For the purposes of this problem, use the following constants and concentrations: R = 2.0 cal/ (mol x °K) ; F = 23,100 cal/(Volt x mol); Absolute zero is -273° C. The energy for pumping ions up an electrochemical gradient can come either from nucleotide triphosphate hydrolysis or by coupling the movement of the ion to that of another ion moving down its electrochemical gradient. a) Compare and contrast the relative amounts of energy generated by ATP hydrolysis to that typically generated by a single sodium ion moving down its electrochemical gradient in a human neuron living at 37°C, with intracellular and extracellular sodium ion concentrations of 12 mM and 150 mM, respectively, and a resting potential of -70 mV.
?G = -RT ln (Nain/Naout)R = 2 cal /(mol*°K)T= 37°C = (37+273) °K = 310KNain = 12mMNaout = 150mM Therefore,?G = (-2*(310) ln (12/150)) = 1566 cal/ molPositive value of ?G (1566 cal/mol) indicates that energy is needed and the reaction is unfavorable.A resting membrane potential of 70 mV (or 0.07V) provides a repulsive force for the movement of sodium ions. The free energy associated with the resting membrane potential is:?G (Resting Membrane Potential) = (Resting Membrane Potential* F) = (-0.07) * 23100 = (-1617) cal/mol?G net = ?G – ?G (Resting Membrane Potential) = (1566-(-1617)) = 3183 cal/mol = 3.183kcal/molThis positive value (3.183 kcal/mol) once again indicates that it is not favorable in terms of energy. For it to occur it has to be coupled to a reaction where there is a net negative free energy observed.
ATP hydrolysis is then used. The energy associated with hydrolysis of one mole of ATP is (-13.6) kcal. If we couple these two reactions we get?G coupled = 3.183+(-13.
6) = (-10.147) kcalb) In the case of the Na+/K+ ATPase, what fraction of the energy released by ATP hydrolysis is used to move Na+ versus K+. Why do you think this movement of K+ up its electrochemical gradient across a membrane is almost never coupled directly to nucleotide triphosphate hydrolysis without also moving another ion along with it? With a mol of ATP having energy of 13.6 kcal and Na having energy of 3.183 kcal/mol, ATP is used to move 4 mols of Na and 9 mols of K+.As seen in the above calculation in (a), it is simply energetically unfavorable to move Na+ and even K+ as they are both positive.
As such, ATP is used to move these ions with a protein embedded in the membrane, the sodium potassium pump. Potassium and Sodium are usually coupled because in a cell. Cells have leaky potassium channels so this balance needs to be maintained by taking in potassium. However, with a cell that has a resting potential of -70mV as the one in this example, an influx of potassium results in a positive potential which again, is unfavorable.
If sodium ions are pumped out while potassium is coming into a cell, a negative potential which is favorable is maintained. This series of steps makes it energetically favorable to move such ions across the membrane and a coupling of ATP with K+ and Na+ possible. c) For the same neuron as in part a, calculate the theoretical maximum of the peak amplitude of the action potential (in mV). Peak amplitude of action potential= ENaENa=VeqVeq=RT/ZF ln (Xout/Xin)R=2 cal /(mol*°K) = 8.
3736 J.K-1.mol-1F = 23,100 cal/ (Volt x mol) = 96650.
4 C.mol-1Therefore, Veq= ((8.3736 J.K-1.mol-1*(310.15))/+(1*96650.
4 C.mol-1) ln (150/12)Veq=67.8 mVThe theoretical maximum of the peak amplitude of the action potential is thus 67.8 mV.
Question 2:You are a molecular neuroscientist consulting for Arthur Daniels Midland. This agricultural conglomerate is looking for new insecticides that target the voltage gated potassium channels, which operate during the action potential. You have isolated an amphipathic, neurotoxic peptide from a poisonous oceanic snail that binds directly to medfly voltage-gated potassium channels at nanomolar concentrations. To study the actions of the toxin, you inject the mRNA of the voltage-gated potassium channel from this insect into Xenopus oocytes, which express the protein and insert the functional channel into the membrane. In a voltage clamp experiment, in the presence of the toxin, the potassium channel maintains its conductance and activates at the usual membrane potential (+15 mV), but takes 2 full seconds to inactivate instead of the usual 1–2 msec. a) Given the properties of the neurotoxin and the general structure of the channel, where are the most likely sites on the channel located for binding the neurotoxin, and why? Neurotoxins that affect voltage-gated potassium channels in neurons will increase the release of acetylcholine at neuromuscular junctions. Voltage gated potassium channels control excitability in the nerves and muscles by controlling the resting membrane potential and by repolarizing the membrane during action potentials. As such, these neurotoxins will be found bound to the nodes of Ranvier and at nerve terminals.
This is how these types of neurotoxins produce their effects of increasing the duration of action potentials and increasing the release of acetylcholine which produces hyper excitability. In general, any neurotoxin that acts on voltage gated potassium channels will block potassium channels and which causes repetitive firing at neurons and prolongs the depolarization.b) Compare (graph) the potassium current across the oocyte membrane as a function of time when you stimulate the oocyte with a voltage pulse (0.5 msec) at +20 mV in the presence and absence of the neurotoxin. In the graph above, in the absence of a neurotoxin, the potassium channel can be seen undergoing a typical action potential; a stimulus, in this case, 20 mV excites the membrane from its resting potential of-75 mV, the threshold is achieved and Na+ channels open allowing Na+ ions to rush in; as the Na+ channel closes, the potassium channels open and depolarization occurs; the action potential slows down when K+ ions decrease in rushing out and finally, hyperpolarization occurs. The resting membrane is re-established as the potassium channels have only one gate which is activated by depolarization and inactivated by repolarization.
When the neurotoxin is present, it triggers a release of transmitters at the neuromuscular junction and end plate to produce action potentials more rapidly. The action potentials go through depolarizations faster with a longer action potential. The rapid release of transmitters stimulates the membrane continuously.
Initial phase of the action potentials is identical, but it is much longer and does not have an after-hyperpolarization. There is a repolarization phase but now the repolarization is due to the process of Na+ inactivation alone. There is no change in the resting potential. c) Assume that the normal sodium equilibrium potential of the insect neurons expressing this K+ channel is +50 mV and that the potassium equilibrium potential of these same neurons is –85 mV, and the resting potential is –75 mV. Graph the membrane potential as a function of time for a normal action potential versus one that has been exposed to the neurotoxin. Explain your reasoning and why this neurotoxin is likely to be deadly to the insect. Normal Action PotentialAction Potential Exposed to NeurotoxinPotassium channels can be voltage gated with rapid kinetics. Neurotoxins that act on potassium channels act as potassium blockers.
These effects would be repetitive firing of the neuron that increases transmitter release as a result of the depolarization of the action potential being prolonged. Dendrotoxins, for example, are neurotoxins that function as potassium blockers and they affect neuromuscular transmission. There is an increase in the number of effective vesicles released in response to a nerve impulse, and repetitive firing of endplate potentials. Blocking potassium channels may also increase acetylcholine release which increases depolarization. This eventually results in an accumulation of acetylcholine at neuromuscular junction and an overstimulation of nerves and muscles that results in paralysis. Question 3:In your laboratory, you are studying the dynamics of axonal transport of an identifiable motor neuron in the abdominal ganglion of a species of giant leech (an annelid). In this (hypothetical) animal, the motor neuron is especially accessible for study because it can be easily visualized within an organ explant, enabling you to do intracellular injections and optical analysis of axonal transport in this neuron in vivo.
The cell body has a diameter of 50 ?m, and its axon terminals are 16 mm away, in the muscle. To study the dynamics of neurofilament (NF) transport in this axon, you have made an mRNA encoding a snail NF protein that is conjugated at its N-terminus to GFP. You inject the GFP-NF mRNA into the neuron on day one. From previous studies in your lab, you know that the mRNA will begin to be translated into protein within 15 minutes of injection and will continue to be translated for approximately 1 hr. After that, no new protein is made, since the injected mRNA is degraded.
Throughout the first day after injection, you observe GFP-NF protein fluorescence as small dots moving through the axon. Approximately 2 hours after injection, the first of these dots arrives at the end of the axon. The concentration of dots increases as you move closer to the cell body, with the peak fluorescence intensity located only 85 ?m from the cell body. In this region, the fluorescence looks more filamentous than it does further down the axon. Two hours later, this peak has moved an additional 50 ?m down the axon.
The size of the peak, however, is reduced, whereas the number of dots and filament-like structures visible along the entire length of the axon has continued to increase, with some filament fragments now visible at 2-3 mm down the axon. a) Based on what you have heard in lecture about intermediate filament transport and assembly, why do you think that you see dots at first and not filaments?Intermediate filaments are composed of different proteins (keratin, neurofilaments, vimentin filaments and nuclear lamins) that create a network of filaments critical to the mechanical support and structure of the cell. The proteins are arranged from non-filamentous to filamentous based on the tissue type as they contain different proteins. The non-filamentous proteins are identified as dots while the filamentous ones are identified as filaments. The “dots” come first because the proteins that resemble those of long fibrillin type filaments are necessary to maintain the structure of the cell and provide stability. b) What is the velocity of the most rapidly moving dots in mm/day? Most rapidly moving dots moved from 0 ?m starting at the cell body then moved 16000 ?m away from it, travelling through the entire axon and reaching the end of the axon within 2 hours.V= (Final Position-Initial position)/ TimeV= (16000?m -0?m)/7200 sec=2.22 ?m/s= 191.
808 mm/dayc) What is the average (most common) velocity of the GFP-NF in mm/day? Some GFP –NF protein moved through the entire axon at 2.22 ?m/s while some moved 85?m away from the cell body and an additional 50?m. The velocity for these dots are: 0.012 ?m/s and 0.
007 ?m/s. As such, the average velocity of GFP-NF protein is:V=(v+u)/2V= (2.22E-6 m/s + 0 m/s)/ 2V= 1.11E-6 m/s= 95.904 mm/dayd) Why doesn’t all the labeled GFP-NF arrive at the tip of the axon at once (i.e., why is there such a difference between the fastest and average velocities?). There must be something preventing the GFP-NF from moving along the axon.
The purpose of the axonal transport is to move mitochondria, lipids, synaptic vesicles, proteins and other organelles to and from the cell body to the axoplasm. These organelles etc. being transported depend on microtubule tracks to be transported along the axon via motor proteins like dynein so it is expected that if there are more microtubules, the more distance the protein (from the given situation) will have to travel.
It will take longer for all of the proteins to reach the axon terminal because of this microtubule organization in addition to the varying sizes in proteins which would result in a longer time to be transported. 4) Acetylcholine is one of the most extensively studied neurotransmitters. It acts on two different subtypes of receptors, nicotinic and muscarinic.
Based on what is known about the mode of action of each of these two classes of receptors, and the physiology of the pre and post synaptic cells, explain why it makes sense that the acetylcholine receptors on skeletal muscle cells at the vertebrate neuromuscular junction, which are innervated by motor neurons, are primarily nicotinic whereas the muscles of the eye’s iris that are innervated by parasympathetic neurons are muscarinic. In skeletal muscle cells, more specifically the motor end plate and neuromuscular junction, nicotinic acetylcholine receptors can be found. The purpose of nicotinic acetylcholine receptors is to trigger rapid neuromuscular transmissions. Muscle cells consist of motor neurons of which when stimulated result in impulses being transported the axon of the motor neuron and signal a muscle contraction. At the neuromuscular junction, acetylcholine transmitters are released from vesicles in axon terminals. When these bind to its nicotinic receptor, it is activated and a conformational change occurs which causes Na+ ions to flow out of the cell and become depolarized then allows K+ to flow into the cell causing a repolarization via an ion pore at the neuromuscular junction; calcium permeability at the pore is also increased.
This excites the neuromuscular junction and causes a muscular contraction.Muscarinic acetylcholine receptors, on the other hand, are slower and play a role in the activity of smooth muscles like that of the iris. As such, they are mostly localized in the parasympathetic nervous system. In fact, a specific subtype of muscarinic receptor known as M-3 muscarinic receptor is found in smooth muscle and will thus yield a response.
5) A monogamous prairie vole male that mates with a female for the first time becomes selectively aggressive toward males and other females, but not toward its newly won mate. In addition, the male continues to show a preference for mating with this female. These behaviors are associated with increased release of vasopressin acting on vasopressin 1A receptors in the diagonal band nucleus of the male prairie vole. a) In situ hybridization, immunocytochemistry, and receptor autoradiography are three methods used to study the distribution of peptidergic neurons and their receptors in the brain. For vasopressin and its V1A receptor, discuss how the results obtained from these methods might differ from one another and how such differences should be interpreted in light of the distributions of the peptide and the receptor relative to their sites of synthesis and physiological action. In situ hybridization can be used to indicate the localization of vasopressin and its V1A receptor by designing a probe against its known DNA or RNA sequence. With in situ hybridization, the peptidergic neuron and its receptors is basically mapped to their respective chromosomal location. MicroRNAs play various roles in brain function and with this technique explains how the brain works.
However, this technique needs to be further optimized in attempt to localize areas of the brain where miRNAs for vasopressin and its V1A receptors might be in low levels. Their expression in its entirety will not be identified. In situ hybridization also requires the tissues to be examined to be thinned and fixed.
Immunocytochemistry also requires tissues to be thinned and fixed but instead of using probes to identify sequences and determine the vasopressin neuropeptide and its receptors, this technique used antibodies. The localization is based on its (vasopressin and its V1A receptor) reacting to the antibody. Immunocytochemistry is more accurate of the two techniques as it identifies localization of the neuropeptide synthesis and release even though it can be multifunctional and exist in various parts of the brain and body.
They might be found in smaller populations, they might be associated with other neuropeptides found in small populations and their synthesis might be episodic. This is not as critically identified via in situ hybridization. However, even immunocytochemistry has its drawbacks-it cannot be done in real time.
Receptor autoradiography can be done in vivo or in vitro by using a radioactive substance bound to vasopressin or V1A and this will determine the distribution of the receptors in living cells; all it requires is the animal being studied to be injected with a radiolabeled hormone. This easily identifies the localization of vasopressin and its V1A receptors and even allows the localization of one region to be compared to that of an adjacent region for reference. Anatomical and functional information are provided simultaneously. Receptor autoradiography thus provides precise anatomical localization and has high sensitivity and can be used for quantification purposes. b) Of the three methods above, why might information obtained from receptor autoradiography be considered the most physiologically relevant? As mentioned above, receptor autoradiography thus provides precise anatomical localization and has high sensitivity and can be used for quantification purposes.
allows the localization of one region to be compared to that of an adjacent region for reference. Anatomical and functional information are provided simultaneously. c) If you wanted to modify the behavior of the animals by injecting an adenovirus containing a gene encoding a dominant negative V1A receptor, which technique would you use to select your target site for injection into the brain and why? I would use Immunocytochemistry. Immunocytochemistry is selective-it is highly sensitive and specific. Adenoviruses will more than likely cause inflammation of the brain tissue and will vary in intensity. With immunocytochemistry, identifying the adenovirus is easier.
Immunocytochemistry would be more effective in keeping track of the adenoviruses that are difficult to detect even by staining and will aid in identifying ones that are present in low numbers.