Edman Degradation - Online Tutor, Practice Problems & Exam Prep

The Edman degradation procedure is a protein sequencing technique that identifies the N-terminal amino acid of a peptide chain through a cycle of three chemical reactions. Initially, the peptide is treated with phenyl isothiocyanate (PITC) under basic conditions, followed by trifluoroacetic acid, which releases the N-terminal amino acid as a derivative. This derivative is then treated with aqueous acid to form a stable PTH amino acid, which can be analyzed using HPLC. This process is repeated until the entire peptide sequence is determined, emphasizing the importance of peptide purification and the sequential nature of the reactions.

Edman Degradation

In this video, we're going to talk about Edman degradation. So, the Edman degradation procedure is really just a protein sequencing technique developed by a scientist named Per Edman way back in the 1960s. And so the Edman degradation procedure can only be used on one single polypeptide chain at a time, which means that before Edman 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 Edman 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 Edman degradation procedure can only be used on one single small peptide chain at a time.

Now, the Edman degradation procedure is really a cycle of three 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 one at a time, the peptide is actually sequenced from the n-terminal end towards the c-terminal end. And that's why Edman degradation procedure is sometimes called n-terminal peptide sequencing. And so because it is a cycle of three different reactions, it makes sense that we need to treat our peptide with three different chemicals. And so we need to treat our peptides sequentially with these three different reagents:

• The first is Phenyl isothiocyanate, which is abbreviated as PITC, and is also known as the Edman reagent.
• The second reagent that we treat our peptide with is trifluoroacetic acid, whose chemical formula is CFC3OC0OH.

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 H3O+ prior to being identified.

So, let's take a look at our example down below of Edman degradation to clear some of this up. 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 R1 through R10. If we want to determine the sequence of this decapeptide, we need to perform Edman degradation. Notice, because Edman degradation is a cycle of three 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. 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.

With the third reagent, we treat the released n-terminal amino acid residue that was released with aqueous acid, or H3O+. This will essentially rearrange our amino acid derivative into a more stable PTH amino acid. This PTH amino acid can be identified with a technique such as HPLC, high-performance liquid chromatography. 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.

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 Edman degradation cycle is complete. One Edman degradation cycle reveals only one amino acid residue. But we want to reveal the entire peptide, which means that we need to perform more Edman degradation cycles. So, we can 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 Edman degradation. Notice we have this blue arrow that's leading back up to the top here and it's leading to the phenylisothiocyanate. So, we can essentially treat our peptide with one less amino acid residue, with phenylisothiocyanate all over again, trifluoroacetic acid. That will pop off the second amino acid residue in our chain, and then we can repeat this process here where we treat it with aqueous acid to identify the second residue and then continue to do more and more Edman cycles.

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'll want to take home the fact that the entire Edman degradation process is actually repeated as a cycle until the full peptide is sequenced. This concludes our lesson on Edman degradation, and we'll be able to get some practice in our next couple of videos. So, I'll see you guys there.

The peptide Leu—Cys—Arg—Ser—Gln is subject to Edman degradation. Products of the 1^st 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

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.

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.

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.

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.

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.

Alright. So here we have practice 1-5, which we know is connected to our previous 4 practice problems, practice 1-1 through 1-4. And so practice 1-5 wants us to deduce the entire sequence of the original nonapeptide using the following hint and the information from the previous four practice problems, practice 1-1 through 1-4. And then it gives us the hint that the sequence of the nonapeptide using the one-letter amino acid codes actually reveals a relevant academic subject. And so, notice here that we have these blank circles where we can provide the sequence of the nonapeptide. And from our previous practice problem in practice 1-4, we were able to determine the entire sequence of the pentapeptide here that was generated from cleaving our original nonapeptide with cyanogen bromide. Cyanogen bromide also produced this tetrapeptide over here, which is the N-terminal fragment. And the N-terminal residue of this fragment is cysteine, and the C-terminal residue is methionine. And again, because this is the N-terminal fragment, we know that the N-terminal residue of this fragment is going to correspond with the N-terminal residue of the nonapeptide. And so we can go ahead and fill in the one-letter amino acid code, for cysteine here, which is just a C. And then, these two residues here, we were not able to determine their identity from our previous four problems. So, we'll leave them as question marks for now, and so there'll be question marks in our nonapeptide above. And then our last residue of this fragment is methionine whose one-letter code is M. And then, of course, the rest of this peptide here corresponds with the pentapeptide that we have below, who we already identified the entire sequence from practice 1-4 in our previous video. So all we need to do is fill in the same residues that we have below. So we have isoleucine, serine, threonine, arginine, and tyrosine. And so what you'll notice is we have 2 residues that have not been identified, but we were able to identify 7 out of the 9 from our previous four problems. And so essentially, these two residues we need to determine from the hint that's provided. And the hint again says that using the one-letter codes, it will reveal a relevant academic subject. So can you guys see it yet? Can you see what amino acids need to be filled in here to reveal this relevant academic subject? That's right. It's got to be histidine and glutamic acid, whose one-letter code is E. And so when we have these two residues as histidine and glutamic acid, notice that the sequence of the nonapeptide reveals the relevant academic subject of chemistry. And so essentially, this here is the answer to our practice problem of practice 1-5. And that concludes this practice video.

Here’s what students ask on this topic:

What is Edman degradation and how does it work?

Edman degradation is a protein sequencing technique that identifies the N-terminal amino acid of a peptide chain. The process involves a cycle of three chemical reactions. First, the peptide is treated with phenyl isothiocyanate (PITC) under basic conditions, which reacts with the N-terminal amino acid. Next, trifluoroacetic acid is used to cleave the N-terminal amino acid as a derivative. Finally, this derivative is treated with aqueous acid to form a stable PTH amino acid, which can be analyzed using high-performance liquid chromatography (HPLC). This cycle is repeated to determine the entire peptide sequence.

Why is Edman degradation limited to small peptides?

Edman degradation is limited to small peptides because the efficiency of the reaction decreases with longer peptide chains. Each cycle of the degradation process removes one N-terminal amino acid, and the accuracy of identifying each amino acid diminishes as the number of cycles increases. This is due to incomplete reactions and side reactions that can occur, leading to a loss of sequence information. Therefore, peptides typically need to be relatively small, usually less than 50 amino acids, to ensure accurate sequencing.

What are the reagents used in Edman degradation?

Edman degradation uses three main reagents in its cycle of reactions. The first reagent is phenyl isothiocyanate (PITC), also known as the Edman reagent, which reacts with the N-terminal amino acid under basic conditions. The second reagent is trifluoroacetic acid (CF₃COOH), which cleaves the N-terminal amino acid as a derivative. The third reagent is aqueous acid (H₃O⁺), which converts the derivative into a stable PTH amino acid for identification using high-performance liquid chromatography (HPLC).

How is the N-terminal amino acid identified in Edman degradation?

In Edman degradation, the N-terminal amino acid is identified through a series of chemical reactions. Initially, the peptide is treated with phenyl isothiocyanate (PITC) to form a derivative of the N-terminal amino acid. This derivative is then cleaved from the peptide using trifluoroacetic acid. The released derivative is treated with aqueous acid to form a stable PTH amino acid. This PTH amino acid is then analyzed using high-performance liquid chromatography (HPLC) to determine its identity. This process is repeated for each subsequent N-terminal amino acid in the peptide sequence.

What is the role of high-performance liquid chromatography (HPLC) in Edman degradation?

High-performance liquid chromatography (HPLC) plays a crucial role in Edman degradation by identifying the PTH amino acid formed after each cycle of the degradation process. Once the N-terminal amino acid is cleaved and converted into a stable PTH derivative, HPLC is used to separate and analyze this derivative. The retention time and specific characteristics of the PTH amino acid in the HPLC system allow for its identification. This information is used to determine the sequence of the peptide, one amino acid at a time.