Use the mass spectrum below & the provided chart with amin...

Question

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.

Mass spectrum showing y and b ion peaks for hexapeptide sequence analysis in biochemistry.

Accepted Answer

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 of a 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, confirm our answer. And so, looking down at our spectrum, notice that we have the relative abundance on the y axis, and the m z ratio on the x axis. And notice again that we have these y ion peaks that are distributed throughout, and we also have a bunch of b ion peaks that are distributed 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 gonna be y ion, a y 6 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 y 6 ion peak, and the highest y ion peak that we have is the y 5 ion. And so the y 5 ion is going to always have the end of the peptide, and it's gonna have 5 residues in it. So it'll have these unknown 5 residues. But it's actually gonna be missing the leucine. And so this y 5 ion, corresponds with the yellow, fragment there, and it's missing the leucine. So that's important to keep in mind, that this Y 5 does not correspond with the Y 6 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 y four 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 an 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 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 it's one letter code v. And, of course, that's going to be the next residue in our sequence. And then we can continue this process, so going from y 3 to y 2, 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 t r p, 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, y 2 here, notice that there's not a y one ion, so we can't really take the difference between, the y two and the next peak. And so, essentially what we need to realize is that this y two 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 gonna 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 know 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 b 6 ion. And so again, we have to notice that we don't have a b 6 ion here, we only have a b 5 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 going to start off with the, I b ion peaks on the left. And so here we have a b two ion is the first b ion that we have, and we know that the b two 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 113.18, so we can subtract off 113 point one eight. And that is going to 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 peak, so we have B 2 here and B 3. 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 b 3 and b 4. So that would be essentially the difference between these two peaks here. And so, if you do 512 0.52 minus 326.34, you'll get an answer of a 186.18. And 186.18 matches with a tryptophan amino acid residue, a 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 gonna do b 4 and b 5 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.

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