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BOGDAN FEDELES: Hi, and welcome
to 5.07 Biochemistry Online,

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00:00:35,540 --> 00:00:41,150
the biochemistry course
on MIT OpenCourseWare.

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I'm Dr. Bogdan Fedeles.

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Let's metabolize some problems.

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Now today we're going to
do a really nice problem.

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This is Problem 1
in Problem Set 2.

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00:00:51,300 --> 00:00:53,340
Now, this is a problem
about elucidating

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00:00:53,340 --> 00:00:56,280
the primary structure
of a protein.

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00:00:56,280 --> 00:00:58,890
Problems like this one
that we're about to discuss

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00:00:58,890 --> 00:01:02,850
are a lot of fun because they're
really biochemistry puzzles.

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00:01:02,850 --> 00:01:06,240
We're given a number of
pieces of data or clues,

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00:01:06,240 --> 00:01:09,510
if you want, and then
we'd have to use,

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00:01:09,510 --> 00:01:12,450
not only our biochemical
sense, but also

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00:01:12,450 --> 00:01:15,390
deduction and elimination
process in order

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00:01:15,390 --> 00:01:17,520
to come up with a final answer.

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00:01:17,520 --> 00:01:20,850
Well, in practice, elucidating
the primary structure

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00:01:20,850 --> 00:01:24,900
of a protein is now a
largely automated process

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00:01:24,900 --> 00:01:28,020
and utilizes high resolution
mass spectrometry.

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00:01:28,020 --> 00:01:30,270
Some of the traditional
chemical methods

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00:01:30,270 --> 00:01:33,930
that we're discussing here
are still occasionally useful.

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00:01:33,930 --> 00:01:37,710
But most importantly, the
logical step-wise process

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00:01:37,710 --> 00:01:41,640
through which we analyze
and use each piece of data

30
00:01:41,640 --> 00:01:44,880
to construct a
big picture result

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00:01:44,880 --> 00:01:49,200
is really representative of
the process by which we make

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00:01:49,200 --> 00:01:50,970
discoveries in biochemistry.

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00:01:50,970 --> 00:01:54,000
Before we begin, let me just
say that this problem assumes

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00:01:54,000 --> 00:01:58,320
familiarity with the structures
and abbreviations of the 20

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00:01:58,320 --> 00:01:59,850
natural amino acids.

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00:01:59,850 --> 00:02:01,620
Feel free to pause
this video and review

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00:02:01,620 --> 00:02:04,830
the relevant chapters in the
book and the lecture notes

38
00:02:04,830 --> 00:02:06,067
before continuing.

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00:02:08,810 --> 00:02:11,630
One important tool that
we have for elucidating

40
00:02:11,630 --> 00:02:15,230
the primary structure of
proteins is proteases.

41
00:02:15,230 --> 00:02:19,700
Proteases are enzymes that can
hydrolyze the peptide bonds

42
00:02:19,700 --> 00:02:22,800
of a polypeptide chain.

43
00:02:22,800 --> 00:02:25,310
Now, proteases that can
cleave in the middle

44
00:02:25,310 --> 00:02:29,720
of a polypeptide chain are
also called endopeptidases.

45
00:02:29,720 --> 00:02:32,060
Now you notice a
lot of these names

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00:02:32,060 --> 00:02:35,920
that end in "ase" denote
enzymes, and peptidase

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00:02:35,920 --> 00:02:38,440
means an enzyme that
acts on a peptide.

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00:02:38,440 --> 00:02:41,300
It's the enzyme that
hydrolyzes the peptide bond.

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00:02:41,300 --> 00:02:43,400
Endo, in this case,
refers to the fact

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00:02:43,400 --> 00:02:47,160
that it acts in the middle
of a polypeptide chain.

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00:02:47,160 --> 00:02:50,800
Now, we're going to be learning
in this problem about trypsin

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00:02:50,800 --> 00:02:52,250
and chymotrypsin.

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00:02:52,250 --> 00:02:54,800
These are proteases that
cleave in the middle

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00:02:54,800 --> 00:02:56,060
of a polypeptide chain.

55
00:02:56,060 --> 00:02:58,340
They are endopeptidases.

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00:02:58,340 --> 00:03:00,380
One important
feature of proteases

57
00:03:00,380 --> 00:03:02,420
is that they are specific.

58
00:03:02,420 --> 00:03:07,610
They don't just cut any which
one peptide bond, but rather

59
00:03:07,610 --> 00:03:11,780
they recognize a specific
sequence of amino acids.

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00:03:11,780 --> 00:03:13,470
In the case of
trypsin, for example,

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00:03:13,470 --> 00:03:16,880
we are told that it cleaves
adjacent to positively

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00:03:16,880 --> 00:03:18,940
charged amino acid.

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00:03:18,940 --> 00:03:21,770
As you know, positively
charged amino acids

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would be lysine or arginine.

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00:03:24,560 --> 00:03:26,710
So trypsin will
always be cutting

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00:03:26,710 --> 00:03:29,510
after arginine or lysine.

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00:03:29,510 --> 00:03:30,850
Let's take a look.

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00:03:30,850 --> 00:03:35,360
Now, here is a polypeptide
chain with amino acid residues,

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00:03:35,360 --> 00:03:39,350
R1, R2, R3, R4.

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00:03:39,350 --> 00:03:42,110
Now, let's look at
this peptide bond

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00:03:42,110 --> 00:03:48,240
right here in the
middle of the chain.

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00:03:48,240 --> 00:03:50,310
Now let's say in
order for trypsin

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00:03:50,310 --> 00:03:53,670
to cut, to hydrolyze
this peptide bond,

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00:03:53,670 --> 00:03:59,700
it means that R2, the amino
acid residue adjacent to it,

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00:03:59,700 --> 00:04:02,040
should be a positively
charged one.

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00:04:02,040 --> 00:04:11,580
So if R2 is lysine or
arginine, then this bond

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00:04:11,580 --> 00:04:14,700
here becomes a good
substrate for trypsin.

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00:04:14,700 --> 00:04:18,370
And what it's going to do, it's
going to use a water molecule--

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00:04:18,370 --> 00:04:19,709
it's going to put here trypsin--

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00:04:22,430 --> 00:04:24,710
and it's going to
hydrolyze forming

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00:04:24,710 --> 00:04:30,150
a carboxyl end and an amine
end of this peptide bond.

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So we're getting this
is a carboxyl end

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00:04:39,080 --> 00:04:53,160
and this is the amine end of
that original peptide bond.

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00:04:53,160 --> 00:04:55,640
So what I want
you to remember is

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00:04:55,640 --> 00:04:59,510
that if we're having a
reaction with trypsin

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00:04:59,510 --> 00:05:04,640
on a polypeptide chain, then
the carboxy end of the peptide

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00:05:04,640 --> 00:05:07,790
that results, which is
this particular amino acid,

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00:05:07,790 --> 00:05:14,540
is going to be one of
either lysine or arginine.

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00:05:14,540 --> 00:05:17,100
So in other words
in a trypsin digest,

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00:05:17,100 --> 00:05:19,760
all the smaller
peptides that we obtain

91
00:05:19,760 --> 00:05:23,990
are going to end in arginine
or lysine except, of course,

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00:05:23,990 --> 00:05:26,180
the very end of the
chain, which might

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00:05:26,180 --> 00:05:29,450
have a different amino
acid at its carboxyl end.

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00:05:29,450 --> 00:05:33,320
Now, this consideration we've
just made about the trypsin

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00:05:33,320 --> 00:05:37,520
digest, in fact, answers
part 1 of the problem, which

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00:05:37,520 --> 00:05:42,110
is asking what is common about
all these peptides generated

97
00:05:42,110 --> 00:05:44,060
by a trypsin digest.

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00:05:44,060 --> 00:05:46,580
And as we've just explained,
all these peptides

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00:05:46,580 --> 00:05:49,700
should be ending with a
positively charged amino acid

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00:05:49,700 --> 00:05:51,440
such as lysine or arginine.

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00:05:51,440 --> 00:05:55,760
This problem also mentions
another protease chymotrypsin,

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00:05:55,760 --> 00:05:58,670
which is a protease that
has a different specificity

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00:05:58,670 --> 00:05:59,820
from trypsin.

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00:05:59,820 --> 00:06:02,540
It actually cleaves
after amino acids

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00:06:02,540 --> 00:06:08,600
that are either very hydrophobic
and large or aromatic.

106
00:06:08,600 --> 00:06:11,790
So let's write that
down and remember.

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00:06:11,790 --> 00:06:20,130
So if we were to do a
digest with chymotrypsin,

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00:06:20,130 --> 00:06:25,560
then our R2, the residue that's
recognized by the protease,

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00:06:25,560 --> 00:06:32,770
is going to have to be either
aromatic, so phenylalanine,

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00:06:32,770 --> 00:06:37,380
tyrosine, or tryptophan,
or something that's

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00:06:37,380 --> 00:06:41,700
large and hydrophobic such
as leucine, isoleucine,

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00:06:41,700 --> 00:06:43,450
or even valine sometimes.

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00:06:43,450 --> 00:06:45,060
Put this in parentheses.

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00:06:45,060 --> 00:06:48,210
So if any of these residues
are at this position,

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00:06:48,210 --> 00:06:50,960
R2, then this
protease, chymotrypsin,

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00:06:50,960 --> 00:06:53,310
is going to generate
two peptides.

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00:06:53,310 --> 00:06:59,670
And once again, the resulting
peptide, R2 at the carboxy end

118
00:06:59,670 --> 00:07:03,300
is going to be one
of these amino acids.

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00:07:03,300 --> 00:07:07,350
That's going to be the signature
of a chymotrypsin digest.

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00:07:07,350 --> 00:07:08,970
Now, keeping these
things in mind,

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00:07:08,970 --> 00:07:11,070
we're ready to tackle
the rest of the problem.

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00:07:14,060 --> 00:07:20,270
Question 2 asks about the use
of DTT, or dithiothreitol.

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00:07:20,270 --> 00:07:23,990
Now, DTT is a commonly
used reducing agent,

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00:07:23,990 --> 00:07:27,840
which can reduce disulfide
bridges in proteins.

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00:07:27,840 --> 00:07:31,160
There are many reasons
why we want to use DTT.

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00:07:31,160 --> 00:07:36,010
For example, when proteins
form disulfide bridges,

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00:07:36,010 --> 00:07:38,990
you may shield
certain amino acids

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00:07:38,990 --> 00:07:42,020
from being accessible
by proteases,

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00:07:42,020 --> 00:07:44,090
and therefore, they're
not going to be cleaved

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00:07:44,090 --> 00:07:46,850
and we'll get a
mixture of peptides.

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00:07:46,850 --> 00:07:50,900
So it makes our analysis and
our results very difficult.

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00:07:50,900 --> 00:07:55,910
Now, disulfide bridges can
also hold two peptides together

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00:07:55,910 --> 00:07:59,610
that have no other covalent
attachment between them.

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00:07:59,610 --> 00:08:02,150
So in that case,
we get one fragment

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00:08:02,150 --> 00:08:05,540
instead of two fragments, and
once again, it complicates

136
00:08:05,540 --> 00:08:08,120
our analysis of the protein.

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00:08:08,120 --> 00:08:10,580
But more importantly,
because we're typically

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00:08:10,580 --> 00:08:14,810
purifying proteins in the
air, in an oxygen atmosphere,

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00:08:14,810 --> 00:08:18,245
proteins can acquire disulfide
bonds, which weren't there

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00:08:18,245 --> 00:08:20,100
in the beginning.

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00:08:20,100 --> 00:08:24,230
So in that case, we can
get very unusual results,

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00:08:24,230 --> 00:08:26,510
unreproducible and artefactual.

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00:08:26,510 --> 00:08:32,450
That's why using DTT can prevent
formation of spurious disulfide

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00:08:32,450 --> 00:08:33,740
bridges.

145
00:08:33,740 --> 00:08:37,039
And finally, DTT is
also useful to tell

146
00:08:37,039 --> 00:08:39,350
if there were any disulfide
bridges in the protein

147
00:08:39,350 --> 00:08:40,470
to begin with.

148
00:08:40,470 --> 00:08:42,440
Because if we're
looking at the analysis

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00:08:42,440 --> 00:08:48,220
before and after using DTT, we
can tell if the results change,

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00:08:48,220 --> 00:08:50,900
and that will tell us whether
disulfide bridges were there

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00:08:50,900 --> 00:08:52,180
to begin with.

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00:08:52,180 --> 00:08:54,680
Now, let's take a look
at the DTT chemistry.

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00:08:58,050 --> 00:09:13,090
Here we have a disulfide
bond or bridge in a protein,

154
00:09:13,090 --> 00:09:16,240
and we're going to
treat this with DTT,

155
00:09:16,240 --> 00:09:19,140
which looks like this.

156
00:09:26,490 --> 00:09:31,440
This is DTT or dithiothreitol.

157
00:09:31,440 --> 00:09:34,800
Now, if we substitute the
SH groups with OH's, you

158
00:09:34,800 --> 00:09:38,340
notice we're going to have
four OH's and four carbons.

159
00:09:38,340 --> 00:09:42,480
That's just an alcohol
derived from a sugar, which

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00:09:42,480 --> 00:09:47,050
is called threose That's why
the threitol part of the name.

161
00:09:47,050 --> 00:09:47,550
All right.

162
00:09:47,550 --> 00:09:50,320
So when we do this
chemical reaction,

163
00:09:50,320 --> 00:09:55,110
this disulfide bridge is going
to transfer between the two

164
00:09:55,110 --> 00:09:57,440
sulfur atoms of DTT.

165
00:09:57,440 --> 00:10:03,780
They're going to form a
intramolecular sulfide bond.

166
00:10:03,780 --> 00:10:12,000
So our cysteines are
going to get reduced,

167
00:10:12,000 --> 00:10:20,590
and from DTT, we're going to
get this intramolecular sulfide.

168
00:10:26,190 --> 00:10:32,910
And because entropical
considerations,

169
00:10:32,910 --> 00:10:34,440
make the formation
of the six-member

170
00:10:34,440 --> 00:10:38,610
ring very, very easy, then
this equilibrium will typically

171
00:10:38,610 --> 00:10:41,520
shift to the right.

172
00:10:41,520 --> 00:10:44,400
Of course, we're also using
vast quantities of DTT

173
00:10:44,400 --> 00:10:48,360
to make sure that the entire
protein becomes reduced,

174
00:10:48,360 --> 00:10:52,290
and then our agent is going
to pick up this [INAUDIBLE]

175
00:10:52,290 --> 00:10:53,400
off the bond.

176
00:10:53,400 --> 00:10:55,560
This sums up the question
2 of the problem.

177
00:10:58,180 --> 00:11:01,600
Next, we're going to
look at distinguishing

178
00:11:01,600 --> 00:11:03,940
between a couple of
different peptides

179
00:11:03,940 --> 00:11:07,660
that we generate during
our analysis of our mystery

180
00:11:07,660 --> 00:11:08,770
protein.

181
00:11:08,770 --> 00:11:13,810
So we're told that we're
isolating by HPLC peptides

182
00:11:13,810 --> 00:11:15,970
of the following composition.

183
00:11:15,970 --> 00:11:20,590
One has tryptophan,
phenylalanine,

184
00:11:20,590 --> 00:11:26,730
valine, aspartate,
lysine, cysteine.

185
00:11:26,730 --> 00:11:28,780
That's peptide one.

186
00:11:28,780 --> 00:11:32,820
Another one has phenylalanine,
serine, cysteine,

187
00:11:32,820 --> 00:11:35,290
and an unknown amino acid.

188
00:11:35,290 --> 00:11:40,251
Finally, the third one
has alanine and lysine.

189
00:11:40,251 --> 00:11:40,750
All right.

190
00:11:40,750 --> 00:11:44,320
Now, obviously, these peptides
have different compositions,

191
00:11:44,320 --> 00:11:49,751
so if you could just put them
through mass spectrometry,

192
00:11:49,751 --> 00:11:51,250
we're going to get
different masses,

193
00:11:51,250 --> 00:11:54,550
so we can tell very
quickly which one is which.

194
00:11:54,550 --> 00:11:57,380
But if we don't have
mass spec available,

195
00:11:57,380 --> 00:12:01,300
we can also tell them apart
only using UV-Vis spectroscopy.

196
00:12:01,300 --> 00:12:05,530
So all you need to remember
is that an amino acid

197
00:12:05,530 --> 00:12:07,040
like tryptophan
that we have here,

198
00:12:07,040 --> 00:12:14,650
W, has a very strong absorption
in the 280 to 300 nanometers.

199
00:12:14,650 --> 00:12:18,220
Whereas, amino acids
like phenylalanine,

200
00:12:18,220 --> 00:12:22,960
present here or here, they
absorb only around 260

201
00:12:22,960 --> 00:12:24,310
nanometers.

202
00:12:24,310 --> 00:12:27,340
Most of the other amino acids
don't absorb in this range

203
00:12:27,340 --> 00:12:28,670
at all.

204
00:12:28,670 --> 00:12:35,458
So if we were to plot the UV-Vis
spectra of these peptides,

205
00:12:35,458 --> 00:12:37,720
this is going to
be our absorption,

206
00:12:37,720 --> 00:12:41,840
and this is going to be lambda
in nanometers, the wavelength.

207
00:12:41,840 --> 00:12:47,140
So let's look in the
range from 200 to 400.

208
00:12:47,140 --> 00:12:48,700
300 is about here.

209
00:12:48,700 --> 00:12:49,930
250 is about here.

210
00:12:52,690 --> 00:12:55,150
And let's label these.

211
00:12:55,150 --> 00:13:03,680
Let's say this peptide is
red, this peptide is blue,

212
00:13:03,680 --> 00:13:06,860
and this one is green.

213
00:13:06,860 --> 00:13:08,460
All right.

214
00:13:08,460 --> 00:13:11,190
So now the red
peptide, as I told you,

215
00:13:11,190 --> 00:13:14,310
contains both tryptophan
and phenylalanine,

216
00:13:14,310 --> 00:13:17,370
so it's going to
absorb both around 260

217
00:13:17,370 --> 00:13:19,770
and both around 280 to 300.

218
00:13:19,770 --> 00:13:24,420
So it's going to have a
pretty big hump in this area

219
00:13:24,420 --> 00:13:26,860
from 250 to 300.

220
00:13:26,860 --> 00:13:29,130
Now, the blue peptide
only has phenylalanine,

221
00:13:29,130 --> 00:13:33,020
so above 280 or so, nanometer
is going to drop off.

222
00:13:33,020 --> 00:13:37,200
So it's going to
look more like this,

223
00:13:37,200 --> 00:13:42,210
whereas, the green one, well,
it has neither phenylalanine

224
00:13:42,210 --> 00:13:45,540
or tryptophan, so
below 240 or so,

225
00:13:45,540 --> 00:13:47,830
it's going to have no
UV absorption at all.

226
00:13:47,830 --> 00:13:51,940
So it's going to
drop off right here.

227
00:13:51,940 --> 00:13:54,540
So basically, if
we're just looking,

228
00:13:54,540 --> 00:13:58,980
say, around 260 nanometers,
we should see a stronger peak

229
00:13:58,980 --> 00:14:01,920
from the red peptide and a
weaker peak from the blue one.

230
00:14:01,920 --> 00:14:04,180
But if we're going
to look around 300,

231
00:14:04,180 --> 00:14:07,320
we should only see
the red peptide.

232
00:14:07,320 --> 00:14:11,160
So by just using the
UV-Vis absorption,

233
00:14:11,160 --> 00:14:13,240
we can tell these
three peptides apart.

234
00:14:16,360 --> 00:14:19,660
Now we're ready to figure out
the structure of our mystery

235
00:14:19,660 --> 00:14:21,430
protein.

236
00:14:21,430 --> 00:14:23,980
This is exactly what the last
part of the problem is asking.

237
00:14:23,980 --> 00:14:27,400
So we're going to take each
one of the clues provided--

238
00:14:27,400 --> 00:14:28,750
A, B, C, and D--

239
00:14:28,750 --> 00:14:30,820
and analyze and see
what information we can

240
00:14:30,820 --> 00:14:32,800
derive from each one of them.

241
00:14:32,800 --> 00:14:35,050
The first piece of
information that we're given

242
00:14:35,050 --> 00:14:37,870
is the result of
a total hydrolysis

243
00:14:37,870 --> 00:14:41,695
of our peptide
with six molar HCl.

244
00:14:41,695 --> 00:14:44,260
So we're told we get
the following amino acid

245
00:14:44,260 --> 00:14:45,338
composition.

246
00:14:48,930 --> 00:14:57,180
We get two phenylalanines, one
methionine, alanine, valine,

247
00:14:57,180 --> 00:15:05,630
two lysines, one serine, two
cysteine, and one aspartate.

248
00:15:05,630 --> 00:15:08,990
Now, recall this is not
a comprehensive list

249
00:15:08,990 --> 00:15:12,020
because the hydrolysis
and six smaller HCl

250
00:15:12,020 --> 00:15:14,690
may destroy some
of the amino acids.

251
00:15:14,690 --> 00:15:18,080
And specifically, we
know for sure something

252
00:15:18,080 --> 00:15:22,997
like tryptophan,
threonine, and tyrosine.

253
00:15:22,997 --> 00:15:24,830
They're going to be
destroyed, and we're not

254
00:15:24,830 --> 00:15:25,920
going to see them here.

255
00:15:25,920 --> 00:15:29,060
So these amino acids may still
be present in our protein,

256
00:15:29,060 --> 00:15:34,880
but we're not going to be able
to see them in this situation.

257
00:15:34,880 --> 00:15:38,440
Now, let's analyze
these amino acids

258
00:15:38,440 --> 00:15:40,500
and see what we can
derive from this.

259
00:15:40,500 --> 00:15:43,700
So first of all,
we have cysteines.

260
00:15:43,700 --> 00:15:45,770
So two cysteines,
it means our protein

261
00:15:45,770 --> 00:15:48,770
can form disulfide bridges.

262
00:15:48,770 --> 00:15:51,440
So from the get
go, two cysteine,

263
00:15:51,440 --> 00:15:54,170
it means we can form an
internal disulfide bridge

264
00:15:54,170 --> 00:15:59,570
or our protein contains a
disulfide bridge between two

265
00:15:59,570 --> 00:16:02,480
otherwise unconnected peptides.

266
00:16:02,480 --> 00:16:05,390
So here are some possibilities.

267
00:16:05,390 --> 00:16:09,020
For example, our
peptide is one chain,

268
00:16:09,020 --> 00:16:14,590
and our cysteines are
not actually connected

269
00:16:14,590 --> 00:16:16,600
by a disulfide bridge.

270
00:16:16,600 --> 00:16:19,660
Another possibility is we
do have a disulfide bridge

271
00:16:19,660 --> 00:16:24,770
between them, like that.

272
00:16:24,770 --> 00:16:29,960
Or yet another possibility is
that we have two pieces, two

273
00:16:29,960 --> 00:16:33,740
polypeptide chains, and
the disulfide bridge

274
00:16:33,740 --> 00:16:37,940
is the only thing that's
holding them together.

275
00:16:37,940 --> 00:16:40,760
Now, in each one of
these cases, obviously,

276
00:16:40,760 --> 00:16:44,300
these polypeptide
chains are oriented.

277
00:16:44,300 --> 00:16:48,582
So in one end, they're going
to have the amine group.

278
00:16:48,582 --> 00:16:50,770
Let me draw it here, NH3.

279
00:16:55,430 --> 00:17:01,680
Here we're going to have
two, and the other end

280
00:17:01,680 --> 00:17:04,735
is going to be the
carboxyl group, COO minus.

281
00:17:07,630 --> 00:17:09,240
And once again, in
this case, we're

282
00:17:09,240 --> 00:17:11,558
going to have two of them.

283
00:17:11,558 --> 00:17:12,359
All right.

284
00:17:12,359 --> 00:17:17,619
Now, how can we narrow down from
these couple of possibilities?

285
00:17:17,619 --> 00:17:22,349
Well, the fact that we're
getting two lysines here

286
00:17:22,349 --> 00:17:25,210
is a very important clue.

287
00:17:25,210 --> 00:17:29,730
Now remember, we're told that
our peptide, this mystery

288
00:17:29,730 --> 00:17:35,190
protein, it actually comes as
a result of a trypsin digest.

289
00:17:35,190 --> 00:17:39,300
And if you recall our
discussion earlier,

290
00:17:39,300 --> 00:17:41,550
we said that upon
trypsin digest,

291
00:17:41,550 --> 00:17:44,130
all the smaller peptides
that were obtained

292
00:17:44,130 --> 00:17:48,270
will end in lysine
or an arginine.

293
00:17:48,270 --> 00:17:50,520
The fact that we have
two lysines here,

294
00:17:50,520 --> 00:17:53,570
it means both of
them they need to be

295
00:17:53,570 --> 00:17:58,500
at the carboxy ends of peptides.

296
00:17:58,500 --> 00:18:01,080
Because if they were in
the middle of a chain,

297
00:18:01,080 --> 00:18:06,090
the trypsin would have
cut that chain in half.

298
00:18:06,090 --> 00:18:16,400
So two lysines means we've
got to have two carboxy ends.

299
00:18:16,400 --> 00:18:22,400
So therefore, this seems to be
the only possibility in which

300
00:18:22,400 --> 00:18:24,170
we can accommodate two lysines.

301
00:18:24,170 --> 00:18:28,980
Basically, the two carboxy
ends that we see here,

302
00:18:28,980 --> 00:18:31,580
each one has to be a lysine.

303
00:18:31,580 --> 00:18:34,250
This possibility only
has one carboxy end,

304
00:18:34,250 --> 00:18:36,710
so the other lysine
that we have to place

305
00:18:36,710 --> 00:18:39,860
will not be at the
end, so would not

306
00:18:39,860 --> 00:18:42,860
be compatible with
a trypsin digest.

307
00:18:42,860 --> 00:18:44,660
Same applies for this case.

308
00:18:44,660 --> 00:18:52,790
So these possibilities are
not consistent with our data.

309
00:18:52,790 --> 00:18:54,920
So we know already
that our protein must

310
00:18:54,920 --> 00:18:57,710
look like this, two
polypeptide chains,

311
00:18:57,710 --> 00:19:00,680
each one ends with
lysine, and there

312
00:19:00,680 --> 00:19:03,660
must be a disulfide bridge.

313
00:19:03,660 --> 00:19:07,430
The second clue we're given
is that the Edman degradation

314
00:19:07,430 --> 00:19:10,430
of the protein yields valine.

315
00:19:10,430 --> 00:19:12,770
Now, as you know,
Edman degradation

316
00:19:12,770 --> 00:19:16,730
is a chemical reaction by
which we can digest the protein

317
00:19:16,730 --> 00:19:19,760
from the N terminus,
from the amino terminus

318
00:19:19,760 --> 00:19:21,920
of a polypeptide chain.

319
00:19:21,920 --> 00:19:26,880
Now as you recall, we have
established in the first part

320
00:19:26,880 --> 00:19:32,450
that we have two amino
termini in our protein.

321
00:19:32,450 --> 00:19:35,990
So the fact that we only get one
amino acid and that is valine,

322
00:19:35,990 --> 00:19:38,390
it says that the
other amino terminus

323
00:19:38,390 --> 00:19:43,760
might be blocked or
somehow unavailable

324
00:19:43,760 --> 00:19:45,540
for the Edman degradation.

325
00:19:45,540 --> 00:19:48,800
So let's update the
structure of our protein

326
00:19:48,800 --> 00:19:50,550
to take into account
the second clue.

327
00:19:55,460 --> 00:19:59,300
So we said we have two
polypeptide chains.

328
00:19:59,300 --> 00:20:02,670
There's a disulfide
bond in between them.

329
00:20:02,670 --> 00:20:06,580
Now, each one of them
ends with a lysine.

330
00:20:06,580 --> 00:20:08,450
This is a carboxy
end, and now we

331
00:20:08,450 --> 00:20:12,020
know from the Edman degradation
that amino end of one of them

332
00:20:12,020 --> 00:20:13,730
has to be valine.

333
00:20:13,730 --> 00:20:17,090
The other one-- we're going
to put a box like that--

334
00:20:17,090 --> 00:20:18,200
is a blocked end.

335
00:20:18,200 --> 00:20:22,110
So the amino terminus
is not available.

336
00:20:22,110 --> 00:20:25,420
So this is as much as we can
tell from these first two

337
00:20:25,420 --> 00:20:27,090
clues.

338
00:20:27,090 --> 00:20:30,510
We're given the products
of the chymotrypsin

339
00:20:30,510 --> 00:20:33,810
digest of our protein
after it was previously

340
00:20:33,810 --> 00:20:35,910
treated with DTT.

341
00:20:35,910 --> 00:20:39,690
So we know first DTT is going
to cleave the disulfide bond,

342
00:20:39,690 --> 00:20:42,230
and then chymotrypsin,
as we talked previously,

343
00:20:42,230 --> 00:20:47,040
is going to cut after large,
hydrophobic, or aromatic

344
00:20:47,040 --> 00:20:48,610
amino acids.

345
00:20:48,610 --> 00:20:52,751
Now, we're told we're getting
five smaller peptides.

346
00:20:52,751 --> 00:20:53,500
Let's take a look.

347
00:20:56,430 --> 00:20:58,500
Here are the five fragments.

348
00:20:58,500 --> 00:21:09,900
One is tryptophan, valine; one
is cysteine, phenylalanine,

349
00:21:09,900 --> 00:21:15,690
another one is aspartate
lysine, another one

350
00:21:15,690 --> 00:21:20,700
is methionine, alanine,
cysteine, and lysine;

351
00:21:20,700 --> 00:21:25,830
and finally, the last one
has serine, phenylalanine,

352
00:21:25,830 --> 00:21:28,920
and something that's
not an amino acid.

353
00:21:28,920 --> 00:21:33,428
It's going to make
it like a small x.

354
00:21:33,428 --> 00:21:35,390
All right.

355
00:21:35,390 --> 00:21:39,280
Now, from what we know
about chymotrypsin digest,

356
00:21:39,280 --> 00:21:43,610
we should be able to orient
these peptides basically

357
00:21:43,610 --> 00:21:46,580
to tell which amino acid
is at the N terminus

358
00:21:46,580 --> 00:21:52,070
and which amino acid is at the
C terminus of each one of them.

359
00:21:52,070 --> 00:21:53,680
Now the first one.

360
00:21:53,680 --> 00:21:56,830
Well, we know from
the second clue

361
00:21:56,830 --> 00:21:58,770
that valine is at
the N terminus,

362
00:21:58,770 --> 00:22:00,630
so that makes it pretty easy.

363
00:22:00,630 --> 00:22:05,770
Then the sequence has to be V-W.
So valine is the N terminus,

364
00:22:05,770 --> 00:22:09,070
W is the C terminus,
and as we said,

365
00:22:09,070 --> 00:22:11,470
tryptophan is one of the
amino acids recognized

366
00:22:11,470 --> 00:22:13,870
by chymotrypsin, so
it's going to end up

367
00:22:13,870 --> 00:22:17,610
at the carboxy terminus.

368
00:22:17,610 --> 00:22:22,540
C-F, that has to go C
and F, phenylalanine,

369
00:22:22,540 --> 00:22:26,950
another chymotrypsin
amino acid that's

370
00:22:26,950 --> 00:22:29,460
left of this carboxy terminus.

371
00:22:29,460 --> 00:22:33,070
D-K. Now, neither
of these amino acids

372
00:22:33,070 --> 00:22:38,680
is recognized by chymotrypsin,
but we remember from clue one

373
00:22:38,680 --> 00:22:41,390
that D-K must be the
carboxy terminus.

374
00:22:41,390 --> 00:22:45,970
So the sequence can only be D-K.

375
00:22:45,970 --> 00:22:49,420
Now, here M, A, C,
and K, none of these

376
00:22:49,420 --> 00:22:52,930
is actually on our recognition
list for chymotrypsin,

377
00:22:52,930 --> 00:22:56,410
but once again, we know that
K must be in the carboxy end.

378
00:22:56,410 --> 00:22:58,840
So for now we're
going to have M, A,

379
00:22:58,840 --> 00:23:04,240
and C in a particular order,
which we cannot establish just

380
00:23:04,240 --> 00:23:07,940
yet and K at the carboxy end.

381
00:23:07,940 --> 00:23:11,390
And finally, we
do an S, F and x.

382
00:23:11,390 --> 00:23:13,570
Well, x is not
even an amino acid,

383
00:23:13,570 --> 00:23:18,760
and F is an amino acid that's
left of the carboxy terminus

384
00:23:18,760 --> 00:23:26,110
by chymotrypsin, so it's
probably x, S, and F. Now,

385
00:23:26,110 --> 00:23:27,580
what about x?

386
00:23:27,580 --> 00:23:41,780
We're told that x is not an
amino acid and is hydrophobic

387
00:23:41,780 --> 00:23:46,865
and it has a molecular
weight of 256 Daltons.

388
00:23:50,381 --> 00:23:50,880
All right.

389
00:23:50,880 --> 00:23:52,860
So in order to
figure out what x is,

390
00:23:52,860 --> 00:23:56,520
we have to read back the
beginning of the problem, which

391
00:23:56,520 --> 00:23:59,460
tells us that we're looking
at a protein that's associated

392
00:23:59,460 --> 00:24:01,140
with a plasma membrane.

393
00:24:01,140 --> 00:24:03,420
And one way for proteins
to associate with a plasma

394
00:24:03,420 --> 00:24:08,370
membrane is to be modified,
to incorporate a fatty acid,

395
00:24:08,370 --> 00:24:10,230
such as palmitate.

396
00:24:10,230 --> 00:24:17,040
So perhaps x, which is blocking
the N terminus of this peptide

397
00:24:17,040 --> 00:24:18,410
is a fatty acid.

398
00:24:21,930 --> 00:24:24,180
Now, the general
formula for fatty acids

399
00:24:24,180 --> 00:24:29,160
is something like this,
CH3 CH2 repeated, say,

400
00:24:29,160 --> 00:24:33,660
n times, and then COOH.

401
00:24:33,660 --> 00:24:37,680
So let's see what would
be the fatty acid that

402
00:24:37,680 --> 00:24:40,890
has the molecular
weight 256 Daltons?

403
00:24:40,890 --> 00:24:45,780
Well, the mass of a
methyl group is 15.

404
00:24:45,780 --> 00:24:49,800
The mass of this carboxylate
is 45, so we have about 60

405
00:24:49,800 --> 00:24:58,290
plus 14n equals 256
or 14n equals 196.

406
00:24:58,290 --> 00:25:02,130
Therefore, n is 14.

407
00:25:02,130 --> 00:25:06,000
So the fatty acid that
would fit these criteria--

408
00:25:06,000 --> 00:25:08,980
it's hydrophobic, it
has a mass of 256--

409
00:25:08,980 --> 00:25:16,770
will be CH3, CH2 14 times
COOH, or the fatty acid

410
00:25:16,770 --> 00:25:18,280
was 16 carbons.

411
00:25:18,280 --> 00:25:19,830
This is palmitic acid.

412
00:25:24,060 --> 00:25:26,580
Now, the answer that
we got, palmitic acid,

413
00:25:26,580 --> 00:25:29,730
was anticipated in the
text of the problem

414
00:25:29,730 --> 00:25:33,380
because it gave us an example
one way to associate proteins

415
00:25:33,380 --> 00:25:36,630
to the plasma
membrane is to form

416
00:25:36,630 --> 00:25:40,960
a covalent linkage with a
fatty acid such as palmitate.

417
00:25:40,960 --> 00:25:44,400
But how does a protein
associate with a plasma

418
00:25:44,400 --> 00:25:50,910
membrane when it has a palmitic
acid residue as part of it?

419
00:25:50,910 --> 00:25:53,580
Well, let's take a
look at a diagram.

420
00:25:53,580 --> 00:25:56,670
Here we have a representation
of the plasma membrane,

421
00:25:56,670 --> 00:25:59,730
where we have the phospholipids
like the hydrophilic head

422
00:25:59,730 --> 00:26:04,410
pointing inside and outside the
cell and the hydrophobic tails

423
00:26:04,410 --> 00:26:09,150
of the fatty acids
lined up to each other.

424
00:26:09,150 --> 00:26:11,880
Now, imagine we have
a protein that's

425
00:26:11,880 --> 00:26:15,450
modified to contain one
of these fatty acids.

426
00:26:15,450 --> 00:26:20,100
Then this fatty acid can
just insert right next

427
00:26:20,100 --> 00:26:23,680
to the phospholipids
of the plasma membrane.

428
00:26:23,680 --> 00:26:26,460
And that way it tethers
the protein right

429
00:26:26,460 --> 00:26:28,500
at the plasma membrane.

430
00:26:28,500 --> 00:26:31,260
The final clue that we're
given in order to figure out

431
00:26:31,260 --> 00:26:33,540
the structure of
our mystery protein

432
00:26:33,540 --> 00:26:37,770
is the digest with an
inorganic agent this time.

433
00:26:37,770 --> 00:26:39,710
It's called cyanogen bromide.

434
00:26:39,710 --> 00:26:44,340
Cyanogen bromide reacts quite
specifically with methionines,

435
00:26:44,340 --> 00:26:47,760
and it cleaves the peptide
bond after methionine

436
00:26:47,760 --> 00:26:51,750
leaving behind an
unnatural amino acid called

437
00:26:51,750 --> 00:26:54,570
a homoserine lactone.

438
00:26:54,570 --> 00:26:56,730
Now, let's take a
look at what happens

439
00:26:56,730 --> 00:26:59,340
when we treat our protein
with cyanogen bromide.

440
00:26:59,340 --> 00:27:09,266
So we're told we're getting
the following peptides, A, K

441
00:27:09,266 --> 00:27:32,470
and W, F, V, D, K, C and F, S,
C, and an unnatural amino acid.

442
00:27:32,470 --> 00:27:37,120
As I just explained to you, the
unnatural amino acid probably

443
00:27:37,120 --> 00:27:40,250
is this homoserine lactone.

444
00:27:45,210 --> 00:27:51,680
So it's most likely,
instead of this amino acid

445
00:27:51,680 --> 00:27:56,170
in the actual sequence,
we had a methionine.

446
00:27:56,170 --> 00:27:59,010
So we know these four residues--

447
00:27:59,010 --> 00:28:01,720
F, S, C, and M--

448
00:28:01,720 --> 00:28:04,260
go together.

449
00:28:04,260 --> 00:28:05,310
All right.

450
00:28:05,310 --> 00:28:08,070
Now, we can start putting
together the clues

451
00:28:08,070 --> 00:28:13,020
from part 3 and 4 and try to
figure out the final structure

452
00:28:13,020 --> 00:28:13,700
of our protein.

453
00:28:17,780 --> 00:28:26,210
So the peptide A, K, we know
has to have K at the C terminus.

454
00:28:26,210 --> 00:28:28,890
Now, from part 3, we
found out that there

455
00:28:28,890 --> 00:28:33,430
was a peptide that
contained M, A, C, and K,

456
00:28:33,430 --> 00:28:37,380
after the chymotrypsin digest.

457
00:28:37,380 --> 00:28:41,553
So now we know K has
to be the carboxy end

458
00:28:41,553 --> 00:28:45,870
and A has to be right
before K, so in order

459
00:28:45,870 --> 00:28:49,080
to get an A-K
peptide, then A has

460
00:28:49,080 --> 00:28:51,810
to be right after
methionine because that's

461
00:28:51,810 --> 00:28:55,035
where cyanogen bromide is going
to cleave the peptide bond.

462
00:28:57,750 --> 00:29:02,010
So the only sequence
that we can have here is

463
00:29:02,010 --> 00:29:05,220
going to be C followed
by M followed by A

464
00:29:05,220 --> 00:29:08,540
followed by K. So when we
treat the cyanogen bromide,

465
00:29:08,540 --> 00:29:11,420
we're going to be cleaving
this bond between M and A,

466
00:29:11,420 --> 00:29:14,460
generating M as a
homoserine lactone.

467
00:29:14,460 --> 00:29:21,730
Now, we also know that M has
to be in this small peptide,

468
00:29:21,730 --> 00:29:25,390
and we know it has to
be C, M as a sequence.

469
00:29:25,390 --> 00:29:31,540
So therefore, S and F have
to be on the amino terminus

470
00:29:31,540 --> 00:29:33,370
of this peptide.

471
00:29:33,370 --> 00:29:36,100
And we also have a
clue from part 3, which

472
00:29:36,100 --> 00:29:43,950
said that S, F, and this
non amino acid moiety

473
00:29:43,950 --> 00:29:49,180
were in the same peptide.

474
00:29:49,180 --> 00:29:51,970
So from here we said that,
well, the sequence there, it's

475
00:29:51,970 --> 00:29:57,280
probably x modifying the amino
S and then F with a carboxy end.

476
00:29:57,280 --> 00:30:04,760
So putting these
three things together,

477
00:30:04,760 --> 00:30:07,400
we can come up with a sequence
for this strand, which

478
00:30:07,400 --> 00:30:16,300
is going to be x, S,
F, then C, M and then A

479
00:30:16,300 --> 00:30:23,150
and K. So this is probably
one of the peptide chains

480
00:30:23,150 --> 00:30:25,670
in our mystery protein.

481
00:30:25,670 --> 00:30:27,350
Then, of course,
the other is going

482
00:30:27,350 --> 00:30:30,600
to be composed of
these amino acids.

483
00:30:30,600 --> 00:30:33,980
Now, we already know something
about the sequence of these.

484
00:30:33,980 --> 00:30:39,130
For example, we know V
goes before W. We also

485
00:30:39,130 --> 00:30:50,600
know D goes before K, and also
know C goes before F. Now, we

486
00:30:50,600 --> 00:30:57,880
know this has to be the carboxy
end of the peptide chain.

487
00:30:57,880 --> 00:31:02,220
Let's write it here, COO minus.

488
00:31:02,220 --> 00:31:06,720
And V has to be the amino end.

489
00:31:10,060 --> 00:31:14,230
So then C-F, there's no other
way, has to be in the middle.

490
00:31:14,230 --> 00:31:19,210
So the only possible
sequence for our second chain

491
00:31:19,210 --> 00:31:27,950
is going to be V,
W, C, F, D, and K.

492
00:31:27,950 --> 00:31:31,250
Now that we've established
the exact sequence

493
00:31:31,250 --> 00:31:33,200
of each one of these
peptide chains,

494
00:31:33,200 --> 00:31:35,450
then we can put together
the final structure

495
00:31:35,450 --> 00:31:36,890
of our mystery protein.

496
00:31:36,890 --> 00:31:42,344
So I'm going to transcribe these
here, the first chain V, W, C,

497
00:31:42,344 --> 00:31:48,980
F, D, and K. And we know
the cysteine is going

498
00:31:48,980 --> 00:31:54,940
to have our disulfide bridge
to the other cysteine, which

499
00:31:54,940 --> 00:32:03,770
goes C, M, A, K, and
then F, and then S.

500
00:32:03,770 --> 00:32:10,370
And now we know the N
terminus of this peptide,

501
00:32:10,370 --> 00:32:19,990
we're going to have our fatty
acid residue CH2 14 times CH3.

502
00:32:22,700 --> 00:32:29,480
So let's just mark, once
again, the carboxy ends here

503
00:32:29,480 --> 00:32:35,260
and the amino end here.

504
00:32:35,260 --> 00:32:41,980
So this is the final
answer for our problem

505
00:32:41,980 --> 00:32:45,700
and the structure of
our mystery peptide.

506
00:32:45,700 --> 00:32:47,590
Well, that's it
for this problem.

507
00:32:47,590 --> 00:32:51,940
I hope you enjoyed this
little protein mystery hunt.

508
00:32:51,940 --> 00:32:55,360
Now, remember that the
strategy that we used here

509
00:32:55,360 --> 00:33:00,850
in which we logically string
together pieces of data

510
00:33:00,850 --> 00:33:04,780
to build a big picture, it's
really the same strategy

511
00:33:04,780 --> 00:33:08,530
that has been used and is
being used right now to advance

512
00:33:08,530 --> 00:33:11,880
our knowledge about living
systems and their underlying

513
00:33:11,880 --> 00:33:14,030
biochemistry.