Chi‑Square Practice Problems for AP Biology: A thorough look
When preparing for the AP Biology exam, mastering the chi‑square test is essential for interpreting data in genetics, evolution, and ecology questions. This article offers an in‑depth look at the chi‑square concept, step‑by‑step solutions, and a variety of practice problems that mirror the style of the exam. By working through these examples, students can build confidence in applying the chi‑square test to real‑world biological scenarios Small thing, real impact..
Introduction
The chi‑square test is a statistical tool used to compare observed data with expected outcomes. In AP Biology, it frequently appears in questions about genetic crosses, phenotypic ratios, and population genetics. The test helps determine whether deviations from expected ratios are due to random chance or indicate a biological phenomenon such as linkage, incomplete dominance, or natural selection.
Key concepts to understand before tackling problems:
- Observed (O): The actual number of individuals counted in each category.
- Expected (E): The number predicted by a theoretical model (e.g., 9:3:3:1 for a dihybrid cross).
- Degrees of freedom (df): Number of categories – 1. For a single‑factor test, df = k – 1.
- Chi‑square statistic: (\chi^2 = \sum \frac{(O-E)^2}{E}).
- Critical value: Retrieved from a chi‑square distribution table at a chosen significance level (usually 0.05).
Steps to Perform a Chi‑Square Test
-
State the null hypothesis (H₀)
Example: The observed phenotypic ratios match the expected Mendelian ratios. -
Collect data
Record the observed counts for each category. -
Calculate expected counts
Multiply the total number of individuals by the theoretical proportion for each category. -
Compute the chi‑square statistic
Use the formula above, summing across all categories. -
Determine degrees of freedom
(df = k - 1), where k is the number of categories It's one of those things that adds up.. -
Find the critical value
Look up the chi‑square value for the chosen significance level and df. -
Make a decision
- If (\chi^2_{\text{calc}} < \chi^2_{\text{crit}}), fail to reject H₀.
- If (\chi^2_{\text{calc}} \ge \chi^2_{\text{crit}}), reject H₀.
Example Problem 1: Classic Dihybrid Cross
Question:
A dihybrid cross between two plants heterozygous for both traits (AaBb × AaBb) yields 360 offspring with the following phenotypic counts:
- 90 show both dominant traits (A‑B‑)
- 90 show dominant A only (A‑bb)
- 90 show dominant B only (aaB‑)
- 90 show both recessive traits (aabb)
Does the observed ratio fit the expected 9:3:3:1 ratio?
Solution:
| Category | Observed (O) | Expected Proportion | Expected (E) | ((O-E)^2/E) |
|---|---|---|---|---|
| A‑B‑ | 90 | 9/16 | 202.227 | |
| aaB‑ | 90 | 3/16 | 67.5 | 0.Consider this: 5 |
| A‑bb | 90 | 3/16 | 67.5 | 0.On the flip side, 227 |
| aabb | 90 | 1/16 | 22. 227 | |
| Total | 360 | 360 | **0. |
- df = 4 – 1 = 3
- Critical value (α = 0.05, df = 3) ≈ 7.815
Since (0.Consider this: 909 < 7. 815), fail to reject H₀. The data are consistent with the expected 9:3:3:1 ratio Not complicated — just consistent..
Example Problem 2: Linkage Analysis
Question:
A cross between two Arabidopsis plants heterozygous for two linked genes, gene X and gene Y, produces 200 progeny with the following counts:
| Phenotype | Observed |
|---|---|
| A‑B‑ | 70 |
| A‑B+ | 20 |
| A+B‑ | 30 |
| A+B+ | 80 |
Assuming independent assortment, the expected ratio would be 1:1:1:1. Does the data suggest linkage?
Solution:
| Category | O | E | ((O-E)^2/E) |
|---|---|---|---|
| A‑B‑ | 70 | 50 | 0.0 |
| A+B‑ | 30 | 50 | 8.Because of that, 8 |
| A‑B+ | 20 | 50 | 18. 0 |
| A+B+ | 80 | 50 | 18.0 |
| Total | 200 | 200 | **44. |
- df = 4 – 1 = 3
- Critical value (α = 0.05, df = 3) ≈ 7.815
Because (44.8 \gg 7.Still, 815), reject H₀. The large discrepancy indicates that the genes are linked, not assorting independently.
Example Problem 3: Population Genetics – Hardy–Weinberg
Question:
In a population of 500 birds, the observed genotype frequencies for a single gene are:
- AA: 260
- Aa: 170
- aa: 70
Test whether the population is in Hardy–Weinberg equilibrium (HWE) at the 0.05 significance level.
Solution:
-
Calculate allele frequencies
(p = \frac{2(260)+170}{2(500)} = \frac{690}{1000} = 0.69)
(q = 1 - p = 0.31) -
Expected genotype counts
- (E_{AA} = p^2 \times 500 = 0.4761 \times 500 = 238.05)
- (E_{Aa} = 2pq \times 500 = 0.4266 \times 500 = 213.30)
- (E_{aa} = q^2 \times 500 = 0.0961 \times 500 = 48.05)
-
Chi‑square statistic
| Genotype | O | E | ((O-E)^2/E) |
|---|---|---|---|
| AA | 260 | 238.Because of that, 30 | 9. Think about it: 05 |
| aa | 70 | 48. 86 | |
| Aa | 170 | 213.On the flip side, 05 | 9. 62 |
| Total | 500 | 500 | **21. |
- df = 3 – 1 = 2
- Critical value (α = 0.05, df = 2) ≈ 5.991
Since (21.991), reject H₀. 40 > 5.The population is not in Hardy–Weinberg equilibrium, suggesting forces such as selection, drift, or non‑random mating are at play.
Practice Problems
Below are ten practice problems, ranging from basic to advanced. Work through each one, then check your answer using the detailed solutions provided at the end.
| # | Problem Statement |
|---|---|
| 1 | A dihybrid cross (AaBb × AaBb) produces 200 progeny: 45 A‑B‑, 55 A‑bb, 50 aaB‑, 50 aabb. |
| 5 | A dihybrid cross (AaBb × AaBb) yields 500 progeny: 140 A‑B‑, 130 A‑bb, 120 aaB‑, 110 aabb. Expected ratio is 3:1. |
| 9 | A dihybrid cross (AaBb × AaBb) produces 600 progeny: 270 A‑B‑, 150 A‑bb, 120 aaB‑, 60 aabb. Test for HWE. |
| 6 | A bacterial colony shows 200 mutants: 150 drug‑resistant, 50 drug‑sensitive. Expected 1:1:1:1 if independent. Which means |
| 2 | A cross of two Drosophila strains yields 120 offspring: 30 show dominant eye color, 90 show recessive. |
| 4 | Two linked genes in a plant cross produce 400 progeny: 120 A‑B‑, 80 A‑B+, 100 A+B‑, 100 A+B+. |
| 3 | In a population of 300 plants, genotype counts are: AA = 180, Aa = 90, aa = 30. Expected ratio 3:1. Test for HWE. Test the 9:3:3:1 ratio. That's why |
| 10 | A population of 400 humans shows genotype frequencies for a disease allele: AA = 200, Aa = 140, aa = 60. |
| 8 | A cross between two heterozygous Arabidopsis plants for two genes (X and Y) yields 300 progeny: 90 X‑Y‑, 60 X‑Y+, 90 X+Y‑, 60 X+Y+. |
| 7 | In a population of 250 fish, genotypes are: AA = 100, Aa = 120, aa = 30. Test for HWE. |
Detailed Solutions
1. Dihybrid 9:3:3:1 Test
- Expected counts: 200 × (9/16) = 112.5; 200 × (3/16) = 37.5; 37.5; 37.5.
- Chi‑square ≈ 7.20.
- df = 3 → critical = 7.815.
- Result: Fail to reject H₀; data fit 9:3:3:1.
2. 3:1 Ratio Test
- Expected: 120 × 0.75 = 90; 120 × 0.25 = 30.
- Chi‑square ≈ 1.33.
- df = 1 → critical = 3.841.
- Result: Fail to reject H₀.
3. HWE Test
- Allele frequencies: p = 0.6, q = 0.4.
- Expected: AA 180, Aa 240, aa 80.
- Chi‑square ≈ 22.50.
- df = 2 → critical = 5.991.
- Result: Reject H₀; not in HWE.
4. Linkage Test
- Expected: 400 × 0.25 = 100 each.
- Chi‑square ≈ 24.00.
- df = 3 → critical = 7.815.
- Result: Reject H₀; genes are linked.
5. Dihybrid 9:3:3:1 Test
- Expected: 500 × 0.5625 = 281.25 for A‑B‑; 93.75 for each of the others.
- Chi‑square ≈ 4.60.
- df = 3 → critical = 7.815.
- Result: Fail to reject H₀.
6. 3:1 Ratio Test
- Expected: 200 × 0.75 = 150; 50.
- Chi‑square = 0.00.
- Result: Fail to reject H₀.
7. HWE Test
- Allele frequencies: p = 0.5, q = 0.5.
- Expected: AA 125, Aa 125, aa 25.
- Chi‑square ≈ 12.80.
- Result: Reject H₀.
8. Linkage Test
- Expected: 300 × 0.25 = 75 each.
- Chi‑square ≈ 27.20.
- Result: Reject H₀; linkage present.
9. Dihybrid 9:3:3:1 Test
- Expected: A‑B‑ 337.5; others 112.5.
- Chi‑square ≈ 6.10.
- Result: Fail to reject H₀.
10. HWE Test
- Allele frequencies: p = 0.75, q = 0.25.
- Expected: AA 225, Aa 150, aa 25.
- Chi‑square ≈ 24.00.
- Result: Reject H₀.
FAQ
Q1: When is chi‑square not appropriate?
A: If expected counts for any category are <5, the chi‑square approximation may be inaccurate. In such cases, use Fisher’s exact test Simple, but easy to overlook..
Q2: How do I remember the critical values?
A: For common df values, memorize the 0.05 critical values:
- df = 1 → 3.841
- df = 2 → 5.991
- df = 3 → 7.815
- df = 4 → 9.488
Q3: Can chi‑square be used for continuous data?
A: Chi‑square compares categorical frequencies. For continuous data, use tests like t‑test or ANOVA Small thing, real impact. No workaround needed..
Q4: What if my calculated chi‑square equals the critical value?
A: Treat it as rejecting the null hypothesis. The boundary is inclusive That's the part that actually makes a difference. That's the whole idea..
Conclusion
Mastering the chi‑square test equips AP Biology students to confidently analyze genetic data, detect linkage, and evaluate population genetics scenarios. By systematically applying the steps outlined—collecting data, computing expected counts, performing the calculation, and interpreting results—students can tackle any chi‑square problem that appears on the exam. Regular practice with diverse problems, as provided above, will sharpen skills and deepen understanding, ensuring readiness for the statistical challenges of AP Biology.
