Download D-Glucose is a carbohydrate which can be classified as which of the

Name: ____Key______
Exam I
Bio200: Cellular Biology
Winter 2014
Multiple Choice Questions
Circle the one best answer for each of the following questions. (2 points each)
1. Phospatidylinositol is what type of molecule?
A. Amino acid
B. Coenzyme
C. Steroid
D. Phospholipid
E. Triglyceride
2. H2O2 and other ROS commonly occur when which enzyme operates incorrectly?
A. Pyruvate Dehydrogenase
B. ATP Synthase
C. Citrate Synthase
D. Phosphatase
E. Complex IV
3. The Cori Cycle does NOT which process or enzyme?
A. Gluconeogenesis
B. Lactate fermentation
C. Glycolysis
D. Oxidative phosphorylation
E. GluT
4. The coenzyme FAD/FADH2 becomes reduced in which process?
A. Transition step
B. Electron transport
C. Chemosmosis
D. Kreb’s cycle
E. None of these
5. The AMPK enzyme becomes active under what physiological conditions?
A. ATP is very scarce
B. ATP is abundant
C. Amino acids are very scarce
D. Amino acids are abundant
E. The plasma membrane is depolarized
6. The shape of animal cells is largely determined by a specialized portion of the cytoskeleton called
A. a lipid raft.
B. a membrane domain.
C. the glycocalyx.
D. the cell wall.
E. the cell cortex.
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7. Complex III contains an Fe3+ ion in it’s active site. That iron atom is likely acting as a
A. coenzyme.
B. prosthetic group.
C. Van der Waals stabilizer.
D. isozyme.
E. disulfide bond.
8. Flippases and Floppases allow the movement of
A. triglycerides across cell membranes.
B. phospholipids between membrane leaflets.
C. fatty acids across cell membranes.
D. steroids between organelles.
E. amino acids across the lysosome membrane.
9. Which of the following proteins could be electrogenic?
A. A Sucrose simple active transporter
B. A water facilitated transporter (aquaporin)
C. A K+ simple active transporter
D. A carbohydrate anaporter
E. A glucose facilitated transporter
10. A pure allosteric activator would be expected to have what effect upon an enzyme?
A. Lower the Vmax and not affect the Km
B. Raise the Vmax and not affect the Km
C. Raise the Vmax and lower the Km
D. Raise the Vmax and raise the Km
E. Lower the Vmax and raise the Km
11. The side chain for the amino acid Serine (Ser) is –CH2-OH. The side chain for the amino acid Lysine
(Lys) is –CH2CH2CH2CH2NH2. The pKa for carboxylic acids is 4 and the pKa for amines is 10. Draw
the chemical structure for N-Lys-Ser-C, including charges found at pH 12. (4 points)
H
O H
H O
│
║
│
│
║
H2N – C – C – N – C – C –O│
CH2
CH2
CH2
CH2
NH2
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│
CH2OH
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12. Shown below are the structures of D-Idose (a hexose) and D-Lyxose (a pentose).
____Idose____
_____Lyxose____
12A. On the lines below the structures, correctly label each sugar as either D-Idose or D-Lyxose.
(2 points)
12B. On the structure of D-Idose, circle the carbon atom whose chirality makes this D-Idose instead
of L-Idose. (1 point)
12C. On the structure of D-Lyxose, circle every carbon atom that is chiral as drawn. (1 point)
12D. Is D-Idose an aldose, ketose, both or neither? (circle one; 1 point)
12E. In the space below, draw the structure of the disaccharide D-Lyxose 16 D-Idose. (3 points)
O
12F. For the disaccharide in 12E, why is the D-Idose not designated as either  or ? (2 points)
Carbon #1 on the Idose is not involved in the glycosidic bond. Thus, it can still open into a linear
structur. There will be a mixture of and  forms.
13. Why do most enzymes have Km values that are considerably larger than the concentration of substrate
that they encounter in a living cell? (3 points)
When the substrate concentration is smaller than the Km value, the velocity changes dramatically with
small changes in substrate concentration.
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14. A prokaryotic cell has the solute concentrations shown below. In this situation, the cell’s membrane
potential is –80 mV; the GNa+,in,chemical = – 1.5 kcal/mol; the GNa+,in,electrical = – 2.0 kcal/mol.
Extracellular:
Cytosol:
Na+
165 mM
2 mM
K+
5 mM
140 mM
H+
10-6 M
10-7 M
NH4+
25 mM
2 mM
Cl165 mM
2 mM
SO42120 mM
25 mM
14A. What is the extracellular pH? ____6______ (1 point)
14B. This organism needs to expel Cl-. What is the G value for the outward movement of Cl-?
(3 points)
-0.5 kcal/mol
14C. The Cl- can be exported by a protein that powers this movement by cutting ATP. The G value
for ATP hydrolysis is -7.6 kcal/mol. Based on your answer to 14B, how many molecules of Cl- could
theoretically be expelled for each ATP that is used? (2 points)
15
14D. The Cl- can also be expelled by a NH4+/Cl- co-transporter. Is this protein a …
antiporter
symporter
simple transporter
potentiator
(circle one, 2 points)
14E. If we provide this prokaryotic cell with a large amount of extracellular glucose, we find that the
extracellular H+ concentration briefly increases to 10-5 M. Why? (3 points)
In aerobic respiration, protons are exported across the plasma membrane by the electron transport
chain.
14F. If these are the only solutes in this system, is the cell hypertonic, hypotonic or isotonic
compared to the surrounding fluid? (circle one; 2 points)
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15. Glycolysis is a spontaneous pathway, with a Go’ value of -17 kcal/mole and produces two
molecules of pyruvate from one molecule of glucose.
15A. How does a G value differ from a Go’ value? (2 points)
Go’ values are determined at standard conditions (1M concentrations) and at pH7.
15B. What can you conclude about the Keq value for glycolysis? (1 point)
_______>1___________
15C. Gluconeogenesis synthesizes a molecule of glucose using two molecules of pyruvate.
Considering that this is the reverse of glycolysis, how is it possible for gluconeogenesis to also be
a spontaneous process? (3 points)
Only seven of the ten steps are the same. Three steps in gluconeogenesis are different from glycolysis.
Those three steps don’t produce ATP in gluconeogenesis even though they consume ATP in glycolysis.
16. Identify the correct subcellular location for each of the following processes or structures in a eukaryote.
(2 points each)
Glycocalyx
Outer leaflet of the plasma membrane
-Oxidation
Mito Matrix
Cytochrome c
Hexokinase
Mito Inner Membrane
Cytosol
17. Why do eukaryotic cells have a slightly lower ATP yield for the aerobic metabolism of glucose than
prokaryotic cells? (3 points)
The electron shuttle is needed to get high energy electrons from glycolysis to the mitochondria. This
effectively swaps an NADH for a FADH2, which has a lower ATP yield.
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18. Name the pathway where each of the following enzymes is found. (1 point each)
Citrate Synthase
Kreb’s Cycle
Fructose-1,6-bisphosphatase
Gluconeogenesis
Lactate Dehydrogenase
Lactate Fermentation
Phosphofructokinase
Glycolysis
Pyruvate Dehydrogenase
Transition Step
Now choose any two enzymes from the above list and show the reaction that each catalyzes
below. Structures are not necessary but be sure to include all coenzymes. (2 points each)
Enzyme: Citrate Synthase
Reaction Catalyzed:
OAA + AcCoA
Citrate + CoA
Enzyme: Fructose-1,6-bisphosphatase
Reaction Catalyzed:
Fructose 1,6-bisphosphate
Fructose-6-phosphate and
one free phosphate
19A. In the electron transport chain, why don’t electrons move from ubiquinone to Complex IV? (2 points)
Ubiquinone does not have a binding site to interact with Complex IV.
19B. Why don’t electrons move from ubiquinone to Complex I? (2 points)
This transfer would be endergonic and there is no energy source to make this possible.
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20. Enolase is an enzyme that catalyzes the conversion of 2-PG  PEP, which is the ninth step of
glycolysis. When this reaction reaches equilibrium, we find that there is twice as much 2-PG as PEP.
20.
To determine the Km and Vmax of Enolase, we will need to make a series of measurements and
build a Michaelis-Menten curve. What should be plotted on each axis of this curve and what
would be reasonable units for each? (4 points total)
x-axis variable: ____[2-PG]_____
units: _____mM_____
y-axis variable: _____vi___________
units: ____nmol/sec____
20B. Based on your answer to 20A, what units will Km and Vmax have? (2 points)
Km ____ mM ____
Vmax ___ nmol/sec ___
20C. In making one of these measurements, we began with 30M 2-PG and 1g of Enolase. If we
allow this reaction to proceed for several hours, what concentration of PEP do you expect to find?
(2 points; include units!)
10M
20D. Tartronate semialdehyde phosphate (TSP) can bind to the active site of Enolase and inhibit the
enzyme. What effect do you predict this will have on the apparent Km and Vmax of the enzyme?
Please briefly explain your answer (2 points)
Since it binds the active site, it is clearly a competitive inhibitor. Thus, it will have no effect on Vmax
but the Km will increase.
20E. When Phosphofructokinase is more active, the rate of the 2-PG  PEP reaction in a living cell
increases dramatically. Describe two reasons why this rate increases (4 points)
Both enzymes are part of the same pathway. As PFK is more active, more 2-PG is available. This
increases the reaction rate in two ways. One, the enzyme velocity increases as we shift to the right on a
Michaelis-Menten plot. Two, we are farther away from equilibrium and thus the extra substrate “pushes”
the reaction to the right (LaChatlier’s Principle)
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21. For oxidative phosphorylation to work efficiently, the mitochondrial inner membrane must be at the
correct fluidity. This is achieved by having the appropriate ratio of saturated, unsaturated and
polyunsaturated fatty acids in the membrane phospholipids. (3 points each)
21A. If the membrane has too many saturated fatty acids in the phospholipids, why does oxidative
phosporylation become inefficient?
The membrane is insufficiently fluid. Thus, proteins in the electron transport chain cannot move
efficiently to transfer their electrons.
21B. If the membrane has too many polyunsaturated fatty acids in the phospholipids, why does
oxidative phosporylation become inefficient?
The membrane is now too fluid and thus also too permeable. The electrons will be passed well but the
transported protons can leak back across the membrane without powering ATP synthase.
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