Biology by Bradford
Unit 5: Heredity. Practice questions with scoring guidelines.
203 original AP-style questions on Unit 5, written to the CED learning objectives: 144 multiple-choice, 59 free-response. Below is a 20-mark practice set built from them, ready to assign as a unit check or homework, or to sit yourself and score against the guidelines. Print it, project it, or build a fresh one.
Topics in this unit
- 5.1 Meiosis 36 questions
- 5.2 Meiosis and Genetic Diversity 31 questions
- 5.3 Mendelian Genetics 49 questions
- 5.4 Non-Mendelian Genetics 96 questions
- 5.5 Environmental Effects on Phenotype 27 questions
In the bank for Unit 5
- 144 multiple-choice
- 59 free-response
Every question is original and tagged to a learning objective and science practice.
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The practice set
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In a species of flowering plant, a true-breeding line with red stems and hairy leaves was crossed with a true-breeding line with green stems and smooth leaves, and all of the F1 plants had red stems and hairy leaves. The F1 plants were self-pollinated. Of 1,600 F2 plants, 1,197 had red stems and hairy leaves and 403 had green stems and smooth leaves; no plant had any other combination of traits. Hypothesis 1 states that a single gene affects both stem color and leaf hairiness. Hypothesis 2 states that stem color and leaf hairiness are controlled by two different genes located very close together on the same chromosome. Which observation would best support hypothesis 2 over hypothesis 1?
- F1 plants testcrossed to plants with green stems and smooth leaves give equal numbers of the two parental phenotypes.
- A chi-square test shows that the 1,197 : 403 ratio of the two F2 phenotypes does not differ significantly from 3 : 1.
- Among the F2 plants with red stems and hairy leaves, homozygous and heterozygous plants occur in a ratio of 1 : 2.
- Among many thousands more F2 plants, a few have red stems with smooth leaves or green stems with hairy leaves.
A diploid animal has a chromosome number of 2n = 8. One primary spermatocyte from this animal completes meiosis I and then meiosis II, producing four sperm. Which statement correctly describes the chromosome content of a cell immediately after meiosis I is complete?
- Each cell contains 4 chromosomes, each made of a single chromatid, because the sister chromatids separated.
- Each cell contains 8 chromosomes, each made of a single chromatid, because only the chromatids separated.
- Each cell contains 4 chromosomes, each still made of two sister chromatids joined at the centromere.
- Each cell contains 8 chromosomes, each of two sister chromatids, as in the parent.
A diploid organism has a diploid chromosome number of 2n = 10. Considering only the independent assortment of homologous pairs during meiosis in each parent and the random fertilization of one gamete by another, and ignoring crossing over, how many chromosomally distinct zygotes can two such parents produce?
- 32
- 64
- 100
- 1,024
In the testes of a species of salamander (2n = 24), some cells divide by mitosis and others undergo meiosis to produce sperm. Cells in different parts of a testis begin dividing at different times, so at any moment dividing cells can be found at every stage of mitosis and of meiosis. Researchers stained testis tissue from 6 males and, for each male, classified 200 cells that were undergoing meiosis according to their stage, counting the two cells formed by each meiosis I division together as one. Table 1 shows the results.
| Stage of meiosis | Meiotic cells at this stage (%) |
|---|---|
| Prophase I | 70.5 ± 3.0 |
| Metaphase I | 9.0 ± 1.4 |
| Anaphase I and telophase I | 4.5 ± 1.0 |
| Prophase II | 3.5 ± 0.9 |
| Metaphase II | 5.0 ± 1.1 |
| Anaphase II and telophase II | 7.5 ± 1.3 |
The following information applies to parts C and D.
In the same tissue, 40 percent of the cells dividing by mitosis were in prophase. Other measurements showed that one complete mitosis, from the start of prophase to the end of telophase, lasts 5 hours and that one complete meiosis, from the start of prophase I to the end of telophase II, lasts 10 days.
In fruit flies (Drosophila melanogaster), gray body (G) is dominant to ebony body (g). A researcher crosses two heterozygous gray-bodied flies and counts the offspring: 172 gray-bodied and 44 ebony-bodied. The researcher performs a chi-square test comparing the observed numbers with those expected for a 3:1 ratio. The critical value of chi-square at p = 0.05 with 1 degree of freedom is 3.84.
| Phenotype | Observed | Expected (3:1) |
|---|---|---|
| Gray body | 172 | 162 |
| Ebony body | 44 | 54 |
In a species of tropical butterfly, adults that develop in the warm wet season have large eyespots (round, ringed markings) on the underside of the wings, and adults that develop in the cool dry season have small eyespots. Researchers used larvae (caterpillars) from a laboratory line that had been bred brother to sister for many generations, so the larvae were nearly identical genetically. Larval development lasts about 20 days. Each of four groups of 30 larvae was kept at 17 °C or at 27 °C for the first 10 days (the early larval stage) and at 17 °C or at 27 °C for the last 10 days (the late larval stage). All of the pupae were then kept at 22 °C, and the area of the largest eyespot was measured on each adult. Table 1 shows the results.
| Group | Early larval stage (°C) | Late larval stage (°C) | Eyespot area (mm²) |
|---|---|---|---|
| 1 | 17 | 17 | 2.1 ± 0.3 |
| 2 | 27 | 27 | 6.8 ± 0.5 |
| 3 | 17 | 27 | 6.5 ± 0.6 |
| 4 | 27 | 17 | 2.4 ± 0.4 |
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Each point is worth 1 point and is credited independently. Accept any one of the listed alternatives per point; ( ) marks optional wording and / separates interchangeable wording.
- D: If a single gene affects both traits, no allele combination can separate them, so red stems would always come with hairy leaves. If two genes lie very close together on one chromosome, a crossover occasionally forms between them in prophase I and produces gametes with a new combination of alleles, so in a large enough sample a few F2 plants with a new combination of traits would appear. Observations A, B and C are expected under both hypotheses: each predicts a 1 : 1 testcross ratio of the parental phenotypes, a 3 : 1 ratio in the F2, and a 1 : 2 ratio of homozygous to heterozygous plants among F2 plants with the dominant phenotype.
- C: Meiosis I separates homologous chromosomes, so each daughter cell has the haploid number (n = 4), but each chromosome still consists of two sister chromatids until they separate in meiosis II.
- D: With 5 homologous pairs, each parent can produce 2⁵ = 32 chromosomally distinct gametes; any one of the 32 eggs can be fertilized by any one of the 32 sperm, giving 32 × 32 = 1,024 possible zygotes.
- Stage of meiosis on the x-axis and mean percentage of meiotic cells on the y-axis, both labeled, with a scale that fits 70.5%, and six bars in order (prophase I to anaphase II and telophase II) drawn at the correct heights.
- Correctly labeled and scaled bar graph with each of the six means plotted accurately in the order the stages occur.
- An error bar on every bar extending the ±2SE value above and below the mean (for example 67.5 to 73.5 for prophase I and 7.6 to 10.4 for metaphase I).
- Error bars of the correct length (±3.0, ±1.4, ±1.0, ±0.9, ±1.1 and ±1.3) drawn on all six bars.
Total for part (A): 2 points
- Most meiotic cells (70.5%) are in prophase I; every other stage holds 9.0% or fewer, and the three stages of meiosis II together hold only 16%.
- Prophase I has far more cells than any other stage, about eight times as many as the next largest (metaphase I, 9.0%); the fewest cells are in prophase II (3.5%).
- Significantly different: the intervals for metaphase I (7.6 to 10.4%) and metaphase II (3.9 to 6.1%) do not overlap.
- Yes, metaphase I has significantly more cells, because 9.0 ± 1.4 and 5.0 ± 1.1 give ±2SE error bars that do not overlap.
- Because the cells began meiosis at different times, a sample taken at one moment catches each cell at a random point in its meiosis; a cell spends more of its meiosis in a long stage, so more cells are caught in that stage, and the fraction of cells at a stage equals the fraction of the total time that the stage takes.
- If all the cells had started together, they would all be at the same stage when the tissue was stained; since their starting times are spread out, each stage holds a share of the cells in proportion to how long cells stay in it, so the longest stage has the most cells.
Total for part (B): 3 points
- Prophase I = 0.705 × 10 days = 7.05 days = 169.2 hours; prophase of mitosis = 0.40 × 5 hours = 2 hours; 169.2 / 2 ≈ 85 times longer (accept 84 to 85).
- 70.5% of 240 hours is about 169 hours, and 40% of 5 hours is 2 hours, so prophase I lasts about 85 times as long.
- In prophase I each chromosome must find its homologous chromosome and pair with it along its whole length, and crossing over (exchange of segments between non-sister chromatids) takes place; in prophase of mitosis homologous chromosomes do not pair or cross over, so less has to happen.
- Pairing of homologous chromosomes (synapsis) and crossing over between them occur only in prophase I, and these extra steps take time that mitotic prophase does not need.
Total for part (C): 2 points
- Metaphase I will increase and metaphase II will fall (to about zero): cells keep arriving at metaphase I from prophase I but cannot leave it, while the cells already past metaphase I finish meiosis and no new cells replace them in meiosis II (a cell that has just left metaphase I passes through metaphase II within about 1.3 days: 4.5 + 3.5 + 5.0 = 13% of 10 days).
- More cells in metaphase I and fewer in metaphase II, because cells pile up at the blocked stage and the cells already in meiosis II complete it without being replaced.
- Testis tissue treated in the same way and for the same 2 days with the solution the chemical is dissolved in, but without the chemical, then stained and classified in the same way; it shows how the stage percentages change over 2 days without the chemical, so any difference from it can be attributed to the chemical.
- Testes from comparable males given no chemical (solvent only) and sampled at the same time, so that changes caused by handling, by the passage of time or by the culture conditions are not mistaken for effects of the chemical.
Total for part (D): 2 points
Total for question 4: 9 points
- The observed offspring numbers do not differ (significantly) from the expected 3:1 (gray:ebony) ratio / any difference is due to chance.
Total for part (A): 1 point
- χ² = (172−162)²/162 + (44−54)²/54 = 100/162 + 100/54 = 0.62 + 1.85 ≈ 2.47 (accept 2.4–2.5).
Total for part (B): 1 point
- Do not reject: the calculated χ² (≈2.47) is less than the critical value (3.84), so the deviation from 3:1 is not significant at p = 0.05.
Total for part (C): 1 point
- A cross of two heterozygotes producing offspring consistent with a 3:1 (dominant:recessive) ratio matches the prediction for a single gene with complete dominance (supported by the chi-square result).
Total for part (D): 1 point
Total for question 5: 4 points
- There is no significant difference: the ±2SE intervals, 1.8 to 2.4 mm² (group 1) and 2.0 to 2.8 mm² (group 4), overlap.
- Not significant, because the error ranges overlap (2.1 ± 0.3 and 2.4 ± 0.4), so the null hypothesis that the two groups have the same mean eyespot area is not rejected.
Total for part (A): 1 point
- Groups with the same late-stage temperature have similar eyespot areas whatever their early-stage temperature (groups 1 and 4, both 17 °C late, 2.1 and 2.4 mm² with overlapping intervals; groups 2 and 3, both 27 °C late, 6.8 and 6.5 mm², intervals 6.3 to 7.3 and 5.9 to 7.1), whereas groups with the same early-stage temperature but different late-stage temperatures differ significantly (group 1, 1.8 to 2.4 mm², against group 3, 5.9 to 7.1 mm²).
- Changing the early-stage temperature while the late stage stays the same does not change eyespot area significantly (group 1 against group 4, group 2 against group 3), but changing the late-stage temperature does (group 1 against group 3, or group 4 against group 2, whose intervals do not overlap), so eyespot size follows the late-stage temperature.
Total for part (B): 1 point
- The temperature during the pupal stage might also affect eyespot size, so it must be controlled: with every pupa at 22 °C, the groups differ only in their early and late larval temperatures, so a difference in eyespot area can be attributed to the larval stages.
- If each pupa had stayed at its late-larval temperature, the late larval stage and the pupal stage would always have had the same temperature, so an effect of the late larval stage could not be separated from an effect of the pupal stage; a common pupal temperature removes this confounding variable.
Total for part (C): 1 point
- Rear larvae at 17 °C and move separate groups to 27 °C on different days of the late larval stage (for example on day 12, 14, 16 or 18), keeping all other conditions the same, then compare the eyespot areas of the groups to find the days on which the warm temperature must be present to give large eyespots.
- Give separate groups a short period at 27 °C (for example 2 days) at different points between day 11 and day 20, with the rest of development at 17 °C, and compare eyespot area among the groups to find which period produces large eyespots.
Total for part (D): 1 point
Total for question 6: 4 points
Other AP Biology units
- Unit 1 Chemistry of Life 196
- Unit 2 Cell Structure and Function 260
- Unit 3 Cellular Energetics 195
- Unit 4 Cell Communication and Cell Cycle 187
- Unit 6 Gene Expression and Regulation 219
- Unit 7 Natural Selection 316
- Unit 8 Ecology 215
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