In a current clamp experiment, you inject enough current into a neuron to bring the voltage from $-80 \mathrm{mV}$ to a steady state of $-60 \mathrm{mV}$. Given a time constant of $5 \mathrm{~ms}$, how long will it take for the voltage to drop to $-71 \mathrm{mV}$ after current injection ends?

Group of answer choices
$0.5 \mathrm{~ms}$
$4 \mathrm{~ms}$
$1.7 \mathrm{~ms}$
$$
6 \mathrm{~m}
$$

Question

In a current clamp experiment, you inject enough current into a neuron to bring the voltage from $-80 \mathrm{mV}$ to a steady state of $-60 \mathrm{mV}$. Given a time constant of $5 \mathrm{~ms}$, how long will it take for the voltage to drop to $-71 \mathrm{mV}$ after current injection ends?

Group of answer choices
$0.5 \mathrm{~ms}$
$4 \mathrm{~ms}$
$1.7 \mathrm{~ms}$
$$
6 \mathrm{~m}
$$

Accepted Answer

***B*** ∻Key Concept∻ ⚹Voltage decay in a neuron post-current injection⚹ ∻Explanation∻ ⚹The voltage decay after current injection in a neuron can be described by the time constant (τ). The time constant is the time it takes for the voltage to change by 63% of the difference between its initial value and its final value. Since the voltage changes from -60 mV to -71 mV, which is a 63% change towards the resting potential (-80 mV), it takes approximately one time constant, which is 5 ms. However, the question asks for the time to reach -71 mV, which is a 37% change (100% - 63%) towards the resting potential. This corresponds to approximately 0.37 of the time constant. Therefore, 0.37 * 5 ms = 1.85 ms, and the closest answer choice to this calculation is 1.7 ms.⚹◊What is the relationship between the time constant and the rate of change in voltage in a current clamp experiment⍭ Generate me a similar question◊

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