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growth Models in Ecology Population growth and

VIEWS: 43 PAGES: 27

									Models in Ecology:
Population growth and
competition
             David Hansen
            Dept. of Biology
      Pacific Lutheran University
                 Linear Growth
                                               Population Growth
    dn/dt = c                     60

Where c is the number                      population
                                  50
of individuals added
each time unit          numbers
                                  40


                                  30

The integrated form               20


                                  10
    Nt = ct + N0                  0
                                       0   2     4       6     8   10   12

                                                        time
Assumptions for linear growth
 Constant number of individuals added each
  time unit
 Number added not proportional to
  population size
             Exponential Growth
                                                        Population growth

   dn/(dt*N) = r                         1200


                                         1000
Where r is the instantaneous                             population

rate of change.                          800




                               numbers
The integrated form                      600


                                         400
   Nt =    N0ert                         200


                                           0
                                                0   2       4     6    8    10   12

                                                                time
  Assumptions for exponential
           growth

 Reproductive rate constant per individual
 Number of individuals reproducing
  proportional to population size
 Unlimited environment
          Realistic growth

 Resources are limited
 Birth rates change
 Death rates change
Deriving equation for changing
     birth and death rates
    given
         dn/dt = rN
    if
         r=b-d
    where:
         b = birth rate
         d = death rate
    then
         dn/dt = (b - d)N
Change in birth and death rates with
        increased numbers
      b0




      d0
                Neq
               numbers
Deriving equation for changing
     birth and death rates
  in
       1. dn/dt = (b - d)N
  substituting for birth and death rates in equation 1,
  the equations for changing birth and death rates
       2. dn/dt = [(b0 - kbN) - (d0 + kdN)]N
  Rearranging and grouping terms
          dn/dt = [b0 - kbN - d0 - kdN]N
          dn/dt = [b0 - d0 - kbN - kdN]N
  And finally
       3. dn/dt = [b0 - d0 - N(kb + kd)]N
Determining values for (kb +
           kd)
For no change in population size, r = 0,
and since r = b + d, then b = d. At this
condition, N = Neq
   thus
          b0 - kbNeq = d0 + kdNeq
   solving for (kb + kd)
          b0 - d0 = kbNeq + kdNeq
          (kb + kd)Neq = (b0 - d0)
    and finally
          (kb + kd) = (b0 - d0)/Neq
Substituting for (kb + kd) in
        equation 3
    dn/dt = [(b0 - d0) - N(kb + kd)]N
becomes

    dn/dt = [(b0 - d0) - N((b0-d0)/Neq)]N
rearranging
    dn/dt = [(b0 - d0)(1 - N/Neq)]N
If b0 - d0 is redefined as r0 (intrinsic rate of growth) and
Neq is defined as K (carrying capacity), then

    dn/dt = r0 N(1 - N/K)
        Plot of logistic growth

                                                             Population growth
the differential equation                           30




                                  population size
                                                    25
 dn/dt = r0N(1 - N/K)                               20
                                                                      Zone of low or
                                                                      no growth
                                                    15


the integrated form                                 10

                                                     5       Zone of rapid growth
                                                     0
 Nt = K/[1 + (K/N0    -1)e-rt ]                          0        5            10      15

                                                                       tim e
  Reproductive strategies derived from a
populations location on logistic growth curve
 r - selection             K - selection
 occurs at low pop.       occurs at high pop.
  density                   density
 little competition       resources limiting
 high no. of offspring    large parental
 low energy/offspring      investment/offspring
 little or no parental    parental care
  care                     slow development
 rapid development        low no. of offspring
 often semelparous        iteroparous
 good colonizers          good competitors
Interactions between two
         species
   Lotka - Volterra Equations
Measurement of intraspecific
and interspecific competition
In the logistic equation, the term N/K measures the
effect on the growth of a population by addition of a
new member of the same species (intraspecific competition).
   dn/dt = rN(1 - N/K)
The effect of a second species on the growth of species one
can be modeled by adding a second term measuring the
effect of adding individuals of a second species (interspecific
competition).

   dn1/dt = r1N1(1 - N1/K1 - N2/K1)
       Measuring ecological
          equivalence
Since the effect of one species on the growth of another
would not likely be identical to a species effect on its
own growth, an equivalence adjustment must be made such
that:
       N1 = 12N2
Where:
         12 measures the effect on species1 by species2.
 and
       N2 = 21N1
Where:
         21 measures the effect on species2 by species1.
.
Lotka - Volterra competition
         equations
Thus for a two species system, the equation for each
species becomes:

for species1

 dn1/dt = r1N1(1 - N1/K1 - 12N2/K1)

for species2

 dn2/dt = r2N2(1 - N2/K2 - 21N1/K2)
Graphical analysis of Lotka-
    Volterra equations
 Searching for equilibrium conditions
Solving equations for zero
         growth
Set equation for each species equal to zero (no growth)
  r1N1(1 - N1/K1 - 12N2/K1) = 0
and
  r2N2(1 - N2/K2 - 21N1/K2) = 0
Dividing by riNi, multiplying through by Ki
and rearranging yields the following pair of
linear equations:
  N1 = K1 - 12N2
and
  N2 = K2 - 21N1
  Finding endpoints for linear
           equations
For each equation, substituting 0 for each Ni gives the end
points on a graph of N1 vs N2, defining an isocline for each
species.
   For species1;
         N1 = K1        In absence of sp2, sp1 will reach
                        carrying capacity
    and
         N2 = K1/12 It will require K1/12 of sp2 to
                        eliminate sp1.
   For species2;
         N2 = K2        In absence of sp1, sp2 will reach
                        carrying capacity
    and
         N1 = K2/21 It will require K2/21 of sp1 to
                        eliminate sp2.
        Graphing isoclines

        K1     dn1/dt = 0

     K2/12
N1
                            dn2/dt = 0




                   K1/21         K2
                    N2
               Graphical solutions
         K1    Sp1 always wins        K2/12   Sp2 always wins

     K2/12                             K1
N1



                       K2   K1/21                 K1/21     K2


          K1   Unstable equilibrium   K2/12   Stable equilibrium

      K2/12
N1                                       K1




                       K1/21    K2                  K2      K1/21
                  N2                                          N2
Conditions for outcomes

  Inequalities K1>K2/12    K1<K2/12


               Unstable      Species 2
 K2>K1/21
              equilibrium      wins


               Species 1      Stable
 K2<K1/21
                 wins       equilibrium
Further analysis of inequalities
Assuming the carrying capacity Ki is equal for both species
and inverting the inequalities, the following conditions occur.

   For unstable equilibrium     For stable equilibrium

           12 > 1                         12 < 1
     and                             and
           21 > 1                         21 < 1
    Biological interpretations
 The ’s measure the ability of a species to
  contain the other species relative to itself.
 If both ’s are <1, then each species has
  greater effect on its own growth than on the
  other species, ie. They are using different
  resources.
 If both ’s are >1, then each species is able
  to exclude the other from resources either
  by out-consuming or defense.
 Principle of competitive
        exclusion
Ecologically equivalent species
cannot coexist. One will go extinct in
the area of competition or switch
resources.
The End

								
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