Peptide Sequencing: Edman Degradation - Online Tutor, Practice Problems & Exam Prep

Edman degradation is a method for sequencing peptides from the N-terminal end using phenylisothiocyanate (PITC). The process involves three main steps: nucleophilic addition, cyclization and cleavage, and rearrangement. In nucleophilic addition, the amino group attacks the carbon of PITC, forming a thiourea derivative. Cyclization leads to the formation of a thioazolinone derivative, which is then cleaved to release the N-terminal amino acid. Finally, rearrangement transforms the thioazolinone into a phenylthiohydantoin (PTH) derivative, allowing for identification through chromatographic techniques.

Peptide Sequencing: Edman Degradation Concept 1

Hey, everyone. In our discussion of peptide sequencing, we now take a look at the Edman degradation. So in our intro to Edman degradation, we're going to say that it is a chemical method to sequence a peptide from the N-terminal end. And we're going to say it uses Phenylisothiocyanate, which we're going to abbreviate as PITC, as the main reagent. If we take a look here, we have Phenylisothiocyanate, PITC.

So our phenyl here is our Ph. And then we have NCS, this whole thing is our Edman reagent. Now, we're going to say each degradation cycle involves 3 reactions and it cleaves 1 amino acid residue from the peptide chain. We're going to say the 3 steps are nucleophilic addition, then we have cyclization and cleavage, and then we have rearrangement. With step 1 here, we have nucleophilic addition.

So, our PITC, we're just going to say PITC just to make it easier to pronounce, reacts with the N-terminal of our amino group to produce our thiourea or urea derivative. So, if we take a look here, we have our PITC here. We're reacting with this amino group here. In this reaction, we're going to create this structure which is our thiourea derivative. Next, we're going to have cyclization and cleavage.

With this, we're going to say our thiourea derivative cyclizes, and this is important, it cyclizes into a thiazolinone derivative. So here it goes right here. We've cyclized this in step 2. We're going to say the N-terminal amino acid is cleaved from the peptide chain. So the peptide chain is PEP-C.

So, as we can see, it's been cut off. It was here, and now it's cut off during the cyclization process. And then finally, we have here step 3, where we use acidic hydrolysis to undergo rearrangement. So, our thiazolinone derivative rearranges to form a more stable, and we're going to say here, phenylthiohydantoin or PTH derivative. So we have this structure here at the end.

Right. So here, if we're talking about these different forms, we've highlighted them here in the boxes, but here they are in their isolated forms. Our thiourea, our thiazolinone, our thiohydantoin is the structure here. Now, we're going to say that our internal amino acid is identified via chromatographic characterizations of the PTH derivative. Alright.

So we're going to go over questions on how we approach Edman degradation, but this gives us a broad overview of the three steps involved.

Peptide Sequencing: Edman Degradation Example 1

Everyone. So here it says complete the structure of the PTH derivative formed by Edman degradation of the following tetrapeptide. So here we have our tetrapeptide, and here we have our thiohydantoin, and we're going to figure out what our derivative will be. The key to this is understanding here we have Tyrosine as the beginning amino acid in this tetrapeptide, and the key is attaching its r group to this structure. When we're talking about Tyrosine, its r group here would be a methylene carbon, so CH₂, and then that would be connected to a phenol.

That is the r group of Tyrosine. Attaching it gives us our PTH derivative, so this would be our final answer.

Peptide Sequencing: Edman Degradation Concept 2

So here, we take a look at step 1, nucleophilic addition, in regards to Edman degradation. We're going to say here, the reaction of the amino group with PITC is analogous to nucleophilic addition to a carbonyl group. So if we're following nucleophilic addition to a carbonyl group, then step 1 would be a nucleophilic attack, and step 2 would just be a proton transfer. If we look at step 1, we're going to say the amino group attacks the carbon of the PITC. So here, the amino group nitrogen has lone pairs.

It will use them to come in and attack this carbon here. It's acting like a carbonyl group, but it can't make 5 bonds. So, the bond connected to nitrogen breaks and goes to nitrogen here, giving us this structure initially. And we're going to say that nitrogen here is making 4 bonds with a positively charged. And then we'd also say that this nitrogen already had a lone pair before the attack but now it gained another lone pair when this pi bond broke.

Remember, when nitrogen makes 2 bonds and it has 2 lone pairs, it's negatively charged. So we've created this structure that has a negative charge and a positive charge, not ideal. So then we go to step 2. We're going to say in step 2 we have our proton transfer. So we’re going to say H⁺ is transferred from the amino group, which is, again, protonated, it's positively charged to the PITC nitrogen atom.

So, here, one of these hydrogens will undergo a proton transfer and migrate to this negatively charged nitrogen. So, as a result of this, now this has 1 H, this has 1 H, both of them still have, as a result, a lone pair. So we've created this structure. So just remember, in step 1 of Edman degradation, we have nucleophilic addition, which can be broken down into two steps, which is our nucleophilic attack for step 1 and our proton transfer for step 2.

Peptide Sequencing: Edman Degradation Example 2

Hey, everyone. In this example question it says, which of the following is an unlikely step in the nucleophilic addition of a peptide to PITC? So we're going to say amino acid group of the N-terminal, amino acid attacks the C atom of PITC. So this is the very first step when it comes to nucleophilic addition, which is our nucleophilic attack. Formation of a resin stabilized ion after nucleophilic attack.

Now, yes. This is also true. Remember, our nitrogen is negatively charged. It's connected to a carbon which is double bonded to an S. There can be some form of resonance here where this could resonate here, kicking this bond up to the sulfur.

So this is true. Proton transfer leading to the formation of a thiourea derivative. Yes. That's also true. So by default, d is the answer.

Nucleophilic attack of the S atom of PITC on the carbonyl group of the N-terminal amino acid. Remember, it starts off with nucleophilic attack. That attack is done by the amino group to the carbon, which is kind of acting like a carbonyl. It's not attacking the sulfur. So, this is an unlikely step when it comes to nucleophilic addition.

So here, our final answer is option d.

Peptide Sequencing: Edman Degradation Concept 3

In reaction 2 of the Edman degradation, we take a look at cyclization and cleavage. So here, we're going to say the formation of the thiazolylidene derivative takes place via a nucleophilic acyl substitution mechanism. Remember, we've seen these types of mechanisms when it comes to our carboxylic acid derivatives. Here, we break it down into our 5 steps: protonation, nucleophilic attack, proton transfer, the loss of our leaving group, and finally, deprotonation. So with step 1, we're going to say here that the carbonyl oxygen of the N-terminal amino acid is protonated.

So here we're going to have an H⁺ and this carbonyl oxygen is going to come out and grab it, becoming protonated. So here we have this structure, and an oxygen here, making 3 bonds, is positively charged now. In step 2, we're going to have an intramolecular nucleophilic attack forming a cyclic tetrahedral intermediate. So now that oxygen's making 3 bonds, which is not ideal, it would want something to come and attack this carbonyl carbon, causing the π bond to break so that oxygen no longer makes 3 bonds. So how do we do this?

Well, we're going to do this in this regard. We're going to say here that our sulfur could come and attack this carbonyl carbon, causing our π bond to break and goes up. But this is caused by this nitrogen making a π bond here, causing this cascading effect. By doing this, we've created our cyclic structure. So if we take a look, we're going to say that this Nitrogen is this Nitrogen, which is connected to this carbon, which is this carbon.

And by following the arrow movement, we see that this is now an OH. We can then say here, if we look, and make this 1, 2, 3, we're going to say here that this would be 4 and 5. So 1, 2, 3, 4, and 5. That's how we made a 5-membered ring. The nitrogen here made a double bond, so here it is right here.

So this would be our structure. Now we've drawn it and bring it down to step 3. So in step 3 here, we're going to say a proton transfer makes the peptide chain a better leaving group. Alright. So here, nitrogen and nitrogen are making 4 bonds so it's positively charged here, and that positive charge carries over to down here.

We're going to say it becomes a better leaving group. We want this nitrogen to become neutral, so we're going to transfer this H⁺ over here. Now, this Nitrogen is making 4 bonds, so it's positively charged, now a better leaving group. Step 4, the OH group pushes its electrons to form the carbonyl and kick out the PEPc chain. So oxygen here decides to push its double bond and when it does, we're going to kick this whole thing out.

So now we have this oxygen which is making 3 bonds again, making it positively charged yet again. Then down here is our structure after the pepc chain is left. Now, we're going to say the carbonyl oxygen is deprotonated to give us our final derivative. So, here we're going to say the nitrogen can use its lone pair to grab this H. Oxygen holds onto the electrons, and now it's just a simple neutral carbonyl oxygen.

So we've made our derivative at the end through these 5 steps. Alright, these are the steps required for reaction 2 of our Edman degradation.

Peptide Sequencing: Edman Degradation Example 3

Hey, everyone. So here it says, which of the following statements about our thiazolinone formation in Edman degradation is incorrect? So here we're going to say it forms through an intramolecular nucleophilic attack. Yes. Remember, that's step 2 of this whole reaction.

So this is true. The tetrahedral intermediate cannot convert into our thiazolidine without a proton transfer step. So yes, a proton transfer step is needed. This is step 3 of the process. The N-terminal amino acid is cleaved from the peptide chain before ring formation takes place.

No. This cannot happen. We have to have the intramolecular attack in order to cyclize, make our ring. And then we can have the loss of our peptide chain later on. So this is incorrect.

And then the S atom from PITC, instead of the N atom, attacks the carbonyl group because it is more nucleophilic. So here, that is true. Sulfur is a larger atom than nitrogen, and because of that, it can act as a better nucleophile. So this is true. So here out of all our options, option C is the one that's incorrect.

Peptide Sequencing: Edman Degradation Concept 4

So here we have rearrangement. We're going to say rearrangement of our thiazolidine to thiohydantoin takes place in 4 steps. So we're going to have step 1 is ring opening, then we're going to have tautomerization, we're going to have rotation, and finally ring closure. So our thiazolinine ring opens via an acid-catalyzed SN2 hydrolysis reaction. So we take a look here, we're going to say that we have this H3O+, which is our acid catalyst.

That's going to allow us to protonate the carbonyl oxygen. We now have a protonated carbonyl oxygen. That opens us up to nucleophilic attack of the carbonyl carbon. This is done by the water coming in and attacking here, causing this pi bond to break and go to the oxygen. So at this point, what we're going to have is we're going to have our OH, which we have already, and then the water that's attached.

Here, this oxygen's making 3 bonds, so it's positively charged. So we bring in a second water molecule to deprotonate it. This would give us 2 OH groups at the end here. At this point now, we can force a ring opening. This can come down to make a double bond, which causes this bond here to break open.

At the same time, we can protonate this sulfur into a thiol group. So remember, this water here accepted an H+, so it exists as H3O+ hanging around. So it can act as a protonating agent. The sulfur can grab an H+. So what we'll have here is we'll have our protonated sulfur which becomes an SH.

And then we'll have this oxygen has made a double bond. It still has its H. It's making 3 bonds so it's positively charged. Now, we don't want to have a positively charged oxygen, so we can use another water molecule. This water molecule, when deprotonated, comes in and deprotonates this carbonyl oxygen. So at the end, we're going to have our carbonyl and then connected to an OH. We created a carboxylic acid. Oh, and don't forget the SH that's here. And here, if we want to add all the lone pairs if we wanted to.

Now, we are going to do tautomerization. We can think of this here as kind of like an enol form. Remember, enol is when you have 2 carbons double bonded connected to an OH. But here instead of having an OH, we have an SH. This can tautomerize to give us our keto form.

So, here, we've changed it to give us a double bond S. Step 3, now we have rotation takes place along the carbon-nitrogen bond. So here we can rotate around this carbon-nitrogen bond. Doing that rotates this structure, so now this part will rotate here and this part will rotate up. So we've created this structure here.

And then step 4 here, this is where we have the ring closure. So here, we're going to say ring closure takes place through an acid-catalyzed intramolecular SN2 reaction. So if we take a look here, here we have our structure. We're going to protonate this carbonyl oxygen. So now it's positively charged the oxygen because it's making 3 bonds.

Then we're going to say here that we can have cyclization of this. Alright. So if we're doing cyclization, let's number this 1, 2, 3, 4, and 5. So this decides to come in and attack here, kicking this pi bond up. So this nitrogen is this nitrogen.

It's connecting to this carbonyl carbon which was this one. And it's connected to 2 OH groups. Now, here we can have a proton transfer. So, we're going to say here, during a proton transfer we take this H and it migrates to either one of those OH groups. We created a water which is a great leaving group.

So, this decides to make a carbonyl group again so the oxygen comes down to make a pi bond kicking out the water molecule. So we get here now is a protonated carbonyl oxygen which becomes deprotonated by this water molecule. So at the end, we're going to make this cyclic structure. So here, let's draw this out. So this would be our final product in this reaction dealing with rearrangement.

Peptide Sequencing: Edman Degradation Example 4

Hey, everyone. So here it says in this example question, acetylation at the N-terminal is a commonly seen post-translational modification. Explain why the following peptide cannot be sequenced using Edman degradation. So here, we have to draw the thioazolinone structure. So here, we have our PITC.

And then here, we have our amino acid with the peptide chain attached to it. Here we put an R-group, symbol R because we're not talking about a specific amino acid. And here's our peptide. Now remember, through a series of processes, we make the structure that we need to draw. Alright.

So here, we're trying to do acetylation at the N-terminal, which is a common post-translational modification. If we do the Edman degradation, remember, we're cleaving off an amino acid. So, we're doing that at the N-terminal end. So it wouldn't be possible for us to do acetylation here because we have the cyclization to create this final structure. Alright.

So we can see here that we're cleaving the internal end, so there is not a chance of us doing acetylation to it. Alright. So this would be our final answer by drawing this structure.

Bradykinin is an important nonapeptide with many biological functions. A student performed a series of experiments to determine its structure and obtained the following results:

1. Complete hydrolysis of the peptide yielded 2 Arg, 2 Phe, 3 Pro, Gly, and Ser.

2. Edman degradation produced PTH-Arg.

3. Treatment with carboxypeptidase A did not produce any fragments.

4. Treatment with chymotrypsin produced Arg, a pentapeptide, and a tripeptide with composition Pro, Ser, Phe.

5. A tripeptide produced during partial acidic hydrolysis was confirmed to be Pro-Gly-Phe.

 

____-____-____-____-____-____-____-____-____

 

Edman: PTH-Arg

____-____-____-____-____-____-____-____-____

 

Carboxypeptidase A = No fragments

____-____-____-____-____-____-____-____-____

 

Chymotrypsin = pentapeptide, tripeptide (Pro, Ser, and Phe), Arg

____-____-____-____-____-____-____-____-____

____-____-____-____-____-____-____-____-____

 

Tripeptide Pro-Gly-Phe

____-____-____-____-____-____-____-____-____

Hey, everyone. So in this practice question it says, bradykinin is an important, yet known peptide with many biological functions. A student performed a series of experiments to determine its structure and obtained the following results. So, it's a complete hydrolysis of the peptide yielded 2 arginines, 2 phenylalanines, 3 prolines, 1 glycine, and 1 serine. Edman degradation produced a PTH-arginine structure. Treatment with carboxypeptidase A did not produce any fragments. Treatment with chymotrypsin produced a pentapeptide and a tripeptide with compositions of proline, serine, and phenylalanine. A tripeptide produced during partial acidic hydrolysis was confirmed to produce proline, glycine, and phenylalanine. Alright. So here, we have our structure. We have all of our amino acids involved. They have to connect together through peptide bonds to make our peptide chain. So, we just have to figure out the order in which we put them. Alright. So first, they tell me that the Edman degradation produced PTH-arginine or arginine, so that has to start it out. We have arginine here. And remember, in this process, we created 2 arginines, so they would go on either end. So we've taken care of these 2 amino acids. Now we have to figure out where the other ones go. It told us that carboxypeptidase A produced no fragments, so we have no change that we know of. It's still arginine on both ends. It then says chymotrypsin created a pentapeptide, a tripeptide which is proline, serine, and, phenylalanine, and then we have arginine. So here we have arginine and still arginine. Remember, chymotrypsin here because we're using that. We know that we'd have a phenylalanine connected to this arginine here. So that takes care of one of these. The other one, we're not exactly sure where that would go. They did say that we created a proline, a serine. Alright. So, remember, we cannot cleave around a proline. And these 3 need to be connected to each other. So in this space here, that's where the 3 of them go. We cannot cleave near a proline, so proline would have to go here in the middle, and serine would have to go here. The other phenylalanine would have to go here as a result because it's connected to this tripeptide. And our pentapeptide, we just have to figure out the way we would connect everything. They're telling me, what do we have left? We've taken out both phenylalanines. We've taken out 1 proline so we have 2 left. We've taken out no glycine. We've taken out a serine. All that's left to draw in are our 2 prolines and 1 glycine. So that's what goes in here. Now, they tell us that we have a tripeptide of proline, glycine, and phenylalanine. What does that tell me? We have arginine here, arginine here, here is our tripeptide, here is the Phenylalanine connected to it. And they're saying that we have another tripeptide which is proline, glycine, and phenylalanine. Here's our Phenylalanine. It has to be the one connected to glycine and then proline. So we've used the proline that we have left and the glycine, meaning we have one more proline left. This is the only open spot, so it has to go here. This will be the sequence of our peptide chain based on the information given to us.

Use the following information to determine the structure of a synthetic heptapeptide:

1. Complete hydrolysis of the peptide yields 2 Thr, Cys, Met, Trp, Lys, and Phe.

2. Edman degradation produces PTH-Phe.

3. Treatment with carboxypeptidase A cleaves off Thr.

4. Cleavage with cyanogen bromide produces a dipeptide and a pentapeptide

5. Treatment with trypsin produces a tetrapeptide and a tripeptide Thr-Cys-Thr.

____-____-____-____-____-____-____

 

Edman: PTH-Phe

____-____-____-____-____-____-____

 

Carboxypeptidase A = Thr

____-____-____-____-____-____-____

 

Cyanogen bromide = dipeptide and a tripeptide

____-____-____-____-____-____-____               or        ____-____-____-____-____-____-____

 

Trypsin = tetrapeptide and tripeptide (Thr-Cys-Thr)

____-____-____-____-____-____-____

 

____-____-____-____-____-____-____

Hey, everyone. So in this practice question, it says, use the following information to determine the structure of a synthetic heptapeptide. Complete hydrolysis of the peptide yields 2 threonines, 1 cysteine, 1 methionine, 1 tryptophan, 1 lysine, and 1 phenylalanine. Edman degradation produces PTH with phenylalanine. Treatment with carboxypeptidase A cleaves off threonine. Cleavage with cyanogen bromide produces a dipeptide and a pentapeptide. And then treatment with trypsin produces a tetrapeptide and a tripeptide of threonine, cysteine, and threonine. Alright. So, we have to figure out the order of these amino acids, so we have to figure out how we fill this in. They tell me that Edman degradation is going to give me PTH phenylalanine. That tells us, since there's only one phenylalanine, it's going to be here on this end because that's where we're going to cleave off and connect the PTH. Remember, the N-terminus end is where we're doing that. And then they're telling me carboxypeptidase A gives us threonine. Remember, that's cleaving off on the C-terminus end. Because of that, threonine has to be on this end. So we figured out through Edman degradation what's on the N-terminus and through carboxypeptidase A what's on the C-terminus. Alright. So, and actually, we're going to write these down here so we can keep track of which ones we're eliminating. Because we don't want to keep going up and down to fill things in. Alright. So, what have we cut out? We've cut out phenylalanine and we've cut out one threonine. So we have this remaining. Now, let's see. They say cyanogen bromide creates a dipeptide and a pentapeptide. Alright. So we have here Phe, Thr, Phe, Thr. Alright. So we have to figure out how exactly this is going to work, dipeptide and a pentapeptide. Alright. So here, what are the possibilities here? Well, one possibility is let's see. Maybe we have we could place methionine. So they tell me that we're going to have threonine connected to cysteine connected to threonine. Alright. So if we look here, here's my threonine. So that means if we want this, cysteine has to be here and threonine has to be here. We skip the cyanogen bromide portion because it can be kind of tricky. Because from this part, we could have methionine here, or we could have methionine here. This one wouldn't work because we need to have these three peptides together, or these three amino acids together to make our tripeptide. So we have Phe here. Right now, what's the last thing? They tell us here we have we've taken care of this. We've taken care of this. See what else is left. We've taken out those three. So all we have left are Met, Tryptophan, and Lysine here, and all we have to do is connect them together. Now, here they're telling us misinformation. We could put phenylalanine here with methionine based on cleavage. So Phe, Met, Thr, Cys, Thr. So all we have left are tryptophan and lysine. And based on the way our enzymes cleave, tryptophan would have to be here and lysine would have to be here. This will represent our final peptide based on the enzymes given.