Question

MORE PRACTICE PROBLEMS 1. A population of Drosophila mauritiana every generation, and generations OCCU 1.000, and its population growth rate is 3.0 per generaron of Drosophila mauritiana reproduces in synchrony at discrete time periods on, and generations occur at two-week intervals. The current population size is a. What kind of growth is this Geometric or exponentials b. What equation should you use? Net - (Ni c. What will the expected size of the population be after six weeks 9. For an annual plant species, there are 100 plants in a population in year 1, and 120 plants in the population in year 2. What is lambda (1) e. A continuously growing population of alligators has a population size of 1,000 and an intrinsic rate of increase of r = 0.05 per year. Assuming that this rate of increase remains the same, about how long should it take for the population to grow larger than 4,000 individuals (Note: The natural logarithm of 2 is about 0.70.) f. A continuously growing population of bears has a population size of 500 and its intrinsic rate of increase is r = 0.07 per year. Assuming that this rate of increase remains the same. about how long should it take for the population to reach 1,000? (Note: The natural logarithm of 2 is about 0.70.) 2. A population of insects triples every year. Initially, there were 40 insects. What is the population growth rate? a. How many insects would there be after 4 years? b. How many insects would there be after 27 years? - Write your answer to this question as an equation - c. The habitat of the insect is degraded such that the population growth rate changes from 3.0 100.75. If there were 100 insects in the population when it's habitat became degraded, how many insects would there be after 3 years B 299 Merit 2019 Discussion 6: Survivorship. Life Tables, and Population Growth

Answer 1

1) (a) This is a geometric growth.

(b) We use .

Ne = xt/2nC

(c) The expected size of the population be after 6 weeks is =
36/2 + 1000
=
33 * 1000
=
27000
.

(d) Here, $\lambda$ is = [Population in year 2]/[population in year 1] = 120/100 = 1.2

(e) Let after n years the population will grow larger than 4,000.

Then we have,

1000*(1+0.05)" > 4000

i.e.,

4000 (1.05)" > 1000

i.e.,

(1.05)" > 4

i.e.,

In(1.05)" > In 4

i.e.,

n* ln(1.05) > In 4

i.e., $n>\frac{\ln4}{\ln(1.05)}$

i.e.,

n > 28.41

Therefore, it should take about 29 years for the population to grow larger than 4000 individuals.

(f) Let after n years the population will grow larger than 1,000.

Then we have,

500*(1 +0.07)" = 1000

i.e., $(1.07)^n=\frac{1000}{500}$

i.e., $(1.07)^n=2$

i.e., $\ln(1.07)^n=\ln2$

i.e., $n*\ln(1.07)=\ln2$

i.e., $n=\frac{\ln2}{\ln(1.07)}$

i.e., $n\approx11$

Therefore, it should take about 11 years for the population to reach 1000.

2) The population growth rate is 3.

(a) The number of insects after 4 years is =
40*31
=
40 * 81
= .

(b) If the number of insects after 27 years is N, then we have,

$N=40*3^{27}$

(c) After 3 years, the number of insects will be = $100*0.75^{3}$$\approx42$.