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ap bio chi square practice problems

ap bio chi square practice problems

3 min read 12-01-2025
ap bio chi square practice problems

The chi-square (χ²) test is a crucial statistical tool in AP Biology, used to determine if observed data significantly differs from expected data. Mastering this test is vital for success on the AP exam. This guide provides practice problems and explanations to help you solidify your understanding. We'll focus on the goodness-of-fit test, the most common type used in introductory biology.

Understanding the Chi-Square Goodness-of-Fit Test

The goodness-of-fit test assesses how well observed data matches a hypothesized distribution. In simpler terms, it helps us determine if deviations from expected results are due to chance or a significant factor influencing the outcome. The test uses the following formula:

χ² = Σ [(Observed – Expected)² / Expected]

Where:

  • Observed: The number of individuals or occurrences in each category from your experiment.
  • Expected: The number of individuals or occurrences expected in each category based on your hypothesis.

The larger the χ² value, the greater the difference between observed and expected results, suggesting a lower probability that the difference is due to chance alone. We compare our calculated χ² value to a critical value from a chi-square distribution table, using degrees of freedom (df = number of categories – 1) and a chosen significance level (usually 0.05). If our calculated χ² exceeds the critical value, we reject the null hypothesis (that there's no significant difference).

AP Bio Chi-Square Practice Problems

Let's tackle some practice problems. Remember to show your work for each step!

Problem 1: Mendelian Genetics

You're performing a monohybrid cross of pea plants, expecting a 3:1 ratio of purple flowers (dominant) to white flowers (recessive). You obtain 72 purple and 28 white flowers. Does this data support the expected Mendelian ratio?

Solution:

  1. State the null hypothesis: The observed phenotypic ratio does not significantly differ from the expected 3:1 ratio.

  2. Calculate expected values: Total plants = 72 + 28 = 100. Expected purple: 100 * (3/4) = 75; Expected white: 100 * (1/4) = 25

  3. Calculate χ²:

    • Purple: [(72 – 75)² / 75] = 0.12
    • White: [(28 – 25)² / 25] = 0.36
    • χ² = 0.12 + 0.36 = 0.48
  4. Determine degrees of freedom: df = 2 (categories) – 1 = 1

  5. Find the critical value: Using a chi-square table with df = 1 and α = 0.05, the critical value is approximately 3.84.

  6. Compare and conclude: Our calculated χ² (0.48) is less than the critical value (3.84). Therefore, we fail to reject the null hypothesis. The observed data supports the expected 3:1 Mendelian ratio.

Problem 2: Hardy-Weinberg Equilibrium

You are studying a population of butterflies with two alleles for wing color: B (brown, dominant) and b (white, recessive). You observe 200 brown butterflies and 50 white butterflies. Does this population appear to be in Hardy-Weinberg equilibrium, assuming the expected allele frequencies are p=0.8 and q=0.2?

Solution:

  1. State the null hypothesis: The population is in Hardy-Weinberg equilibrium.

  2. Calculate expected genotype frequencies:

    • BB: p² = (0.8)² = 0.64
    • Bb: 2pq = 2*(0.8)*(0.2) = 0.32
    • bb: q² = (0.2)² = 0.04
  3. Calculate expected number of individuals: Total butterflies = 250

    • BB: 250 * 0.64 = 160
    • Bb: 250 * 0.32 = 80
    • bb: 250 * 0.04 = 10
  4. Since we can’t distinguish between BB and Bb visually we need to combine those two phenotypic categories: Observed Brown butterflies = 200 Expected Brown butterflies = 160 + 80 = 240 Observed White butterflies = 50 Expected White butterflies = 10

  5. Calculate χ²:

    • Brown: [(200 – 240)² / 240] ≈ 6.67
    • White: [(50 – 10)² / 10] = 160
    • χ² = 6.67 + 160 = 166.67
  6. Determine degrees of freedom: df = 2 – 1 = 1

  7. Find the critical value: The critical value remains 3.84 (α = 0.05, df = 1)

  8. Compare and conclude: Our calculated χ² (166.67) is far greater than the critical value (3.84). We reject the null hypothesis. The population is not in Hardy-Weinberg equilibrium.

These examples demonstrate how to apply the chi-square test in biological contexts. Remember to practice more problems and familiarize yourself with chi-square tables for different degrees of freedom and significance levels. Good luck with your AP Biology studies!

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