QofQuimica: You may recall from your neurobiology lessons that the axons of neurons are myelinated, and that myelin speeds the transmission of action potentials. However, you may not have ever thought about why this is true. This post explains an application of neurotransmission concepts to biology (cell membranes). So for all of you who are wondering why you have to study physics, and what physics has to do with anything in the "real world" for a pre-health student, here is one great example. It will hopefully also increase your intuitive understanding of the principles behind myelination and neurotransmission.

Lindyhopper: My remembrance is that myelin increases the resistance of the membrane, and therefore reduces its capacitance. If we compare a myelinated axon with an unmyelinated axon, the unmyelinated axon has a larger capacitance, so more charge must be deposited on the membrane to change the potential across the membrane. Thus, the current must flow for a longer time to produce a given depolarization. I’m trying to come up with a quick, simple analogy that might possible be helpful to explain this. I was thinking, perhaps, of comparing the unmyelinated axon to a soaker hose & the myelinated to a normal garden hose. But this has problems. The ions charging the capacitor do not normally flow through the membrane. I’m also not sure if the water flowing through the soaker hose flows significantly more slowly.

Nutmeg: If you’re looking for an anology to understand membrane capacitance, here’s the one that I always use for wave propagation. Imagine that you have one huge, heavy, chain and one long, unassembled fiberglass tent pole (the kind you use for a dome tent, where you have lengths about a foot long, connected by a single elastic cord that runs through the whole length) of equal length. The tent pole will sag a bit, yes, but the chain sags completely. Now imagine trying to hold one end and getting the other end to respond to a quick snap–basically, you’re trying to whip it. With the chain, it will take enormous effort to get any degree of propagation because there are so many places where the movement gets transferred. With the tent pole, each segment is relatively stiff, and movement of one section sends (propagates) the energy much further than does a single chain link.

I usually think of these sorts of matters in terms of stiffness when trying to propagate a wave. You end up with a continuum, from a heavy steel chain, to a cotton jump rope, to a stiff nylon rope, to a bamboo pole, to a steel rod. The things that matter are the number of joints (more jointed things will be slower and take more nergy), the weight (the heavier it is, the more energy it needs), and the stiffness. In the case of the axon, the myelin makes the axon have fewer "joints." So, instead of a constant in-out flux down the entire length of the axon membrane, the local depolarization at the nodes of Ranvier can cause brief ionic diffusion along the length of the axon only at discrete points. The "weight" in this situation is analogous to the amount of charge required to propagate the signal. The decreased capacitance of the myelinated axon means that less charge is needed to propagate an action potential, as Lindyhopper stated. Likewise, the heavy chain needs more work to get the same pulse size as the light chain.

Finally, there is the stiffness. In the world of ropes and rods, a stiff region will behave very similarly to neighboring regions. Something floppy, however, can easily bend, and it doesn’t transfer energy as quickly or efficiently. The internodal regions, in this example, are very "stiff". They do not have local variations because they can’t communicate with the extracellular space very well. Thus, when one area changes in polarity, since the membrane cannot pass ions to or from the extracellular space, the adjacent intracellular regions will pass ions instead. You’ve reduced what are known as the degrees of freedom–the number of possible modes that can be adopted by the axon–such that charge will go up or down the length of the axon, but not in and out of the cell.

Or, if all that was too complicated, imagine a chain of people playing telephone. In each case, the message will be sent a distance of one kilometer. If you place the people at a distance of two meters apart and have them relay a message, one to the next, it will take significantly longer to send a message one kilometer than it would if you had spaced the people 25 meters apart. Analogously, having fewer nodes means that the message will be transmitted over the same distance in a shorter amount of time.

Some people still have problems converting cm^3 and mm^3 to m^3 for problems such as density and volume.

1) When converting from cm^3 to m^3 move 6 decimal places to the left. For example if the volume of a box is 120 cm^3, then converting it, we will get .000120 m^3 or 1.2 x 10^-4 m^3

2) When converting from mm^3 to m^3 move 9 decimal places to the left. For example, if the volume is 120 mm^3, the conveting it, we will get .000000120 m^3.

**REASONING**

1 m^3 = 1m x 1m x 1m = 100 cm x 100 cm x 100cm = 1000mm x 1000mm x 1000 mm

Recent MCATs have included several questions about simple machines — devices that alter the strength of the input force and the distance through which force is applied. You should be familiar with the force multiplication concept, and somewhat familiar with some of the machines that use it.

The basic concept is this: you can change force, but you can’t get something for nothing, i.e., energy is not created or destroyed by any of these machines. The energy of a force is, more properly stated, the work done by that force; it follows that the work done by an input force is the same as that done on the object that is being moved (unless there’s friction, in which case energy is lost; we’ll neglect friction for the rest of this discussion). As work = force x distance x cos(theta), if we increase force we have to decrease distance equivalently, and vice versa. (For this discussion we’ll consider forces that do work in the direction in which they are applied, i.e., cos(theta) = 1, and W = Fd.) The amount by which force is increased by a system, and therefore distance decreased, is the force multiplier.

Because of the fixed relationship between force and distance, it is often easier to consider distance (which we can see) than force, and then work backwards. For example, if you can see that the thing you’re moving travels, say, three times farther than the force that moves it, then the force on the object is three times less than the input; if you can see that it travels only a quarter as far, then the force exerted on it is four times as great, and so on.

The simple machines (and other devices with force multipliers) follow, including all six classic simple machines (even those that aren’t likely to appear in the test; I note which these are). Do not memorize these; instead, be moderately familiar with them, especially the more common ones.

• Pulley: A single pulley or a system thereof is not necessarily a simple machine, but it can be. For MCAT-style pulley systems, output force is input force multiplied by the number of ropes that are effectively pulling on the object to be moved; equivalently, by the number of ropes that must (not just might) shorten in order for the resistance to move. Another hint: if none of the pulleys is free to move, then the force multiplier is 1: force and distance remain the same. Of course, if the force multiplier is not 1, i.e., you get greater force on the object than you put in, you will get less distance by an equivalent factor. Appears on the MCAT.
• Inclined plane: the force multiplier (relative to straight lifting) is the sine of the angle of the plane with the horizontal. Distance is divided by the same factor. Consider the problem another way: If you push a block up an incline, you push along the hypotenuse, but lift only the amount of its height; the ratio of these two is sin(theta) Because distance is less than what you input, force must be greater by the same factor — the force multiplier. Common on the MCAT.
• Lever: the force multiplier is the ratio of distance from fulcrum to object, to distance from fulcrum to applied force. (The "fulcrum" of a lever is its pivot point.) Look at the picture to make sure you know whether distance goes up or down (hence force down or up). This treatment works even when the force and the object are on the same side of the fulcrum. I saw a lever — a claw hammer pulling a nail — and a question about its force multiplier and another about its fulcrum, on the August 2004 MCAT, so you can’t ignore this one.
• Hydraulic jack: not usually considered to be a simple machine like the aforementioned, but effectively the same. Force multiplier is the ratio of the areas of the tops of the two pistons. Recall that force changes, not pressure — pressure is the same (at the same height) everywhere in a body of liquid. Assuming that the change in height is negligible, it also doesn’t matter what liquid is used. Appears on the MCAT.
• Wedge: similar to an inclined plane; the force multiplier is length divided by width; distance is divided by the same factor. Unlikely on the MCAT.
• Wheel and axle: the force multiplier is ratio of radii of the wheel and the axle. To figure whether to divide or multiply the two, look at the picture to see whether the distance the object travels is greater or less than the distance over which the force is applied; that will tell you whether the force is decreased or increased, respectively. Unlikely on the MCAT.
• Gears (technically not a separate type of simple machine, but they work the same as the others): the force multiplier is the ratio of the numbers of teeth on the two wheels. Again, look at the picture to see which increases, force or distance. Unlikely on the MCAT.
• Screw: an inclined plane wrapped around a rod. Unlikely on the MCAT. For completeness: the force multiplier is 2 x pi x radius x (turns/unit length). Don’t worry about this one at all.

It is possible to combine two or more of the types: for example, a cam is an inclined plane combined with a wheel and axle. But don’t worry: such a problem is very unlikely to appear on the MCAT, and if it did it would necessarily be a very simple setup.

The only conversion you will have to do is of prefixes, e.g., micro-whatsis to kilo-whatsis. No kilograms to slugs, or liters to hogsheads, or meters to chains, or…

For those who have trouble converting units, I suggest the following method:

First, forget the conversion factors listed in whatever reference material you’re using. They’re correct, of course, but if you’re having trouble they’re what’re mixing you up. Instead, just think in terms of big and small, as shown here. Also, eschew multiplying by negative exponents — divide instead. Finally, think of the units as variables, and cancel as you would with variables.

An example problem will show the method best. Lets say you want to know the speed of light, c = 3×10^8 m/s, but in cm/ms.

Write down 3×10^8 m/s on the left. Write down the units you want, cm/ms, on the right. Now multiple and divide to get rid of the m and the s, and to get cm and ms in the right spots:

Code:

3x10^8 m x _____cm x _____s_ = _______cm       s        m         ms          ms

Now fill in the numbers, which you should know, just focusing on which unit is bigger, and remembering that it takes more of the smaller ones to make a big one:

Code:

3x10^8 m x 10^2_cm x ___1_s_ = _______cm       s      1 m    10^3 ms          ms

Check to ensure that the old units cancel out, that you wind up with the new units, that you have more of the smaller units in each fraction (the fractions must each equal 1); now do the math:

Code:

3x10^8 m x 10^2_cm x ___1_s_ = 3x10^7_cm       s      1 m    10^3 ms          ms

Finally, never use a squared or cubed conversion factor: use the normal conversion factor, and square or cube it at the end. As another example, find the density of osmium in kg/m^3:

Code:

22.4 g___ x ___1_kg x (____cm)^3 = _______kg_     cm^3        g    (    m )^3          m^3

Code:

22.4 g___ x ___1_kg x (100_cm)^3 = 3x10^7_kg_     cm^3   1000 g    (  1 m )^3          m^3

Code:

22.4 g___ x ___1_kg x 10^6_cm^3 = 2.24x10^4_kg_     cm^3   1000 g       1 m^3              m^3

Notice that we filled in the cm/m conversion, which we know well, and squared at the end. Notice also that there’s no need to use scientific notation for many numbers — the notation is a tool, not an end in itself.