Can we explain the Isle Royale wolf-moose data as a purely predator-prey prey system? How much of the wolf and moose abundances can be explained with one of the simplest predator-prey models we have? Here we will try to find out.

When we use mechanistic models in science, we are following the scientific method we may have first been exposed to in K-12 education:

  1. observe a pattern in nature.
  2. propose an explanation (hypothesis) that can generate predictions.
  3. gather data to test the predictions.
  4. re-evaluate the hypothesis in light of the data.

Here are our data.

Natural populations undergo fluctuations. Here we smooth the data over different proportions of the time series to reveal different patterns of squiggliness.

Natural populations undergo fluctuations. Here we smooth the data over different proportions of the time series to reveal different patterns of squiggliness.

Next, let’s describe a model of predator and prey interaction. See your tetbook for more information.

Lotka-Volterra prey-dependent predation

A mechanistic model has parameters and state variables. State variables are the quantities that vary through time, such as population size. In contrast, parameters are constants that determine how fast the state variables change. In our exponential growth equation, \(dN/dt=rN\), \(N\) is the state variable, and \(r\) is the parameter.

Here are the rate equations of the Lotka-Volterra predator-prey model, where \(N\) is the prey population, and \(P\) is the predator population. The Lotka-Volterra predator-prey model is among the simplest predator-prey models there are.

\[\frac{dN}{dt} = rN-aNP\] \[\frac{dP}{dt} = baNP -mP\] In the above equations,

  1. How would you characterize the growth of the prey in the absence of predators? (Pretend \(P=0\))
  2. How would you characterize the growth of the predator in the absence of prey? (Pretend \(N=0\)).

This model is famous for its oscillations. Do our data show evidence of oscillations? Yes it does. Look at the figures below to estimate how often they undergo a cycle.

Use this phase plane plot to count how many times the populations undergo a complete counterclockwise cycle.

Use this phase plane plot to count how many times the populations undergo a complete counterclockwise cycle.

Equilbria

One thing we can learn from simple models are the equilibria they predict. An equilibrium is a point when the populations stop bouncing around. At these points, the populations stop changing - their growth rates equal zero. By setting growth rates equal to zero, we can use the model to make predictions about what those equilibria are.

Another name for an equilibrium is an attractor. We call it an attractor because the population appears to be attracted to it. If some event pushes the abundance of a population away from an equilibrium, the equilibrium seems to attract the population back to it.

Here we solve for the equilibria of the Lotka-Volterra predator-prey model. We set the prey population growth rate equal to zero, and then use algebra to simplify the expression. This is one way we learn the logical consequences of prey growth being zero.

\[0 = rN-aNP\] If we divide both sides by \(N\), we get \[0 = r-aP\] and rearranging we get \[P=r/a\] This means that the prey population growth rate will be zero when the predator population size is \(r/a\). This happens to be the intrinsic rate of increase divided by the attack rate.

If the prey population stops changing when \(P=r/a\), do you think the predator population will stop growth too?

Next, we set the predator population growth rate equal to zero, and then use algebra to simplify the expression.

\[0 = baNP - mP\]

  1. Simplfy this expression on your own. What do you get? What will \(N\) be equal to?

If the prey population stops changing when \(P=r/a\) and the predator population stops chaning wehn \(N=m/(ba)\), this means that at these two abundances, the populations will no longer fluctuate. These are the expected equilibria for these two populations under this set of assumptions. Something different happens when \(N\) and \(P\) are not at these values.

We abbreviate these equilibria as \(N^*\) and \(P^*\), which we refer to as “N star” and “P star”.

If our populations are regulated approximately according to these rules, we might expect that the average abundances of these populations are approximately equal to these values.

  1. Find the average or mean population size of the wolf population in our data set.
  2. Find the average or mean population size of the moose population in our data set.

Next we find values for the model parameters.

Intrinsic rate of increase, r

One way to think about the intrinsic rate of increase of a population, \(r\), is that it is the maximum per capita growth rate. A crude way to estimate that is to find the maximum observed value in our data.

  1. Estimate \(r\). Return to your Isle Royale spreadsheet and find the maximum \(r\) or ln(lambda) for the moose population across all years. You can do that visually, or you can use “=MAX(D2:D53)”. Let that maximum be \(r\).

Attack rate, a

It is worth noting that \(aN\) is the rate at which the average wolf kills moose. We will estimate the attack rate, \(a\), by realizing that the predator equilibrium equals \(r/a\), or \[a = r/P^*\] where our estimate of \(P^*\) is the average size of the wolf population.

  1. Estimate \(a\) using your above estimates of \(r\) and \(P^*\).

Predator mortality rate, m

One way to think about the predator mortality rate, \(m\), is that it is the maximum rate per capita rate of decline. Put another way, it is the minimum per capita population growth rate, and this will be a negative number.

  1. Estimate \(m\). Return to your Isle Royale spreadsheet and find the minimum \(r\) or ln(lambda) for the wolf population across all years. You can do that visually, or you can use “= ABS( MIN(D2:D53) )”. Let that the absolute value of that minimum be \(m\). Remember that \(m\) by itself is a positive number….

Conversion efficiency, b

The predator conversion efficiency is the prey-dependent birth rate of predators. Thus, for each prey killed, an average predator will produce and successfully rear \(b\) offspring. Given that only a few females in a population produce pups, and that number is averaged across the entire wolf population, that number is relatively small. However, a single moose is pretty big and so goes a long way.

Here we use our previous estimates of \(a\), \(m\), and \(N^*\) to estimate \(b\), where \[N^* = \frac{m}{ba}\] and \[b=\frac{m}{aN^*}\]

  1. Use the above expression to estimate \(b\).

Predicting population dynamics

In this next section, you will use a spreadsheet (predprey.xlsx) that has the Lotka-Volterra model built-in, so that you can explore the dynamics of this model.

Open the spreadsheet and add the parameters you just derived above. You will find cells in the spreadsheet for each parameter.

  1. Once you do this, you should get about 3-6 cycles over 52 years, and the minimum and maximum population sizes should differ tremendously from the observed data. If you get something wackier than that, you’ve gone astray. If so, ask for help.

Once you are confident that you got something reasonably correct, assess the predicted dynamics.

  1. In what ways are the model dynamics similar to the observed data? Consider the number of cycles, the size of the peaks and the troughs, and the averages.
  2. In what ways are the model dynamics different to the observed data?
  3. How could you test the effect of the parvovirus that began to affect the wolves in 1981?

Write down your answers to these questions, and we will go over them.

For the next exam,