Concepts of Biology (BIOL116)  Dr. S.G. Saupe; Biology Department, College of St. Benedict/St. John's University, Collegeville, MN 56321; ssaupe@csbsju.edu; http://www.employees.csbsju.edu/ssaupe/ 
SURFACETOVOLUME RATIOS IN BIOLOGY
These exercises are designed to introduce you to the concept of surfacetovolume ratios (S/V) and their importance in 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.
EXERCISE 1. INFLUENCE OF SIZE ON S/V RATIOS. 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 surfacetovolume 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 (mm^{2}) 
Volume (mm^{3}) 
Surface/volume ratio 
1 

2 

3 

4 

5 

6 

7 

8 

9 

10 
Questions and Analysis:
EXERCISE 2. 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.
Shape  Dimensions (mm)  Volume (mm^{3})  Surface Area (mm^{2})  S/V ratio  Volume of environment within 1.0 mm 
Sphere  1.2 diameter  
Cube  1 x 1 x1  
Filament  0.1 x 0.1 x 100 
note: volume of a sphere = 4/3 (p)r^{3} = 4.189r^{3} surface area of sphere = 4(p)r^{2} = 12.57 r^{2}
Questions & Analysis:
EXERCISE 3. 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 (mm^{2})  volume (mm^{3})  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:
EXERCISE 4. 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 (mm^{2})  volume (mm^{3})  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:
EXERCISE 5. 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 (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 gelatinlike material called agar. The agar has an acidsensitive 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.
Small Cell 
Large Cell 

Colored Portion Before Feeding (initial)  diameter (mm)  
radius (mm)  
height (mm)  
surface area (mm^{2})  
volume (mm^{3})  
S/V ratio  
Colored Portion After Feeding (final)  diameter (mm)  
radius (mm)  
height (mm)  
surface area (mm^{2})  
volume (mm^{3})  
Percent of volume (cell) fed 
volume of a cylinder = pr^{2}l where
p = 3.14
surface area of cylinder = 2p
r^{2} + 2 prh
percent cell fed = (initial volume  final volume)/initial volume x 100
Analysis/Questions:
EXERCISE 6. 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. (cm^{3}) x specific heat of water (4.2 joules/cm^{3} 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).
Small 
Large 

1. height (cm)  
2. diameter (cm)  
3. radius (cm)  
4. volume (cm^{3})  8 
80 
5. surface area (cm^{2})  
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/cm^{3}) 
Analysis/Questions:
ADDITIONAL QUESTIONS:
References:
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