# Week_9_Answers - STUDY QUESTIONS(Lectures 29-33 1 Consider...

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STUDY QUESTIONS (Lectures 29-33) 1) Consider four populations: A , B , C , and D . The growth rate of each population (d N /d t ) is a function of population abundance ( N ). Suppose that each population has the functional relationships given by the following graphs: 10 20 30 40 50 N dt dN 10 20 30 40 50 N dt dN 10 20 30 40 50 N dt dN 10 20 30 40 50 N dt dN AB CD Fill in the following table describing whether these populations increase (INC), decrease (DEC) or remain constant (CON) when population abundance is at the number given in the rows: population size A B C D 0 CON CON CON CON 10 INC INC DEC DEC 20 INC INC DEC CON 30 INC INC DEC INC 40 INC CON DEC CON 50 INC DEC DEC DEC The general differential equation for single population dynamics is: N r dt dN N = For which of the above populations is r N (the per-capita growth rate) constant? What does a constant per-capita growth rate mean? Both populations A and C possess a constant per-capita growth rate. Note the differential equations describing population growth for A and C both have the form d N /d t = r N (where r is positive for population A and negative for population C). The per-capita growth rate is given by the slope of the lines in each graph. A constant per-capita growth rate means that individual contributions to population growth (or decay) do not change with population density. As a consequence the population exhibits exponential growth (population A) or exponential decay (population B) over time. (You should be able to draw N as a function of t for different starting population densities for both these populations). Do note that population growth is density-dependent for all four populations.

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2) In 1935, 102 Cane toads were introduced into Australia to control the cane beetle (a pest of sugar cane). By 2005, these cane toads hit 200 million. Assuming exponential growth, calculate the continuous per-capita growth rate. Let N 0 be the number of toads in 1935 ( N 0 =102). After 70 years, we have 2 × 10 8 toads (i.e., N 70 =2 × 10 8 ). Assuming exponential growth, we find the continuous per-capita growth by solving the following equation for r : N t = e rt N 0 2 × 10 8 = e r (70) (102) 1.96 × 10 8 = e r (70) ln(1.96 × 10 8 ) = r (70) r = 0.207 Let’s say that the Cane toad population is 210 million currently. What can you say about an assumption of exponential growth? How would account for this new data point? It is now 2007 (72 years after t =0). Under the exponential growth model, we would predict: N 72 = e 0.207 (72) (102) = 3.03 × 10 8 However, we have currently nowhere near 303 million toads. Indeed, there are many fewer toads than expected. One possibility is that the population is not growing exponentially. Rather the toads are approaching a carrying capacity and their growth might be better described as logistic. This new data point then becomes part of the “S-shaped” curve that is characteristic of logistic growth.
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## This note was uploaded on 04/20/2008 for the course BIOL 180 taught by Professor Freeman during the Winter '07 term at University of Washington.

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Week_9_Answers - STUDY QUESTIONS(Lectures 29-33 1 Consider...

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