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BOGDAN FEDELES: Hi, and welcome to 5.07 Biochemistry Online,
the biochemistry course on MIT OpenCourseWare.
I'm Dr. Bogdan Fedeles.
Let's metabolize some problems.
Now today we're going to do a really nice problem.
This is Problem 1 in Problem Set 2.
Now, this is a problem about elucidating
the primary structure of a protein.
Problems like this one that we're about to discuss
are a lot of fun because they're really biochemistry puzzles.
We're given a number of pieces of data or clues,
if you want, and then we'd have to use,
not only our biochemical sense, but also
deduction and elimination process in order
to come up with a final answer.
Well, in practice, elucidating the primary structure
of a protein is now a largely automated process
and utilizes high resolution mass spectrometry.
Some of the traditional chemical methods
that we're discussing here are still occasionally useful.
But most importantly, the logical step-wise process
through which we analyze and use each piece of data
to construct a big picture result
is really representative of the process by which we make
discoveries in biochemistry.
Before we begin, let me just say that this problem assumes
familiarity with the structures and abbreviations of the 20
natural amino acids.
Feel free to pause this video and review
the relevant chapters in the book and the lecture notes
before continuing.
One important tool that we have for elucidating
the primary structure of proteins is proteases.
Proteases are enzymes that can hydrolyze the peptide bonds
of a polypeptide chain.
Now, proteases that can cleave in the middle
of a polypeptide chain are also called endopeptidases.
Now you notice a lot of these names
that end in "ase" denote enzymes, and peptidase
means an enzyme that acts on a peptide.
It's the enzyme that hydrolyzes the peptide bond.
Endo, in this case, refers to the fact
that it acts in the middle of a polypeptide chain.
Now, we're going to be learning in this problem about trypsin
and chymotrypsin.
These are proteases that cleave in the middle
of a polypeptide chain.
They are endopeptidases.
One important feature of proteases
is that they are specific.
They don't just cut any which one peptide bond, but rather
they recognize a specific sequence of amino acids.
In the case of trypsin, for example,
we are told that it cleaves adjacent to positively
charged amino acid.
As you know, positively charged amino acids
would be lysine or arginine.
So trypsin will always be cutting
after arginine or lysine.
Let's take a look.
Now, here is a polypeptide chain with amino acid residues,
R1, R2, R3, R4.
Now, let's look at this peptide bond
right here in the middle of the chain.
Now let's say in order for trypsin
to cut, to hydrolyze this peptide bond,
it means that R2, the amino acid residue adjacent to it,
should be a positively charged one.
So if R2 is lysine or arginine, then this bond
here becomes a good substrate for trypsin.
And what it's going to do, it's going to use a water molecule--
it's going to put here trypsin--
and it's going to hydrolyze forming
a carboxyl end and an amine end of this peptide bond.
So we're getting this is a carboxyl end
and this is the amine end of that original peptide bond.
So what I want you to remember is
that if we're having a reaction with trypsin
on a polypeptide chain, then the carboxy end of the peptide
that results, which is this particular amino acid,
is going to be one of either lysine or arginine.
So in other words in a trypsin digest,
all the smaller peptides that we obtain
are going to end in arginine or lysine except, of course,
the very end of the chain, which might
have a different amino acid at its carboxyl end.
Now, this consideration we've just made about the trypsin
digest, in fact, answers part 1 of the problem, which
is asking what is common about all these peptides generated
by a trypsin digest.
And as we've just explained, all these peptides
should be ending with a positively charged amino acid
such as lysine or arginine.
This problem also mentions another protease chymotrypsin,
which is a protease that has a different specificity
from trypsin.
It actually cleaves after amino acids
that are either very hydrophobic and large or aromatic.
So let's write that down and remember.
So if we were to do a digest with chymotrypsin,
then our R2, the residue that's recognized by the protease,
is going to have to be either aromatic, so phenylalanine,
tyrosine, or tryptophan, or something that's
large and hydrophobic such as leucine, isoleucine,
or even valine sometimes.
Put this in parentheses.
So if any of these residues are at this position,
R2, then this protease, chymotrypsin,
is going to generate two peptides.
And once again, the resulting peptide, R2 at the carboxy end
is going to be one of these amino acids.
That's going to be the signature of a chymotrypsin digest.
Now, keeping these things in mind,
we're ready to tackle the rest of the problem.
Question 2 asks about the use of DTT, or dithiothreitol.
Now, DTT is a commonly used reducing agent,
which can reduce disulfide bridges in proteins.
There are many reasons why we want to use DTT.
For example, when proteins form disulfide bridges,
you may shield certain amino acids
from being accessible by proteases,
and therefore, they're not going to be cleaved
and we'll get a mixture of peptides.
So it makes our analysis and our results very difficult.
Now, disulfide bridges can also hold two peptides together
that have no other covalent attachment between them.
So in that case, we get one fragment
instead of two fragments, and once again, it complicates
our analysis of the protein.
But more importantly, because we're typically
purifying proteins in the air, in an oxygen atmosphere,
proteins can acquire disulfide bonds, which weren't there
in the beginning.
So in that case, we can get very unusual results,
unreproducible and artefactual.
That's why using DTT can prevent formation of spurious disulfide
bridges.
And finally, DTT is also useful to tell
if there were any disulfide bridges in the protein
to begin with.
Because if we're looking at the analysis
before and after using DTT, we can tell if the results change,
and that will tell us whether disulfide bridges were there
to begin with.
Now, let's take a look at the DTT chemistry.
Here we have a disulfide bond or bridge in a protein,
and we're going to treat this with DTT,
which looks like this.
This is DTT or dithiothreitol.
Now, if we substitute the SH groups with OH's, you
notice we're going to have four OH's and four carbons.
That's just an alcohol derived from a sugar, which
is called threose That's why the threitol part of the name.
All right.
So when we do this chemical reaction,
this disulfide bridge is going to transfer between the two
sulfur atoms of DTT.
They're going to form a intramolecular sulfide bond.
So our cysteines are going to get reduced,
and from DTT, we're going to get this intramolecular sulfide.
And because entropical considerations,
make the formation of the six-member
ring very, very easy, then this equilibrium will typically
shift to the right.
Of course, we're also using vast quantities of DTT
to make sure that the entire protein becomes reduced,
and then our agent is going to pick up this [INAUDIBLE]
off the bond.
This sums up the question 2 of the problem.
Next, we're going to look at distinguishing
between a couple of different peptides
that we generate during our analysis of our mystery
protein.
So we're told that we're isolating by HPLC peptides
of the following composition.
One has tryptophan, phenylalanine,
valine, aspartate, lysine, cysteine.
That's peptide one.
Another one has phenylalanine, serine, cysteine,
and an unknown amino acid.
Finally, the third one has alanine and lysine.
All right.
Now, obviously, these peptides have different compositions,
so if you could just put them through mass spectrometry,
we're going to get different masses,
so we can tell very quickly which one is which.
But if we don't have mass spec available,
we can also tell them apart only using UV-Vis spectroscopy.
So all you need to remember is that an amino acid
like tryptophan that we have here,
W, has a very strong absorption in the 280 to 300 nanometers.
Whereas, amino acids like phenylalanine,
present here or here, they absorb only around 260
nanometers.
Most of the other amino acids don't absorb in this range
at all.
So if we were to plot the UV-Vis spectra of these peptides,
this is going to be our absorption,
and this is going to be lambda in nanometers, the wavelength.
So let's look in the range from 200 to 400.
300 is about here.
250 is about here.
And let's label these.
Let's say this peptide is red, this peptide is blue,
and this one is green.
All right.
So now the red peptide, as I told you,
contains both tryptophan and phenylalanine,
so it's going to absorb both around 260
and both around 280 to 300.
So it's going to have a pretty big hump in this area
from 250 to 300.
Now, the blue peptide only has phenylalanine,
so above 280 or so, nanometer is going to drop off.
So it's going to look more like this,
whereas, the green one, well, it has neither phenylalanine
or tryptophan, so below 240 or so,
it's going to have no UV absorption at all.
So it's going to drop off right here.
So basically, if we're just looking,
say, around 260 nanometers, we should see a stronger peak
from the red peptide and a weaker peak from the blue one.
But if we're going to look around 300,
we should only see the red peptide.
So by just using the UV-Vis absorption,
we can tell these three peptides apart.
Now we're ready to figure out the structure of our mystery
protein.
This is exactly what the last part of the problem is asking.
So we're going to take each one of the clues provided--
A, B, C, and D--
and analyze and see what information we can
derive from each one of them.
The first piece of information that we're given
is the result of a total hydrolysis
of our peptide with six molar HCl.
So we're told we get the following amino acid
composition.
We get two phenylalanines, one methionine, alanine, valine,
two lysines, one serine, two cysteine, and one aspartate.
Now, recall this is not a comprehensive list
because the hydrolysis and six smaller HCl
may destroy some of the amino acids.
And specifically, we know for sure something
like tryptophan, threonine, and tyrosine.
They're going to be destroyed, and we're not
going to see them here.
So these amino acids may still be present in our protein,
but we're not going to be able to see them in this situation.
Now, let's analyze these amino acids
and see what we can derive from this.
So first of all, we have cysteines.
So two cysteines, it means our protein
can form disulfide bridges.
So from the get go, two cysteine,
it means we can form an internal disulfide bridge
or our protein contains a disulfide bridge between two
otherwise unconnected peptides.
So here are some possibilities.
For example, our peptide is one chain,
and our cysteines are not actually connected
by a disulfide bridge.
Another possibility is we do have a disulfide bridge
between them, like that.
Or yet another possibility is that we have two pieces, two
polypeptide chains, and the disulfide bridge
is the only thing that's holding them together.
Now, in each one of these cases, obviously,
these polypeptide chains are oriented.
So in one end, they're going to have the amine group.
Let me draw it here, NH3.
Here we're going to have two, and the other end
is going to be the carboxyl group, COO minus.
And once again, in this case, we're
going to have two of them.
All right.
Now, how can we narrow down from these couple of possibilities?
Well, the fact that we're getting two lysines here
is a very important clue.
Now remember, we're told that our peptide, this mystery
protein, it actually comes as a result of a trypsin digest.
And if you recall our discussion earlier,
we said that upon trypsin digest,
all the smaller peptides that were obtained
will end in lysine or an arginine.
The fact that we have two lysines here,
it means both of them they need to be
at the carboxy ends of peptides.
Because if they were in the middle of a chain,
the trypsin would have cut that chain in half.
So two lysines means we've got to have two carboxy ends.
So therefore, this seems to be the only possibility in which
we can accommodate two lysines.
Basically, the two carboxy ends that we see here,
each one has to be a lysine.
This possibility only has one carboxy end,
so the other lysine that we have to place
will not be at the end, so would not
be compatible with a trypsin digest.
Same applies for this case.
So these possibilities are not consistent with our data.
So we know already that our protein must
look like this, two polypeptide chains,
each one ends with lysine, and there
must be a disulfide bridge.
The second clue we're given is that the Edman degradation
of the protein yields valine.
Now, as you know, Edman degradation
is a chemical reaction by which we can digest the protein
from the N terminus, from the amino terminus
of a polypeptide chain.
Now as you recall, we have established in the first part
that we have two amino termini in our protein.
So the fact that we only get one amino acid and that is valine,
it says that the other amino terminus
might be blocked or somehow unavailable
for the Edman degradation.
So let's update the structure of our protein
to take into account the second clue.
So we said we have two polypeptide chains.
There's a disulfide bond in between them.
Now, each one of them ends with a lysine.
This is a carboxy end, and now we
know from the Edman degradation that amino end of one of them
has to be valine.
The other one-- we're going to put a box like that--
is a blocked end.
So the amino terminus is not available.
So this is as much as we can tell from these first two
clues.
We're given the products of the chymotrypsin
digest of our protein after it was previously
treated with DTT.
So we know first DTT is going to cleave the disulfide bond,
and then chymotrypsin, as we talked previously,
is going to cut after large, hydrophobic, or aromatic
amino acids.
Now, we're told we're getting five smaller peptides.
Let's take a look.
Here are the five fragments.
One is tryptophan, valine; one is cysteine, phenylalanine,
another one is aspartate lysine, another one
is methionine, alanine, cysteine, and lysine;
and finally, the last one has serine, phenylalanine,
and something that's not an amino acid.
It's going to make it like a small x.
All right.
Now, from what we know about chymotrypsin digest,
we should be able to orient these peptides basically
to tell which amino acid is at the N terminus
and which amino acid is at the C terminus of each one of them.
Now the first one.
Well, we know from the second clue
that valine is at the N terminus,
so that makes it pretty easy.
Then the sequence has to be V-W. So valine is the N terminus,
W is the C terminus, and as we said,
tryptophan is one of the amino acids recognized
by chymotrypsin, so it's going to end up
at the carboxy terminus.
C-F, that has to go C and F, phenylalanine,
another chymotrypsin amino acid that's
left of this carboxy terminus.
D-K. Now, neither of these amino acids
is recognized by chymotrypsin, but we remember from clue one
that D-K must be the carboxy terminus.
So the sequence can only be D-K.
Now, here M, A, C, and K, none of these
is actually on our recognition list for chymotrypsin,
but once again, we know that K must be in the carboxy end.
So for now we're going to have M, A,
and C in a particular order, which we cannot establish just
yet and K at the carboxy end.
And finally, we do an S, F and x.
Well, x is not even an amino acid,
and F is an amino acid that's left of the carboxy terminus
by chymotrypsin, so it's probably x, S, and F. Now,
what about x?
We're told that x is not an amino acid and is hydrophobic
and it has a molecular weight of 256 Daltons.
All right.
So in order to figure out what x is,
we have to read back the beginning of the problem, which
tells us that we're looking at a protein that's associated
with a plasma membrane.
And one way for proteins to associate with a plasma
membrane is to be modified, to incorporate a fatty acid,
such as palmitate.
So perhaps x, which is blocking the N terminus of this peptide
is a fatty acid.
Now, the general formula for fatty acids
is something like this, CH3 CH2 repeated, say,
n times, and then COOH.
So let's see what would be the fatty acid that
has the molecular weight 256 Daltons?
Well, the mass of a methyl group is 15.
The mass of this carboxylate is 45, so we have about 60
plus 14n equals 256 or 14n equals 196.
Therefore, n is 14.
So the fatty acid that would fit these criteria--
it's hydrophobic, it has a mass of 256--
will be CH3, CH2 14 times COOH, or the fatty acid
was 16 carbons.
This is palmitic acid.
Now, the answer that we got, palmitic acid,
was anticipated in the text of the problem
because it gave us an example one way to associate proteins
to the plasma membrane is to form
a covalent linkage with a fatty acid such as palmitate.
But how does a protein associate with a plasma
membrane when it has a palmitic acid residue as part of it?
Well, let's take a look at a diagram.
Here we have a representation of the plasma membrane,
where we have the phospholipids like the hydrophilic head
pointing inside and outside the cell and the hydrophobic tails
of the fatty acids lined up to each other.
Now, imagine we have a protein that's
modified to contain one of these fatty acids.
Then this fatty acid can just insert right next
to the phospholipids of the plasma membrane.
And that way it tethers the protein right
at the plasma membrane.
The final clue that we're given in order to figure out
the structure of our mystery protein
is the digest with an inorganic agent this time.
It's called cyanogen bromide.
Cyanogen bromide reacts quite specifically with methionines,
and it cleaves the peptide bond after methionine
leaving behind an unnatural amino acid called
a homoserine lactone.
Now, let's take a look at what happens
when we treat our protein with cyanogen bromide.
So we're told we're getting the following peptides, A, K
and W, F, V, D, K, C and F, S, C, and an unnatural amino acid.
As I just explained to you, the unnatural amino acid probably
is this homoserine lactone.
So it's most likely, instead of this amino acid
in the actual sequence, we had a methionine.
So we know these four residues--
F, S, C, and M--
go together.
All right.
Now, we can start putting together the clues
from part 3 and 4 and try to figure out the final structure
of our protein.
So the peptide A, K, we know has to have K at the C terminus.
Now, from part 3, we found out that there
was a peptide that contained M, A, C, and K,
after the chymotrypsin digest.
So now we know K has to be the carboxy end
and A has to be right before K, so in order
to get an A-K peptide, then A has
to be right after methionine because that's
where cyanogen bromide is going to cleave the peptide bond.
So the only sequence that we can have here is
going to be C followed by M followed by A
followed by K. So when we treat the cyanogen bromide,
we're going to be cleaving this bond between M and A,
generating M as a homoserine lactone.
Now, we also know that M has to be in this small peptide,
and we know it has to be C, M as a sequence.
So therefore, S and F have to be on the amino terminus
of this peptide.
And we also have a clue from part 3, which
said that S, F, and this non amino acid moiety
were in the same peptide.
So from here we said that, well, the sequence there, it's
probably x modifying the amino S and then F with a carboxy end.
So putting these three things together,
we can come up with a sequence for this strand, which
is going to be x, S, F, then C, M and then A
and K. So this is probably one of the peptide chains
in our mystery protein.
Then, of course, the other is going
to be composed of these amino acids.
Now, we already know something about the sequence of these.
For example, we know V goes before W. We also
know D goes before K, and also know C goes before F. Now, we
know this has to be the carboxy end of the peptide chain.
Let's write it here, COO minus.
And V has to be the amino end.
So then C-F, there's no other way, has to be in the middle.
So the only possible sequence for our second chain
is going to be V, W, C, F, D, and K.
Now that we've established the exact sequence
of each one of these peptide chains,
then we can put together the final structure
of our mystery protein.
So I'm going to transcribe these here, the first chain V, W, C,
F, D, and K. And we know the cysteine is going
to have our disulfide bridge to the other cysteine, which
goes C, M, A, K, and then F, and then S.
And now we know the N terminus of this peptide,
we're going to have our fatty acid residue CH2 14 times CH3.
So let's just mark, once again, the carboxy ends here
and the amino end here.
So this is the final answer for our problem
and the structure of our mystery peptide.
Well, that's it for this problem.
I hope you enjoyed this little protein mystery hunt.
Now, remember that the strategy that we used here
in which we logically string together pieces of data
to build a big picture, it's really the same strategy
that has been used and is being used right now to advance
our knowledge about living systems and their underlying
biochemistry.