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AP Biology: Unit 5, Mendelian Inheritance (Drill 18)

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About This Drill

AP Biology: Unit 5, Mendelian Inheritance (Drill 18) 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 Mendel's laws to predict inheritance patterns with this AP Biology drill. You will use Punnett squares to calculate genotype and phenotype ratios, distinguish the law of segregation from the law of independent assortment, and analyze crosses involving both monohybrid and dihybrid scenarios.

Passage

Gregor Mendel's experiments with pea plants established the foundational principles of inheritance. Mendel's first law, the Law of Segregation, states that each organism carries two alleles for each trait, and these alleles separate during gamete formation so each gamete carries only one allele per gene. Mendel's second law, the Law of Independent Assortment, states that alleles of different genes assort into gametes independently of one another, provided the genes are located on different chromosomes. In standard Mendelian genetics, one allele is dominant (uppercase, e.g., A) and the other is recessive (lowercase, e.g., a). Dominant alleles mask recessive alleles in heterozygous genotype (Aa). Pea Plant Cross Data A genetics student crossed two pea plants heterozygous for seed color (Y = yellow, y = green) and seed shape (R = round, r = wrinkled). Cross: YyRr x YyRr Expected phenotypic ratio (dihybrid): 9 yellow round : 3 yellow wrinkled : 3 green round : 1 green wrinkled The student grew 320 offspring:
  • Yellow round: 181
  • Yellow wrinkled: 57
  • Green round: 62
  • Green wrinkled: 20
  • Total: 320

Questions & Explanations

Question 1. In the cross YyRr x YyRr, how many different genotypes are possible in the offspring, and what fraction of offspring is expected to be homozygous dominant for both traits (YYRR)?

  • A) 4 possible genotypes; 1/4 of offspring are YYRR
  • B) 16 possible genotypes; 1/9 of offspring are YYRR
  • C) 9 possible genotypes; 1/16 of offspring are YYRR ✓
  • D) 9 possible genotypes; 1/4 of offspring are YYRR

Explanation: Correct: (C) In a dihybrid cross YyRr x YyRr, each gene independently produces 3 genotypic classes (e.g., YY, Yy, yy for the Y gene), but the two genes combine to give 3 x 3 = 9 possible genotypes total. To find P(YYRR), apply the product rule: P(YY) x P(RR) = 1/4 x 1/4 = 1/16. This illustrates how independent assortment allows us to calculate probabilities for two genes by multiplying individual probabilities for each gene.

Question 2. The student's observed results are 181:57:62:20. The expected ratio is 9:3:3:1. Do these data support Mendel's laws, and what approach helps evaluate this statistically?

  • A) The data contradict Mendel's laws because the observed numbers are not in an exact 9:3:3:1 ratio.
  • B) The data are consistent with Mendel's laws; deviations from an exact 9:3:3:1 ratio are expected due to random sampling variation, and a chi-square test could determine whether deviations are statistically significant. ✓
  • C) The data support Mendel's laws only if the yellow round class shows exactly 180 plants.
  • D) The data contradict Mendel's laws because the green wrinkled class (20) is exactly as expected, indicating artificial selection rather than random assortment, independent assortment, or sampling error in the observed cross.

Explanation: Correct: (B) In any finite sample, observed counts will deviate from expected ratios due to random chance. Expected counts for 320 offspring at a 9:3:3:1 ratio: yellow round = 180, yellow wrinkled = 60, green round = 60, green wrinkled = 20. The observed values (181, 57, 62, 20) are very close to these expectations. A chi-square goodness-of-fit test compares observed to expected counts; if p > 0.05 we fail to reject the null hypothesis that the data fit a 9:3:3:1 ratio.

Question 3. A true-breeding YYRR plant is crossed with a true-breeding yyrr plant. All F1 offspring are YyRr. The F1 plants self-fertilize. Which of the following correctly describes the F2 generation?

  • A) The F2 generation will show a 9:3:3:1 ratio of yellow round : yellow wrinkled : green round : green wrinkled, demonstrating independent assortment. ✓
  • B) All F2 plants will be yellow round because dominant alleles were established in F1.
  • C) The F2 generation will show a 1:1:1:1 ratio of yellow round : yellow wrinkled : green round : green wrinkled.
  • D) The F2 generation will show only yellow round and green wrinkled offspring in a 3:1 ratio, because recombination cannot produce new phenotypic combinations.

Explanation: Correct: (A) This is a classic Mendelian dihybrid experiment. When F1 plants (YyRr) self-fertilize, independent assortment of the Y and R genes produces four phenotypic classes in a 9:3:3:1 ratio. The 9:3:3:1 ratio was one of Mendel's key results supporting the law of independent assortment. Option C (1:1:1:1) would result from a testcross (YyRr x yyrr), not from selfing. Option D describes the result expected only if the two genes were completely linked with no crossing over.

Question 4. A student proposes that seed color (Y gene) and seed shape (R gene) are located on the same chromosome. Which observation would be most inconsistent with this proposal, assuming no crossing over?

  • A) Yellow round offspring appearing in the F2 generation of a dihybrid cross
  • B) All F1 offspring of a cross between YYRR and yyrr parents being yellow and round
  • C) Green round offspring appearing in lower frequency than yellow round offspring in the F2 generation
  • D) A 9:3:3:1 phenotypic ratio appearing in the F2 generation of a dihybrid cross (YyRr x YyRr) ✓

Explanation: Correct: (D) If the Y and R genes were completely linked with no crossing over, gametes from a YyRr parent could only be YR or yr. Self-fertilization would yield offspring of only two phenotypes: yellow round (Y_R_) and green wrinkled (yyrr) in a 3:1 ratio. The appearance of yellow wrinkled and green round recombinant classes and the resulting 9:3:3:1 ratio are the hallmarks of independent assortment and directly contradict complete linkage. Option A (yellow round offspring) is entirely consistent with linkage. Option B (all F1 yellow round) is also consistent with either model.

Question 5. A cross between two pea plants produces 75 yellow and 25 green offspring. Which of the following best identifies the most likely parental genotypes?

  • A) YY x YY, because all offspring would be yellow if both parents are homozygous dominant
  • B) Yy x Yy, because a 3:1 (yellow:green) ratio results from a cross of two heterozygous parents ✓
  • C) YY x yy, because crossing a homozygous dominant with a homozygous recessive produces a mix of yellow and green offspring
  • D) Yy x yy, because a testcross produces green offspring at a significant frequency

Explanation: Correct: (B) The observed 75:25 ratio is very close to 3:1. A 3:1 phenotypic ratio is the expected result of a monohybrid cross between two heterozygous parents (Yy x Yy). Offspring genotypes are 1 YY : 2 Yy : 1 yy; since Y is dominant, YY and Yy are yellow (3/4) and only yy is green (1/4). A YY x yy cross (C) produces only yellow F1 offspring, all heterozygous Yy, with no green offspring. A Yy x yy testcross (D) produces a 1:1 ratio of yellow to green, not 3:1. A YY x YY cross (A) produces only yellow offspring.