Due Feb 1,2019 Biology 363 population growth exercise 2019 A pest management specialist is hired to monitor an apartment complex for cockroaches. Research shows that if 2 cockroaches invade an apartment the population, on average, follows the growth pattern below Month Averape populatiom 0 2.0 1 4. 2 10. 325.6 4 59,8 139 6 324 745 8 1676 3597 10 7056 1 11979 12 17070 a) Using only the first and last populations calculate rm per month for this population assuming exponential growth. b) Calculate DN, month and rm per month for successive generations in this series (Excel can be used here). Does this indicate that the growth is exponential or logistic? Explain your interpretation. c) Add a column for lm from part a) where you assumed exponential growth. Plot, on the same graph, rm as a function of population size for both the exponential and logistic series of rHow do these curves differ from one another? Why is the logistic rm at N-0 larger than the rm at N-0 for exponential growth d) Using only the graph estimate the carrying capacity for this population. Let N,-population size at time t N, original population size t = time during which population is growing r_ = intrinsic rate of natural increase (max rate at which the population can grow; also the value entered into, and returned by, the growth equations) a intrinsic rate of natural increase at a specific point on the growth curve e base of natural logarithms 00finite rate of increase (or multiplication rate) bbirth rate per capita (births per 1000 population) ddeath rate per capita (deaths per 1000 population) If growth is exponential and continuous, the change in size of the population is proportional to the size of the population or where rm b-d Integrating equation (1) we get N,N taking the natural logarithms of both sides of the equation we get In N,In No 4) Equation 4 is in the form of a linear equation so that if you plot in N versus time (t) we can calculate our value of r. from the slope of the semilog plot. You will require about 130 counters and one die. Each counter represents one member of the population. Start the game with 4 counters (individuals) to represent the initial population size. The die is thrown once for each individual present at the beginning of the round i.e. 4 times in round one). If the die shows 1 -3 spots that particular individual is assumed to have reproduced by binary fission and one new member is added to the population. If the die shows 4 – 6 spots the individual is assumed not to have reproduced. Subsequent rounds follow the same rules, the die being thrown once for each individual present when the round starts. The number of times you throw the die per round steadily increases. Follow the population for 7 rounds and record your data in table 1 In N’- 1 n N° + r-,t (4) Equation 4 is in the form of a linear equation so that if you plot in N versus time (t) we can calculate our value of r from the slope of the semilog plot. You will require about 130 counters and one die. Each counter represents one member of the population. Start the game with 4 counters (individuals) to represent the initial population size. The die is thrown once for each individual present at the beginning of the round (i.e. 4 times in round one). If the die shows 1 -3 spots that particular individual is assumed to have reproduced by binary fission and one new member is added to the population. If the die shows 4 6 spots the individual is assumed not to have reproduced. Subsequent rounds follow the same rules, the die being thrown once for each individual present when the round starts. The number of times you throw the die per round steadily increases. Follow the population for 7 rounds and record your data in table 1 You may expect the value of r to be approximately 0.5 because r b -d (birth rate 0.5 and death rate 0). However, in this particular model of exponential growth assuming continuous growth r is calculated as follows. Rearranging equation (3) we get N. Ift-1 then taking the logarithms of both sides we find r,,-1 n λ (5) and the inverse relationship is λ In our case λ 1,5 sorm-In 1.5-0.4055 Another useful but seldom used, equation for exponential growth is NN Analysis and Theory. We can predict what on the average the future population size will be. The probability of reproduction per individual per round is 0.5, although the realized value may be higher or lower than this. The probability of death is 0. Thus, the population should increase 1.5 times each round on average. Our predicted average growth of the population will therefore, be 4, 6,9, 13.5 etc. Calculate the series and enter your results in Table in the Excel Spreadsheet provided. The actual class average will also be provided for each round and you can also enter these results in Table Which is the usual differential equation describing logistic growth. When we integrate this (a difficult operation) we get To find the value of a (the other terms in the equation are known) we rearrange 9 as follows. N, taking the logarithms of both sides a-r-t = 1n (K-N) N, when t -0, N, N and a- (rx0)- In (K-N) N, substituting this back in equation (9) we get (10) Equation (10) can also be written in a more proper mathematical form as 1 N Equation () can also be written in a linear form (12) Equation (12) can also be written in a more proper mathematical form as (13)