Chemical Cleavage of Bonds - Online Tutor, Practice Problems & Exam Prep

Cyanogen bromide (CNBr) cleaves peptide bonds on the C-terminal side of methionine residues, aiding in protein analysis. Hydrazine identifies the C-terminal amino acid by forming aminoacylhydrazides, leaving the C-terminal residue free for distinction. Beta mercaptoethanol reduces disulfide bonds, while iodoacetate permanently carboxymethylates cysteine residues, preventing reformation of disulfides, crucial for Edman degradation sequencing. Understanding these chemical agents is essential for effective protein characterization and analysis in biochemistry.

Chemical Cleavage of Bonds

In this video, we're going to begin our discussion on the chemical cleavage of bonds. So it turns out that there are actually many different chemicals that biochemists can use to cleave the bonds within proteins, and you're definitely not expected to know all of those chemicals and the bonds that they cleave. So we're only going to focus on the more common chemicals that your professor's likely going to want you guys to know. And so the first chemical that we're going to cover is cyanogen bromide, which can be abbreviated with CNBr and cleaves very specific peptide bonds. The specific peptide bonds that it cleaves are the peptide bonds that are on the carboxyl side of methionine amino acid residues.

Down below in our example, it's asking us where will cyanogen bromide cleave the peptide? And what you'll notice is we have cyanogen bromide here, and we know that it can be abbreviated with CNBr. Something that I came up with that helps me remember exactly where cyanogen bromide cleaves is if we take this B here in the abbreviation for cyanogen bromide, and we rotate it 90 degrees counterclockwise, we can get this B to kind of look like an M. And so when we do that, it's literally telling us where cyanogen bromide cleaves. It cleaves next to methionine residues. And so hopefully that'll help you guys remember where cyanogen bromide cleaves.

Notice, over here on the left what we have is a tetrapeptide or a peptide with 4 amino acid residues represented by these 4 circles here. The residues are alanine, methionine, cysteine, and histidine. And of course, it has an N-terminal with a free amino group and it has a C-terminal on the opposite end with a free carboxyl group. Notice up above what we have is the cyanogen bromide structure, where we have a carbon triple-bonded to a nitrogen and that carbon is bonded to a bromine. And so this is cyanogen bromide. Again, it's asking where is cyanogen bromide going to cleave the peptide? Now we know, by rotating this B here in this abbreviation, that it cleaves next to methionine residues. And so we've got 2 methionine, we've got 2 peptide bonds around the methionine residue. We've got this peptide bond here and this peptide bond over there. Now, because cyanogen bromide starts with the letter C here, that helps me remember that it cleaves closest to the C-terminal end. And so essentially, it's going to cleave the peptide bond that's closest to the C-terminal end, which is going to be this peptide bond right here.

And so, what we can do is we can draw a little zigzag line through here to represent this is the peptide bond being cleaved, and we can write 'cleaved' underneath it. And so essentially, cyanogen bromide is going to split our peptide in half so that we have an alanine-methionine fragment on one side and a cysteine-histidine fragment on the other side. And so, we'll be able to get some more practice with the chemical cleavage of bonds using cyanogen bromide in our next practice video. So, I'll see you guys there.

What fragments result from treating the following peptide with CNBr? 

So this practice problem asks what fragments result from treating the following peptide with cyanogen bromide or CNBr. And what you'll see is that we've got a peptide down below and, of course, we know that our peptide is going to have an N-terminal end that's going to be on the far left here. So, aspartic acid is the N-terminal residue, and then it's going to have a C-terminal end on the opposite end. And so that means that asparagine is going to be the C-terminal residue. And so, from our previous lesson, we know that every time we see cyanogen bromide, we know that it's literally telling us where it cleaves. And so with cyanogen bromide, we know that if we take this B here and rotate it 90 degrees counterclockwise, we can kind of get it to look like an M. And so, essentially, it's telling us that it cleaves next to methionine residues. And so, looking at our peptide, we need to find the methionine residues. And you'll see that there are 2 methionine residues in our peptide. But on either side of the methionine residues, there are peptide bonds. And so which of these peptide bonds is going to be cleaved? Well, we can remember from our previous lessons that every time we see the C in cyanogen bromide, that can remind us that it cleaves closer to the C-terminal end. And so it's going to cleave the peptide bond closest to the C-terminal end. And so if we get rid of these colors here, we can say, alright. Well, it's this peptide bond here, next to the methionine closest to the C-terminal end that will be cleaved, and this peptide bond over here that will be cleaved. And so, what that's going to generate are 3 different peptide fragments. And so, essentially, that's what we are going to draw for our answer. So we're going to have an aspartic acid, a methionine residue, a leucine, a phenylalanine, a methionine residue, and then we're going to have an arginine, an alanine, a tyrosine, a glycine, and an asparagine fragment. And so these are the 3 fragments that are going to be generated upon cleavage of our peptide with cyanogen bromide. So, essentially, this part over here is our answer, and this concludes our practice problem. So, I'll see you guys in our next video.

Chemical Cleavage Of Bonds

So the next chemical that we're going to talk about that cleaves the chemical bonds within proteins is the chemical agent hydrazine. And hydrazine is used to initiate a process called hydrazinolysis. And hydrazinolysis is used to identify the C-terminal amino acid residue of a peptide. And so the way that it works is that when a peptide is treated with hydrazine, it actually forms aminoacylhydrazides with every single amino acid residue, except for the C-terminal amino acid residue. So the C-terminal amino acid residue is the only amino acid residue that will not form an aminoacyl hydrazide. And so what that means is that the C-terminal amino acid residue is going to be chemically this free C-terminal amino acid residue can be pretty easily distinguished from all of the other residues since it's chemically different. And it can be pretty easily identified. And that's exactly why we use hydrazine and hydrazine analysis in the first place, is to identify the C-terminal amino acid residue. So let's take a look at our example of hydrazine analysis down below. And what you'll notice is we're starting here with a tetrapeptide or a peptide with 4 amino acid residues in it. And we can quickly tell that it's a tetrapeptide just by looking at the number of R groups that it has. So notice that it has 1, 2, 3, 4 different R groups. And, of course, what you'll notice is that this N over here on the left is the N-terminal end because it has a free amino group, and this end over here on the right is the C-terminal end because it has a free carboxylate group. And so if we take our tetrapeptide here and we treat it with hydrazine, so notice here we have the chemical structure of hydrazine being shown. Essentially, hydrazine is going to react with our peptide in a way that every single C-terminal residue So notice that the C-terminal residue over here is the only residue that does not form an aminoacyl hydrazide. And so literally, you can see that there's this chemical group that's attached to the amino end of all of the residues that makes it an aminoacyl hydrazide, but the C-terminal residue does not have that same group. It actually has its normal carboxylate group. And so that is what distinguishes the C-terminal amino acid residue from all of the other amino acid residues in the peptide. And since they're so chemically different, that allows us to easily identify and distinguish the C-terminal amino acid residue. So, one way that helps me remember how hydrazine and hydrazine analysis works, is that I, associate that the fact that Z is the last letter of the alphabet, and that hydrazine has a Z in it, and it's one of the only chemicals that has a Z in it that we're going to talk about. And so hydrazine is used to identify the last residue of a peptide or the C-terminal residue of a peptide. And so if you can make that association, then you'll be able to remember how hydrazine and hydrazinolysis is used to identify C-terminal amino acid residues. And so this concludes our lesson on hydrazinolysis, and we'll be able to get some practice in our next video. So I'll see you guys there.

Sketch the approximate result of the following reaction.

So this practice problem says to sketch the approximate result of the following reaction. And notice on the left here what we have is a pentapeptide or a peptide with 5 amino acid residues represented by these 5 different shapes, and these lines in between them represent peptide bonds. So on the far left, what we have is the N-terminal, represented here by a triangle, and on the opposite end, we have our C-terminal with this blue rectangle. And so notice here we're treating our pentapeptide with this chemical, and this chemical is actually hydrazine. We know that hydrazine will initiate hydrazinolysis, which will form aminoacyl hydrazides with every single amino acid residue, except for the C-terminal amino acid residue. And so when we're drawing the approximate result here, essentially all we need to do is draw each residue with an aminoacyl hydrazide group extending off it on its amino end. So that will exist for all of our residues here. So that was our first residue, which is done. Here's our second residue here, with an aminoacyl hydrazide. So that one's done. Then we have a green circle with an aminoacyl hydrazide. And, of course, we have our last but not least, this shape here, this oval shape with an aminoacyl hydrazide. And so all of these are our aminoacyl hydrazides. So these are our aminoacyl hydrazides. And the only residue that will not form an aminoacyl hydrazide is our C-terminal residue here, which is going to be this blue rectangle. And so instead of having this hydrazide group on the end, it's going to actually have just its normal carboxylate group. And so this is the C-terminal residue. We can mark that in blue. And so essentially what we have here are really two sets of results. We have the aminoacyl hydrazides, which are all 4 of these residues here, and then we have the C-terminal residue, which is chemically different. And really, these are the results, approximate results of treating this pentapeptide here with hydrazine. So, remember that this allows us to easily identify the C-terminal residue, because they are so chemically different from the treatment with hydrazine. And so this concludes this practice problem, and I'll see you guys in our next video.

Chemical Cleavage Of Bonds

So the next set of chemicals that we're going to talk about, which are used to cleave specific bonds within proteins, are beta-mercaptoethanol and iodoacetate. These are used in conjunction with one another. Now, recall from our previous lesson videos that beta-mercaptoethanol is specifically used to cleave disulfide bonds. Let's take a look at our example down below to refresh our memories on how beta-mercaptoethanol cleaves disulfide bonds. Notice on the far left here what we have is a single cysteine residue. A cysteine residue is when we have two cysteines covalently connected via a disulfide bond, which is shown in red. With the addition of beta-mercaptoethanol, whose structure is shown, the disulfide bond is reduced and cleaved, resulting in two separate cysteine molecules, each on its own and not covalently linked via a disulfide bond since beta-mercaptoethanol broke that bond. However, under the appropriate conditions, these two cysteine molecules can actually reform that disulfide bond and form the cysteine again. As you can see through the removal of beta-mercaptoethanol and under oxidizing conditions, we can oxidize these two cysteines to reform the disulfide bond.

It turns out that we want to get rid of disulfide bonds when we're looking to sequence a protein because disulfide bonds interfere with the sequencing procedure. Therefore, it's important that we break the disulfide bonds prior to Edmond degradation sequencing. Recall that we mentioned Edmond degradation sequencing when we did the overview of direct protein sequencing in our previous lesson videos when we covered the map for these next couple of lesson videos. We'll cover Edmond degradation sequencing in a lot more detail in another video later in our course. Disulfide bonds interfere with the sequencing procedure, and we must break the disulfide bond so that they do not interfere. Using beta-mercaptoethanol alone may not be sufficient to permanently break the disulfides as there is the possibility for them to reform the disulfide bond. That's exactly where iodoacetate comes into play.

When we use beta-mercaptoethanol in conjunction with iodoacetate, it actually permanently breaks the disulfide bonds of cysteine residues. Iodoacetate works by carboxymethylating cysteinesulfhydryl groups, which prevents the reformation of the disulfide bond. The carboxymethylation is depicted below where the iodoacetate structure is shown. Iodoacetate reacts with each of these sulfhydryl groups, carboxymethylating them, as shown on the far right. We have carboxymethylated cysteines, and these are not able to reform the disulfide bond present on the far left. This one-way transformation allows us to proceed with Edmond degradation sequencing without any interference from disulfide bonds. This concludes our lesson on beta-mercaptoethanol and iodoacetate, and we'll be able to get some practice in our next couple of videos. I'll see you guys there.

Which of the following statements about cystine is correct?

• A. Cystine is formed through a peptide linkage between two cysteines.
• B. Cystine forms when the -CH₂-SH group is oxidized to form a -CH₂-S-S-CH₂- disulfide bridge between 2 cysteines.
• C. Cystine is formed by the oxidation of the carboxylic acid groups on two cysteines.
• D. Two cystines are released when a -CH₂-S-S-CH₂- disulfide bridge is reduced to -CH₂-SH.

Chemical Cleavage Of Bonds

Hey, guys. In this video, we're going to do a quick recap on the chemical agents that we covered in our most recent lesson videos. And so, none of the information in this video is new information. It's all review from our older videos. And really, this entire section is just here to help refresh your memories and to give you a place to consolidate all of the information that you've learned. And so, what we're going to do is fill in the blanks in the chart below to recap the effects that these chemicals have on proteins. And so, in the left-hand column in our chart below, what we have is the chemical agent. And then in the right-hand column, what we have is the result that the chemical agent has on the protein. And notice that our chemical agents here are, our chart is actually color-coded. And so, the first chemical that we have is 1-Fluoro-2,4-Dinitrobenzene, which is also abbreviated as FDNB, and known as Sanger's reagent. And so, the reason that Sanger's reagent or FDNB here is color-coded with blue is because it's the only reagent in this list that actually doesn't cleave any chemical bond of a protein. And so, we know that FDNB, essentially what it does is it reacts to covalently label the free N-terminal amino acid residues of all polypeptide chains. And so, FDNB alone doesn't actually cleave any bonds. It just covalently labels and terminal amino acid residues. But we know from our previous lesson videos that when we use FDNB along with 6 molar hydrochloric acid, that allows us to reveal 2 different things from different topics about our proteins. The first is it reveals the n-terminal residues. So, we can write n-terminal residues, and the second thing that it reveals is the number of subunits. And so remember that none of this information here is new information, all of it is review. And this area here is for you guys to refresh your memories and take notes and consolidate the information. And so, the next chemical that we have is 6 molar hydrochloric acid, or 6 molar HCl. And we know that this is used to induce complete amino acid hydrolysis to cleave all of the peptide bonds found in a protein and to release all of the amino acid residues as free amino acids. And we know that if we follow up the use of 6 molar hydrochloric acid with techniques such as HPLC inear chain of DNA, these include transcription factors, activators or repressors. or mass spectrometry , that we're actually able to reveal the amino acid composition. Alright. So, our next chemical that we have is Cyanogen bromide. And we know that cyanogen bromide can be abbreviated with CNBr. And we know that, by taking this B and rotating it 90 degrees counterclockwise, that it tells us literally where it cleaves. So cyanogen bromide cleaves next to methionine residues. And so what we'll see here is that it cleaves the peptide bonds on the C-terminal side, and we know it's the C-terminal side because cyanogen bromide starts with the letter C here. So that helps remind us that it cleaves the C-terminal side of Methionine residues. And, the next chemical that we have is Hydrazine, and hydrazine structure is shown right here, NH_2NH_2. And we know that hydrazine, because it has the Z in it, and Z is the last letter of the alphabet, hydrazine is used to initiate hydrazinelysis and identify the last residue of a protein. And so, it identifies the C-terminal amino acid residue of a protein. Now, the last but not least, what we have is beta-mercaptoethanol and iodoacetate. And this one is kind of easy because we've already covered beta-mercaptoethanol plenty of times—it disrupts disulfide bonds. And so the iodoacetate here is really just used to carboxymethylate the sulfhydryl groups to prevent the reformation of the disulfide bonds. So essentially, beta-mercaptoethanol and iodoacetate permanently break the disulfide bonds. And we know that the disulfide bonds interfere with sequencing and that we're going to need to break any disulfide bonds before we move forward with Edman degradation sequencing. And so this here is our quick recap of the chemicals, and we'll be able to get some more practice in our next couple of videos. So I'll see you guys there.

Given the amino acid sequence DAHGHFLKAGMT, what reagent would cleave the last residue, T, from the peptide? That is, one of the products after treatment will be the free amino acid: T.

• A. Trypsin.
• B. Phenyl isothiocyanate.
• C. Cyanogen bromide.
• D. Iodoacetate.
• E. Chymotrypsin.

Here’s what students ask on this topic:

What is the role of cyanogen bromide in protein analysis?

Cyanogen bromide (CNBr) is a chemical reagent used in protein analysis to cleave peptide bonds specifically on the C-terminal side of methionine residues. This selective cleavage is useful for breaking down large proteins into smaller, more manageable fragments, which can then be analyzed through various techniques such as mass spectrometry or Edman degradation. The specificity of CNBr for methionine residues helps in mapping the protein's primary structure and identifying the sequence of amino acids.

How does hydrazine help in identifying the C-terminal amino acid of a peptide?

Hydrazine is used in a process called hydrazinolysis to identify the C-terminal amino acid of a peptide. When a peptide is treated with hydrazine, it forms aminoacylhydrazides with every amino acid residue except the C-terminal one. This leaves the C-terminal amino acid chemically distinct and free, making it easily identifiable. The unique chemical properties of the C-terminal residue allow for its differentiation from other residues, facilitating the determination of the peptide's C-terminal end.

Why are beta mercaptoethanol and iodoacetate used together in protein sequencing?

Beta mercaptoethanol and iodoacetate are used together in protein sequencing to permanently break disulfide bonds. Beta mercaptoethanol reduces disulfide bonds, converting them into free cysteine residues. However, these bonds can reform under certain conditions. To prevent this, iodoacetate is used to carboxymethylate the sulfhydryl groups of cysteine residues, ensuring that the disulfide bonds cannot reform. This permanent disruption of disulfide bonds is crucial for accurate protein sequencing, particularly in Edman degradation.

What is the function of 1-Fluoro-2,4-dinitrobenzene (FDNB) in protein analysis?

1-Fluoro-2,4-dinitrobenzene (FDNB), also known as Sanger's reagent, is used in protein analysis to label the free N-terminal amino acid residues of polypeptide chains. FDNB reacts covalently with these N-terminal residues, allowing for their identification. When used in conjunction with 6 M hydrochloric acid, FDNB helps reveal the N-terminal residues and the number of subunits in a protein. This labeling is essential for determining the sequence and structure of proteins.

How does 6 M hydrochloric acid aid in protein analysis?

6 M hydrochloric acid (HCl) is used in protein analysis to induce complete amino acid hydrolysis. It cleaves all peptide bonds within a protein, releasing all amino acid residues as free amino acids. This process is essential for determining the amino acid composition of a protein. Following hydrolysis, techniques such as high-performance liquid chromatography (HPLC) or mass spectrometry can be used to analyze the free amino acids, providing detailed information about the protein's primary structure.