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Biochemistry 2000, Fall 2009
Midterm 1 – 75 min
1 of 7
Student Name : _________________________
Student ID : _________________________
2008-10-9
Instructions:
Write neatly and clearly. Cross out with a single line any material you do not wish to have
marked. Marks will be deducted for incorrect statements in certain cases. Students must
work independently and may not knowingly utilize resource materials or share resource
materials with other students. Students may use pens, pencils, erasers and calculators only.
Electronic devices including cell phones, personal information managers and audio devices
are prohibited.
Question
Mark
Total
Marks
1
4
2
10
3
3
4
10
5
11.5
6
3
7
6
8
4
9
10
10
2
Total
63.5
Biochemistry 2000, Fall 2009
(1)
Midterm 1 – 75 min
2 of 7
In the space provided, give unique definitions or descriptions of the following
biochemical terms and phrases: (4 marks)
(a) Hydrophobic effect – tendency of nonpolar compounds to minimize their contact
with water
(b) Tertiary structure – spatial arrangement of secondary structures and side chains in
a polypeptide
(c) Domain – structural independent folding unit with properties of small globular
protein
(d) Protein denaturation – unfolding of a protein’s secondary, tertiary and quaternary
structure induced by external factors (denaturants) (such as pH, detergents, organics,
chaotropic salts, heat)
(2)
Draw the following structures: (10 marks)
(a) A Trp-Pro dipeptide with a cis peptide bond.
(1 mark for Trp, 1 mark for Pro, 2 marks for cis peptide bond – 4 in total)
O
+
N
H3N
O
O
HN
(b) A parallel β-sheet composed of two β-strands, each of four residues in length.
Represent each of the side chains with the generic symbol R. Indicate all H-bonds
between main chain atoms.
1 mark for parallel
1 mark for 2 strands
1 mark for 4 residues
1 mark for backbone
2 mark for all H-bonds
(6 in total)
Biochemistry 2000, Fall 2009
Midterm 1 – 75 min
3 of 7
(3)
The following is a fragment of a primary sequence alignment of the protein Cbf5 from
several organisms. Classify each of the primary sequence positions as I (invariant), C
(conservative substitutions) or NC (non-conservative substitutions). (3 marks)
(4)
Ile
Ile
Ile
Ile
Glu
Asp
Glu
Glu
Leu
Val
Val
Ile
I
C
C
Tyr
Gly
Gly
Asp
Asp
Thr
Lys
Gln
NC
NC
Glu
Glu
Glu
Glu
0.5 mark for each
I
Bacterial cells synthesize a protein called porin (MW 31.5 kDa) which is located in the
hydrophobic outer cell membrane. Structure determination of this protein revealed that
it consist of 16-stranded antiparallel β-barrel. Each β-strand transverses the outer
membrane (thickness 45 Å). Porins act as channels to allow flow of water and small
solutes through the inside of the β-barrel thus crossing the outer cell membrane. (10
marks)
(a) Calculate the minimum number of amino acid residues necessary for one β-strand
to transverse the membrane completely.
7 Å corresponds to two residues, i.e. 45 Å / 7 Å = ~6.5 x 2 residues = 13
(2 marks: one for 7 Å per 2 residues, 1 for answer)
(b) Estimate the percentage of the porin protein that is involved in membrane-spanning
β-strands. Use an average amino acid residue weight of 110 Da.
Amino acids spanning membrane: 16 x 13 = 208
Molecular weight of 208 residues: 208 x 110 Da = 22880 Da = 22.88 kDa
Percentage of porin in membrane spanning b-strands:
22.88 kDa / 31.500 kDa * 100% = 72.6 %
(2 marks)
(d) Draw a 2D topology diagram of porin. (2 marks)
(c) Which class of amino acids do you expect on the outside of the β-barrel and which
do you expect on the inside? Why?
Outside: nonpolar residues as these contact the hydrophobic membrane
Inside: polar charged and polar uncharged residues as the inner part of the porin forms
a channel for water and must be hydrophilic
(4 marks: 1 for each answer, 1 for each explanation)
Biochemistry 2000, Fall 2009
(5)
Midterm 1 – 75 min
4 of 7
Provide short answers or fill in the blanks for the following questions: (11 marks)
(a) What is the net charge of the Tyr-Glu-Met-Arg tetrapeptide at pH 10? Shortly justify
your answer.
α-amine (9.4)
charge = 0
Tyr R-group(10.1)
charge = 0 / -0.5 (both OK)
Glu R-group(4.2)
charge = -1
0.5 marks for each correct
Arg R-group (12.5)
charge = +1
partial charge
α-carboxyl (2.2)
charge = -1
Net charge = -1 to -1.5
(b) An α-helix contains _3.6__ residues per turn and has a pitch of __5.4 Å_____. In
the α-helix, the N-H of the backbone from residue 7 forms hydrogen bond with the
_backbone C=O____ of the residue number ___3____. (1 mark each – 4 total)
(c) What type of electrostatic interaction most likely occurs between the side chains of
the following amino acid pairs?
(i) Lys and Tyr (pH 9.0)
(ii) Glu and His (pH 5.0)
(iii) Gln and Asn (pH 7.0)
(iv) Asp and Arg (pH 6.0)
(v) Met and Phe (pH 4.0)
(6)
dipole-ion
(1 mark each – 5 total)
salt bridge
dipole-dipole (or hydrogen bond)
salt bridge
van der Waal’s
The following is the titration of an unknown tripeptide with strong acid. What can you
conclude about the identity of residues that comprise the tripeptide? (3 marks)
12
10
8
pH
6
4
2
0.5
1.5
2.5
3.5
4.5
H+ equivalents added
2 Arg:
Arg pKa = 12.5, two titratable groups (x axis 0.5 and 1.5) )
1 His
His pKa = 6.0, one titratable group (x axis 3.5)
(amino terminus, pKa = 9.4, x axis 2.5)
(carboxy terminus, pKa = 2.2, x axis 4.5)
2 marks
1 mark
Biochemistry 2000, Fall 2009
(7)
5 of 7
Draw a Ramachandran plot, label the axis and the regions of secondary structure.
Define both angles. (6 marks)
(1 for each axis – labels and numbers/ -2 if axis are switched, 1 for each region (β/α), 2
for definitions of phi & psi)
Phi (Φ)
Psi (Ψ)
(8)
Midterm 1 – 75 min
C N – Cα C
N Cα – C N
Match the following folds to the structures (4 marks)
(A) α + β domain
(I)
(A) – (I)
(B) – (IV)
(C) – (II)
(D) – (III)
(B) α/β domain
(II)
(C) β domain
(III)
(D) α domain
(IV)
Biochemistry 2000, Fall 2009
6 of 7
You have a mixture of 3 proteins (see table below) that have been subjected to the
following experiments: (10 marks)
Protein Name
A
B
C
Quaternary Structure
Heterotrimer
Homodimer
Monomer
Molecular Mass (kDa)
2 x 20, 1 x 42
56 (total)
34
pI
6
7
8
Sample
(a) Show the result of an SDS-PAGE experiment on the
picture provided to you on the right. Identify the protein
that corresponds to each band.
Molecular
Weight
Marker
(9)
Midterm 1 – 75 min
(4 marks, one for each band, -1 for each wrong band)
70 kD
50 kD
A (42 kDa)
40 kD
C (34 kDa)
30 kD
B (2x 28 kDa)
20 kD
A (2x 20 kDa)
(b) Draw a reasonable chromatogram for a cation exchange chromatography at pH 7.0.
Assume the sample was loaded onto the column in a low salt solution and bound
protein was eluted using a high salt solution. For each peak in the chromatogram,
identify the corresponding protein.
•
•
Protein A (pI < pH, i.e. more
deprotonated): negatively
charged, not binding
Protein B (pI = pH): no charge,
not binding, elutes together
with A in low salt
Protein C (pI > pH, i.e. more
protonated): positively carged,
does bind, elutes only with
high salt
High-salt gradient
1.00
0.75
A 280
•
0.50
A+ B
C
0.25
20
40
60
80
Volume, mL
(3 marks, 1 for each peak, 1 for labels
Marks deducted for wrong peaks: 3 peaks (two in high-salt) – 1 mark,
3 peaks (two in low-salt) – 2 marks)
100
Biochemistry 2000, Fall 2009
Midterm 1 – 75 min
7 of 7
(c) Draw a reasonable chromatogram for a size exclusion chromatography experiment.
For each peak, identify the corresponding protein and its size.
1.00
82 kD
56 kD
34 kD
0.75
0.50
0.25
20
40
60
80
100
Volume, mL
(3 marks for each peak and its identity, 0 marks if order wrong or subunits)
The unfolding of the α helix of a polypeptide to a
randomly coiled conformation is accompanied by a large
Poly(Lys)
Poly(Glu)
decrease in a property called its specific rotation, a
measure of a solution’s capacity to rotate planepolarized light. Polyglutamate, a polypeptide made up of
only L-Glu residues, has the α-helical conformation at
pH 3. When the pH is raised to 7, there is a large
decrease in the specific rotation of the solution. Similarly,
polylysine (L-Lys residues) is an α helix at pH 10, but
0 2 4 6 8 10 12 14
when the pH is lowered to 7 the specific rotation also
pH
decreases, as shown by the right graph.
What is the explanation for the effect of the pH changes on the conformations of
poly(Glu) and poly(Lys)? Why does the transition occur over such a narrow range of
pH? (2 marks)
At pH > 6, the carboxyl groups of poly(Glu) are fully deprotonated; repulsion among
negatively charged carboxylate groups leads to unfolding of the α helix. However, at
pH < 4.2 (pKa of Glu), the side chains are protonated and thus uncharged. Thus
poly(Glu) can form a helix with the side chains sticking out into solution and interacting
with the surrounding water. Similarly, at pH < 9, the amino groups of poly(Lys) are fully
protonated; repulsion among these positively charged groups also leads to unfolding of
the α helix. Above pH 10.5 (pKa of Lys), the side chains are uncharged and a helix can
form. The transition occurs over a narrow range because deprotonation of only a few of
the Glutamates or protonation of only a few lysines creates charges of the same type
within the helix. This small number of charges causes large repulsion and thus rapid
unfolding of the α helix. (2 marks – 0.5 for change in ionization state, 1 for repulsion
of equally charged residues, 0.5 for rapid transition caused by only a few charges)
Specific rotation
(10)