Some real numbers
Before we get into the psychadelic equations, let's do some simpler math with ions and channels.
A typical cell volume is about 10-10 liters, and the concentration of K+ ions is 140mM (millimoles per liter). So, about how many K+ ions are in a typical cell?
- millimoles to moles : 140 millimoles = about .1 moles (rounding is OK)
- moles of K+ : (.1 moles/liter) *( 10-10 liters / cell) = 10-11 moles/cell
- particles per mole: there are about 6 * 1023 particles in a mole
- so ... : (10-11 moles/cell) * (6*1023 ions/mole) = ...
I think I have the answer: 6*1012 K+, or about 6 trillion
Six trillion -- let's see, about 1000 times the population of the earth...
Now let's see how many of those 6 trillion ions could be leaving the cell at any given time:
A typical cell has 10,000 K+ channels, and each channel can let 100,000 ions through per second that it is open. However, the typical channel is only open for 1 millisecond out of every second. So, how many K+ can leave at a time?
- ions per channel opening: 100,000 * 1/1000 = 100 every time a channel opens
- ... if each channel opens once a second ... : 10,000 channels * 100 each ions per opening = 1 million
I think I have the answer: 1 million ions per second
A million per second -- if ions were people, that would be a tenth of New York City every second -- sounds like a stampede to me!
What PERCENT of the cells K+ ions can typically leave in 1 second?
- I need a hint: 1 million out of 6 trillion
- ... using scientific notation... : 1* 106 / 6 * 1012
- ... to divide, subtract the exponents... : 1/6 * 10-6
I think I have the answer: 0.000017%
So on the one hand, ions are rushing out at the pace of a million a second... on the other hand, that's only about one hundred thousandth of a percent of the number of ions present! This is a very tiny percentage!! And the movement of that tiny percentage is what causes the tenth-of-a-volt membrane potential.
Here's another way to think about it: water can gush over a dam at a rate of hundreds or thousands of gallons a minute, yet the level of the water above and below the dam doesn't change perceptibly -- because there are millions of gallons of water involved. And, despite the fact that the flow is only a small percentage of the total water, it can still do a significant amount of work as it falls.
Finally, let's imagine for a moment that all of the K+ ions could continue to leak out of the cell at the rate of 1 million per second.
How long would it take to completely empty the cell of K+ ions at this rate?
- how many seconds?: 6 trillion ions / 1 million ions per second = 6 * 10^12 ions / 10^6 ions per second = 6 * 10^6 seconds needed
- how long is that? : 6,000,000 seconds * (1 hour/3600 seconds) = ...
I think I have the answer: 1,667 hours, or about 69 days
This last calculation is a fantasy, because in fact the cell will not continue to empty out at the same rate. Remember the 2 opposing gradients? As diffusion moves ions out of the cell, a voltage gradient builds up which pushes them back in. Eventually the two forces even out and ... voila, equilibrium.
Copyright University of Maryland, 2007
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