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AP Biology Unit 8 practice questions

Try 10 multiple-choice questions from Unit 8 here. Pick an answer to see why it's right or wrong, then take the timed test above for a full AP-style set.

  1. Question 1

    Nutrient runoff from a farm causes an algal bloom in a nearby pond. The density of algal cells increases from 1,200 cells/mL on Day 1 to 96,000 cells/mL on Day 5. What is the average rate of change in algal cell density per day during this period?

    19,200 cells/mL per day

    23,700 cells/mL per day

    24,000 cells/mL per day

    94,800 cells/mL per day

    Show answer

    Answer: (B)

    Calculating the rate of change requires dividing the change in population (94,800 cells/mL) by the time interval (4 days).

  2. Question 2

    A researcher measures mercury concentrations in a marine food web affected by industrial pollution. Phytoplankton contain 0.05 ppm of mercury, while sharks at the top of the food chain contain 32.0 ppm. How many times greater is the mercury concentration in the sharks compared to the phytoplankton?

    6.4 times greater

    64 times greater

    640 times greater

    6,400 times greater

    Show answer

    Answer: (C)

    Calculating the ratio requires dividing the concentration in the top predator (32.0 ppm) by the concentration in the producer (0.05 ppm).

  3. Question 3

    Deforestation for agriculture reduced a tropical forest fragment from 800 hectares in 1990 to 320 hectares in 2020. What is the percent change in the area of the forest fragment over this 30-year period?

    A decrease of 40%

    A decrease of 60%

    A decrease of 150%

    A decrease of 250%

    Show answer

    Answer: (B)

    Calculating percent change requires finding the difference (-480) and dividing by the initial value (800), then converting to a percentage.

  4. Question 4

    Atmospheric data indicates that CO2 levels increased from 350 ppm in 1980 to 410 ppm in 2020 due to human activities. What is the average annual rate of increase in CO2 concentration during this period?

    1.5 ppm per year

    8.8 ppm per year

    10.3 ppm per year

    60.0 ppm per year

    Show answer

    Answer: (A)

    Calculating the rate requires dividing the total change in concentration (60 ppm) by the time interval (40 years).

  5. Question 5

    A population of 900 deer lives in an ecosystem with a carrying capacity (K) of 1000. If the maximum per capita growth rate (r_max) is 1.0, calculate the population growth rate (dN/dt) considering density-dependent negative feedback.

    90 individuals/year, reflecting resource limits

    100 individuals/year, reflecting resource limits

    900 individuals/year, reflecting exponential growth

    1000 individuals/year, reflecting exponential growth

    Show answer

    Answer: (A)

    Calculates logistic growth correctly: 1.0 * 900 * ((1000-900)/1000) = 90.

  6. Question 6

    A student places 100 isopods in a choice chamber with wet and dry sides to test humidity preference. After 10 minutes, 70 isopods are on the wet side and 30 are on the dry side. Assuming no preference (null hypothesis), the expected count is 50 per side. The critical value for p=0.05p=0.05 with 1 degree of freedom is 3.84. What is the chi-square value and conclusion?

    8.0; fail to reject the null hypothesis

    16.0; reject the null hypothesis

    16.0; fail to reject the null hypothesis

    32.0; reject the null hypothesis

    Show answer

    Answer: (B)

    Calculating the chi-square statistic involves summing the squared differences divided by expected values for both groups: (70−50)250+(30−50)250=40050+40050=16.0\frac{(70-50)^2}{50} + \frac{(30-50)^2}{50} = \frac{400}{50} + \frac{400}{50} = 16.0. Since 16.0 exceeds the critical value of 3.84, the null hypothesis is rejected.

  7. Question 7

    In a phototropism experiment, 200 seedlings are exposed to unidirectional light. Observed results show 140 bending toward light and 60 growing straight. Under the null hypothesis of random growth, expected counts are 100 each. The critical value at p=0.05p=0.05 is 3.84. Which choice correctly describes the statistical outcome?

    Chi-square is 16.0; light significantly affects growth direction

    Chi-square is 32.0; light significantly affects growth direction

    Chi-square is 32.0; light has no significant effect on growth

    Chi-square is 64.0; light has no significant effect on growth

    Show answer

    Answer: (B)

    The chi-square calculation is (140−100)2100+(60−100)2100=1600100+1600100=16+16=32.0\frac{(140-100)^2}{100} + \frac{(60-100)^2}{100} = \frac{1600}{100} + \frac{1600}{100} = 16 + 16 = 32.0. Because 32.0 is greater than 3.84, the null hypothesis is rejected, indicating a significant response to light.

  8. Question 8

    Researchers test phototaxis in 60 fruit flies, observing 40 move to light and 20 move to darkness. If the null hypothesis states no preference, the expected count is 30 per group. With 1 degree of freedom, the critical value is 3.84. What is the result of the chi-square test?

    3.33; fail to reject the null hypothesis

    6.67; fail to reject the null hypothesis

    6.67; reject the null hypothesis

    13.33; reject the null hypothesis

    Show answer

    Answer: (C)

    The calculation is (40−30)230+(20−30)230=10030+10030=3.33+3.33=6.67\frac{(40-30)^2}{30} + \frac{(20-30)^2}{30} = \frac{100}{30} + \frac{100}{30} = 3.33 + 3.33 = 6.67. Since 6.67 > 3.84, the null hypothesis is rejected, showing a significant preference.

  9. Question 9

    Ninety fish are released into a tank with equal areas of sand, gravel, and mud. Observed counts are 50 on sand, 30 on gravel, and 10 on mud. The null hypothesis predicts 30 fish per area. The critical value for 2 degrees of freedom at p=0.05p=0.05 is 5.99. What is the chi-square value and biological interpretation?

    13.33; fish show no significant substrate preference

    26.67; fish show a significant substrate preference

    26.67; fish show no significant substrate preference

    40.00; fish show a significant substrate preference

    Show answer

    Answer: (B)

    The calculation sums contributions from all three groups: Sand (40030=13.33\frac{400}{30}=13.33) + Gravel (00) + Mud (40030=13.33\frac{400}{30}=13.33) = 26.67. This exceeds 5.99, indicating a significant preference.

  10. Question 10

    Fifty caterpillars are offered corn or tomato leaves. Thirty choose corn and 20 choose tomato. The null hypothesis predicts equal preference (25 each). The critical value at p=0.05p=0.05 is 3.84. Which statement correctly reports the chi-square value and conclusion?

    1.0; fail to reject the null hypothesis

    2.0; fail to reject the null hypothesis

    2.0; reject the null hypothesis

    4.0; reject the null hypothesis

    Show answer

    Answer: (B)

    The calculation is (30−25)225+(20−25)225=2525+2525=2.0\frac{(30-25)^2}{25} + \frac{(20-25)^2}{25} = \frac{25}{25} + \frac{25}{25} = 2.0. Since 2.0 is less than 3.84, the null hypothesis cannot be rejected, meaning no significant preference is shown.

What's on the AP Biology Unit 8 practice test

The standard version takes about 60 minutes and follows the timing and directions of the AP exam, sized to one unit:

  • 20 Multiple Choice Questions (30 minutes)
  • 1 FRQ – FRQ 1 – Interpreting and Evaluating Experimental Results (Long) (15 minutes)
  • 1 FRQ – FRQ 3 – Scientific Investigation (Short) (15 minutes)

Questions are drawn from 3469 multiple-choice questions and 101 FRQs written for Unit 8, Ecology.

Practice AP Biology Unit 8 FRQs

Answer the unit's free-response questions one at a time and get a score on every part.

Review Unit 8 before you start

Other AP Biology unit tests