At rest, neurons are negatively charged with respect to the extracellular fluid. The magnitude of this electrical difference is referred to as Vm, the cell's membrane potential. An action potential is a positive change in membrane voltage initiated, carried, and terminated by different electrical currents across the membrane. In the 1940s and 1950s, Andrew Huxley and Alan Hodgkin used an apparatus called a voltage clamp to study the changes in electrical currents throughout the course of an action potential. As an action potential progresses down an axon, the membrane depolarizes and then repolarizes. Vm moves from –60 mV to more than 0 mV and then returns to –60 mV within milliseconds, making it difficult to observe the current's pattern. An experimenter can set the voltage clamp to hold the axon at a certain voltage, called the command potential, freezing the action potential at a moment in time. A voltmeter measures Vm and that value is compared to the command potential. If the values match, the machine does nothing. If they do not, the apparatus will supply a current to the axon sufficient to change the measured Vm to the command potential.Whatever current the axon's membrane naturally allows in or out will be countered with equal and opposite current from the apparatus. If the experimenters know what current they have applied, they know what current the membrane has passed. By performing repeated experiments, they measure the membrane currents that control the action potential.Figure 1 shows Hodgkin and Huxley's voltage clamp. A voltmeter measures Vm; one of its electrodes is placed within a section of axon, and the other is grounded in the extracellular fluid. The measured value is compared to the command potential and the voltage clamp amplifier supplies the current necessary to equalize the two values. The amount of current delivered is calculated by measuring the voltage drop over a series of two resistors.Figure 1 Voltage clamp apparatus, connected to axonQuestion 17Given R1 = 3 Ω and R2 = 1 Ω, and that the potential between d and f is –25 mV, how many electrons pass through point f in 4 seconds and in which direction? (The charge of an electron is –1.6 × 10–19 C.)A.3.54 × 10–28 away from the axonB.2.04 × 1016 toward the axonC.1.56 × 1017 away from the axonD.1.84 × 10–27 toward the axon
Question
At rest, neurons are negatively charged with respect to the extracellular fluid. The magnitude of this electrical difference is referred to as Vm, the cell's membrane potential. An action potential is a positive change in membrane voltage initiated, carried, and terminated by different electrical currents across the membrane. In the 1940s and 1950s, Andrew Huxley and Alan Hodgkin used an apparatus called a voltage clamp to study the changes in electrical currents throughout the course of an action potential. As an action potential progresses down an axon, the membrane depolarizes and then repolarizes. Vm moves from –60 mV to more than 0 mV and then returns to –60 mV within milliseconds, making it difficult to observe the current's pattern. An experimenter can set the voltage clamp to hold the axon at a certain voltage, called the command potential, freezing the action potential at a moment in time. A voltmeter measures Vm and that value is compared to the command potential. If the values match, the machine does nothing. If they do not, the apparatus will supply a current to the axon sufficient to change the measured Vm to the command potential.Whatever current the axon's membrane naturally allows in or out will be countered with equal and opposite current from the apparatus. If the experimenters know what current they have applied, they know what current the membrane has passed. By performing repeated experiments, they measure the membrane currents that control the action potential.Figure 1 shows Hodgkin and Huxley's voltage clamp. A voltmeter measures Vm; one of its electrodes is placed within a section of axon, and the other is grounded in the extracellular fluid. The measured value is compared to the command potential and the voltage clamp amplifier supplies the current necessary to equalize the two values. The amount of current delivered is calculated by measuring the voltage drop over a series of two resistors.Figure 1 Voltage clamp apparatus, connected to axonQuestion 17Given R1 = 3 Ω and R2 = 1 Ω, and that the potential between d and f is –25 mV, how many electrons pass through point f in 4 seconds and in which direction? (The charge of an electron is –1.6 × 10–19 C.)A.3.54 × 10–28 away from the axonB.2.04 × 1016 toward the axonC.1.56 × 1017 away from the axonD.1.84 × 10–27 toward the axon
Solution
To solve this problem, we first need to convert the potential difference from millivolts to volts. So, -25 mV becomes -0.025 V.
Next, we use Ohm's law (V = IR) to find the current (I) flowing through the circuit. Since the potential difference is across both resistors, we add them together to get the total resistance (R). So, R = R1 + R2 = 3 Ω + 1 Ω = 4 Ω.
Now, we can calculate the current: I = V/R = -0.025 V / 4 Ω = -0.00625 A.
The negative sign indicates that the current is flowing in the opposite direction to the conventional current flow (from positive to negative), which means it is flowing towards the axon.
Next, we need to find the total charge (Q) that has passed through point f in 4 seconds. We use the formula Q = It, where t is the time in seconds. So, Q = -0.00625 A * 4 s = -0.025 C.
Finally, we find the number of electrons by dividing the total charge by the charge of a single electron (-1.6 × 10^-19 C). So, the number of electrons is Q / e = -0.025 C / -1.6 × 10^-19 C = 1.56 × 10^17.
So, the answer is C. 1.56 × 10^17 electrons pass through point f in 4 seconds, moving towards the axon.
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