Questlon 2. (2 pts) You have constructed a model system to study ionic permeabil
ID: 193103 • Letter: Q
Question
Questlon 2. (2 pts) You have constructed a model system to study ionic permeability, consisting of two compartments separated by a lipid bilayer. The bilayer contains K' channels and Na channels. The ionic concentrations in the right and left compartments are: k: left = 400 mM, right = 10 mM Na": left 50 mM, right = 460 mM At room temperature, what are the equilibrium potentials for K' and Na" in this system? Express each membrane potential as referring to the charge of the left compartment relative to the right (e.g. a positive membrane potential means that the left compartment is more positive than the right compartment). Question 3. You are studying the electrical properties of a neuron in a dish. The extracellular solution bathing the cell contains 145 mM Na' and 5.6 mM K. You determine that under these conditions, the Na' equilibrium potential is +35 mV and the K' equilibrium potential is -60 mV. All measurements are conducted at room temperature. 0 a. (2 pts) What are the intracellular Na" and K' concentrations? b. (2 pts) If you held the neuron at +20 mV, in which direction would the net flow of each ion be? c. (2 pts) If you held the neuron at +50 mV, in which direction would the net flow of each ion be?Explanation / Answer
Ans. As per the nernst equation for equilibrium potential we know that
V = RT/zFln[K+]out/[K+]in
= 2.303 RT/zFlog[K+]out/[K+]in
= 2.303x8.314x298/96485 log 400/10
= 94.72 mV
Hence the equilibrium potential for K+ would be 94.72mV.
For Na+ it would be
V = 2.303 RT/zFlog[Na+]out/[Na+]in
= 2.303 x8.314x298/96485 log 50/460
= -56.98 mV
Therefore for Na+ ions it would be -56.98 mV
Ans 3 a. For Na+ ions the intracellular concentration would be
35 = 2.303 x 8.314 x 298 / 96485 log 145/ x
Hence x = 37.103 mM
Therefore the intracellular concentration of Sodium ions would be 37.103 mM
For K+ ions the intracellular concentration would be
-60 = 2.303 x 8.314 x 298 / 96485 log 5.6/x
Therefore x = 57.93 mM
Hence the potassium ion concentration would be 57.93 mM.
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