C2006/F2402 -- Key to Exam #3
1. A-1. (2 pts) G1 to S. Growth factors control the transition from G1 to S by controlling the synthesis of G1 (start) cyclin. Cells that lack this cyclin arrest at the G1/S border or "start" checkpoint. Cells that lack the G2 cyclin (part of active MPF) arrest at the transition from G2 to M. GF's do not affect the product of G2 cyclin or arrest at the G2/M checkpoint. Cells in cleavage generally arrest at G2/M if they stop at all, but cells past the cleavage stage generally arrest at G1/S.
A-2. (2 pts) Cyclin. The only two items listed that are synthesized at a particular point in the cycle are cyclin and histones. Only cyclin must be made to enable the G1 to S transition. The "start" cyclin is made before the G1/S checkpoint, and gradually builds to a critical value. Once there is enough cyclin, it activates CDK to enable passage of the G1/S checkpoint, and then the cyclin is degraded after the cell passes the checkpoint. Histones are made after the cell enters S, after the transition is already made.
A-3. (3 pts; -1 for each additional wrong answer.) Map kinases, start kinases, and ras are activated to enable the G1 to S transition. (MPF is activated, but that's at the G2 to M transition; histones and lamins are modified at the time of entry into M.)
Order of activation/synthesis of items: GF --> ras --> MAP kinases --> cyclin --> CDK --> histones --> MPF --> lamins
Explanation: GF binds to its receptor --> activates ras (activated by GTP displacing GDP) --> MAP kinases are activated --> kinases phosphorylate and activate TFs such as jun, fos, myc --> transcription of gene for cyclin --> cyclin made --> CDK/cyclin complex forms --> complex activated by additional factors --> TF (E2F) activated --> transcription of genes for histones (& enzymes for replication of DNA) --> histones made --> finish S, G2 --> activate MPF (G2 cyclin + CDK) --> modify lamins, enter M.
B. Rb-E2F complex declines, because the complex falls apart after Rb is phosphorylated. Active CDK phosphorylates Rb, E2F and Rb-phosphate are released, E2F acts as a transcription factor to transcribe the genes for histones, enzymes of DNA synthesis, etc.
2. 10 pts, 2 pts each part. One point for each answer and one for a suitable corresponding backup statement.
A. Both. The blastopore is the point where cells invaginate. It becomes the anus in deuterostomes and the mouth in protostomes. Both types must invaginate (at the blastopore) to form a three layered structure.
B. The PNS develops from ectoderm. Both parts of the nervous system, PNS and CNS, and the skin develop from the ectoderm, the outer of the 3 germ layers.
C. Determination occurs before differentiation. Determination refers to the time the final state of the cell is set (or the process by which the cell's fate is set up); differentiation refers to the final state of specialization that is actually reached.
D. Either one (for any one case of embryonic induction) or both (for the process of embryonic induction overall). The process of embryonic induction involves cell-cell contact (by juxtacrines) in some cases and use of diffusible signals (by paracrines) in other cases. An individual act of induction can involve either one.
E. Necrosis. Cell growth and cell death are both required for proper development -- some cells must multiply and some must be removed to shape tissues and organs. The type of cell death involved in development is apoptosis or programmed cell death -- it is triggered by specific signals and results in controlled destruction of the cell from within by an active process. Specific enzymes must be made and/or activated for apoptosis to occur. Necrosis is "unprogrammed" cell death that results from damage. The cell's contents are released as it dies (which does not happen in apoptosis), and the release triggers inflammation.
3. A-1. (Ans. 2 pts; backup 1 pt.) Recruitment. PKB is in the cytoplasm, while the PIP3 is in the membrane. The enzyme must move to where the activator is, not the other way around.
A-2. (Ans. 2 pts; backup 2 pts.) Receptor must be a separate protein from PKB. The receptor must be a transmembrane protein (since it is both a receptor and a TK). The ligand (signal) binding site must be extracellular and the TK must be intracellular. PKB is entirely intracellular, so it cannot carry out the receptor or ligand binding part. (Intracellular receptors, such as the ones for steroids, are generally not TK's.)
A-3. (Ans. 2 pts; backup 2 pts.) Act before PKB in the pathway. (The usual terminology is to say it acts "upstream" of PKB, meaning at an earlier step in the pathway.) The receptor TK is responsible for receiving a signal from outside the cell. The PKB is inside the cell. So the signal must go first to the TK and then to the PKB. The pathway must be: signal/ligand --> binding to extracellular domain of receptor TK --> change in intracellular domain of TK --> (1 or more steps) activation of PKB.
B. (2 pts each answer; 4 pts for explanation.) B-1. Higher PIP3 than normal; B-2. Larger eyes than normal. Explanation: If PTEN levels are low, PIP3 will not be degraded as it is normally. It will continue to be made, but its breakdown will be slower than usual. So the level of PIP3 will go up. (2 pts) Activity of PKB is dependent on the level of PIP3. If there is more PIP3 than usual, the activity of PKB will be higher than usual (assuming the PIP3 is the limiting factor). High levels of active PKB lead to production of extra large eyes, presumably by overstimulation of growth of the eye cells. (2 pts). Note that either more stimulation of PIP3 production or less degredation of the PIP3 will have the same effect -- too much activation of PKB, too much cell growth, and extra large eyes.
C. (Ans 2 pts; explanation 4 pts.) PTEN is a tumor suppressor. A tumor suppressor is a protein that prevents excessive growth; when it is inactive or missing, too much growth (leading to a tumor) can occur. (2 pts for what a tumor suppressor or oncogene or proto-oncogene is.) PTEN prevents high levels of PIP3 from building up, by degrading the PIP3. If PTEN is inactive, PIP3 levels get too high, the PIP3 over activates PKB, and excessive growth occurs. (2 pts.)
4.
The proteins we discussed in class are found in this sarcomere, but
they’re not labeled. What is
labeled is some of the additional proteins that have been identified in muscles.
Circle all correct answers below.
No explanation is needed. (12 points) (2
point each, but you lost points if you included incorrect answers, or if you
omitted one of the answers in F.}
A.
The role of nebulin is most likely to (attach the DHP and ryanodine
receptors) (strengthen the actin filaments)
(help the actin filaments to shorten)
(bind individual myosin proteins together in a bundle)
(bind ATP to actin) (bind
Ca++ to actin) nebulin is shown alongside actin
B.
The thin filaments are attached to (myomesin)
(titin) (nebulin) The
thin filaments are made of actin, reverts to previous question.
C.
The one protein that must physically shorten during muscle contraction is
(titin) (nebulin) (actin)
(myosin) (troponin) (tropomyosin)(myomesin) (actinin) (desmin) The
way titin is drawn, it would block actin from sliding past myosin, so it must
shorten to allow the vertical z-line to come towards the center of the sarcomere.
D.
From the location of tropomodulin, it seems most likely that it (was in
modern St. Tropez where it was first identified, using muscles of beach bums)
(anchors tropomyosin in place, so that it doesn’t slide into the
grooves between the actin chains) (binds
to troponin, at least when the muscle is maximally contracted)
(caps F-actin so that it is a constant length and doesn’t polymerize
further) (determines whether
calcium will bind to troponin or to tropomyosin) If
additional globular actin molecules attached to the actin in thin filaments, the
actin on one side would hit the actin on the other side, and the sarcomere
wouldn't be able to shorten.
E.
A single muscle cell has multiple units of sarcomeres near each other,
but the final sarcomere must be bound to the cell membrane.
The protein that binds the sarcomere to the membrane is most likely to be
(myomesin) (myosin) (desmin)
Must be something at the end of the final sarcomere.
F.
Myomesin may be needed to (stabilize thick filaments) (allow
cross-bridge formation) (move thin filaments) (bind myosin molecules to each
other) Myomesin is shown in the center of the thick
filament, which is made of a bunch of myosin molecules bound together, so it is
likely to stabilize this bundle.
5.
You have an isolated muscle in an in vitro solution, and obtain the following
experimental results:
A.
When you stimulate the motor neuron, the muscle contracts.
You add caffeine to the solution, and the muscle contraction becomes even
stronger.
B. You don’t stimulate the motor
neuron, but you add a high concentration of caffeine to the solution, and the
muscle contracts spontaneously, even without neuronal stimulation.
C. To the solution in part B, you
now add curare, which blocks nicotinic receptors, and the muscle is not
affected; it stays contracted.
Caffeine
is most likely to act on (acetylcholinesterase)
(the receptor for acetylcholine) (the
receptor for norepinephrine) (the
muscarinic receptor) (ATPase)
(axon varicosities) (the
ryanodine receptor) (3}
(Ca++ channels in the presynaptic cell) (sarcoplasm)
Circle the one best answer.
Explain
your answer.
Your answer should include an explanation of the molecular events that
occur when the muscle contracts spontaneously in the presence of caffeine, and
an explanation of how the results above led to your conclusion.
(18 points)
Caffeine stimulates the ryanodine receptor, such that the Ca++ channel opens (2} and Ca++ flows out of the sarcoplasmic reticulum, into the sarcoplasm (2}. Calcium binds troponin (2} which moves tropomyosin away from its blocking position (2}, and actin-myosin crossbridges form, as actin slides past myosin (2}.
In Experiment A, the presence of caffeine increases neuronally stimulated contraction. From this, one might think that caffeine works either on the neuron, (to increase the frequency of action potentials, or the release of neurotransmitter), or it might act on the muscle itself. Experiment B, shows that the the neuron is not necessary. Caffeine stimulates contraction even if the neuron is not present. So the choices now would be that it affects muscle, either by mimicking the natural neurotransmitter (acetylcholine) at its receptor, or by working someplace downstream of the receptor. The results of Experiment C indicates that when nicotinic receptors are blocked, which are the receptors found at the neuromuscular junction, then the muscle still responds to caffeine with contraction, so caffeine must not be stimulating the receptor directly. Of the choices given, ryanodine receptor is most likely. (5 points}
6.
In 1774, Captain James Cook recorded the consequences of eating a certain
food he’d discovered:
“About three o’clock in the morning we found
ourselves seized with an extraordinary weakness and numbness all over our limbs.
I had almost lost the sense of feeling; nor could I distinguish between
light and heavy bodies of such as I had strength to move, a quart pot full of
water and a feather being the same in my hand….”
More
recently, doctors have described the time course of response to this toxin: First, tingling and numbness in the lips and tongue, then in
the face and fingers. Then muscle
weakness and difficulty contracting the muscles needed to move or speak.
Later, heart beat becomes irregular and the person loses consciousness
and goes into a coma.
A.
( 5 points) The time course
described here allows you to conclude that in response to the toxin, (more IPSPs
are produced in all neurons) (glial cells are the first to respond) (the central
nervous system is affected before the peripheral nervous system)
(afferent neurons are affected before efferent neurons)
(motor neurons are affected before sensory neurons)
(sympathetic neurons are affected before parasympathetic neurons)
(the autonomic nervous system is not affected at all)
(myelination protects neurons from this toxin, since myelinated neurons
are not affected at all) Circle all answers that are supported by the
information given. No
explanation needed.
The
initial response is tingling and numbness, suggesting that sensory signals are
not getting through. Sensory neurons are afferent neurons. Only
later are the efferent motor neurons affected.
B.
(15 points) A person who has
eaten this food is brought to the emergency room soon afterwards.
You insert an electrode into a neuron in her pinky finger, and find that
the resting potential is –70 mV. You
scratch the side of her pinky, and you see that the neuron you’re studying has
been stimulated, since an action potential moves down its axon.
But the rising phase of the action potential is less steep than usual (ie,
a slower rise) and the amplitude, or height, of the action potential is lower
than usual. However, the width of
the action potential is the same as usual.
This suggests that the channels that are affected are probably (K+ leak)
(Na+ leak) (Ca++ leak)
(voltage-gated K+) (voltage-gated
Na+) (voltage-gated Ca++) (ligand-gated K+) (ligand-gated Na+) (ligand-gated
Ca++) Choose the one best answer.
Explain your answer. Your
explanation should include why you selected this channel and what is responsible
for the changed appearance of the action potential.
Note: Tetrodotoxin, found in the pufferfish, has these effects. See James Cook's log
In neurons, the rising phase of the action potential occurs as voltage-gated Na+ channels open. Since the membrane depolarizes more slowly than usual, there must be a problem with these channels, either there are fewer of these channels able to open, or perhaps they are opening more slowly than usual. Na+ enters more slowly, so the slope is less steep, and not as much Na+ can enter before the gates shut, so the amount of depolarization (or the amplitude) is less. (5}
Voltage-gated K+ channels are responsible for the repolarization of the membrane, as K+ rushes out and carries out positive charge, the membrane becomes more negative inside again. Since the action potential is normal width, that is, the membrane repolarizes at the normal time, this implies that K+ channels are opening at the normal time after threshold, so they must be okay. (5}
Ligand-gated channels are not likely to be involved. These would be expected if a neurotransmitter were responsible for initiating the signal, but since it is pressure that initiated the signal, a ligand was not involved .