--- title: "AP Biology Unit 5 Practice Test" description: "A timed AP Biology Unit 5 (Heredity) practice test in the AP format, drawn from 1,912 multiple-choice questions and 62 FRQs written for the unit. The standard version takes about 60 minutes." canonical: "https://fiveable.me/ap-bio/unit-test/5" type: "unit-test" subject: "AP Biology" unit: "Unit 5" --- # AP Biology Unit 5 Practice Test ## Overview A timed AP Biology Unit 5 (Heredity) practice test in the AP format, drawn from 1,912 multiple-choice questions and 62 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=5): 11 questions - [FRQ 2 – Interpreting and Evaluating Experimental Results with Graphing (Long)](/ap-bio/frqs/graphing?unit=5): 6 questions - [FRQ 3 – Scientific Investigation (Short)](/ap-bio/frqs/scientific-investigation?unit=5): 13 questions - [FRQ 4 – Conceptual Analysis (Short)](/ap-bio/frqs/conceptual-analysis?unit=5): 12 questions - [FRQ 5 – Analyze Model or Visual Representation (Short)](/ap-bio/frqs/model-analysis?unit=5): 4 questions - [FRQ 6 – Analyze Data (Short)](/ap-bio/frqs/data-analysis?unit=5): 16 questions ## Practice Questions ### Question 1 Researchers study the phagocytosis of two types of bacteria, Type A and Type B, by macrophages. They hypothesize that macrophages show no preference and engulf both types equally. In an experiment, macrophages engulf a total of 200 bacteria: 120 of Type A and 80 of Type B. Using a chi-square test with $$p=0.05$$ (critical value $$= 3.84$$ for $$df=1$$), what is the correct statistical conclusion? - (A) Reject $$H_0$$ because the calculated $$\chi^2$$ value is $$8.0$$ - (B) Reject $$H_0$$ because the calculated $$\chi^2$$ value is $$4.0$$ - (C) Fail to reject $$H_0$$ because the calculated $$\chi^2$$ value is $$8.0$$ - (D) Fail to reject $$H_0$$ because the calculated $$\chi^2$$ value is $$4.0$$ **Answer: (A)** The calculated chi-square value is 8.0, which exceeds the critical value of 3.84, leading to a rejection of the null hypothesis. ### Question 2 A geneticist crosses two heterozygous mice with a mutation affecting receptor-mediated endocytosis. The expected phenotypic ratio of offspring is 3 normal to 1 affected. The researcher observes 280 normal and 120 affected offspring in a sample of 400. Using a chi-square test ($$\alpha=0.05$$, critical value $$= 3.84$$), which statement best describes the result? - (A) The calculated $$\chi^2$$ is $$5.33$$, so the null hypothesis is rejected - (B) The calculated $$\chi^2$$ is $$5.33$$, so the null hypothesis is accepted - (C) The calculated $$\chi^2$$ is $$2.67$$, so the null hypothesis is rejected - (D) The calculated $$\chi^2$$ is $$2.67$$, so the null hypothesis is accepted **Answer: (A)** The expected counts are 300 normal and 100 affected, so $$\chi^2 = (280-300)^2/300 + (120-100)^2/100 = 1.33 + 4 = 5.33$$. This exceeds the critical value of 3.84 (df = 1), so the null hypothesis of a 3:1 ratio is rejected. ### Question 3 Neuroscientists hypothesize that exocytosis of synaptic vesicles occurs uniformly across three active zones (A, B, and C) in a neuron. They observe 30 events at Zone A, 40 at Zone B, and 20 at Zone C (Total = 90). Using a chi-square goodness-of-fit test with $$df=2$$ and a critical value of $$5.99$$, what is the outcome? - (A) $$\chi^2 = 6.67$$; the distribution is not uniform - (B) $$\chi^2 = 6.67$$; the distribution is uniform - (C) $$\chi^2 = 3.33$$; the distribution is not uniform - (D) $$\chi^2 = 3.33$$; the distribution is uniform **Answer: (A)** The calculated chi-square value of 6.67 exceeds the critical value of 5.99, leading to the rejection of the null hypothesis of uniform distribution. ### Question 4 A cell biologist measures pinocytosis rates in four different cell lines, hypothesizing that fluid uptake is equal across all lines. The observed vesicle counts are 15, 25, 25, and 15 (Total = 80). Using a chi-square test ($$df=3$$, critical value $$= 7.82$$), what conclusion is supported by the data? - (A) Fail to reject $$H_0$$ because $$\chi^2 = 5.0$$ - (B) Fail to reject $$H_0$$ because $$\chi^2 = 2.5$$ - (C) Reject $$H_0$$ because $$\chi^2 = 5.0$$ - (D) Reject $$H_0$$ because $$\chi^2 = 2.5$$ **Answer: (A)** The calculated chi-square value is 5.0, which is less than the critical value of 7.82, so there is insufficient evidence to reject the null hypothesis. ### Question 5 Researchers treat cells with a drug and observe the type of endocytosis used: clathrin-mediated vs. caveolae-mediated. In untreated cells, the ratio is 60:40. In the treated sample ($$n=100$$), they observe 50 clathrin-mediated and 50 caveolae-mediated events. Using a chi-square test ($$df=1$$, critical value $$= 3.84$$), does the drug significantly alter the ratio? - (A) Yes, because $$\chi^2 = 4.17$$ - (B) Yes, because $$\chi^2 = 8.33$$ - (C) No, because $$\chi^2 = 4.17$$ - (D) No, because $$\chi^2 = 8.33$$ **Answer: (A)** The calculated chi-square value of 4.17 exceeds the critical value of 3.84, indicating a significant difference from the expected 60:40 ratio. ### Question 6 A protein is hypothesized to be sorted equally into four secretory vesicle types (1:1:1:1). Observed counts for 80 proteins are: 20 in Type I, 20 in Type II, 30 in Type III, and 10 in Type IV. Using a chi-square test ($$df=3$$, critical value $$= 7.82$$), what is the calculated statistic and conclusion? - (A) $$\chi^2 = 10.0$$; reject the null hypothesis - (B) $$\chi^2 = 10.0$$; fail to reject the null hypothesis - (C) $$\chi^2 = 5.0$$; reject the null hypothesis - (D) $$\chi^2 = 5.0$$; fail to reject the null hypothesis **Answer: (A)** The calculated chi-square value is 10.0, which is greater than the critical value of 7.82, so the null hypothesis of equal sorting is rejected. ### Question 7 A mutation is expected to cause cells to secrete only 20% of a specific enzyme via exocytosis, retaining 80%. In a sample of 500 cells, 120 secrete the enzyme and 380 retain it. Using a chi-square test ($$df=1$$, critical value $$= 3.84$$), which conclusion is valid? - (A) Reject $$H_0$$ because $$\chi^2 = 5.0$$ - (B) Reject $$H_0$$ because $$\chi^2 = 4.0$$ - (C) Fail to reject $$H_0$$ because $$\chi^2 = 5.0$$ - (D) Fail to reject $$H_0$$ because $$\chi^2 = 4.0$$ **Answer: (A)** The calculated chi-square value is 5.0, which exceeds the critical value of 3.84, indicating the observed secretion rate differs significantly from the expectation. ### Question 8 A virus enters cells via endocytosis. Researchers test three viral strains (X, Y, Z) to see if entry rates differ. Observed entry events are: X=50, Y=60, Z=40 (Total = 150). Assuming a null hypothesis of equal entry rates, what is the chi-square value and decision ($$df=2$$, critical value $$= 5.99$$)? - (A) $$\chi^2 = 4.0$$; fail to reject $$H_0$$ - (B) $$\chi^2 = 4.0$$; reject $$H_0$$ - (C) $$\chi^2 = 8.0$$; fail to reject $$H_0$$ - (D) $$\chi^2 = 8.0$$; reject $$H_0$$ **Answer: (A)** The calculated chi-square value is 4.0, which is less than the critical value of 5.99, so there is insufficient evidence to reject the null hypothesis. ### Question 9 A protein normally folds in the Endoplasmic Reticulum (ER). When expressed in the cytosol, 80% of the protein aggregates (misfolds). In the ER, only 5% aggregates. A chi-square test yields $$p < 0.001$$. The null hypothesis states that the ER environment does not influence protein folding efficiency. Which conclusion is supported? - (A) Reject the null hypothesis; the low p-value indicates the ER provides a specialized environment essential for correct folding. - (B) Reject the null hypothesis; the low p-value indicates protein folding efficiency depends on chaperone proteins present in the ER. - (C) Fail to reject the null hypothesis; the p-value indicates the difference in aggregation rates is due to random variation rather than ER influence. - (D) Reject the null hypothesis; the low p-value indicates the cytosol environment actively promotes protein aggregation rather than the ER preventing it. **Answer: (A)** Interpreting the p-value ($$p < 0.001$$) against a standard alpha ($$0.05$$) leads to rejecting the null hypothesis. This supports the biological concept that the ER membrane compartmentalizes specific folding factors absent in the cytosol. ### Question 10 A biologist observes 200 cells stained for lysosomal acidity to test for a membrane transport defect. Based on wild-type efficiency, 160 cells should show high acidity and 40 low acidity. The observed counts are 140 high acidity and 60 low acidity. Using a chi-square test with a critical value of 3.84 ($$\alpha = 0.05$$), what is the correct statistical conclusion regarding lysosomal compartmentalization? - (A) Reject $$H_0$$; the $$\chi^2$$ value of 12.5 indicates a significant loss of compartmentalization. - (B) Reject $$H_0$$; the $$\chi^2$$ value of 10.0 indicates a significant loss of compartmentalization. - (C) Fail to reject $$H_0$$; the $$\chi^2$$ value of 12.5 implies normal compartmentalization function. - (D) Fail to reject $$H_0$$; the $$\chi^2$$ value of 2.5 implies normal compartmentalization function. **Answer: (A)** Calculating the test statistic requires summing the squared differences divided by expected values: $$(140-160)^2/160 + (60-40)^2/40 = 2.5 + 10 = 12.5$$. Since 12.5 exceeds the critical value of 3.84, the null hypothesis is rejected, indicating the defect significantly alters the internal environment. ## Review the Unit - [5.1 Meiosis](/ap-bio/unit-5/meiosis/study-guide/FC0aTuODYikjJuhlBO1Z) - [5.2 Meiosis and Genetic Diversity](/ap-bio/unit-5/meiosis-genetic-diversity/study-guide/YZOFYsQw4twZNkp5WM3l) - [5.3 Mendelian Genetics](/ap-bio/unit-5/mendelian-genetics/study-guide/SdlMbZYAD4sxuXuRygPv) - [5.4 Non-Mendelian Genetics](/ap-bio/unit-5/non-mendelian-genetics/study-guide/5oRHoGlMbML8IgtaHaHs) - [5.5 Environmental Effects on Phenotype](/ap-bio/unit-5/environmental-effects-on-phenotype/study-guide/hLZNliseyo0zAayZWnah) - [Chromosomal Inheritance Review](/ap-bio/unit-5/chromosomal-inheritance/study-guide/PzK71wcPD3xAmEId5SWv) ## 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 6: Gene Expression and Regulation](/ap-bio/unit-test/6) - [Unit 7: Natural Selection](/ap-bio/unit-test/7) - [Unit 8: Ecology](/ap-bio/unit-test/8) ## 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`)