Plant Physiology (Biology 327)  - Dr. Stephen G. Saupe;  College of St. Benedict/ St. John's University;  Biology Department; Collegeville, MN  56321; (320) 363 - 2782; (320) 363 - 3202, fax;    ssaupe@csbsju.edu

Surface-to-Volume Ratios in Biology

These exercises are designed to introduce you to the concept of surface-to-volume ratios (S/V) and their importance in plant biology.   S/V ratio refers to the amount of surface a structure has relative to its size. Or stated in a more gruesome manner, the amount of "skin" compared to the amount of "guts". To calculate the S/V ratio, simply divide the surface area by the volume.    We will first examine the effect of size, shape, flattening an object, elongating an object on surface-to-volume ratios.  Then, we will do some hands-on activities to examine the importance of S/V ratios in two areas, cell size and metabolic rate.

I.  INFLUENCE OF SIZE ON S/V RATIOS.

The purpose of this exercise is to see how the S/V changes as an object gets larger.  We will use a cube to serve as a model cell (or organism). Cubes are especially nice because surface area (length x width x number of sides) and volume (length x width x height) calculations are easy to perform. To calculate the surface-to-volume ratio divide the surface area by the volume. Complete the table below for a series of cubes of varying size:

 Length of a side (mm) Surface Area (mm2) Volume (mm3) Surface/volume ratio 1 2 3 4 5 6 7 8 9 10

Questions and Analysis:

1. Which cube has the greatest surface area? volume? S/V ratio?
2. What happens to the surface area as the cubes get larger? What happens to the volume as the cubes get larger? What happens to the S/V ratio as the cubes get larger?
3. Proportionately, which grows faster - surface area or volume? Explain.
4. Which cube has the most surface area in proportion to its volume?
5. If you cut a cube in half, how does the volume, surface area and S/V ratio of one of the resultant halves compare to the original?
6. As the linear dimension of the cube triples, the surface area increase by the (square or cube) of the linear dimension, and the volume increases by the (square or cube) of the linear dimension.
7. Plot, all on one graph, the following: s/v ratio vs. cube size (length in mm); volume vs. cube size (length in mm); and surface area vs. cube size (length in mm).

II.  S/V RATIOS IN FLATTENED OBJECTS
:

In this exercise we will explore how flattening an object impacts the surface to volume ratio. Consider a box that is 8 x 8 x 8 mm on a side. Then, imagine that we can flatten the box making it thinner and thinner while maintaining the original volume. What will happen to the surface area, and s/v ratio as the box is flattened? Complete the table below.

 Box No. Height (mm) Length (mm) Width (mm) Surface area (mm2) volume (mm3) s/v ratio 1 8 8 8 2 4 16 8 3 2 16 16 4 1 32 16 5 0.5 32 32

Questions/Analysis:

1. Explain why leaves are thin and flat.
2. Why do elephants have large, flat ears?
3. Explain why desert plants are leafless.

III.  S/V RATIOS IN ELONGATED OBJECTS:
In this exercise we will explore how elongating an object impacts the surface to volume ratio. Consider a box that is 8 x 8 x 8 mm on a side. Then, imagine that we pull on the ends to make it longer and longer while maintaining the original volume. What will happen to the surface area, and s/v ratio as the box is flattened? Complete the table below.

 Box No. Length (mm) Height (mm) Width (mm) Surface area (mm2) volume (mm3) s/v ratio 1 8 8 8 2 16 4 8 3 32 4 4 4 64 2 4 5 128 2 2

Questions/Analysis:

1. Explain the shape of blood vessels.
2. Explain why roots have "hairs".
3. Explain why plants are dendritic.

IV.  SHAPE AND S/V RATIOS

In this exercise we will explore the impact of shape on surface to volume ratios. The three shapes given below have approximately the same volume. For each, calculate the volume, surface area and S/V ratio and complete the table.  The last column in the table, "Volume of environment within 1.0 mm of the object" is particularly important.  Since the materials that an organism exchanges with its environment comes from its immediate surroundings, the greater this volume, the more material that can be exchanged.

 Shape Dimensions (mm) Volume (mm3) Surface Area (mm2) S/V ratio Volume of environment within 1.0 mm of object Sphere 1.2 diameter Cube 1 x 1 x1 Filament 0.1 x 0.1 x 100

note: volume of a sphere = 4/3 (π)r3 = 4.189r3 surface area of sphere = 4(π)r2 = 12.57 r2

Questions & Analysis:

1. Make a sketch, to scale, of the three objects.
2. Which shape has the greatest surface area? volume? s/v ratio?
3. If you had to select a package with the greatest volume and smallest surface area, what shape would it be?
4. Explain why the shape of animals is basically "spherical", whereas plants and fungi are "filamentous".

V. WHY ARE CELLS SMALL?

The typical eukaryotic cell is rather small - approximately 100
mm in diameter. This exercise is designed to help provide an explanation why cells are not normally larger.

Obtain 2 cell models, one small and one large. Measure the length and diameter of each and then record your data in the table below. Place each cell in a bowl containing clear vinegar or dilute HCL (0.1N; BE CAREFUL!). Allow to sit for a few minutes, or until most of the blue color is gone from the smallest cell. Remove the models with a plastic spoon (CAUTION: don't get the vinegar on your hands!!!!) and place it on a piece of paper towel. Then, measure the size of the colored area remaining and record these data in the table below. Complete the calculations.

Theory: The cell models are made of a gelatin-like material called agar. The agar has an acid-sensitive dye incorporated into it. The dye turns from blue to yellow (clear in the presence of acid). The uptake of acid, and hence colorless areas of the cell models represents the uptake of food/nutrients by the cell. From this, we can calculate the percent of each cell that was fed during the incubation period.

 Table 5.  Effect of cell size on feeding rates Small Cell Large Cell Colored Portion Before Feeding (initial) diameter (mm) radius (mm) height (mm) surface area (mm2) volume (mm3) S/V ratio Colored Portion After Feeding (final) diameter (mm) radius (mm) height (mm) surface area (mm2) volume (mm3) Percent of volume (cell) fed

volume of a cylinder = πr2l where p = 3.14
surface area of cylinder = 2
π r2 + 2 πrh
percent cell fed = (initial volume - final volume)/initial volume x 100

Analysis/Questions:

1. In which cell did the greatest portion get fed? Explain.
2. Which cell would be in greatest danger of starving? Explain.
3. Suggest some reasons why cells are small.
4. If a cell divides in half, how does the volume, surface area and S/V ratio of each new cell compare to that of the original?
5. Explain why the rate of cell growth slows as a cell gets larger.
6. Explain why cells divide when they get large.

VI.  WHY DO MICE HAVE GREATER METABOLIC RATES THAN ELEPHANTS?
It is well known that there is an inverse relationship between body size and metabolic rate. The purpose of this exercise is to determine the reason for this relationship.

Procedure:

1. Pack ice around the containers in the pan to a depth of about two inches. Then add cold water to bring the water level up to the top of the ice.
2. Turn on the hot water tap and let it run until the temperature reaches 55-65 (measure with your thermometer).
3. Measure 8 mL of hot water with a pipet and place it into the 30 mL flat-bottomed vial.
4. Place the thermometer in the hot water. At first the temperature will rise. When it begins to fall again, measure the temperature. If it is below 40 C, start again with hotter water.
5. When the temperature reaches 40 C, record the time. Exactly two minutes later record the temperature again. Record your data in the table below.
6. Measure 80 mL of hot water with a graduate cylinder and transfer it to the larger (100 mL) beaker. Then repeat steps 4 & 5.
7. Calculate the initial and final heat content (rows 8 and 10) by multiplying the temperature (rows 7 and 9) by the volume of water (row 4) by the specific heat of water (4.2 joules/cm3 C). [Note - specific heat is an indication of the amount of heat it takes to change the temperature of a substance.] Thus, this equation is:
8. Heat content (joules) = temp. (C) x vol. (cm3) x specific heat of water (4.2 joules/cm3 C)

Calculate the total heat loss during the two minute period (row 11) by subtracting the final heat content (row 10) from the initial heat content (row 8).

9. Calculate the relative heat loss (joules lost/cm3) for each volume (row 13) by dividing the total heat loss by the volume of the vessel (row 4).
10. Lastly, measure the height (row 1) and diameter (row 2) of the water in each container and then calculate radius (row 3), volume (Row 4), surface area (row 5) and S/V ratio (row 6) for both containers.
 Table 6.  Effect of size on the rate of cooling Small Large 1. height (cm) 2. diameter (cm) 3. radius (cm) 4. volume (cm3) 8 80 5. surface area (cm2) 6. S/V ratio 7. initial temp (C) 40 40 8. initial heat content (J) 9. final temp (C) 10. final heat content (J) 11. total heat loss (Joules/2 minute) 12. total heat loss (Joules/min) 13. relative heat loss (kJ/min/cm3)

Analysis/Questions:

1. Which vessel had the greatest volume? surface area? S/V ratio?
2. Which vessel lost more total heat? Did you expect that? Explain.
3. Which vessel lost the most heat relative to its size? Did you expect that? Explain.
4. Which vessel showed the greatest temperature change?
5. To maintain a constant temperature in the vessels, which would have required the most total heat input? Which would have required the most heat proportional to its size?
6. Who will loose more total heat in a given period, an infant or an adult? A mouse or an elephant? Explain.
7. Who will loose more heat relative to its volume, an infant or an adult? A mouse or an elephant? Explain.
8. Who will need to eat the most food, an infant or an adult? A mouse or an elephant? Explain. Who will need to eat the most food relative to size? Explain.
9. Explain why small animals have a higher metabolic rate than large animals
10. Explain why shrews are voracious feeders.

Other Plant Applications:

1. Explain why plants are essentially a cluster of filaments, whereas animals are blobs. In other words, why is a thin, elongated rectangle a good model for a plant, but a square a good model for an animal?
2. Explain how S/V ratios relate to the form of plants that have evolved in mesic (moderate), xeric (dry) and hydric (aquatic) environments.
3. Explain why the cells of the spongy mesophyll layer are irregular in shape whereas those of the palisade layer are more rectangular.
4. Describe the trends that have occurred in s/v ration during the evolution of plants from algae to mosses to ferns to angiosperms.
5. Obtain the leaf of a mesophytic plant. Record the scientific name and family of this species. Calculate the surface area of the leaf (ignore the edges of the leaf). Then, calculate the dimension of a cube that would have the same surface area. Construct a cube of this size from paper.

1. Explain why cells divide when they get large.
2. Explain why the rate of cell growth slows as cells get larger.
3. Explain why cats can fall off of tall buildings and survive. Why do people splat?
5. Describe the scientific inaccuracy in the episode of Goldilocks and the porridge.
6. Explain why lungs, gills and intestines have the shape they do.
7. Describe the shape of a radiator? Explain why it has this shape.
8. Medieval churches were often built in the shape of a crucifix. Explain why.
9. Mice have large eyes relative to size and elephants small ones. Explain why. Are large eyes better than small ones?
10. The earth is geologically active (has a molten core) but the moon is apparently no longer geologically active. Explain why using S/V ratios.
11. Based on size relationships, explain why a human is smarter than a mouse.
12. Shrews have a reputation for being "mean". In other words, they must feed constantly. Explain why.
13. Why are there few small animals in the arctic?
14. Why would a mouse-sized human lack intelligence?

Useful Equations:

 Shape Equation for Volume Equation for Surface Area cube l x w x h l  x w x 6 box (filament) l x w x h determine SA for each side then add sphere 4/3 (π) r3 = 4.189 r3 4 (π) r2 = 12.57  r2 cylinder π r2 l 2 π  r2 + 2 π r h = 2 π r (r + h)

Supplies Required:

• Cell Exercise:  Small tray with 2-3 paper towels, spoon, jar of vinegar, ruler, pair of gloves (1/group); cell models (1 of each size per group; 0.04% bromocresol green, 2% agar)
• Heat/Metabolism Demo:  2 pans with 8 and 80 ml beakers glued into the bottom; 2 thermometers; hot plate; flask with water; 100 ml graduate cylinder; 10 ml pipet and pump; ice; hot pad/glove

References:

• Barrett, D. 1983. Body size and temperature: an extended approach. J. Biol. Educ. 71:78.
• Cohen, A, AB Moreh, R Chayoth. 1999. Hands-on Method for Teaching the Concept of the Ratio Between Surface Area & Volume. The American Biology Teacher 61: 691 - 696.
• Diamond, Jared. 1989. How cats survive falls from New York skyscrapers. Natural History, August, pp 20 - 26.
• Haldane, J.B.S. On Being the Right Size.
• Gould, S. 1977. Size and shape. In Ever Since Darwin. Norton, NY.
• Moog, F. Gulliver was a bad biologist. Scientific American.
• Stanek, Jr., J. A. (1983) Why don't cells grow larger? American Biology Teacher 45:393-395.
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Last updated:  01/07/2009     � Copyright  by SG Saupe