# MathBench>Cellular Processes

## 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?

(To make this problem interactive, turn on javascript!)

• 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?

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• 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?

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• 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?

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• 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.