Questions for Exam 3 - Columbia University



C2006/F2402 ’10 Key to Exam #3

Each answer and each explanation was 2 pts unless indicated otherwise. Several errors were noticed in the original key. The correct answers are blocked in blue. If you notice any other errors, please email Dr. M.

Links to papers on electric fish & genetically engineered mice (with altered taste) are on the last page.

1. A-1. You would expect to find voltage-gated K+ channels in the electrocytes of (fresh water) electric fish.

A-2. You would expect to find ungated K+ channels in the electrocytes of (both) types of electric fish.

Explanation: (1). Only freshwater fish generate action potentials in their electrocyte membranes. Therefore only freshwater fish will need voltage-gated K+ channels – they’ll need the channels to repolarize after an AP. (2). Both types will need ungated K+ channels (leak channels) as these are needed to generate a RMP. You need a baseline RMP to generate an EPSP (in both types of fish). You need voltage-gated channels for EPSPs to lead to an AP (freshwater fish only). You need voltage-gated Na+ channels for the depolarization phase and voltage-gated K+ channels for the repolarization phase of the AP.

B-1. The esterase should be found in (the synaptic cleft).

B-2. If you inhibit the esterase, the shock delivered by the electric ray will probably be (stronger or longer).

Explanation: Esterase is an enzyme that splits the NT (active) into acetate and choline (both inactive as NTs). The enzyme is found in the synaptic cleft, which is where the NT is. If the enzyme is inhibited, the average NT molecule will last (remain in the cleft) longer. Therefore there will be more NT to bind to the receptors on the post-synaptic side, and the EPSP (& subsequent shock) generated in the electrocyte will be larger.

C-1. The electrocyte is probably derived from which kind of muscle? (skeletal)

C-2. Assume all vertebrates (such as fish and humans) have a similar body plan. Then the cell bodies of the neurons that innervate the electrocytes in fishes should be (in the CNS).

Explanation: The receptor here is a nicotinic acetyl choline receptor. That type of receptor is only found on skeletal muscle.

The neurons that innervate skeletal muscle have their bodies in the spinal cord (CNS) and their axons in the PNS – the axons extend from the CNS to the target muscle. (If you thought the electrocyte was derived from smooth muscle, the cell body would be in a ganglion of the PS. One pt for this.)

2. A. The gene for this enzyme was probably transcribed in the (nucleus).

B. To reach its normal location, the protein would have to pass through (the OMM) (the IMM).

Explanation: Mitochondria contain DNA, so they can make some of their own proteins. However the mito. DNA is very small, and only codes for a few proteins. (Mito. DNA codes for a few highly hydrophobic transmembrane proteins in the IMM. Not for any matrix proteins.) Most of the proteins found in mitochondria are encoded in the nucleus. The genes are transcribed in the nucleus, and the mRNA is translated in the cytoplasm. The ribosomes translating the mRNA are not attached to the ER. The resulting proteins are released into the cytosol, and are then transported into the matrix, crossing both the OMM & IMM.

3. A. How should receptor cells in normal mice respond to sweet compounds?

A-1. The plasma membranes in the receptor cells should (depolarize).

A-2. Voltage-gated Na+ channels in the receptor cells should (neither).

A-3. The level of Ca++ in the cytosol of the receptor cells should _increase__.

Explanation (any two of the three was sufficient):

A-1. Cations are ions with a positive charge; monovalent ions have a charge of one. A channel for monovalent cations will allow both K+ and Na+ to pass through. When the M5 channels open, Na+ will enter the cell and K+ will leave. Both will go down their concentration gradients, but they won’t move to the same extent. Since Na+ is very far from its equilibrium potential, lots of Na+ will enter the cell. Since K+ is very close to its equilibrium potential, not much K+ will exit. Therefore the cell will become more positive inside = depolarized.

A-2. There should not be any voltage-gated Na+ channels in the receptor cells since the cells do not fire action potentials. The receptor cells should have only ligand-gated channels that generate graded (receptor) potentials, and the amount of NT released should be proportional to the receptor potential. The second cell in the afferent pathway, the sensory neuron, should contain voltage-gated Na+ channels because it is the first cell in the pathway that can fire an AP.

A-3. Ca++ levels must be elevated to stimulate exocytosis and release of NT.

B-1. In the axon of the neuron, the interval between spikes should ________decrease______________

B-2. In the axon of the neuron, the size of the graded potential should _______stay the same at 0 (or STS)

B-3. In the dendrites of the neuron, the size of the graded potential should _____increase______________

Explanation (any 2 of the 3 was sufficient):

B-1. More stimulus means more NT released from the receptor cell, meaning more APs in the sensory neuron per sec. A faster rate of firing of APs means less time between spikes.

B-2. There is no graded potential in the axon.

B-3. A graded potential is found on the dendrites, on the post synaptic side of a synapse. The more NT from the receptor cells, the more channels open at the synapse in the sensory cells, and the more depolarization – bigger EPSP.

C-1. The receptor proteins for sweetness are probably (metabotropic) .

C-2. Which of the following proteins probably bind(s) to sweet compounds? (a GPCR).

C-3. Of the following choices, the M5 channel is most likely to be gated by (Ca++).

C-4. If sweetness detection uses a 2nd messenger, the 2nd messenger is most likely to be (IP3).

Explanation:

C-1. The sweet ‘tastant’ binds to the extracellular domain of the T1R protein = GPCR. If the T1R were an ionotropic (direct) receptor, it would be a channel itself, and no additional channel would be required.

C-2. The activated GPCR activates a G protein. The tastant binds to the GPCR on the outside of the cell; all the other substances are intracellular and inaccessible to the ‘first messenger.’

C-4. Since PLC is required, you assume the G protein in turn activates PLC. The PLC splits PIP2 in the membrane, and releases IP3 (& DAG). You know you have activated the IP3 pathway since PLC is required. (The cAMP pathway would require AC.)

C-3. Since IP3 usually opens channels for Ca++ (Ca2+) in the ER, the [Ca++] in the cytoplasm – really the cytosol -- should rise, and that could gate (open) the M5 channel. Alternatively, IP3 itself, or DAG could bind to and open the M5 channel. (These are not listed as options. See * below.) Opening of the channel should depolarize the cell, and that should lead to release of NT. (Perhaps depolarization would open voltage-gated Ca++ channels, leading to a rise in Ca++ and exocytosis.) If the G protein itself bound to and opened the channel, you shouldn’t need PLC. There are more complex possible pathways possible, but these are the obvious ones.

* Of the choices listed, Ca++ is the only reasonable candidate. However Ca++ itself causes exocytosis, and so a rise in Ca++ might be expected to cause release of NT – in which case you wouldn’t need the M5 channel at all. However both the channel and PLC are required. Therefore it seems likely that the IP3 or DAG or some factor ‘downstream’ from them -- something further in the pathway, other than Ca++ -- opens the channel, that leads to depolarization, and that leads to a rise in Ca++ and subsequent NT release. Current research will reveal what actually gates the channel. (It does not appear to be Ca++, although that is a reasonable suspect.)

4. A. If the mice are missing the M5 channel, sensory neurons should fire when the mice drink water containing (lemon juice).

B. If the mice are missing T1R receptors, they will NOT be able to distinguish (water & sugar solution).

Explanation: A. The M5 channel is required for detecting sweet and bitter tastes but not sour (lemon juice). The B. T1R receptors are required to detect sweetness. Mice who have no T1R receptors should not be able to detect sweetness – therefore they shouldn’t be able to tell the difference between a tasteless solution and a sweet one. They can still detect sour and bitter, so they should still be able to tell water from Cyx solution (or from lemon juice).

No points were awarded for the explanation. Anything reasonable was accepted, as this was just a way to be sure you understood the experimental set up. However points were deducted for errors.

C-1. The simplest explanation of these results is that the two types of receptor proteins (T1R & T2R) --

a. use different signaling pathways in their respective receptor cells. No, they use the same pathway – both use the M5 channel and PLCß2.

b. are in cells that synapse on neurons connecting to different cells in the brain. This fits the ‘labeled line’ theory – each receptor is hooked up (by a chain of neurons) to a particular part of the brain. All signals coming in are the same (APs) but it’s where they go in the brain that matters. If the ‘lines’ go to the sweetness center, the mouse registers sweetness. Similarly for bitterness.

c. cause their respective receptor cells to release different NTs. No. They use the same pathway in the same cells.

d. are in cells that synapse on the same neurons. No. If so, how would mice know which receptor was tripped?

C-2. You would expect these genetically engineered mice to:

a. Have a preference for (Cyx solution) AND b. Have an aversion to (neither).

Explanation: Both receptors use the same pathways, so either type of receptor should work in either type of cell. (Presumably they both use the same G protein.) If the T2R receptor binds its ligand, the cell with the receptor will send a signal to the brain. If that cell is the one that usually has T1R receptors, the signal will go to the sweetness center. The mouse will drink Cyx solution, which is normally considered to be bitter, but the mouse’s brain will register sweetness. Therefore the mouse will prefer Cyx solution, just as it normally prefers sugar solution.

If the mouse has no T1R receptors, it will find sugar solution to be tasteless. The mouse will not avoid either type of solution – it will think the Cyx solution is sweet and the sugar solution is tasteless.

5. A-1. Which of the following should be higher in females than in males? (number of spikes per min. in axon).

A-2. When the steroid is added to female fish, which of the following should increase? (time for pacemaker potential to reach threshold).

Explanation: When steroid is added, the pacemaker potential should decrease – its slope should be flatter. Therefore it will take the cells longer to reach threshold (which will be at the same value), and the rate of firing spikes will decrease – the time between spikes will increase.

B-1. The effect of the steroid should be (blocked by inhibitors of transcription) & (blocked by inhibitors of translation). 1 pt each.

B-2. The steroid receptor is most likely to be synthesized on ribosomes that are (in the cytoplasm).

Explanation: Steroids generally bind to receptors that are TFs and the complex affects transcription, leading to changes in translation. If the complex stimulates transcription, causing synthesis of, say, a protein that slows opening of ‘funny channels’, then blocking transcription or translation of the mRNA should reduce the effect. When transcription or translation is blocked, less protein should be made in response to steroid, and there would be less interference with the opening of channels & less effect on the pacemaker potential.*

The receptor should be inside the cell in the nucleus or cytoplasm, so it should be made on free ribosomes. Note that the question asks about the site of synthesis of the receptor protein, not the protein made in response to steroid (which might be a membrane protein made on the ER).

* If the steroid inhibits transcription, say to prevent synthesis of channel protein, then the steroid effect will be mimicked (in part) by blocking transcription or translation. In the female fish, the steroid could inhibit synthesis of a protein to reduce the number of ‘funny channels’ or increase synthesis of a protein that interferes with the opening of the ‘funny channels.’ Either one would decrease the slope of the pacemaker potential.

6.A. The ratio of nerve endings to internal sphincter muscle cells should be (>1) (about 1) (1.

Since a single nerve ending has multiple varicosities that can contact more than one cell, at least according to the class handout, there might be multiple cells innervated by a single nerve ending, and the ratio would be ................
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