2. Cells are capable of generating different resting membrane potentials, which
ID: 190082 • Letter: 2
Question
2. Cells are capable of generating different resting membrane potentials, which are essential to secondary active transport mechanisms, excitability, and generation of action potentials. The resting membrane potential of a cell is determined by which ions it actively pumps across the membrane (and in which direction) and to what extent the ions are allowed to "leak" back across the membrane down their electrochemical potential gradient. Suppose a muscle cell has a resting membrane potential that is primarily determined by the transport of K', Na', and Cl. The cell actively pumps Na* and Cl out and pumps K in. Passive leakage (permeability) of Na* across the membrane is very low, while the permeability of K is about 10 times that of Na The following are the concentrations of these ions inside and outside of the cell lon Extracellular Fluid (mM Cytoplasm (mM) [Na ] 122 140 [C11 122 The resting membrane potential for this cell at 37°C is experimentally found to be -86 mV Determine the permeability of Cl of the cell relative to that of Na". At this temperature, the quantity RT/F is equal to 26.72 mV and the membrane potential equation is Nat Nat In In this equation, o subscripts indicates outside the cell (extracellular) and i subscripts indicate inside the cell (intracellular). Note that this is the same equation given at the end of the corresponding slide deck, but uses slightly different notation as is arranged differently (there is no minus sign in from of VmExplanation / Answer
Ans. The above equation is an example of Goldman equation which determines the resting memebrane potential and considers the permeability of cells as well.
We have been given that permeability of K+ is 10 times of Na+ . At resting phase permeability of K+ is 1 hence permeability of Na
Vm = 2.303 RT/zflog (PK+CK+i + PNa+CNa+i + PCl-CCl-o / PK+CK+o + PNa+CNa+o + PCl-CCl-i)
-86 = 61.5 log(1*140+0.1*9+x2 / 1*2.5 + 0.1*122 + x122)
-1.39 = log (140+0.9+2x/2.5+12.2+122x)
Hence by considering the above equation the permeability of Cl- relative to Na+ would be -7.33 to 0.1.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.