MCDB151_HW2_key - MCDB 151/251 - HOMEWORK #2 Due Friday...

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MCDB 151/251 - HOMEWORK #2 Name____________________ Section_____________ Due Friday 10/16/09 in Lecture Ion: Extracellular Concentration: Intracellular Concentration: K + 10 mM 120 mM 1) A cell containing only Na + channels and K + channels has the following intracellular and extracellular ionic composition: Na + 130 mM 20 mM Cl - 150 mM 120 mM a) What are the equilibrium potentials for K + , Na + , and Cl - ? (3 pts) Use the Nernst equation: E equil. = 58 log 10 [ion] out Z [ion] in E Na = 58 log (130/20) = +47 mV E K = 58 log (10/120) = -63 mV E Cl = (58/-1) log (150/120) = -6 mV b) At what potential will the Na current through open Na channels be zero? (1 pt) When voltage = equilibrium potential for sodium = +47 mV (it would only be 0mV if the extracellular and intracellular concentrations were equal) c) You perform voltage clamp experiments on this cell by holding the membrane potential at -90 mV, then the currents from the whole cell are recorded for voltage pulses to -20 mV, +20 mV and +60 mV (voltage pulses are indicated below). Draw the currents that you would expect from this cell. Label your currents corresponding to each of the voltage pulses (-20, +20 and +60 mV). (6 pts) How do you do problem 1C?
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First, ask which channels are present in a cell. You are told that only Na and K channels are in this cell. No blockers have been added, so you should expect both voltage-activated Na currents and voltage-activated K currents. Remember that Na currents are fast. They activate fast, then inactivate. K currents are slower: the K currents activate, but do no inactivate. You know that the Na and K currents each will be described by the equation: I = G (V-E equil ) It is easiest to separately figure out the Na and K currents, then to draw them summed together. You will need to calculate the equilibrium potential for each ion (Remember that you
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MCDB151_HW2_key - MCDB 151/251 - HOMEWORK #2 Due Friday...

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