INTERSPECIFIC COMPETITION
Table 1. Different type of interactions between species and the effect on density, where – (indicate a reduction), 0 (no change) and + (indicate an increase), in the density of a particular species due to the presence of the other species.
Interactions Competition Predator-Prey Parasite-Host Commensalism Mutualism Protocooperation
Species 1 + (predator) + (parasite) + + +
Species 2 - (prey) - (host) 0 (host) + +
Theoretical Models We already discussed the regulation of population growth in a density-dependent manner using the logistic growth model. The population growth rate of a species – Species 1 – using the logistic growth model dN1 = r1N1 (K1 – N1) dt K1 and for a second species; Species 2
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dN2 = r2N2 (K2 – N2) dt K2 where the subscripts are used to indicate values for species 1 and species 2 N1 t r1 K1
= = = =
population size for species 1 time instantaneous growth rate for species 1 carrying capacity for species 1
Also the same as N2, t, r2 dan K2 for species 2 Assume, that a second species competes with species 1 for some resource, such as food and space and that Each individual of species 2 uses an amount of α of this resource. The value of α indicates the relative per capita effect of species 2 on species 1, and is often between 0 and 1 for species of similar size. For example; if there are 100 individuals of species 2 and each uses 0.5 as much of the resources necessary for species 1 as an individual of species 1 (α = 0.5), then the resource available for species 1 is reduced to the amount it would be if 50 more individuals of species 1 were present. The effect of the competitor on the population growth rate of species 1 can be incorporated into logistic equation: 2
dN1 = r1N1 (K1 - N1 - αN2) dt K1 αN2 is the product of the per capita effect of species 2 on species 1 and N2 is the number of species 2 Therefore, the remaining carrying capacity available for species 1 is a function of both the 1) number of species 1 present and 2) the resources used by individuals of species 2. If α = 0, then the second species does not affect the growth of species 1, If α = 1, then an individual of species 2 is equivalent in its resource usage to an individual of species 1. If α > 1, then this may indicate that each individual of species 2 consumes resources much faster than each individual of species 1 or that there is direct aggression between the two species and that species 2 is much larger or a more successful aggressor.
Similarly, the rate of population growth of species 2 is affected by the numbers and resource utilization of species 1, therefore dN2 = r2N2 (K2 – N2 - βN1) dt K2 where β indicates the effect of species 1 on the resources for species 2. 3
The values α and β are called the competition coefficients and indicate the relative per capita effect of species 2 on species 1 and the relative per capita effect of species 1 on species 2, respectively. Note that α and β are not necessarily equal but that in many cases they are probably of similar magnitude, particularly when the competing species are related. These equations, giving the population growth rates of two competing species, are often called the Lotka-Volterra competition equations after the two ecologists, A. J. Lotka and V. Volterra, who first used them. The Lotka-Volterra equations can be used to predict the eventual outcome of interspecific competition. Overall, there are four possible outcomes: 1) species 1 wins, 2) species 2 wins, 3) either species I or species 2 wins depending upon the starting composition, or 4) the two species coexist. Let us begin by determining the conditions for the stable coexistence of two species. As before, we know that if dN1/dt = 0, then there is an equilibrium for species 1.
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The only nontrivial way (as earlier, r1 = 0 or N1 = 0 are trivial situations) in which this can happen is if the numerator
K1 — N1 — αN2 = 0. This equation can also be written as N1 = K1 — αN2 indicating that there is no change in species 1 numbers for the values of N1 that satisfy this equation. The solution to this equation is a straight line, with N1 = K1 when N2 = 0 and N2 = K1/α when N1 = 0. This line for species 1 Let us begin by determining the conditions for the stable coexistence of two species. As before, we know that if If
dN1 = 0 so that, dt
then, there is an equilibrium for species 1. The only nontrivial way (as earlier, r1 = 0 or N1 = 0 are trivial situations) in which this can happen is if the numerator K1 – N1 - αN2 = 0 This equation can also be written as N1 = K1 - αN2
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Indicating that there is no change in species 1 numbers for the values of N1 that satisfy the equation. The solution to this equation is a straight line, with N1 = K1 when N2 = 0 and N2 = K1 apabila N1 = 0 α
This line for species 1 (the values for which dN1=0) dt and the analogous one for species 2 are called zero isoclines. To clarify let us give the constant K1 and α particular values and calculate the line of points that satisfies the above equation. If we assume that K1 = 100 and α = 0.5, then N1 = 100 – 0.5 N2 If N2 = 0, then N1 = 100, the carrying capacity for species 1 and if N1 = 0, then N2 = 100/0.5 = 200. This isocline is drawn in Figure 1a, where the number of species 1 is on the horizontal axis and the number of species 2 is on the vertical axis. This isocline also divides the area into two regions in which dN1/dt is either positive (the shaded region below the line) or negative (the region above the line). For example, if the numbers of N1 and N2 are at some combination below the line, then dN1/dt is positive and the numbers of N1 will increase. This
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occurs because if we look at the numerator in the population growth equation, N1 + αN2 < K1 making the numerator and the equation positive. On the hand, if N1 + αN2 > K1 then the numerator is negative, and the number of species 1 declines as indicated by the arrows above the line. However, we have only given the equilibrium number of species 1. In order for the numbers of both species to remain constant, then dN2/dt must also equal zero. Analogously to species 1, this occurs when K2 – N2 - βN1 = 0 or N2 = K2 - βN1 Again, the solution to this equation is a straight line, in which N2 = K2 when N1 = 0 and N1 = K2 when N2 = 0 β Let us assume that K2 = 80 and β = 0.4, so that, N2 = 80 – 0.4N1 If N1 = 0, then N2 = 80, the carrying capacity for species 2, and 7
if N2 = 0, then N1 = 80/0.4 = 200. This isocline is drawn in Figure 1b, and it divides the graph into regions in which dN2 < 0 above the line dt dN2 > 0 the shaded region below the line. dt In order for there to be no change in the number of either species, then both must equal zero. dN1 = 0 and dN2 = 0 dt dt In the numerical case discussed above, this is the point where the two isoclines cross. If we superimpose the two parts of Figure 1, then as given in Figure 2, the two lines cross and divide the total area into four parts. The change in numbers for both species together is given by the arrows (or vectors). In the unshaded region both species decline in numbers, and in the darkly shaded region both species increase. The numbers of the two species change so that from any point in the total area they converge on a single value, the stable equilibrium. For example, if we let r1 = 0.1 and r2 = 0.1, α = 0.5, β = 0.4 8
and assume that initially we have 30 individuals of species 1 and 120 species 2 (still assuming K1 = 100; K2 = 80), then we can calculate the expected change in population numbers for both species. This combination of N1 and N2 is in the upper shaded region of Figure 2, so we would expect N1 to increase and N2 to decline. For species 1, the change is dN1 = r1N1 (K1 – N1 - αN2) dt K1 dN1 = (0.1)(30) X (100 – 30 – (0.5)(120) dt 100 = 0.3 For species 2, the change is dN2 = (0.1)(120) X (80 – 120 – (0.4)(30) dt 80 = - 7.8 Therefore, the numbers are expected to change form 30 and 120 of species 1 and 2 to 30.3 and 112.2, respectively, after one time unit.
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What are the numbers of the two species at the stable equilibrium? These numbers can be obtained by solving the two equations that were set equal to zero above. In other words, the values of the two unknowns, N1 and N2, that satisfy equations N1 = K1 - αN2 N2 = K2 - βN1 are the equilibrium numbers. If we substitute the values used above for the carrying capacities and competition coefficients, we can solve these equations. Inserting K1 = 100, K2 = 80, α = 0.5, β = 0.4, then N1 = 100 – 0.5N2 N2 = 80 – 0.4N1 Let us substitute the value of N from the first equation into the second so that N2 = 80 – 0.4 (100 – 0.5N2) N2 = 50 This value can be inserted in the first equation so that N1 = 100 – 0.5 (50) N1 = 75 Therefore, the equilibrium numbers of species 1 and 2 in this case are 75 and 50 respectively. This is the point where the two isoclines cross, indicated by a closed circle in Figure 2. Notice
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that the total number of individual of both species at equilibrium, 125, is higher than the carrying capacity of either of the species when they are present by themselves.
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