Edman Degradation - Video Tutorials & Practice Problems
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Edman Degradation
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Video transcript
In this video, we're gonna talk about Edmond degradation. So the Edmond degradation procedure is really just a protein sequencing technique developed by a scientist named Per Edmond way back in 19 sixties. And so the Edmond degradation procedure can only be used on one single polypeptide chain at a time, which means that before Edmond degradation, we're going to need to use our protein purification techniques to isolate one particular peptide of interest. And so this peptide actually needs to be relatively small. Now, later in our course, when we talk about Edmond degradation reaction efficiency, we'll explain exactly why the peptide needs to be small. But for now, all I want you guys to know is that Edmond degradation procedure can only be used on one single small peptide chain at a time. Now, the Edmond degradation procedure is really a cycle of 3 different chemical reactions. And these three reactions remove 1 n terminal amino acid residue at a time, and then identifies that n terminal amino acid residue upon removal. And because it is the n terminal amino acid residue that's removed 1 at a time, the probe the the peptide is actually sequenced from the n terminal end towards the c terminal end. And that's why Edmond degradation procedure is sometimes called n terminal peptide sequencing. And so because it is a cycle of 3 different reactions, it makes sense that we need to treat our peptide with 3 different chemicals. And so we need to treat our peptides sequentially with these 3 different, reagents. The first is Phenyl isothiocyanate, which is abbreviated with PITC, and is also known as the Edmond reagent. The second reagent that we treat our peptide with is trifluoroacetic acid, whose chemical formula is CF 3c00h. And as we'll see down below in our example, after treating our peptide with the first two chemicals, the n terminal amino acid residue pops off of the rest of the chain, and is released as an amino acid derivative. And so with this third reagent here, the released n terminal amino acid derivative is treated with aqueous acid or h three o plus prior to being identified. And so let's take a look at our example down below of Edmond degradation to clear some of this up. And what you'll notice here is we're starting with a decapeptide or a peptide with 10 amino acid residues, and we have no idea what the sequence is. Notice that the r groups are labeled r one through r 10. And so if we want to determine the sequence of this decapeptide, we need to perform Edmond degradation. And so notice, because Edmond degradation is a cycle of 3 different chemical reactions, and we need to treat it first with phenyl isothiocyanate, that is exactly what we're seeing down below, treatment with phenyl isothiocyanate, which we know is abbreviated with PITC. And this first reaction needs to occur under basic conditions, where the pH is approximately 9. Now after phenyl isothiocyanate, we treat our peptide with our second reagent, which is trifluoroacetic acid, whose chemical formula is shown here. And then after treatment of our peptide with the first two chemicals, notice that the n terminal amino acid residue is no longer attached to the chain because it pops off as a released amino acid intermediate or derivative. And so that means that the rest of the chain is left, with having one less amino acid residue, because, again, the N terminal amino acid residue popped off. And that would be this first one here. And so, with the 3rd reagent, we treat the released N terminal amino acid residue that was released with aqueous acid, or H three O plus. And that will essentially rearrange our amino acid derivative into a more stable PTH amino acid. And this PTH amino acid can be identified with a technique such as HPLC, high performance liquid chromatography. And so notice down below that specifically what we have in this example is a PTH alanine, because the R group here is just a methyl group, which is indicating alanine. And so after we identify this first, PTH amino acid as PTH alanine, that means that the first residue in our chain up above was alanine and that our first Edmond degradation cycle is complete. And so notice that one Edmond degradation cycle reveals only one amino acid residue. But we want to reveal the entire peptide, which means that we need to perform more Edmond degradation cycles. And so what we can do is take our peptide with one less amino acid residue right here, and essentially, return the remainder of our peptide back to conditions in step number 1, to begin the next cycle of the Edmond degradation. And so notice we have this blue arrow that's leading back up to the top here, and it's leading to, the phenoisothiocyanate. So we can essentially treat our peptide with one less amino acid residue, with phenoisothiocyanate all over again, trifluoroacetic acid. That will pop off the second amino acid residue in our chain, and then we can essentially repeat this process here where we treat it with aqueous acid to identify the second residue, and then continue to do more and more Edmond cycles. And so what's important to note here is that the PTH amino acid is actually the final product that is analyzed to identify the N terminal amino acid residue. So this is very important. This PTH amino acid is the final product. And of course, you wanna take home the fact that the entire Edmond degradation process is actually repeated as a cycle until the full peptide is sequenced. And so this, concludes our lesson on Edmond degradation, and we'll be able to get some practice in our next couple of videos. So I'll see you guys there.
2
Problem
Problem
The peptide Leu—Cys—Arg—Ser—Gln is subject to Edman degradation. Products of the 1st cycle include:
A
PTH—Leu, PTH—Cys, PTH—Arg, PTH—Ser, and PTH—Gln
B
PTH—Leu—Cys—Arg—Ser—Gln
C
PTH—Gln and Leu—Cys—Arg—Ser
D
PTH—Leu and Cys—Arg—Ser—Gln
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Problem
Problem
Suppose you isolated a nonapeptide (9 amino acid residues) from a patient's blood. Reaction of the nonapeptide with FDNB followed by acid hydrolysis produces a DNP-product with a sulfhydryl R-group, indicating that:
A
C-terminal residue of the nonapeptide is Cysteine.
B
N-terminal residue of the nonapeptide is Cysteine.
C
N-terminal residue of the nonapeptide is Tyrosine.
D
C-terminal residue of the nonapeptide is Threonine.
E
C-terminal residue of the nonapeptide is Tyrosine or Glycine.
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Problem
Problem
Treatment of the nonapeptide from the previous problem (1-1) with CNBr produces a tetrapeptide containing the N-terminal amino acid and a pentapeptide. After one round of Edman degradation on the pentapeptide, a product is produced that contained a nonpolar, aliphatic R group, meaning that the pentapeptide has:
A
N-terminal I.
B
N-terminal S.
C
C-terminal E.
D
N-terminal H.
E
N-terminal M.
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Problem
Problem
The second and third rounds of Edman degradation on the same pentapeptide from the problem above (1-2) produced products with aliphatic alcohol groups, meaning that the pentapeptide had:
A
S and H.
B
I and Y.
C
M and C.
D
T and S.
E
M and Y.
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Problem
Problem
Hydrazinolysis of the same pentapeptide from the problems above produced modified amino acids & a free α-amino acid with an aromatic-alcohol group. Combining the info from these problems, the pentapeptide is most likely:
A
STIRY.
B
HTSMY.
C
TISMY.
D
ISTRY.
E
ISMRY.
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Problem
Problem
Deduce the entire sequence of the original nonapeptide using the following hint and the information from the previous four practice problems (1-1 through 1-4). Hint: The sequence of the nonapeptide using the one-letter amino acid codes reveals a relevant academic subject.