Key Old quiz questions about synapses

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Old quiz questions about synapses KEY Note that not all the material in this question smay have been covered in the Winter term 2024 E x = RT zF ∗ ln [ X out ] [ X ¿ ] R= 8.314 J o K -1 mol -1 T is in o K 0 o C = 273 o K F= 96,492 Coulombs mol -1 If temp is not specified E x = (60/z) * log 10 [out]/[in] At 37º C E x = (61.5/z) * log 10 [out]/[in] At 20º C E x = (58/z) * log 10 [out]/[in] For two permeant ions: g a g b = E b − V m V m − E a or V m  = g a g b ∗ E a + E b g a g b + 1 For more than 2 permeant ions V m  = g a ∗ E a + g b E b + ... + g z E z g a + g b + ... + g z E rev = RT F ∗ ln ( [ Cation A out ] ∗ P A + [ Cation B out ] ∗ P B + [ AnionC ¿ ] ∗ P C + … [ Cation A ¿ ] ∗ P A + [ Cation B ¿ ] ∗ P B + [ AnionC out ] ∗ P C + … ) ix=gx(Vm-Ex) g= 1/R V=IR R=  V/  I Q=charge in Coulombs= CV I= dQ/dt icap= C dV/dt f ( t ) = A ∗ exp ( t ) + C If V Final =0, V(t) = V 0 * exp (-t/  ) If V 0 =0, V(t) = V Final *(1- exp (-t/  ) ) V(x) = V 0 * exp (-x/  ) λ = √ g i g m or λ = √ r m r i V(t) = V 0 * exp (-t/  ) V(t) = V Final *(1- exp (-t/  ) )  m = R m *C m For geometrical relationships, use d for diameter (2* radius) rather than r for radius to avoid confusion with resistance For a spherical cell body, Surface area =  d 2 so g m ∝ d 2 and c m ∝ d 2 and r m ∝ 1 d 2 For an axon, Surface area = circumference* length =  dl so g m ∝ d and c m ∝ d and r m ∝ 1 d For an axon, cross sectional area =  d 2 /4 so r i ∝ 1 d 2 therefore λ = √ r m r i ∝ √ d 2 d ∝ √ d INa=gNa(Vm-ENa) IK=gK(Vm-EK) g Na ¿ g Na ( max ) ∗ m 3 h g K ¿ g K ( max ) ∗ n 4 m= average evoked response/average mini m= n*p 0! = 1 Number (x) e − x 0 1 1 0.37 2 0.14 3 0.05 4 0.02

Old quiz questions about synapses KEY Poisson Probability y ( X = k ) = e − m ∗ m k k ! Coupling coefficient = V Cell2 /V cell1

Old quiz questions about synapses KEY 1) (10 points, 2 each) Pharmacological tools are exceptionally useful in understanding molecular mechanisms in neuroscience. Fill in the missing information Agent Action of the agent and Molecular target Muscarine Agonist on metabotropic Ach R TTX or STX Blocker of voltage sensitive sodium channels CNQX Antagonist of AMPA type ionotropic glutamate receptors Naloxone Antagonist of multiple types of opiate receptors (If you said a specific type of opiate receptor, you also got full credit) ATP Endogenous Agonist of P2X receptors 2) (5 pts) Are the following receptors ionotropic or metabotropic AMPA __________Ionotropic___________ Muscarinic ____________Metabotropic_________ Nicotinic __________Ionotropic___________ GABA B __________metabotropic___________ mGluR ____________Metabotropic_________

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Old quiz questions about synapses KEY 3) 14 pts) You identify a new neurotransmitter and call it Jerrionin. In order to characterize the actions of this neurotransmitter, you conduct an experiment in which you stimulate the Jerrionin-containing neuron while recording the postsynaptic current using voltage clamp. You get the results shown in the left figure: Holding Potential (mV) a) (4 pts). Use the peak amplitude of each trace to plot the I-V curve for Jerrionin responses on the graph to the right. Make sure to plot the points to the appropriate scale and to label your axes. b) (4pts) What is the reversal potential for the Jerrionin receptor? Explain how you got this value. -60 mV, which is the potential the line fit to the points takes on a value of 0 nA c) If the cell is bathed with a physiologically normal solution (E Na = +60 mV, E K = -80 mV, E Cl = -60 mV, E Mg = 0 mV, E Ca = +100 mV), there are two possible correct answers for which ion has the greatest conductance through the open Jerrionin channel. Explain both possibilities. Possibility 1- If it is permeable exclusively to Cl, Erev = ECl = -60 mV Possibility 2 – If it is permeable to two ions, the most permeable one must be potassium, as -60 is closer to EK than to any of the other equilibrium potentials Holding Potential Current (nA ) -100 -80 -60 -40 -20 0 20 40 60 -80 -60 -40 -20 0 20 40 60 80 0 1 2 3 4 5 6 7 8 9 10 11 12 -20 -10 0 10 20 30 40 50 60 70 80 - 8 0 m V Time (ms) Current (nA)

Old quiz questions about synapses KEY 4) (10 pts) You are a homicide detective convinced that a neuropharmacologist has murdered her abusive husband using her knowledge of the action of drugs on the nervous system. Explain how the action of each of these substance might have caused death, by indicating its molecular target and action. Drug Molecular target Action on target (agonist, antagonist, allosteric modulator or other) NMDA NMDA type glutamate receptor Agonist Bicuculine GABA A receptor Antagonist Botulinum toxin Snare proteins Other (it cleaves them) Valium GABA A receptor Allosteric modulator Atropine Muscarinc AChR Antagonist 5) (6 pts) Put the following steps in the correct order (For example, if they are already in the correct order, your answer would be A, B, C, D, E, F). A. Neurotransmitter release B. Ca 2+ entry into the presynaptic terminal C. Axonal action potential D. Fusion of synaptic vesicle with the presynaptic plasma membrane E. Opening of voltage-gated Ca 2+ channels F. Depolarization of the presynaptic terminal C, F, E, B, D, A 6) (4 pts) Under normal conditions, every time you stimulate neuron A you record a 5 mV excitatory postsynaptic potential in neuron B. You add a drug that blocks all voltage dependent calcium channels, and yet the response of neuron B is unchanged. What can you conclude about this synapse? It must be an electrical synapse, since release of neurotransmitter requires calcium entry through voltage gated calcium channels

Old quiz questions about synapses KEY 7) (14 pts) To study neurotransmitter release, you make intracellular recordings from a mouse neuromuscular junction under conditions where Poisson statistics apply. You stimulate the motor neuron axon 150 times at 1 Hz. a. (2 pts) What manipulation to the solution bathing the NMJ did you need to take to produce a synapse that follows Poisson statistics? You had to lower the probability of transmitter release so it is much lower than normal. The easiest way to do this is by dropping the extracellular calcium, but adding an inhibitor of calcium channels at a concentration that blocks most but not all would also work. b. (4 pts) 55 of the stimuli evoked no response. What is the mean quantal content at this neuromuscular junction? Show your work The mean quantal content (m) = ln (trials/failures) = ln (150/55) = ln(2.72) = 1.0 c. (4 pts) The average postsynaptic response to nerve stimulation is 2 mV. Given your answer to part (b) estimate the average mini amplitude at this synapse, and explain how you got this answer. (If you are unable to solve part b, just pick any value you wish for mean quantal content and use it to answer this question). m = Mean response/mean mini so mean mini = mean evoked response/m = 2 mV/1 = 2 mV d. (4 pts) Given your answer to part (b), on how many of the 150 trials would you expect to release exactly 2 quanta? Show your work. (If you are unable to solve part b, just pick any value you wish for mean quantal content and use it to answer this question). P(2 quanta) = e − m ∗ m k k ! = e − 1 ∗ 1 2 2 = 0.18 Number of two quantal events = probability time number of trials =.18 * 150 = 28

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Old quiz questions about synapses KEY 8) (8 points, 2 each) For each box, give the name of one the Nobel Laureate who did the indicated work. If you name more than one and any are wrong, you will get no points. Discovery Name of Nobel Laureate (last name) Identified molecules of the core vesicle fusion machinery Scheckman or Sudhoff or Rothman Determined that acetylcholine is a chemical neurotransmitter Loewi or Dale Demonstrated that neurotransmitter release is quantal Katz Made the first single channel recordings from a nicotinic acetylcholine receptor Neher or Sakmann

Old quiz questions about synapses KEY 9) (22 points) One neurotransmitter that is highly expressed in the nervous system of all animals is GABA. a. (2 points) All the ionotropic GABA receptors are permeable to the same ion. What is it? Cl - (Chloride) b. In the vast majority of mammalian neurons, GABA acting on its ionotropic receptors is inhibitory, but in some neurons, it is excitatory. i. (2 points) At what developmental stage is GABA excitatory? Prior to and just after birth ii. (2 points) What is the key ion concentration differences between the neurons for which the ionotropic GABA receptors are excitatory and those for which it is inhibitory? Excitatory responses are produced when E Cl is positive to threshold. To make this occur, intracellular Cl is much higher than in adulthood (indeed Cl out and Cl in are nearly equal so E Cl is near 0 mV). c. You study a GABAergic synapse at which resting potential of the postsynaptic cell is -50 mV, and the reversal potential of GABA is -80 mV. You have a stimulating electrode in the presynaptic neuron and a recording electrode in the postsynaptic neuron. When you stimulate the presynaptic neuron to produce an action potential with normal levels of calcium present, the postsynaptic neuron hyperpolarizes from -50 mV to -78 mV ±0.1 mV on every trial . i. (4 points) When you remove all extracellular calcium, you get no evoked postsynaptic responses when you stimulate the presynaptic GABA releasing cell, but you do see random spontaneous hyperpolarizations all of which are of very similar amplitude, -1 ± 0.1 mV. What is the cause of these spontaneous hyperpolarizations? Each response is due to the spontaneous release of one GABA filled vesicle. ii. ( 4 points) When you stimulate the presynaptic neuron with a much lower extracellular calcium concentration than normal 10 times in a row, you get responses (in mV) as shown in the table. The 10 trials are representative of all responses from this synapse. Explain the two ways you could potentially estimate mean quantal content from these data and the data above. For this part, you do not need to calculate the mean quantal content. Approach 1- Sum all the responses and divide by the number of responses to get the mean evoked response, then divide by average mini size Approach 2- Count the number of failures (responses that are 0) and use the Poisson equation to calculate m iii. (4 points) Calculate mean quantal content using one of the two approaches you described in c(ii). [Note: because of the small sample size, the two approaches will give similar, but not identical values.] Approach 1 – Sum of all responses from 10 trials is -11 mV, so m = -1.1 mV/ -1 = 1.1 Approach 2- There are 3 failures- From the Poisson equation, m = ln(10/3) = 1.2 d. (4 points) You return the synapse to normal calcium, then add a drug that doubles the amount of GABA released on each trial. This results in double the number of GABA receptors being activated. Under this condition the neuron hyperpolarizes to -79mV on every trial. Suggest a reason why doubling the amount of open receptor channels only increased the hyperpolarization by 1 mV from the level seen without the drug. No matter how many GABA receptors you can activate, you cannot go more negative than E Cl (-80 mV). The original response got you almost all the way there, so twice as many active channels get you only a tiny bit closer. This is referred to as non-linear summation. Trial 1 Trial 2 Trial 3 Trial 4 Trial 5 -2 0 -1 -1 -2 Trial 6 Trial 7 Trial 8 Trial 9 Trial 10 0 -1 0 -1 -3

Old quiz questions about synapses KEY 10) (16 points) The process of neurotransmitter release consists of many steps in a sequence. a. (10 points, 2 each) You determine the location of the following proteins at a mouse neuromuscular junction in which the presynaptic and postsynaptic cells are at their resting potential, and no vesicles are currently endocytosing or exocytosing. Indicate one of the six options as a likely place to find each protein. Note that for some molecules, more than one location is correct and that some of the six options may not apply to any of these molecules. Protein Location – 1) Presynaptic axon 2) Presynaptic terminal, plasma membrane 3) Presynaptic vesicles 4) Synaptic cleft 5) Postsynaptic cell, plasma membrane 6) Postsynaptic vesicles Voltage gated sodium channel 1, 2 and 5 are all correct answers Synaptobrevin 3 Voltage gated calcium channel 2 (5 is also correct in some species and so was given credit) Nicotinic receptor 5 Acetylcholinesterase 4 b. (4 points) Every chemical synapse has a delay of a millisecond or longer from the time the presynaptic action potential reaches the nerve terminal. In principle, the rate limiting step that accounts for the delay might be binding of calcium to the calcium sensor on vesicles, the fusion process once calcium is bound, the diffusion of the transmitter across the synaptic cleft or the opening of the postsynaptic receptors. However, in reality, it is none of these steps. What is the rate limiting step, and what evidence led to this conclusion? The rate of opening of the voltage gated calcium channels is very sluggish.so it requires over one millisecond until enough have opened to allow the amount of calcium needed to release transmitter to enter the terminal. The evidence is that when you step to the reversal potential of calcium channels (to open all of them, but not let any calcium in) when you then step to a potential with a large negative driving force so Ca can flow in instantly, postsynaptic responses are seen within 200 microseconds. c. (2 points) The core fusion reaction requires only 3 proteins, but when only these proteins are present, the rate of fusion is very slow. What protein is responsible for the huge increase in fusion rate when bound to calcium? Synaptotagmin is the vesicular Ca sensor. d.

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Old quiz questions about synapses KEY 11) (14 pts) The protein GCaMP3 is frequently expressed in neurons to assess their level of activity, even though it does not respond directly to changes in membrane potential. a. (3 points) What parameter of cell function does GCaMP3 directly measure? Intracellular Free calcium b. (3 points) Explain why the parameter that GCAMP3 directly measures is usually tightly correlated with the spike activity. Spiking opens voltage gated calcium channels in the cell body, so intracellular calcium usually rises when cells are spiking c. (3 points) What kind of signal does GCaMP3 produce when it is active? An increase in fluorescence emission (green) when illuminated by the appropriate wavelength of light (blue) d. (3 points) You would like to examine activation of cortical GABAergic inhibitory interneurons using GCaMP3. Explain one promoter you could use to express GCaMP3 protein exclusively in GABAergic neurons? The promotor that makes the enzyme that synthesizes GABA (GAD) e. (2 points) There is inhibitory GABAergic input from other classes of neurons onto most GABAergic inhibitory interneurons. However, when you activate these inputs, you see no change in the signal from GCaMP3. Explain why activating GABAA receptors does not produce a signal when these neurons are sitting at their resting potential. E Cl is usually negative to rest, so activating these receptors will cause a hyperpolarization. This does not elecit any change in intracellular calcium, so there is no signal.

Old quiz questions about synapses KEY 12) You have discovered a human family that has members with a sex linked recessive mutation that results in subtle coordination deficits during all voluntary movements. a. (2 points) What features would you see in the pedigree that would lead you to conclude it was sex linked? Most or all of the affected individuals are male. If the mutant allele is common enough, there will also be some affected females. In this case, the father must have shown the mutant phenotype, but the mother might or might not b. (2 points) Can you conclude that the mutant gene codes for a non-functional protein? Explain. No. All you can include is that one copy of a non-mutant protein is sufficient to give normal function. IN this case, the mutant protein is functional, but if the person is homozygous for it, their motor system functions differently from normal (see answer to part j and k) Sequencing genomes from members of the family shows that the mutation is in the calcium channel found in the presynaptic terminals of motor neurons. It is a missense mutation that changes a single base. You decide to explore the effects of this mutation in a mouse model, because the amino acid sequence of the mouse channel is very similar to the human channel. c. ( 2 points) The mutation is to one of the arginines in the first S4 domain. If your goal was to use site directed mutagenesis to remove this positive charge but not otherwise cause dramatic structural changes, name one amino acid that might be a good choice and one that might be a bad choice. A good choice would be something that was uncharged, (rather than positively or negatively charged), and not too hydrophobic – A good choice might be S or A, but N, Q or T would not be bad choices. A bad choice would be something that is highly hydrophobic (V,L, I, F,M), induces bends (P or W) or retains a positive charge (K) or reverses the charge (D, E or H) G and C would also be problematic, as G is so small that it often disrupts structure, and C can form ectopic disulfide bonds. d. (4 points) You successfully make a plasmid that has the full-length mouse calcium channel sequence, except that you changed the arginine at the position homologous to the human mutation to match the change in the human mutation. Describe a way that you could use an in vitro expression system to test whether voltage dependent activation of this channel was normal or altered. [You can assume that the plasmid has any promotor sequences you will need to be able to use your approach]. Option 1 – Use a plasmid with a CMV promotor to drive ubiquitous expression Transfect into cells in tissue culture (perhaps HEK293 cells) and perform whole cell recording. Option 2- Sue a plasmid which has a viral promotor that allows you to make RNA in vitro, and then inject the RNA into Xenopus oocytes and carry out two electrode voltage clamping With either approach, you would hold the cells the normal resting potential of neurons and then give steps to more depolarized levels up to the reversal potential of these channels

Old quiz questions about synapses KEY e. (2 points) The experiment you did in part d demonstrated that the mutant channel activated much more slowly than the wild type channel, but that the voltage dependence as revealed by the g-V curve was unchanged. How do you produce a g-V curve from electrophysiological data? Step 1 – Apply any drugs needed to block endogenous channels, so you see only the calcium channel current Step 2 – Determine the reversal potential (the potential at which there is no current, even though there are inward currnts for voltage steps a few mV positive to or negative to it. Step 3 – Measure the peak current in response to each voltage step Step 4 – Calculate conductance at each potential as I/(Vm-Erev) and then plot as a function of the step potential f. (6 points) It turns out that a kinetic model that accurately describes Ca channel gating is identical to the model that Hodgkin and Huxley used for voltage gated sodium channels, but with different values for the parameters. i. (2 points) What are the parameters of the Hodgkin Huxley model for sodium channels? m, h and g max The equation is g = g max * m 3 h ii. (2 points) The result of your analysis was that in the wild type calcium channel the time constant for activation of a single gating sensor moving to its preferred positon at 0 mV is 3 msec, and in the mutant 5 msec. Is activation faster or slower in the mutant? The time constant indicates how long it takes to get to 67% of the final level, so a shorter time constant indicates it will get there fast, so activation is faster for the 3 msec time constant iii. (2 points) How long would you expect to have to wait until all wild type gating sensors had moved to the final positon appropriate for 0 mV? Explain. The individual sensors follow in expoential time course, so after 5 time constants the function will be within 1% of its final value, which we consider done. That will occur at 15 msec for sild type g. ( 4 points) You decide to test whether adding mCherry to the N terminal end of the calcium channel protein alters it function. i. (2 points) if you made this fusion protein successfully, would the mCherry be extracellular or in the cytosol. Explain. Intracellular- All proteins in the family have an even number of transmembrane domains with both N and C terminals intracellular ii. (2 points) Give one reason you might prefer to make a transgenic animal with that is a fusion with mCherry rather than with just the wild type calcium channel. It fluorescence red under green light, and so allows you to observe which cells are expressing your transgene in living animals. h. You carry out in vitro electrophysiological experiments that demonstrate that your mCherry tagged proteins have properties identical to the untagged proteins. You therefore decide to go forward with making a knock-in mouse replacing the wild type calcium channel sequence with the mutant sequence fused at its N terminal to mCherry. i. (2 points) Describe how you would do this using a Cas9 based strategy.

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Old quiz questions about synapses KEY  Design a guide RNA that will base pair to a unique sequence in the calcium channel gene that is near a PAM sequence. It will also need to have all the sequences necessary to bind Cas9 and allow it to cut the DNA. Ideally this will target a sequence very near the coding sequence for the arginine of interest.  Design and Synthesize a double stranded DNA target sequence o It should begin with genomic sequence of the calcium channel gene just prior to the start of translation followed by o the coding sequence for mCHerry fused in frame without a stop to the initiator methionine of the calcium channel followed by o Additional genomic sequence for the calcium channel up to and slightly beyond the arginine of interest  Inject the guide RNA and the target DNA along with Cas9 into newly fertilized embryos, and then implant the embryos into a pseudopregnant host  Sequence DNA of newborn pups to see if any have made the desired modification.  The Cas9 + plus guide RNA should have created double stranded breaks, with some repaired by swapping in your target sequence- Because the target sequence is long, you might have to sampe wuite a few pups before you had one with the desired genotype. ii. (1 point) Given what you know about calcium channel protein structure, why might making a knock in with a N terminal mCherry fusion be easier than making one with a C terminal fusion? There are a very large number of amino acids between the arginines of the first repeat and the C terminal (the last 1/3 of repeat 1 and all of repeats 2,3 and 4) so the targeting DNA construct would be very long and the chances of successful homologous recombination low. i. You successfully carry out all the molecular steps and discover that all mice expressing the N terminal fusion protein die immediately after birth because they cannot breath, whether the sequence at the arginine of interest is wild type or mutant. i. (2 points) Just before birth, you dissect out a muscle from a knock-in mouse expressing the calcium channel-mCherry fusion protein and rapidly fix it. When you then process it for freeze fracture electron microscopy, you discover that although you can very occasionally see a “dimple” indicative of a vesicle fusion site, but there are no double rows of small particles adjacent to any of the dimples. What can you conclude about how the fusion protein is different that results in cell death? The double rows of particles are the calcium channels at their normal location, so if they are not present, the mCherry must have caused the calcium channels to fail to correctly localize. ii. (2 points) If your hypothesis form part I was correct, explain a way you could have used fluorescence microscopy to reach the correct conclusion, and any limitations on whether this approach would be successful given the em results You can visualize the location of the mCherry with fluoresce microscopy. The the mCherry casued it to fail to localize to terminals at all, this would be easy to observe. However, if the mCherry got to the terminals, and was only mis localized to slightly farther away from its normal location, the resolution of light microscopy might be too low to see a difference. Suerp-resolution light microscopes methods would potentially be of use, but mCherry is not easy to switch in the way Betzig did with Kaede.

Old quiz questions about synapses KEY j. (2 points) You spend another year doing molecular biology and genetics, and make a new line of knock-in mice at which mCherry is fused to the C terminal. These mice have a normal lifespan. The mice with the arginine mutation are less coordinated, mimicking the human disorder. When you record from neuromuscular junctions of the mutant mice, you find that the synaptic delay is 2 msec longer in the mutants. Explain why this was the expected result given your in vitro recording results described in parts e and f. The kinetics of the opening of calcium channel is the rate limiting step of release, and so the casue of the synaptic delay. If they open more slowly in the mutant, as indicated in parts w and f, the delay will be longer. k. (2 points) Would you expect the rate at which minis occur in the mutant to be different from the wild type? Explain. Mini frequency is independent of calcium influx through voltage gated calcium channels, which are all closed at the resting potential, , so it should be unchanged.

Old quiz questions about synapses KEY 13) (10 points) David Julius shared the 2021 Nobel Prize in Physiology or Medicine for discovering a receptor important in somatosensation. Among the other receptors he discovered were an ionotropic serotonin receptor (5-HT 3 R) and ionotropic and metabotropic ATP receptors. Furthermore, the ionotropic glutamate receptors were identified in other labs using the “expression cloning” methods he had pioneered. a. ( 2 points) The 5-HT 3 R is a pentamer, with four transmembrane segments per subunit. Name two other receptors that belong to the same gene family. Nicotinic AChR, GABA A -R and ionotropic glycine R are all possible answers. These are cys-loop receptors b. You are studying a neuron with typical ion gradients and a resting potential of -60 mV. The spike threshold when the neuron had been at resting potential for a while is -45 mV. i. (2 points) The 5-HT 3 R is a non-selective cation channel equally permeable to Na and K. Is it an excitatory or inhibitory receptor? Explain. Since ENa is about +60 mV and EK about -80 to -100 mV, the reversal potential Erev will be half way between them (between -20 and -10 mV). This is very positive to threshold, so these are excitatory receptors. ii. (2 points) Based on your answer to part b-I, if you voltage clamped a neuron expressing 5-HT 3 Rs to +40 mV, and then applied serotonin, would the current generated be an inward current or an outward current? Explain. Calculate the driving force Vm – Erev It will be positive (+20 to + 30 depending on Erev). A positive driving force produces a positive current, since g can only be 0 or a positive number. c. You are studying a non-neuronal cell expressing an ionotropic glutamate receptor, but you are not sure whether it is an AMPA receptor, kainate receptor, or NMDA receptor. i. ( 2 points) You make a whole cell recording from this cell, using no pharmacological tools other than applying glutamate together with glycine to the cell to cause a response. Explain one thing you could do that would allow you to conclude whether it was an NMDA-R or a non-NMDA-R. [Note- Some people said apply glutamate alone, or to apply an antagonist, but these are in consistent with the instruction, which said “using no pharmacological tools other than applying glutamate together with glycine” so such answer received at most half credit. ] Assuming a physiologically normal [Mg] was in the extracellular solution, you could hold the cell at a variety of potentials and then plot the I-V curve NMDA R will give a J shape- AMPA or Kainate receptors will give linear or inwardly rectifying responses. If there was no Mg in the solution, you could still do a useful experiment. Hold the cell at +40 mV (where NMDA- R will carry current whether or not Mg is present) and just observe the duration of the response when the two transmitters are only applied very briefly NMDA-R responses to a very brief application of transmitter are long lasting (over 100 ms) while non-NMDA-R are rapidly decaying (back to baseline within 20 ms, and some much faster than that) ii. ( 2 points) It turns out that the cell is expressing a calcium-permeable AMPA receptor. What does this physiological observation tell you about the molecular composition of the glutamate receptors expressed in this cell? Explain. None of the expressed receptors have an R at the Q/R site, since that would suppress Ca permeation. Therefore the mature protein encoded by the GluA2 subunit is not present. You also know at least one of GluA1, GluA3 or GluA4 (or a Q/to R mutant of A2) is being expressed. Any combination of these would give the same results.

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Old quiz questions about synapses KEY 14) ( 2 each) Fill in the table with the name (last name only is fine) of an individual who earned a Nobel Prize for studies related to synaptic function using the indicated organism . [Note that there may be more than one correct answer for some organisms] Organism Name (and what they used it for, which is not needed for full credit) Frog Katz (skeletal NMJ) or Loewi (heart) Squid Katz (giant synapse) Genetically modified mice Sudhoff (testing the role of Synaptotagmin) Yeast Scheckman (discovery of components of fusion complex) 15) ( 12 points) You are studying a neuron that expresses a novel metabotropic receptor (Receptor X) whose downstream signaling pathway causes voltage-gated calcium channels to become phosphorylated. Phosphorylated voltage-gated calcium channels spend less time open when the cell is depolarized. a. (3 points) If the calcium channels are in nerve terminals, what does activating this receptor do to the amount of neurotransmitter released by a presynaptic action potential? Explain. Less will be released. Normal calcium channels allow the calcium that triggers release to enter the cell. If they spend less time open, less calcium will enter to trigger exocytosis. b. (3 points) You make a whole cell recording from a neuron, so you can replace all the intracellular GTP with a GTP analog that cannot be hydrolyzed. Would you expect the effect on calcium currents of briefly activating receptor X with serotonin to be changed from normal? Explain based on your understanding of the most relevant step in the signaling cascade activated by Receptor X. The suppression of current would be much longer lasting, because normally hydrolysis of the GTP shuts down the cascade, but this is prevented when the non-hydrolyzable analog is present. c. (3 points) The normal affinity of Receptor X for the neurotransmitter that activates it (Neurotransmitter Y) is 50 nM. If you made a knock-in mouse with a modified receptor X that had an affinity of 0.1 mM, would the endogenous levels of Neurotransmitter Y still be able to transmit a signal using this receptor? Explain. No. Affinity is a measure of how well the receptor binds its ligand and has evolved to match the endogenous levels of it. This mutant receptor would require 2,000 times more of Neurotransmitter Y to signal, which would surely never be present. d. (3 points ) An intermediate step in the pathway by which Receptor X brings about its effects is the release of calcium from intracellular stores to activate the kinase that phosphorylates the Ca channels.

Old quiz questions about synapses KEY How will filling the cell with a fast calcium chelator (for example BAPTA) affect transmitter release? BAPTA will directly prevent most or all transmitter release, by capturing any calcium that comes in through the open calcium channels. [The kinase activity would also stay low, even if the receptor were active. However, this effect doesn’t matter for answering the question posed]

Old quiz questions about synapses KEY 16) (10 points) You are a biochemist skilled at making subcellular fractions to understand brain function. a. (2 points) Two Nobel laureates used this approach to make major contributions to our understanding of synaptic function. Name one of the two. If you name two people, you will only receive 1 point, even if both are correct answers. Either Rothman or Sudhoff is acceptable The electric organ of the marine ray Torpedo califonica, has massive nicotinic cholinergic synapses. You obtain one, and make the following subcellular fractions i) postsynaptic plasma membranes only ii) presynaptic plasma membranes only and iii) Synaptic vesicles only b. (2 points) To which fraction would α-bungarotoxin bind? Explain. The postsynaptic fraction. This toxin has a very high affinity for the nicotinic AChR. c. ( 2 points) The key synaptic molecular Acetylcholinesterase is not present in any of the three fractions you isolated. Explain what this molecule does, and where it is localized. AChE breaks down ACh, to end the synaptic responses. It is in the extracellular space between the cells (the synaptic cleft) and not in any membrane. d. Prior to isolating these fractions, you genetically engineered the ray so that all synaptic vesicles express a transmembrane protein that is non-fluorescent until the vesicles release their contents, at which point the molecule glows bright green. When you mix these labelled vesicles with presynaptic plasma membranes in test solution 1, the rate of fluorescence appearance is very slow. Test solution 1 is isosmotic with the vesicles and contains only NaCl and a buffer to set pH to 7.0. i. (2 points) Explain what you could add to your test solution 1 to greatly accelerate the appearance of fluorescence, and why it would do so. Add calcium- which binds to Synaptotagmin and greatly accelerates the basal rate of fusion [An acceptable alternate answer was to add any of the other molecules that enhance docking of vesicles, as long as there was a reasonable explanation to accompany this type of answer] ii. (2 points) Prior to mixing your synaptic vesicles with plasma membranes, you exposed them to a mixture of all the known botulinum toxins. Would the rate of appearance of fluorescence be greater than, the same as or less than the accelerated rate seen when you add the correct material described in part d-i? Explain. There would be no fusion at all. These toxins cleave the snare proteins that allow fusion.

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Old quiz questions about synapses KEY 17) (10 points) Quantal analysis of the type pioneered by Bernard Katz has been used to study many different synapses. a. ( 2 points) The vertebrate skeletal neuromuscular junction has hundreds of release sites in each presynaptic terminal contacting a muscle fiber. Explain why quantal analysis at this synapse is always done with lower than normal extracellular calcium. At normal levels, the response is so large that non-linear summation prevents one from seeing different levels of responses as the number of released quanta varies from trial to trial. b. The typical presynaptic axon ending on neurons in the submandibular ganglion of rats makes about 25 contacts, all on the cell body, since these neurons have no dendrites. When you stimulate the nerve four times in a row you see the responses shown in the figure. i. ( 2 points) The arrow at the end of the stimulus artifact indicates the instant the presynaptic action potential reached the nerve terminal. Approximately how long is the synaptic delay, and what is the major cause of it? From looking at the time scale, the delay is in the range of 1.5-2 ms . Almost all of the delay is due to the very sluggish opening of the voltage gated Ca channels in the presynaptic membrane. ii. (2 points) There are no spontaneous miniature EPSPs present in these recordings. What peak amplitude do you expect miniature EPSPs to have? Explain. 1.5 mV. The smallest nerve evoked response is often due to release of one quantum (one vesicle worth of transmitter) Consistent with this idea, the other responses that are not failures are integral multiples of this (3 mV = 2X, 6 mV = 4X) iii. ( 2 points) Based on a very large sample of additional responses, the mean quantal content for this synapse (m) equals 3.0. Are these 4 traces representative of the overall sample if release was done under conditions that follow the rules for a Poisson process? Explain. Estimate m based on these traces and see if it is close to 3. Approach 1 m = mean EPSP/mean mini = [(6+3+1.5+0)/4]/1.5 = 2.65/1.5 = 1.75 Approach 2 m = ln (#trials/# failures) = ln (4/1) = 1.39 Neither is close to 3, so these traces are not representative. However in such a small sample, they could occur c. ( 2 points) Some synapses in the CNS have only a single presynaptic release site in the entire synaptic contact area with their target neuron. Explain why it would be necessary to use a binomial model, rather than a Poisson model to assess the statistics of release at the latter type of synapses. Poisson statistics assume that n (the number of possible release sites) is large. In this case n is only 1.

Old quiz questions about synapses KEY 18) (8 points) You perform an experiment in which you inject substances at a variety of concentrations directly into a region of the brain you know controls locomotion. You plot a dose-response curve of your results for agonists of two different receptors a. (2 points) What is the EC 50 for Ligand X ? [Be sure to give the units. Show work to be eligible for partial credit] The maximum is 20, so the EC 50 is the concentration the dotted curve is at 10. That appears to be at the first tick mark corresponding to 2 * 10 -10 M which can also be expressed as 200 pM or 0.2 nM b. (2 points) Does Ligand X or Ligand Y have higher potency? Explain. Using the same logic as A, the EC 50 for Ligand Y is about 5 µM. High potentcy means a low EC 50 , so Ligand X has much higher potency. c. (2 points) Does Ligand X or Ligand Y have the greater Hill coefficient? Explain. Ligand Y. The greater the Hill coefficient, the narrower the range over which the response begins to change from not detectable to responding maximally. It only requires about a 100fold range for Ligand Y to do this, while it takes more than a 1-million fold range for ligand X. d. (2 points) One of the two ligands is acting on nicotinic ACh receptors and the other is acting on a metabotropic peptide receptor. Is Ligand X or Ligand Y the nicotinic agonsit? Explain. Ligand Y. The potency of ligand X is in the picomolar range, and this is impossible for ionotropic receptors like the nicotinic receptr, as they could not rapidly stop responding if they had the high affinity requred to produce pM potency.

Old quiz questions about synapses KEY 19) (10 points) Many different types of channels are involved in synaptic communications. a. (5 points) Fill in the missing information in this table (0.5 points each) Receptor or Channel Major Ion that passes through the indicated channel type Number of transmembrane domains per subunit Number of subunits in the functional protein Known antagonist IP3 receptor Calcium GIRK channel Potassium 4 GABA A receptor Chloride 4 NMDA receptor 4 APV (also called AP5) or MK801 or PCP P2X receptor 2 3 Connexon 12 b. The most widely expressed ionotropic receptors in the mammalian brain are AMPA receptors. i. (3 points) Explain one way that electrophysiological experiments can easily determine whether a synaptic response to glutamate is caused by AMPA receptors or NMDA receptors. Approach 1 – Without pharmacology -Voltage clamp the cell at -80 mV. If there is a response, it is due to AMPA receptors since NMDA receptors are completely blocked by Mg at this potential. If there is no response, also voltage clamp at a potential significantly positive to the NMDA-R reversal potential of +5 mV (typically about +50 mV) where the NMDA Rs are completely unblocked and will carry easily detected currents Approach 2- Without pharmacology - Look at the duration of the response to a very brief glutamate application (when glycine is continuously present)- NMDAR responses will be long-lasting (over 50 ms) while AMPA responses are very brief (<10 ms) Approach 3 – Use pharmacological tools- See if an AMPA antagonist (CNQX) or an NMDA antagonist (APV) makes the current at +50 mV disappear. ii. (2 points) What is one line of evidence presented in the MCDB 322 lectures that the majority of the AMPA receptors in the mammalian brain are heteromers of at least two different AMPA receptor subunits? The majority of AMPA receptors have linear I-V relations. Only channel with a mix of GluA2 and a second subunit (GluA1 or A3 or A4) have this property. OR The majority of AMPA receptors are calcium impermeable. Only channel with a mix of GluA2 and a second subunit (GluA1 or A3 or A4) have this property.

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Old quiz questions about synapses KEY 20) (10 points) You characterized the synapses between 3 neurons by giving stimuli that evoked action potentials in each neuron (one at a time) and recording the responses of the other two neurons. a) (2 points) One potential explanation of the reciprocal effects of cell A and Cell B is that they have an electrical synapse between them. An alternative explanation is that there are reciprocal excitatory chemical synapses. Briefly describe one experiment that could determine which is correct. Possibility 1 - Repeat the stimulation in the absence of extracellular calcium. If the synapse is electrical, the response will persist. If it is chemical, it will not. Possibility 2 - Repeat the stimulation in the presence of a calcium channel blocker. If the synapse is electrical, the response will persist. If it is chemical, it will not. Possibility 3 – Pass hyperpolarizing current into cell A and record the response in Cell B. If the synapses is chemical, you should see no response. If the synapse is electrical, you should see a hyperpolarization in cell B b) (2 points You hypothesize that the responses of Cell C to stimulating Cell A is due activation of GABA A receptors. Name one antagonist you could use to test your hypothesis and what you would expect to see if the hypothesis is correct. Add an antagonist of GABA A receptors, such as picrotoxin or bicuculline. If the hypothesis is correct, the response to stimulating Cell A should be attenuated or abolished. c) The hypothesis in part b is incorrect. However, when you apply 1 µM GABA, Cell C hyperpolarizes from -50 mV to -80 mV. You later learn that the GABA response was due to metabotropic GABA B receptors that work via G q . i. (2 points) The response is prevented if you fill Neuron C with a high concentration of BAPTA before applying GABA. Provide a plausible explanation of this result based on known Gq signaling pathways. Both kinases activated by the Gq pathway (CamKII and PKC) require elevated cytosolic Ca to become active. BAPTA is a calcium chelator that prevents calcium from rising in response to release of Ca from the ER by IP3. ii. (2 points) Based on the amplitude of the response to GABA, what is the major ion passing through the channels that GABA activates? Explain. It must be potassium, as activation with GABA is taking the cell to -80 mV, which is negative to the equilibrium potential of all other ions. E Na = +60 mV, E K = -90 mV, E Ca = +120 mV, E Cl = -60 mV Neuron stimulated Response in Cell A Response in Cell B Response in Cell C A Action Potential Hyperpolarized by 5 mV B Action Potential Hyperpolarized by 5 mV C No response No response

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Old quiz questions about synapses KEY 21) (12 points) Questions about the mechanism of neurotransmitter release a. (3 points) Botulinum toxins have a subunit that allows them to enter presynaptic terminals and once inside cells, they are very slowly degraded, so it can take weeks to recover from BOTOX poisoning. What is the mechanism by which these toxins cause muscle paralysis? They cleave all three SNARE molecules (Synaptobrevin, Syntaxin, and SNAP25) and thus prevent them from forming the 4 helix SNARE complex required to allow vesicle fusion. b. (3 points) The Rothman lab tested whether they had identified all the required components of the core vesicle fusion machinery by mixing two types of synthetic vesicles. For one, the only protein on the surface was synaptobrevin. For the other, both syntaxin and Snap-25 were on the surface. In these experiments, neither vesicle type alone had significant fluorescence, but fluorescence increased as vesiclse fused. Explain why. The vesicles with the synaptobrevin had both a fluorescent molecule, and a second molecule that acted as a quencher, so they were barely fluorescent. The second set of vesicles were filled only with a salt solution, so they were non- fluorescent. When the two types of vesicles fused, the volume increased, allowing greater distance between quencher and the fluorescent molecule, so more fluorescence gets to the detector. c. (6 points) You measure both EPPs and mEPPs at a frog neuromuscular junction while using a concentration of a toxin that blocks 50% of the presynaptic calcium current and results in a decrease of “m” from 160 to 10 . i. (2 points) Will mEPP amplitude be larger, smaller or the same when toxin is present? Explain. mEPPs will be unchanged – When a vesicle is released spontaneously, it will have the normal amount of ACh and interact with the normal number of ACh receptors. ii. (2 points) If you give 100 superthreshold stimuli to the presynaptic axon without toxin, and then 100 more with toxin, will the % of responses that are failures be expected to be larger, smaller or the same when toxin is present? Explain. The same (0%). For m=10, the Poisson equation predicts that proportion of trials that will be failures is e -10 which is 4.5 * 10 -5 , so for 100 trials you expect 0.005 failures (stated another way you expect one failure for every 22,222 trials). For m=160 the predicted number of failures is far smaller (3*10 -68 ). [Note- People who said the proportion of failures would increase in toxin got 1 point, since had m decreased further (to <4.5), failures would indeed have become more common in a sample of 100 traces] iii. (2 points) What property of synaptotagmin is responsible for the fact that a relatively modest decrease in calcium current caused such a large decrease in “m”? Synaptotagmin has 5 calcium binding sites, most or all of which must be occupied to trigger release. This is the reason that in the steep part of the concentration-response relation, vesicle release is proportional to the 4 th power of calcium concentration. Thus a 16fold decrease in release is what was expected for a 2fold decrease in calcium current [Note: When we graded the page, we gave 1 point for saying Synaptotagmin was the calcium sensor. To receive a second point the answer had to indicate there are multiple calcium binding sites on synaptotagmin]

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Old quiz questions about synapses KEY 22) (4 points, 1 each) For each of the indicated Nobel Laureates indicate the contribution that was most relevant to their Prize. Name Contribution Earl Sutherland Discovered the cyclic AMP is a second messenger Bernard Katz Worked out the quantal nature of chemical synaptic transmission and other aspects of the release of neurotransmitter Al Gilman Discovered G proteins Thomas Sudhof Discovered many of the proteins involved in rapid vesicle fusion and tested their roles 23) (6 points) The voltage clamp was developed to help understand the mechanism of action potentials, but it has proved to have great advantages for studying synaptic potentials as well. d. (3 points) One advantage is that it eliminates non-linear summation of postsynaptic responses as the amount of transmitter increases. Explain why non-linear summation occurs when you do not use a voltage clamp. When you open channels, they bring the cell closer to their reversal potential and thus decrease the driving force present when more channel open, so the amplitude of response to constant increments of transmitter gets smaller and smaller. e. (2 points) Another advantage is that the voltage clamp immediately allows one to see whether there is rectification in the current voltage relation. If you were given a plot of currents obtained under voltage clamp in response to a fixed amount of ligand, how would you decide whether the channels rectify? If the response is not a straight line, the channels rectify. [The rectification is referred to as inward if, for driving forces of equal magnitude, but opposite sign, the inward currents are greater. Outward rectification is the opposite] f. (1 points) Name one type of ligand gated channels that show rectification, and whether it is an outward rectifier or an inward rectifier. NMDA receptors outwardly rectify

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Old quiz questions about synapses KEY AMPA receptors with only Qs at the QR site show inward recetificatoin [Although you were not taught this, P2X receptors, 5HT3 receptors and many of the neuronal nicotinic receptors also show inward rectification]

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Key Old quiz questions about synapses

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