Biochemistry
MITES 2010
Name:
Problem Set 1 Solutions
1. (3) Name the three domains of life.
Bacteria, Archaea, Eukaryota.
2. (2) Does an archaea cell have a nucleus? No
3. (8) Find the functional groups in the following molecules:
O
O
HO!C !CH2CH2CH—C—OH
NH2
CH3
SH
CH2CH3
C
H
CH3CH2CH CH3
C
H
4.(7) Identify and label all of the functional groups on the following molecules.
MITES
Biochemistry
5. (2) What is the smallest R-group of "-amino acids? H
6. (4) Transcribe the DNA strand:
AATCGGA
TTAGCCT
7.
a.
AGTAGCC
TCATCGG
*BONUS* Which of the molecules has a higher mass?
(14) Sketch titration curves for the following acids. Be sure to label the axes and the pKa(s). At
the starting point and at each endpoint, identify the dominant species. At each midpoint, identify
the two species that have equal concentrations. The table on page 12 may be useful.
(5) Formic acid
1: x-axis (labeling and amounts)
1: y-axis (labeling and pKa(s))
3: species identification
b.
(9) Citric acid
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Biochemistry
1: x-axis (labeling and amounts)
1: y-axis (labeling and pKa(s))
3: species identification
8.(14) What will be the pH of buffers made by combining the following components:
a. (2) 0.7 moles HClO and 1.0 moles ClO! in 500 mL water
b. (2) 0.2 moles sodium citrate (NaC6H7O7) and 0.6 moles disodium citrate (Na2C6H6O7) in 1 L
water
c. (2) 0.6 moles sodium citrate (NaC6H7O7) and 0.6 moles disodium citrate (Na2C6H6O7) in 1 L
water
d. (4) 0.3 moles citric acid (C6H8O7) and 0.4 moles disodium citrate (Na2C6H6O7) in 1 L water
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Biochemistry
We can simplify the problem by
realizing that in order to reach
equilibrium, 0.3 moles of citric acid
(C6H8O7) and 0.3 moles of disodium
citrate (Na2C6H6O7) will be
consumed to form 0.6 moles of
sodium citrate (NaC6H7O7).
Combining this with the remaining
0.1 moles of disodium citrate
(Na2C6H6O7), we have a simple
C6H7O7# /C6H7O72# buffer.
e. (4) 0.4 moles citric acid (C6H8O7) and 0.4 moles trisodium citrate (Na3C6H5O7) in 1 L water
This is similar to part d. We can
simplify the problem by realizing that in
the process of equilibrating, some of
those protons on citric acid (C6H8O7)
will be transferred to trisodium citrate
(Na3C6H5O7). At equilibrium, we will
essentially have a simple
C6H7O7# /C6H7O72# buffer with equal
amounts of both components.
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MITES
Biochemistry
9.(3) Choose one of the following properties (melting point, boiling point, specific heat, or heat of
vaporization) and explain how hydrogen bonding leads to a high value of this property in water.
Hydrogen bonds between molecules must be overcome in order to either move water molecules
farther apart (as in a phase change) or make the molecules move faster (as in a temperature
change). This requires additional energy in the form of either higher temperatures or higher
heat input.
10. (10) Will the following pair of molecules interact by hydrogen bonding? If not, state why not. If
so, show how this interaction would occur.
OH
OH
CH3 CH CH3
f.
+
CH3 CH CH3
CH3 CH2 CH3
isopropanol
isopropanol
g.
Yes, hydrogen bonds will form as
shown below
CH3 CH CH3
H O
H O
CH3 CH CH3
O
h.
C
CH3 CH3
acetone
+
j.
propane
propane
OH
CH3 CH2 CH3
C
CH3 CH3
acetone
i.
O
CH3
OH
+
CH3 CH CH3
isopropanol
CH3 CH2 CH3
No, propane will not interact with
itself by hydrogen bonding because it
only contains nonpolar C-C and C-H
bonds, and because the
electronegativity of carbon is too low
to form hydrogen bonds.
O
No, isopropanol will not interact with
itself by hydrogen bonding because
the only hydrogen atoms are bound to
carbon, whose electronegativity is too
low to form hydrogen bonds.
C
CH3 CH3
acetone
+
O
H
O C
CH3
CH3 CH CH3
Yes, hydrogen bonds will form as shown above
5
propane
+
CH3 CH CH3
isopropanol
No, propane and isopropanol will not
interact by hydrogen bonding. While
isopropanol has a hydrogen atom
bound to a highly electronegative
atom, there is not another highly
electronegative atom in propane to
interact with the hydrogen atom.
MITES
Biochemistry
11. 1 mole glycine + 0.5 moles NaOH. This is a half-equivalent of NaOH. This means we’re at the first
midpoint on the titration curve. At this point, pH = pKa1=2.34
1 mole glycine + (0.5 + 1.0) moles NaOH. This means we’re at the second midpoint on the titration
curve. At this point, pH = pKa2=9.60
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Biochemistry
12.
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Biochemistry
13. (5) Identify each of the molecules below as hydrophilic, hydrophobic, or amphipathic
O
CH3(CH2)16
O
C O CH2
O P O CH2
O
CH3(CH2)8CH=CH(CH2)7
O
C O CH2
m.
phosphatidylserine
+
O
C
l.
O
amphipathic
squalene
hydrophobic
8
H
OH
OH
H
H
CH2 CH2 CH NH3
k.
CH2OH
O
OH
fructose 6-phosphate
hydrophilic
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Biochemistry
O
O
n.
O
C CH2 C C
O
O
oxaloacetate
hydrophilic
O
C OH
o.
arachidonic acid
amphipathic
14.
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