Biology154
Molecular and Cellular Neurobiology
FINAL EXAM
Fall 2004-2005
December 8, 2004
Name:
Student ID number:
Student email:
Section TA Name:
Section Day & Time:
DO NOT OPEN YOUR EXAM UNTIL YOU ARE TOLD TO DO SO!
You have 3 hours to complete this exam.
This is an open book exam. All notes, lecture slides, textbooks are allowed.
Computers are not allowed.
Please make sure that your answers are brief and specific. If incorrect information
is included in your answer, you can lose points even if the correct information is also
there.
If you write legibly the graders will look more kindly on your answer.
NOTE: Some questions may have multiple correct answers; as long as yours is
reasonable, logical, and consistent, you can get full credit. Be aware of point values
for each question and budget time accordingly. Some questions may be challenging
but don’t panic; grading will be on a curve. Good luck!
Please limit your answers to the space provided.
At the conclusion of this exam, please sign the honor code statement on the next page.
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Honor code statement:
In signing, I acknowledge my responsibility in the maintenance of the honor code.
In preparing for and in completing this exam, I have adhered fully to the
requirements of the honor code of Stanford University
__________________
Signature
Point allocation (for graders only)
Q 1 ______________/20
Q 5 ______________/20
Q 2 ______________/20
Q 6 ______________/12
Q 3 ______________/20
Q 7 ______________/12
Q 4 ______________/12
Q 8 ______________/12
Total ______________/128
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QUESTION 1 (20 pts)
1. Comparing the brain representation of the senses of vision, olfaction and taste,
what is a common principle? (1 pt) Which two senses are more similar to each other
and why? (3 pts)
examples of possible answers
common: e.g. a spatial map is used to encode information.
similar senses: olfaction and taste, b/c both use spatially distinct maps
as opposed to a gradient fashion seen in the visual system
0 point for talking about signal transduction in sensory neurons
(note "brain representation"!)
2. Charles likes to knock out genes in mice and test the effects of their taste
preference (as measured by relative licking rate in two bottle choice assay as shown
in the figure below). When he knocked out the gene encoding a signal transduction
component expressed in taste cells, phospholipase C (PLC2), he found the
following results.
(wild-type (white) vs. knockout (dark gray). Hint: AceK/Sac have the similar taste
as glucose; PROP/Den/SOA tastes bitter).
What are the two ways of explaining the result of this experiment? (6 pts)
1. Sweet+umami+bitter and sour+salty are encoded by distinct sets
of cells, former requring PLCB2.
2. All 5 modalities are encoded by same cells but only sweet, umami
and bitter transduction requires PLCB2.
6 points if explicitly said that sweet, umami and bitter require PLCB2
BUT salty and sour do not.
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3. To further understand the organizational principle, Charles introduced into the
PLC2 knockout mice a transgene expressing PLC2 under the control of a bitter
receptor promoter and tested the taste choices (rescue: light grey bars in the figure
above). What conclusions can you make from this rescue experiment? (3 pts)
2 points for mentioning that bitter is encoded in different CELLS than
sweet or umami, just 1 point if said bitter is independently coded from
sweet or umami.
1 point for mentioning that many bitter receptors are likely co-expressed
with each other (note that the promoter of one bitter receptor gene rescued
all three bitter tastes tested)
4. In a separate experiment (unrelated to the above figure), Charles identified a
human taste receptor for a novel ligand X that does not taste anything to mice (i.e.
the mice drink equally from pure water and X-containing water). He generated a
first strain of transgenic mice that express this human receptor using a bitter
receptor promoter. Mice strongly prefer pure water over X-containing water. He
then generated a second strain of transgenic mice that express receptor X using a
sweet receptor promoter. Mice prefer X-containing water than pure water. What
can you learn from these two experiments? (4 pts). For fun, Charles then created
mice that have one copy each of the above two transgenes. Which water do you
think the mice will choose -- pure water or X-containing water? Explain your
prediction. (3 pts)
It is the activation of sweet or bitter taste cells, rather than specific receptors that determine the
taste behavior. Or the signal transduction pathway for the ligand X receptor is the same as
sweet and bitter receptors. Other logical answers were accepted. For fun: All logical answers
were accepted. For example the mice might choose the pure water because evolutionarily bitter
has been associated with death. Even though X-containing water signals both sweet and bitter,
the bitter “wins” out and the mice avoid the X-containing water.
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QUESTION 2 (20 pts)
Inspired by Roger Sperry’s eye rotation experiments, you decided to search for the
“chemoaffinity tags” that might mediate cell-cell recognition during synapse
formation. You decided to use a simple organism, C. elegans to explore this
question.
In wild type animals, you found there was one neuron A, which forms synapses on
muscle B. The axon of A also contacts muscles C and D, but does not form synaptic
connections onto either C or D. (Refer to Figure I.) You hypothesize that there are
molecules on A that can distinguish B, C and D in froming synaptic connections. In
preliminary experiments, you found defects in synaptic patterning in two types of
animals. The results are described and shown below:
(Figure II.) In animals II, cell B is displaced so that B does not contact A and A
forms synapses on D.
(Figure III.) In animal III, cell B is eliminated and A forms synapses on D.
Stars represent synapses.
1. What does this tell you about B’s role in forming a presynaptic compartment in
cell A? (1 pt) What type of cue does cell B most likely possess (diffusible or
membrane attached)? Why? (2 pt)
Cell B is likely to play an inductive role or maintenance role in the formation of a
presynaptic compartment. Cell B likely possesses a membrane-attached cue because
when cell B is displaced from cell A (II), a presynaptic compartment is not formed near
cell B.
2. In a genetic screen, you uncovered a number of mutants in which A forms
synapses on D instead of B. What neuronal developmental processes could have
been affected by these mutations? List three. (6 pts)
Many answers were accepted as long as they were logical such as:
axon guidance of A (so it misses B)
cell fate determination of A or B (so that the correct membrane tags are not expressed)
synapse formation of A or B (mutations in membrane molecules themselves)
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cell fate in D (gain of function, mutation results in cell fate transformation)
synaptic specificity in D (gain of function membrane molecule)
3. You went ahead and cloned two of these mutants, x and y. You found that X and
Y encode two membrane molecules that have characteristics of adhesion molecules.
What are the key experiments to further understand the function of X and Y in A-B
synapse formation? List three. (6 pts)
Any of the following or other logical and relevant experiments:
Determine the cellular and/or subcellular localization of X and Y
Determine cell autonomy using cell specific rescue (In an X (or Y) mutant background,
express X (or Y) specifically in A (or B) to see if the synapse is formed.)
Assay whether X and Y interact through yeast two-hybrid, immunoprecipitation, or cellaggregation.
Structure function analysis of proteins X and Y
Note: Rescue experiments that were not cell specific were not accepted. Loss-of-function
experiments that were not cell specific were not accepted.
4. Assume that X functions in neuron A and Y functions in muscle B, how can you
show that the potential X-Y interaction is sufficient to initiate synapse formation? (3
pts)
Full credit was given to experiments that ectopically expressed X and Y in cells not A and
not B (that do not normally form synapses) to see if synapses were formed. 2 points were
given for experiments that just expressed Y in cell C.
5. In the absence of B, A forms synapses with D but not C. What did we learn from
this result about how specific synapses are formed? (2 pts)
There is a hierarchy of synapse formation. If the wild type synaptic partner is not present
then synapses are formed specifically with a secondary target.
QUESTION 3 (20 pts)
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You are investigating the cellular mechanisms of temperature sensation in mammals.
Based on previous studies, you hypothesize that TRP channels may mediate temperature
sensitivity. To test this hypothesis you clone many of the rat TRP channels and express
each one individually in HEK293T cells, which lack TRP channels. You can then
characterize each channel’s response to changes in temperature.
You record from a TRP-expressing cell using a whole-cell patch clamp configuration (the
electrode is continuous with the inside of the cell). Below are shown the results of an
experiment where you recorded the current (I) at a range of different temperatures over
time. You conducted this experiment while holding the membrane potential at +100mV or
–80mV. The x axis represents time.
(a) Is this heat sensitive or cold sensitive channel? (2 pts)
The channel opens at “colder” temperatures around 20 C, so the answer we were looking for
was cold-sensitive. However, if you put heat sensitive and said that the channel closes at colder
temperatures, that was also acceptable.
(b) Briefly explain why the TRP current is positive when the cell is held at +100mV, but
negative at –80mV. (5 pts)
At +100 mV, this is above the equilibrium potential for all cations, so therefore, cations would
be moving OUT of the cell, and cations moving out of the cell is a positive current (Lecture
Synapse I Slide 16 illustrates this). (2.5 pts)
At -80 mV, this is below the equilibrium potential for all cations, so therefore ions would be
moving IN to the cell, and cations moving into the cell is a negative current. (2.5 pts)
(c) How would an increase in extracellular K+ affect the membrane potential under
normal conditions? (3 pts) How would an increase in extracellular Cl- affect
membrane potential? (3 pts)
At resting membrane potential, increasing the extracellular K+ concentration would decrease
the K+ gradient moving out of the cell, therefore more K+ ions would remain in the cell, and
this would depolarize the membrane potential (more + in cell, make it larger).
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At resting membrane potential, increasing the Cl- concentration would increase the Cl- gradient
moving into the cell, therefore more Cl- ions would be in the cell, hyperpolarizing the membrane
potential (more - in cell, making it smaller).
(d) Lastly, you want to determine which cations pass through this TRP channel. Describe
an experiment to differentiate between the permeability of the channel to Ca2+, Na+ and K+.
(6 pts)
Many experiments were possible. You could express the TRP channel in Xenopus oocytes and
clamp the voltage at the Ca2+, Na+ and K+ equilibrium potential and see at what which one of
them there is no current passing through the TRP channel. You could express the TRP channel,
and then only have one of the ions present in solution and see which one would pass a current
when the TRP channel is activated. You could look at calcium imaging to see if Ca2+ is passing
through the channel. 2 pts for each discriminating between each of the ions.
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QUESTION 4 (12 pts)
(a) Among primates, humans, apes, old world monkeys (OWM), and the Howler
monkey (a species of new world monkeys) have full trichromatic vision. This means
that they have three distinct loci for opsin genes. On the other hand, all new world
monkeys (NWM) except Howler do not possess full trichromatic vision:
heterozygous females can be fully trichromatic, but all males are dichromatic. Full
trichromatic vision is believed to have evolved through a gene duplication event of
opsin genes. Based on this, label (with an X) in the phylogenetic tree shown below
where you think the gene duplication event(s) took place that led to full trichromatic
vision during evolution. (4 pts)
see attached (arrows are used instead of X's)
2 points for each “X”
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Olfactory receptor (OR) genes are encoded by the largest gene family in the
mammalian genomes. As shown in the figure below, a large proportion of OR genes
are pseudogenes. For example, ~51% of human OR genes are pseudogenes.
(b) What is a pseudogene? (3 pts)
A sequence of DNA similar to a gene but nonfunctional; probably the remnant of a oncefunctional gene that accumulated mutations.
-1 point for saying no protein/mRNA is expressed (it's just non-functional,
could be expressed-- see olfaction lecture)
(c) Explain briefly the relationship between the proportion of OR pseudogens and
full trichromatic vision. (3 pts)
Full trichromatic vision = higher proportion of OR pseudogenes
(d) During evolution, which do you think happened first, change in the proportion
of OR pseudogenes, or emergence of full trichromatic vision? Why? (2 pts)
Full trichromatic vision. Presumably acquisition of full trichromatic vision allowed
deterioration of the sense of smell, resulting in the increase of OR pseudogenes. If
deterioration of the sense of smell had occurred first, it would have had a catastrophic
effect on the species' survival
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QUESTION 5 (20 pts)
1. Ciliary and rhabdom photoreceptors represent two evolutionarily distinct cell
types involved in light detection. As result of their independent origins, they employ
a number of molecularly distinct mechanisms to convert the photoisomerization of
11 cis-retinal into a change in membrane potential. Outline three such differences
(4 pts).
There are quite a few differences (all summarized on one slide in my lecture): among
them: different G proteins (Gq vs transducin), hyperpolarizing versus depolarizing, TRP
channel versus CNG channel, PLC versus PDE etc…
2 points for first correct answer
1 point for each following correct answer
2. As the transduction cascades in these two photoreceptor types are so
different, you wonder how the electrophysiological properties of the two
photoreceptor types might differ. You begin by recording the dark noise
signal in Drosophila photoreceptors (just as Denis Baylor did with
vertebrate rods), and compare it to the single photon response. You
observe the following:
What does this result tell you about the discrete noise component seen in
normal (WT) flies? (2 pts) How does this differ from what Baylor observed
in vertebrate rods? (1 pt)
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The key difference is that the discrete noise component is smaller than the single photon
response. Therefore the noise source must reflect spontaneous activation of a
downstream component, rather than thermal isomerization of retinal leading to
rhodopsin activation. In vertebrate rods, discrete noise amplitude is the same as a single
photon response.
This discrete noise in flies is G protein dependent; since rhodopsin activation is excluded,
spontaneous G protein activation is the most likely source.
In this question you were asked to compare the amplitude of the discrete dark noise (WT
dark) to the single photon response (WT light)
0.5 points for mentioning that the amplitudes are of different sizes
1 point for mentioning that thus, the discrete dark noise in flies is not caused by the
isomerization of 11-cis retinal
0.5 points for suggesting a potential downstream component that could be activated in
the WT dark noise
In the second question you were asked to compare WT dark & WT light for flies vs. WT
dark and WT light for vertebrates.
2 points for saying that in vertebrates, dark noise is the same as a single photon response
(isomerization)
0 points for making a comparison between frequency of dark noise ( you were asked to
consider all 4 situations, not just 2)
0 points for talking about the Baylor experiment where light saturation eliminated all
noise in vertebrate rods
3. Gaq1 is a mutation in the promoter region of the alpha subunit of a
particular G protein, Gq, that only causes the amount Gq protein to be
reduced by approximately 100-fold compared to the normal level. All other
transduction components are unaffected. The lowest trace in this figure
records the dark noise signal in Gaq1 mutant photoreceptors. What does
this observation tell you about the source of the discrete dark noise? (2
pts).
In this mutant, there are lower levels of Gq protein and lower amounts of dark noise.
This means the dark noise probably comes from spontaneous activity of the G-protein.
Encouraged by this work, you go on examine the light-induced responses
of normal and Gaq1 photoreceptors, administering brief light flashes to
each, and recording the responses of a large population of photoreceptors
(panels A and B), as well as the responses of single photoreceptors (panels
C and D, displaying only single photon responses). You observe the
following:
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What are the two key differences between the light-induced responses of
normal (WT) and Gaq1 mutant photoreceptors? (2 pts) Suggest a molecular
explanation for these differences that takes into account the nature of the
Gaq1 mutation? (6 pts) Suggest an experimental test of one aspect of your
model. (2 pts)
Amplitude of the response is greatly reduced, and the response itself is greatly prolonged.
Mechanism: amplitude is determined in part by how many G proteins each receptor can
activate before being shut off: if there aren’t as many Gq molecules around, each
activated rhodopsin is going to activate fewer G proteins, reducing the size of the peak.
Explaining the increased duration of the response requires knowing that inactivation
mechanisms are activation dependent. That is, when the G protein level is reduced, a
photo-activated rhodopsin might not be able to find even one G protein for some time.
Until it does, it can’t be switched off – so it’s a bit like a time bomb just waiting for a G
protein to show up.
The source of the dark noise must be spontaneous activation of Gaq or something
upstream, but not isomerization of retinal (2 points).
In terms of experiments, it sort of depends on the model proposed, and on the style of
experiment (genetic, cell biological, imaging a fret signal, biochemical, whatever). I
think that the only thing that absolutely shouldn’t be accepted is a Gq rescue experiment
in the mutant, since we already know what the answer is. Almost anything else could be
appropriate, depending on how it’s argued.
Mutant has lower amplitude and slower termination of population response (2 points). As
Gaq is less abundant in the mutant, each rhodopsin activates fewer G-proteins,
generating a smaller evoked current (3 points). Termination of the light-evoked response
is delayed because some rhodopsins are delayed in activating Gaq. Since rhodopsin
inactivation depends on rhodopsin first being activated, these late activations contribute
the population response (3 points).
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QUESTION 6 (12 pts)
You are a graduate student in Eric Kandel’s lab studying the molecular basis of
learning in Aplysia. You use a reduced cell culture preparation in which a sensory
neuron (SN) synapses onto two different motor neurons (MN).
You find a peptide, FMRFamide (FMRFa), which modulates the strength of the SNMN synapse. Shown below is data from an experiment in which you puff FMRFa
or serotonin (5HT) locally onto the individual synapse depicted. 24 hours later you
record the change in EPSP amplitude in motor neurons (Y axis).
1. Focus on panel A. What is the effect of 5HT on synaptic strength? What is the
effect of FMRFa? (2 pts) Briefly describe four general mechanisms by which
FMRFa may be affecting synaptic strength. (4 pts)
5-HT strengthens/facilitates the synapse (1point). FMRFa weakens/inhibits it
point).
(1
Any plausible mechanisms get credit (4 points): 1) decrease release probability 2)
decrease # release sites 3) decrease # vesicles 4) decrease receptor sensitivity 5)
decrease # functional receptors
2. You also puff 5HT and FMRFa simultaneously onto different synapses (Panel B).
EPSP amplitude is measured as before. You note that FMRFa application at one
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SN-MN syanpse changes the response of the other synapse to 5HT. You guess that
FMRFa binding to its receptor on SNs initiates these synaptic changes. You have at
your disposal antibodies to the FMRFa receptor which can bind to and inactivate
the receptor. You inject them into the SN. But when you repeat the experiments
depicted in Panel A and B, the results are unchanged. Propose a model for FMRFa
function consistent with these results. (4 pts) Design an experiment to test your
model. (2 pts)
FMRFa binds to its receptor on the motor neuron (2 points). This causes the motor
neuron to send a retrograde signal to the sensory neuron (1 point). This signal inhibits
the response of other synapses to 5HT (1 point). You did not receive points for suggesting
that FMRFa binds to intracellular signaling machinery (e.g. CamKII) in the SN. As a
peptide FMRFa is unlikely to enter the cell.
The easiest experiment is to inject FMRFa receptor blocking antibodies into motor
neurons and measure plasticity. Another possibility is to transfer medium containing
retrograde messenger to naïve dissociated prep. (2 points)
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QUESTION 7 (12 pts)
1. List 2 lines of evidence that supports that synaptic transmission requires calcium.
(4 pts)
Many answers were possible; however, it must tie calcium to synaptic transmission.
Saying that calcium is an important signaling molecule, or saying that calcium is
important for LTP would not qualify.
1. If you hold calcium at its equilibrium potential, a post-synaptic potential does not
occur.
2. An increase in calcium concentrations is sufficient for neurotransmitter release. 3.
Synaptotagmin is a calcium sensor, and modulates neurotransmission.
2. While searching the mouse genome, you identify this very interesting protein which has
homology to a C. elegans protein, abc-1. Abc-1 has been previously shown to be involved
in synapse function. You wish to determine if this putative mouse homolog, mabc-1, has
any effect on synapses. You decide to go ahead and make the loss-of-function mouse
mutant. The mouse does not exhibit gross morphological defects of motor neurons or
muscles. To characterize this mouse mutant, you decide to inject aldicarb, an
acetylcholinesterase inhibitor, into the mouse muscle. This causes hypercontraction of the
muscle in wild type animals, but in mabc-1 knockout mice, you don’t observe
hypercontraction of the muscle. Interpret this result. List 5 possible explanations where
mabc-1 could be acting in synaptic transmission. (5 pts)
Acetylcholinesterase normally degrades acetylcholine in the synaptic cleft. When
aldicarb, an acetylcholinesterase inhibitor is applied, that causes an excess of
acetycholine at the synapse, which overstimulates the neuromuscular junction and causes
hypercontraction of the muscle.
This mouse is resistant to aldicarb, meaning that some part of the synaptic transmission
pathway is affected and synaptic transmission is not occurring, or is weakened.
Essentially, any part of the synaptic transmission pathway could be affected. Here are
some examples.
1.
2.
3.
4.
5.
6.
Could be post synaptic receptor.
No production of neurotransmitter.
Transmitter is produced, but not been localized correctly.
Could have problems in fusion or exocytosis..
Could be a problem with the calcium sensor.
Could have a problem with endocytosis
Reuptake of acetycholine was not accepted because acetycholine is DEGRADED, and is
not re-uptaken.
3. You decide to do further characterization by injecting levimasole, an acetylcholine
receptor agonist, into the mouse muscle. This also normally causes hypercontraction of the
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muscle in wild type animals. However, you also don’t observe hypercontraction of the
muscle in your mabc-1 knockout mouse. Where is mabc-1 most likely acting in the process
of synaptic transmission? Design one experiment, which could test your model. (4 pts)
Because the mouse is resistant to levimasole, that tells you that it is most likely acting on
the post synaptic site, because if the problem was presynaptically or in the synaptic cleft,
the animal should be affected by levimasole.. It could be a post synaptic receptor, or
something that is associated with a post synaptic receptor. You could test if it is an
ionotropic post synaptic receptor by expressing it in Xenopus oocytes, and see if current
is passed when you throw on different neurotransmitters. If it is a metabotropic
receptor, you could see if the G protein associated with muscarinic acetycholine
receptors, Gi, associates with your gene. If it is a protein that is associated with the post
synaptic receptor, you could do a co-immunoprecipitation and see if the protein
associates with the post synaptic receptor.
QUESTION 8 (12 pts)
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1. We learned that the mammalian SCN (suprachiasmatic nucleus) has an intrinsic
rhythm that is responsible for mediating the circadian rhythm in mice. Explain
how the SCN is entrained to natural light/dark cycles. (2 sentence max) (2 pts)
2 points for explaining the role of melanopsin in RGCs in light entrainment.
1 point for good explanation that neglected to mention melanopsin
2. Organisms as diverse as mammals, insects, plants, and yeast have circadian
rhythms that can be entrained to natural light/dark cycles; however, pulses of light
during the dark half of a cycle can phase shift circadian rhythms. When Drosophila
is entrained to natural light/dark cycles, it is most active at dawn (morning) and
dusk (evening). Below is a histogram of locomotor activity for Drosophila. If wildtype or mutant flies are exposed to a pulse of light at the given point (asterisk),
answer the following questions about how behavior will change.
For each of the following flies, at what point (A, B, or C) would you expect the peak
of the next period of activity to occur? Why? (2 pts each)
wild-type:
cry (cryptochrome mutant):
tim (timeless null mutant):
wild-type: C because the pulse of light causes a phase delay. The pulse causes
degradation of tim which resets the evening rhythm, effectively bringing it back to the
beginning of night. This will result in a shift whereby evening onset is later and “night”
finishes later.
Cry: B: cry(baby) mutants are insensitive to light pulses and therefore do not shift their
rhythm in response to a pulse.
1 point for A, B, or C, one point for a correct explanation. If A was called a phase delay,
0 points.
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3. A group of scientists interested in circadian rhythms showed that subsets of
neurons in Drosophila control morning and evening activity. Below is a figure that
shows locomotor behavior of light-dark (LD) -adapted flies over a 24 hour period in
DD (constant darkness) conditions. White represents subjective day and grey
represents subjective night.
Explain how ablation of both C & P subsets of neurons affects the circadian rhythm.
(1 pt)
C & P are important for setting circadian rhythem and C & P ablation causes
arrhythmic flies.
4. When only the P subset of neurons is ablated, the evening part of the circadian
rhythm is restored (histogram on the far right). Further experiments showed that
ablation of the C neurons resulted in restoration of the morning rhythm. Draw a
diagram of the expected locomotor activity of flies in the C ablation experiment.
What does this result imply about the role of C and P neurons in establishing the
circadian rhythm in Drosophila? (2 pts)
1 point: Diagram of a peak only in the morning.
1 point: C controls evening rhythm, P controls morning rhythm.
5. Ablation of C and P neurons is accomplished by inducing cell death of these
neurons. You can imagine that the loss of these neurons may have unexpected
effects on development of the brain or brain function. In the course of making these
genetic cell ablation constructs, it was found that many genes involved in the
circadian clock are expressed in C and P neurons. Describe a more elegant set of
genetic experiments that would directly test the importance of a circadian rhythm in
these neurons. (3 pts)
3 points: Shibire expression in P and C subsets of neurons will eliminate circadian
rhythms in a temperature-sensitive fashion.
Other cre-lox, Flp, Frt also accepted – the idea was to eliminate circadian rhythm genes
in only the P and C neurons.
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2 points: - MARCM- this would result in a subset of neurons being homozygous and in
some cases, it may be possible for a small number to be sufficient for the behavior. If a
small number of neurons were sufficient, these animals would not display a phenotype.
- LOF (loss of function) and rescue in C and P neurons – LOF in the whole animal may
mess up a lot of things.
1 point: microarray, antibodies, in situ hybridization – these experiments will only
provide correlation data.
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