Neurons A and B synapse on a postsynaptic cell ( V m = –70 mV, threshold = –65 m
ID: 166245 • Letter: N
Question
Neurons A and B synapse on a postsynaptic cell (Vm = –70 mV, threshold = –65 mV). Neuron A produces a 6 mV PSP due to the closing of potassium channels. Neuron B produces an 8 mV PSP due to the opening of chloride channels (ECl- = –78 mV). Neuron B's synapse is closer to the axon hillock than neuron A's. If the PSPs from neurons A and B arrive simultaneously, the postsynaptic cell will __________.
hyperpolarize
NOTE: This question is from my mastering A & P for my textbook. They told me my inital answer (it will fire an action potential) is wrong and that the corrrect answer is hyperpolarize however i do not understand how.
To figure this question out, i added the 6mV and 8mV to get 14mV that arrive at the same time. Given the information in the question that the membrane potential of the cell is -70mV and the threshold for an action potential is -65mV, i calculated that -70mV + 6mV + 8mV = -56mV which means it has depolarized the membrane (made it more positive than the existing membrane potential (-70mV) and has made it more positive than the required threshold for an action potential. So i do not understand how the answer could be hyperpolarized because that would mean that the membrane was made more negative than the inital membrane potential (-70mV). Please explain!!
fire an action potential not change its membrane potential depolarize but not fire an action potentialhyperpolarize
NOTE: This question is from my mastering A & P for my textbook. They told me my inital answer (it will fire an action potential) is wrong and that the corrrect answer is hyperpolarize however i do not understand how.
To figure this question out, i added the 6mV and 8mV to get 14mV that arrive at the same time. Given the information in the question that the membrane potential of the cell is -70mV and the threshold for an action potential is -65mV, i calculated that -70mV + 6mV + 8mV = -56mV which means it has depolarized the membrane (made it more positive than the existing membrane potential (-70mV) and has made it more positive than the required threshold for an action potential. So i do not understand how the answer could be hyperpolarized because that would mean that the membrane was made more negative than the inital membrane potential (-70mV). Please explain!!
Explanation / Answer
During nervous excitation, many steps take place. Firstly when the neuron is at rest, its membrane has nearly -70mV of potential. When a stimulus arrives, the opening of sodium channel gates takes place and influx of sodium ions occurs. This causes the membrane to acquire a net positive charge of +35mV due to over-concentration of sodium ions inside the neuron. This state is termed as depolarization and action potential is generated.
Further, when the action has been completed in the neuron, the refraction of the action potential takes place. During this phase, the calcium and potassium channels open up and calcium and potassium ions move outside the neuron making the neuron membrane acquire a net negative charge as compared to outside. This charge is nearly -70mV in nature and this phase is termed as repolarization.
However, there should again occur closing of these positive channels but the potassium channels are kept open for a slightly longer duration which permits net movement of more potassium ions outside causing generation of a more negative charge of -80mV. This state is termed as hyperpolarization.
Now, according to the information, two neurons A and B are placed in viscinity and they generate membrane potentials of 6mV and 8mV respectively due to opening of potassium and chloride ions. Here, this should be carefully noted that these two neurons are residing in close proximity of a synapse. Under these conditions, the nerve potential of one neuron will drive the nerve potential of second neuron.
Consider that the neuron A is depolarized and generates an action potential of -70mV, it will transmit the information during the refractory phase to the next neuron. The next neuron B which is already hyper-polarized with membrane potential of 8mV will undergo depolarization as this stimulus will arive and perform the same cycle of impulse generation. This is because when neuron A will be hyperpolarized to a potential of -80mV, this potential will add up to the membrane potential of neuron B (+8mV) and give a net potential of -72mV to it (following law of summation of nerve potential). This will immediately generate a state of hyperpolarization in neuron B and it will be bound to undergo depolarization due to arrival fo this stimulus.
Thus, correct answers is choice 4, hyperpolarization will take place.