The chloride ion that conducts a distrubution of $55 \mathrm{mM}$ outside the cell and $35 \mathrm{mM}$ inside the cell. The system at $37 \mathrm{C}$. Threshold is $-45 \mathrm{mV}$ if one sets the membrane voltage to $-65 \mathrm{mV}$ the driving force (D) on the in conducted by this channel will be

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Sure, let's address the question about the driving force on chloride ions given the provided conditions. ∻Answer∻ ⚹ The driving force on the chloride ion can be calculated using the Nernst equation and the given membrane voltage. ⚹ ∻Solution∻ ‖ a ⋮ Calculate the Nernst potential ($E_{Cl^-}$) for chloride ions using the Nernst equation: $E_{Cl^-} = \frac{RT}{zF} \ln \left( \frac{[Cl^-]_{out}}{[Cl^-]_{in}} ight)$ ‖ ‖ b ⋮ Given values: $R = 8.314 \, ext{J/mol·K}$, $T = 310 \, ext{K}$ (37°C), $z = -1$ (charge of chloride ion), $F = 96485 \, ext{C/mol}$, $[Cl^-]_{out} = 55 \, ext{mM}$, $[Cl^-]_{in} = 35 \, ext{mM}$ ‖ ‖ c ⋮ Substitute the values into the Nernst equation: $E_{Cl^-} = \frac{8.314 imes 310}{-1 imes 96485} \ln \left( \frac{55}{35} ight)$ ‖ ‖ d ⋮ Simplify the equation: $E_{Cl^-} = -0.0267 \ln \left( 1.571 ight)$ ‖ ‖ e ⋮ Calculate the natural logarithm: $\ln(1.571) \approx 0.451$ ‖ ‖ f ⋮ Final calculation: $E_{Cl^-} = -0.0267 imes 0.451 \approx -0.012 \, ext{V} \, ext{or} \, -12 \, ext{mV}$ ‖ ‖ g ⋮ Determine the driving force ($D$) using the formula: $D = V_m - E_{Cl^-}$, where $V_m = -65 \, ext{mV}$ ‖ ‖ h ⋮ Substitute the values: $D = -65 \, ext{mV} - (-12 \, ext{mV}) = -65 \, ext{mV} + 12 \, ext{mV} = -53 \, ext{mV}$ ‖ ∻Key Concept∻ ⚹ Driving force on ions ⚹ ∻Explanation∻ ⚹ The driving force on an ion is the difference between the membrane potential and the ion's equilibrium potential. In this case, the driving force on chloride ions is -53 mV, indicating the direction and magnitude of ion movement across the membrane. ⚹◊What is the Nernst equation and how is it used to calculate membrane potential?⍭ Generate me a similar question◊

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