Edman Degradation - Video Tutorials & Practice Problems
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Edman Degradation
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in this video, we're going to talk about Edmund Degradation. So the Edmund Degradation procedure is really just a protein sequencing technique developed by a scientist named Pair Edmund way back in the sixties. And so the Edmund Degradation procedure can Onley be used on one single poly peptide chain at a time. Which means that before Edmund Degradation, we're going to need to use our protein purification techniques toe 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 Edmund 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 the Edmund degradation procedure can only be used on one single small peptide chain at a time. Now, the Edmund Degradation Procedure is really a cycle of three different chemical reactions, and these three reactions remove one and terminal amino acid residue at a time and then identifies that in terminal amino acid residue upon removal and because it is the end terminal amino acid residue that's removed one at a time, the peptide is actually sequence from the n terminal end towards the C terminal end. And that's why Edmund degradation procedures sometimes called end 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 re agents. The first is Final is a diocesan eight, which is abbreviated with P I. T. C and is also known as the Edmund Re Agent. The second re agent that we treat our peptide with is Try Floro acetic acid, whose chemical formula is C F three C 00 h. And as we'll see down below. In our example, after treating our peptide with the first two chemicals, the end 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 re agent here, the released and terminal amino acid derivative is treated with acquis acid, or H 30 plus prior to being identified. And so let's take a look at our example down below of Edmund degradation to clear some of this up and what you'll notice here is we're starting with a deck, a peptide or a peptide with 10 amino acid residues. And we have no idea what the sequences notice that the are groups are labeled are one through our 10. And so if we want to determine the sequence of this deck a peptide we need to perform Edmund degradation and so notice, uh, because Edmund degradation is a cycle of three different chemical reactions and we need to treat it first with fennel is a diocesan. It That is exactly what we're seeing down below. Treatment with fennel is a diocesan. It which we know is abbreviated with P I. T. C. And this first reaction needs to occur under basic conditions. Where the pH is approximately nine now after final is a diocesan eight. We treat our peptide with our second re agent, which is Try Floro acetic acid, whose chemical formulas shown here and then after treatment of our peptide with the first two chemicals noticed that the end terminal amino acid residue is no longer attached 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 end terminal amino acid residue popped off. And that would be this first one here. And so, uh, with the third re agent, we treat the released and terminal amino acid residue that was released with acquis acid, or H 30 plus, and that will essentially rearrange our amino acid derivative into um or stable p th amino acid. And this p th amino acid can be identified with the techniques such as H PLC, high performance liquid chromatography and so notice down below that. Specifically, what we have in this example is a P. T. H. Allen in because the our group here is just a metal group which is indicating Allan E. And so after we identify this first uh, p th amino acid as p t. A challenging, that means that the first residue in our chain of above was a winning and that our first Edmund degradation cycle is complete. And so notice that one Edmund degradation cycle reveals Onley one amino acid residue. But we want to reveal the entire sequence of our peptide, which means that we need to perform Mawr Edmund 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 and step number one, uh, to begin the next cycle of the Edmund degradation. And so notice we have this blue era that's leading back up to the top here, and it's leading to, uh, the female is a diocesan eight, so we could essentially treat our peptide with one less amino acid residue with fennel is a diocesan it all over again. Try floor acetic 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 a kiss acid to identify the second residue and then continue to do Maura Maura Edmund cycles. And so what's important to note here is that the p th amino acid is actually the final product that is analyzed to identify the n terminal amino acid residue. So that's very important. This p th amino acid is the final product, and of course you want to take home the fact that the entire Edmund degradation process is actually repeated as a cycle until the full peptide is sequenced. And so this concludes our lesson on Edmund 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
3
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.
4
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.
5
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.
6
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.
7
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.
