Download 7.06 Cell Biology EXAM #1 KEY February 28, 2006

Name: _____________KEY__________________
7.06 Cell Biology
EXAM #1 KEY
February 28, 2006
This is an OPEN BOOK exam, and you are allowed access to books, a calculator,
and notes
BUT NOT computers or any other types of electronic devices.
Please write your answers in PEN (not pencil) to the questions in the space allotted.
Please write only on the FRONT SIDE of each sheet, as we will Xerox all of the
exams and thus only grade writing on the front.
And be sure to put your NAME ON EACH PAGE in case they become separated!
There are EIGHT pages including this cover sheet.
Question 1. 23 pts __________
Question 2. 27 pts __________
Question 3. 21 pts __________
Question 4.
9 pts __________
Question 5. 20 pts __________
TOTAL
100 pts __________
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Question 1. (23 pts)
You are studying transepithelial transport of glucose from the intestinal lumen to the
extracellular medium surrounding the basal surface in cells that are growing in culture.
You are using a system identical in set-up to that described for MDCK cells in the
textbook, copied below:
(a, 3 pts) If you performed immunofluorescence on permeabilized intestinal epithelial
cells using an antibody against GLUT2, where would you find the fluorescent signal?
You would see signal only in the basolateral surface of the membrane, as that is the only
portion of the cell in which GLUT2 protein is found. GLUT2 is the uniporter that allows
glucose to flow down its concentration gradient, from the inside of the cells to the blood.
(b, 5 pts) If you performed the same experiment as above on intestinal epithelial cells
that had been depleted of occludin and claudin, where would you see the fluorescent
signal? Justify your answer.
You would see the signal all over the plasma membrane (both the basolateral and apical
sides), because these two proteins are essential for proper formation of tight junctions.
One of the very important functions of tight junctions is to disallow proteins from
diffusing freely between the basolateral and the apical sides of the membrane.
(c, 8 pts) You are trying to design a new artificial transporter that would transport
glucose from the intestinal lumen into intestinal epithelial cells. You have the ability to
create novel transporters and target them to the apical membrane. You make the two new
transporters listed below, and to test them you use recombinant DNA techniques learned
in 7.02 and 7.03 to express these in the MDCK cells. For each new transporter below,
state whether it could allow for movement of glucose across the apical membrane at all,
and if yes, whether you think that glucose would move into or out of cells.
(i) A 2 Ca2+/1 glucose symporter
Yes, it would allow movement of glucose inside of the cell. Glucose is higher inside these
cells than outside, and calcium is higher outside than inside. Thus the free energy
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released from moving calcium down its concentration gradient could be used to drive the
movement of glucose into the cells (up its concentration gradient).
(ii) A 2 Ca2+/1 glucose antiporter
Yes, it would allow movement of glucose outside of the cell. Glucose is higher inside
these cells than outside, and calcium is higher outside than inside. Thus it would be
impossible for this antiporter to move calcium outside (against its gradient) and glucose
inside (against its gradient). It would be possible, however, for the antiporter to move
calcium inside (down its gradient) and glucose outside (down its gradient).
(d, 7 pts) You expose wild-type mice to a modified version of ouabain. Ouabain inhibits
the Na+/K+ ATPase in all body cells, but your version inhibits the Na+/K+ ATPase only
in intestinal epithelial cells. You think this might be useful as an antidiabetic drug.
Predict the following:
(i) What would happen over time to glucose uptake from the intestinal lumen to the blood
and why?
Glucose uptake would cease. The uptake of glucose from the lumen is powered by the
Na+ concentration gradient, because sodium flowing down its gradient (from the lumen
to inside cells) releases free energy that is used to power the movement of glucose up its
gradient (from the lumen to inside cells). The Na+/glucose symporter is the protein that
allows for this movement. If the Na+/K+ ATPase were inhibited, then the concentration
gradients of Na+ and K+ would eventually be dissipated, and so there would no longer
be a concentration gradient of Na+. Thus the Na+/glucose symporter would no longer
be powered.
(ii) What would happen over time to the magnitude of the membrane potential of the
intestinal epithelial cells?
It would decrease in magnitude. If the Na+/K+ ATPase were inhibited, then the
concentration gradients of Na+ and K+ would eventually be dissipated, and so there
would no longer be a concentration gradient of K+. Thus the movement of K+ through
the open resting potassium channels would cease. This movement of K+ is what creates
the membrane potential of cells, so therefore the magnitude of the membrane potential
would decrease.
Question 2. (27 pts)
You are studying the TGF-β pathway in mouse cells growing in culture. Describe how
the results obtained from each of the two following experiments would be different,
depending on whether or not TGF-β had been added the cells for an hour. Also
describe how you would interpret that difference in results (i.e. what the different results
in the presence or absence of TGF-β tell you about the effect that TGF-β has).
(a, 7 pts) You perform immunofluorescence on cells using an antibody against Smad-3.
(Assume you performed the immuofluorescence correctly, using standard procedures.)
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Result without TGF-β:
You would see staining in the cytoplasm.
Result + TGF-β:
You would see staining in the nucleus.
Interpretation of difference in results:
TGF-beta pathway activation leads to the translocation of the the transcription factor
Smad-3 into the nucleus. This is because TGF-beta binding to its receptor results in
receptor dimerization and activation, and subsequent phosphorylation of Smad-3.
Phosphorylated Smad-3 complexes with Smad-4, and Smad-3’s now exposed NLS
sequences allow it to be translocated into the nucleus.
(b, 7 pts) You dissolve the cells in a nonionic detergent, remove all remaining membrane
fragments by centrifugation, and run the cell lysate over a gel filtration column. You then
do Western blotting on each fraction that comes off the column using an antibody against
Smad-3.
Result without TGF-β:
Smad-3 would elute in a relatively late fraction, so you would see a band corresponding
to Smad-3 in a lane of your Western in which you loaded a relatively late fraction.
Result + TGF-β:
Smad-3 would elute in a relatively early fraction, so you would see a band corresponding
to Smad-3 in a lane of your Western in which you loaded a relatively early fraction.
Interpretation of difference in results:
TGF-beta pathway activation leads Smad-3 forming a complex. Without TGF-beta, it is
a monomer. This is because TGF-beta binding to its receptor results in receptor
dimerization and activation, and subsequent phosphorylation of Smad-3. Phosphorylated
Smad-3 complexes with Smad-4 and another monomer of Smad-3.
(c, 4 pts) You overexpress in wild-type cells a mutant version of Smad-3 in which its
three critical serine residues at the C- terminus – the ones normally phosphorylated by the
TGF-β receptor - were replaced with aspartic acid. These cells (in contrast to wild-type
cells) rapidly stop growing in culture whether or not TGF-β is added to the cells. Explain
this result.
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This is a constitutively active mutant version of Smad-3, because you are replacing the
amino acid that is normally phosphorylated with an amino acid that has a negatively
charged R group that mimics phosphorylation. Thus this Smad-3 will always be
activated, and will always form complexes and translocate into the nucleus, where it will
turn on genes that inhibit the cell cycle. Thus cells are always inhibited from growing
rapidly in culture.
(d, 3 pts) Would the result from part (c) change if the cells had not also been expressing
endogenous wild-type Smad-3? If so, how would it change?
No. This is a mutant that is constitutively active, which is why you see the mutant
phenotype even though there is wild-type Smad-3 present in addition to mutant Smad-3.
(e, 6 pts) Would cells in culture rapidly stop growing if you over-expressed (in wild-type
cells) a mutant version of Smad-3 that lacked an NLS…
…in the absence of TGF-β? …in the presence of TGF-β? Explain.
No in the absence and no in the presence. If Smad-3 lacked an NLS, it could never go
into the nucleus, and thus it could never turn genes off that inhibited the cell cycle. Thus
the cells would always proliferate. This is a dominant negative mutation, which is why
you see the mutant phenotype even though there is wild-type Smad-3 present in addition
to mutant Smad-3. Mutant Smad-3 will bind to any wild-type Smad-3, and prevent the
entire complex from translocating into the nucleus.
Question 3. (21 pts)
You form a liposome in buffer #1 that contains 5mM KCl and ATP at pH7. You then
place the formed liposome into buffer #2 that contains 50mM KCl and ATP also at
pH7. For each subpart below: Describe what would happen very shortly after liposome
formation, and then eventually (after a long period of time).
(a, 7 pts) There are no proteins incorporated into the liposome.
(i) Will K+ move across the membrane, and if so, in which direction does it move?
No. Ions cannot pass through the membrane unless there is a channel, pump, or
transporter that allows movement of the ion.
(ii) What happens to the concentration gradient across the liposome bilayer over time?
Nothing. Ions cannot pass through the membrane unless there is a channel, pump, or
transporter that allows movement of the ion. If no ions can flow, then the concentration
gradient cannot change.
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(iii) How does the magnitude of the membrane potential across the lipid bilayer change
over time?
It doesn’t. It remains at zero the whole time. Ions cannot pass through the membrane
unless there is a channel, pump, or transporter that allows movement of the ion. Ions
traveling through channels is what creates a membrane potential in the first place.
(b, 7 pts) You now repeat the experiment from the introduction, but using liposomes into
which the H+/K+ ATPase (isolated from stomach epithelial cells) is incorporated in the
same orientation as it is in the cell, with the ATP-binding site inside the liposome.
(i) Will K+ move across the membrane, and if so, in which direction does it move?
Yes, inward. The H+/K+ ATPase brings K+ into the cell and pumps H+ out of the cell.
(ii) What happens to the K+ and H+ concentration gradients across the liposome bilayer
over time?
The H+/K+ ATPase brings K+ into the cell and pumps H+ out of the cell. Thus, an H+
gradient will be created so that H+ is higher outside the liposome. The K+ gradient will
initially dissipate (as it starts off higher outside than inside), but once the concentration
gradient of K+ reaches zero, the pump keeps pumping, thus creating a new gradient
where K+ becomes higher inside.
(iii) How does the magnitude of the membrane potential across the lipid bilayer change
over time?
It does not change. No net charge is flowing across the membrane. The outside and
inside of the liposome remain neutral.
(c, 7 pts) You now repeat the experiment from the introduction, but using liposomes into
which only a K+ channel is incorporated.
(i) Will K+ move across the membrane, and if so, in which direction does it move?
Yes, inward. K+ begins higher inside than outside. When the channels are first open,
K+ will want to flow down its concentration gradient (from the outside to the inside).
This results in the inside of the liposome becoming positive. Thus, eventually, K+ will
want to stop flowing inside because it is positive and thus repelled by the + charge inside
the liposome. Thus eventually there will be no net flow of K+ because an equilibrium has
been reached (where the membrane potential across the liposome membrane is equal to
the Nernst potential for potassium).
(ii) What happens to the concentration gradient across the liposome bilayer over time?
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Very little. It may decrease slightly because K+ is flowing inward, but then it reaches an
equilibrium, and the number of ions that must flow to create the membrane potential at
that equilibrium is miniscule compared to the number of ions that create the
concentration gradient.
(iii) How does the magnitude of the membrane potential across the lipid bilayer change
over time?
At first, the inside of the liposome will be becoming more and more positive, but
eventually the membrane potential will stabilize at the Nernst potential for K+. K+
begins higher inside than outside. When the channels are first open, K+ will want to
flow down its concentration gradient (from the outside to the inside). This results in the
inside of the liposome becoming positive. Thus, eventually, K+ will want to stop flowing
inside because it is positive and thus repelled by the + charge inside the lipsome. Thus
eventually there will be no net flow of K+ because an equilibrium has been reached
(where the membrane potential across the liposome membrane is equal to the Nernst
potential for potassium).
Question 4. (9 pts)
You are studying muscle contraction. When muscles receive the signals from neurons to
contract, calcium is released into the cytosol from an organelle called the sarcoplasmic
reticulum (SR), which stores calcium. The sarcoplasmic reticulum contains abundant
amounts of a P-class Ca2+ ATPase.
(a, 5 pts) What would be the effect over time on muscle contraction if you added a drug
that inhibited the Ca2+ ATPase in the SR, and why?
Initially, contractions would be continual. This is because the SR calcium pump pumps
calcium from the cytosol into the lumen of the SR. If you inhibit this pump, when calcium
rushes into the cytosol, it will not be cleared from the cytosol back into the SR. Thus the
cell will feel like it is constantly being stimulated to contract. However, eventually the SR
would be depleted of calcium, and thus the muscle cell would lose the ability to contract.
This is because the calcium that initially flows into the cytosol can be cleared into the
extracellular medium by the calcium pump that is present in the plasma membrane of all
cells.
(b, 4 pts) There are calcium channels in the membrane of the SR. Do you think that
those channels are non-gated or gated? Justify your answer.
Gated. If the channels were non-gated, then calcium would always be flowing into the
cytosol, because calcium wants to flow down its concentration gradient (from the SR to
the cytosol).
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Question 5. (20 pts)
Adipocytes (fat cells) store fatty acids as triacylglycerol – a molecule of glycerol (HOCH2 – CHOH- CH2OH) esterified to three long chain fatty acids. Lipase, the enzyme that
hydrolyzes triacylglycerols (TAGs) to one molecule of glycerol and three long chain fatty
acids, is found mainly in adipocytes. It is phosphorylated by Protein Kinase A (PK-A)
and is catalytically active only when phosphorylated.
You are studying a line of cultured adipocytes that express the G-protein-coupled
receptor for adrenaline that is coupled to Gsα.
For parts (a) – (d) below, state whether the following conditions would result in either:
constitutively high levels of TAGs,
constitutively low levels of TAGs,
high levels of TAGs in the absence of adrenaline but low levels in its presence,
OR
low levels of TAGs in the absence of adrenaline but high levels in its presence.
Explain your choice.
(a, 5 pts) In addition to the wild-type β-adrenergic receptor, your cells express a mutant
β-adrenergic receptor (containing three amino acid changes in the cytosol-facing Cterminal segment) that is a GEF for Gsα regardless of the presence or absence of
adrenaline.
Constitutively low levels of TAGs. If the receptor always allows for the exchange of GDP
for GTP, then G alpha will always be bound to GTP. Thus it will always activate
adenylate cyclase, leading to constitutively high cAMP levels. Thus PKA will always be
active and lipase will always be phosphorylated and thereby activated. Thus TAGs will
always be cleaved.
(b, 5 pts) A mutation in the catalytic domain of PK-A that prevents dissociation from the
regulatory domains even if 3’-5’-cAMP is present. (No wild-type PKA is present.)
Constitutively high levels of TAGs. If the catalytic subunit of PKA can never dissociate
from the regulatory subunit, then PKA can never be an active kinase. Thus it will never
phosphorylate and activate lipase, so TAGs will never be hydrolyzed.
(c, 5 pts) A mutation in IP (the inhibitory subunit of phosphoprotein phosphatase) that
prevents its phosphorylation by PK-A. (No wild-type IP is present.)
Constitutively high levels of TAGs. PP is a phosphatase that takes off the phosphate
groups put on substrate proteins by PKA. (PKA and PP are opposites of each other.) PP
is inhibited by phosphorylated IP. If IP cannot be phosphorylated, then IP will never
inhibit PP, and the phosphatase will always be active. Thus the phosphatase will always
remove any phosphorylation put on substrate proteins by PKA. Thus the substrates will
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never phosphorylated. Thus lipase will never be active and will never be able to cleave
TAGs.
(d, 5 pts) You incubate the cells in growth medium to which non-hydrolyzable GTP has
been added.
Normal levels (i.e. high in the absence of adrenaline and low in its presence). GTP of
any form cannot cross the plasma membrane, because it is charged, and charged
molecules can only cross the membrane if there is a channel, pump, or transporter to
move that molecule.
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