Suppose a cell is a bathing solution of 1 mM NaCl across the membrane there is a

Question

Suppose a cell is a bathing solution of 1 mM NaCl across the membrane there is an electric potential of -59 mV. If the cell is allowed to go to equilibrium for both ions while maintaining a membrane potential of -59 mV, the concentrations inside the cell will be (draw yourself a picture, it may help) [Na+] = 0.1 mM and [CI-] = 0.1 mM [Na+] = 0.1 mM and [Cl-] = 10 mM [Na+] = 1 mM and [CI-] = 1 mM [Na+] = 100 mM and [CI-] = 0.01 mM [Na+] = 10 mM and [Cl-] = 0.1 mM You are observing a cell at equilibrium with respect to the potassium cation (K+). The cell is in a bathing solution of 1 mM potassium and you measure the membrane potential of the cell to be -118 mV. What is the internal potassium concentration? 0.1 mM 10 mM 100 mM 1 mM 0.01 mM

Explanation / Answer

Q.No 6

The correct answer is [Na+] = 0.1 mM and [Cl-] = 10 mM

In a resting state, the membrane potential of a cell is called the Resting Membrane Potential and is equal to -70mV. The negative sign indicates the cell is negative inside which is surrounding the extracellular fluid.

Q.No 7

The correct answer is Option C : 100 mM.

The resting membrane potential of a cell is around -70 mV (most neurons). [K+] is an intracellular cation with a normal intracellular concentration of 140 mM.

By nerst equation, V= 62* log(K+ outside of cells / K+ inside of cells)

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