Bio98 Winter 2015
Midterm 3 (Version A)
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01. I purify a protein, “X” from cells. I then add NADH, FADH2 or
FADH2+Rotenone to purified “X” and measure the amount of
QH2 that is produced after 60 mins. My data is shown in the
graph. What is the BEST inference from these data ?
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 ? A) Rotenone is required for QH2 production by “X”
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 ? B) Rotenone has no effect on QH2 production by “X”
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C) “X” is most likely Complex II
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D) “X” is most likely Complex III or Complex IV
02. I culture plant leaf cells in conditions that causes complete inhibition of both
Photosystem I and II. In addition, these growth conditions causes the stroma of
chloroplasts to contain plenty of ATP and NADPH. What is the level of carbon fixation
in these cells compared to cells grown in normal conditions (in which PSI and PSII function
normally)?
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 ? A) Same as cells grown in normal conditions; cells have adequate ATP and NADPH
for carbon fixation
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 ? B) Higher than cells grown in normal conditions; cells have plenty of ATP and
NADPH
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C) Lower than cells grown in normal conditions; stroma will not have high pH or
reduced Ferredoxin
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D) In light conditions, there will be more carbon fixation than cells grown in normal
conditions; ATP and NADPH are already present
03. I add a drug that causes membranes in the cell to become permeable to protons.
What effect will this drug have on plant cells ?
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 ? A) Mitochondria will have no PMF; no synthesis of ATP. Photosynthesis normal
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 ? B) No PMF; No ATP production in mitochondria or chloroplasts
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C) Chloroplasts will have no PMF; no fixation of CO2. Mitochondrial ETC is normal
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D) e- transport will stop in ETC of mitochondria; photosynthesis will be normal
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04. I have mutant cells that make no PFK2 enzyme. I measure rates of oxygen
consumption in these mutant cells and compare them to wild type cells. What is the
MOST LIKELY outcome of this experiment ?
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 ? A) Mutant cells will have the same O2 consumption as wild type; PFK2 does not
affect the ETC
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 ? B) Mutant cells will have increased O2 consumption; they will need to increase ETC
to compensate for lack of PFK2
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C) Mutant cells will have decreased O2 consumption; decreased glycolysis leads to
decreased TCA and ETC
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D) Mutant cells will have increased O2 consumption; decreased PFK2 will increase
glycolysis, which will increase ETC
08. The reaction Glc  Glc-6-PO4 is catalyzed by the enzyme Hexokinase. I start a
reaction with Glc, ATP and Hexokinase, and allow the reaction to reach equilibrium. At
equilibrium, I have 200 µmoles of Glucose and 300 mmoles of Glc-6-PO4. I now replace
25 µmoles of the Glucose with 25 µmoles of radiolabeled Glucose (so that the
equilibrium is not disturbed). After letting the reaction run several minutes, will I see
radiolabeled Glc-6-PO4? Will the total [Glc-6-PO4] change? Explain your answer.
[4 points]
09. Two enzymes, Xam1 and Xam2 catalyze the same reaction, and have the same Vmax.
However, Xam1 has a Km of 34 nmolar, while Xam2 has a Km of 79 nmolar. I set up
identical reactions containing 34 nmolar substrate and either Xam1 or Xam2 and
measure the amount of product formed after 30 seconds. Would I expect to see more
product using Xam1 or Xam2? Or will they both produce the same amount of product
in 30 secs? Explain your answer.
[4 points]
10. I have a rats with either wild type or mutant
Pyruvate dehydrogenase (PDH). I measure the
enzyme kinetics of Liver PDH from these
animals in conditions of low or high blood
glucose. My data is shown (assume that the mutant
responds identically to wild type with all allosteric
regulators). Based on all this, what is the MOST
LIKELY problem with the mutant PDH ?
[4 points]
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11. I take equal numbers of liver cells and incubate them with radiolabeled Glucose
either in buffer only, or buffer+Glucagon. I measure the level of radiolabeled
pyruvate in the cells after 30 mins. Cells that are incubated with buffer+Glucagon
convert ~5% of the Glucose to pyruvate. How much of the Glucose would have been
converted to pyruvate in the cells incubated with buffer only?
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A) 5%. Same numbers of cells used, and cells have the same rate of
conversion.
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B) >5%. Glycolysis would not be inhibited by Glucagon.
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C) <5%. Glucagon makes cAMP in liver cells which would inhibit
Glycolysis.
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D) 0%. Since there is no hormone, there is no signal for the cells to
utilize the Glucose.
12. I inject equal amounts of radiolabeled glucose into the thigh
muscles of sleeping mice, awake mice, or mice running on a
treadmill. Two hours later, I measure the amount of
radiolabeled glucose present in the blood. My data is shown
in the graph. What is the BEST explanation for my results?
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A) The running mice used more Glucose
for energy.
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B) The sleeping mouse has a very low metabolic rate.
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C) Running mouse muscles made lactate that was converted to
Glucose in liver and released back in the blood.
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D) The running mouse secreted the highest amount of Glucagon.
The glucagon acted on liver cells to increase Gluconeogenesis.
13. I add radiolabelled Glyceraldehyde 3-Phosphate (GA-3-P) to wildtype and mutant
cells and follow the radiolabel over time. My data is shown below (white rows are
WT, and shaded rows are MUTANT). What is the BEST, COMPLETE explanation for
these results?
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