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AP Biology: Unit 5, Probability & Chi-Square in Genetics (Drill 20)

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About This Drill

AP Biology: Unit 5, Probability & Chi-Square in Genetics (Drill 20) is a practice drill. It contains 5 original questions created by Brian Stewart, a Barron's test prep author with over 20 years of tutoring experience.

Practice applying probability rules and chi-square analysis to genetics problems with this AP Biology drill. You will use the product rule and sum rule to calculate probabilities for genetic crosses, distinguish expected from observed ratios, and interpret chi-square test results to draw conclusions about inheritance patterns.

Passage

Probability is essential for predicting outcomes of genetic crosses. Two fundamental rules apply: Product Rule (AND rule): The probability that two independent events both occur equals the product of their individual probabilities. Example: P(boy AND type A blood) = P(boy) x P(type A) = 0.5 x 0.25 = 0.125 Sum Rule (OR rule): The probability that at least one of two mutually exclusive events occurs equals the sum of their individual probabilities. Chi-Square (χ²) Goodness-of-Fit Test: Used to determine whether observed data differ from expected ratios due to chance alone. Formula: χ² = Σ (O − E)² / E where O = observed count, E = expected count. Critical Values (df = number of classes - 1)
dfp = 0.05 critical value
13.84
25.99
37.82
49.49
If calculated χ² exceeds the critical value at p = 0.05, reject the null hypothesis. If below, fail to reject. Student Experiment: A Pp x Pp cross produced 400 offspring: 315 purple and 85 white plants.

Questions & Explanations

Question 1. For a monohybrid cross Aa x Aa, what is the probability of obtaining offspring with genotype Aa?

  • A) 1/4
  • B) 1/2 ✓
  • C) 3/4
  • D) 1

Explanation: Correct: (B) In a monohybrid cross Aa x Aa, the Punnett square gives genotypic classes: 1 AA : 2 Aa : 1 aa. Therefore P(Aa) = 2/4 = 1/2. Using the product rule as an alternative approach: P(Aa) = P(A from parent 1 AND a from parent 2) + P(a from parent 1 AND A from parent 2) = (1/2)(1/2) + (1/2)(1/2) = 1/4 + 1/4 = 1/2. Option C (3/4) is the frequency of the dominant phenotype, not the Aa genotype.

Question 2. Two genes on different chromosomes: Ee x Ee and Ff x Ff. Using the product rule, what is the probability of offspring with genotype EeFF?

  • A) 1/8 ✓
  • B) 1/4
  • C) 3/16
  • D) 1/16

Explanation: Correct: (A) Because the two genes are on different chromosomes, their inheritance is independent and probabilities can be multiplied. P(Ee) from Ee x Ee = 2/4 = 1/2. P(FF) from Ff x Ff = 1/4. Product rule: P(Ee AND FF) = 1/2 x 1/4 = 1/8. Option D (1/16) would be the probability of EEFF x FF, not EeFF. Option C (3/16) represents the probability of the E_ dominant phenotype combined with FF.

Question 3. In the student's experiment, a Pp x Pp cross produces 315 purple and 85 white plants out of 400 total. What are the expected numbers under a 3:1 ratio, and what is the calculated chi-square value?

  • A) Expected: 280 purple, 120 white; chi-square = 14.58
  • B) Expected: 300 purple, 100 white; chi-square = 0.75 (purple term only)
  • C) Expected: 300 purple, 100 white; chi-square = 2.25 (white term only)
  • D) Expected: 300 purple, 100 white; chi-square = (315-300)^2/300 + (85-100)^2/100 = 0.75 + 2.25 = 3.00 ✓

Explanation: Correct: (D) Under a 3:1 ratio with 400 total offspring: expected purple = 3/4 x 400 = 300, expected white = 1/4 x 400 = 100. Chi-square = (315-300)^2/300 + (85-100)^2/100 = 225/300 + 225/100 = 0.75 + 2.25 = 3.00. Notice that both classes must be included in the calculation; Options B and C each show only one term. The chi-square statistic sums the squared deviations for all classes.

Question 4. Using the critical value table, what conclusion should the student draw from chi-square = 3.00 for the flower color experiment?

  • A) Reject the null hypothesis; the data do not fit a 3:1 ratio because chi-square = 3.00 exceeds the critical value.
  • B) Fail to reject the null hypothesis; the data are consistent with a 3:1 ratio because chi-square = 3.00 is below the critical value of 3.84 for 1 degree of freedom. ✓
  • C) Reject the null hypothesis; the white class (85 observed vs. 100 expected) shows a deviation too large to be due to chance.
  • D) Fail to reject the null hypothesis, but only because the sample size of 400 is too small to draw any meaningful conclusion.

Explanation: Correct: (B) With 2 phenotypic classes (purple and white), df = 2 - 1 = 1. The critical value at p = 0.05 with 1 df is 3.84. The calculated chi-square = 3.00 is below this critical value. We fail to reject the null hypothesis: the data are consistent with a 3:1 ratio. The phrase "fail to reject" does not mean we have proven the 3:1 ratio is correct; it means we have insufficient statistical evidence to conclude the data deviate from the expected ratio. Option A incorrectly states that 3.00 exceeds 3.84.

Question 5. A different student obtained 270 purple and 130 white plants out of 400. The calculated chi-square is 10.67. What conclusion is appropriate?

  • A) Fail to reject the null hypothesis; 10.67 is below the critical value, so the data fit a 3:1 ratio.
  • B) Reject the null hypothesis; chi-square = 10.67 exceeds the critical value, confirming the trait must be X-linked.
  • C) Reject the null hypothesis; chi-square = 10.67 exceeds the critical value of 3.84 (1 df), so the data do not fit a 3:1 ratio. Possible explanations include selection against one phenotype, incorrect parental genotypes, or a non-Mendelian inheritance mechanism. ✓
  • D) Fail to reject the null hypothesis; the 3-to-1 ratio only applies when exactly 300 purple and 100 white plants are observed, and any other sample size or deviation from those exact counts would automatically disprove the expected Mendelian inheritance pattern.

Explanation: Correct: (C) With 1 df, the critical value is 3.84. Chi-square = 10.67 far exceeds this value, so we reject the null hypothesis. The observed 270:130 ratio does not fit the expected 3:1 ratio, and this deviation is unlikely due to chance alone. The chi-square test identifies that the deviation is statistically significant; it does not identify the biological cause. Option B incorrectly infers X-linkage from a chi-square result; that conclusion would require different data, such as differences in phenotype frequencies between sexes. A chi-square test cannot distinguish X-linkage from other explanations.