Edman Degradation Reaction Efficiency - Online Tutor, Practice Problems & Exam Prep

Edmond degradation is effective for small peptides, typically under 50 amino acids, due to a reaction efficiency of 99% per cycle. However, the 1% failure rate accumulates, complicating sequencing as cycles increase. Cumulative yield, crucial for accurate protein sequencing, is calculated using the equation: $e^n$, where e is reaction efficiency and n is the number of cycles. A cumulative yield above 60% is ideal for reliable results, emphasizing the need for efficient peptide cleavage methods.

Edman Degradation Reaction Efficiency

In this video, we're going to talk about Edman degradation reaction efficiency. So in our previous videos, we've said that Edman degradation can only be used on small peptides. And here, we're just reinforcing that same idea with a little bit more detail by saying that Edman degradation is only practical for small peptides with less than about 50 amino acid residues. Now, that might seem like a really limiting factor about Edman degradation since most proteins in nature have way more than just 50 amino acid residues. But that's exactly why most proteins need to be cleaved down into smaller peptide fragments with less than 50 amino acid residues each in order for those proteins to be sequenced via Edman degradation. But the question here is actually why? Why is it that Edman degradation is only practical for small peptides with less than 50 amino acid residues? Well, it turns out the answer has to do with the reaction efficiency per cycle for Edman degradation. And the reaction efficiency per cycle for most modern Edman degradation sequinators is about 99%, which means that in each Edman degradation cycle, 99% of the reactions or 99% of the peptides are going to react successfully. And so let's face it, a 99% reaction efficiency is a really high success rate, and if you guys were to get a 99% on your next test, I wouldn't be complaining. You wouldn't be complaining. We'd all be pretty happy and satisfied. But with Edman degradation, 99% reaction efficiency means that in each Edman degradation cycle, there's still going to be about 1% of the reactions or 1% of the peptides that are going to fail to release their N-terminal amino acid residue in the correct cycle. And so although 99% reaction efficiency seems really, really high, we have to remember that this is the reaction efficiency per cycle. And one Edman degradation cycle reveals only 1 amino acid residue. And so if we have 50 amino acid residues in the protein, then we need 50 Edman degradation cycles. And again, with each Edman degradation cycle, 1% of the peptides fail to release their amino acid and so we get an accumulation of 1% of failed products with each cycle and so this accumulation of failed products with each cycle is really the reason why Edman degradation is limited to small peptides with less than 50 amino acid residues. So just to clear up that idea, let's take a look at our example down below. And notice in this example, we have a pool of identical peptides on the left-hand side and these are decapeptides because they have 10 amino acid residues. And they have an amino N on the far left and a carboxyl N on the far right. And notice that the N-terminal amino acid residue is highlighted here in gold. And that's because after one round or one cycle of Edman degradation, this N-terminal amino acid residue highlighted in gold is the one that's going to pop off of the chain and be identified as a PTH amino acid residue. And so, notice that the first cycle of Edman degradation can be initiated with Phenyl isothiocyanate or PITC to initiate the first reaction. And then we can treat it with trifluoroacetic acid or CF₃COOH to initiate the second reaction. And then, of course, we treat the released amino acid derivative with aqueous acid or H₃O⁺ to initiate the third reaction that generates that PTH amino acid final product that we identify. And so we know this from our previous lessons, and the result is that most of the peptides are going to release their N-terminal amino acid residue, and we can see that down below. Most of the peptides here have indeed released their N-terminal amino acid residue and that's indicated by these check marks here. But notice that not all of the peptides release their N-terminal amino acid residue. So this peptide here failed to release its N-terminal amino acid residue. Now in this diagram, it might seem that 1 out of 4 or 25% of the peptides are going to fail to release their amino acid residue. But in reality, it's only 1% of the peptides that fail. And so, it's not as bad as 25%, but even with 1% that's still capable of limiting Edman degradation to small peptides. And the real reason is that notice that this peptide here is saying, whoops. Guess I'll just release it in the next cycle. And that's really the issue here. The fact that the next cycle is supposed to identify the second amino acid residue, not the first amino acid residue. And so, if this peptide releases its N-terminal residue in the next cycle, it's really just contaminating the second cycle with unwanted PTH amino acids. And so essentially, these unwanted PTH amino acid side products will accumulate with each Edman degradation cycle. And so if you have enough Edman degradation cycles, you'll get a lot of side products that accumulate and ultimately, these side products are going to obscure the results and make it really, really difficult are going to require more Edman degradation cycles. And with more Edman degradation cycles, there are going to be more side products that accumulate. And again, more side products accumulating means that the results are going to be obscured and more difficult to interpret. And so again, we know that most proteins in nature exist as being naturally long, so they have lots and lots of amino acid residues and they range from having several hundred up to several thousands of amino acid residues. But that means if we were to try to sequence those large proteins with Edman degradation, we would need several hundred to several thousand Edman degradation cycles. And, again, that's a lot of Edman degradation cycles. And that's exactly why the solution to sequencing long proteins in nature is to cleave down those large proteins into smaller fragments before Edman degradation. And that's exactly why we talked about all of those protein cleavage techniques in our previous lessons, such as, amino acid hydrolysis, chemical cleavage, and peptidases. And so, this here, concludes our lesson on Edman degradation reaction efficiency. And in our next lesson video, we're going to talk about cumulative yield and how the reaction efficiency can be used to calculate the cumulative yield. So I'll see you guys in that video.

Edman Degradation Reaction Efficiency

In this video, we're going to talk about how to use the Edman degradation reaction efficiency to calculate the cumulative yield, which is something that your professor is likely going to want you guys to do. We already know that the cumulative yield can be calculated from the reaction efficiency. The cumulative yield is the relative amount of a very specific product that's obtained in a chemical reaction. With Edman degradation, the very specific final product that's obtained and analyzed is the PTH amino acid.

Notice with this equation shown below, it expresses the relationship between the Edman degradation reaction efficiency, the number of Edman degradation cycles, and the cumulative yield. The reaction efficiency per cycle raised to the power of the number of Edman degradation cycles is equal to the cumulative yield. For context, accurate protein sequencing typically requires a high cumulative yield, usually greater than about 60%. A cumulative yield of 60% suggests that 60% of the products of that Edman cycle are the correct PTH amino acid. The remainder of the percentage, essentially 40%, are going to be unwanted PTH amino acid side products.

Recall from our previous lesson video that these unwanted PTH amino acid side products can obscure the results. The goal is to keep this percentage of unwanted PTH amino acid side products as low as possible and to keep the cumulative yield percentage as high as possible. A biochemist can expect accurate protein sequencing with any combination of reaction efficiency and number of Edman degradation cycles that gives a cumulative yield greater than 60%. However, any combination of reaction efficiency and number of Edman cycles that yields a cumulative yield lower than 60% should be approached with caution because inaccurate protein sequencing is at risk.

That's something important to keep in mind. Our degradation procedure has a reaction efficiency of 99%, where 1% of each reaction cycle produces unwanted PTH amino acid side products. Calculate the total cumulative yield of the correct PTH amino acid immediately after the 50th Edman degradation cycle. Since it's asking us to calculate the cumulative yield, all we need to do is use our equation from above. Given the reaction efficiency of 99%, and as a decimal, this is 0.99.

If we take the reaction efficiency and raise it to the power of the number of Edman degradation cycles, which is 50, then we can get our cumulative yield. The cumulative yield, abbreviated as cy, can be calculated using: cy 50 = 0.99 50 If you calculate this, you'll get an answer of 0.605, or 60.5% when converted to a percentage. This exceeds the 60% threshold, suggesting that with a reaction efficiency of 99% at the 50th Edman degradation cycle, we can expect accurate protein sequencing. Note that 60.5% is just barely above the threshold. Adding one more Edman degradation cycle, changing the cycle number to 51, the cumulative yield becomes approximately 59.9%, below our 60% threshold.

What this means is that with 51 Edman cycles, we're starting to risk accurate protein sequencing. It's possible that we may not get accurate protein sequencing with the 51st cycle. Typically, biochemists use a threshold of 60% to ensure the correct PTH amino acid is in high abundance. This concludes our lesson on how to use reaction efficiency to calculate the cumulative yield. In our next couple of practice videos, we'll be able to get more practice. I'll see you guys there.

Assuming 98% reaction efficiency, calculate the total cumulative yield of the correct PTH-amino acid at the 50^th Edman degradation cycle.

So this practice problem says, assuming 98% reaction efficiency, calculate the total cumulative yield of the correct PTH amino acid at the 50th Edman degradation cycle. And so the only difference between this practice problem and the previous example from our previous lesson video, is really the 1% difference in reaction efficiency. Here we have a 98% reaction efficiency, and in our previous example problem we had a 99% reaction efficiency. So, we'll be able to compare how drastic of an effect, 1% difference in reaction efficiency has on the cumulative yield. And so, recall from our previous lesson video that the relationship between the reaction efficiency per cycle, the number of Edman degradation cycles, and the cumulative yield is expressed through this equation shown here. And so all we need to do to calculate the cumulative yield is to plug into this equation. Notice we're given a reaction efficiency of 98%, but 98% expressed as a decimal is 0.98. And so if we raise 0.98 to the number of Edman degradation cycles, we'll have to raise it to 50, since there are 50 degradation cycles. And so this is going to be equal to our cumulative yield. So if we take our calculators and do 0.9850, we'll get an answer of about 0.3642. And so this is the cumulative yield in a decimal format. But if we want to convert it into a percentage, all we need to do is multiply it by 100%. And this is going to be equal to 36.42%. And so, essentially, what this percentage means, this is the cumulative yield here, what this percentage means is that with a 98% reaction efficiency per cycle, at the 50th Edman degradation cycle, only about 36.42% of the products are actually going to be the correct PTH amino acid, which means that most of the products are not the correct PTH amino acid, and they are unwanted side products. And so that's going to make it really difficult to detect the correct PTH amino acid with this reaction efficiency. In our previous example problem with a 99% reaction efficiency, we had a much larger cumulative yield, where the correct PTH amino acid was in, the majority. It had the majority of the percentage. But here, it has only 36.42%. So, we can see that a difference in reaction efficiency of just 1% has a very large impact on the cumulative yield, especially when the number of Edman degradation cycles is relatively high. And so, the answer to this practice problem is 36.42% for the cumulative yield, and that concludes this practice. So I'll see you guys in our next video.

A) A peptide with the primary structure Lys-Arg-Pro-Leu-Ile-Asp-Gly-Ala is sequenced by the Edman degradation procedure. If each Edman cycle is 93% efficient, what percentage of the PTH-amino acids in the fourth Edman cycle will be PTH-Leu?

B) What percentage of the PTH-amino acids in the eighth Edman cycle will be PTH-Ala?

Alright. So here's a 2-part practice problem broken up into part a up above and part b down below. Part a says a peptide with the primary structure shown here is sequenced by the Edman degradation procedure. If each Edman cycle is 93% efficient, what percentage of the PTH amino acids in the 4th Edman degradation cycle will be PTH Leucine? We know the 4th Edman degradation cycle will reveal the 4th amino acid residue from the N-terminal end, which is leucine. Although this practice problem might seem like it's asking for something different than our previous practice problem, it's really asking for the same exact thing, just worded differently. This percentage of the correct PTH amino acid in that Edman cycle is really just asking us for the cumulative yield. We know that our equation, where we take the reaction efficiency per cycle and raise it to the number of Edman cycles, will reveal the cumulative yield. All we need to do is plug into this equation. We're told that the reaction efficiency per cycle is 93%, and 93% as a decimal is 0.93. We then raise that to the number of Edman cycles, which will be 4. If you take your calculators and do 0.93 4 , you'll get an answer of 0.748. This is our cumulative yield in decimal format. But if we want to convert it into a percentage, we just multiply it by 100%, and that gives us 74.8%. This is the cumulative yield and the answer to Part a. Recall from our previous lessons that a cumulative yield above 60% allows for accurate protein sequencing. Since 74.8 is greater than 60%, that means we'll get accurate protein sequencing up through the 4th Edman cycle and sequencing up through leucine. Now that we've got part a, let's move on to part b. Part b asks, what percentage of the PTH amino acids in the 8th Edman cycle will be PTH alanine? Looking at our peptide, notice that the 8th residue here is alanine. We are looking for the percentage of the correct PTH amino acid. All we need to do is use our formula again, except change the number of Edman cycles. We're still working with a reaction efficiency of 93%, which is 0.93 in decimal format. The number of admin cycles now changes to 8. If we take our calculators and do 0.93 8 , that gives us an answer of about 0.56. This is our cumulative yield in decimal format. If we want a percentage, we multiply it by 100%, and that gives us 56%. This is the answer for part b, the percentage of the correct PTH amino acid. Since this cumulative yield is less than 60%, it indicates that accurate protein sequencing through the 8th Edman cycle, sequencing through the 8th residue here, will not be very accurate. One of the major takeaways from this practice problem is that this 93% efficiency, reaction efficiency, seems very low, especially since we said in our previous lesson videos that 99% was already low. But through this practice problem, we learn that we're able to sacrifice reaction efficiency if we limit the number of Edman degradation cycles. When we only had 4 Edman degradation cycles, we were able to get accurate protein sequencing with a high cumulative yield above 60%. But once we increase the number of Edman cycles, we notice that we had to sacrifice our cumulative yield and accurate protein sequencing. This highlights the fact that not only reaction efficiency but also the number of Edman cycles will affect the cumulative yield and accurate protein sequencing. This concludes this practice problem, and I'll see you guys in our next video.

Here’s what students ask on this topic:

Why is Edman degradation limited to small peptides?

Edman degradation is limited to small peptides, typically those with fewer than 50 amino acids, due to its reaction efficiency of 99% per cycle. While 99% efficiency is high, the 1% failure rate accumulates with each cycle. For a peptide with 50 amino acids, this means 50 cycles are needed, leading to a significant accumulation of failed reactions. These failed reactions produce side products that obscure results, making it difficult to accurately sequence longer peptides. Therefore, Edman degradation is most effective for small peptides.

How is cumulative yield calculated in Edman degradation?

Cumulative yield in Edman degradation is calculated using the equation: $e^n$, where $e$ is the reaction efficiency per cycle and $n$ is the number of cycles. For example, with a reaction efficiency of 99% (0.99) and 50 cycles, the cumulative yield is ${0.99}^{50}$, which equals approximately 60.5%. This means that 60.5% of the products are the correct PTH amino acids, while the rest are side products.

What is the significance of a 60% cumulative yield in Edman degradation?

A cumulative yield of 60% is significant in Edman degradation because it indicates a high likelihood of accurate protein sequencing. A yield above 60% means that the majority of the products are the correct PTH amino acids, minimizing the impact of side products. If the cumulative yield falls below 60%, the accuracy of the sequencing is at risk due to the increased presence of unwanted side products. Therefore, maintaining a cumulative yield above 60% is crucial for reliable results.

What are the steps involved in an Edman degradation cycle?

An Edman degradation cycle involves three main steps: First, the N-terminal amino acid is labeled with phenyl isothiocyanate (PITC). Second, the labeled amino acid is cleaved from the peptide using trifluoroacetic acid (CF₃COOH). Finally, the cleaved amino acid derivative is converted to a more stable form, typically a phenylthiohydantoin (PTH) amino acid, using aqueous acid (H₃O⁺). This PTH amino acid is then identified, revealing the sequence of the peptide one amino acid at a time.

Why do side products accumulate in Edman degradation?

Side products accumulate in Edman degradation due to the 1% failure rate per cycle. In each cycle, 1% of the peptides fail to release their N-terminal amino acid correctly. These failed peptides release their amino acids in subsequent cycles, contaminating the results with unwanted PTH amino acids. As the number of cycles increases, the accumulation of these side products becomes significant, obscuring the results and making accurate sequencing more difficult.