C2006/F2402 '09 Key to Exam #1
Each answer and each explanation are 2 pts, unless it says otherwise.
1. A. Integrin. Adhesive junctions that connect to the ECM
(hemidesmosomes & focal adhesions) use integrin as a TM protein to connect MF or
IF inside the cell to the ECM proteins on the outside. Cadherins connect one
cell to another; the other proteins have nothing to do with adhesive junctions.
By circling the correct answer, you told us that 'integrin
binds to the ECM.' To get credit for the explanation you had to add some
additional true fact about integrin or adhesive junctions involving the ECM.
Similar reasoning applies to the grading of all the questions -- to get credit
for the explanation you have to add additional information -- you can't simply
restate your circled answer.
B. Collagen. Collagen is the only extracellular protein listed. All the other proteins are intracellular, so collagen is the only reasonable candidate on the list.
C. Mitochondria. Laminin is extracellular, so platelets shouldn't contain any. Platelets have no nuclei. That rules out lamin and nucleoli. Platelets have no junctions to other cells or the ECM. That rules out everything else but mitochondria. Since platelets do have cytoplasm, they should contain all the usual cytoplasmic organelles, such as mitochondria, lysosomes, peroxisomes, etc., unlike RBC.
2. A. The TM protein should be in the plasma membrane, with its carboxyl end on
the extracellular side of the membrane. This follows from the mechanism of
fusion of vesicles and the plasma membrane during exoctyosis. What's in
the lumen of the vesicle ends up on the outside of the cell. For a diagram, see
texts or
https://eapbiofield.wikispaces.com/file/view/399px-Exocytosis_types.svg.png
B-1. They used the EM. Components of the cytoskeleton are too small to see in the light microscope unless they are tagged with fluorescent material. (You might be able to infer the state of the cytoskeleton from the overall shape of the cell, but you wouldn't be able to tell the state of the MFs and MTs without an EM.)
B-2. 3 (but maybe 4 if you want to see both MTs and MFs at the same time). You need two different, primary, unlabeled antibodies, one to tubulin and one to actin. Antibodies are specific -- each Ab fits to and binds to a specific target antigen. So an Ab to tubulin won't bind actin and vice versa. If both primary antibodies are made in the same species, they will have the same constant end, and you can use the same secondary, fluorescently labeled antibody, to bind to and visualize them.
If you want to distinguish the MFs from the MTs in one picture, you might want to use two different secondary antibodies that have different colored fluorescent tags. (MTs and MFs are different in size, so you could probably tell them apart even if they were same color.) If you want to use two different secondary antibodies, than either the two primary antibodies must have different constant regions (i. e. be made in different organisms) or each of the two secondary antibodies must react with a different part of the constant region.
3. A. Polymerize more than normal. Taxol binds to MTs (polymerized tubulin) and prevents depolymerization of MT. This shifts the equilibrium between monomers of tubulin (really alpha/beta dimers)and polymers of tubulin (MT). There are always some monomers in the platelets, since the MTs are dynamic structures that constantly add and lose monomers. When MTs are stabilized, monomers can't fall off, but they can add, so more polymerization occurs.
B. Remain; neither.
B-1. You would expect the ring of MTs to remain, since MTs
are stabilized by taxol (see A). This is what actually happens.
B-2. If the state of the MT ring is responsible for the shape
of the platelets, you would expect that the platelets would not deform in the
presence of taxol. But the platelets do deform anyway, indicating that the MT
state does not control the shape of the platelets.
Some students reasoned that since taxol promotes polymerization, and the platelets deform in the presence of taxol, deformation must depend on polymerization (not depolymerization). Partial credit was given for this explanation, although it is incorrect, because students who picked it recognized that the results rule out depolymerization as a solution. It is reasonable to suggest that during activation or cooling the MTs from the ring depolymerize and then re-polymerize in a different formation to support a star shape. (This is what was assumed to be the case before these experiments were done.) However, a rearrangement of tubulin could not occur in the presence of taxol. So polymerization can't be the explanation.
C. Actin, both, activity of a myosin kinase.
Explanation of C-1 & C-2 (not required): Myosin is the motor molecule of actin.
Myosin binds to actin, but not to tublin or IFs. If myosin is
phosphorylated reversibly, there must be an enzyme to add the phosphate at the
expense of ATP (a kinase) and a different enzyme to remove the phosphate by
hydrolysis (a phosphatase.
Explanation of C-3: When platelets are warmed, phosphate is
removed from myosin, and the dephosphorylated myosin no longer binds actin. (We
know from the information given that more phosphorylation means more binding to
actin, not vice versa.) It takes a phosphatase, not a kinase, to remove the
phosphate.
D. More MFs; cytochalasin D. (If you said
'less MFs' the correct answer to D-2 is phalloidin.) The experimenters who did
this work actually counted MFs, and they found more MFs in the activated state.
Cytochalasin would stabilize monomers and reduce net polymerization of actin
into MFs; phalloidin would do the reverse.
Explanations: (2 pts for explaining each of the following.)
(1) We know that myosin binds to MFs, and that more myosin
binds in the activated state. The number of MFs could stay the same, and the
state of the myosin alone could change. However the problem says there are more
MFs in one state than the other. In that case, it makes more sense to have more
MFs when there is more binding of myosin, not the other way around. You can also
argue that there is more surface area in a star shape than in a disc, and more
MFs would be needed to maintain the shape.
(2). There are more MFs, but the amount of actin per cell
doesn't change. The activation is fast and reversible, which indicates that what
changes is the polymerization of actin, not the amount of actin.
Note: If you thought both MFs and MTs were involved, that was okay, as long as you explained the MF part properly. If you missed both (1) & (2) above, part credit was given for explaining how the (correct) drug worked.
Problem 3 is based on a Letter to Nature, 313, 3 January 1985, pp. 70-72. (Nachmias, Kavaler & Jacubowitz) A pdf is at http://www.nature.com/nature/journal/v313/n5997/pdf/313070a0.pdf
Click here for a copy of page 3, with the information for problem 4.
4. A. In expt 1, you are measuring secondary active transport
(symport) across the plasma membrane. Na+ is flowing down its
gradient, and the energy from the movement of Na+ is being used to
transport serotonin across the plasma membrane, against its gradient, using a
transporter protein. It is symport because both the Na+ and the
serotonin are moving in the same direction -- into the platelets.
This transport requires a protein transporter or carrier, but
the process is not called 'carrier mediated transport' because that term is
reserved for transport of a substance down its gradient, using a transport
protein, without additional input of energy. This case is called secondary
active transport because energy from an ion gradient is being used to push the
substance to be transported (serotonin) up its gradient. No primary active
transport is involved here, but a [Na+] gradient is established in
normal cells and platelets by primary active transport, using the Na+/K+
pump.
B. For this part, each ans = 1 pt;
explanation = 3 pts. (8 total)
In expts. 2 & 3 you are measuring transport into the organelles. Transport is
occurring by antiport with H+s (secondary active transport). First,
primary active transport is used to pump H+ into the organelles and
make them acid. Once the [H+] gradient is established, with high [H+]
on the inside, serotonin can be exchanged for H+. (The lag, which
should be in expt. 2, is the time it takes for the H+ gradient to be
established at the expense of ATP.) The exchange is an example of antiport -- H+s
leave the organelle, going down their gradient, and serotonin goes in, up its
gradient.
Why is uptake important? Once exocytosis occurs to release
serotonin, the organelles (storage vesicles) reform by endocytosis, and
serotonin is transported back into the cytoplasm by co-transport with Na+.
Antiport with H+ is used to refill the organelles with serotonin for
another round of exocytosis. A similar process occurs at during nerve impulse
transmission at synapses (ends of neuron axons). Too much or too little uptake
can cause problems with brain function and behavior.
If you were in 833 Mudd, and thought the lag was in expt. 3,
then it was difficult to come up with a satisfactory explanation for the results
of experiments 2 & 3. The results implied that H+ transport was
producing ATP (for primary active transport of serotonin into the organelles)
and not the reverse. Therefore you received credit if you chose 'primary active
transport' and 'uniport' and explained it clearly. (This applies to students in
833 Mudd only.)
C-1. Na+ should be higher inside. To get efflux, Na+ must flow down its gradient out of the platelets and drive serotonin with it. In this case serotonin may be moving down its gradient, not up, but the conformational changes that allows the transport protein to bind and move serotonin across the plasma membrane are dependent on binding of Na+. In other words, the action of this transporter is reversible -- it can be used to push serotonin in or out, and the direction depends on the direction of the Na+ gradient. (In real life it is more complicated, and K+ is involved too.) In the revolving door analogy, the transporter must bind first Na+ and then serotonin in order to 'turn' and move the serotonin to the other side of the membrane. It won't 'turn' unless it binds them both. (A similar argument applies to H+ /serotonin exchange; see below.)
C-2. Raising the pH should increase efflux. (MDMA is a basic substance, an amine, and probably does have this effect, although it has others as well.) The H+s flow down their gradient, pushing serotonin in the opposite direction. (Same basic idea as with Na+ and serotonin above -- transporter is reversible, and direction it 'pushes' depends on the ion gradient. However this is antiport with H+ instead of symport with Na+.) Normally, the organelle is acid inside, and H+s flow out (causing uptake of serotonin) If the organelle is basic, H+s are higher outside, and will flow in (causing efflux of serotonin).
Your answers to C were graded based on your answers to A & B. Wherever possible, you were given credit if the two parts fit together properly and made sense.
Problem 4 is based on an article by Rudnick et al, J. Biol. Chem., 255: 3638-3641 (April 25 1980). If link doesn't work, go to http://www.jbc.org/ and use the search function. That will give you a link to the pdf.
5. A. LDL uptake is by endocytosis; Apo B and LDLR separate in endosomes; You need a peripheral protein, such as clathrin, to provide a coat and help the membrane to bend and invaginate.
B-1. Ab goes to lysosomes and is degraded.
B-2. LDLR is both endocytosed and exocytosed (it is recycled
back to the plasma membrane)
B-3. Ab and LDLR separate inside the cell, probably in
endosomes, just like LDLR and LDL.
B-4. The curve reaches a plateau when Ab degradation is
balanced by Ab uptake.
Ab must bind to the LDLR and be internalized by RME. Once the Ab gets inside the cells, it is degraded in lysosomes. Since the amount of degraded material increases continuously, after a short lag, it means that the receptors are being recycled -- they aren't being degraded, but are being sent back to the plasma membrane (exocytosed) to bring in more Ab for degradation. Therefore the Ab and the LDLR must separate inside the cell, leaving the Ab in the cell and recycling the LDLR. It's the same old receptors that are being used over and over, not new ones, since inhibiting protein synthesis doesn't affect degradation. The curve plateaus when the amount of radioactive Ab coming into the cells = the amount of radioactive degradation product leaving the cells (by some unspecified process).
The curve shown (fig. 4) is a curve of X (Ab) inside vs time, which is like curve #1 on handout 4C. It is not a curve of X uptake vs. [X] = Ab uptake vs [Ab]. That would be equivalent to curve #2. A curve of X inside vs time always reaches a plateau because the system eventually reaches a steady state where the rate of X going out equals the rate of X going in. It doesn't matter if you have low levels of Ab (and lots of empty receptors) or a high levels of Ab that saturate all the receptors. The results would look the same. You aren't checking the maximum rate of uptake. If you want to find out what that is, and see what it takes to saturate the receptors, you have to measure Ab uptake as a function of [Ab]. The authors did that, but it is a different experiment and the results are shown in a different figure.
C. Explanation: 4 pts. The rare patients have
LDL receptors that bind Ab but not ApoB.
You know the receptors bind Ab, because Ab must be
internalized to be degraded. If the receptors were missing, or not internalized,
there would be no degradation. So the rare patients must have receptors with a
normal binding site for Ab. (2 pts)
You know the receptors don't bind ApoB because the patients
have hypercholesterolemia -- they aren't taking up LDL well from the blood. ApoB
is the part of LDL that normally binds to the receptor. So the rare patients
must have altered receptors -- the part that binds ApoB must be altered, so they
don't internalize LDL.(2pts)
Problem 5 is based on a paper by Beisiegel et al from the Brown and Goldstein lab in J. Biol. Chem, 256: 11923 (1980). Fig. 4 is from this paper. (I rewrote the legend.) If link doesn't work, go to http://www.jbc.org/ and use the search function. That will give you a link to the pdf.
