1 Which of structures below stands for D-ribose?
Answer = (b)
2 Hydrophobic interactions most likely occur between the R groups of which of the following
a) arginine and histine
b) leucine and methionine
c) phenylalanie and threonine
d) valine and tyrosine
3 The three-dimensional conformation of a protein may be strongly influence by amino acid
residues that are very far apart in sequence. This relationship is in contrast to secondary
structure, where the amino residues are:
a) always side by side
b) generally near each other in sequence
c) generally on different polypeptide strands
d) generally near the polypeptide chain’s amino terminus or carboxyl terminus
e) restricted to only about seven of the twenty standard amino acids found in proteins
4 Myoglobin mainly consists of .
a) α helix b) β sheet c) β turn d) a) and b)
5 Amino acid analysis of an oligopeptide seven residues long gave
Asp Leu Arg Met Phe Tyr
The following facts were observed:
a. Trypsin treatment had no apparent effect.
b. The phenylthiohydantoin released by Edman degradation was
c. Brief chymotrypsin treatment yielded several products including a dipeptide and
a tetrapeptide. The amino acid composition of the tetrapeptide was Leu, Arg, and Met.
d. Cyanogen bromide treatment yielded a dipeptide, a tetrapeptide, and free Arg.
What is the amino acid sequence of this heptapeptide?
a) N- Arg Asp Tyr Met Leu Met Phe
b) N- Arg Asp Tyr Met Met Leu Phe
c) N- Phe Asp Tyr Met Leu Met Arg
d) N- Phe Asp Tyr Met Met Leu Phe
e) N- Tyr Asp Arg Met Leu Met Phe
6 functions to prevent intramolecular and intermolecular aggregates of unfolded
a) protein precursor b) ligand c) structural motif d) molecular chaperone
7 The H+
concentration of 0.5M NaOH is : the H+
concentration of 0.5M acetic
acid is ( assume the decrease in the concentration of acetic acid due to
dissociation is negligible: K’ of acetic acid = 1.74 x 10-5
a) 2 x 10-14
M 0.5 M
b) 0.5 M 2.95 x 10-3
c) 2 x 10-14
M 9.33 x 10-4
d) 2 x 10-14
M 2.95 x 10-3
e) 0.5 M 9.33 x 10-4
8 In the binding of oxygen to hemoglobin, the relationship between the concentration of
oxygen and the fraction of binding sites occupied can best be described as .
c) linear with a positive slope
d) linear with a negative slope
9 A fetus obtains oxygen from maternal circulation via the placenta. This biological process
can be realized because
a) Fetal hemoglobin (HbF) has a higher oxygen affinity than maternal hemoglobin (HbA)
b) The concentration of BPG is higher in fetal erythrocytes.
c) BPG could bind more tightly to deoxyHbA than to deoxyHbF.
d) a) and b)
e) a) and c)
10 Water molecules bind to one another by means of hydrogen bonds. Additionally, water is
able to form hydrogen bonds with each type of functional group except?
b) Amino groups (N-H)
c) Hydroxyl groups (O-H)
d) Carbonyl groups (C O)
e) None of above
11 Two monosaccharides combine to form disaccharidesor polysaccharides. This involves the
OH group of one monosaccharide forming an ester linkage with the anomeric carbon of the
second monosaccharide. This type of bond is called:
c) N-glycosidic bond
d) O-glycosidic bond
12 Proteins must fold into their final structures to be functional. Which of the following
statements are not true?
a) Folding begins with disulfide bond formation;
b) Structural motifs such as αα or ββ act as seeds around which the rest of the protein
c) Helper proteins called chaperones may assist protein folding;
d) Hydrophobic amino acids are buried in the interior.
13 Which of the following is correct with respect to the amino composition of all proteins?
a) Proteins with the same molecular weight have the same amino acid composition
b) Proteins contain at least one each of the twenty different standard amino acids.
c) Larger proteins have a more uniform distribution of amino acids than smaller proteins.
d) Proteins with different functions usually differ significantly in their amino acid
e) The average molecular weight of an amino acid in a protein increases with the size of the
14. The first step in two-dimensional gel electrophoresis generates a series of protein bands by
isoelectric focusing. In the second step, this gel is turned at a right angle, placed on another gel, and
a second electric current is applied. In this second step:
a) the individual bands undergo a second, more intense iselectric focusing.
b) the individual bands become stained so that the isoelectric focus patterns can be
c) the individual bands become visualized by interacting with protein-specific antibodies in
the second gel.
d) the proteins in the bands separate more completely because the second electric current is
in the opposite polarity to the first current.
e) proteins with similar isoelectric points become further separated according to their
molecular weights by an SDS gel.
15. Compare the following sequences taken from four different proteins, and select the answer
that best characterizes their relationships.
a) Protein 4 is the protein that shows the greatest overall homology to protein 1.
b) The portions of amino acid sequence shown suggest that these proteins are completely
c) Based only on sequences in column B, protein 4 reveals the greatest evolutionary
d) Comparing proteins 1 and 2 in column A reveals that these two proteins have diverged the
most throughout evolution.
e) Proteins 2 and 3 show a greater evolutionary distance than proteins 1 and 4.
16. Experiments on denaturation and renaturation after the reduction and reoxidation of the -
-S-S- bonds in the enzyme ribonuclease (RNase) have shown that:
a) the primary sequenced of RNase is sufficient to determine the formation of a specific
secondary and tertiary structure.
b) the enzyme, dissolved in water, is thermodynamically stable relative to the mixture of
amino acids whose residies are contained in RNase.
c) native ribonuclease does not have a unique secondary and tertiary structure.
d) the completely unfolded enzyme, with all –S-S- bonds broken, is still enzymatically
e) the folding of denatured RNase into the native, active conformation, requires the input of
energy in the form of heat.
17. A sequence of amino acids in a certain protein is found to be:
-Ser-Gly-Pro-Gly-. The sequence is most likely to be part of a(n):
a) beta turn
b) parallel beta-pleated sheet
c) alpha helix
d) anti-parallel beta pleated sheet
18. Which of the following amino acids has a side chain that can be modified by glycosylation?
19. In the β-pleated sheet conformation
a) there are hydrogen bonds perpendicular to the direction of the polypeptide chain
b) the polypeptide chain is almost fully extended
c) the polypeptide chains may be hydrogen bonded together in a parallel or
d) both a) and c)
e) a), b), and c)
20. Answer = (b)
1. (a) Calculate the pH at which the gamma carboxyl group of glutamic acid is two-
thirds dissociated. The pKa of the gamma carboxyl group is 4.3. Show your work.
The pKa of γ-carboxyl group of glutamic acid is 4.3
pH = pKa + log _[COO-
When the carboxyl group is 2/3 dissociated
] + [COOH] = [COO-
(i.e., 2/3 of the total is in the COO-
From this equation we find that [COOH] = [COO-
]/2 and, upon substituting into
Henderson-Hasselbach equation we find,
pH = pK + log 2
or, pH = 4.6
1. (b) Draw the structures of the following amino acids showing the charges on the
ionizable side chains as they would exist at pH 1. Give the three letter code and the one
letter code below each.
1 letter code
3 letter code
1. (c) The pKa of an ionizable group is affected by what is in the vicinity of that group.
How might the presence of calcium ions affect the pKa’s of the side chains of aspartic
and glutamic acid?
) may bind to carboxylate ions. Therefore, the presence of calcium
may lower the apparent pKa’s of aspartic and glutamic acid.
2. (a) Protein purification schemes typically involve several separation procedures. For
the two techniques listed below describe in some detail the basis for protein separation by
(i) Gel filtration chromatography
Answer 2(a) (i)
Gel filtration or molecular sieve chromatography separates proteins on the basis
of size. A porous matrix is employed such as beads of dextrans, agarose, or
polyacrylamide. A column of hydrated beads contains two aqueous volumes, the volume
within the beads or the internal volume, and the volume outside of the beads, the external
volume. The degree to which a protein can enter the internal volume is dependent on the
protein’s size. Large proteins cannot enter the internal volume at all, whereas smaller
proteins may be partially or completely accessible to the internal volume. Separation of
proteins of different sizes occurs because for every volume element added to the column,
the proteins are displaced that volume element down the column. Large proteins can
move only in the external volume whereas smaller proteins move in the external and
internal volumes. Molecular sieve chromatography is likely to be used at a later stage in
protein purification. It may be performed under conditions favoring the native state of the
protein in which case the elution profile is dependent on the native molecular weight of
the protein. Alternatively, denaturing conditions may be employed allowing individual
polypeptide chains to be separated.
(ii) affinity chromatography
Answer 2 (a) (ii)
Affinity chromatography exploits the fact that proteins have complementary
surfaces that exhibit incredible specificity. These surfaces might be a substrate binding
site, a cofactor binding site, or a protein-protein interaction surface as examples. The
principle is to attach a molecule, that interacts specifically with a particular protein, to an
inert, insoluble matrix. A mixture containing the protein of interest is added to the affinity
matrix under conditions favoring complementary interactions. The matrix is washed to
remove unbound protein and then subjected to conditions that disrupt the complementary
interactions. These conditions may be high salt, or high ionic strength, a change in pH, or
high concentration of cofactor or substrate. In favorable cases, affinity chromatography
may be a one-step purification scheme. Generally, affinity chromatography is employed
at a later stage at which the protein of interest has been partially purified and therefore
represents a high percentage of the total protein.
2. (b) How are protein purification procedures monitored? Take an enzyme purification
procedure as an example and describe the typical features of a purification table and how
each feature might be measured
A typical purification table would include the following properties:
Volume; protein concentration (mg/ml)
Total protein (vol x [protein]); biological activity (e.g. enzyme activity) in
units/ml; total biological activity (volume x e.u./ml)
Specific activity (e.u./mg protein)
Measurement required = biological activity, e.g. by enzyme assay, ELISA
Measurement required = protein assay (e.g. by 280/260 nm measurement, dye
assay (Bradford), Lowry, Biuret
3. (a) Draw the structure of α-D-fructofuranose in the Haworth projection. Indicate
which is the anomeric carbon atom.
3. (b) Chitin is a homopolymer of N-acetyl glucosamine residues. Knowing the physical
properties of chitin and that it serves as the exoskeleton of insects, what would you
predict about the nature of the linkage between the N-acetyl glucosamine units? Justify
Because the chitin exoskeleton of insects has hard, brittle properties one would
expect the structure to be similar to cellulose. The β-linkage of cellulose allows for the
formation of relatively linear polysaccharide chains which can H-bond both intra-chain
and between chains forming a closely layered insoluble polymer. This chitin is expected
to have the same kind of beta-(1-4) linkage (see below).
4. (a) Myoglobin and the subunits of hemoglobin are polypeptides of similar size and
structure. Compare the expected ratio of nonpolar to polar amino acid residues in
myoglobin and in hemoglobin.
Hemoglobin subunits have a high proportion of nonpolar residues in the contact
regions between subunits. Analogous regions of myoglobin are exposed to solvent and
are therefore largely polar. Consequently, hemoglobin has a higher ratio of nonpolar to
polar residues than does myoglobin.
4. (b) When oxygen binds to an iron residue at the center of one of the four heme groups
of hemoglobin, the iron atom moves approximately 0.6 angstroms, coming to rest in the
plane of the porphyrin ring. Explain how this small change can trigger conformational
changes in the polypeptide chain of this particular subunit and further result in substantial
movements of the other subunits with respect to each other. What effects do these
changes have with respect to the binding of other oxygen molecules to this hemoglobin?
When oxygen binds to the iron atom in one of the hemes, not only does the iron
atom move, but so too does the proximal histidine which is coordinated to the Fe atom on
the opposite side to which oxygen binds. The movement of this histidine residue in turn
triggers a movement of the alpha helix of which the histidine is a part (F8 helix). This
movement is transmitted through the polypeptide chain of this globin and eventually
results in the breaking of one set of interchain non-covalent linkages (H-bonds and salt
bridges) and the formation of a new set of intersubunit interactions. This change3 in
subunit interactions initiates a change in conformation of the non-oxygenated subunits.
The net result of these conformational changes is to greatly increase the affinity for
oxygen of these hemoglobin subunits.
4. (c) It only takes 1 mole of 2,3-bisphosphoglycerate (BPG) per mole of hemoglobin to
alter the oxygen binding properties of the hemoglobin substantially. Describe how this
one mole of BPG can affect all 4 subunits of hemoglobin at the same time. Does the
presence of BPG increase or decrease affinity for oxygen?
BPG binds in the central cavity between the 4 globin chains of deoxyhemoglobin
and can thus affect all 4 chains at once. The multiple negative charges on BPG interact
with positively charged groups on the globin chains (N-termini and lysine and histidine
side chains) and thereby stabilizes the deoxy (T) state. This in turn decreases the affinity
of the hemoglobin for oxygen. The central cavity of the (R ) state is too narrow to contain
BPG. (see V&V p. 234)
5. (a) Treatment of a polypeptide by dithiothreitol yields two polypeptides that have the
following amino acid sequences:
Chymotrypsin-catalyzed hydrolysis of the intact polypeptide yields polypeptide
fragments with the following amino acid compositions:
3. (Ala, Phe)
4 (Asn, Cys2 Met, Tyr)
5. (Cys, Gly, Lys)
6. (Cys2, Leu, Trp2, Val)
7. (Ile, Phe, Val)
Chymotrypsin cleaves on the C-terminal side of aromatic residues. Indicate the
positions of the disulfide bonds in the original polypeptide. Describe your reasoning.
5. (b) A new protein of unknown structure has been purified. Gel filtration
chromatography reveals that the native protein has a molecular weight of 240,000.
Chromatography in the presence of 6 M guanidine hydrochloride yields only a peak for a
protein of Mr 60,000. Chromatography in the presence of 6 M guanidine hydrochloride
and 10 mM β-mercaptoethanol yields peaks for proteins of Mr 34,000 and 26,000.
Explain what can be determined about the structure of this protein from these data.
Guanidine hydrochloride is a powerful denaturing agent that disrupts tertiary and
quaternary structure by disrupting hydrogen bonds. In the presence of 6M guanidine-HCl,
only a 60kD species is observed, indicating that the native protein may in fact be a
tetramer of four 60 kD subunits. Upon denaturation in the presence of β-mercaptoethanol
( a disulfide reducing agent), two protein peaks are observed, one at 34 kD and one at 26
kD. This indicates that the 60 kD subunit is a heterodimer of two chains, 34 kD and 26
kD, held together by a disulfide bond.
6. (a) The dielectric constant of water is very high compared with liquid hydrocarbons.
Describe the role that this high dielectric constant plays in the solubility of ionic solids.
6. (b) Changes in the dielectric constant of the surroundings can also affect the ionization
of the side chains of amino acids. Consider the side chains of the amino acid histidine.
Will the pKa of the side chain be increased or decreased if the side chain moves from a
region of higher to lower dielectric constant? Explain why.
In a region of lower dielectric constant than water there will be a reduction in the
ability of a proton and the ionized side chain to remain apart. This is because they will
lack the insulating effect of having a layer of water molecules surrounding them. With
less tendency to ionize the side chains will have an effectively higher pKa.