Growth Rates and Competition
I. Introduction.
In your study of density dependant effects and population growth rates in the lecture section of the course, you realize that growth is actually much more dynamic than implied by taking size at two points in time. In this lab we will follow the growth rate of populations through time. You will compare growth rates of a species when alone and when in competition with another species. The results from the competition part of the exercise are exactly comparable to the results you saw in class from Gause's classical experiments.
II Methods.
You will use two aquatic plants the fern Azolla and the duckweed Lemna. Both the fern and the angiosperm are composed of small isolated plants that float on the water surface. While both can reproduce sexually, they also reproduce asexually by fission, and it is this process that we will follow. Azolla exists symbiotically with a blue-green cyanophyte. Do you think this would have any effects on its ability to grow? If so, under what conditions?
All experiments will be carried out in plastic cups containing freshwater to which a small amount of fertilizer has been added. In the growth experiments one or two plants will be added to a cup and the density of plants will be followed over time.
Here are some things you will need to consider before you set up your study.
a)How will you define density?
b)How will you measure density?
c) How often will you need to take measurements and how long will the experiment run?
In the competition experiments a few individuals of each species will be added and the numbers of each will be followed through time. The data you collect will be the same as in the simple growth study, but keeping the data for the two species separately.
III Growth rate
There are several means to convert increase in density to a useable value for growth rate. One that is frequently seen is the specific growth rate - the number of individuals produced by each individual in the population over some unit of time.
Ind Ind-1 t-1
Depending on the particular organism under study, only females may be considered [i.e. number of females per female per unit time] and the time unit may vary from hours to years. Specific growth is often represented as µ [mu] and its units are t-1 since IndInd-1 cancels out.
µ = (ln N2 - ln N1) * (T2 - T1)-1
Where N2 is the number of individuals at time 2 (T2)
Where N1 is the number of individuals at time 1 (T1)
and ln signifies the natural log
Note that this method uses only two data points density at the start and density at the end. Is that likely to be a good measure? What kind of things might make a two variable estimate incorrect or misleading? How could you solve such problems?
Other ways of expressing growth rate are:
Doubling time - i.e. the length of time it takes for the population to double from
a size N to a size 2N
DT = ln 2 µ-1
As an exercise, derive this relationship from that given for the determination of µ
A third means of expressing growth rate, particularly for micro-organisms is:
Divisions per day - i.e. the number of times an individual divides in a day (this assumes T is in days)
Divisions per day = DT-1
Can you see why this is so?
Realize that in a population that is at equilibrium, net growth rate equals zero. On the other hand, a population that is very small relative to the ability to support it (the "carrying capacity") may be growing very quickly. When populations are growing maximally they often exhibit exponential growth, that is, the size of the population at any given time is an exponential function of the population size at some earlier time. For example:
NT = N0 * R0T
Where:
NT is the number of individuals in the population at time T
N0 is the number of individuals in the population at time zero
R0 is the specific growth rate - often in number of individuals per individual per generation
T is the length of a generation
What do you think “generation” implies? How would you measure this?
For populations without discrete generations - i.e. continuous reproduction or overlapping the expression used is:
Nt = N0 * ert
Where:
e is the base of natural logarithms (2.7183.....)
t is the time elapsed
r is the instantaneous growth rate
What do you think “instantaneous growth rate” is? What are its units?
If required resources of any sort (food, space, shelter) become limiting, or if mortality (from disease, predation etc.) increases, than the population will grow at less than its maximum rate (rm).
When a population begins growth in an environment that is originally well supplied with all necessary resources, the growth will, after a "lag” phase where growth rate is low as the individuals adjust to the new environment, grow exponentially until resources begin to limit numbers and growth rate decreases approaching zero. This sequence generates the familiar sigmoid growth curve described by many models including the well known logistic.
dn/dt = rN * {(K-N)/K}
Where:
dn/dt is the actual growth rate at a point in time
N is the size of the population at that point in time
(so actually it should have a subscript – but we can leave it out for now)
K is a “constant” the maximum possible population size or carrying capacity
R is the intrinsic growth rate for that species
IV. Competition:
In addition to obtaining the growth curve for each of the species, you will also be able to carry out a competition experiment using the same organisms. If you start a cup with a few individuals of both species present, they will both begin to grow exponentially, just as they do in the single species cups. But, at some point some resource, space, nutrients etc., will begin to limit the growth rate. At that point the differential abilities of each species will begin to come into play. You know from the Lotka-Volterra competition equations that:
dN1/dt = r1N1 *{(K1-N1-N212)/K1}
dN2/dt = r2N2 *{(K2-N2-N121)/K2}
Where:
the i's are the equilibrium numbers of species 1 and 2
the Ki's are the carrying capacities of species 1 and 2
the ij's are the competition coefficients of each species on the other
You can get the K's and the r's from the single species cups and so can determine the competition coefficients for each species:
12 = {(K1-1)/2)}
21 = {(K2-2)/1)}
Where the 's are the equilibrium densities in the presence of the other species.
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