Question

please give an explanation for question 31 thank you!

31. GIVEN: Anion Q has an equilibrium potential of -60 mV. a. (2 pts) Is Q in greater concentration inside or outside the cell? Explain your answer. Q is greater consentration Instre of the cell. This is because Q is a ano is trying to enter the cell, where to making it regate b. (2 pts) If the cell is at rest (with a membrane potential of -70 mV), would the electrochemical Q move it into or out of the cell? Explain your answer. The electrechend force will move it outside the cell Because the electried fore it many it outsie chod te clemeal force +1 - 70 +60=-10 mV ? C. (2 pts) In this resting cell, if channels for Q suddenly opened, how would membrane potential (Would it become more negative, less negative or not change?) Explain. It would become less nentive since it will be moving outside

Answer 1

1) Since the equilibrium potential of anion Q is -60mV, this means that at -60mV there is no net flow of anion Q across the membrane. The equilibrium potential can be calculate using the Nernst potential.

Equation for Nernst potential is V_eq = (RT/zF) ln([X]_out/[X]_in)

where:

• R is the universal gas constant
• T is the temperature in Kelvin
• z is the valence of the ionic species (+ve for cations and -ve for anions)
• F is the Faraday's constant.
• [X]_out is the concentration of the ionic species X outside the cell
• [X]_in is the concentration of the ionic species X inside the cell
• ln is natural log

Now, here, z for anion Q will be negative (-). Therefore, all the other terms (except z, which is already –ve) in the equation should be positive as the L.H.S. (left hand side), which is the equilibrium potential we know is -60mV. All the terms except the ion concentrations inside and outside the cell, are constants so, for now, we can disregard them. Since the natural log of any number less than 1 but greater than 0 will be negative, the ratio of concentration outside to concentration inside has to be greater than 1. This can only be the case if [X]_out > [X]_in. Therefore, from this we can infer that the concentration of the ion is greater outside the cell as compared to the inside the cell.

2) Since the equilibrium voltage for anion Q is -60mV it means that at -60mV, net flow of ions is zero. When the resting membrane voltage is -70mV, it means that the voltage inside the membrane is more negative as compared to the outside of the membrane. Electrochemical force on an ion is influenced by two main factors: its concentration gradient across the membrane as well as the potential across the membrane. Since the negative potential is greater inside the cell, it will act on the anion to try to move it outside the cell (so as to try and reach the equilibrium potential). Therefore, the electrochemical force will try to move anion Q outside the cell.

You can also solve for this using the Nernst equation, Where V is -70mV (L.H.S.) and the only way that V can go from -60mV to -70mV is if [X]_out increases even more (as everything else is constant). Which means the force acting on the ion will try to push it outside to increase [X]_out.

3) If the channels are opened, the ions will flow through the channels to try and reach the equilibrium potential (here the assumption is that there are no other ions or ion channels). Since the equilibrium potential for anion Q is -60mV, and the membrane potential is -70mV, the ions will try and flow outside the cell to reduce the negative membrane potential, to -60mV, thereby causing the net flow of ions to become zero. Therefore, the membrane potential will become less negative.