final-f99-key - Name: KEY Page 1 of 14 Instructions:...

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Unformatted text preview: Name: KEY Page 1 of 14 Instructions: --Write your name on all the pages --Make sure that all 13 pages are attached. The last page is blank and is for you to use as a scratch pad or in case you mess up another section of your exam. Leave it attached! --Any math needed to solve a problem should be relatively simple. If not, please give your answer by showing how you would make the calculation: e.g., showing "(10+10)/4" is as good an answer as "5" (writing down an appropriate equation and clearly defining the variables, as well as indicating their values, if known, will also suffice). GENERAL ECOLOGY PCB 4044 FALL 1999 FINAL PAGE POINTS POSSIBLE SCORE 2 11 _____ 3 15 _____ 4 10 _____ 5 10 _____ 6 6 _____ 7 13 _____ 8 12 _____ 9 21 _____ 10 18 _____ 11 16 _____ 12 18 _____ 13 0 _____ _____ _________ TOTAL 150 _____ Name: KEY Page 2 of 14 1. (3 points) Two populations exhibit no density-dependence, have stable age distributions, and have the same net reproductive rates (R ), yet they increase in size at different per capita rates (i.e., their r ’s are not the same). Explain how the net reproductive rates can be the same, but the per capita growth rates can be different. Generation times differ, leading to a lower rate of increase for the population with the longer generation time. R is expressed per generation, whereas r (or λ ) are expressed per unit time. E.g., r = ln(R )/G 2. As part of your independent project, you monitored the demographic rates of salamanders in Devil’s Millhoppper. Here is the life table you came up with: Age class (yrs.) l x m x 0 1.00 0 1 0.90 0 2 0.45 2 3 0.15 1 4 0.10 1 5 0.00 0 a. (3 points) Assuming these estimates apply to all years, how will the population change in size once it achieves a stable age distribution? Circle the best answer: decrease remain the same increase b. (5 points) Using the life table and given the following abundances of age classes in 1999, please calculate the abundances of all age classes in the following year (2000): Age class 1999 2000 0 400 ( 50 x 2) + (33 x 1) + (67 x 1) = 200 1 100 400 x (0.9/1.0) = 360 2 100 100 x (0.45/0.9) = 50 3 100 100 x (0.15/0.45) = 33 4 100 100 x (0.10/0.15) = 67 Name: KEY Page 3 of 14 3. (9 points) List the three major elements of a well designed field experiment, and briefly (in a sentence) explain why each is important: Replication: units treated in similar ways will vary; replication allow you to estimate how much variation arises within a treatment group and thus assess whether the differences among groups are greater than expected. Increasing the number of replicates increases the statistical power. Controls: a control provides a basis for comparison that allows the experimenter to isolate the factor of interest from other factors that might be associated with a particular treatment. (A control is usually NOT "a treatment that is not unmanipulated")....
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This note was uploaded on 06/04/2011 for the course PCB 4043 taught by Professor Osenberg during the Spring '10 term at University of Florida.

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final-f99-key - Name: KEY Page 1 of 14 Instructions:...

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