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Lecture 5

BIO220H1 Lecture Notes - Lecture 5: Exponential Growth, Logistic Function, Inflection Point

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John Stinchcombe

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Lecture 5: Ecology and Evolution of Harvested Population
Humans harvesting biodiversity
We need to acquire energy from natural/man systems
Some of the energy from these systems are being diverted from us (for our
Taking energy out of naturally occurring energy ows
In agriculture completely modify environment
oSet it up to suit us
oEnergy from sun converted into crops/domesticated animals for food
We need to consume for energy so some harvesting is inevitable for living
organisms to survive
What does this do to ecology and evolution of harvested species?
And because it’s inevitable can we prevent bad results
1. Simple models of species growth and harvesting strategies
a. Apply population growth models to understand optimal harvesting
2. How harvesting a*ects demographic of harvested species
3. Next lecture: How harvesting a*ects evolution of species that we manage and
we depend on
(1) The general rule
Aim is to harvest out the recruitment and leave the stock alone
oImagine population of species we want to manage
oThat population will be growing at some rate per year
oOnly want to harvest out growth and leave population to regenerate
oRecruitment = new additions to populations
Maturing o*spring
oCurrent population
So if can take out new additions only and leave stock (current population) the
way it is have potential to have self-perpetuating resources
Simple model for how populations grow
Logistic growth equation
oChange in number of individuals per unit time (dN/dt) is equal to product
to 3 terms:
R: Intrinsic population growth rate
How fast species can begin reproducing and regenerate
N: Number of individuals currently in population
Term in brackets
odN = change in abundance
odt = change in time
oso dN/dt is a rate (rate of change of population over time)
oK is the carrying capacity

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Graph of equation:
oCurrent population size as function of time
oEarly phases: Exponential growth
oAs more and more individuals = reach asymptote at K
oHave inection point
Below K/2, have exponential growth
Above K, have slowing of growth rate = fewer individuals produced
per unit time
Clicker Question
Which of statements are correct?
(1) As N → K, dN/dt →0 (correct)
(2) As N→0, dN/dt → exponential growth (As N is close to 0, then get rN which is
exponential growth) (correct)
(3) If r<0, dN/dt < 0 (correct)
Density-dependent growth rate can limit population size
Graph shows number of birds against time
As islands are colonized by terns
oTern carrying capacity on islands limited by nesting space
At >rst have exponential growth, then levels o* (Bird Island)
Ram island has highest carrying capacity
oStill in exponential growth phase
Using logistic curves to predict response of harvesting
Time of harvest on horizontal axis
Size of population in terms of K on vertical axis
Each vertical line represents a harvest season
At top arrows, if harvest up to there, then next generation population will grow
back to where it was
oHarvest down a bit, then it regrows
oBut since population close to K, overall rate of population growth is not
Don’t harvest a lot, it recovers but not a lot of recruitment because already close
to K
Same at bottom arrows
oRate of growth is quick but staggers up
How much we should harvest down to get maximum rate of growth?
oAt 50% of K slope of graph is the steepest (inection point)
If solve equation for maximum growth rate, will get K/2
So ideal situation:
oIf population is at K
oHarvest it down to K/2
Brings population down to density where it’s rate of growth is
Keeps population at region where rate of population growth is the
Bene>cial to us:
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