bio 365 HW1 - N t+1 =200*R 20 = 200*(1.2) 20 = 7667.52 (7)...

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BIO 356, Summer 2010 Homework 1 Juhee Kang 106977136 Mating in a population of semalparous frogs occurs each spring. The probability that an egg develops and that the resulting frog survives until the next spring to reproduce is estimated to be 0.012. Given that a surviving adult female frog lays an estimated 200 eggs (and assuming an even sex ratio) (1) Annual fecundity 200 * 0.012 = 2.4 2.4/2 = 1.2 f = 1.2 (2) Survivorship S=0 (3) Average growth rate R=s+f=1.2 (4) Is this population shrinking, growing, or staying the same (and how do you know this)? The population is growing because R is greater than 1. (5) If you capture and mark 50 frogs and later catch 40 frogs, of which 10 are marked, what is the estimate for the current frog population size (assuming a closed population with unbiased capture rates)? N=m*C t /C m = 50*40/10 = 200 (6) How many would you expect there to be in 20 years (assuming deterministic geometric growth)?
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Unformatted text preview: N t+1 =200*R 20 = 200*(1.2) 20 = 7667.52 (7) How long does it take for the population to increase 10-fold? 10 = 1.2 t ln(10) = t ln(1.2) t = 12.629 12 years and 7month (8) If the number of eggs laid per adult female frog were 10% higher, what percent higher would the projected size after 20 years be? 200 eggs : 1.2 20 *200 = 7667.52 200 + 200*.1 = 220 eggs : f=110*.012 = 1.32 220eggs : 220*1.32 20 =56741.56 56741.56/7667.52 * 100 = 740.025% (9) If there are 520 frogs after 10 years, what is the average annual growth rate and was it larger, smaller, or the same as the estimate from question 3? 520 = 200*R 10 R = 1.1 It is smaller than the annual growth rate in Question 3. (10) If the frogs were iteroparous, but still had the same annual fecundity as in question 1, how would this affect the growth rate? f = 1.2 s = 200 * 0.012 = 2.4 R = f+s = 3.6 Therefore, R is greater than the R in question 1....
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bio 365 HW1 - N t+1 =200*R 20 = 200*(1.2) 20 = 7667.52 (7)...

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