#1. The protein pictured below is bovine insulin. Determine the
number and the size of the fragments that would be generated
upon treatment with the following:
-look for the cleavage points
-is the disulfide bond broken (DTT present) or unbroken (no DTT)
Without DTT
Trypsin
Chymotrypsin
BrCN
1, 7 and 43 amino acids
1, 1, 1, 4, 5, 10, 29
51
Reagent (source)
Trypsin (bovine pancrease)
Chymotrypsin (bovine pancrease)
Staphylococcus V8 protease
Pepsin (porcine pancrease)
Cyanogen bromide (chemical)(CnBr)
With DTT
1,7, 21 and 22
1,1, 1, 2, 4, 5, 8, 14, 15
21 and 30
Specificity
Lys, Arg (C)
Phe, Trp, Tyr (C)
Glu, Asp (C)
Phe, Trp, Tyr (N)
Met (C)
#2. A peptide was cleaved into smaller fragments using chymotrypsin and CnBr.
The fragments were separated by HPLC and then sequenced by Edman
degradation. Determine the sequence of the original peptide from the following
fragment sequences.
Cleavage with chymotrypsin
EAGPDGMECAF
GPF
EAAMCKW
HR
IAHTY
Cleavage with CnBr
ECAFHR
CKWEAGPDGM
IAHTYGPFEAAM
-align the peptides
IAHTYGPFEAAMCKW EAGPDGMECAFHR
Reagent (source)
Trypsin (bovine pancrease)
Chymotrypsin (bovine pancrease)
Staphylococcus V8 protease
Pepsin (porcine pancrease)
Cyanogen bromide (chemical)(CnBr)
Specificity
Lys, Arg (C)
Phe, Trp, Tyr (C)
Glu, Asp (C)
Phe, Trp, Tyr (N)
Met (C)
#3. The following three peptides are subjected to anion-exchange
chromatography at pH 7.5 using a NaCl gradient to elute the peptides. There
are no disulfide bridges. What is the charge of each peptide at pH 7.4 and in
what order will they elute from the column? Use Table 3.2 to find the pKa
values. (1) MARKER
(2) SADDLE
(3) CRACKED
pKa
Form at pH 7.4
Charge
 amino group
9.3
HA
+1
M
pKa
Form at pH 7.4
Charge
 amino group
9.2
HA
+1
pKa
Form at pH 7.4
Charge
 amino group
C
10.7
8.4
HA
HA
+1
0
S
A
R
K
E
12.5 10.5 4.1
HA
HA A+1
+1 -1
A
D
3.9
A-1
D
3.9
A-1
L
R
12.5
HA
+1
A
C
8.4
HA
0
K
10.5
HA
+1
R
12.5
HA
+1
E
4.1
A-1
E
4.1
A-1
 carboxyl group
1.8
A-1
= +2
 carboxyl group
2.1
A-1
= -3
D  carboxyl group
3.9
2.0
AA-1
-1
=O
The order of elution from an anion exchange column (binds negatively charged proteins) is
MARKER, CRACKED, SADDLE or 1,3,2.
#4. A smallpeptide was found to contain equimolar amounts of the following amino acids:
arginine, glutamic acid, glycine, lysine, methionine and phenylalanine. Individual samples of
the peptide were treated with the following agents with the results noted:
a) trypsin: arginine and a pentapeptide
Trypsin cuts on the C side of arginine thereforethe arginine must be the N-terminal residue.
arg (glu, gly, lys, met, phe)
b) cyanogen bromide: two tripeptides
Cleavage occurs on the C side of met so met must must be the third residue in order for two
tripeptides to form.
arg, ?, met, ?, ?, ?
c) S. aureus V8 protease: lysine and a pentapeptide
Cuts at glu and asp on the C-side so lys must be C-terminal with a glu preceding it.
arg, ?, met, ?, glu, lys
d) chymotrypsin: a dipeptide and a tetrapeptide, with the latter showing absorbance at
260nm.
If phe is second then we have a dipeptide containg phe and if it is fourth then the tetrapeptide
contains phe. Since the tetrapeptide absorbs light at 260nm then it must contain the phe
because the aromatic amino acids absorbs UV light.
What is the primary structure of the peptide? Explain each piece of evidence given and the
reasoning that led to your answer.
arg, gly, met, phe, glu, lys
#5 You have 2 globular proteins. Protein X has 300 amino acids and protein Y has
400 amino acids. Which one will emerge first using gel filtration chromatography?
Gel filtration separates on the basis of size. The gel filtration resin consists of very
small beads with tiny holes. The protein equilibrates between inside and outside the
beads. The smaller proteins spend more time inside the bead and therefore move
more slowly down the column. Thus, larger protein Y will emerge first and X second.
#6 Below is an SDS gel in the presence of mercaptoethanol. Label the
band of the highest and lowest molecular weight band and explain.
The band closest to the top is the largest protein and the
one closest to the bottom is the smallest. In SDS the
protein unfolds and is like a big piece of spaghetti
covered with SDS and therefore has a bunch of negative
charge. Because it is longer and more stringer than the
smaller protein it cannot move as fast through the
agarose gel. The top is the - charge and bottom is the +
charge. This way the protein moves toward the bottom
because the protein is covered with SDS and therefore
has a - charge.
#7
a) Reverse-phase chromatography (HPLC) separates peptides based on
molecular weight. F
b) One method used to prevent disulfide bond interference with protein
sequencing procedures is to reduce disulfide bridges and prevent their
reformation with the addition of iodoacetic acid. T
c) High molecular weight proteins will migrate farther during gel
electrophoresis (SDS-PAGE). F
d) -sheet protein structures can be stabilized by hydrogen bonding
between distant residues on the same polypeptide. T
e) -sheets are a type of secondary structure and are found in every
protein. F
f) In the α-helix, the hydrogen bonds that stabilize the helix occur mainly
between electronegative atoms of the R groups. F
g) Negatively charged peptides flow through a cation-exchange
chromatography column without binding. T
h) The enzyme hexokinase (972 amino acids) will elute from a gel filtration
column before myoglobin (155 amino acids). T
i) Separation of proteins in the first dimension of 2D gel electrophoresis is
based on a protein’s molecular weight. F