BSC 2011 (Periods 3 and 5; Spring 1997)

Discussion for April 18^th

SPECIES INTERACTIONS

NOTE: You might find that Populus will again come in handy when evaluating this week's material (e.g., to explore the effects of interspecific competition, or to further examine effects of age-structure). If you want to download Populus, check out the instructions associated with the week of April 7. However, this is not necessary, and you should be able to answer the following questions without using Populus. However, to get a more complete understanding of the material, feel free to "play around" with Populus.

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.

SEE FIGURES BELOW FOR SOLUTIONS

Part a and b:

Part c:

Part d:

Note for Parts c and d: The "blips and dips" in the figures on the right (e.g., for species 1's dynamics in part d and species 2's dynamics in part c) might at first seen unexpected, but are 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. Thus Species 1 "overshoots" its equilibrium density -- because Species 2 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.

Also notice that I've given the equilibrium solutions for N₁ and N₂ in these figures. They can be found by setting dN₁/dt =0 and dN₂/dt=0 and solving these two equations simultaneously. A little algebra will reveal that N₁(equilibrium)=83.3 and N₂(equilibrium)=33.3.

2. A community consists of 4 species: Sp. 1 = an algae; Sp. 2 = another algae; Sp. 3 = a copepod; Sp. 4 = a fish. Species 1 and 2 compete for nutrients (each algal cell of Species 1 reduces the per capita growth rate of Species 2 by .1/day and Species 1 by .2/day; each algal cell of Species 2 reduces the per capita growth rate of Species 1 by .05/day and Species 2 by .15/day). Species 1 is consumed by the copepod at a rate of .03/copepod/day; Species 2 is not eaten by the copepod. The fish eats the copepod, and each fish imposes a mortality rate of .04/day. Using this information, please write down four 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).

dN₁/N₁dt = r₁ - a₁₁N₁ - a₁₂N₂ - a₁₃N₃

= r₁ - 0.2N₁ - 0.05N₂ - 0.03N₃

dN₂/N₂dt = r₂ - a₂₂N₂ - a₂₁N₁

= r₂ - 0.15N₂ - 0.1N₁

dN₃/N₃dt = r₃ - a₃₄N₄ + a₃₁N₁

= r₃ - 0.04N₄ + a₃₁N₁

dN₄/N₄dt = r₄ + a₄₃N₃

Note: a₄₃ and a₃₁ are positive terms because they denote the effect of a prey species on the predator population. These terms are assumed to consist of two components: the feeding rate (f) and the conversion efficiency (c), which tells how many prey must be eaten to produce one new predator. In such a case, we could write:
a₄₃ = f₄₃c₄₃ = .04c₄₃
a₃₁ = f₃₁c₃₁ = .03c₃₁



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)

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

R₀= SUM (l_xb_x)= 1.4 offspring/individual/generation

c. Estimate generation time.

G = SUM (xl_xb_x)/SUM (l_xb_x) = 3.6/1.4 = 2.6

d. Estimate r, and LAMBDA.

r= (ln R₀)/G = (ln 1.4)/2.6 = .13/year

LAMBDA = 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= N₁₀LAMBDA^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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