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PCB 4043C Study Guide - Midterm Guide: Predator Satiation, Rich Mix, Ectotherm

9 pages88 viewsSpring 2014

Course Code
PCB 4043C
Study Guide

of 9
Name: KEY
Page 1 of 9
--Write your name on all the pages
--Make sure that all 9 pages are attached.
--You may not use calculators or other electronic aids. The math needed to solve a problem
should be relatively simple. If you can't do the math, just show what needs to be
calculated (e.g., "answer = .5 x .3")
--If you get hung up on a problem, skip it; return to it after you’ve answered the “easy”
PCB 3034/4044
Spring 2008
1 2 _____
2 12 _____
3 15 _____
4 8 _____
5 11 _____
6 13 _____
7 10 _____
8 17 _____
9 12 _____
_____ _________
TOTAL 100 _____
(2 points):
1) Are you interested in being an undergraduate TA for this course next year? ( YES, NO )
2) From where/whom did you take Intro Bio (Ecology): e.g., “UF, Dr. Palmer”: ________
Name: KEY
Page 2 of 9
2-1. a. (4 points) Rank the following animals in order of lowest to highest production
efficiency. Also indicate lowest and highest with words.
__4___worm that eats decomposing plant material (Highest)
__3___fish that eats other fish
__1___goose that grazes on wetland grasses (Lowest)
___2__chicken that eats grain
b. (4 points) Explain your ranking for each organism in relation to its (1) activity
level, (2) assimilation efficiency of food resources and (3) thermoregulation.
Goose: Lowest production efficiency because it has a high activity level (grazing, flying, escaping
predators, migrating in search of food resources),a high cost of homeostasis since it is an
endotherm/homeotherm, and it is consuming grasses, a food resource that has relatively low
assimilation efficiency (60-70%).
Chicken: Like the goose, it has a high activity level (finding grain, flying, escaping predators, “the
pecking order”) and is an endotherm. Unlike the goose, it is consuming a food resource with a
relatively high assimilation efficiency (80%). As a result, it has higher production efficiency than the
Fish: lower activity level than birds and low cost of homeostasis due to
ectotherm/poikilotherm/heterotherm thermoregulation. High assimilation efficiency of meat (60-
Worm: Highest production efficiency because of low activity (surrounded by food resources) and
medium-high assimilation efficiency since it is consuming a rich mix of plant material and microbes.
It is an ectotherm so the energetic costs of homeostasis are relatively low.
2-2. (4 points) Explain why there are no wood eating hummingbirds.
High energetic cost of activity and homeostasis, low assimilation efficiency of wood.
Name: KEY
Page 3 of 9
3-1. Assume that trophic efficiency for phytoplankton-animal energy transfer is 10%,
animal-animal energy transfer is 50%.
a) (5 points) How many kg of phytoplankton would it take to produce 70 kg of
human biomass if that phytoplankton is first eaten by sardines, the sardines
are eaten by tuna, and the tuna is eaten by humans?
70 g human / 0.5 = 140 g tuna, 140 g tuna / 0.5 = 280 g sardines, 280 g sardines/ 0.1 = 2,800
g phytoplankton
b) (5 points) How many kg of sardines would it take to produce 70 kg of human
biomass if the humans ate the sardines directly? How many kg of
phytoplankton if the humans ate the phytoplankton directly?
70 g human / 0.5 = 140 g sardines
70 g human / 0.1 = 700 g phytoplankton
3-2. (5 points) On a recent trip to the north, you are startled by a rat scurrying across the
snowy surface of a parking lot; you do not see it at first because its creamy white fur
blends almost perfectly with the snow. How would you determine if the rat’s white
fur is an acclimation to the snowy conditions?
An acclimation is a reversible physiological change that helps to maintain the functioning of
the organism in changed environmental conditions. Watch the rat over time and see if its coat
color changes as the snow melts and the dark tarmac is revealed or some other activity that
determines whether coat color is a reversible change that is related to the environment.
Name: KEY
Page 4 of 9
4-1. A student doing an independent project (Zoo 4905) monitored Daphnia growth in a
lake, and based on their performance was able to describe how the per capita growth
rate varied as a function of density. Her data are shown below.
Daphnia Density (no. / 10 ml)
Per Capita Growth Rate, dN/Ndt
(per week)
a) (4 points) Given enough time, to what density will the population converge if
initiated at the following densities:
Final Density
0 = 1 0
0 = 3 11
0 = 7 11
0 = 14 11
b) (4 points) Give a biologically plausible explanation for this hump-shaped
Possible answers: The increase arises from a positively density-dependent
factor, such as social facilitation of reproduction, feeding facilitation (Daphnia
increase food quality by feeding), or predator satiation (or handling time).
The decrease arises from a negatively density-dependent factor, such as
competition for food (it also could represent the increase incidence of disease
or an increase in predation caused by a habitat or diet shift in predators that
target Daphnia as their density increases).
Name: KEY
Page 5 of 9
5-1. (3 points) Dr. I.M. Green followed a cohort of plants through time, and at each point
in time, recorded the number of survivors. (note the logarithmic scale).
age (or time)
Number still alive
Which of the following statements is correct (circle the correct one)?
i. Younger plants have a higher annual survival probability than older plants
ii. Older plants have a higher annual survival probability than younger plants
iii. Young and old plants survive with the same probability
iv. The answer can’t be obtained from the data available
5-2. The dynamics of three species can be described using the following equations:
dN1/N1dt = +1.0 - 0.1N1 - 0.2N2 - 0.1N3
dN2/N2dt = +0.6 – 0.2N1
dN3/N3dt = -0.1 + 0.1N1
For each of the following, choose (i.e., circle) the best option from the those available
within parentheses:
b) (2 points) Species 1 and 2:
( compete, are predator and prey; are mutualists, do not interact directly).
c) (2 points) Species 1 and 3:
( compete, are predator and prey; are mutualists, do not interact directly).
d) (2 points) Species 1 and 2 interact:
( indirectly only, directly only, directly and indirectly)
e) (2 points) The indirect effect between Species 2 and 3 results in a:
( negative, positive, neutral ) effect of Species 2 on 3.
Name: KEY
Page 6 of 9
6-1. The following life cycle diagram is based upon a time interval of one year.
Eggs Medium
Juveniles Adults
0.6 0.5 0.3 0.1
0.3 0.1 0.5
a. (5 points) What proportion of individuals in each life stage survive from one year
to the next?
Eggs 0.6
Small Juveniles 0.9
Medium Juveniles 0.3
Large Juveniles 0.1
Adults 0.5
b. (5 points) If there were 100 individuals of each stage on April 1, 2008, how many
individuals of each stage should there be one year later (on April 1, 2009)?
Eggs 2000
Stage 1 Juveniles 90
Stage 2 Juveniles 50
Stage 3 Juveniles 40
Adults 60
c. (3 points) Given enough time (and assuming the transition values remain
constant), this population will grow:
(pick one: lognormally, exponentially, logistically, inverse demogogically )
Name: KEY
Page 7 of 9
7-1. (4 points) To get coexistence between two competitors in the Lotka-Volterra
competition model, which of the following must be true? (circle all choices that are
true and put an X through the false ones)
i) α12α21 < 1;
ii) the strength of intraspecific competition must be greater than interspecific
iii) the strength of interspecific competition should exceed intraspecific
iv) there must be some form of resource partitioning
7-2. (6 points) Circle the best option from the choices given in bold (1 point each):
a) (1 point) an effective bio-control agent should have a (low; moderate; high)
conversion efficiency.
b) (1 point) an effective bio-control agent should have a (low; moderate; high)
attack rate (or feeding rate).
c) (1 point) The concept of Maximum Sustained Yield via a fixed quota system (i.e.,
a constant amount harvested) is a ( good; bad ) way to achieve sustainability.
d) (1 point) Based on the Lotka-Volterra competition model (and the Competitive
Exclusion Principle), two competing species (can; cannot) stably coexist on one
limiting resource.
e) (1 point) Exploitative competition arises when two species:
( both consume a limited resource; eat each other; defend mating territories ).
f) (1 point) A population with a growth rate or r = 0.1 will grow _____a population
with a growth rate of λ=1.1: (slower than, the same as, faster than )
Name: KEY
Page 8 of 9
8-1. (9 points) Explain (you must use words, although you also can supplement this with
equations) the conditions that must be satisfied for a disease that has infected only a
few hosts thus far to spread to a greater number of hosts. Include a discussion of the
role of three key factors (e.g., think about variables in models that we explored) that
play an important role in this process.
An infected host must infect >1 new host before the host dies or recovers.
This is more likely if: 1) there are many susceptible hosts; 2) the transmission
rate is high; 3) the host remains infective for a long time (it does not die or
becoming immune quickly).
8-2. (4 points). To be effective, does a vaccination need to be given to all of the
susceptible hosts? Justify your answer.
No. A successful vaccination program only requires that the pool of
susceptible hosts be reduced below that needed for an epidemic (i.e.,
#Susceptible hosts must be reduced below the ratio of transmission rate to
‘recovery’ rate).
8-3. Two species of plant are both limited by nitrogen. When grown in isolation, little
blue stem exhibits positive growth until it drives nitrogen concentrations down to 0.1
mg/kg of soil (i.e., at [N]<0.1, it shows negative growth). Ragweed, when grown in
isolation, depletes N to 0.3 mg/.
(4 points) What will be the outcome of competition when the two species are grown
together and Nitrogen is limiting? (e.g., do the two species coexist, or does one
species win; if so, which one?)
Little blue stem will outcompete ragweed (it has a lower R*).
Name: KEY
Page 9 of 9
9-1. Two species compete for resources. Their interaction can be described using the
Lotka-Volterra Competition Equations. The following data are available:
in the absence of Species A, Species B reaches an equilibrium density of 40 /
in the absence of Species B, Species A reaches an equilibrium density of 50 /
The effect of 10 individuals of Species B on dNA/NAdt is the same as the effect of
5 individuals of Species A.:
The effect of 1 individual of Species A on dNB/NBdt is the same as the effect of 1
individual of Species B.
a) (6 points) Label a phase plane diagram and the no-growth isoclines for this
system. Indicate the numeric value for x- and y-intercepts of the isoclines.
0 20406080100
Density of Species B
Density of Species A
Sp B's iso cline SpA's isoclin e
b) (6 points) Draw the dynamics (i.e. how the N’s change through time) for a system
that starts with NA=NB=10 on the phase plane above AND the figure below.
sp A sp B
Tim e
end of exam

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