BIOLOGY 1M03 Lecture Notes - Lecture 18: Logistic Function, Exponential Growth, Housefly

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Population ecology: mark recapture (discrete and continuous pop) Look at fraction of recaptured of individuals (marked vs unmarked) should reflect total caught in first place and actual size. N is the estimate of the total population size (solve for this). In our example: 50/300 = 100/n n = 600. General growth equation: assuming no immigration or emigration (just births and deaths, dn/dt = (b-d) x n. N = # individuals at current time: e. g. , n = 5000, b = 20 births/1000 ind*yr. D = 10 deaths/1000 ind*yr dt = 1 year dn/dt = [(20-10)/1000*1yr] x 5000 dn/dt = +50. Practise question- year 2 n = 5050 dn/dt. A) 6000: +50, +50. 5, +60, +60. 6, what if n = 6(cid:1004)(cid:1004)(cid:1004), dn/dt = [(cid:894)(cid:1006)(cid:1004)-10)/1000*1yr] x 6000 = 60. Intrinsic rates of increase: e. coli rmax = 60/individual*day, h. sapiens rmax= 0. 0003/individual*day 0. 11/individual*year. At a point in time under real conditions populations grow at the per capita growth rate r (not rmax)

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