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Could someone check if I answered these questions correctly? (Biology)

  1. Feb 9, 2014 #1
    Information Given:
    Resting Membrane Potential: -72.88 mV
    Na+ Nernst Potential: 69.91 mV
    K+ Nernst Potential: -96.88 mV
    Cl- Nernst Potential: -21.10 mV


    1. At the resting potential, what is the direction of net movement for Na+, K+, and Cl-.

    My Answer:
    The Na+ ions are trying to get inside the cell since the Na+ ions would like to bring resting membrane potential closer to their own Nerst Potential of 69.91 and thus depolarize the cell. The K+ ions are trying to get outside the cell since the Na+ ions would like to bring resting membrane potential closer to their own Nerst Potential of -96.88 and thus hyperpolarize the cell to. The Cl+ ions are trying to get outside the cell since the Cl+ ions would like to bring resting membrane potential closer to their own Nerst Potential of -21.10 and thus depolarize the cell.


    2. A commonly used strategy to depolarize the neurons is to add K+ in the bath. What would be the resting membrane potential of this cell if you add K+ so that the total extra-cellular K+ becomes 30 mM?

    My Answer:
    Extracellular K+ Concentration: 30 mM
    Intracellular K+ Concentration: 155 mM

    New Nernst Potential of K+: (61/1)(Log(30/155)) = -43.51 mV -> Thus, the K+ ions will try to get inside of the cell to bring the cell’s resting membrane potential value of -72.88 mV to a value closer to -43.31 mV and consequently depolarize the cell.

    New Resting Membrane Potential: 61log(((1x30)+(0.04x140)+(0.01x40))/((1x155)+(0.04x10)+(0.01x120))) = -38.95 mV


    3. Suppose this cell is now poisoned by a spider toxin that blocks the K+ channel (PK becomes 0.0), will the cell be depolarized or hyperpolarized? Why?

    My Answer:
    The cell will depolarize since the internal K+ ions within the cell will accumulate and become more concentrated within the cell and won’t be able to leave the cell since the K+ channel has been blocked by the spider toxin. The cell membrane will deviate from its approximate value of -70 mV to a more positive value.

    New Resting Membrane Potential when PK=0: 61log(((0)+(0.04x140)+(0.01x40))/((0)+(0.04x10)+(0.01x120))) = 35.02 mV
     
  2. jcsd
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