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
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:
- Explain why leaves are thin and flat.
- Why do elephants have large, flat ears?
- 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:
- Explain the shape of blood vessels.
- Explain why roots have "hairs".
- 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
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 = 2π
r2 + 2 πrh
percent cell fed = (initial volume - final volume)/initial volume x 100
Analysis/Questions:
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:
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).
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:
- Which vessel had the greatest volume? surface area? S/V ratio?
- Which vessel lost more total heat? Did you expect that? Explain.
- Which vessel lost the most heat relative to its size? Did you expect that? Explain.
- Which vessel showed the greatest temperature change?
- 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?
- Who will loose more total heat in a given period, an infant or an adult? A mouse or an elephant? Explain.
- Who will loose more heat relative to its volume, an infant or an adult? A mouse or an elephant? Explain.
- 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.
- Explain why small animals have a higher metabolic rate than large animals
- Explain why shrews are voracious feeders.
Other Plant Applications:
Additional Questions:
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:
References:
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Last updated:
01/07/2009 � Copyright by SG
Saupe