A plant cell whose intracellular (sap) concentration for sodium, potassium, and
ID: 54620 • Letter: A
Question
A plant cell whose intracellular (sap) concentration for sodium, potassium, and chloride is shown in above table is immersed in an external solution whose temperature is 24 degrees C and whose composition is also given in the table. The composition of the sap is constant over time, as is the difference in potential between inside and outside of this cell; measurements with an intracellular electrode indicate the inside of the cell is at a potential of –86 mV with respect to the outside. For each of the ions shown in the table determine if its transport requires an active transport mechanism to maintain a steady state and, if so, in which direction across the membrane (inward or outward) it is transported.
b) Calculate the “driving force” (Em – Eion) for the 3 ions at the cell’s resting potential. From the relationship between ionic current and driving force do your results agree with the convention that inward current is negative and outward current is positive? Calculate ICl at rest assuming the resting chloride conductance is 0.1 nS (note: nS = nanosiemens).
c) If a population of stretch-activated K+-selective channels suddenly became open at rest what happens to the cell’s membrane potential, membrane resistance, and membrane capacitance? (Assume no change in cell size or shape).
Ion Sap (mmol/L) External (mmol/L) sodium 15 0.1 potassium 28.7 1.0 chloride 38 1.3Explanation / Answer
To maintain steady state concentration the ions from the driving force, which can be calculated by using the formula
VDf = Vm - Veq; Where Vdf - Driving force; Vm - membrane potential, Veq - Equilibrium potential of ion of unrest.
The Vm for the given cell is -86 mV. Veq can be calculated using Nernst equation.
Veq = RT/zF ln(Xin/Xout)
Where R-gas constant, T- Temperature in Kelvin, z- valancy of the ion, F- faraday constant. Xin - concentration of ion inside the cell and X out - concentration of ion outside the cell.
R= 8.314 J.K-1.mol-1; Given Temperature = 24 degree C; K = 24°C + 273.15 = 297.15; z for Na = 1, K = 1; Cl= -1
F= 96485 C.mol-1
Veq for Sodium = -0.12 V
Veq for potassium = -0.085 V
Veq for chlorine = 0.086 V
Driving force for each ion
Vdf of sodium = -86 mv - Leads to influx into cell
Vdf of potassium = 764 mV - leads outside the cell
Vdf of chlorine = -172 - leads to effluxes outside the cell.
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