AP Biology · Topic 8.3
Population Ecology Practice
Part of Ecology.(SYI-2.B)
Practice questions
29
Sample questions
5 of 29 — sign in to practice the rest with adaptive difficulty and mastery tracking.
Sample 1difficulty 2/5
The S-shaped logistic growth curve plateaus at the population's
- Acheck_circle
Carrying capacity (K)
- B
Initial size (N0)
- C
Intrinsic growth rate (r)
- D
Per capita death rate (d)
Why
Logistic: dN/dt = rN(K-N)/K. As N approaches K, growth slows to zero.
- A
Sample 2difficulty 2/5
A bacterial culture was sampled hourly for 8 hours after inoculation in fresh media. Cell density was plotted on a semi-log scale.
What does the lag phase represent biologically?
- Acheck_circle
Cells are synthesizing enzymes and adjusting to the new medium before rapid division.
- B
DNA replication is inhibited.
- C
The population has reached carrying capacity.
- D
Cells are dying due to nutrient deprivation.
Why
During lag, cells acclimate, synthesize ribosomes/enzymes, and prepare to divide. The population doesn't yet grow exponentially.
- A
Sample 3difficulty 2/5
The age-structure pyramid shown best predicts what for the population?
- A
Stable or declining population
- B
Negative growth in 50 years
- Ccheck_circle
Rapid future growth (expanding population)
- D
Equal growth and death rates
Why
A broad base with each younger cohort larger than the next indicates a high reproductive potential and predicts rapid expansion.
- A
Sample 4difficulty 2/5
At which population size does the logistic growth curve show the maximum dN/dt?
- A
N = K
- B
N just below K
- Ccheck_circle
N = K/2
- D
N = 0
Why
In the logistic model dN/dt = rN(1 - N/K), the maximum growth rate occurs at N = K/2, the inflection point of the S-curve.
- A
Sample 5difficulty 2/5
Researchers captured 100 fish in a lake, marked them, and released them. One week later they captured 80 fish, of which 20 were marked.
Using N = (M x C)/R, what is the estimated population size?
- Acheck_circle
400 fish
- B
1600 fish
- C
100 fish
- D
200 fish
Why
N = (100 x 80)/20 = 8000/20 = 400 fish.
- A