Download 1 Bi/CNS/NB 150 Problem Set 4 Due: Tuesday, Nov. 17, at 4:30 pm

Bi/CNS/NB 150
Problem Set 4
Due: Tuesday, Nov. 17, at 4:30 pm
Instructions:
1) Drop off in the Bi 150 box outside Baxter 331 or e-mail to the head TA (jcolas).
2) Submit with this cover page.
3) Use a separate sheet of paper for each problem.
4) Type all answers if possible.
5) Use complete, grammatically correct sentences.
6) Include your name and the page number on every page.
7) Note that late problem sets receive a 10% deduction for every day past the due date.
Name:
Time and date submitted:
Total pages (including cover page):
Comments:
Problem 1 grade:
Problem 1 comments:
Problem 2 grade:
Problem 2 comments:
Problem 3 grade:
Problem 3 comments:
Total grade:
1
Problem 1 (1.5 points): G-protein-coupled receptors
Problem 1.A (0.6 points): Mutants and toxins
1.A.a. G-protein-coupled receptors (GPCRs) activate G proteins by reducing the
strength of GDP binding. This results in rapid dissociation of bound GDP that is then
replaced by GTP, which is present in the cytosol in much higher concentrations than
GDP. What consequences would result from a mutation in the αs subunit of a G protein
that caused its affinity for GDP to be reduced without significantly changing its affinity for
GTP?
The mutant G protein would be activated almost continuously, as GDP would
spontaneously dissociate and allow GTP to bind even in the absence of an activated Gprotein-coupled receptor. (0.2 pt)
1.A.b. Compare the effects of this mutation with the effects of the cholera toxin.
The consequences for the aforementioned mutation would be similar to those caused by
the cholera toxin, which modifies the α subunit of Gs so that it cannot hydrolyze GTP to
shut itself off. In contrast to the cholera toxin, however, the mutant G protein would not
stay permanently activated. Rather, it would switch itself off normally but then instantly
become activated again as GDP dissociated and GTP re-bound. (0.2 pt)
1.A.c. Describe how the mechanism of the pertussis toxin differs from the mechanism
of the cholera toxin on GPCRs. Describe how several subsequent actions cause
prolonged neuronal activation.
The pertussis toxin alters the α subunit of Gi, a different type of G protein. The toxin
locks the G protein into its inactive GDP-bound state. This prevents activation of K+
channels, inhibition of Ca2+ channels, and inhibition of adenylyl cyclase. All of these
effects of the pertussis toxin would excite neurons. (0.2 pt)
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Problem 1.B (0.4 points): Pathways
1.B.a. You have ascribed a G-protein pathway to two events. When you stimulate the
presynaptic neuron while activating the GPCR on the presynaptic neuron’s membrane,
more neurotransmitter is released. You observe phosphorylation of a K+ channel in the
presynaptic neuron. Explain your hypothesis for linking these two results.
Synaptic action at presynaptic terminals can activate second messengers through a
GPCR pathway that activates protein kinases, which can regulate presynaptic K+ and
Ca2+ channels. Phosphorylation of a K+ channel in this case likely closes the channel
since an increase in neurotransmitter release is observed. Activation of voltage-gated
Ca2+ channels may also result in an increase of intracellular Ca2+, which can interact
with synaptotagmin to increase neurotransmitter release. (0.2 pt)
1.B.b. You have ascribed another G-protein pathway to two events. When you
stimulate the pathway for several days, the target neuron expresses 36 genes more
strongly than in the absence of stimulation. You find that the receptor activates Gq.
Explain your hypothesis for linking these two results.
Gq activates PLCß with concomitant production of DAG and IP3 leading to activation of
PKC and Ca2+ signaling. Increased Ca2+ in the cytosol activates a kinase, which
phosphorylates CREB. pCREB is a transcription factor and thus can induce synthesis
of new proteins that alter the cell’s gene expression. (0.2 pt)
Problem 1.C (0.5 points): Inwardly rectifying potassium channels
The phenotype of Down syndrome may be a consequence of overexpressed genes in
an extra copy of chromosome 21. One such gene is Kcnj6/Girk2, which encodes Gprotein-coupled inwardly rectifying K+ channel subunit 2 (GIRK2). The DS mouse
model, Ts65Dn, overexpresses GIRK2 throughout the brain and particularly in the
hippocampus. Increased expression of GIRK2-containing channels has functional
consequences that likely affect the balance between excitatory and inhibitory neuronal
transmission.
1.C.a. Beginning with appearance of GABA near GABAB receptors, describe how GIRK
current is activated in a neuron.
Agonists bind to the G-protein-coupled receptor, which is GABAB in this case. The
activation leads to the exchange of GDP for GTP in the α subunit followed by
dissociation of the β and γ subunits from the α subunit of the heterotrimeric Gi protein
complex. The βγ complex can then activate GIRK channels. (Note that GABAB
receptors are unique among GPCRs in that signaling requires the dimerization of two
different GABAB subunits, B1 and B2.) (0.1 pt)
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1.C.b. What role do GIRK channels play in neuronal function?
Synaptic activation of GABAB receptors in the dendrites of neurons produces a slow
inhibitory postsynaptic potential that is carried by GIRK channels. Activation of GIRK
channels results in K+ conductance that holds the membrane potential near EK, thus
reducing membrane excitability and inhibiting firing of action potentials. (0.1 pt)
1.C.c. RGS4 encodes a GTPase-activating protein. The current trace below comes
from a voltage-clamp experiment conducted at a holding potential of -80 mV with
extracellular and intracellular K+ concentrations of 30 mM and 140 mM, respectively.
What effect do RGS proteins have on the activity of GIRK channels? Redraw the trace
below and then superimpose the trace produced if the cell were to express RGS4.
RGS proteins accelerate the hydrolysis of GTP. Therefore, RGS proteins terminate
action of GABAB-activated GIRK currents. The trace returns to baseline more quickly.
(There should be less maximal current as well, but this is not required for full credit.)
(0.15 pt)
1.C.d. Describe the effect on GIRK activity if a non-hydrolyzable GTP analog, GTPγS,
were injected into the cell. What effect would this have on the activity of GIRK
channels? Redraw the trace above and then superimpose the trace produced if GTPγS
were present. Also consider changes that could occur before GABA is applied.
GTPγS exchanges for GDP on the Ga subunit. Because GTPγS cannot be hydrolyzed,
Ga remains activated much longer after the receptor is activated. Perhaps GTPγS
would also allow the βγ complex to constitutively activate GIRK channels. The trace
returns to baseline more slowly. (The baseline should start lower as well, but this is not
required for full credit.) (0.15 pt)
4
Problem 2 (1.5 points): Visual system
Problem 2.A (0.6 points): Impairments
Consider each of the following lesions and mutations:
1)
2)
3)
4)
5)
6)
Selective and complete loss of rod cells in both retinae
Bilateral and complete lesions of visual area V4
Lesion of the lower bank of the calcarine sulcus in the left hemisphere
Selective loss of function of parvocellular ganglion cells in both retinae
Selective and complete loss of cone cells in both retinae
Complete sagittal transection of the optic chiasm cutting only those fibers that
would normally cross there
7) Bilateral lesions in posterior parietal cortex
8) Mutation of retinal cGMP-gated channels causing selective impermeability to Ca2+
9) Selective loss of function of magnocellular ganglion cells in both retinae
10) Mutation of retinal cGMP-gated channels causing selective impermeability to Na+
11) Lesion of the optic radiation fibers that curve into the temporal lobe in the left
hemisphere
Match each of the following visual impairments with a single lesion or mutation in
the visual system from the list above that could be responsible for the impairment.
No lesion or mutation should be provided as an answer twice. Explain how the
given lesion or mutation produces the specified visual impairment.
2.A.a. Complete loss of vision only in the temporal halves of both visual fields
6) Complete sagittal transection of the optic chiasm cutting only those fibers that would
normally cross there
Fibers crossing over in the optic chiasm receive inputs from the nasal half of the retina.
The nasal half of the retina receives stimuli from temporal visual field.
(0.05 pt for each match and 0.05 pt for each explanation)
2.A.b. Complete loss of vision only in the superior quadrant of the visual field on the
right side and with the central visual field unaffected
3) Lesion of the lower bank of the calcarine sulcus in the left hemisphere
The lower bank carries information from the upper visual hemifield. Visual cortex also
encodes contralateral representations. There is an overrepresentation of the fovea due
to cortical magnification.
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2.A.c. Inability to perceive a pattern of bright and dark bars that have both low
spatial frequency and high temporal frequency
9) Selective loss of function of magnocellular ganglion cells in both retinae
Magnocellular (M) cells are able to respond to stimuli with low spatial resolution and high
temporal resolution. Parvocellular (P) cells would not be malfunctioning because color
contrast and high spatial frequencies are unaffected.
2.A.d. Inability to see or dream in color
2) Bilateral and complete lesions of visual area V4
V4 is a relatively high-level region of visual cortex along the ventral stream that is
necessary for perception of color.
2.A.e. Difficulty to adapt to different light intensities quickly
8) Mutation of retinal cGMP-gated channels causing selective impermeability to Ca2+
Ca2+ inhibits guanylyl cyclase. Lowering the Ca2+ concentration speeds up the
inactivation of visual pigments, such that the effectiveness of a given light flash for
activating cGMP phosphodiesterase is reduced.
2.A.f. Inability to reach for visual objects normally
7) Bilateral lesions in posterior parietal cortex
Posterior parietal cortex is part of the dorsal “where” pathway that processes the spatial
locations of objects.
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Problem 2.B (0.9 points): Simple and complex cells
2.B.a. A simple cell responds to bars of light with a specific orientation. Draw a network
of on- or off-center ganglion-cell inputs to a simple cell that would enable it to respond to
this stimulus. Draw the receptive fields of each ganglion cell needed and the spatial
relationships of these receptive fields. Explain how your model works.
A simple cell would receive convergent excitatory inputs from three or more on-center
cells that together represent light falling along a straight line in the retina. (0.3 pt)
2.B.b. A complex cell also responds to bars of light with a specific orientation.
However, unlike the simple cell, the receptive field of a complex cell is generally large
enough that the position of bars of light in the receptive field is not critical to evoke the
response in the neuron. These properties of complex cells are derived from their
inputs from multiple simple cells. Draw a network of multiple simple cells with different
receptive fields that all connect to a complex cell and enable it to have a larger
receptive field with orientation tuning. Explain how your model works.
A complex cell receives convergent excitatory inputs from simple cells, each of which
has a receptive field with similar organization and orientation tuning. (0.3 pt)
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2.B.c. Furthermore, a complex cell can respond selectively to unidirectional movement
across its receptive field in a specific orientation (e.g., a horizontal bar of light moving
upwards). This could also be explained by the connectivity between multiple simple
cells and one complex cell. In the network shown below, three simple cells with
different receptive fields form synapses onto a complex cell at different locations along
a single dendrite. This complex cell is most responsive to a bar of light moving from
the receptive field of simple cell 1 through the receptive field of simple cell 2 to the
receptive field of simple cell 3 but not as responsive to a bar of light moving in the
opposite direction. Explain how this directional specificity might be achieved.
This is a case of temporal summation. If a bar of light is moving from the receptive field
of simple cell 1 through the receptive field of simple cell 2 to the receptive field of
simple cell 3, it evokes action potentials in cells 1, 2, and 3 sequentially. As a result,
EPSPs are generated at different locations along the dendrite of the complex cell.
These three EPSPs travel along the dendrite towards the axon hillock of the complex
cell, where they are summed and potentially converted into an action potential. Cell 1
is farthest from the axon hillock of the complex cell, and cell 3 is closest to it. A 1-2-3
stimulus may result in nearly simultaneous arrival of three EPSPs, which leads to
greater summation. In contrast, a 3-2-1 stimulus would produce lesser summation.
(0.3 pt)
8
Problem 3 (1 point): Olfactory system
Problem 3.A (0.2 points): Anatomy
Beginning with the nose, describe the major excitatory synaptic connections in the main
olfactory pathway of mammals.
ORNs in the olfactory epithelium within the nose synapse onto mitral and tufted cells in
the glomeruli within the olfactory bulb as well as periglomerular neurons. These
neurons in turn project to the anterior olfactory nucleus, piriform cortex, the amygdala,
entorhinal cortex, and the olfactory tubercle.
Problem 3.B (0.8 points): Olfactory receptor neurons
3.B.a. Describe the intracellular signaling pathway in olfactory receptor neurons
(ORNs) from odorant stimulation to action potentials. Explain the role of each
molecule involved in this signaling cascade.
Binding of an odorant causes the ORN to interact with Gαolf, the α subunit of a
heterotrimeric G protein. This interaction releases a GTP-coupled Gαolf, which
stimulates adenylyl cyclase. Adenylyl cyclase induces the synthesis of cAMP from
ATP. cAMP-gated cation channels on the membranes of ORNs are opened with
increased concentrations of cAMP. The cation influx depolarizes the membrane
and induces action potentials with sufficient depolarization. (0.2 pt)
3.B.b. Individual odorant receptors show relatively low affinity for a range of
ligands. What would be the selective advantage of this broad spectrum of ligand
binding if two different concentrations of the same odorant were applied
sequentially?
If instead ORNs had very high affinity and were fully activated by low concentration of
odorants, they would not exhibit differential responses to higher concentrations of
odorants because activation would already be saturated. Thus, low-affinity binding is
important for supporting differential responses to different concentrations of odorants.
(0.2 pt)
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3.B.c. What would be the selective advantage of this broad spectrum of ligand
binding if two similar odorants were applied sequentially?
Humans have approximately 350 different odorant receptors, whereas mice have
approximately 1000. If each receptor exhibits very specific and strong binding to
odorant molecules, the number of unique odorant receptors limits the number of
odorants that can be differentiated. The broad spectrum of ligand binding can facilitate
the representation of odorant identity because a slight change in odor can be
represented by a change in the firing pattern across multiple neurons. This highdimensional representation enables differentiation of similar odorants. (0.2 pt)
3.B.d. The olfactory systems of most animals feature sensitivity on the order of only a
few odorant molecules. Yet, we become less sensitive to an odor after being exposed to
it for an extended duration. Using your knowledge of GPCR pathways, describe a
possible mechanism in ORNs that might contribute to this phenomenon.
1) Inhibition of cAMP synthesis due to CaMKII-dependent phosphorylation of adenylyl
cyclase
2) Reduced sensitivity of cAMP-gated channels to cAMP due to Ca2+-dependent
phosphorylation of cAMP-gated channels
3) Inactivation of odorant receptors by a cAMP-dependent protein kinase
4) Closing of cAMP-gated cation channels due to increased Ca2+ concentration and
thus inhibition of cAMP
(0.2 pt)
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