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Notes to Instructors
Chapter 40 Basic Principles of Animal Form and Function
What is the focus of this activity?
All organisms, whether single-celled or complex and multicelled, must be able to
maintain homeostasis. For each cell in the organism to remain alive, it must be
What is this particular activity designed to do?
Exercise 40.1 How does an organism’s structure help it maintain homeostasis?
Activity 40.1 263
Answers
Activity 40.1 How does an organism’s structure help it
maintain homeostasis?
1. To remain alive, an organism must be able to maintain homeostasis of its internal
environment relative to the external environment. What behaviors, structure(s) or
system(s) are of primary importance in maintaining homeostasis in the following
situations in amoeba versus mammal?
Situation Amoeba Mammal
a. Change in
environmental: pH
Behavioral: Amoebas can
move to avoid areas that
have unfavorable pH or
temperature.
Behavioral: Mammals can
avoid areas that have
unfavorable pH or
temperature.
(Continued on next page)
264 Activity 40.1
Situation Amoeba Mammal
b. Reception of stimuli,
for example: light
Amoebas do not have
specific subcellular
organelles designed to
As multicellular organisms,
mammals have specialized
parts of the body for various
c. Response to stimuli Amoebas are negatively
phototropic. When touched
Both the eye and touch
receptors respond to stimuli
2. Cells must be bathed continuously in an aqueous medium to take in oxygen and
nutrients and get rid of waste products via diffusion. Diffusion is efficient over only
short distances. In fact, diffusion is efficient only for a distance of about three cell
diameters maximum (approx. 200 to 300 m). Note the following times required to
diffuse specific distances:
Diffusion Distance (m) Time Required for Diffusion
1 0.5 msec
10 50 msec
100 5000 msec (5 sec)
1,000 (1 mm) 500,000 msec (8.3 min)
10,000 (1 cm) 50,000,000 msec (14 hr)
Activity 40.1 265
a. Graph the data in the table and calculate the mathematical relationship between
the increase in distance and the corresponding increase in time required to diffuse
that distance. You can do this by hand or you can use a software program with
graphing capability to do this. Include your graph and your calculations below.
In general, for each 10-fold increase in distance, a 100-fold increase in time for
Diffusion Distance vs. Time
60000000
50000000
40000000
y = 0.5x2
R2 = 1
Diffusion
Distance vs.
Time
b. How much time would be required for oxygen to diffuse 5 m? 200 m?
The difference between 1 and 5 m is 5. We square this value (5 ⫻5 ⫽25) to
3. What happens to the surface-area-to-volume (SA/V) ratio of a three-dimensional
object (such as a cell) as its linear dimension increases? For example, how does the
SA/Vratio of a sphere or cube change as the linear dimensions increase? (Formulas
for a sphere: surface area = 4r2; volume = 4/3r3.) (Also see Activity 7.1.)
As the linear dimensions of an object increase by two times, the surface area
266 Activity 40.1
Propose the effect(s) the physical properties of diffusion are likely to have on the
structure and function of epithelia and epithelial cells and digestive and circulatory
systems as animals become larger and larger. In your answer, consider how SA/V ratios
change as organisms become larger and the effect(s) this change is likely to impose on the
structure and function of organisms.
See the description of various types of epithelial cells in Chapter 40 of Campbell Biology,
9th edition. Also, refer to Figure 40.5. In general, barrier-type cells are columnar and have
40.1 Test Your Understanding
