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BS11 Midterm 2 (1999)
NAME:_______________________
Question 1. 12 points.
A. (4 pts) Briefly explain the difference in melting points between trans-oleic acid
(18:1,∆9)(44.5C) and cis-oleic acid (18:1,∆9) (13.4C).
The cis double bond puts a kink or bend in the hydrocarbon chain, while the trans-oleic
acid is more or less fully extended. The trans configuration allows better packing in a
crystal than the cis arrangement, and hence has a higher melting point.
B. (4 pts) Would you expect to see an amphipathic helix in the transmembrane domain
of a single-pass transmembrane protein or a channel protein? Why?
Answer:
A channel protein. Amphipathic helices have one side that is polar or charged, which
would be unfavourable in the nonpolar membrane. In a channel protein this charged or
polar side of the helix face the channel, and other polar residues, and away from the
nonpolar lipids.
C. (4 pts) How many amino acids would be the minimum to cross the lipid bilayer?
Show your calculations for full credit.
The minimum residues required to cross the lipid bilayer of 3-5 nm thickness, given 15 Å
per aa in a helix
3/1.5 or 5/1.5
=20-35 aa
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BS11 Midterm 2 (1999)
NAME:_______________________
Question 2. 23 points.
A. (8 pts) The difference in pH between the internal and external surfaces of the inner
mitochondrial membrane is 1 pH unit (external side acidic). If the membrane potential is
–0.18V(negative inside), calculate ∆G for the transport of H+ from outside to inside the
matrix.
Answer:
∆G = 2.3RT(pHout - pHin) + zF∆Ψ
=2.3(0.6kcal/mol)(-1) + (1)(23kcal/V*mol)(-0.18V)
= -1.38 kcal/mol – 4.14 kcal/mol
= -5.52 kcal/mol
B. (15 pts). Assume it takes 3 moles of H+ to synthesize 1 mole of ATP. What is the
largest ratio of [ATP]/[ADP][Pi] under which ATP synthesis can occur? (∆G0ATP = 7.5
kcal/mol)
Answer:
∆GH+ = ∆G0ATP + RT ln[ATP]/[ADP][PI]
(5.52 x3) kcal = 7.5 kcal + 0.6 ln [ATP]/[ADP][PI]
[ATP]/[ADP][PI] = e9.06/0.6
= 3.6 x10 6/1
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BS11 Midterm 2 (1999)
NAME:_______________________
Question 3. 25 points
Inhibitors of mitochondrial function have provided extremely useful tools for analyzing
how mitochondria work. The figure below shows three distinct patterns of oxygen
electrode traces obtained using a variety of inhibitors. In all experiments, mitochondria
were added to a phosphate-buffered solution containing succinate as the sole source of
electrons for the respiratory chain. After a short interval ADP and then an inhibitor were
added, as indicated in the figure. The rate of O2 consumption at various times during the
experiment is shown by downward sloping lines, with faster rates shown by steeper lines.
A. (10 pts) From the description of the inhibitors in the list that follows, assign each
inhibitor to one of the O2 traces in the figure. All these inhibitors stop ATP synthesis.
Inhibitor
1. Dinitrophenol (DNP)
Action
Trace (Answer)
Makes membranes permeable to protons
___b____
2. Malonate
Prevents oxidation of succinate
___a____
3. Cyanide
Inhibits cytochrome oxidase
___a____
4. Atractylate
Inhibits ADP-ATP exchange
___c____
5. Oligomycin
Inhibits ATP synthase
___c____
a.
b.
mitochondria
ADP
inhibitor
mitochondria
ADP
inhibitor
c.
mitochondria
ADP
inhibitor
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BS11 Midterm 2 (1999)
NAME:_______________________
B. (15 pts) Using the same experimental protocol, sketch the O2 traces that you would
expect for the SEQUENTIAL addition of the pairs of inhibitors listed below.
1. DNP followed by cyanide
mitochondria
ADP
DNP
cyanide
2. DNP followed by oligomycin
mitochondria
ADP
DNP
oligomycin
3. Oligomycin followed by DNP
mitochondria
ADP
oligomycin
DNP
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BS11 Midterm 2 (1999)
NAME:_______________________
Question 4. 20 points.
A student just started to study protein translocation with the eventual goal of isolating the
individual components involved. His first task was to isolate the different subcellular
fractions that are required for this process. Unfortunately, his labels fell off and he could
not tell which fraction was what. Since he had the in vitro translation system (IVT) ready
to go, he decided to use that to help him figure out what the fractions were.
Shown below are the pertinent reactions. The IVT contains all the components necessary
for cell-free translation, an mRNA that encodes a protein known to translocate to the ER,
and 35S-labelled methionine to visualize the translated polypeptide. Translation was
carried out in the absence (lane 1) or presence of fraction C (lane 2), fraction E (lane 3),
or both (lanes 4-7). Endo-glycosidase H (endo H) or protease was added after translation
was complete. The samples were analyzed by SDS-PAGE, followed by exposure of the
dried gel to X-ray film. Results are shown below.
1.
2.
3.
4.
5.
IVT
IVT + fraction C
IVT + fraction E
IVT + fraction C + fraction E
IVT + fraction C + fraction E
+ detergent + endo H
6. IVT + fraction C + fraction E
+ protease
7. IVT + fraction C + fraction E
+ detergent + protease
1
2
3
4
5
6
7
a) 5 pts. What is fraction C? Briefly explain.
The cytoplasmic fraction containing SRP. Addition to the IVT causes translational
arrest seen in lane 2.
b) 5 pts. What is fraction E? What are the components in this fraction that our student
would be able to isolate?
This is the ER, or microsomal fraction. It contains SR, Sec61p (translocon/channel),
glycosylase and signal peptidase. BiP, calnexin and other proteins that assists
protein folding are also present, though the experiment does not implicate them.
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BS11 Midterm 2 (1999)
NAME:_______________________
c) 5 pts. What is the purpose of the endoglycosidase H treatment? What accounts for
the size of the band in lane 5?
Endonuclease removes N-linked sugars that are added in the ER. The polypeptide is
smaller than that in lanes 3 and 4 because sugars and the signal peptide have been
removed.
d) 5 pts. Based on the data, can you tell whether the test protein is lumenal or
transmembrane? Briefly explain.
The protein is probably transmembrane. A lumenal protein will not be digested with
protease alone, since it is protected inside the microsome. There are probably
cytoplasmically exposed regions to which the protease have access, leaving
transmembrane and ER fragments seen in lane 6.
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BS11 Midterm 2 (1999)
NAME:_______________________
Question 5. 20 points.
In most organisms, the citric acid cycle begins with the reaction of acetyl-SCoA with
oxaloacetate, as shown on the next page. However, in Bacillus logotenens, a bacterium
found near ocean vents, the initial reaction occurs between acetyl-SCoA and pyruvate
(shown below).
CH3
C=0
C00-
Draw a cycle for the oxidation of acetate to 2 CO2 that might occur in this organism. The
rules for making and breaking C-C bonds and for oxidation reactions are the same in B.
logotenens as in other organisms.
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