During an action potential, Na^+ ions move into the cell at a rate of about 3x10^-7 mol/m^2*s.
How much power must be produced by the "active Na^+ pumping" system to produce this flow against a +30-mV potential difference? Assume that the axon is 40cm long and 30um in diameter.
Please help me out I am lost. Thanks
Physics please help - drwls, Sunday, February 10, 2013 at 12:52pm
I am not familiar with the terminology. How can a "potential" be an event or a time interval?
Power = (current) x (voltage)
The voltage is apparently 30*10^-3 V
Convert the 3x10^-7 mol/(m^2*s) to
coulombs/(m^2*s)
1 mole of Na+ ions contains 6.02*10^23 ions with a charge of 9.63*10^4 Coulombs. Your ion flux is 2.89*10^-2 Amps/m^2
Multiply that by the appropriate axon area.
Please help am still stuck on this one I cant get the right answer!
To calculate the power required by the "active Na+ pumping" system, we need to follow a few steps:
1. Firstly, convert the given molar flow rate of Na+ ions (3x10^-7 mol/m^2*s) to electric current by using Faraday's constant.
1 mole of Na+ ions contains 6.02x10^23 ions with a charge of 9.63x10^4 Coulombs. Therefore, the ion flux can be calculated as follows:
Ion Flux = (3x10^-7 mol/m^2*s) x (6.02x10^23 ions/mol) x (9.63x10^4 C/ion)
2. Next, calculate the area of the axon that the ions are flowing through. The axon diameter is given as 30 µm, so the radius would be half of that (15 µm or 15x10^-6 m). The cross-sectional area of the axon can be calculated as follows:
Axon Area = π(radius)^2 = π(15x10^-6 m)^2
3. Now multiply the ion flux by the axon area to get the electric current flowing through the axon:
Current = Ion Flux x Axon Area
4. Finally, calculate the power using the formula: Power = Current x Voltage
The voltage difference across the axon is given as +30 mV, which can be converted to volts by multiplying it by 10^-3.
Power = Current x Voltage = Current x (+30x10^-3 V)
By following these steps and plugging in the given values, you should be able to calculate the power required by the "active Na+ pumping" system.