Exam 2

Question 1

8. Consider two synapses located at different sites on the dendritic tree of a neuron. If both synapses cause identical potential changes at the site at which they are located, which synapse is most effective in determining the generation of action potentials of the neuron: the synapse closest to or furthest from the axon hillock? Why?

Answer 1

The synapse closest to the axon hillock will be the most effective synapse.

The reason is that some of the positive charges that entered the cell from the synapse farther from the axon hillock will leak out of the cell through potassium channels as it travels from the synaptic site to the hillock. The closer the synapse, the less current will leak out of the cell and thus there will be more current to depolarize the hillock, the site at which action potentials are initiated.

Question 2

8. In testing the effects of a drug on synaptic transmission at a synapse, you find that the drug lowers the amplitude of the EPSP but that the drug does not alter the amplitude of the miniature endplate potentials. What can you conclude from this finding about where and how the drug is acting?

Answer 2

The drug must be acting on the presynaptic terminal to limit the number of vesicles that release transmitter in response to an action potential. The smaller number of quanta released (i.e., the smaller amount of transmitter released) results in a smaller EPSP. The receptors on the postsynaptic cell are unaffected by the drug, as shown by normal mepps that are recorded. In other words, the spontaneous release of a quantum of transmitter causes the same mepp when the drug is present or when it is absent.

Question 3

8. Write the 5 basic steps in chemical transmission as given on page 88 in Chapter 8.

Answer 3

1.The nerve terminal (of the motor neuron) is invaded by the action potential and the membrane potential of the terminal approaches the sodium equilibrium potential.
2. Depolarization of the nerve terminal opens voltage gated Ca++ channels that are located in
the axon terminal. Ca++ ions enter the nerve ending. Voltage gated Ca++ channels have not been mentioned previously, but they are very similar to the voltage gated K+ channels discussed previously. Like K+ channels, they have an activation gate that is closed at rest but is opened by membrane depolarization.
3. The entry of Ca++ into the terminal causes a sudden increase in a normally very low level of ACh secretion. The mechanism of ACh secretion is the fusion of the membrane of a synaptic vesicle with the membrane of the presynaptic terminal. ACh is stored in the vesicles.
4. The transmitter is released from the vesicles and diffuses across the gap that separates the nerve terminal from the postsynaptic membrane. The gap at most neuromuscular junctions is about 50 mμ (micrometers or 500 Å).
5. ACh binds to protein receptor molecules embedded in the postsynaptic membrane of the muscle fiber under the nerve terminal. This binding of ACh to a receptor causes the ion channel in the receptor to open. Opening of the ion channel leads to a conductance change in the membrane to Na+ and K+ ions. Ions flowing through these synaptically activated channels cause a local potential change in the muscle fiber. This synaptic potential is given a special name at the neuromuscular junction: it is called the end-plate potential and is abbreviated epp. There is a delay of about 0.5 msec between the arrival of the action potential in the motor neuron terminal and the onset of the depolarization of the epp in the muscle fiber. (One difference between the epp and the epsps occurring in most neurons is that the epp is so large that the muscle fiber membrane potential is always depolarized past threshold for an action potential. Thus, each motor neuron action potential gives rise to a corresponding muscle fiber action potential. In the case of the epp, there is no integration!
6. The action of ACh is brief, lasting 2-3 msec. The epp is terminated because the ACh released undergoes hydrolysis (to acetate and choline) by the enzyme acetylcholinesterase (also called an esterase for short) located in the basal lamina in the synaptic cleft.

Question 4

8. What is the neurotransmitter used at the neuromuscular junction?

Answer 4

ACh is the neurotransmitter used at every neuromuscular junction.

Question 5

8. What is an end-plate potential and how does it differ from an EPSP (or does it differ)?

Answer 5

The end-plate potential, or epp, is the depolarization at the endplate caused by the release of ACh from the axonal terminals of the motor neuron that innervates the muscle.

It is exactly the same thing as and excitatory postsynaptic potential (EPSP), except that the depolarization is called an epp in muscle while the depolarization in postsynaptic neurons is called an EPSP.

Question 6

8. What exactly is the neuromuscular junction and what are its major parts?

Answer 6

The neuromuscular junction is the synapse that a motor neuron makes with a muscle fiber.

The major parts are the synaptic boutons, the axonal endings of the motor neuron, the endplate, which is the specialized region of the muscle fiber at the junction. The enplate has numerous junctional folds, the infolding of the muscle membrane at the endplate that contains the ACh receptors. In addition, there is the basal lamina, a basement membrane located between the bouton endings and the endplate. The basal lamina contains acetylcholine esterase, the enzyme that hydrolyzes ACh into acetate and choline.

Question 7

8. What factors determine whether a synapse is excitatory or inhibitory?

Answer 7

The critical factor is the response of the ligand-gated channel to the release of transmitter. If the channel is permeable to chloride, it allows the influx of negative charges, which causes the membrane potential to hyperpolarize and is inhibitory. If the channel is permeable to sodium and potassium, the influx of sodium and efflux of potassium drives the membrane potential to the reversal potential, which is depolarizing and typically ranges from about -17-0 mV.
​
Usually, inhibitory presynaptic fibers release either glycine or GABA while excitatory fibers release glutamate or acetylcholine. However, there are synapses in the brain in which a particular transmitter is inhibitory at one synapse but excitatory at another because the receptors at each synapse are different. The classic example is ACh. ACh is excitatory at all neuromuscular junctions, because the receptors at the neuromuscular junction are all nicotinic ACh receptors, but ACh is inhibitory at the heart because the heart has muscarinic ACh receptors.

Question 8

8. Explain briefly how an IPSP can summate with an EPSP. What is the result of such a sumation. In other words, how does the postsynaptic response caused by summation of the EPSP and IPSP differ from the synaptic response that would have been evoked only by the excitatory input or the inhibitory input to the cell?

Answer 8

If one of the inputs is excitatory and the other inhibitory, the charges from the inhibitory inputs subtract the charges from the excitatory input.

Since the positive charges entering the neuron are what cause the membrane potential to depolarize, the subtraction of positive charges by the negative charges would cause the membrane potential to depolarize less than it would have if the only synaptic input was from the excitatory input.

Similarly, the positive charges injected into the cell from the excitatory synapse make the negative charges from the inhibitory synapse less effective.

That is, if only the inhibitory inputs were activated, the result would be a hyperpolarization but if the excitatory inputs were also activated, the result would either be a much smaller hyperpolarization or even a depolarization.

Question 9

8. What is the difference between spatial and temporal summation?

Answer 9

Spatial summation refers to the summation of the influx of charges from two or more inputs located on different parts of the dendrite or cell body.

Temporal summation refers to the effects of sequential inputs from the same input fiber.

Question 10

8. What is meant by the summation of synaptic events?

Answer 10

The addition of charges that enter a postsynaptic cell due to the release of transmitter from one or more presynaptic neurons.

In other words, the positive charges that enter through ligand gated channels sum and cause a larger response than that evoked by one input alone

Question 11

8. What is an inhibitory postsynaptic potential (IPSP)?

Answer 11

The hyperpolization (increase in membane negativity) of the postsynaptic cell in response to the release of transmitter from a presynaptic neuron at a synaptic site.  
It is called inhibitory because the hyperpolariztion of the membrane tends to drive the membrane potential farther from threshold and tends to prevent the genertion of an action potential.

Question 12

10. Who was Otto Loewi and what did he imagine in his famous dream? Describe the experiment he conducted and what did the results of this experiment on the heart demonstrate?

Answer 12

Otto Loewi was a Nobel prize winning neuroscientist who had a dream in which he conducted the definitive experiment to show that neurons could communicate with their targets via chemical transmission. When he awoke, however, he had completely forgotten how to conduct the experiment. The next night the dream reoccurred, but this time he woke up and wrote down the experiment so that he would be sure to have it in the morning. The next day he conducted his now famous experiment.

​​What he did was to isolate two frog hearts and place each in a different container filled with Ringer’s solution. Heart 1 was placed in container 1, and heart 2 in container 2. The vagus nerve in heart 1 was intact and its heart rate was slowed when the vagus nerve was stimulated. The fluids in container 1 flowed into container 2 through a tube that connected the two containers. A few moments after vagal stimulation, the rate of heart 1 slowed but the key result was that the rate of the heart 2 also slowed. Clearly stimulation of the vagus to the first heart caused the release of some substance into the surrounding fluid which, when it reached the second heart, produced the slowing of its rate. This showed, for the first time, that nerves communicate with other cells by chemical transmission

Question 13

10. What is the major excitatory transmitter in the brain? What are the major inhibitory transmitters in the brain?

Answer 13

The major excitatory transmitter in the brain is the amino acid, glutamate, while the main inhibitory transmitters are the amino acid, glycine, and gamma amino butyric acid (GABA).

Question 14

10. What is Sarin?

Answer 14

Sarin is a very powerful cholinesterase inhibitor that is used in some forms of nerve gas. It is deadly because it would block the normal breathing cycle by preventing a person from exhaling and death could occur by asphyxiation.

Question 15

14. What is the mechanism by which the depolarization induced suppression of inhibition is produced?

Answer 15

The key feature is that CB1 receptors are present only on the terminals of inhibitory neurons in the cortex and hippocampus, and are not on the terminals of excitatory neurons. So if a cell in the cortex or hippocampus is, at one point in time, strongly stimulated by only by excitatory inputs the cell will depolarize strongly. The strong depolarization will open voltage gated Ca++ channels, and the calcium influx will trigger the cascade that synthesizes endocannabinoids from the lipid membrane. The endocannabinoids will diffuse out the cell and bind to the CB1 receptors on the inhibitory terminals, which are not active at that moment. The binding of the endocannbinoids activates the receptor which then closes some or most of the voltage gated calcium channels on the terminal of the inhibitory axon.
​A few seconds later, both the excitatory and inhibitory axons become active. However, since the voltage gated Ca++ channels on the inhibitory terminal are closed, little or no inhibitory transmitter is released. There are no such constraints on the excitatory terminal, since the excitatory terminals do not have CB1 receptors. Hence, the cell is now strongly excited by only the excitatory innervation, because the inhibitory innervation is suppressed by the previous depolarization of the postsynaptic cell.

Question 16

14. What is depolarization induced suppression of inhibition?

Answer 16

Depolarization in both hippocampal and cortical cells can suppress or weaken the inhibitory inputs to the cell. Thus, if an initial signal activates both the excitatory and inhibitory inputs at the same time, a small excitatory response will be evoked because the inhibition partially cancels the excitation. If following that initial signal, a second signal evokes only the excitatory input and strongly depolarizes the cell, the strong depolarization will generate a series of events that will then suppress the inhibitory inputs for several seconds. If during that period of several seconds, while the inhibitory inputs are suppressed, both the excitatory and inhibitory inputs are again driven, the the excitatory inputs to have a greater effect (evoke a larger depolarization) than they did before the inhibitory inputs were suppressed.

Question 17

14. What effect does the binding of an endocannabinoid to a CB1 receptor have on synaptic strength and why does a change in synaptic strength occur?

Answer 17

The CBI receptor is a metabotropic receptor located on the axon terminal. When the CB 1 receptor is activated, the beta-gamma complex migrates to bind to voltage gated Ca++ channels and acts to close the channels. When an action potential invades the axon terminal, there are now fewer voltage gated Ca++ channels available to open, and hence there is a smaller influx of Ca++. The reduced amount of Ca++ influx results in a reduced amount of transmitter released, and thus the post-synaptic response is smaller than it was before the Ca++ channels were closed, thereby weakening the synapse.

Question 18

14. What are CB1 receptors and where are they located?

Answer 18

CB1 receptors are metabotropic receptors that are located in the terminal of the presynaptic axon.

Question 19

14. How do endocannbinoids reach their receptors?

Answer 19

Endocannabinoids are synthesized in the post-synaptic cell and are lipid soluble. Since they are lipid soluble, they diffuse through the membrane of the postsynaptic cell and bind onto CB1 receptors on the axon terminal of the presynaptic cell. This is a mechanism by which neurons can communicate backwards across synapses to modulate their inputs, and thereby adjust synaptic strength.

Question 20

14. How are endocannabinoids produced?

Answer 20

Through a cascade activated by the influx of Ca++ through voltage gated calcium channels in the cell body. The net result is that the endocannabinoids are synthesized from lipids in the cell membrane.

Question 21

14. In Kandel’s experiments on the sea slug, Aplysia, he found that serotonin secreted from the sensory neuron causes a change in the duration of the action potential in the sensory axon. Explain why that change in the duration of the action potential occurs, and how that change in action potential duration changes the synaptic strength of the synapse that the sensory neuron makes with a motor neurons.

Answer 21

The activation of the facilitating neuron by touching the siphon releases serotonin onto the axon terminal of the sensory neuron from the siphon, which innervates a motor neuron that causes gill contraction. The serotonin activated a metabotropic receptor. The activated G protein then turned on adenylate cyclase which phosphorylates some of the voltage gated K+ channels in the terminal, the S-type of K+ channels (the S standing for serotonin type of K+ channel). The phosphorylation then closes the S-type of K+ channel.
​When an action potential invades the terminal, its duration is lengthened because there are now fewer voltage gated K+ that opened and thus it now takes longer to repolarize the membrane, thereby increasing the duration of the action potential.
​Since depolarization opens voltage gated Ca++ channels that cause the release of transmitter, the prolonged duration of the action potential causes a prolonged release of Ca++, and thus an enhanced amount of transmitter released. The enhanced transmitter then causes a larger EPSP in the motor neuron than had been evoked before the S-type of K+ channels were closed.

Question 22

14. Give an example of the effects of the beta-gamma component of the G-protein when a metabotropic receptor is activated.

Answer 22

One example is the activation of muscarinic ACh channels in the sympathetic ganglion. The binding of ACh on these receptors activates the beta-gamma component of the G protein which then binds to and closes of K+ channels. The net affect of closing these K+ channels is to cause a subthreshold depolarization of the membrane potential, thereby making the cell more excitable, and thereby causing the cell to fire more action potentials in response to ACh released on nicotinic ACh receptors.

Question 23

14. Give an example of the effects of the alpha component of the G-protein when a metabotropic receptor is activated.

Answer 23

One example is the activation of adenylate cyclase by the activated alpha component.

Question 24

14. How is adenylate cyclase activated by a G protein and what is the cascade that is evoked when the cyclase is activated?

Answer 24

The activated alpha component of the G protein migrates along the membrane and binds to adenylate cyclase, which is also embedded in the membrane. The binding of the alpha component activates the cyclase which then converts ATP in the axoplasm into cyclic AMP. The cyclic AMP then binds onto the inhibitory subunits of protein kinase A, causing the inhibitory subunit to disassociate from the catalytic subunit. The catalytic subunits then phosphorylates an ion channel, causing the channel to close (or open in some cases) and remain in the closed (or open) state as long as the subunit is bound. If the channel is a K+ channel, the opening of the channel will cause the cell to hyperpolarize. If it closes a K+ channel, the cell will depolarize (because the cell now loses fewer positive charges).

Question 25

14. How does the activation of a metabotropic receptor generate an amplified response?

Answer 25

The response can be amplified at several stages. The first amplificaton is that more than one G protein is activated when the receptor binds a transmitter. In the case of the activation of adenylate cyclase, the activated enzyme then converts many ATPs into many cyclic AMPs. The cyclic AMPs, in turn, can activate many molecules of protein kinase A, and thus can phosphorylate many ion channels.

Question 26

14. Explain how metabotropic receptors work.

Answer 26

Metabotropic receptors do NOT have an ion pore but rather are coupled to a G-protein. That protein has three subunits, an alpha, a beta and a gamma subunit. The beta-gamma subunits operate jointly as a dimmer. The protein is normally is coupled to a molecule of guanosine diphosphate (GDP). When transmitter binds to the receptor, the receptor dislodges the GDP and take on a molecule of guanosine triphosphate (GTP). The GTP activates the protein causing the alpha subunit to dissociate from the beta-gamma unit. The dissociated units of the protein then exert effects downstream from the receptor.

Question 27

12. MICE
    Explain what the investigators did to inactivate the fear memory that had been previously established in mice and what they subsequently did to reactivate the fear memory?

Answer 27

The investigators showed that in the mice that experienced the pairing of light with shock while bar pressing, the synaptic connections that auditory neurons made on amygdala neurons were strengthened , as shown by LTP at those synapses. The rationale was that if the pairing of light and shock strengthened synapses, and thus was responsible for the fear response evoked by light while the mice were bar pressing, then weakening those synapses by inducing LTD should inactivate the fear response.
​Since the light guides were permanently implanted into the amygdala, the investigators took the mice that had previously been conditioned to freeze in response to light, and subjected them to low frequency light stimulation at the amygdala, a protocol that induces LTD. The mice in which LTD was induced, were again placed in the cage to bar press for a liquid reward. While they were bar pressing, the light was shown on the amygdala, but now the mice did NOT freeze and stop bar pressing, although the same mice did freeze in response to light before they were subjected to the LTD protocol.
​The next day, the same mice were given an LTP protocol; that is light was shown at a high rate on the amygdala, which produced LTP. When the mice were then placed back into the cage to press a bar for a liquid reward, they now froze and stopped bar pressing when the light was presented. This showed that the fear memory had been reactivated.
The next day they were given the LTD protocol for a second time, and when placed in the cage and were pressing the bar for a liquid reward, they again did NOT freeze when light was shown on the amygdala. This showed that the LTD protocol once again had inactivated the fear memory. The fear memory could then be re-activated by giving the LTP protocol again. Thus the fear memory could be inactivated by inducing LTD at those synapses and reactivated by inducing LTP

Question 28

12. MICE

Why was there only an outward current when the cell was clamped at +40 mV?

Answer 28

When the membrane is at +40 mV, it is close to the sodium equilibrium potential. Thus when the AMPA and NMDA receptors open, there is very little driving force on sodium and therefore very little inward sodium current.
​On the other hand, there is a large driving force on potassium. The concentration force drives potassium out of the cells and since the inside of the cell is now positive, both the concentration and electrical forces drive the positively charged potassium ions out of the cell. In other words, there is a very large driving force on potassium that causes a large outward flow of potassium when either the AMPA or NMDA or both channels are open. It is for this reason that there is a large outward current when both receptors are open, and a smaller outward current after the AMPA receptors have closed but the NMDA receptors are still open. That is the pure NMDA current.

Question 29

12. MICE

How did the investigators measure the NMDA current?

Answer 29

They next clamped the membrane potential at +40 mV. At this depolarized potential, the Mg++ plug is expelled from the pores of the NMDA receptors. There is an important feature about NMDA receptors that has not yet been discussed; namely, that NMDA receptors bind glutamate more tightly than do AMPA receptors. This means that when glutamate is released, the gates on both AMPA and NMDA receptors open and current flows through the pores of both receptors. However, once the glutamate is removed from the synaptic cleft, it first diffuses off of the AMPA receptors, which then close, and several milliseconds later, the glutamate diffuses off of the NMDA receptors. In other words, there is a period following the closing of the AMPA receptors when current is still flowing through the NMDA receptors because glutamate is still bound to the NMDA receptors. The current flowing during that period, after the AMPA receptors have closed is the pure NMDA current, as indicated in the above figure by the horizontal arrow.
​So what the investigators did is to first measure the pure AMPA current when the membrane was clamped at -60 mV (blue graph in the figure above). The time when the current flowed is when the AMPA channels were open, and that open time is independent of membrane potential (the binding and diffusion of glutamate from the AMPA receptors is the same regardless of membrane potential; it is NOT voltage dependent). They then clamped the cell at +40 mV and measured the maximum current during the period AFTER the AMPA channels had closed but while the NMDA receptors were still open (the maximum pure NMDA current as indicated by the purple vertical double-headed arrow in the figure above). They then took the ratio of the maximum pure AMPA current (measured when the membrane was clamped at -60 mV) and the maximum pure NMDA current (measured when the membrane potential was clamped at +40 mV), and then compared the ratio obtained from amygdala neurons in unpaired mice to the ratio obtained in paired mice.
​In unpaired mice, the ratios were about 2; the AMPA current was about twice as large as the NMDA current. In paired mice, however, the ratios were about 4; the AMPA current was about four times as large as the NMDA current. The reason that the AMPA current was so much larger in paired mice is because the synapses in paired mice had undergone LTP, and thus had more AMPA receptors than in the unpaired mice, whose synapses did not have LTP and thus had less AMPA receptors.

Question 30

12. MICE

Why was there a large inward current when the cell was clamped at -60 mV?

Answer 30

Although no potassium flows through the open AMPA receptors, because the membrane is clamped at Ek, there is a very large driving force on sodium, which then causes in a large inward sodium current to flow through the AMPA receptors.
That current record (above figure) shows both the maximum amplitude of the current (vertical arrow) and the time course of the current (how long the AMPA receptor was open, as indicated by the time between the two dashed lines). This is the pure AMPA current as shown in the above figure.

Question 31

12. MICE

How did the investigators measure the AMPA current independent of the NMDA current?

Answer 31

Since they wanted to evaluate currents through AMPA receptors the investigators voltage clamped the amygdala neurons. They first clamped the amygdala cell at -60 mV, which is the potassium equilibrium potential in these cells. They then stimulated the auditory fibers with light, which evoked action potentials in those fibers and caused them to release glutamate onto the amygdala neuron.
​The important point, however, is that no current can flow through the NMDA receptors because the membrane potential is clamped at -60 mV, which prevents the Mg++ ion from being expelled from the pore of the NMDA receptor. Thus when glutamate is released, the only receptors passing current are the AMPA receptors, even though the gates on both the AMPA and NMDA receptors are opened by glutamate.

Question 32

12. MICE
    Why did the investigators assume that LTP was generated in the synapses that the fibers from auditory cortex and medial geniculate made with amygdala neurons in mice that had undergone the paired condition?

Answer 32

The basic assumption is that pairing light with shock strengthens the synapses made by the auditory fibers, which are activated by light, and the amygdala neurons with which they make synaptic connections. To show the strengthening, the investigators compared the ratio of the currents through AMPA and NMDA receptors in the mice that experienced the paired condition and the mice that experienced the unpaired condition. Since synapses that express LTP have additional AMPA receptors inserted into the synapse, it follows that mice that experienced the paired condition should have a substantially higher ratio of AMPA/NMDA currents than do the same synapses in mice that experienced the unpaired condition.
​

Question 33

12. What is the difference between the paired condition and the unpaired condition?

Answer 33

In the paired condition, the tone or light was presented simultaneously with the shock, and thus the mice learned to associate the tone and shock. In the unpaired condition, the tone (or light) was presented, but the shock was presented a minute after the tone (or light). Because of the long time separation between the tone (or light) and shock, the mice did not associate the tone (or light) with the shock. In the subsequent tests, the mice did NOT freeze when the tone (or light) was presented alone.

Question 34

12. How did the investigators who conducted this study create a conditioned fear response to tones and how did they do it by stimulating the brain with light?

Answer 34

They trained mice to press a lever for a liquid reward. After the mice had learned the task, they would present a tone and a foot shock while the mice were pressing the bar. The shock was painful and the mice immediately stopped pressing the bar and froze. After one or two trials, they learned to associate the tone with the shock. The freezing indicates fear, fear of the pain from the shock.. When they were tested the next day, they immediately froze and stopped bar pressing when the tone came on even though no shock was given.

In a subsequent experiment, mice were again trained to press a bar for a liquid reward. After they learned the task, a shock was presented while they were bar pressing but now the auditory fibers that innervated the amygdala were activated with blue light while the shock was presented. Presumably, the activation of the auditory fibers by light was equivalent to the activation of the same auditory fibers by the tone. When the mice were tested the next day, the mice froze when the blue light was shown on the amygdala, even though no shock was presented. This showed that the mice had learned to associate the activation of the auditory fibers that innervated the amygdala with shock, in the same way other mice had learned to associate a tone with shock.

Question 35

12. Which parts of the brain were transfected with channel rhodopsins in this study?

Answer 35

The auditory portions of the brain in the cortex and a lower auditory center, both of which send axonal projections to the amygdala.

Question 36

12. What is “Optogenetics” and how is it used to study the brain?

Answer 36

Optogentics refers to the technique of using light to stimulate neurons transfected with channel rhodopsins. The “Opto” refers to the use of light for stimulation while “genetics” refers to the transfection of the neuron’s DNA with the channel rhodopsin gene.

Question 37

12. How is channel rhodopsin inserted into a restricted region of the brain?

Answer 37

The gene for channel rhodopsin has been cloned. The channel rhodopsin gene and a gene for a red fluorescent protein (d-tomato) are fused to a promoter specific for neurons (synapsin). This engineered gene is inserted into a virus, which is then injected into a specific region of the brain. The channel rhodopsin and the red fluorescent protein are then expressed in the transfected neurons.

Question 38

12. How does channel rhodopsin work?

Answer 38

The rhodopsins capture blue light, which then triggers the opening of a channel that is permeable to sodium. Thus activation of the channel rhodopsin causes sodium to enter and depolarize the cell.

Question 39

12. What is a channel rhodopsin and from which organism did it come from?

Answer 39

Channel rhodopsins are light sensitive rhodopsins used by algae to orient to light.

Question 40

11. THOUGHT
    A scientist finds that he can evoke long term potentiation (LTP) in a postsynaptic cell by tetanically stimulating the presynaptic fiber. In another experiment on the same synapse, the scientist applies a drug that selectively blocks only NMDA receptors and finds that LTP can no longer be generated by tetanic stimulation at that synapse. In yet another experiment on the same synapse, he applies a drug that selectively blocks only AMPA receptors and again finds that LTP can no longer be generated by tetanic stimulation.

Explain why blocking NMDA receptors prevented LTP from being generated.

Explain why blocking AMPA receptors prevented LTP from being generated.

Answer 40

If NMDA receptors are blocked, calcium cannot flow through the pores of those receptors, and thus CaM-Kinase II will not be activated.

Even though calcium is released at that synapse, there would be no depolarization because the AMPA receptors are blocked, and thus there would be no depolarization to expel the magnesium block in the NMDA receptors. Since current could not flow through the NMDA receptors, there would be no calcium entering the cell and CaM-kinse II would not be activated and thus there would be no LTP generated

Question 41

11. Who was Donald Hebb and state the Hebb Postulate.

Answer 41

Donald Hebb was a Canadian neuroscientist who, in 1948, proposed a cellular mechanism for learning and memory. That mechanism, now known as the Hebb Postulate, states, in essence, that cells that fire together, wire together.

Question 42

11. How can associative strengthening be produced?

Answer 42

The high activity in synapse 1 causes a strong depolarization at that synapse (with a concomitant high influx of calcium through NMDA receptors). The moderate activity at synapse 2 opens AMPA receptors but is not strong enough to open NMDA receptors. However, some of that positive current from synapse 1 then spreads to moderately active synapse 2, which sums with the positive current entering through the AMPA receptors. If the summed current at is large, it can expel the magnesium plug from the NMDA receptors in synapse 2. Since synapse 2 is moderately active, glutamate is present which opens the gates on the NMDA receptors. Now the NMDA receptors are conductive, allowing for the entry of calcium into synapse 2. If there is a large calcium influx, the calcium will activate CaM-Kinase II and allow it to autophosphorylate, which then will generate a synaptic strengthening at synapse 2, a strengthening that would not have occurred if the modeate activity at synapse 2 were not associated with high activity in neighboring synapse 1.

Question 43

11. What is the difference between “specific strengthening” of a synapse and “associative strengthening”?

Answer 43

Specific strengthening refers to the strengthening of one synapse due to a high rate of presynaptic activity at that synapse. Associative strengthening refers to a synapse which is only moderately active (not active enough to induce strengthening by LTP) but is moderately active at exactly the same time as a nearby synapse that is very active. The simultaneous activity of the two synapses enables the moderately active synapse to be strengthened, but only if it is active at the same time as its highly active neighbor.

Question 44

11. What is calcineurin?
12. How is it activated?
13. Why does calcineurin have a much greater effect with a small influx of Ca++ ​at the synapse than with a large Ca++ influx?

Answer 44

Calcineurin is a phosphatase (an enzyme that de-phosphorylates proteins) that is concentrated in the postsynaptic density, just under the receptors in the postsynaptic cell.

Calcineurin is activated by Ca++/calmodulin.

When calcineurin is activated, it can dephosphorylate proteins and is more efficatious than the CaM-Kinase II, because only one or a few subunits on the kinase are open (the efficacy of the kinase is low).
​​With a large calcium influx, more calcineurin will be activated, but the CaM-Kinase II will also be far more efficatious because many more subunits will be activated on each molecule. The important point is that once the calcium is removed from the postsynaptic cell, the effects of calcineurin are over (because there is no longer any Ca++/calmodulin to activate it). However, the CaM-Kinase II is still fully active because of the autophosphorylation that resulted from the initially high calcium influx. Thus, the kinase can now exert its phosphorylating effects without competition from the dephyosphorylating effects of calcineurin.

Question 45

11. Explain what happens to this kinase when there is a small calcium influx into the cell and how does that differ from the effects of high calcium levels in the cell?

Answer 45

With a low calcium influx, one or only a few subunits will be opened on the full kinase because there is only a small amount of ca++/calmodulin present. Thus, the enzyme is only partially activated (only one or a few subunits can phosphorylate another protein) and relatively ineffective. In addition, once the calcium influx has been removed from the interior of the postsynaptic cell, the subunits close and the kinase is rendered completely non-functional. In other words, with a small calcium influx, the activity of the kinase is low and only acts for a short time, while calcium is present.

​With a high calcium influx, much more ca++/calmodulin is present, which opens many of the subunits on the kinase. Under these conditions, some or many neighboring subunits will open, thereby allowing those subunits to be autophosphorylated. The important point is that due to the autophosphylation, those subunits remain active even after all of the calcium in the cell has been removed. Thus, the larger number of active subunits makes the kinase far more effective than with a low calcium signal, and the kinase can exert its effects for a much longer period of time.

Question 46

11. What is the structure of CaM-Kinase II?

Answer 46

The enzyme is composed of 12 subunits, where each subunit has both a catalytic domain, and an autoinhibitory domain. The catalytic domain contains a substrate binding site, the S-site, that can bind to and phosphorylate a substrate, and a T-site on the autoinhibitory domain. The full-length form of the kinase has almost no catalytic activity under basal conditions because the autoinhibitory domain of each subunit binds to its own catalytic domain, thereby closing the subunit and inhibiting its catalytic activity. Binding of ca++/calmodulin to the autoinhibitory region opens the subunit and renders it active. The T site on the autoinhibitory domain can be phosphorylated by the active S site on the catalytic domain, if and only if, a neighboring subunit is open due to the binding of ca++/calmodulin.
​

Question 47

11. What is CaM-Kinase II and where is it located in the synapse?

Answer 47

CaM-Kinase II is an enzyme activated by calmodulin, but only after the calmodulin has bound calcium. It is located in the postsynaptic density, just below the synaptic region of the postsynaptic cell.
​

Question 48

11. Define long term potentiation (LTP) and how does it differ from long term depression (LTD).

Answer 48

LTP is the strengthening of a synapse due to a high rate of activity in the presynaptic neuron. LTD is exactly the opposite, that is it is the weakening of a synapse due to a low rate of activity in the presynaptic neuron.

Question 49

11. What ions are AMPA receptors permeable to and what ions are NMDA receptors permeable to?

Answer 49

AMPA receptors are permeable to both sodium and potassium and when open, drive the membrane potential to the reversal potential, which is about -2.0 or 0 mV. The pore of the NMDA receptor is larger, and is permeable to sodium, potassium and most importantly to calcium.

Question 50

11. What is required to open NMDA receptors that allow the influx of positive charges and thus membrane depolarization?

Answer 50

NMDA receptors have two requirements.

The first is that the receptor has to bind glutamate, which opens the conductive pore in the receptor.

The second is that the membrane has be depolarized by about 20 mV. The more positive membrane potential then expels the magnesium ion that normally plugs the pore of the receptor. Thus, the receptor can conduct current only when it binds glutama

Question 51

11. What is required to open AMPA receptors and allow the influx of positive charges and thus membrane depolarization?

Answer 51

AMPA receptors open when they bind glutamate.

Question 52

10. What effect would a cholinesterase inhibitor have on the respiratory muscles that control breathing?

Answer 52

It would greatly prolong the postsynaptic response in the diaphram ​muscles, thereby preventing the muscles from relaxing

Question 53

10. There is a class of compounds that act as cholinesterase inhibitors. What changes occur in the endplate response in a motor neuron when a cholinesterase inhibitor is applied to the neuromuscular junction?

Answer 53

The duration of endplate potential is greatly prolonged because the ​ACh is not broken down in the synaptic cleft, and thus can exert its effects for a ​much longer time than when it is cleared quickly from the cleft due to the ​hydrolysis of ACh.

Question 54

10. Are other neurotransmitters cleared in the same way as ACh?

Answer 54

No.

ACh is the only transmitter that is broken down into its constituents by an enzyme. All other neurotransmitters are not broken down, but simply transported back into the axon terminal by specific transporters.

Question 55

10. How is ACh cleared from the synaptic cleft?

Answer 55

Once ACh is released, it is broken down into choline and acetate by the acetylchoine esterase in the basal lamina. The acetate diffuses into the extracellular fluid and blood while the choline is transported back into the axon terminal by choline transporters and is used to resynthsize ACh.

Question 56

10. Where and how is ACh synthesized in the neuron?

Answer 56

The synthesis of acetylcholine occurs in the cytoplasm of the axon terminal by the enzyme, acetylcholine​transferase (ChAT). ChAT catalyzes the reaction of acetyl co-enzyme A and choline to form ACh .

Once synthesized in the cytoplasm of the axon, ACh is transported into the vesicles. The transport is accomplished by proteins in the membrane of the vesicles that carry the ACh from the cytoplasm and actively ​pump it into the interior of the vesicle.

Question 57

10. Why does curare block ACh receptors at the neuromuscular junction but not in the heart?

Answer 57

Because there are different receptors at the two synapses. The receptors at the neuromuscular junction are nicotinic ACh receptors, for which the agonist is nicotine and the antagonist is curare (which blocks the synapse), while the agonist at the heart is muscarine (hence they are called muscarinic ACh receptors) while the antagonist is atropine.

Question 58

10. How are these compounds used to distinguish receptor types that bind the same transmitter?

Answer 58

If the receptor at one synapse behaves differently to a compound that ​is an agonist or antagonist for a receptor at another synapse, the receptors ​must be different, even though they bind the same natural neurotransmitter.

Question 59

10. What is meant by an agonist of a receptor and what is an antagonist??

Answer 59

An agonist is a drug or compound that produces the same effect on ​the postsynaptic receptor as the natural neurotransmitter.

​An antagonist is a drug or compound that when applied to the postsynaptic ​receptors will bind to the receptor but will not open the receptor, and thus ​block the action of the natural neurotransmitter.

Question 60

10. What are the three criteria for establishing that a chemical is a neurotransmitter?

Answer 60

1) nerve stimulation must cause the release of the substance and the release must be shown to come from the nerve
2) the effects of nerve stimulation are mimicked by applying the suspected transmitter directly to the postsynaptic cell

3) nerve-evoked and extrinsic application responses must display the same pharmacology, i.e. response to various drugs.
​

Question 61

9. ​​Ligand gated channels that evoke inhibition are permeable to Cl-. What is the reversal potential that results from the binding of an inhibitory neurotransmitter?
   ​

Answer 61

The reversal potential would be the chloride equilibrium potential, since the channel is permeable to only one ion, Cl-. You would have to know the concentrations of chloride inside and outside the cell to calculate Ek, and thus the reversal potential for that channel

Question 62

9. THOUGHT
   Consider this problem: A nerve cell has a resting potential of -70 mV and a threshold of -55 mV. The reversal potential for one synapse on this neuron is -60 mV. In which direction does positive current cross the membrane during the action of the transmitter? Is the synapse excitatory or inhibitory? Why? What if there is a second synapse on the same neuron that has a reversal potential of -30 mV. If the same conductance change occurs at each synapse, which synapse will produce the largest EPSP?

Answer 62

The first synapse would be depolarizing and would generate an​EPSP. That is, it would allow for an influx of positive charge and drive the membrane potential to -60, which is more depolarized than the resting potential. Notice that the membrane would never reach threshold, -55 mV, because the reversal potential is -60, which is below threshold. So while the evoked response would be a depolarizing EPSP, that EPSP would actually be inhibitory because it would in effect clamp the membrane potential at the reversal potential of -60 mV and prevent the membrane potential from reaching threshold.

The second synapse would also be depolarizing and would generate an EPSP. However, the membrane would never get to the reversal potential of -30 mV because the membrane would reach threshold, -55 mV, and evoke an action potential

Question 63

9. THOUGHT
   An animal is discovered that uses ACh as a transmitter at some synapses in its nervous system. The ACh receptors of that animal are ligand gated receptors. However, their ACh receptors differ from those in other vertebrates in that when they bind ACh, the channel allows HCO3- as well as Cl- (instead of Na+ and K+) to go through equally well (the channel is equally permeable to HCO3- and Cl). Assuming that EHCO3 = -90 mV, and ECl = +10 mV, and the resting potential is -20 mV.

What is the reversal potential for ACh at these synapses?

Answer 63

The reversal potential is -90 + (+10) /2= -40 mV. Thus, when the channels open, they would drive the membrane potential to a value more NEGATIVE than the resting potential and would evoke an IPSP!

Question 64

9. Explain how the voltage clamp was used to evaluate the reversal potential at the neuromuscular junction.

Answer 64

By voltage clamping the muscle fiber at various holding or clamped potentials, one can follow the current coming into or out of the cell. The point is that when one clamps the membrane potential at -75 mV, Ek, there is a large influx of positive current, all of which is carried by Na+. As the membrane potential is clamped at progressively less negative values, the influx of positive current becomes progressively smaller. The reason is that driving force on Na+ is decreasing, causing a progressively smaller influx of Na+, while the driving force on K+ is increasing, causing a progressively larger efflux of positive charge carried by K+. When the membrane is clamped or held at a certain potential, there is no influx of positive current because at that potential, the reversal potential, the influx of Na+ is equal to the efflux of K+. At even a slightly more positive membrane potential, the current reverses; that is, now the influx of Na+ is smaller than the efflux of K+ (more positive current leaves the cell carried by K+ than enters the cell carried by Na+.

Question 65

9. What is the definition of a reversal potential?

9. What is the reversal potential for ACh at the neuromuscular junction?

Answer 65

The reversal potential is the potential to which the membrane is driven due to the opening of a ligand gated channel. The value is midway between the equilibrium potentials of the ions to which the channel is permeable.
​
Since nicotinic ACh channels at the neuromuscular junction are equally permeable to sodium and potassium, the reversal potential is equal to the sum of the equilibrium potential for sodium and the equilibrium potential for potassium divided by 2. It is about -17mV at the frog neuromuscular junction.

Question 66

9. List two major differences between a voltage-gated ion channel and a ligand-gated ion channel.

Answer 66

Obviously, voltage gated ion channels are opened by the charge on the inside of the neuron while ligand gated channels are opened by a chemical neurotransmitter. In addition, voltage gated channels, at least the type we have discussed in class so far, are located all along the axon or, in the case of voltage gated Ca++ channels at the axon terminal, while ligand gated channels are located in the dendrites and cell body, at the site of the synaptic connection with the presynaptic axon.

Question 67

8. What is an excitatory postsynaptic potential (EPSP)?

Answer 67

The depolarization of the postsynaptic cell in response to the release of transmitter a the presynaptic neuron at a synaptic site.

It is called excitatory because the depolarization of the membrane tends to drive the membrane potential closer or even to threshold for an action potential.

Question 68

10. What is Sarin?

Answer 68

Sarin is a very powerful cholinesterase inhibitor that is used in some forms of nerve gas. It is deadly because it would block the normal breathing cycle by preventing a person from exhaling and death could occur by asphyxiation.