At 50 mV, what ion would exert the greatest force… would it be the number closest to 50 or farthest?
The correct answer and explanation is:
To determine which ion would exert the greatest force at 50 mV, we need to consider the relationship between the electrochemical gradient and the membrane potential. The force exerted by an ion is influenced by its concentration gradient across the membrane and its charge. This is often quantified using the Nernst equation, which calculates the equilibrium potential for an ion based on its concentration inside and outside the cell.
The Nernst equation is given as:
$E = \frac{RT}{zF} \ln\left(\frac{[ion]{out}}{[ion]{in}}\right)$
Where:
- $E$ is the equilibrium potential (or Nernst potential) of the ion.
- $R$ is the gas constant.
- $T$ is the temperature in Kelvin.
- $z$ is the charge of the ion.
- $F$ is the Faraday constant.
- $[ion]{out}$ and $[ion]{in}$ represent the concentration of the ion outside and inside the cell, respectively.
At a given membrane potential, ions that are closest to their equilibrium potential will exert the greatest force. When the membrane potential is far from the equilibrium potential of an ion, there is a strong electrochemical gradient that drives the ion to move across the membrane, resulting in a greater force.
At 50 mV, the ion whose equilibrium potential is closest to this value will exert the greatest force. The ion’s force is determined by how far the membrane potential is from the ion’s equilibrium potential. The further the membrane potential is from an ion’s equilibrium potential, the greater the electrochemical driving force for that ion.
Therefore, it is not necessarily the ion closest to 50 mV in terms of absolute value that will exert the greatest force, but rather the one whose equilibrium potential is closest to 50 mV. This creates a larger electrochemical gradient and a greater driving force for that ion to move across the membrane.