--- title: "AP Biology Unit 8 Practice Test" description: "A timed AP Biology Unit 8 (Ecology) practice test in the AP format, drawn from 3,469 multiple-choice questions and 101 FRQs written for the unit. The standard version takes about 60 minutes." canonical: "https://fiveable.me/ap-bio/unit-test/8" type: "unit-test" subject: "AP Biology" unit: "Unit 8" --- # AP Biology Unit 8 Practice Test ## Overview A timed AP Biology Unit 8 (Ecology) practice test in the AP format, drawn from 3,469 multiple-choice questions and 101 FRQs written for the unit. The standard version takes about 60 minutes. ## Test Format - 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) ## Unit FRQs by Type - [FRQ 1 – Interpreting and Evaluating Experimental Results (Long)](/ap-bio/frqs/experimental-results?unit=8): 16 questions - [FRQ 2 – Interpreting and Evaluating Experimental Results with Graphing (Long)](/ap-bio/frqs/graphing?unit=8): 13 questions - [FRQ 3 – Scientific Investigation (Short)](/ap-bio/frqs/scientific-investigation?unit=8): 21 questions - [FRQ 4 – Conceptual Analysis (Short)](/ap-bio/frqs/conceptual-analysis?unit=8): 16 questions - [FRQ 5 – Analyze Model or Visual Representation (Short)](/ap-bio/frqs/model-analysis?unit=8): 19 questions - [FRQ 6 – Analyze Data (Short)](/ap-bio/frqs/data-analysis?unit=8): 16 questions ## Practice Questions ### 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? - (A) 19,200 cells/mL per day - (B) 23,700 cells/mL per day - (C) 24,000 cells/mL per day - (D) 94,800 cells/mL per day **Answer: (B)** Calculating the rate of change requires dividing the change in population (94,800 cells/mL) by the time interval (4 days). ### 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? - (A) 6.4 times greater - (B) 64 times greater - (C) 640 times greater - (D) 6,400 times greater **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). ### 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) A decrease of 40% - (B) A decrease of 60% - (C) A decrease of 150% - (D) A decrease of 250% **Answer: (B)** Calculating percent change requires finding the difference (-480) and dividing by the initial value (800), then converting to a percentage. ### 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? - (A) 1.5 ppm per year - (B) 8.8 ppm per year - (C) 10.3 ppm per year - (D) 60.0 ppm per year **Answer: (A)** Calculating the rate requires dividing the total change in concentration (60 ppm) by the time interval (40 years). ### Question 5 A study measured the cooling effect of evaporation on skin temperature. The mean difference between skin treated with water versus oil was 4.0°C, with a standard error of the difference equal to 1.5°C. Using a t-critical value of 2.0 for a 95% confidence level, does the data support the claim that water provides significantly more cooling than oil? - (A) Yes, because the 95% confidence interval of 1.0°C to 7.0°C excludes zero. - (B) Yes, because the 95% confidence interval of 2.5°C to 5.5°C excludes zero. - (C) No, because the 95% confidence interval of 1.0°C to 7.0°C includes zero. - (D) No, because the 95% confidence interval of 2.5°C to 5.5°C includes zero. **Answer: (A)** Calculates the interval for the difference (4.0 ± 2.0×1.5 = 4.0 ± 3.0) as 1.0 to 7.0 and correctly identifies that excluding zero implies a significant effect. ### Question 6 African elephants have large, thin ears that they flap frequently, whereas woolly mammoths adapted to the Arctic had very small ears. Which reasoning best justifies the claim that large ears are an adaptation for thermoregulation in hot environments? - (A) Large ears increase the organism's overall surface area-to-volume ratio, facilitating rapid heat loss to the environment. - (B) Large ears increase the organism's overall surface area-to-volume ratio, facilitating rapid heat gain from the environment. - (C) Large ears decrease the organism's overall surface area-to-volume ratio, preventing excessive heat loss to the environment. - (D) Large ears increase the organism's overall surface area-to-volume ratio, minimizing water loss through evaporation. **Answer: (A)** Connecting the physical trait to thermodynamic principles justifies the claim. The large, thin structure maximizes the surface area exposed to the air relative to the blood volume inside, allowing heat to dissipate efficiently. ### Question 7 Organism A has a mass of $$10 \text{ kg}$$ and a metabolic rate of $$2000 \text{ mL O}_2/\text{hr}$$. Organism B has a mass of $$5 \text{ kg}$$ and a metabolic rate of $$1500 \text{ mL O}_2/\text{hr}$$. Calculate the ratio of the mass-specific metabolic rate of Organism B to that of Organism A. - (A) 0.67 - (B) 0.75 - (C) 1.50 - (D) 2.00 **Answer: (C)** Specific rate A = $$2000/10 = 200$$. Specific rate B = $$1500/5 = 300$$. The ratio of B to A is $$300/200 = 1.50$$. ### Question 8 A biologist observes that a 30-gram mouse consumes much more food per gram of body mass than a 3,000-kilogram elephant. Which explanation best accounts for this difference in metabolic rate? - (A) The mouse has a higher surface area-to-volume ratio, requiring a faster metabolism to maintain body temperature. - (B) The mouse has a lower surface area-to-volume ratio, requiring a faster metabolism to maintain body temperature. - (C) The mouse has a higher surface area-to-volume ratio, allowing a slower metabolism to conserve energy reserves. - (D) The mouse has a lower surface area-to-volume ratio, allowing a slower metabolism to conserve energy reserves. **Answer: (A)** Connecting body size to thermodynamics reveals that small animals lose heat rapidly due to a high SA:V ratio, necessitating a higher metabolic rate per unit mass to generate sufficient replacement heat. ### Question 9 A researcher compares a mouse and an elephant exposed to the same sudden drop in environmental temperature. Which of the following best predicts the metabolic response required for each organism to maintain body temperature? - (A) The mouse consumes more oxygen per gram of tissue because it has a higher surface area-to-volume ratio. - (B) The mouse consumes less oxygen per gram of tissue because it has a higher surface area-to-volume ratio. - (C) The mouse consumes more oxygen per gram of tissue because it has a lower surface area-to-volume ratio. - (D) The mouse consumes less oxygen per gram of tissue because it has a lower surface area-to-volume ratio. **Answer: (A)** Connecting body size to thermodynamics reveals that the smaller mouse has a much larger surface area relative to its mass, causing it to lose heat rapidly. To maintain homeostasis, it must generate heat through a significantly higher metabolic rate per gram than the larger elephant. ### Question 10 A 5-gram shrew must consume approximately three times its body weight in food daily to survive, while a 5,000-kilogram elephant consumes only a small fraction of its body weight daily. Which question best guides an investigation into the cause of this metabolic difference? - (A) Does a high surface area-to-volume ratio increase the specific rate of heat loss? - (B) Does a high surface area-to-volume ratio decrease the specific rate of heat loss? - (C) Does a high surface area-to-volume ratio increase the efficiency of food digestion? - (D) Does a high surface area-to-volume ratio decrease the efficiency of food digestion? **Answer: (A)** Connecting the small size of the shrew to a high SA:V ratio explains the rapid heat loss that necessitates a high metabolic rate. ## Review the Unit - [8.1 Responses to the Environment](/ap-bio/unit-8/responses-environment/study-guide/RTwUxFq3Kcs28woJXlQX) - [8.2 Energy Flow Through Ecosystems](/ap-bio/unit-8/energy-flow-through-ecosystems/study-guide/A1PeQD1Zy3BIMu1zSMzd) - [8.3 Population Ecology](/ap-bio/unit-8/population-ecology/study-guide/JiYkhCa7zQ0XPgs6OpbK) - [8.4 Effect of Density of Populations](/ap-bio/unit-8/effect-density-populations/study-guide/Zn70P0oeUAlNnbvUEjW3) - [8.5 Community Ecology](/ap-bio/unit-8/community-ecology/study-guide/GhiVt7Egu8crmrHtQXXc) - [8.6 Biodiversity](/ap-bio/unit-8/biodiversity/study-guide/UQxfkl91v4pCcoar2qMD) - [8.7 Disruptions to Ecosystems](/ap-bio/unit-8/disruptions-ecosystems/study-guide/ra0njykAUxN9gf0swqKV) ## Other Unit Tests - [Unit 1: Chemistry of Life](/ap-bio/unit-test/1) - [Unit 2: Cells](/ap-bio/unit-test/2) - [Unit 3: Cellular Energetics](/ap-bio/unit-test/3) - [Unit 4: Cell Communication and Cell Cycle](/ap-bio/unit-test/4) - [Unit 5: Heredity](/ap-bio/unit-test/5) - [Unit 6: Gene Expression and Regulation](/ap-bio/unit-test/6) - [Unit 7: Natural Selection](/ap-bio/unit-test/7) ## About This Document Canonical Fiveable pages are available as Markdown at the same path plus `.md`. - [llms.txt](https://fiveable.me/llms.txt): index of Fiveable's sections and URL patterns - [llms-full.txt](https://fiveable.me/llms-full.txt): complete subject and unit listing - [MCP server](https://fiveable.me/mcp): call Fiveable as tools instead of fetching pages (`https://fiveable.me/api/mcp`) - [MCP server for AP teachers](https://fiveable.me/mcp/teachers): a teacher's classes, assignments and AP-rubric grading (`https://fiveable.me/api/mcp/teacher`)