Bio 481 - Problem Set 3

 

General Note: a^b stands for a raised to the b power.

  1. A certain species of parthenogenetic lizard has a maximum life span of four years. Two-year-old lizards (those between 2 and 3 years of age) produce two babies per year, and three-year-old lizards prdouce three babies per year. In 1918 there were 50 babies (individuals between 0 and 1 years old), 10 between ages 1 and 2, 5 between 2 and 3, and 2 between 3 and 4. What are the values of Nx(t), where N is the number of individuals, x is age category, and t is time, for x=0,1,2,3 and t=1918? What are the values of mx (maternity factor = number of babies produced per individual in category x) for x=0,1,2,3?

    This should be pretty intuitive, with no real work involved:
    and

  2. For the population of lizards in problem 1, the number of babies in 1919 was 16, the number of yearlings (one year old individuals) was five and the number of three-year-olds was two. Estimate the values of sxy (probabilities of surviving from category x to y) for all four age categories. Why might these values not be correct?

    For survival from age class 0 to 1, we have 50 individuals -> 15 individuals (s01=.3). For survival from age class 1 to 2, 10 individuals -> 5 individuals (s12=.5). For survivial from age class 2 to 3, 5 individuals -> 2 individuals (s23=.4). These values are as "correct" as we can get from this information, but there might be some problems with these values in the model. For example, we would have to assume that our method of aging individuals is 100% accurate. Also, we would have to have sampled the population at the same time of year in both years. Also the time at which we would sample should be right at the end of the breeding season (since we are interested in the survivorship of those lizards that are reproducing). Finally, there may be stochastic variation that leads to different survivorship values from year to year (not to mention that this model doesn't take any density dependence into consideration).

  3. How many lizards in eage age category would you expect in 1920?

    You can do this two ways. One way would be to write down the difference equations for each age class and then iterate it for one year. The other method is to take the projection matrix and multiply it by the population at 1919. This is what I am going to do here.

    N1920 = A*N1919 where A is our projection matrix.

  4. Write down the population projection matrix for the above lizard example.

    We already wrote this down in problem 3. Here it is again: A =

  5. Use the projection matrix to obtain the value of N(1921) for this lizard population (N without a subscript will stand for the total population in these exercises).

    It is important to note that matrices are not commutative (i.e. A*B = B*A is false). However they are associative (i.e (A*B)*C = A*(B*C) ). We know that N(1921) = A*N(1920). Since N(1920) = A*N(1919), N(1921) = A*(A*N(1919)). Using the property of associativity, N(1921) = A2*N(1919).
    Here is the solution:

  6. What is the age distribution vector (the values of Nx for all age categories) in the year 1940, for this lizard population?

    1940-1919=21. Therefore, N(1940)=A21*N(1919).


  7. Consider a model population in which m0=0, m1=2, m2=3, s12=0.5, and s23=0.2. Begin with the following age-distribution vector

    and project the population 10 times. Compute the values of Bx (the proportion of individuals in category x) for x=0,1,2, and t=1,2,3,4,5,6,7,8,9,10. Plot B0(t), B1(t), and B2(t) against time. Plot the total number of individuals, N(t), against time.

    Our projection matrix, C, is:
    Nt=Ct*N0: for t=1 to 10.
    Bi(t)=Ni(t)/[N1(t)+N2(t)+N3(t)]
    for t=0 to 10.

    Here you can see that the population is coming close to stable age distribution (proportion of individuals in class i is becoming constant).


  8. Repeat exercise 7, beginning with the age-distribution vector,

    What is the difference between the trajectories here and those in exercise 7?

    I'm not going to go through all the work again, but the major difference is that the population starts off at nearly the stable age distribution. Here is a graph of the total population through time to show this (close to exponential growth):



  9. Using the projection data from exercise 8, estimate the value of lambda, where: N(t+1)=lambda*N(t).

    We can assume that the population is at (or very near) SAD by time 9. You can also show this by look at the fact that the proportion of individuals in each age class is constant by then. Therefore, N(10)/N(9) will give us lambda. Since the population is at SAD, we can look at the proportion of a single age class rather than the entire population if we wish.
    N(10)=114.18. N(9)=101.48. lamba = N(10)/N(9) = 1.125.


  10. Consider a rodent population with the follwing projection matrix,

    Begin with 10 individuals in the first age category and project the population 150 time units. Plot the total population over time. What is the stable age distribution? What is the value of lambda?

    Here is the plot of the population over time:

    At time 150, the population vector is: . This gives a SAD of (where Bi=Ni/(Ntotal)): Lambda = Ntotal(150)/Ntotal(149) = 1.506.


  11. Repeat exercise 10 but reduce the survivorship of the last two age categories to .1. Discuss the result.

    This doesn't have much affect on the overall population growth. It lowers the population at 150 years by less than an order of magnitude (which is quite small when we are dealing with a population in the 27th order of magnitude). It also has little affect on the SAD, changing only the last two proportions significantly. Here's the SAD:
    Lambda is also very similar. Now it's 1.489.


  12. Repeat exercise 10 but reduce the survivorship of the first two age categories to .1. Discuss the result. (hint. To compute lambda you may have to start with a very large number of individuals in the population, rather than the 10 with which you started in exercise 10).

    This has a much larger negative effect on the population growth. The new lambda is 0.837. The SAD is also changed, although not that much. Now the first two age classes contain a greater proportion of the population than before. The SAD is:


  13. Bristleberry trees occus in four distinguishable forms: seed, seedling, small tree, large tree. In 1974 there were 5000 viable seeds, 500 seedlings, 100 small trees and 10 large trees. In 1975 there were 6000 seeds, 500 seedlings, 100 small trees and 11 large trees. Let x=0 represent seedlings, x=2 small trees and x=3 large trees. What are the values for Nx(1974) for all stage categories? What are the values for Nx(1975)?

    Well, if you have made it this far, you are probably pretty excited to see a question that is this easy and short. Here are the population vectors:


  14. For the bristleberry trees above, in 1974 a random sample of 2000 seeds was marked. By repeatedly sampling the seeds we know that 20 of them germinated by 1975 and the rest rotted. Also randomly sampled seedlings were marked. By 1975 20 had died, five had become small trees, and the rest were still seedlings. In addition 50 small trees were marked and it was found that two had become large trees, five had died, and the rest were still small. All 11 large trees were marked; one had died and the rest were still alive. Estimate the transition probabilities for the stage projection matrix.

    We are assuming that the trees cannot "shrink" (go to a smaller age class). Here is the transition matrix as it is right now with 'b' for unknown fecundity values:


  15. For the population in exercise 14, it has been found that each adult tree produces an average of 200 seeds per year, and each small tree produces 10 seeds per year. What will be the overall size of this population in 10 years?

    Now we can fill in the rest of the transition matrix: . Knowing this, the population in 10 years (1985) will be equal to the transition matrix to the 10th power times our N(1975) vector. The total population at this time is: 7921.9. Note: you will get a different answer if you project the population 11 years from N(1974) because the transition from 1974 to 1975 does not obey our transition matrix.


  16. It has been discovered that the rodent population of exercise 10 contains important behavioral constraints on survivorship. In particular the third age category exhibits strong density dependence. Model density dependence with the simple function, [a(1+Nx)]-1 for the second age category, letting a=.001 (that is, multiply the original survivorship by this function). Repeat exercise 10 with this density dependent addition to the matrix. Graph the result. Experiment with different values of a. Add density dependence to other age categories and see what happens.

    You may notice that this density dependence function is the same general function as the saturating function I was talking about with the saturating predation example on problem b) of the previous assignment. The transition matrix is:


  17. Suppose density dependence is itself more complicated such that the function [a(1+Nx - bNx2)]-1 is a reasonable model for density dependence of the second age category (that is, multiply the original survivorship by this function). Let b=.0001. Graph the population projection for a couple of thousand time units. Experiment with different values of b. (be careful not to divide by zero in this exercise).


  18. For the bristleberry population of exercise 14, devise a density-dependent model for one or more of the stages and project the population.


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