Biology Reference
In-Depth Information
slowing of growth is caused by the increasing concentration of ethanol in
the medium, alcohol being one of the products of anaerobic respiration
or fermentation, as any brewer or vintner can attest. The concentration of
ethanol rises until it reaches levels toxic to new yeast cells, resulting in
the observed decrease in the growth rate.
It is evident from this graph that the ability of the environment to
support the growth of the yeast diminishes as the population increases—
a reality we must modify our models given by Eqs. (1-1) and (1-2) to
reflect. Modification of the continuous model is discussed in detail; the
discrete case is left as an exercise.
Given that an environment can sustain only so many organisms,
we need to modify the model so the net per capita growth rate r
depends on the size of the population. In terms of an equation, we
could say:
dP
dt
¼
r
ð
P
ð
t
ÞÞ
P
ð
t
Þ;
(1-9)
emphasizing that r now is not constant but depends on P.
Specifically, we now assume:
1. The environment can sustain a maximum population of the species,
reflecting its carrying capacity, K.
2. The smaller the population, the higher the per capita rate of popu-
lation growth. In general, as long as the population remains smaller
than the carrying capacity K, the population will grow, but the
closer to K the population gets, the slower the growth rate
will be.
3. If the population ever exceeds K (e.g., by immigration), then the
population will diminish and approach K; that is, the net per capita
rate of change should be negative for P
>
K.
We want to modify our model to reflect the simplest case—that the
environment accommodates zero growth when the population is K and the
maximal per capita growth rate when the population is near zero. Suppose
the highest per capita growth rate is a
0. If we want to graph per capita
growth rate versus population size (see Figure 1-5), we want no growth
when the population is K,so(K, 0) must be a point on our graph. We also
want the maximum growth rate to be at the hypothetical population of 0,
so (0, a) is another point on our graph. Letting (x
1
, y
1
)
>
¼
(K, 0) and (x
2
, y
2
)
¼
(0, a), the slope of the line passing through these two points is
y
2
y
1
a
0
a
K
:
m
¼
x
1
¼
K
¼
(1-10)
x
2
0
The graph of this line is depicted in Figure 1-5, and its equation is
