Biology 212: Cell Biology
In-class quiz #2
You will have 50 minutes to complete this 120-point quiz. The point value of each question is shown in brackets.
Good luck!
In signing your name below, you agree to complete this exam without aiding or receiving aid from any of your
classmates and without consulting your notes, textbook, etc.
______________Greg Crowther [revised key]_________
1. Use the attached excerpts from Triglia et al. (1997) to help you answer the following questions. [5 points
apiece]
A. Fig. 3A and 3B look quite similar. What information does Fig. 3B provide (regarding the proteins on the gel)
that Fig. 3A does not?
Fig. 3A used Coomassie blue, which stains all proteins, whereas Fig. 3B used an antibody that specifically binds to
the enzyme dihydropteroate synthase (DHPS). Therefore, while Fig. 3A simply shows that only one band of protein
was left after the purification process, Fig. 3B shows that that protein was in fact DHPS.
B. In two well-crafted sentences, explain what an inhibition constant (Ki) is.
An inhibition constant indicates the affinity with which the enzyme binds to a particular inhibitor (just as the K m
reflects an enzyme's affinity for its substrate). A high Ki indicates a low affinity, whereas a low Ki indicates a high
affinity.
C. What effect does changing dihydropteroate synthase (DHPS)'s 437th amino acid from alanine (abbreviated A) to
glycine (abbreviated G) have on the Ki of this enzyme for sulfadoxine? Briefly justify your answer by citing
specific evidence from the figures and/or tables.
Changing amino acid #437 from A to G raises the Ki about 10-fold. Table 1 shows that DHPS from isolates D10
and 3D7 are identical except that D10 has an A at position 437 and 3D7 has a G there; therefore any differences in
Ki between the two enzymes must be due to that one amino acid change. Table 2 shows that the D10 version of the
enzyme has a Ki of 0.14 μM for the D10 version of the enzyme and 1.39 μM for the 3D7 version of the enzyme.
D. Let's say you're studying the "Tak9/96-C" version of DHPS. Assume that the initial concentration of paminobenzoic acid (a substrate of DHPS) is 9 mM and that there is an excess of 6-hydroxomethyldihydropteroate
(DHPS's other substrate). Further assume there is no inhibitor or product present. What is the approximate initial
rate (v) of the reaction catalyzed by DHPS? Briefly explain your reasoning and/or calculations.
The Km of the Tak9/96 enzyme is 88.9 nM; in other words, the reaction rate will be half-maximal when the substrate
concentration is 88.9 nM. 9 mM is 5 orders of magnitude above this Km,; in terms of the Michaelis-Menten curve
(Fig. 3.16), the enzyme will be working way up on the flat part of the curve at V max because there's so much
substrate around. Therefore the initial reaction rate will be the V max, which for this enzyme is 33.0 mU/mg protein
according to Table 2. [This question is based in part on review sheet question #36.]
E. In continuing your study of the Tak9/96-C DHPS, you do an experiment in which you make a Michaelis-Menten
curve for the enzyme (i) without any inhibitors present and (ii) in the presence of both sulfadoxine and a new drug
called doxasulfine (i.e., both are present at the same time). Your data look like this (see next page):
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Biology 212: Cell Biology
In-class quiz #2
initial
reaction
rate
(v)
(i) no inhibitors
(ii) sulfadoxine + doxasulfine
[substrate]
Based on this graph and what you know about sulfadoxine, is doxasulfine a competitive inhibitor or a
noncompetitive inhibitor of DHPS? Briefly explain your reasoning.
You know from reading the article that sulfadoxine is a competitive inhibitor. Competitive inhibitors raise the Km
but don't affect the Vmax, whereas noncompetitive inhibitors lower Vmax but don't affect Km. In the above graph, both
the Km and Vmax have been altered. Since the lowered Vmax couldn't be due to the sulfadoxine, it must be due to the
other inhibitor (doxasufine). Therefore doxasulfine must be a noncompetitive inhibitor. [This question is based in
part on review sheet question #38.]
2. Choose the single best answer to each question. [2.5 points apiece]
A. According to Chang & Lee (the authors of the Great Moment paper presented by Jasmine, Kelli, Mel, and
Monica), snake venom inhibits neuromuscular function by
i. breaking down acetylcholine in the synapse
ii. inhibiting acetylcholine release from presynaptic neurons
iii. blocking acetylcholine's access to its receptors on muscle cells
iv. both (i) and (ii)
v. both (ii) and (iii) -- CORRECT
B. To find out whether the CFTR protein really is an ion channel, Bear et al. (the authors of the Great Moment paper
presented by Ben, Evan, Jesse, and Zack)
i. mutated the gene for CFTR and tested whether cells had trouble depolarizing
ii. made a hydrophobicity plot for the protein
iii. purified the protein and observed its behavior when placed in phospholipid vesicles -- CORRECT
iv. performed gel electrophoresis to see whether its subunits were the right size to be ion channel
components
v. repeatedly asked it, "Are you an ion channel?" until it confessed
C. In the mouse cell-human cell fusion experiments of Frye & Edidin (the authors of the Great Moment paper
presented by Kerri, Kevin, Louis, and Stephanie), the only thing that slowed proteins' movement from one part of
the cell membrane to another was
i. inhibition of protein synthesis
ii. inhibition of ATP synthesis
iii. inhibition of glutamine-dependent pathways
iv. addition of saturated lipids into membrane
v. lowered temperature -- CORRECT
D. The main question explored by Quistorff et al. (the authors of the Great Moment paper presented by Erin, Jake,
Maggie, and Victoria) in their study of exercising calf muscle was
i. Does glycolysis continue after exercise if ADP and P i remain high? -- CORRECT
ii. Does [ADP] change during exercise?
iii. Do muscles fatigue more quickly when blood flow to them is cut off?
iv. Do muscles consume ATP during exercise?
v. Can PCr be detected by NMR spectroscopy?
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Biology 212: Cell Biology
In-class quiz #2
E. Samples being prepared for scanning electron microscopy (SEM) are subjected to all of the following steps
except [based on review sheet question #24]
i. dissected free of other tissue
ii. placed in fixing solution
iii. dehydrated
iv. sliced into thin slices -- CORRECT
v. coated with a thin layer of metal
F. The amount of cellular space devoted to intracellular organelles (nuclei, mitochondria, ER, etc.) is lowest in
which of the following cells? [based on review sheet question #26]
i. skeletal muscle cells
ii. pancreatic cells
iii. red blood cells -- CORRECT-iv. tracheal cells
v. cells of the small intestine
G. Recall that ΔG = ΔG˚' + 2.303RTlog10([B]/[A]). Under which conditions would the reaction citrate  isocitrate
have the highest (most positive) ΔG? [based on review sheet question #32]
i. [isocitrate] = 100 mM, [citrate] = 1 mM -- CORRECT
ii. [isocitrate] = 10 mM, [citrate] = 1 mM
iii. [isocitrate] = 1 mM, [citrate] = 1 mM
iv. [isocitrate] = 0.1 mM, [citrate] = 1 mM
v. [isocitrate] = 1 M, [citrate] = 1 M
H. Which of the following is a correctly written formula? [based on review sheet questions #30 and 36]
i. ΔG = ΔH - ΔS
ii. ΔH = TΔS + ΔG -- CORRECT
iii. v = Vmax * (Km/([S] + Km))
iv. v = Vmax / ([S] + Km)
v. Km = Vmax/2
Note that (v) is incorrect because the Km is not Vmax/2; the Km is the substrate concentration at which the reaction
rate is Vmax/2.
I. Glycolysis differs from the Krebs cycle in all of the following ways except [based on review sheet questions #5255]
i. the Krebs cycle produces GTP and glycolysis produces ATP
ii. the Krebs cycle produces FADH2 and glycolysis does not
iii. the Krebs cycle makes use of Coenzyme A and glycolysis does not
iv. the Krebs cycle includes 6-carbon (C6)intermediates and glycolysis does not -- CORRECT
v. the Krebs cycle takes place in the mitochondria and glycolysis does not
K. A competitive inhibitor [based on review sheet questions #36 and 38]
i. binds to an enzyme's active site -- CORRECT
ii. binds to an enzyme at a location other than the active site
iii. binds to the substrate to prevent it from binding to the enzyme
iv. lowers the activation energy of a reaction
v. acts allosterically on an enzyme
L. To alter membrane fluidity, cells can adjust the number of double bonds in their [based on review sheet question
#41]
i. cholesterol molecules
ii. phospholipids -- CORRECT
iii. integral proteins
iv. free fatty acids
v. lipid-bound proteins
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Biology 212: Cell Biology
In-class quiz #2
M. A person's blood type -- O, A, B, or AB -- is determined by [based on review sheet questions #44-45]
i. glycolipids in the membranes of red blood cells (RBCs) -- CORRECT
ii. the presence or absence of Band 3 in the membranes of RBCs
iii. the presence or absence of glycophorin A in the membranes of RBCs
iv. the ratio of saturated lipids to unsaturated lipids in the membranes of RBCc
v. the person's final grade in Cell Biology
N. The "threshold" of a neuron is [based on review sheet question #48]
i. the amount of calcium that must enter the axon terminal in order to trigger release of synaptic vesicles
ii.the membrane potential at which voltage-gated K+ channels open
iii. the neurotransmitter concentration at which ligand-gated Na+ channels open
iv. the membrane potential at which voltage-gated Cl- channels open
v. usually around -55 mV -- CORRECT
(Note that I wouldn't expect you to remember that choice (v) is true, but you should have been able to rule out the
other 4 choices.)
O. Why does facilitated diffusion of glucose into muscle cells increase when blood insulin levels are high? [based
on review sheet question #50]
i. high insulin levels increase the synthesis of new glucose transporters
ii. high insulin levels phosphorylate the glucose transporters, making them more active
iii. high insulin levels cause transporter-carrying vesicles to fuse with the cell membrane -- CORRECT
iv. high insulin levels increase active transport of Na+ and K+, which then powers glucose uptake
v. high insulin levels make the cell membrane less hydrophobic
P. In the glycolytic pathway, 1 mole of NADH is produced for every ____ mole(s) of glucose broken down.
[based on review sheet question #52]
i. 0
ii. 0.5 -- CORRECT
iii. 1
iv. 2
v. 4
Q. The main reason yeast makes ethanol from pyruvate under anaerobic conditions is to [based on review sheet
question #54]
i. synthesize new glucose in times of starvation
ii. produce a 2-carbon molecule that can enter the Krebs cycle
iii. produce NADH for the electron transport chain
iv. regenerate the NAD+ needed for earlier step of the glycolytic pathway -- CORRECT
v. maintain an acidic cellular environment
R. Hydrogen bonds can form between [based on review sheet question #17 and Quiz 1 questions 6A and 6C]
i. O atoms and C atoms
ii. N atoms and C atoms
iii. N atoms and H atoms -- CORRECT
iv. H atoms and H atoms
v. H atoms and their dads
S. Which of the following lists contains only amino acids with nonpolar R groups? [based on review sheet question
#16 and Quiz 1 question 8]
i. valine, lysine, tyrosine
ii. methionine, proline, threonine
iii. asparagine, isoleucine, valine
iv. leucine, alanine, tryptophan -- CORRECT
v. phenylalanine, alanine, glutamine
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Biology 212: Cell Biology
In-class quiz #2
T. Noncovalent bonds and interactions are important determinants of most proteins' __________ structure. [based
on review sheet question #17 and Quiz 1 question 1F]
i. quaternary
ii. tertiary and quaternary -- OK
iii. secondary, tertiary, and quaternary -- CORRECT (hydrogen bonds are noncovalent bonds)
iv. secondary and quaternary
v. primary, secondary, tertiary, and quaternary
U. Some neurons can be a meter or more in length. This is an exception to the general rule that [based on review
sheet question #9]
i. cells must be small because diffusion is slow over long distances -- CORRECT
ii. cells must be small in order to maintain a low surface area to volume ratio
iii. neurons must be short so that the signal can make it from one end of the axon to the other
iv. neurons must be short so that their neurotransmitters don't have as far to travel at synapses
iii. neurons must be short so that they can be insulated by myelin
3. True/false questions for hip young students…. State whether each of the following sentences is ROCKIN'
(true) or BOGUS (false). If it is bogus, explain in one well-crafted sentence why it is bogus. [3 points apiece]
A. Under normal cellular conditions, a chemical reaction that has a ΔG > 0 can proceed if it is coupled to the
reaction ADP + Pi  ATP.
Bogus! A chemical reaction that has a ΔG > 0 can proceed if it is coupled to the reverse reaction (ATP  ADP +
Pi). ATP synthesis does not proceed spontaneously, but ATP hydrolysis does. [This question is based on review
sheet questions #32-33.]
B. The fatty acid cycle is a typical anabolic pathway in that it breaks complex organic compounds into smaller ones,
reduces NAD+ and FAD, and leads to the production of ATP.
Bogus! The fatty acid cycle is a typical catabolic pathway for all of these reasons. [This question is based on
review sheet questions #39 and 55.]
C. If a particular enzyme becomes less active when one of its tyrosine residues becomes phosphorylated, it is an
example of enzyme activity being regulated via covalent modification.
Rockin'! (See p. 113 of your text.) [This question is based on review sheet question #40.]
D. You suspect that amino acids #55 to 77 of a newly discovered polypeptide is one subunit of an ion channel.
Therefore you predict that this portion of the polypeptide will have the 3D structure of an α-helix with hydrophobic
residues facing the lipid bilayer and hydrophilic residues facing the pore of the presumed channel.
Rockin'! (See Fig. 4.20 in your text.) [This question is based in part on review sheet question #42.]
E. Na+ ions don't pass through bacterial KscA channels to any great extent because they're too small to
simultaneously interact with all of the electromagnetic O atoms lining the narrow part of the channels.
Rockin'! (See p. 157 of your text.) [This question is based on review sheet question #49.]
F. In the graph below of free energy versus reaction progress, the solid line and dotted line could represent the same
chemical reaction with and without the benefit of an enzyme catalyst, respectively.
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Biology 212: Cell Biology
In-class quiz #2
Bogus! The free energy of the starting materials and products differ for the two lines. Enzymes cannot change the
free energy of substrates or products; they can only change the activation energy (difference in free energy between
the substrate and the high-energy intermediate). [This question is based on review sheet questions #34-35.]
free
energy
(G)
reaction progress 
4. Miscellaneous short-answer questions. [4 points apiece]
A. Briefly describe the structure of a nucleotide. How does it differ from a nucleic acid?
A nucleotide consists of a phosphate group, a 5-carbon sugar (ribose or deoxyribose), and a nitrogen-containing
base (adenine, cytosine, guanine, thymine, or uracil). Nucleotides are the monomers from which nucleic acid
polymers are made. [This question is based on review sheet question #28.]
B. Give an example of a membrane with a low protein-to-lipid ratio and an example of a membrane with a high
protein-to-lipid ratio. Briefly explain why these membranes have such different ratios.
Myelin has a low protein-to-lipid ratio because it electrically insulates axons, which doesn't require proteins.
Red blood cells (RBCs) have a high protein-to-lipid ratio because the proteins help maintain the integrity of the
membrane, which is deformed and scraped by the capillaries through which the RBCs pass. Inner mitochondrial
membranes have a high protein-to-lipid ratio because they are packed with the respiratory enzymes of the electron
transport chain. Intestinal villi have a high protein-to-lipid ratio because they have numerous protein transporters
for moving nutrients out of the intestine and, ultimately, into the blood. (Any of these 3 examples is fine.)
[This question is based on review sheet question #43.]
C. Na+ ions can pass from one side of a neuron's membrane to the other via three different membrane proteins.
What are they, and in which direction (inside to outside or outside to inside) does the Na+ go in each case?
i. Voltage-gated Na+ channels -- Na+ flows down its electrochemical gradient into the cell.
ii. Ligand-gated Na+ channels -- Na+ again flows down its electrochemical gradient into the cell.
iii. Na+/K+ pump -- Na+ is pumped against its electrochemical gradient out of the cell.
[This question is based on review sheet question #48.]
D. What are van der Waals forces?
Van der Waals forces are weak attractive forces between atoms or molecules that are near each other but not
covalently bonded to each other. The attraction occurs between a transient partial positive charge on one atom or
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Biology 212: Cell Biology
In-class quiz #2
molecule and a transient partial negative charge on another atom or molecule. [This question is based on review
sheet question #12.]
E. If 1 molecule of stearic acid (an 18-carbon fatty acid) was completely oxidized to CO2, how many molecules of
FADH2 would be produced? Show how you came up with the number that you got. Feel free to refer to the
diagram on the next page.
The fatty acid cycle produces 9 molecules of acetyl CoA from an 18-carbon fatty acid. For each acetyl CoA
produced, 1 FADH2 is produced (except for the 9th and final acetyl CoA, but we didn't talk about that), so that's 9
(or 8) FADH2 so far. Then each acetyl CoA goes into the Krebs cycle and gets oxidized to CO 2. For each of the 9
acetyl CoA molecules that go in, 1 FADH2 is produced, so that's 9 more FADH2 for a total of 18 (or 17). [This
question is based on review sheet question #55.]
F. On the diagram above, circle the names of all amino acids from which net production of glucose can occur.
All should be circled except the Phenylalanine/Tyrosine/Leucine/Lysine/Tryptophan group and the
Leucine/Isoleucine/Tryptophan group. [This question is based in part on review sheet question #56.]
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