a mass spectrum, we're going to want to analyze it from the right of the spectrum to the left of the spectrum. And we know that when we analyze it in that way, we're revealing the sequence from the N-terminal end to the C-terminal end. And so, really that's the major key for this problem. And when we're starting on the right, what we can see is that this peak here is representing our unfragmented peptide, so it has the largest m/z ratio. Now the next peak, notice it has a glycine residue above it. And the glycine residue above is just indicating that the difference in mass or m/z ratios between these two peaks corresponds with a glycine amino acid residue. And so because that's the case, essentially what we can do is say that our N-terminal residue is a glycine residue, or G. All we need to do is just continuously read the residues that are above the peaks from the right to the left. The next peak that we have here is valine, so it is V. The next one that we have is leucine/isoleucine. So is it leucine or is it isoleucine? And the answer is we do not know, and that's because mass spectrometry cannot differentiate between leucine and isoleucine, as they have the same exact mass because they're isomers. And so, the difference in mass between these two peaks here could be either a leucine or an isoleucine. For that reason, we can put 'L/I' here for leucine or isoleucine. We can continue, we have a valine that comes next, then we have another valine that comes. We have an alanine that's next, so A. We have another alanine. Then we have a serine, so that's S. A glycine residue, which is G, and an aspartic acid residue, which is D. Then we have another serine, which is S, another glycine, which is G. You guys can see the point here, we're just reading it from right to left. Then we have alanine, we have serine is next, another serine, and then we have another leucine/isoleucine. So again, we can put 'L/I' here. We have a serine, a tyrosine, which is Y, and a proline, which is P, and then last but not least, the C-terminal residue is an alanine. By reading the spectrum from the right to the left, we've revealed our protein sequence from the N-terminal end to the C-terminal end. And really, that concludes this practice problem, so pretty easy, pretty straightforward. Just keep that in mind, analyze from the right to the left. Alright. That concludes this practice, and I'll see you guys in our next video.

Mass Spectrum

So now that we understand mass spectra a little bit better and we know that we can use them to reveal a protein's primary structure, in this video, we're going to talk about how a protein actually fragments during mass spectrometry. And we're going to define b and y ions and how they contribute to the mass spectrometry spectra. So we already know that mass spectrometry ionization fragments most of our protein molecules only one time at a single peptide bond. And so it turns out that when a peptide bond is actually fragmented during mass spectrometry ionization, it actually leads to the generation of 2 prominent sets of ions, and those 2 prominent sets of ions are the b ions and the y ions. Now the b ions always contain the n terminal amino acid residue, and those b ions are read from the left of our spectrum to the right of our spectrum, so from left to the right. Now the y ions, on the other hand, they always contain the c terminal amino acid residue and those peaks are always read from the right of our spectrum to the left of our spectrum, so in this direction here. And so remember that that's the same direction that we tend to analyze our spectrum to reveal the sequence of our protein. And so what that tells us is that the y ions here must be key for revealing the sequence of our protein. Now another way that helps me distinguish between b and y ions is that because b is the second letter of the alphabet, it's towards the beginning of the alphabet. And so b ions always contain the beginning amino acid residue of our protein or the n terminal amino acid residue of our protein. And they're always read from the beginning of our spectrum to the end of our spectrum, so from left to right. Now for the y ions, because y is the second to last letter of the alphabet towards the end of the alphabet. That helps me remember that y ions always contain the amino acid residue on the end of our protein, or the c terminal, end of our protein. And those y ion peaks are read from the end of our spectrum to the beginning of our spectrum, and so they're read from right to left. And so if you can remember that, that'll help you distinguish between the b and the y ions. So let's take a look at our example down below to clear up this idea of b and y ions. And so, what you'll notice is that we have the same exact pentapeptide from our previous lesson video. And we know that during mass spectrometry ionization, most of our protein molecules are going to be fragmented. And so you can see that we have fragmentation as our first step here. And we know that when that fragmentation occurs, most of the molecules are only going to be fragmented one time at a single peptide bond. And when that single peptide bond is fragmented, it's going to generate 2 prominent sets of ions. And so let's take a look at our example and imagine that this peptide bond here is the one that's being cleaved. Well, essentially, that's gonna generate 2 prominent sets of ions, b ion and a y ion. And so the b ion, remember, always contains the beginning amino acid residue or the n terminal amino acid residue. And so that fragment is going to be this entire fragment here, and that's the b ion that's generated from this single peptide bond that's broken. Now the y ion always contains the amino acid residue at the end or the c terminal residue. And so that's gonna be this residue right here. And so down below, you can see that we have the same exact, fragmentation here. You can see that we have the Y ion here, and we have the B ion over here. Now, through the color coding here, we can see that this, y ion that's generated from this peptide bond fragmentation is just gonna be glycine, so we can put in glycine here. And because this y ion only has one amino acid residue, it's called the y₁ fragment. And because the b ion has 4 amino acid residues in it, it's called the b₄ fragment. And so you can essentially see that by fragmenting this one peptide bond, we generated these 2 ions, b ion and y ion. And that was just from cleaving this one peptide bond that's being shown, here. And so, what you can see is that if we were to cleave a different peptide bond, say it wasn't this peptide bond, instead it was this peptide bond that was cleaved. Well, essentially, we're still gonna get 2 prominent sets of ions, the b ion with the n terminal amino acid residue, and the y ion with the c terminal amino acid residue. And that's exactly what we see down below, the y ion and the b ion over here. So we can fill in the residues that are missing. So the blue corresponds with valines, the pink with histidines, and the green with alanines. So, you can see the pattern here, and you can see that we need to fill in the blanks for the rest of these fragmentations. So if we were to fragment our next peptide bond over here, we would get valine and histidine right here. And then, for the y ions, we would get alanine, asparagine, and glycine. So we can fill those in, alanine, asparagine, and glycine. And then, of course, for our last peptide bond that gets fragmented, we're going to get, the, b ion of just a valine shown here, and then we're going to get the y ion of this entire fragment here, so we can fill that in as well. So we have histidine, alanine, asparagine, and glycine. All right. So now we've generated all of the B and Y ion fragments that could result from breakage of any of these peptide bonds. And so because b ions and y ions are both ions, they both have a charge on them, and they're both gonna be deflected in an electric field, and they're gonna be detected by the detector and show up on a mass spectrum. And so it's important to note that, intermixed b and y ions can actually show up on a mass spectrum. But the y ions are actually more stable than the b ions, and they're structurally more stable because the b ions are very short-lived and they tend to fragment further into other ions. But the y ions, because they're more stable, they tend to have higher intensity and abundance, and they're often going to be the most prominent peaks in the spectrum. And so essentially, because they are the most prominent peaks in the spectrum, we tend to have a preference to analyze the y ions in a spectrum. And so because we have that preference, the b ions over here, essentially, we can cross off since we're not really going to be analyzing them very much moving forward. But the y ions, again, because they're the more prominent peaks in the spectrum, these are the ones that we're typically going to focus on in a mass spectrum. And so you can see that in our mass spectrum up here, you can see that we only have y ions being shown, and so you can see the y₁ ion is being shown here. We have the y₂ ion being shown here, the y₃ ion showing here, and the y₄ ion showing right here. And then, of course, the y₅ ion is going to be the unfragmented ion that is the original peptide. And so that's something important to keep in mind that these y ions are going to be the most prominent peaks. Now, again, this spectrum is only showing the y ions, but in a real spectrum of a protein, there is going to be intermixed b and y ions showing. So there would be some b ions that would be smaller, less prominent peaks if this were an actual spectrum. So that's another important factor to keep in mind. And so because we focus on the y ions and the y ions are read from right to left, from the right to the left of our spectrum, what this means is that in most cases, it is safe to assume that the mass spectrum is analyzing the y ions and is read from right to left, especially moving forward in our course. And so, essentially what you can see is that that is why we read our spectrum from the right to the left to reveal our amino acid sequence. So this here concludes our lesson on the difference between B and Y ions, and we'll be able to get some practice in our practice video. So I'll see you guys there.

Upon fragmentation of a peptide bond during mass spectrometry, what ions can be detected on the spectrum?

• A. b ions.
• B. y ions.
• C. b & y ions.
• D. Ions are deflected but not detected.

Use the mass spectrum below & the indicated y-ion peaks (red) to reveal the sequence of the peptide.

So this practice problem says to use the mass spectrum below and the indicated y ion peaks in red to reveal the sequence of the peptide. And down below, we have a spot to provide the peptide sequence with all of these blanks here. We're given a chart here with the 20 standard amino acids and the mass of the residue in grams per mole for those 20 standard amino acids. And then down below, we're given a mass spectrum where we have the relative abundance on the y-axis and the m/z ratio on the x-axis. Notice we have all of these different peaks, and up above, we were provided that the y ion peaks are the ones that are in red. All of these black peaks that we see are likely to be b ion peaks. Because these are y ion peaks, we know that we're going to be analyzing our spectrum from the right of our spectrum to the left of our spectrum, in order to reveal our peptide sequence from the N-terminal end to the C-terminal end. Remember that this peak on the far right represents the unfragmented peptide. And so it has an MZ ratio of 1,298.6. And so, looking at the neighboring y ion peak, which is this peak here at 1,199.52. If we wanted to determine our first N-terminal residue, we just take the difference between these two peaks. So, if you take your calculator and do 1,298.6 minus 1,199.52, you get an answer of 99.08. And that mass difference between these two peaks, is specific to our valine residue, 99.08. So this number here matches with the 99.08. So that tells us that this difference in mass is referring to a valine amino acid residue. And so, up here in our peptide sequence, we can put a V for valine. So now we can move on to our next set of peaks. And so essentially what we have is 1142. We are going to take the difference between these two peaks. And so 1,199.52 minus 1,142.44 comes out to 57.08. Looking at our chart, notice that 57.08 matches with glycine. And so that tells us that glycine here is going to be our next residue, and we can put a G up here for that. Essentially, this whole practice problem is going to be taking the difference between all of these peaks. So, just to fast forward this a bit, what we have is, the difference between these two peaks, is going to be 163.18, which is specific to tyrosine, so tyrosine is here, 163.18. So we can put tyrosine here, and we can put tyrosine up here with a one letter code of Y. Then we can move on to our next peak here, so 880. The difference between these two comes out to, again, 99.08, which we already said was specific for a valine residue. Down here, we can see that. And so we can put valine here and another V up here. Then moving on to our next peak, what we have is 765.1. So the difference between these two peaks is just 115.08. And so that matches up with aspartic acid, which is up here, or aspartate, 115.08. And so we can put in aspartic acid next down here, ASP, and that is the one-letter code D. Again, if we continue to do this, we would take the difference between this peak and the next peak which is going to come out to 115.08 again. So this is essentially, the same aspartic acid, so ASP, so we have another aspartic acid. Then we can move on to the next one, which is the difference between these two peaks, and that will give us 101.08, which is specific to threonine in our chart. So you can see threonine over here has a mass of 101.08, and so, that is THR and just T. Then, this peak right here, the difference between these two, comes out to 128.14, which is specific to glutamine, GLN, or just, Q for our one-letter code. And then continuing forward here, we have, the difference between these two peaks comes out to 147.18, which is specific for Phenylalanine. So we can see Phenylalanine over here 147.18. Its three-letter code is PHE, and its one-letter code is an F. And then, what we have is the difference between these 2 small peaks here. The difference between these two comes out to essentially 186.54, so 186.54, which is specific to our tryptophan residue, so TRP is next, and tryptophan's one-letter code is W. And then, of course, this last residue here, 87.08, is specific to our serine, which is right here, 87.08. And that's just the mass of this peak and the difference of this peak from the peak to the zero mark on the x-axis. So, Serine's one-letter code is just an S. And so essentially what we've done is we've analyzed our mass spectrum from the right to the left looking at the y ions and we've revealed the sequence of our peptide, which is up above. So that concludes this practice problem, and I'll see you guys in our next one.

In your mass-spectrometry of a pure protein with an m/z of 1,582, you found peaks of y ions with the following m/z ratios of 1,582, 1396 and 1283. The mass in Daltons for the possible relevant amino acids are provided: Y (163), N (114), W (186), D (115), G (57), L (113) and M (131). From this data, it is obvious that the C-terminal amino acid residue of the 1,582 fragment is:

• A. G.
• B. L.
• C. N.
• D. W.
• E. Can't be determined.

Use the mass spectrum below & the provided chart with amino acid masses to determine the sequence of a hexapeptide (6 amino acid residues). In the mass spectrum, y ion peaks are indicated with 'y' while b ion peaks are indicated with 'b.' The N-terminal residue is given as Leu and the C-terminal residue is given as Lys. Determine the remaining amino acid sequence using either the y ions or the b ions.

Alright. So this practice problem wants us to use the mass spectrum on the left and the chart over here on the right to determine the sequence hexapeptide or a peptide with 6 amino acid residues in it. And then it tells us that in the mass spectrum below, the Y ion peaks are indicated with a Y at the top, and the B ion peaks are indicated with a B at the top. And then it says that the N terminal amino acid residue is given to us as leucine, whose one-letter code is L, so we can see down below in our peptide sequence that the N terminal residue is indeed leucine. And we're also told that the C terminal residue is lysine, whose one-letter code is K. And then it says to determine the remaining amino acid sequence, using either the Y ion peaks or the B ion peaks in the mass spectrum. Now up until this point, we've only used the Y ion peaks to determine the sequence from a mass spectrum. And in this video, we're gonna continue to do the same, and we'll use the Y ion peaks to reveal the sequence of the hexapeptide. But then after we do that, we can talk about how we could have used the B ion peaks to essentially do the same thing, and we can use these B ions to confirm our answer. And so, looking down at our spectrum, notice that we have ion peaks that are distributed throughout, and we also have a bunch of B ion peaks that are distributed throughout, and we also have a bunch of B ion peaks that are throughout as well. Now, recall from our previous lessons that because Y ions has a Y in it and Y is the second to last letter of the alphabet towards the end of the alphabet, That means that Y ions always have the end of the polypeptide chain, so they always have the C terminal amino acid residue. And so an unfragmented hexapeptide is going to be a Y ion with 6 residues, so it's going to be Y ion, a Y6 ion, and that's because there are 6 residues in our hexapeptide that's unfragmented. But notice looking down below at our spectrum that there is no Y6 ion peak, and the highest Y ion peak that we have is the Y5 ion. And so the Y5 ion is going to always have the end of the peptide, and it's going to have 5 residues in it, so it'll have these unknown 5 residues. But it's actually going to be missing the leucine, and so this Y5 ion, corresponds with the yellow, fragment there, and it's missing the leucine. So that's important to keep in mind, that this Y5 does not correspond with the Y6 unfragmented peptide. And so, we have to work with the information that's being provided, so we have to take that into account. And so we know that, because these are Y ions, that we read the spectrum from the right of the spectrum to the left of the spectrum, and we can take the difference between corresponding peaks to reveal amino acid residues. So that means that we can take the difference between this peak and this peak right here to reveal the amino acid residue, right? No, not right. So recall that we need to take the difference between the appropriate sets of peaks. But because this is a B ion peak, it's not an appropriate match to a Y ion peak. And so the next Y ion peak would be the Y4 ion here. And so we have to take the difference between these two peaks here. And so if you do 584.6 minus 470.52, you'll get an answer of 114.08. And we know that this mass will correspond to the mass of an amino acid residue. And so looking at our answer options over here, notice that 114.08 matches with asparagine, and asparagine's 3 letter amino acid code is ASN, and its one letter amino acid code is N. And so, essentially, what we can see is because, Y ions are read from right to left, and it reveals amino acid residues from the N terminal to the C terminal N, then that means that this first residue here is going to be our next N terminal amino acid residue. So it'll be asparagine at this position here. Alright. So now we're gonna just continue to do the same thing, so we can take the next set of Y ion ion peaks and take the difference between them, so between these 2. And so if you do 470.52 minus 371.44, you'll get an answer of 99.08. And, of course, that matches with an amino acid mass, which is valine. And so, what we can do is put valine's 3 letter code here, which is VAL, and its one letter code V. And, of course, that's gonna be the next residue in our sequence. And then we can continue this process, so going from Y3 to Y2, taking the difference between the 2. And, that comes out to 186.18. And so looking at our chart over here on the right, notice that it matches with answer option for tryptophan. And tryptophan's 3 letter code is TRP, and its one letter code is W. And so up above, we can say that tryptophan W is the next residue. And so moving on to our last residue, Y2, here, notice that there's not a Y1 ion, so we can't really take the difference between, the Y2 and the next peak. And so, essentially what we need to realize is that this Y2 ion is going to contain our it's always going to contain the N terminal, the C terminal residue here, and it's also going to contain this, missing residue here since it has 2, residues in it, a Y ion with 2 residues. And so, we know that this mass right here of a 185.26, which we can rewrite up here, contains both of these residues. So if we subtract off the mass of the lysine residue, we'll get the mass of this missing residue. And so looking at our lysine over here on the right, notice that lysine's mass is a 128.18, so we can subtract that off. And so this comes out to 57.08, And this is the mass of a, glycine residue. And so what that means is that glycine is identified as the next residue in our chain. And so, essentially, what we've done is we've confirmed the sequence of our peptide. We've revealed the sequence of our peptide using the Y ions. And this is the answer to our practice problem, and we're essentially done here. So with the rest of this, video, what we're going to do, if you guys are interested, we're going to talk about how we're going to use the B ions to essentially, determine the same exact answer and to confirm our answer. And so, what we can do is, we now realize that because the B ions have a B in them, and B is the second letter of the alphabet towards the beginning of the alphabet, that tells us that B ions always contain the beginning of the polypeptide chain, so it always has the N terminal residue. And the unfragmented, hexapeptide here is also called a B6 ion. And so again, we have to notice that we don't have a B6 ion here, we only have a B5 ion. And also, because, this is a B ions that we're working with, we read the spectrum from the beginning of the spectrum to the end of the spectrum. So we're gonna start off with the, ion B ion peaks on the left. And so here we have a B2 ion as the first B ion that we have, and we know that the B2 ion is going to contain the first two amino acid residues. And so what we can do is, we can realize that this first Oh, I'm sorry. These first two amino acid residues here has a mass of 227.26. So we can rewrite that. So, essentially, we can take this up here and say, alright, well, we have 227.26, and we can subtract off the mass of the known residue here, leucine. And so notice that leucine's mass here is a 113.18, so we can subtract off a 113.18. And that is gonna be equal to a 114.08, And this corresponds with the mass of an asparagine residue, ASN, and whose one letter code is N. And, of course, that's the next residue here. So, up above in blue for the B ions, we can say that we've confirmed that this residue here is an asparagine. And so now what we can do is take the difference between the next, B ion peaks, so we have B2 here and B3. And so if we take the difference between these two peaks here, which I'll just rewrite down below since there's not a lot of space in there. So, it comes out to 99.08, and this mass here corresponds with a valine amino acid residue. And so the next residue is going to be valine or V. And, we can see up above that, indeed, it is valine. That's the next residue. And so again, we can continue to do this. So now we can do the difference between B3 and B4. So that would be essentially the difference between these two peaks here. And so, if you do 512.52 minus 326.34, you'll get an answer of a 186.18. And 186.18 matches with a tryptophan amino acid residue, 186.18. So the next residue here is gonna be tryptophan or W. And so looking up, we can see that W is indeed the next residue. And, of course, we can continue to do this. So now we're going to do B4 and B5 here, so the difference between these two peaks. And so 569.6 minus 512.52 comes out to 57.08, and that is the mass of a glycine amino acid residue. And so glycine is the next residue, and we can see up above that indeed, glycine is the next residue. So we can see here how we were able to use the B ions to essentially confirm the same exact answer that we got using the Y ions. And so, this here shows you how biochemists can use both the Y ion and the B ion peaks that are obtained in a mass spectrum to help them confirm the sequence of a peptide. And so this here concludes our practice problem, and I'll see you guys in our next video.

Here’s what students ask on this topic:

What is a mass spectrum in mass spectrometry?

A mass spectrum is a graphical representation of mass spectrometry data. It plots the relative abundance of detected ions on the y-axis against their mass-to-charge (m/z) ratios on the x-axis. Each peak in the spectrum corresponds to a specific ion, with the height of the peak indicating its relative abundance. By analyzing the differences in m/z ratios between peaks, one can deduce the primary structure of proteins, including the sequence and composition of amino acids. This analysis is crucial for understanding protein fragmentation and identifying specific residues.

How are b ions and y ions different in mass spectrometry?

In mass spectrometry, b ions and y ions are fragments generated from the cleavage of peptide bonds. B ions contain the N-terminal amino acid residue and are read from left to right in the spectrum. In contrast, y ions contain the C-terminal amino acid residue and are read from right to left. Y ions are generally more stable and prominent in the spectrum, making them key for determining the protein sequence. Understanding the distinction between these ions is essential for interpreting mass spectrometry data accurately.

How can mass spectrometry reveal the primary structure of a protein?

Mass spectrometry reveals the primary structure of a protein by analyzing the mass-to-charge (m/z) ratios of peptide fragments. When a protein is ionized and fragmented, it produces a spectrum with peaks corresponding to different fragments. By examining the differences in m/z ratios between these peaks, one can determine the sequence of amino acids. This process involves identifying specific residues based on their unique molecular weights and reading the spectrum from right to left to reveal the sequence from the N-terminal to the C-terminal end.

Why are y ions more prominent in mass spectrometry spectra?

Y ions are more prominent in mass spectrometry spectra because they are more stable than b ions. This stability results in higher intensity and abundance peaks for y ions. Y ions contain the C-terminal amino acid residue and are read from right to left in the spectrum. Their prominence makes them crucial for determining the protein sequence, as they provide clearer and more reliable data compared to the less stable b ions, which tend to fragment further.

What are the limitations of mass spectrometry in differentiating amino acids?

One significant limitation of mass spectrometry is its inability to differentiate between isomeric amino acids, such as leucine and isoleucine. These amino acids have the same molecular mass but differ in their structural arrangement. Since mass spectrometry relies on mass-to-charge (m/z) ratios, it cannot distinguish between molecules with identical masses. This limitation can complicate the analysis of protein sequences, as it introduces ambiguity in identifying specific residues.