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AP Biology: Unit 5, Pedigrees & Genetic Analysis (Drill 21)

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About This Drill

AP Biology: Unit 5, Pedigrees & Genetic Analysis (Drill 21) 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 reading and interpreting pedigrees to determine inheritance patterns with this AP Biology drill. You will identify autosomal vs. X-linked patterns, distinguish dominant from recessive inheritance, determine carrier genotypes, and calculate the probability that future offspring will be affected by a genetic condition.

Passage

A pedigree traces inheritance of a trait across multiple generations. Standard symbols: squares = males, circles = females; filled symbols = affected individuals; horizontal lines between individuals = mating; vertical lines = offspring. Key Patterns: Autosomal Recessive: Unaffected parents can have affected children; affects males and females equally; trait often skips generations; affected individuals are homozygous recessive (aa). Autosomal Dominant: At least one parent of an affected child is usually affected; trait appears every generation; affects males and females equally. X-Linked Recessive: Affects more males than females; carrier females pass trait to sons; affected fathers cannot pass to sons (daughters of affected father are carriers); trait often skips generations through female carriers. Recombination Frequency: Genes on the same chromosome can be separated by crossing over during meiosis I. Recombination frequency = (recombinant offspring / total offspring) x 100%. Genes with recombination frequency close to 50% behave as if unlinked. Genes with recombination frequency below 50% are linked. Pedigree for Analysis: Generation I: Unaffected father (I-1) x Unaffected mother (I-2) have four children. Generation II: Son II-1 (affected), Son II-2 (unaffected), Daughter II-3 (unaffected), Daughter II-4 (affected). Generation III: II-2 x unaffected female (II-5) have Son III-1 (affected) and Daughter III-2 (unaffected). Both sexes are affected in roughly equal proportions in the pedigree.

Questions & Explanations

Question 1. In the pedigree, two unaffected parents (I-1 and I-2) have an affected son (II-1) and an affected daughter (II-4). Which inheritance pattern is most consistent with these observations?

  • A) Autosomal dominant, because the trait appears in the first generation of offspring.
  • B) X-linked dominant, because both sexes are affected and the trait skips Generation I.
  • C) Autosomal recessive, because unaffected parents produce affected offspring of both sexes, indicating both parents are carriers of a recessive allele. ✓
  • D) X-linked recessive, because more males than females are affected in the pedigree.

Explanation: Correct: (C) The most informative clue is that two unaffected parents produce affected offspring. In autosomal dominant inheritance, at least one parent is typically affected. Since both parents are unaffected, autosomal dominant is essentially ruled out. The fact that both a son and a daughter are affected makes X-linked recessive less likely, since X-linked recessive conditions affect males at much higher rates. Autosomal recessive fits perfectly: both parents are heterozygous carriers (Aa), unaffected themselves, with a 1/4 chance of producing affected (aa) offspring of either sex.

Question 2. If the condition is autosomal recessive, what are the most likely genotypes of the unaffected parents (I-1 and I-2)?

  • A) Both I-1 and I-2 are Aa (heterozygous carriers), because only a carrier x carrier cross can produce affected (aa) offspring. ✓
  • B) I-1 = AA (homozygous dominant) and I-2 = Aa (heterozygous carrier).
  • C) Both I-1 and I-2 are AA (homozygous dominant), because they are both unaffected.
  • D) I-1 = Aa and I-2 = aa, but I-2 is unaffected because the recessive allele has incomplete penetrance.

Explanation: Correct: (A) For unaffected parents to produce an aa child, each parent must contribute one recessive allele. Since both parents are unaffected (not aa themselves), their genotypes must be Aa (heterozygous carriers). The cross Aa x Aa produces 1 AA : 2 Aa : 1 aa, giving a 1/4 probability of affected offspring. Option B (AA x Aa) cannot produce aa offspring, this cross yields 1 AA : 1 Aa, with no homozygous recessive children possible.

Question 3. The unaffected son II-2 (both parents are Aa carriers) marries an unaffected woman and has an affected son (III-1). Before knowing about his affected son, what was the probability that II-2 is a carrier (genotype Aa), given that he is unaffected?

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

Explanation: Correct: (B) From the cross Aa x Aa, offspring genotypes are 1/4 AA, 2/4 Aa, 1/4 aa. Among unaffected offspring (those who are AA or Aa), the conditional probability of being a carrier is: P(Aa | unaffected) = P(Aa) / P(unaffected) = (2/4) / (3/4) = 2/3. Among unaffected children of two carriers, 2 out of every 3 are expected to be carriers (Aa) and 1 out of 3 is expected to be homozygous dominant (AA). This conditional probability calculation is important in genetic counseling.

Question 4. A geneticist analyzes two genes in fruit flies on the same chromosome. A testcross (AaBb x aabb) produces 400 offspring: 170 AaBb, 170 aabb, 30 Aabb, and 30 aaBb. What is the recombination frequency between Gene A and Gene B?

  • A) 15% ✓
  • B) 30%
  • C) 50%
  • D) 7.5%

Explanation: Correct: (A) The parental (non-recombinant) types are AaBb (170) and aabb (170). The recombinant types, produced by crossing over during meiosis, are Aabb (30) and aaBb (30). Recombination frequency = (recombinant offspring / total offspring) x 100% = (30 + 30) / 400 x 100% = 60/400 x 100% = 15%. Genes A and B are therefore approximately 15 centimorgans (map units) apart. A recombination frequency less than 50% confirms the genes are linked on the same chromosome.

Question 5. A woman is a confirmed carrier (Aa) of an autosomal recessive condition. Her partner is unaffected; his mother was a carrier (Aa) and his father was homozygous dominant (AA). What is the probability their first child will be affected (aa)?

  • A) 1/4, because when one parent is confirmed to be a carrier the risk to offspring is always 1/4.
  • B) 1/6, because the probability the man is a carrier given he is unaffected is 2/3 and the probability of an affected child if both parents are Aa is 1/4, so P = 2/3 x 1/4 = 1/6.
  • C) 1/4, because the woman is definitely a carrier so the risk does not depend on the man's genotype.
  • D) 1/8, because the man has a 1/2 probability of being a carrier (from an Aa x AA cross), and if both parents are carriers the probability of an affected child is 1/4; combined probability = 1/2 x 1/4 = 1/8. ✓

Explanation: Correct: (D) This is a two-step probability calculation. The man's mother was Aa and his father was AA. From the cross Aa x AA, offspring are 1/2 AA and 1/2 Aa. Since the man is unaffected and his father is AA (not a carrier), both AA and Aa are consistent with being unaffected, and we do not need to apply a conditional adjustment, being unaffected provides no new information when the father is AA. Therefore P(man is Aa) = 1/2. P(affected child | both parents Aa) = 1/4. Combined: 1/2 x 1/4 = 1/8. Option B incorrectly applies the 2/3 conditional probability, which would apply if the man's father were also Aa (as in Q3 above), not AA.