LAB ORIENTATION AND THE SURFACE TO VOLUME RATIO IN ANIMALS - 1
Lab Orientation and the Surface Area to
Volume Ratio in Animals
by
Antoine Morin and Gabriel Blouin-Demers
Lab Orientation
Details of your activities during the lab orientation component of your first lab will be explained to you at
the beginning of your lab on January 17, 18, 19 or 20, 2017.
Introduction
Many aspects of protozoan and animal shapes, body plans and behaviours can be explained by the necessity
that oxygen, water and nutrient requirements are met at the cellular level. Diffusion is the most important
overriding mechanism that supplies these requirements and it is involved in the function of various
structures, tissues and organs. During this lab period we will study some of the factors that influence
diffusion rates by using some very simple techniques. After that you will examine a variety of different
organisms and identify the strategies that they use to enhance gas exchange and how they reduce water loss
from evaporation in the terrestrial environment
Metabolic requirements of animals
An organism's oxygen and energy requirements are proportional to its size and the level of cellular and
muscular activity that they exhibit. At rest, the metabolic rate is proportional to the mass to the power of
0.75 (M0.75). Large animals therefore have higher overall requirements then the smaller ones, but when this is
expressed per unit of mass, the larger animals requirements are below those of smaller ones. For any
particular body weight, single celled organisms require less energy than cold-blooded animals, while warmblooded animals have the highest energy demands.
To meet these requirements by diffusion large animals generally rely on large surface areas for exchange.
However, for all organisms, as they increase in size their mass will increase more rapidly than their surface
area. This means that their needs will increase more rapidly with the increase in size than can be met by
diffusion. This would imply that there will be some sort of shortfall unless other adaptations or
modifications develop to improve diffusion in the larger animals.
© Antoine Morin and Gabriel Blouin-Demers,
Department of Biology, University of Ottawa
2 - ANIMAL FORM AND FUNCTION
Figure 1. Effect of body mass (M, in kg) on basal metabolic rate (R, in Watts) of unicellular and multicellular organisms. R=0.018M0.75 for
unicellular organisms, R=0.14M0.75 for poikilotherms, R=4.1M0.75 for homeotherms. (Modified from Hemmingsen 1960)
The surface area to volume ratio in organisms is a measure of how easily oxygen and nutrients can be
taken up and how well metabolic wastes can be eliminated. For two organisms with different sizes (but with
a similar body plan) to balance their metabolic demands with supply (diffusion rate), means that their
surface area to volume ratio must be similar. This can be done by changing the shape of the organism and
moving away from a spherical shape which has the lowest surface area to volume ratio. Changes in body
shape that result in an advantageous surface area to volume ratio do not always work for large animals that
typically use morphological, physiological or behavioural adaptations to enhance diffusion rates.
Factors that influence diffusion
Diffusion rates depend primarily on three factors: the surface area (S) across which the exchange occurs,
the permeability of the membrane or tissue through which diffusion will occur (always inversely
proportional to its thickness L) and the difference in the concentration (ΔC) between the two sides of the
membrane or tissue.
SΔC
D ∝ ____
L
Adaptations to minimize or maximize diffusion
To maximize oxygen diffusion and minimize heat and water loss by evaporation in the terrestrial
environment, different types of animals have adaptations that relate to these three factors. For example, in
animals without respiratory and circulatory systems, we usually see a change in the general shape of the body
between small and large organisms. Small animals are spherical or cylindrical while large forms are flattened
to increase their surface area to volume ratio. In terrestrial animals, the outer body covering is impermeable
BIO 2135 - Winter 2017
LAB ORIENTATION AND THE SURFACE TO VOLUME RATIO IN ANIMALS - 3
to water in order to reduce the loss of water by evaporation. Finally, mixing at the exchange surfaces is used
to maintain the concentration gradient and maximize oxygen diffusion at the respiratory surfaces by
exposing the oxygen poor internal fluids (or rich in metabolic wastes), to oxygen rich water at the external
surface of the membrane. You will examine some organisms that represent different phyla and you will be
looking for modifications and adaptations that permit them to either maximize or minimize exchange with
their surrounding environment.
Measurements
For each organism you will estimate the external surface area, its volume and the surface area to volume
ratio. To do this you will assume that the surface area and volume of each animal approximate those of
simple geometric shapes. For example, the surface area and volume of an earthworm is similar to that of a
cylinder with the same length and diameter of the worm. For flat organisms where their thickness is hard
to measure we will make the simple assumption that their thickness is 2% of their longest dimension. For
animals with more complex and interesting shapes (such as an undergraduate student), estimate the surface
area and volume of the limbs and torso as if they were cylinders and the head as if it were a sphere.
Formulas used to calculate surface areas and volumes
Surface area
Volume
Sphere
S = 4 π R2
V = -4-- πR 3
3
Cylinder
S = 2 πR + 2 πR L
2
V = π R 2L
Parallelepiped
(rectangle)
S = 2 L H + 2 L P + 2HP
V = LHP
R=radius, L=length, H=width, P=depth or thickness
Protozoans
Look at a small protozoan, Tetrahymena, and a large one, Spirostomum. Do these organisms have a high surface
area to volume ratio? How do they improve their diffusion rates? How does the shape of the large protist
differ from the small one? How does this change in shape affect diffusion?
Flat worms
Examine a free living flatworm, such as a planarian and a parasitic flatworm, such as a liver fluke. These
animals depend upon diffusion across their skin to obtain oxygen that they require. How does their shape
favour this exchange? Do you think that this type of permeable epidermis would allow these organisms to
live in a terrestrial environment?
© Antoine Morin and Gabriel Blouin-Demers,
Department of Biology, University of Ottawa
4 - ANIMAL FORM AND FUNCTION
Earthworms
Worms breathe by diffusion across their epidermis. Is their surface area to volume ratio higher than that of
the flatworms? Earthworms have a circulatory system with a respiratory pigment (haemoglobin) that
increases the efficiency of exchange with their environment.
Sea urchins
Look closely at the oral surface of the sea urchin. These animals depend on diffusion across the fine
membranes of their numerous tube feet to meet their oxygen requirements. In fact their body is filled with a
network of tubes filled with water, without any cells, that enhances gas diffusion throughout the body. As
effective as this may be, these animals cannot sustain high levels of muscular activity that require large
amounts of oxygen.
Fish
Fish use special structures (gills), for respiratory gas exchange. Gills are made of many folded sheets and
their surface area for exchange is much larger than it looks. Fish use a circulatory system and respiratory
pigments to enhance oxygen exchange at the gill surface.
Turtles
Turtles, like all other animals that live in the terrestrial environment, must find a balance between the
exposed respiratory surface available for oxygen uptake and carbon dioxide release, and an impermeable
skin that prevents dehydration. Note that the epidermis which is covered by scales or a shell prevents any
form of diffusive exchange. The respiratory surfaces are the lungs which are located in the thoracic cavity,
and they make contact with the outside by respiratory passages which among other things reduce water loss.
Humans
We have, like other warm-blooded animals, high metabolic needs. Our respiratory surfaces and circulatory
system, which pumps blood in and out of the lungs, optimize gas exchange. Our epidermis is essentially
impermeable, and other parts of our bodies are covered with hairs that prevent heat loss (how?). Since our
surface area to volume ratio is relatively small, it helps us maintain our body temperature.
Graphs to hand in
The surface area (in mm2) and volume (in mm3) data for nine species and instructions for the
report will be available on Dr. Houseman's web page in an Excel file and pdf file respectively.
Graphs should be designed so that they may be projected onto a screen (caption included below the figure).
Text and symbols should be large enough to be read from the back of the laboratory. The hand in date for
the graphs (Version 1) is Jan. 30, 2017 before 4:00 pm in the document drop off area near the front
doors of the BioSciences Complex. Place your report in the slot of the box labeled with your lab section.
BIO 2135 - Winter 2017
LAB ORIENTATION AND THE SURFACE TO VOLUME RATIO IN ANIMALS - 5
Surface area vs volume
The first graph you will draw in Excel is that of surface area as a function of volume. You will appreciate
quickly that there is a scaling problem. The surface area and volume of humans is quite a bit larger than
those measured in the other animals, so that it becomes impossible to compare them on the same graph.
Logarithm base 10 (Log10) of the surface area vs Log10 of the volume
Use a transformation to solve this problem of scale. In Excel, create two new columns with formulas to
calculate Log10 of the surface area and Log10 of the volume. This transformation effectively reduces the
values and has an even greater effect as the numbers increase. It allows us to easily compare widely different
values on a common scale.
Here is the formula in Excel for Log10:
=LOG10(cell)
Be aware that you do not have to type in the formula for each of the rows of data! In Excel, you will type
the formula into the cell referring to the first data row, and then copy and paste it into the cells that apply to
the other data rows. This is one simple operation.
The second graph you are asked to draw is a graph of the Log10 of the surface area as a function of the
Log10 of the volume. Much nicer, isn't it? Please adjust the maximum and minimum axis values by double
clicking on each axis and changing the values under the scale tab to +10 and -10 respectively.
Use Excel to calculate the slope (m) of this relationship and to obtain the standard error (SE) of the slope.
This formula must be entered as a matrix formula. To begin, type the formula for the slope into a cell near
but not within your data:
=LINEST(known_y's,known_x's,constant,statistics), for example
=LINEST(D2:D500,E2:E500,TRUE,TRUE)
Press the enter key. Select the cell containing the formula and the cell immediately below it. Press the F2
function key. Press and hold the Control+Shift keys, followed by the Enter key. The value in the upper cell
will be the slope of the line (m in Excel) and the value in the lower cell will be the standard error of the
slope (SE in Excel). Use the help function for LINEST if you have difficulties.
© Antoine Morin and Gabriel Blouin-Demers,
Department of Biology, University of Ottawa
6 - ANIMAL FORM AND FUNCTION
For large sample sizes (n > 100), as in our case, the standard error may be used to calculate the confidence
interval. Use the following formula (a slight variation on the confidence interval formula found at the end
of your lab manual) to calculate the 95% confidence interval of the slope:
Lower limit of the 95% confidence interval of the slope = m - 1.96(SE)
Upper limit of the 95% confidence interval of the slope = m + 1.96(SE)
If the 95% confidence interval about m does not include 1, you may conclude that the slope is significantly
smaller than 1. If that is the case, what do you conclude about the increase in the surface area as a function
of the increase in the volume?
Log10 of the surface area to volume ratio vs Log10 of the volume
At this time create a new column in which you will calculate the surface area to volume ratio using the
following formula:
=surface/volume
Create another column to calculate the Log10 of the surface area to volume ratio. If you are beginning to feel
at ease in Excel, these last two steps could have been completed in one step using the following formula:
=LOG10(surface/volume)
Remember to copy and paste this formula into the cells referring to all of the other data rows.
Make a graph (your third graph) of the Log10 of the surface area to volume ratio as a function of the
Log10 of the volume. Adjust the maximum and minimum axis values by double clicking on the x-axis and
changing the values under the scale tab to +10 and -10. Do the same for the y-axis but change the
maximum and minimum values to +5 and -5 respectively. Calculate the slope of this relationship, as well as
the 95% confidence interval as completed above. If the 95% confidence interval does not include 0, what
are you able to conclude about the decrease in the surface area to volume ratio as a function of animal size
(measured by volume in our case)? What are the implications of this decrease for very large animals?
BIO 2135 - Winter 2017