AP Biology Unit 5 practice questions
Try 10 multiple-choice questions from Unit 5 here. Pick an answer to see why it's right or wrong, then take the timed test above for a full AP-style set.
Question 1
A plant heterozygous for two genes is testcrossed with a plant homozygous recessive for both genes. The offspring have observed counts of 22, 28, 24, and 26 for the four phenotype combinations. The null hypothesis predicts a 1:1:1:1 ratio under independent assortment. Using a critical value of 7.82 (df=3, p=0.05), what is the chi-square value and conclusion?
Chi-square = 0.8; fail to reject the null hypothesis, data is consistent with independent assortment.
Chi-square = 0.8; reject the null hypothesis, data is inconsistent with independent assortment.
Chi-square = 3.2; reject the null hypothesis, data is inconsistent with independent assortment.
Chi-square = 0.2; fail to reject the null hypothesis, data is consistent with independent assortment.
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Answer: (A)
Expected is 25 each. Chi-square = (3^2+3^2+1^2+1^2)/25 = (9+9+1+1)/25 = 20/25 = 0.8. Since 0.8 < 7.82, the null hypothesis is not rejected.
Question 2
In a drought study, 100 broad-leaf plants (high SA:V) and 100 needle-leaf plants (low SA:V) are observed. 20 broad-leaf and 60 needle-leaf plants survive. If leaf shape did not affect survival, 40 survivors were expected in each group. Calculate and interpret (, critical value=3.84).
; reject , low SA:V ratios reduce water loss.
; reject , high SA:V ratios reduce water loss.
; fail to reject , shape is unrelated to survival.
; fail to reject , low SA:V increases water loss.
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Answer: (A)
Calculates . Rejection indicates a significant difference; needle-leaf plants (low SA:V) survived better, consistent with reduced transpiration.
Question 3
Researchers treat a bacterial culture with lysozyme, an enzyme that degrades peptidoglycan cell walls. Based on the enzyme concentration, they expect 60% of the cells to lyse (burst) and 40% to remain intact. In a sample of 200 treated cells, 130 lysed and 70 remained intact. Using a critical value of 3.84 (, ), which choice correctly describes the chi-square () analysis?
; the observed results are not significantly different from the expected lysis rate
; the observed results are significantly different from the expected lysis rate
; the observed results are not significantly different from the expected lysis rate
; the observed results are significantly different from the expected lysis rate
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Answer: (A)
Calculating requires determining expected counts ( lysed, intact) and summing for each category (). Since , the null hypothesis cannot be rejected.
Question 4
A testcross is performed to see if genes for two metabolic enzymes assort independently (1:1:1:1 ratio). The observed offspring counts are 45, 45, 35, and 35. Using a critical value of (), what is the conclusion?
, fail to reject null; genes assort independently.
, fail to reject null; genes assort independently.
, reject null; genes are genetically linked.
, reject null; genes are genetically linked.
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Answer: (A)
Connecting the low statistic to the null hypothesis confirms independent assortment. Expected counts are 40 each; , which is less than , so the deviation is not significant.
Question 5
Two genes encoding enzymes for the Krebs cycle are tested for linkage. A chi-square test for independence is performed, resulting in a p-value of . Using a significance level of , what is the correct biological conclusion?
Reject ; the genes are likely linked
Reject ; the genes assort independently
Fail to reject ; the genes are likely linked
Fail to reject ; the genes assort independently
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Answer: (A)
Interpreting the p-value requires comparing it to alpha. Since , the null hypothesis of independent assortment is rejected, suggesting the genes do not assort independently and are likely linked.
Question 6
Researchers investigate whether mitochondrial density (High vs. Low) is independent of tissue type (Muscle vs. Skin). Observed counts are: Muscle/High=, Muscle/Low=, Skin/High=, Skin/Low= (). Using a critical value of (, ), what is the correct statistical conclusion?
Reject the null hypothesis because the chi-square value of exceeds the critical value of
Fail to reject the null hypothesis because the chi-square value of exceeds the critical value of
Reject the null hypothesis because the chi-square value of exceeds the critical value of
Fail to reject the null hypothesis because the chi-square value of is less than the critical value of
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Answer: (A)
Expected counts for all cells are (based on row/column totals of ). Summing for four cells () yields , which is significant.
Question 7
Researchers investigate if testosterone signaling influences aggression in male fish. They observe 50 intact males and 50 castrated males. Intact males show 40 high-aggression and 10 low-aggression behaviors. Castrated males show 10 high-aggression and 40 low-aggression behaviors. Using a critical value of 3.84 (p=0.05), what is the chi-square result?
Chi-square = 36.0; reject the null hypothesis, linking testosterone to aggression levels
Chi-square = 36.0; fail to reject the null hypothesis, finding no link to aggression
Chi-square = 18.0; reject the null hypothesis, linking testosterone to aggression levels
Chi-square = 18.0; fail to reject the null hypothesis, finding no link to aggression
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Answer: (A)
Expected count for all cells is 25. Calculation: (40-25)²/25 * 4 = (225/25) * 4 = 9 * 4 = 36.0. Since 36.0 > 3.84, the null is rejected, indicating a link.
Question 8
Yeast cells exposed to a mating factor are expected to form projections called "shmoos" at a rate of 70%. In a sample of 100 cells, 60 form shmoos and 40 do not. Using a critical value of 3.84 (), what is the conclusion?
Chi-square is 4.76, suggesting the mating factor response differs from 70%
Chi-square is 4.76, suggesting the mating factor response matches 70%
Chi-square is 2.38, suggesting the mating factor response differs from 70%
Chi-square is 2.38, suggesting the mating factor response matches 70%
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Answer: (A)
Expected counts are 70 shmoo and 30 no shmoo. . Since , the null hypothesis is rejected.
Question 9
In a species with 2n = 46, three cell types are examined. Type A cells have 46 chromosomes, Type B cells have 23 chromosomes, and Type C cells have 46 chromosomes. Which conclusion is supported by the data regarding the cell types?
Types A and C are somatic cells produced by mitosis, while Type B is a gamete produced by meiosis.
Types A and B are somatic cells produced by mitosis, while Type C is a gamete produced by meiosis.
Types B and C are somatic cells produced by mitosis, while Type A is a gamete produced by meiosis.
Types A and C are somatic cells produced by meiosis, while Type B is a gamete produced by mitosis.
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Answer: (A)
Identifying the data points (46 vs 23) allows classification of cell types. Mitosis preserves the diploid number (46), characterizing somatic cells, while meiosis reduces it to haploid (23), characterizing gametes.
Question 10
A biologist observes a germ cell undergoing meiosis to produce gametes. Which statement best describes how the chromosome number is reduced from diploid to haploid during this process?
Homologous chromosomes separate during meiosis I, reducing the chromosome count by half.
Sister chromatids separate during meiosis I, reducing the chromosome count by half.
Homologous chromosomes separate during meiosis II, reducing the chromosome count by half.
Sister chromatids separate during meiosis II, reducing the chromosome count by half.
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Answer: (A)
Identifying the reduction division requires recognizing that ploidy changes when homologous pairs separate. Meiosis I is the reductional division where the cell transitions from diploid (2n) to haploid (n).
What's on the AP Biology Unit 5 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 1884 multiple-choice questions and 62 FRQs written for Unit 5, Heredity.
Practice AP Biology Unit 5 FRQs
Answer the unit's free-response questions one at a time and get a score on every part.
- FRQ 1 – Interpreting and Evaluating Experimental Results (Long) (11 questions)
- FRQ 2 – Interpreting and Evaluating Experimental Results with Graphing (Long) (6 questions)
- FRQ 3 – Scientific Investigation (Short) (13 questions)
- FRQ 4 – Conceptual Analysis (Short) (12 questions)
- FRQ 5 – Analyze Model or Visual Representation (Short) (4 questions)
- FRQ 6 – Analyze Data (Short) (16 questions)
Review Unit 5 before you start
- 5.1 Meiosis
- 5.2 Meiosis and Genetic Diversity
- 5.3 Mendelian Genetics
- 5.4 Non-Mendelian Genetics
- 5.5 Environmental Effects on Phenotype
- Chromosomal Inheritance Review