2006F Bio205H5: Tutorial 4
Competition
weeks of Nov 6th and 13th
due 2 weeks after tutorial
7% of final grade
The purpose of this tutorial is to learn how to model interactions between
species. We will focus on interspecific competition to understand the effects that
two populations can have on each other.
Background:
1. Competition
Recall the density-dependent model of population growth: it incorporates a term
(carrying capacity, or K) that demonstrates the effect that resource limitation has
on population growth. The resource limitation arises from intraspecific
competition – as the population increases in size, the per-capita amount of
resources decreases, resulting in competition among members of the population.
Here, we will examine the effects of 2 species competing for a limiting pool of
resources. As we have discussed in class, there are two possible outcomes of
interspecific competition: coexistence or competitive exclusion of the inferior
competitor. Recall the equation for density-dependent growth:
dN
⎛K−N ⎞
= rN ⎜
⎟ (eq.1)
dt
⎝ K ⎠
We can extend this basic model to look at population growth of 2 species that
are competing for a pool of resources. If N1 = population density (or size) of
species 1 and N2 = population density (or size) of species 2, the feedback term
in eq. 1 can be modified to incorporate the effects of both populations on the
carrying capacity of species 1:
⎛ K1 − ( N1 + N 2 ) ⎞
⎜
⎟
K1
⎝
⎠
The above feedback term considers that the two populations have the same
effect on each other as they have on themselves (i.e., an individual of species 2
exerts the same effect on an individual of species 1 as an individual of species 1
does on a conspecific, and vice versa). This is often not the case. If the two
species do not overlap completely in their resource requirements, it is possible
that conspecifics have a greater competitive effect on each other than do
members of the other species. Conversely, competition among conspecifics may
be less than that between members of different species. Consider two species of
squirrel - the red squirrel and the fox squirrel - that occupy the same habitats in
forests in the eastern US. Both eat the same types of food (more or less), but
the fox squirrel is much larger and thus has greater food requirements. A red
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squirrel with many fox squirrel neighbours will experience greater competition for
food resources than a red squirrel with many red squirrel neighbours. The
feedback term in the growth equation needs to reflect the effect of interspecific
competition relative to intraspecific competition. Thus, a scaling factor (the
coefficient of competition) must be applied. Two coefficients of competition
must be considered: the effect that species 2 has on species 1 (α), and the
effect that species 1 has on species 2 (β). To illustrate, let’s look at the possible
values of α. If interspecific effects are the same as intraspecific effects (an
individual of species 2 has the same effect on an individual of species 1 as does
a member of its own species), α = 1. If individuals of species 2 have less effect
on an individual of species 1 than members of its own species, α<1. If
individuals of species 2 have a more severe effect on an individual of species 1
than members of its own species, α>1. Thus, α quantifies the per-capita
reduction in population size of species 1 caused by species 2. Eq.1 can be
modified to show the effects of both intra- and interspecific competition on the
population growth rate of species 1:
⎛ K − ( N1 + α N 2 ) ⎞
dN1
= r1 N1 ⎜ 1
⎟ (eq. 2)
dt
K
1
⎝
⎠
The change in size of population 1 over time depends on its own intrinsic
growth rate, its population size at a given time, the carrying capacity, and the
degree of resource competition exerted by population 2. This equation can be
rearranged to look at the per-capita growth rate:
dN1
r
= r1 − 1 ( N1 + α N 2 ) (eq. 3)
N1dt
K1
Now, we must consider the effects of species 1 on species 2. It is important to
note that the coefficient of competition of species 2 on species 1 (α) is not
necessarily equal to that of species 1 on species 2 (β). We can re-write eq.’s 2
and 3 from the perspective of species 2:
⎛ K − ( N 2 + β N1 ) ⎞
dN 2
= r2 N 2 ⎜ 2
⎟ (eq. 4)
dt
K
2
⎝
⎠
and
dN 2
r
= r2 − 2 ( N 2 + β N1 ) (eq. 5)
N 2 dt
K2
Equations 2 and 4 are known as Lotka-Volterra models of competitive
interaction. The problem with these equations is that they are difficult to solve –
the dynamics of the 2 populations are interdependent: population 1 has an effect
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on population 2, which then has an immediate effect on 1, etc. How do we get
around this? Computers allow for instantaneous numerical integration, so that
instantaneous changes in sizes of the 2 populations can be determined.
Coexistence or displacement:
We can use Populus to determine the effects of both inter- and intraspecific
competition on two populations living in the same habitat and exploiting at least
some of the same resources. Using the default parameters in Populus, consider 2
populations. Species 1 starts with a density of 10 individuals; the intrinsic growth
rate is 0.9; the carrying capacity is 500 individuals; and the coefficient of
competition of species 2 on 1 is 0.6. Species 2 starts with a density of 20
individuals; the intrinsic growth rate is 0.5; the carrying capacity is 700
individuals; and the coefficient of competition of species 1 on 2 is 0.7. What will
happen to the 2 populations? Will one drive the other to extinction, or will they
coexist indefinitely?
Fig. 1. Equilibrium coexistence of populations 1 & 2.
Fig. 1 shows population size (or density) of the 2 populations over time. Initially,
species 1 is more abundant, due to its greater intrinsic rate of increase.
However, species 1 has a lower carrying capacity than species 2, and thus
quickly starts to experience the effects of both inter- and intraspecific
competition for resources. Species 2 has a slower growth rate and a higher
carrying capacity; thus, it does not suffer as quickly from the effects of both
inter-and intraspecific competition. These 2 species are able to coexist at stable
(unchanging) population densities, but note that both exist at levels less than the
carrying capacity of either species alone. Now, consider a situation where one
species drives another to extinction. In this case the population parameters are:
species 1 starts with a density of 10 individuals; the intrinsic growth rate is 1.2;
the carrying capacity is 440 individuals; and the coefficient of competition of
species 2 on 1 is 0.9. Species 2 starts with a density of 20 individuals; the
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intrinsic growth rate is 0.5; the carrying capacity is 580 individuals; and the
coefficient of competition of species 1 on 2 is 0.9:
Fig. 2. Equilibrium displacement of population 1 by population 2.
Species 1 eventually goes extinct (rapid growth rate plus low carrying capacity
means that it suffers greatly from both intra- and interspecific competition). The
equilibrium population densities are 0 for population 1, and carrying capacity
(580) for population 2 (this makes sense, because population 1 is extinct!).
Technically, it is not correct to call this extinction (presumably, species 1 has not
been wiped off the face of the earth!). The term for this type of competitive
interaction is displacement (species 1 cannot coexist with species 2).
Isocline analyses:
From the discussion above, we can see that stable equilibria (when the growth
rate of the 2 populations is zero) define the outcomes of competitive
interactions. Thus, it is useful to quantify the population densities at which the
growth rate of population 1 and 2 are zero. Obviously, the two populations affect
each other, so the population density at which 1 stops growing depends on the
density of population 2 and vice versa. Think about this for a minute: if species
2 was not present, population 1 would stop growing at its carrying capacity.
However, as you start adding individuals of species 2 to the habitat, the density
at which population 1 stops growing will be at a level less than carrying capacity
(because those members of species 2 are using up resources that population 1
needs). As the density of population 2 varies, the density at which population 1
stops growing will vary. How can we figure out this relationship? To do this, we
need to plot density of population 2 versus density of population 1. Going back
to eq.2, we can set the equation equal to zero:
⎛ K − ( N1 + α N 2 ) ⎞
dN1
= 0 = r1 N1 ⎜ 1
⎟ (eq. 6)
dt
K1
⎝
⎠
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What is the relationship between the terms of the equation that allows the
growth rate to be zero? We can ignore the trivial case where either r = 0 (no
intrinsic growth) or N = 0 (no initial population). If we leave these out, then zero
population growth occurs when the feedback term is equal to 0:
0=
K1 − ( N1 + α N 2 )
K1
(eq. 7)
We can rearrange this equation to see the relationship between N1 and N2:
N2 =
K1
α
−
1
α
N1 (eq. 8)
This is a straight-line plot in the form y = a + bx, where y = N2, x = N1, the yintercept = K1/α, the slope = -1/α, and the x-intercept = K1. The line defines the
zero-net-growth isocline for population 1. Isoclines are lines on a graph
that show all the combinations of x and y that satisfy a particular condition (in
this case, zero population growth, or dN = 0 ):
dt
K1
α
N2
N1
K1
Thus, the red line (slope = -1/α) defines the various combination of N2 and N1
that force population 1 to stop growing (dN/dt = 0). The green arrows show the
area of the graph where a combination of N2 and N1 would allow N1 to increase
in size (positive growth; birth rate > death rate). The red arrows show the area
of the graph where a combination of N2 and N1 would force N1 to decrease in size
(negative growth; birth rate < death rate). How about the zero-net-growth
isocline for population 2? We can do the same type of algebraic rearrangement
of eq. 4 by setting the growth rate of population 2 to 0:
⎛ K − ( N 2 + β N1 ) ⎞
dN 2
= 0 = r2 N 2 ⎜ 2
⎟ (eq. 9)
dt
K2
⎝
⎠
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The zero-net-growth isocline for population 2 is defined as:
N 2 = K 2 − β N1 (eq.10)
K2
N2
N1
K2
β
In this case, the blue line is the zero-net-growth isocline (slope = -β) for
population 2. The green arrows show the area of the graph where a
combination of N2 and N1 would allow N2 to increase in size (positive growth;
birth rate > death rate). The red arrows show the area of the graph where a
combination of N2 and N1 would force N2 to decrease in size (negative growth;
birth rate < death rate). What happens if we superimpose the two isoclines of
population 1 and 2? Consider the following 2 isoclines:
Case I:
K2
K1
α
N2
N1
K1
K2
β
What does this tell us about the dynamics of these 2 populations? The secret is
in the comparisons of the intercepts. The x-intercept for population 1 is its
carrying capacity (K1); thus, it is a measure of intraspecific competition
for population 1 (the number of population-1 individuals needed to stop
growth of population 1 when population 2 is absent). The x-intercept of
population 2 is a measure of interspecific competition of population 1 on
population 2 (K2/β); it is the number of population-1 individuals that cause
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population 2 to stop growing (at the theoretical limit of N2= 0; in any case, when
there is virtually no effect of population 2 on itself). Because K1 < K 2 β ,
population 1 stops growing due to intraspecific effects on itself before it can
reach the density at which it would stop population 2 from growing. Now
compare the y-intercepts. The y-intercept for population 2 is its carrying
capacity (K2); thus, it is a measure of intraspecific competition for
population 2 (the number of population-1 individuals needed to stop growth of
population 2 when population 1 is absent). The y-intercept of population 1
(K1/α) is a measure of interspecific competition of population 2 on
population 1; it is the number of population-2 individuals that cause population
1 to stop growing (at the theoretical limit of N1= 0; in any case, when there is
virtually no effect of population 1 on itself). Because K1 α < K 2 , population 2 can
continue to grow to its carrying capacity after population 1 has stopped growing
due to the effects of population 2. Thus, species 2 is the better competitor, and
will displace species 1 from this habitat. The yellow area of the graph is where
the densities of the 2 populations are such that both can continue to grow; the
green area of the graph is where the densities are such that population 2 can
keep growing but population 1 must shrink; and the pink area is where both
populations would have to shrink. The black dot indicates the stable equilibrium
point – the population densities at which the net growth for both species will be
zero. In this case, population 1 has been excluded (its density is 0), and
population 2 will stabilize at its carrying capacity (K2). Now consider a situation
where the two isoclines cross:
Case II:
K1
α
K2
N2
N1
K1
K2
β
We can see that effect of intraspecific competition in each population is greater
than the effect that either population has on the other: on the x-axis, population
1 reaches its carrying capacity before it limits population 2; on the y-axis,
population 2 reaches its carrying capacity before it limits population 1. In this
case, the stable equilibrium population densities will be reached where the 2
isoclines cross. In the yellow area, both populations grow; in the green area,
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population 2 grows, but population 1 shrinks; in the pink area, population 1
grows, but population 2 shrinks; and in the blue area, both must shrink. The
black dot indicates the stable equilibrium point – the population densities at
which the net growth for both species will be zero.
It now becomes apparent that there are only 4 possible outcomes of a
competitive interaction, depending on the relative location of the intercepts of
the 2 zero-net-growth isoclines:
1. If K2 > K1/α, and K2/β > K1: population 2 will exclude
population 1 and the stable equilibrium point will be at N1=0 and
N2=K2. This is the situation in Case I (isocline of pop 2 is higher
and to the right of the isocline of population 1).
2. If K1 > K2/β, and K1/α > K2: population 1 will exclude
population 2 and the stable equilibrium point will be at N2=0 and
N1=K1. This is the opposite of the situation in Case I (isocline of
pop 1 is higher and to the right of the isocline of population 2).
3. If K1/α > K2, and K2/β > K1: each species inhibits itself more than it
inhibits the other species. The two isoclines will cross and the
stable equilibrium population densities will be reached at this point
of intersection. This is the situation in Case II (stable coexistence).
These are the only stable outcomes, but a fourth unstable outcome is possible:
4. If K1 > K2/β, and K2 > K1/α: each species inhibits the other species
more than it inhibits itself. The two isoclines will cross, but stable
coexistence is not possible. Eventually, one species will be excluded
and the equilibrium densities will be either N1=0 and N2=K2 or
N1=K1 and N2=0. This is the situation in Case III below:
K2
Case III:
K1
α
N2
N1
K2
β
K1
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In the yellow area, both populations grow; in the green area, population 1 grows
to carrying capacity but population 2 shrinks; in the pink area, population 2
grows to carrying capacity but population 1 shrinks; and in the blue area, both
must shrink. The black dots indicate the two possible stable equilibrium points –
K1 or K2.
In situations where one isocline is higher and to the right of the other isocline,
the 2 isoclines do not cross and it is obvious that one population will
competitively exclude the other. In situations where the 2 isoclines cross (Case
II and Case III), how can we quickly tell if the outcome will be stable coexistence
or exclusion? Recall that stable coexistence occurs when:
K2 <
This can only occur if:
K1
α
and
K1 <
K2
β
(eq.11 & 12)
αβ < 1 (eq.13)
This makes sense: two species can coexist if they don’t have much impact on
each other. If two species exert a severe impact on each other (i.e., the
coefficients of competition are large), they are less likely to coexist stably and
one will end up excluding the other. By extension of eq.s 11 & 12, the
conditions for stable coexistence are:
α<
K1 1
<
K2 β
(eq.14)
For example, if α = 0.6, stable coexistence can only occur if β < 1.67, and the
ratio of K1/K2 is greater than 0.6.
Isocline analyses in Populus: Phase-plane graphs
Recall that Figs 1 and 2 are graphs generated in Populus showing the changes in
the sizes of the populations over time. Now we can look at the isoclines of the 2
populations using a phase-plane graph:
Fig 3. Phase-plane plot of the relationship shown in Fig 1.
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In this figure, the 2 axes are now the densities (or sizes) of the 2 populations.
The red line is the zero-net-growth isocline for population 1 and the blue line is
the zero-net-growth isocline for population 2. The x-intercept for population 1 is
its carrying capacity (500). The y-intercept for population 2 is its carrying
capacity (700). The x-intercept for population 2 = K2/β = 700/0.7 = 1000. The yintercept for population 1 is K1/α = 500/0.6 ≈ 833. Because both populations
reach their carrying capacity before they inhibit the growth of the other
population, the populations will coexist.
The green line is the phase-plane plot of the 2 populations. This tracks the
relative densities of the 2 populations. At the beginning (starting at the origin)
population 1 increases quickly relative to population 2 and the line veers right.
As population 2 increases in size, population 1 starts to decrease in size and the
line veers left. The phase plot stabilizes at the equilibrium densities of the 2
populations. From figure 1, we know that dN/dt = 0 for species 1 when its
population density is a bit more than 100 individuals, and that dN/dt = 0 for
species 2 when its population density is around 600 individuals. In this case, it is
the point where the 2 zero-net-growth isoclines meet (where the lines cross, and
the green arrow points). What are the exact densities of population 1 and 2 at
this point? We can use the following equations to solve for these values:
K − α K2
Nˆ 1 = 1
1 − αβ
and
K − β K1
Nˆ 2 = 2
1 − αβ
(eqs.11 & 12)
Thus, the equilibrium density of population 1 is:
K − α K 2 500 − (0.6 × 700)
Nˆ 1 = 1
=
= 138
1 − αβ
1 − (0.6 × 0.7)
and the equilibrium density of population 2 is:
K − β K1 700 − (0.7 × 500)
Nˆ 2 = 2
=
= 603
1 − αβ
1 − (0.6 × 0.7)
Now consider the situation described in Fig 2 (competitive exclusion of species 1
by species 2). What does the phase-plane plot look like?
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Fig 4. Phase-plane plot of the relationship shown in Fig 2.
The plot indicates that equilibrium is achieved when N1 is 0 (population 1 has
been displaced) and when N2 = 580 (the carrying capacity of population 2).
Assignment
All simulations should be run using the Multi-Species: Lotka-Volterra Competition
Model. Remember to run all simulations until steady state!!!
Part A.
1. Run a simulation in Populus using the Multi-Species: Lotka-Volterra
Competition Model starting with the following parameters: for both
populations, N0 = 30, r = 0.5, K = 700, and both α and β are equal to 1.
i. Show the time trajectory (N vs. t ) and phase-plane (N2 vs. N1) plots.
[4 marks]
ii. Consider why the graphs look the way they do and what these graphs tell
you about the populations (in terms of growth patterns, equilibrium
densities, etc.). (max. ½ page). [4 marks]
2. Now change parameters so that the starting density of N2 is different from
the starting density of N1 (but all other parameters are the same as in part
A1).
(Hint: You should start by keeping the same initial N1 as in Part A1 and
slowly vary N2 to see how it affects the simulation)
i. Include one representative set of graphs (one time trajectory and one
phase-plane). You may be penalized for including redundant or
irrelevant graphs. [4 marks]
ii. Explain what happens to the population dynamics when the starting
density of one population is larger than the other? Refer to your graphs
to illustrate your points (max. ½ page) [4 marks]
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Part B.
1. Start a simulation with K1 = 700, K2 = 400, α = 1 and β = 0.5. For both
populations, N0 = 30 and r = 0.5.
i. Show the time trajectory and phase-plane plots. [4 marks]
ii. Describe what α and β mean in general and with regards to the
populations in your simulation (max. ½ page). [6 marks]
iii. Calculate the equilibrium densities of the two populations (show full
calculations). [2 marks]
2. Experiment with the effects of changing starting densities and intrinsic growth
rates on population dynamics. You should do this methodically by starting
with the simulation parameters from part B1 and only changing one
variable at a time. First, while observing the time trajectory and phaseplane graphs created in part B1, try slowly increasing and decreasing the
initial density of species 1. Then, after returning to your initial
parameters, see what happens when you slowly increase and decrease the
r value for species 1.
i. Show one representative set of graphs (one time trajectory and one
phase-plane) that demonstrate what happens when you change N0 and
one set that demonstrate what happens when you change r. [8 marks]
ii. Compare the new graphs to the original simulation (in part B1) as well as
to each other. Interpret what is happening in the time trajectory and
phase-plane plots (Hint: What changes? What doesn’t?) (max. ¾ page).
[10 marks]
Part C.
1. Starting with the parameters used in part B1, try varying the carrying
capacities and coefficients of competition to produce the four different
outcomes described in the background material.
i. Show the time trajectory and phase-plane graphs for each scenario. [16
marks]
ii. Explain what is happening in each case (max. ¼ page per scenario).
[8 marks]
2. Starting with the parameters used to create the example of an unstable
equilibrium in part C1, examine the effects of changing starting densities
and intrinsic growth rates on population dynamics as you did in part B2.
i. Show one representative set of graphs (one time trajectory and one
phase-plane) that demonstrate what happens when you change N0 and
one set that demonstrate what happens when you change r. You may be
penalized for including redundant or irrelevant graphs. [8 marks]
ii. Compare the new graphs to the original unstable simulation (in part C1)
as well as to each other. Interpret what is happening in the time
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trajectory and phase-plane plots and explain how these simulations differ
from the ones you ran in part B2 (max. ¾ page). [8 marks]
***All answers should be typed and double-spaced using a standard
12pt font with one inch margins. Remember to include important values
where appropriate and refer to your figures to illustrate your points. Your
answers should be clear and concise and written in paragraph form using full,
grammatically correct sentences. Organization, writing style, grammar and
spelling count.
Total = 100 marks
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