1 MAXE · 171 OIB
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UMICH Department of Ecology & Evolutionary Biology
EST. 1817
ARTES · SCIENTIA · VERITAS
BIO 171 — Exam 1
P O P U L AT I O N E CO LO G Y, E VO LU T I O N A RY B I O LO G Y & V I R O LO G Y
INSTITUTION University of Michigan COURSE CODE BIO 171
PROGRAM Ecology & Evolutionary Biology ACADEMIC YEAR
EXAM TITLE BIO 171 Exam 1 — Population TOTAL QUESTIONS 110 Questions
Ecology & Virology
COURSE TITLE Introductory Biology: Ecology & FORMAT Multiple Choice — Select the
Evolutionary Biology Single Best Answer
EXAMINATION INSTRUCTIONS
▸ Select the single best answer for each question unless "Choose all that apply" is specified.
▸ Questions cover population ecology, population growth models, virology, experimental design, and
evolutionary biology.
▸ Content is aligned with University of Michigan BIO 171 course objectives.
▸ Correct answers and detailed rationales appear below each question.
, BIO 171 EXAM 1 — POPULATION ECOLOGY, GROWTH
Questions 1 – 110
MODELS, EXPERIMENTAL DESIGN & VIROLOGY
1. A population of prairie dogs shows logistic growth. As the population approaches
carrying capacity (K), the number of individuals added to the population per unit of time
________.
A. Increases exponentially
B. Remains constant
C. Decreases
D. Fluctuates randomly
CORRECT ANSWER C — Decreases
RATIONALE In logistic growth, as N approaches K, the term (K-N)/K approaches zero, reducing
the population growth rate. The number of individuals added per unit time slows
as the population nears carrying capacity because density-dependent factors
(competition for resources, space) increasingly limit growth. Exponential addition
(option A) only occurs when the population is far below K.
2. Density dependent factors are influenced by the number of organisms in a population.
A. True
B. False
C. Only in exponential growth models
D. Only when resources are unlimited
CORRECT ANSWER A — True
RATIONALE Density dependent factors are defined as factors whose effects on population
growth vary with population density. Examples include competition for resources,
disease transmission, and predation. As population density increases, these
factors exert stronger effects on birth and death rates, producing logistic growth
patterns.
,3. If a population has 500 individuals in it in 2010, and the per capita birth rate is 0.3 and the
per capita death rate is 0.2, is the population growing or shrinking?
A. Growing
B. Shrinking
C. Staying the same size
D. It is not possible to say whether it is growing or shrinking without knowing whether there
is migration.
CORRECT ANSWER D — It is not possible to say whether it is growing or shrinking without
knowing whether there is migration.
RATIONALE r = (b-d) + (i-e). With b=0.3 and d=0.2, the birth-death component suggests
growth. However, if emigration exceeds immigration by more than 0.1, the
population could still be shrinking. All four vital rates (birth, death, immigration,
emigration) must be known to determine population trajectory. This is a key
concept in population ecology.
4. In the population model dN/dt = rmax·Nt, "rmax" refers to:
A. The current population growth rate at any given time
B. Per capita population growth rate when b is as high as possible and d is as low as possible
C. The carrying capacity of the environment
D. The geometric growth rate lambda
CORRECT ANSWER B — Per capita population growth rate when b is as high as possible and d is
as low as possible
RATIONALE rmax (the intrinsic rate of increase) is the per capita population growth rate under
ideal conditions — when birth rate is maximized and death rate is minimized. It
represents the maximum potential growth rate for a population. Carrying
capacity (K) is the maximum population size, not a growth rate. Lambda (λ) is the
geometric growth rate.
, 5. A population of starlings reproduces once per year. The annual geometric growth rate of
the population is 4. The population had 80 individuals in 1982. How many individuals will
there be in 1984?
A. 320
B. 640
C. 1280
D. 2560
CORRECT ANSWER C — 1280
RATIONALE Using Nt = λ^t·N0: From 1982 to 1984 is 2 years. N1984 = 4² × 80 = 16 × 80 = 1280.
Each year the population multiplies by 4: Year 1: 80×4=320. Year 2: 320×4=1280.
This tests application of the geometric growth equation with discrete breeding
seasons.
6. As a population approaches its carrying capacity, how does its growth change?
A. The growth rate accelerates as N approaches K.
B. The growth rate remains constant regardless of N.
C. The growth rate slows as N approaches K.
D. The growth rate oscillates unpredictably.
CORRECT ANSWER C — The growth rate slows as N approaches K.
RATIONALE In logistic growth, as N approaches K, (K-N)/K becomes smaller, reducing dN/dt.
When N is small relative to K, growth is approximately exponential. As N
increases, density-dependent factors increasingly limit population growth until N
stabilizes near K. This produces the characteristic S-shaped logistic curve.