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filtration occurs in one step, when the filtrate is formed from blood at high pressure in the glomerulus (Fig. 2.2). Filtration in earthworms, in contrast, is a two-step process. First, coelomic fluid is produced by filtration directly from the blood, analogously to the deposition of filtrate in the glomerulus. Then coelomic fluid is filtered across the nephridiostome into the tubule. Both filtration steps are pressure driven.2 The first step is driven by the worm's blood pressure. The annelid heart can sustain blood pressures that, while modest compared with ours (roughly 20-50 percent of the average systolic pressure of humans), are high by invertebrate standards. The second step is driven by

2. Among some annelid worms, differences in osmotic pressure between the coelom and blood also drive filtration.

coelomic pressure, which is imparted to the coelomic fluid by the tunic of locomotory muscles that invest the worm's body. Once in the tubule, salts and water are recovered from the filtrate by reabsorption, and the filtrate's composition is otherwise modified by secretion, just as it is in the vertebrate nephron. The urine then passes out the tubule through the nephridiopore to the outside.

The structures of annelid nephridia correlate with particular habitats, just as the fish nephron reflects the animal's "proper" physiological place. Marine polychaetes, for example, have very small and poorly vascularized nephridiostomes, and the nephri-diostomes are closed, not open, imposing a barrier to the production of filtrate by coelomic pressure (Fig. 7.2). The overall result, as one would expect, is a nephridium that produces only small quantities of filtrate. Nephridiostomes of freshwater polychaetes and freshwater oligochaetes, on the other hand, are large, well perfused, and open, apparently "designed" to produce large quantities of filtrate (Fig 7.1).

Freshwater annelids take this pattern a couple of steps further. In most annelids, the nephridiostome sits a.

nephridiostome b.

nephridiostome solenocytes

Figure 7.2 "Low filtration" nephridia of annelids. a: A "closed" nephridiostome from the polychaete Phyllodoce paretti. The end of the nephridiostome is a closed tubule, in which filtration is driven by currents set up by the solenocytes (flagellated cells that drive fluid down the tubule). b: The nephridium of Phyllodoce paretti is formed from a cluster of these tubules, each topped with a cluster of solenocytes. [From Goodrich (1945)]

solenocytes nephridial canal

Figure 7.2 "Low filtration" nephridia of annelids. a: A "closed" nephridiostome from the polychaete Phyllodoce paretti. The end of the nephridiostome is a closed tubule, in which filtration is driven by currents set up by the solenocytes (flagellated cells that drive fluid down the tubule). b: The nephridium of Phyllodoce paretti is formed from a cluster of these tubules, each topped with a cluster of solenocytes. [From Goodrich (1945)]

in the same segment as the rest of nephridium, and the septa between segments are incomplete, so that coelomic fluid can flow from one segment to another (Fig. 7.3). This arrangements limits the extent to which coelomic pressure in any one segment can rise above that in an adjacent segment. Freshwater oligo-chaetes, however, typically have complete septa, a design that allows large pressure differences to develop between segments, as high as a few hundred pascals (Fig. 7.3). Furthermore, the nephridiostome for one nephridium pokes through the septum into the adjacent segment. The pressure differences between segments provide an additional motivating force for production of filtrate. The bottom line, though, is that nephridia of freshwater oligochaete worms are constructed to produce large quantities of filtrate, as one would expect.

What about the nephridia of earthworms? They, in fact, have much in common with the nephridia of freshwater oligochaetes, and very little in the way of the structural adaptations one would expect to see in an animal living on land. Functionally, earthworms produce urine as if they were freshwater animals, with losses of water ranging from 60 to 90 percent of body weight per day (in contrast, our own urinary losses of water range from 5 to 10 percent of body weight per day). There appears to be considerable recovery of salts as well, on the order of 90 percent of the salts in the original filtrate. Marine polychaete worms, with their more appropriately constructed nephridia, have urine production rates that are only about 5-35 percent of body weight per day, with little recovery of salts.

It is hard to escape the conclusion that earthworms are essentially aquatic oligochaetes, poorly equipped physiologically for life on land. Yet there they are. Earthworms still face the water balance problems of a terrestrial animal, and their successful habitation on land implies something is doing the physiological work necessary for a terrestrial existence. This raises an obvious question: if earthworms' internal physiology equips them so poorly for this task, then what is doing the work? You probably will have guessed by now that

Figure 7.3 The effect of the septum on filtration across the nephridiostome. a:When the septum is incomplete, pressure in one segment can do work moving coelomic fluid from one segment to another. Consequently, little energy is left to drive filtration. b: When the septum is complete, more energy is available to drive filtration.

Figure 7.3 The effect of the septum on filtration across the nephridiostome. a:When the septum is incomplete, pressure in one segment can do work moving coelomic fluid from one segment to another. Consequently, little energy is left to drive filtration. b: When the septum is complete, more energy is available to drive filtration.

my answer to this question lies in the structures built by earthworms as they tunnel through soils.

Before I am able to convince you of this, I must first develop two arguments. First (and fairly straightforward), I need to show how respiratory gases like oxygen behave in air and in water, because this characteristic illuminates why animals go to the trouble of coming onto land in the first place. Second (and somewhat more involved), you need to know something about the physical forces that move water around in the porous medium of soils. Once those pieces are in place, we will be ready to look at how the burrows and other structural modifications earthworms make to soil help do their physiological work for them.

Life on Land: Why Bother?

If moving onto land is so challenging physiologically (Box 7A), it is tempting to ask why organisms bothered to make the move in the first place. This is one of biology's Big Questions, and, not surprisingly, several possible answers have been proposed, some of which make more sense than others. Explanations have included: easier access to light for powering photosynthesis, escape from aquatic predators, more readily exploitable food stores, and so on. Among the more sensible answers, in my opinion, is simply that oxygen from water is scarce and expensive to obtain, much more so than oxygen obtained from air. Just how scarce and expensive it can be is illustrated with a simple calculation. Suppose an animal wants to metabolize and extract the energy from one gram of sugar. How much oxygen does it need to do this? And how much air, or water, is required to provide that much oxygen?

The first question is easy to answer. We have already seen (Chapter 2) that six molecules of oxygen are required to oxidize one molecule of glucose. If you work out the respective molecular weights of oxygen and glucose, you will see that about 1.1 grams of oxygen are required to oxidize one gram of glucose.3 The amount of air you need to get 1.1 g of oxygen is also easy to calculate: you need a little over 3.6 liters of air. Remember this number.

The availability of oxygen in water is limited by its solubility, which describes how much oxygen can be dissolved in water before it starts coming out as gas bubbles. The quantity of oxygen in solution is governed by Henry's law, which states:

3. The molecular weight of glucose is 180 g mol- , so 1 g of glucose is 5.56 mmol. Glucose oxidation will require six times that molar quantity of oxygen, or 33.3 mmol, or (since the molecular weight of oxygen gas is 32 g mol-1) 1.07 g. With gases, one mole will occupy roughly 22.41. The volume occupied by 33.3 mmol of oxygen will therefore be 746 ml. Air is a mixture of gases, of which 21 percent is oxygen. The volume of air required to hold 746 ml of oxygen is therefore 3.56 l.

where [O2]s is the concentration of oxygen in solution (in moles per liter), pO2 is the oxygen partial pressure (in kilopascals, or kPa), and aO2 is oxygen's Bunsen solubility coefficient (mol l-1 kPa-1). In water at 20oC oxygen's solubility is 13.7 X 10-6 mol l-1 kPa-1. At sea level, where atmospheric pressure is about 101 kPa, the oxygen partial pressure will be about 21 percent of that, or about 21.3 kPa. Water exposed to atmospheric air will therefore have an oxygen concentration of about 292^moll-1 (1 ,amol = 10-6 mol). To extract 1.1 grams of oxygen (which is about 0.037 moles, or 37,000 ,amol), you will need 126 liters of water. This is roughly 35 times the volume needed (3.6 liters) of air. This is only the best case: it gets worse if the water is warmer than 20oC, or if its access to the air is somehow impeded, or if the altitude is higher than sea level, or if you are sharing the water with lots of other things that are consuming oxygen.

Not only is oxygen in water scarce, it is expensive to extract. Whenever an animal extracts oxygen from either water or air, it must move the fluid past a gas exchange organ, either a gill or lung. This means doing work on the fluid to pump it past the gas exchanger: obviously, the less fluid that has to be pumped, the lower the energy costs for pumping will be. Air is the clear winner here, because only about 3 percent as much of it must be pumped to extract the same quantity of oxygen from water. Air is also easier to pump, because it is about a thousand times less dense than water, and less viscous. The bottom line is that the costs of breathing are considerably less for animals that breathe air (roughly 0.5-0.8 percent of total energy expenditure) than it is for animals that breathe water (5-20 percent of total energy expenditures). The much lower "overhead" means that air breathers have more energy left over to make babies.

So that—breathing easy—is the first piece of our puzzle. The aquatic ancestors of earthworms probably made the transition from water to terrestrial soils because it enabled them to tap into the more abundant and easily exploitable oxygen there.

Water in Soils

The second piece of the puzzle is how water moves in soils. This is rather more involved than the straightforward matter of oxygen, but at root it is really fairly simple. You just have to remember two things. First, for a parcel of water to move anywhere, it must be pushed or pulled there—that is, some force must act on it to start it moving and keep it moving. Second, if a parcel of water is standing still, this does not mean there are no forces acting on it. It does mean that the net force acting on it is nil—that is, every force acting to push it one way is balanced by an equal force pushing it the other way.

Soil scientists have been concerned for a long time with how water moves in soils, and they have developed a comprehensive body of theory to help them with this problem. We need to understand the elements of this theory, which is centered around a quantity known as the water potential. As the name implies, the water potential is a measure of the potential energy that can do work on water (namely move it around). The water potential, symbolized by the Greek letter W (psi), is equivalent to a pressure, conveniently enumerated with units of pascals (equivalent to newtons per square meter).

There are several kinds of forces that can act to move water, and the water potential accounts for all these forces. These forces are: the pressure potential, Wp; the gravity potential, Wg; the osmotic potential, Wo; and the matric potential, Wm. These add to give the water potential, W:

Of the four components of the water potential, the pressure potential is the easiest to grasp: it is simply the hydrostatic pressure acting on a parcel of water. Water in a hose pipe has a pressure potential, for example. Pressure potentials can be positive (forcing water away) or negative (drawing water in, that is, a suction pressure). For an earthworm, pressure potential is important because the heart and muscles of the body can elevate the pressure of its internal water by several thousand pascals (kilopascals).

The gravity potential is also straightforward.4 Water has mass, and therefore gravity will pull it downward. Gravity potential is proportional to the height of a parcel of water, h, with respect to some reference level, hO.

vention, the osmotic potential is always negative. In other words, it is always represented as a suction pressure, the magnitude of which depends upon solute concentration:

where p is the water's density (roughly 1,000 kg m-3) andg is gravitational acceleration (9.8 m s-2). The reference level, ho, usually is the height (in meters) at which water would come to rest independent of any other force acting on it. For example, if a parcel of water resting on the ground was lifted 1 m, it will have a gravity potential of 9,800 Pa, 9.8 kPa, with respect to ground level. Gravity potential is also handy for determining such practical things as pumping requirements for ground water. For example, if the water table is 10 m below the ground (h - ho = -10 m), getting it up to the surface where it can water a crop requires a pump capable of developing 1,000 kg m-3 X 9.8 m s-2 X -10 m = -98,000 Pa or -98 kPa of suction pressure.

The osmotic potential reflects the effect of solute concentration on movement of water. The osmotic potential represents a pressure that draws water toward parcels with higher concentrations of solute. By con

4. Some of this discussion may be confusing without some explanation of the "convenient" units of pascals. The pascal is a so-called derived unit, which distinguishes it from the fundamental units of length, time, mass, and temperature. The pascal is a measure of pressure, the quotient of a force, in newtons (N), acting over an area, in square meters. Thus, 1 Pa = (1 N) h-(1 m2) = 1 N m-2. The newton itself is a derived unit, the product of a mass (in kg) and an acceleration (in m s-2), which in fundamental units would be written as 1 N = (1 kg) X (1m s-2) = 1 kg m s-2. The pascal in its fundamental units would therefore be 1 Pa = (1 kg m s-2) X (1m-2) = 1 kgm-1 s-2. This should clarify how a water potential arises from the formulas given in the text. Taking the gravity potential as an example (equation 7.3), is the product of a water density, p (in kg

m ), a gravitational acceleration, g (inms ), and a height, h (in m). The units of this product are (kgm-3) X (ms-2) X (m) = kg m-1 s-2, the same as the derived unit of the pascal.

where C is the molar concentration of solute (mol m-3), R is the universal gas constant (8.314 J mol-1 K-1), and Tis the absolute temperature (in kelvins).5

The matric potential is the trickiest of the components of the water potential, and it cannot be encapsulated in a neat formula the way the other components can be. It is also the most important to the story I am building, so we must give it a fair bit of attention.

Anyone who has ever soaked up a spill with a paper towel has seen matric potentials at work. Water is drawn up because the towel exerts a force on the water, the so-called matric force, sometimes called capillary action. The matric force arises from an interaction between electrical charges on surfaces, in this example charges on the cellulose threads of the paper towel and similar but opposite charges on water molecules. Just as opposite poles of a magnet exert a mutual attractive force on one another, so too do these opposite charges attract one another.

This forceful interaction is visible to the naked eye in a glass containing water. Most of the water's surface is flat, but near the sides of the glass the water creeps up slightly onto the glass wall, forming a curved surface called a meniscus. The meniscus arises from a balance of three forces. Pulling upwards are the electrostatic

5. Similarly, expressing osmotic potential as pascals requires some legerdemain with units. The product of concentration (in molm-3), gas constant (in J mol-1 K-1) and temperature (inK) results in units of (molm-3) X (Jmol-1K-1) X (K) = Jm-3 (equation 7.4). The joule is also a derived unit, being the product of force (in newtons) and distance (in meters), that is 1 J =

1 N m. In fundamental units, the joule is therefore 1 J = (1 kg m s-2) X (1m) = 1 kg m2 s-2. The osmotic potential is therefore equivalent to 1 Jm-3 = (1 kgm2 s-2) X (1 m-3) = -1 -2

1 kg m s , which, as was shown in footnote 4, is equivalent to the pascal.

forces I described just above, but this time between the glass surface and the water molecules. The stronger this attraction is, the more wettable the surface is said to be. The meniscus is formed by the interaction of this force with the other two. Gravity obviously pulls the water down with a force proportional to its density. In addition, the water's surface tension exerts a force parallel to the water surface that pulls inwardly toward the center of the glass. The balance of all three gives the meniscus its gracefully curved surface.

When the multiple forces that give rise to a meniscus are translated to a porous medium containing a matrix of small pores and channels, the result is the matric force. Commonly, matric forces are much stronger than those that draw water up the side of a glass: even a cheap paper towel can lift water ten or fifteen centimeters. This indicates that the paper towel exerts a more forceful upward pull than the glass, which lifts water only by a few millimeters. The stronger force in the paper towel arises, in part, because the very small pores force the menisci to be more tightly curved there. Forcing curvature onto a water surface magnifies the surface tension forces: it follows that the smaller the pores in the matrix, the stronger the matric forces will be. Different grades of paper towels prove my point: "better" paper towels are woven more tightly and will draw water into them more strongly than will the more loosely woven "inferior" brands. Soils are like paper towels in that they are a matrix of small pores, small air spaces bounded by the mineral grains and small organic particles that make up the soil. Consequently, water in these small spaces between the soil grains, or micropores, will experience strong matric forces.

Understanding matric potentials in soils is fairly straightforward as long as you remember two simple rules. First, matric potentials require that a boundary, or interface, exist between water and air. It is the interaction between the wetting forces and surface tension that makes up the matric potential. If there is no air-water interface, there is no surface tension, and hence no matric potential. Second, creation of new surface

0 soil water content matric potential

Figure 7.4 Typical relationship between a soil's matric potential and its water content (quantity of water held in a particular quantity of soil—say kilograms of water per kilogram of dry soil). As the soil water content decreases (moves toward zero), the matric potential becomes more and more negative (that is, the suction pressure becomes greater and greater).

area requires work to be done, because surface tension opposes the increase of surface area. Just think of a stretched rubber sheet: the elastic forces in the sheet act similarly to surface tension, and you cannot stretch a rubber sheet without doing work on it.

Let us now apply these rules to understanding a particular behavior of the matric potential: its variation with soil water content, the quantity of water held by a particular quantity of soil (Fig. 7.4). When soil is satu-rated—that is, when all the pores spaces in a soil are completely filled with water—the matric potential is nil. In wet but less than saturated soil, matric potential is weak, and it strengthens only slightly as the soil dries. At a critical water content, however, matric potential strengthens considerably, dropping rapidly to hundreds of thousands or millions of pascals below zero.

Let us now use our rules to explain this behavior. The total volume of the micropores in a soil forms a void space, which can be occupied either by air or water (Fig. 7.5). The nil matric potential in saturated soil is easily explained by our first rule. Water completely fills the void space, there is no air-water interface anywhere within the void space, no surface tension forces there, and hence no matric potential. As a soil dries,

Figure 7.4 Typical relationship between a soil's matric potential and its water content (quantity of water held in a particular quantity of soil—say kilograms of water per kilogram of dry soil). As the soil water content decreases (moves toward zero), the matric potential becomes more and more negative (that is, the suction pressure becomes greater and greater).

Figure 7.5 Different regimes of matric forces in wet and dry soils. a: In very wet soil, many of the micropores are filled with water, and matric forces are weak. b: In dry soils, the small amount of water is distributed among many micropores, and matric forces are strong.

though, some fraction of the void space comes to be occupied by air, and we now need to fall back on our second rule.

The existence of a critical water content implies that there are two mechanisms for the creation of new interfacial surface, one of which dominates in soils wetter than the critical water content, and the other dominating in drier soils. When soil is wetter than the critical water content, water is distributed through the void space in fairly large parcels. The pores that do contain an air-water interface are relatively few in number and the interfaces gravitate toward pores of larger diameter. As the wet soil dries, the water retreats deeper into these pores, and the interfaces become more tightly curved as they retreat (Fig. 7.5).

The increase of the interfacial surface area results mostly from this increased curvature of already existing interfaces. Hence, matric potential changes slowly as soil dries, because interfacial surface area varies only slightly as the water retreats deeper into the pores. Below the critical water content, though, these large parcels of water break up into many smaller parcels. This creates new interfacial surface at a much greater rate than the simple shrinkage of a few large parcels.6 The rapid increase of interfacial surface is reflected in the dramatic strengthening of the matric potential (Fig. 7.4) as the soil dries.

We can also use our rules to explain how matric potentials behave in different soil types (Fig. 7.6). Clays have higher critical water contents than do more coarsely divided soils, like loams or sands, and their water potentials drop to much lower values. The reason, obviously, is that pore spaces are smaller across the board in finely divided clays than in the coarser sands and loams. A parcel of water will break up when the forces holding it together (its cohesiveness) are weaker than the forces pulling it apart (the surface tension). When water is distributed into small pores, as in clays, surface tension forces pull more strongly than they will in more coarsely divided soils, and parcels of water will break up at higher water contents than they do in sands. Clays, therefore, can hold a lot of water, but this abundant water will be difficult to obtain because it is held by very strong matric forces.

Handy Features of the Water Potential The water potential is such a powerful tool because it enables one to analyze the movements of water and their energetic consequences. Suppose, for example,

6. You can convince yourself that this argument is true with an easy calculation. Suppose you have one liter of water occupying a single spherical mass. One liter occupies a volume of

1,000 cm3, so the sphere has a radius r = 3j3V/4n = 6.2 cm. The

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