BSC 2011 (Periods 3 and 5; Spring 1999)

a. Draw the phase plane diagram associated with this system (i.e., plot N₂ vs. N₁), labelling the isoclines, the starting point, and the resulting dynamics.

b. Now, use this information to plot how the numbers of Sp. 1 and Sp. 2 change through time (i.e., plot N₁ and N₂ vs. time). [Note -- don't worry about the precise rates of change or specific numbers of rodents at every point in time -- we'd have to tell you r₁ and r₂ in order for you to even attempt this -- here we're just interested in the qualitative patterns, and the initial densities and final densities.]

[NOTE -- in this example, the alphas are both less then 1, but there is no coexistence. That's because the result we discussed in class (i.e., that ALPHA₁₂xALPHA₂₁<1), is a necessary but not sufficient condition for coexistence.]

c. Repeat a and b, but change the Carrying Capacity for Species 2 to 100 (i.e., K₂=100). Contrast the dynamics for this two species system with the dynamics that would have resulted had each species been put in the field alone (i.e., without the other species).

d. Repeat the first part of c, but start the system off with 200 individuals of each rodent species.

For this problem, we are describing the dynamics of the system using the competition equations Dr. Osenberg gave in class:

per capita growth for Species 1:
dN₁/N₁dt = r₁(K₁-N₁-alpha₁₂*N₂)
                                 K₁
per captia growth for Species 2:
dN2/N₂dt = r₂₍K₂-N₂-alpha₂₁*N₁)
                                 K₂

To draw the isoclines for Species 1 and 2 in part A, we need to figure out the x and y intercepts for each isocline.  The x intercept for the Species 1's isocline is its carrying capacity or 100.  The y intercept for Species 1's isocline is K₁/alpha₁₂ or 200 (plug in 100/0.5).  Now you can simply "connect the dots" to draw the isocline or no net growth line for Species 1.  The y intercept for Species 2's isocline is its carrying capacity or 250.  The x intercept for Species 2's isocline is K₂/alpha₂₁ or 312.5 (plug in 250/0.8).  Now you can simply "connect the dots" to draw the isocline or no net growth line for Species 2.  See the graphs below for more examples.

SEE FIGURES BELOW FOR SOLUTIONS

Part a and b:

When we innoculate the system in part A with 15 individuals from each species, you can see from the above graphs that initially, both species grow.  Then, as Species 1 approaches its isocline, growth begins to slow.  When the population crosses Species 1's isocline, Species 1 is no longer growing, but Species 2 still grows (at this point, the population is still be below Species 2's isocline).  From this point on, Species 1 declines until it goes extinct.  Species 2 continues to grow until the population reaches the carrying capacity of Species 2.  Then, Species 2 reaches an equilibrium.

Part c:

In part C, note that when we change the carrying capacity of Species 2, the isoclines of the two species cross.  We now have an equilibrium point which allows coexistence (in fact, this is a stable equilibrium point or an attractor).  Also, we can see that Species 1 reaches a higher equilibrium value than Species 2.  This is due to the competition coefficients-alpha₁₂ (the effect of species 2 on species 1) is lower that alpha₂₁ (the effect of species 1 on species 2).  Therefore, when all other conditions are the same, Species 1 is a better competitor than Species 2, and more of Species 1 persists in the system.

The "blip" in the figure on the right for Species 2 might at first seem unexpected, but is analogous to the initial rise seen in part A for Species 1. In part C, this occurs because the total competitive environment experienced by Species 2 (during the initial increase) is still "low" enough that dN₂/dt>0. Because Species 1 is not yet at it's equilibrium, the total competitive effect is still less than at the eventual equililbrium. Notice that the system soon settles down to the equilibrium.

Part d:

 Here we have introduced more individuals into the environment than the environment can support.  Both species decline until they reach their respective isoclines.  Again, the isoclines still cross and we have coexistence (and we still have a stable equilibrium). And, Species 1 has a higher equilibrium density than Species 2 (again, due to the difference in the competition coefficients).  In this case, there is a slight dip in the right figure for species 1.  This is again analogous to the initial rise seen in part A for Species 1. This occurs because the total competitive environment experienced by Species 1 (during the initial decrease) is still "high" enough that dN₁/dt<0. Thus Species 1 "overshoots" its equilibrium density -- notice that the system soon settles down to the equilibrium.

Also notice that I've given the equilibrium solutions for N₁ and N₂ in these figures. They can be found by setting dN₁/N₁dt =0 and dN₂/N₂dt=0 and solving these two equations simultaneously.

In other words, from  dN₁/N₁dt = r₁(K₁-N₁-alpha₁₂*N₂)
                                                                     K₁
set (K₁-N₁-alpha₁₂*N₂)=0 and from dN₂/N₂dt = r₂₍K₂-N₂-alpha₂₁*N₁)
                                                                                              K₂
set (K₂-N₂-alpha₂₁*N₁)=0.  Then solve these two equations simultaneously.

A little algebra will reveal that N₁(equilibrium)=83.3 and N₂(equilibrium)=33.3.

I am listing the variables and their values for the generalized Lotka-Volterra models that can be obtained from each bulleted statement.  Note the signs of the interaction terms, and remember the notation (i.e.,  a₁₂ is the effect of species 2 on species 1).  Below (at the end of question 2), I will show the full equation for each species.
 
• Species 1 and 2 compete for nutrients (each algal cell of Species 1 reduces the per capita growth rate of Species 2 by .01/day and Species 1 by .02/day; each algal cell of Species 2 reduces the per capita growth rate of Species 1 by .005/day and Species 2 by .015/day).

a₁₁=-0.02, a₂₁=-0.01, a₁₂=-0.005, a₂₂=-0.015
 
• In the absence of copepods, and at low densities (very close to zero) of both algal species, Sp. 1 has a per capita growth rate of  .9/day and Sp. 2 has a per capita growth rate of .75/day.


r₁=0.9, r₂=0.75
 

• Species 1 is consumed by the copepod at a rate of .03/copepod/day;

a₁₃=-0.03

• Species 2 is not eaten by the copepod.


a₂₃=0
 

• The copepod grows at a per capita rate of .2/day when Species 1 occurs at an abundance of 100 cells and there are no fish, and a rate of .08/day when Sp. 1 occurs at an abundance of 50 cells and there are no fish.


By plotting dN₃/N₃dt vs. N₁ and calculating the slope of this relationship, you can find the effect of species 1 on species 3 (a₃₁) .  The slope of that line is 0.002 (0.2-0.08/100-50), so a₃₁=0.002.
 

• The fish eats the copepod, and each fish imposes a mortality rate of .04/day on the copepods.

a₃₄=-0.04

• In the absence of copepods, fish die (via starvation) at a per capita of .1/day.  When copepod abundance is 20, fish have a per capita growth rate of 0.

 

 

Similar to above, by plotting dN₄/N₄dt vs. N₃ and calculating the slope of this relationship, you can find the effect of species 3 on species 4 (a₄₃).  The slope of that line is 0.005 (0-[-0.1]/20-0), so a₄₃=0.005.

Using this information, please write down four equations (e.g., using the generalized form of the Lotka-Volterra equations) to describe the dynamics of this system (i.e., write equations for dN₁/N₁dt, dN₂/N₂dt, dN₃/N₃dt, and dN₄/N₄dt as functions of N₁, N₂, N₃, and N₄).  Use actual numbers instead of variables whenever possible for the interaction coefficients.  Some of the interaction coefficients may not be estimable from the information given.  In these cases, just use a variable instead of an actual number.
 

dN₁/N₁dt = r₁ + a₁₁N₁ + a₁₂N₂ + a₁₃N₃
                    = 0.9 + (-0.02)N₁ + (-0.005)N₂ + (-0.03)N₃

dN₂/N₂dt = r₂ + a₂₂N₂ + a₂₁N₁
                 = 0.75 + (-0.015)N₂ + (-0.01)N₁

dN₃/N₃dt = r₃ + a₃₄N₄ + a₃₁N₁
                = r₃ + (-0.04)N₄ + (0.002)N₁

dN₄/N₄dt = r₄ + a₄₃N₃
                = r₄ + (0.005)N₃
 

3. Dr. Frank N. Stein developed a life table for the three-toed yellow-dotted swamp frog. This frog is very rare and it took Dr. Stein many years to collect these data. This endangered frog used to thrive in wetlands near Gainesville, but went locally extinct in the late 1960's. Dr. Stein recently introduced 20 frogs (10 newborns, and 10 one year olds) into a swamp near Gainesville in an attempt to re-establish a local population. Dr. Stein has not been feeling (or behaving) well lately, and he has asked you to continue his studies (he's in seclusion in his laboratory). You must project the population size of the frogs over the next 100 years, and estimate various population parameters. Here's the relevant information:

a. Project the population size through time (i.e., say how many individuals of each age class there will be after 1, 2, 3, .... 10 years; in "year 0" there are 10 "0 year" frogs and 10 "1-year old" frogs).
 

Time	n0	n1	n2	n3	n4	newborns calculated as:
	(14)
0	10	10
1	11.2	7	5.6			2 x 5.6
2	13.6	7.8	4	2.8		2 x (4 + 2.8)
3	15.8	9.52	4.5	2	1.4	2 x (4.5 + 2 + 1.4)
4	17.2	11.1	5.4	2.2	1	2 x (5.4 + 2.2 + 1)
5	20.2	12	6.3	2.7	1.1	2 x (6.3 + 2.7 + 1.1)
6	23	14.1	6.9	3.2	1.4	2 x (6.9 + 3.2 + 1.4)
7	26.2	16.1	8.1	3.4	1.6	2 x (8.1 + 3.4 + 1.6)
8	29.8	18.3	9.2	4	1.7	2 x (9.2 + 4 + 1.7)
9	34.2	20.9	10.5	4.6	2	2 x (10.5 + 4.6 + 2)
10	38.8	23.9	11.9	5.2	2.3	2 x (11.9 + 5.2 + 2.3)

It is pretty straight forward to figure out the survivorship of the 10 newborns in the above table.  To calculate how many newborns made it to become one year olds, multiple 10 by the survivorship of one year olds (0.7), and you can see that 7 of the original 10 newborns became one year olds.  Next, to figure how many of the 10 newborns became two year olds, multiple 10 by the survivorship of two year olds (0.4), and you can see that 4 of the original 10 newborns became two year olds.  You can do this for the rest of the ageclasses.
To figure out the fate of the original 10 one year olds, you can first need to calculate how many frogs survive between each age class (i.e., from age zero to one = 0.7 [.7/1.0], from age one to two = 0.56 [0.4/0.7], from age two to three = 0.50 [0.2/0.4], from age three to four = 0.50 [0.1/0.2], from age four to five = 0 [0/0.1]).  Then you need to multiple the number of individuals in an age class in one time step (i.e., time=1, n₂=5.6)  by the mortality rate for that time step (mortality for age two to three = 0.5) to get the number of individuals in the next age class during the next time step (i.e., time=2, n₃=2.8). You can do this for every age class and time step to fill out the above table.

b. What is Ro? Will the population increase or decrease in size?

R₀= S (l_xb_x)= 1.4 offspring/individual/generation
c. Estimate generation time.
G = S (xl_xb_x)/ S (l_xb_x) = 3.6/1.4 = 2.6
d. Estimate r, and l.
r= (ln R₀)/G = (ln 1.4)/2.6 = .13/year

l = e^r = 1.14

Therefore, the population increases 14% per year.
 

e. Assume the population age structure has stabilized. Use the number of frogs in year 10, and information from parts b,c, and d to estimate how many frogs there will be 100 years after the frogs were introduced into the swamp.
N₁₀₀= N₁₀e^rt= N₁₀l^t
=82.1 (1.14)⁹⁰= approximately 10,000,000 frogs!

slight rounding errors can result in large differences in your final result.

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