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Biology/Chemistry 302-01 Biochemistry
Fall 2006
Exam 1
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This exam consists of 5 questions on 5 pages including this one. Please check to
make sure you have a complete copy.
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Please write clearly. If I can’t read your answer, I can’t give you credit.
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Please show all work. Please indicate any values you use that are not given (for
example, pKa’s).
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Remember to include units and significant figures where appropriate.
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Please ask if you are unsure of what is being asked in any question. I will happily
provide any information that I believe I can fairly offer without comprising the
integrity of the exam.
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You may use a calculator.
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Please do not share answers or calculators.
1. (15 points)
a. Suggest at least two ways in which proline is unique among amino acids.
Proline’s ring structure constrains its phi angles severely compared to other amino
acids.
Proline’s secondary amine results in no H-bonding capability of its amide bond in
peptides.
b. In what type(s) of secondary structure(s) would proline be most likely found? Why?
Proline’s lack of hydrogen bonding and constrained angles make it both a helix disrupter
and a β-strand disrupter, so it is most often found in loops and turns.
c. Myohemerythrin is an Fe-binding protein. The figure below shows the iron bound in
the interior of the protein. What would be the effect of a mutation that placed a proline at
point A in the figure?
A
A mutation that places a proline at position A would disrupt the helical nature of the
peptide region, unfolding the helix. Without the helical structure to appropriately place
the sidechains, the iron-binding capabilities would likely be disrupted.
2. (5 points)
Turns and loops often participate in interactions between two proteins. Why?
Turns and loops connect the strands of α-helices or β-sheets, which fold into compact
structures. Turns and loops then necessarily lie on the surface of proteins where they are
available to interact with other proteins.
3. (15 points)
a. Draw the structure of the peptide indicated below at pH 0.
READY
pKa 10
H2N
pKa 12.5
NH2+
OH
NH
pKa 9
O
O
H
N
H
N
+H3N
OH
N
H
O
N
H
O
O
OH
O
OH
pKa 4
pKa 4
b. What is the average charge on the peptide at pH 4?
(N-end) + (R) + (E) + (D) + (Y) + (C-end) = avg. charge
(+1) + (+1) + (-0.5) + (-0.5) + (0) + (-1) = 0
O
pKa 2
4. (10 points)
a. In the model that follows of phospholipase A, label at least two instances of each type
of secondary structure and label any elements of tertiary or quaternary structure you find.
turns
β-sheet
β-sheet
loop
domains
α-helix
α-helix
loop
b. If the structure of phospholipase A showed isolated acidic or basic residues in an
interior region of the protein, what would it suggest about that region?
Polar residues on the interior of the protein suggest that region is involved in the
function of the protein, for example in the enzymatic activity or substrate binding of the
protein.
5. (15 points)
a. Carbon dioxide is dissolved in blood (pH 7.4) to form a mixture of carbonic acid
(H2CO3, pKa 6.37) and bicarbonate (HCO3-, pKa 10.25). What fraction will be present as
carbonic acid? Would you expect a significant amount of carbonate (CO32-)?
pH = pKa + log [A-]
[HA]
7.4 = 6.37 + log [HCO3-]
[H2CO3]
10(1.03) = [HCO3-]
[H2CO3]
10.7 = [HCO3-]
[H2CO3]
10.7 [H2CO3] = [HCO3-]
[H2CO3] + [HCO3-] = 1
[H2CO3] + 10.7 [H2CO3] = 1
[H2CO3] = 0.085
8.5 % is H2CO3
We don’t expect any significant amount of CO32- because the pH is more than 2 pH units
below the pKa of bicarbonate.
b. Since it is difficult to measure how much CO2 you bubble into a solution, it is easier to
make a carbonic acid buffer by starting with sodium bicarbonate instead. How would
you make a 100 mM carbonate buffer at pH 7.4 starting with sodium bicarbonate and
solutions of HCl or NaOH.
At pH 7.4, a carbonic acid buffer will have 8.5% H2CO3 (from above), so if starting with
all sodium bicarbonate, we must convert some HCO3- to H2CO3 using the HCl solution.
To make the solution, measure 100 mmol of sodium bicarbonate, add 8.5 mmol HCl, and
dilute the solution to 1 L.