Biochemistry
Lecture 6
January 24, 2013
Lecture 6: UV Absorption of Amino Acids & Protein Structure
Assigned reading: Horton - 3.2B, 3.9-3.10.
Key Terms:
 UV Absorption
 Peptide/protein nomenclature
 Edman degradation
 Cyanogen bromide cleavage
Nelson 5e - box 3.1, 3.4.


OLI: Quiz: 2
Chymotrypsin & trypsin cleavage
Protein sequence determination.
+
6A. UV Absorption Properties of Amino Acids
H3N
Three aromatic amino acids absorb light in the ultraviolet range (UV).
O
O
N
H
+
H3N
O
O
OH
+
H3N
O
The extinction coefficients (or molar absorption coefficients)
of these amino acids are:
5
Extinction Coefficient (MAX)
5,500 M-1cm-1 (280 nm)
1,490 M-1cm-1 (274 nm~280nm)
220 M-1cm-1 (257 nm)
The amount of light absorbed by a solution of concentration
[X] is given by the Beer-Lambert Law:
4
3
2
1
0
I
A  log O   [ X ] l
I
where
 A is the absorbance of the sample;
 I0 is the intensity of the incident light;
 I is the intensity of the light that leaves the
sample.
6
Absorbance
Amino acid
Trp
Tyr
Phe
O
230 240 250 260 270 280 290 300 310 320 330
Wavelength (nm)

ε is the molar extinction coefficient at a
specific wavelength, e.g. at max;
 [X] is the concentration of the absorbing
species
 l is the path length (usually 1 cm).
Therefore, given a known extinction coefficient it is possible to measure the concentration of a protein.
Calculation of molar extinction coefficients: If a molecule contains a mixture of N different chromophores,
the molar extinction coefficient can generally be calculated as the sum of the molar extinction coefficient for
each absorbing group in the protein:
N
 Protein    i
i 1
Therefore, the molar extinction coefficient for a protein can be calculated from its amino acid composition.
Example: A protein has two Tryptophan (Trp) residues and one Tyrosine (Tyr) what is its extinction
coefficient?
1
Biochemistry
Lecture 6
January 24, 2013
6B. Protein Structural Hierarchy:
1. Primary structure (1°): The amino acid sequence.
2. Secondary structure (2°): Configuration of mainchain.
3. Tertiary structure (3°): Side chain packing in the 3-D structure.
4. Quaternary structure (4°): Association of subunits.
O
6C. Primary Structure
H3C
CH3
O
H
CH3
O
H
N
H3N
O
N
O
+
H
H
H
O
O
H
O
O
CH3
H 3N
+
O
O
H
N
O
N
H
O
O
H 3C
CH3
Example: Draw the structure of Gly-Ala-Ala
i) draw mainchain atoms first.
ii) add sidechains
Sequencing Proteins
Protein Sequencing: Edman Degradation
We will focus on N-terminal sequencing using Edman
degradation coupled with fragmentation of the peptide in the
case of larger proteins.
Edman Degradation: Cleaves single amino acids from the
amino-terminus. Producing:
i) The PTH derivative of the released amino acid can be
identified.
ii) A peptide that is one residue shorter is produced, this can
be treated with PITC to obtain the next residue, etc.
iii) Errors accumulate because the release of the PTH-amino
acid is not 100%, limiting the length of sequence to ~75
residues:
AGCTWGCAYPF
AGCTWGCAYPF
100%
Ala (A)
GCTWGCAYPF
CTWGCAYPF
GCTWGCAYPF
CTWGCAYPF
AGCTWGCAYPF
GCTWGCAYPF
CTWGCAYPF
AGCTWGCAYPF
GCTWGCAYPF
GCTWGCAYPF
AGCTWGCAYPF
AGCTWGCAYPF
(underlined residue released at each cycle)
2
75%
Gly(G)
25%
Ala(A)
GCTWGCAYPF
S
C
N
O
R1
8
H2N
N
(N) amino acids
R2
O
PITC
S
N
R1
O
N
O
PTC polypeptide
R2
O
N
N
S
50%
Cys(C)
50%
Gly(G)
R1
N
H
O
H2N
(N-1) amino acids
R2
Specific PTH-amino acid
Biochemistry
Lecture 6
January 24, 2013
Fragmentation: Because it is not possible to sequence proteins larger than ~75 residues it is necessary to
fragment the protein to extend the sequence information. After cleavage, the individual peptide fragments are
separated from each other and each is subject to N-terminal sequencing using the Edman degradation
method.
a. Cyanogen bromide (CNBr) cleaves the peptide bond after Methionine residues. You do not need to know
the mechanism, just the specificity of the reaction.
CNBr
Ser  Met  Gly  Ala  Phe  Arg  Leu  Ile   Ser  Met

Gly  Ala  Phe  Arg  Leu  Ile
b. Chymotrypsin hydrolyzes the peptide bonds that follow large hydrophobic residues, e.g. Phenylalanine,
Tyrosine, Tryptophan. You should remember this cleavage pattern.
Chymotrypsin
Ser  Met  Gly  Ala  Phe  Arg  Leu  Ile  Ser  Met  Gly  Ala  Phe

Arg  Leu  Ile
c. Trypsin hydrolyzes the peptide bonds that follow positively charged residues, e.g. Lysine and Arginine.
You should remember this cleavage pattern as well.
Trypsin
Ser  Met  Gly  Ala  Phe  Arg  Leu  Ile  Ser  Met  Gly  Ala  Phe  Arg
 Leu  Ile
Sequence Strategies:
If only two fragments are produced by the cleavage reaction, then it is straightforward to reconstruct the
sequence using the known amino terminal sequence of the original protein to determine which fragment is
first.
i) Obtain sequence past cleavage point (X)
ii) Cleave, purify fragments, sequence 2nd fragment.
However if the original protein is cleaved into three or more fragments, or they cannot be sequenced in their
entire length, then it is not possible to determine the correct order of fragments using a single cleavage agent.
Multiple overlapping fragments have to be used to determine the correct ordering:
K
frag. A
K
Y
Y
K
Y
Y
K
frag. B
Y
Y
K
K
Y
K
frag. D
Y
Y
Y
frag. E
frag. D
Y
K
frag. F
K
Y
frag. C
Y
Y
frag. E
overlapping sequences
3
Original protein
frag. B
K
Cleavage with
trypsin [Lys(K)/Arg(R)]
Two possible solutions
using trypsin fragments
(assuming amino-terminal seq. is known).
K
K
Y
frag. A
frag. C
K
Y
K
Original protein
K
K
frag. F
Cleavage with
chymotrypsin [Tyr(Y)/Phe(F)/Trp(W)]
Overlapping sequences
from second cleavage
reagent indicates correct
order of trypsin fragments.
Biochemistry
Lecture 6
January 24, 2013
An example: Ala-Gly-Met-Ser-Thr-Gly-Val-Val-Lys-Gly-Ser-Ala-Phe-Leu
Ala
Gly
Met
Ser
Thr
Gly
Val
Val
Lys
Gly
Ser
Ala
Phe
Leu
Here I have assumed that 7 cycles of Edman degradation are possible; note ~75 are more typical.
A. The first 7 cycles of Edman sequencing on the intact peptide released the following amino acids, in this
order: Ala, Gly, Met, Ser, Thr, Gly, Val, therefore , the initial, amino-terminal sequence of the peptide is:
Ala
Gly
Met
Ser
Thr
Gly
Val
Val
Lys
Gly
Ser
Ala
Phe
Leu
B. A new sample of the intact peptide was treated with CNBr. The two peptides (CNBr-1, CNBr-2) that
were produced were isolated and each was subject to sequencing, giving the following sequences (The
last three residues (Ala-Phe-Leu )could not be determined due to limitation in the Edman degradation to
7 steps, these are highlighted gray).
Ala
Ala
Gly
Gly
Met
Ser
Thr
Met
Gly
Ser
Thr
Val
Val
Lys
Gly
Val
Val
Gly
Lys
Ser
Ala
Gly
Phe
Ser
Ala
Leu
Phe
Leu
C. A new sample of the peptide was treated with Trypsin. The two peptides (Trp1, Trp2) that were produced
were isolated and each was sequenced.
Al a
Al a
Gl y
Gl y
Met
Met
Ser
Ser
Thr
Thr
Gl y
Gl y
Val
Val
Val
Val
Lys
Gl y
Ser
Gl y
Lys
Al a
Ser
Phe
Al a
Leu
Phe
Sequence alignment strategy: Find overlaps between fragments obtained with different cleavage reagents
& use these overlaps to correctly order the peptides obtained from one sequencing reaction.
The overlaps can be readily identified by finding a cleavage site in a peptide that would be cut by another
cleavage reagent and then identifying the correct fragment based on the expected amino-terminal sequence.
A) The original amino-terminal sequence contains a Methionine (Met), so I’d look for a CNBr fragment
beginning with Ser-Thr-Gly-Val, the information from that CNBr fragment extends the sequence:
Ala
Gly
Met
Ser
Thr
Gly
Val
Ser
Thr
Gly
Val
Val
Lys
Gly
Ser
Ala
Phe
Leu
B) The lysine indicates that there should be a trypsin fragment beginning with Gly, the sequence of that
trypsin fragment allows the completion of the sequence:
Ala
4
Gly
Met
Ser
Thr
Gly
Val
Ser
Thr
Gly
Val
Val
Lys
Gly
Ser
Ala
Phe
Leu
Gly
Ser
Ala
Phe
Leu
Leu