AP Biology Heredity — Worked Answer Explanations

Unit 5 · 12 questions explained

Below is a complete answer key for our AP Biology Heredity practice questions. For each question you'll find the correct choice, a full written explanation of how to get there, and — for every wrong answer — a short note on exactly why it's tempting and where it goes wrong. Reading these straight through is one of the fastest ways to find the gaps in a unit before exam day.

Prefer to test yourself first? Take the timed Heredity practice test and come back here to review, or head back to the Heredity unit overview.

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Question 1 · Easy
During meiosis, homologous chromosomes separate from each other during which phase?
- A
  Meiosis I, anaphase ICorrect
- B
  Meiosis II, anaphase II
  Why not B: Anaphase II separates sister chromatids, not homologous chromosomes.
- C
  Meiosis I, prophase I
  Why not C: Prophase I is when homologs pair (synapsis) and crossing over occurs, not when they separate.
- D
  Meiosis II, metaphase II
  Why not D: Metaphase II aligns single-chromatid chromosomes at the equatorial plate; separation of homologs already occurred in meiosis I.
Explanation
Meiosis I is the reductional division: homologous chromosome pairs (bivalents) align at the metaphase plate during metaphase I and are pulled toward opposite poles during anaphase I by shortening of kinetochore microtubules. This reduces the chromosome number from diploid (2n) to haploid (n). In contrast, meiosis II is equational — it separates sister chromatids, similar to mitosis.

Key takeaway
Homologous chromosomes separate at anaphase I of meiosis I, reducing ploidy from 2n to n.
Question 2 · Easy
In pea plants, round seed shape (R) is dominant over wrinkled (r), and yellow seed color (Y) is dominant over green (y). Two plants both heterozygous for both traits are crossed. What fraction of the offspring is expected to have round yellow seeds?
- A
  $\frac{1}{4}$
  Why not A: One-quarter represents a single phenotypic class in a monohybrid cross, not the dominant-dominant class in a dihybrid.
- B
  $\frac{9}{16}$ Correct
- C
  $\frac{3}{16}$
  Why not C: Three-sixteenths is the proportion for either dominant–recessive phenotypic classes, not the dominant–dominant class.
- D
  $\frac{1}{2}$
  Why not D: One-half is the dominant phenotype proportion in a monohybrid cross (Aa × Aa), not a dihybrid cross.
Explanation
In a dihybrid cross (RrYy × RrYy), each gene assorts independently (Law of Independent Assortment). The probability of round seeds (R_) = 3/4 and the probability of yellow seeds (Y_) = 3/4. Multiplying these independent probabilities: 3/4 × 3/4 = 9/16. The classic 9:3:3:1 phenotypic ratio confirms: 9/16 round yellow, 3/16 round green, 3/16 wrinkled yellow, 1/16 wrinkled green.

Key takeaway
In a dihybrid cross between double heterozygotes, 9/16 of offspring display both dominant phenotypes.
Question 3 · Easy
In humans, red-green color blindness is an X-linked recessive trait. A color-blind woman and a man with normal color vision have children. What is the expected phenotypic ratio for their sons?
- A
  All sons have normal color vision
  Why not A: Sons receive their only X chromosome from their mother; a color-blind mother (X^c X^c) gives all sons the X^c allele.
- B
  All sons are color blindCorrect
- C
  Half of the sons are color blind, half have normal vision
  Why not C: This would be true if the mother were a carrier (X^c X^+), not homozygous color blind.
- D
  No sons are color blind because the father's normal vision allele is passed on
  Why not D: Sons inherit their Y chromosome from their father; they do not receive the father's X chromosome.
Explanation
X-linked recessive traits follow the inheritance pattern where sons get their sole X chromosome from their mother. A color-blind woman is homozygous X^cX^c. The color-blind father's genotype is X^cY. All sons receive X^c from their mother and Y from their father, giving them genotype X^cY — they are all color blind (hemizygous, so the recessive allele is expressed). All daughters receive X^c from both parents and are therefore all carriers or color blind depending on father's X, but the question asks only about sons.

Key takeaway
Sons of a color-blind mother are all color blind because they receive her only X chromosome, which carries the recessive allele.
Question 4 · Easy
A plant breeder crosses two true-breeding plants: one with red flowers and one with white flowers. All F₁ plants have pink flowers. When F₁ plants are self-fertilized, the F₂ ratio is 1 red : 2 pink : 1 white. This pattern of inheritance is best explained by which of the following?
- A
  Complete dominance of the red allele over the white allele
  Why not A: Complete dominance would produce all red F₁ and a 3 red : 1 white F₂, not the 1:2:1 ratio observed.
- B
  Incomplete dominance, where the heterozygote shows an intermediate phenotypeCorrect
- C
  Codominance, where both alleles are fully expressed in the heterozygote
  Why not C: Codominance would show both red and white patches (not a blended pink) in F₁ heterozygotes.
- D
  Epistasis between two separate genes controlling flower color
  Why not D: Epistasis involves two separate genes; this cross produces a 1:2:1 ratio consistent with a single gene with two alleles showing incomplete dominance.
Explanation
Incomplete dominance occurs when neither allele is fully dominant, producing an intermediate phenotype in heterozygotes. Here, one allele (R) produces red pigment and the other (W or r) produces none, so heterozygotes (RW or Rr) produce half the normal pigment concentration, appearing pink. Self-fertilization of pink F₁ (Rr × Rr) yields: 1 RR (red) : 2 Rr (pink) : 1 rr (white) — a 1:2:1 phenotypic ratio that mirrors the genotypic ratio. This is the hallmark of incomplete dominance.

Key takeaway
Incomplete dominance produces an intermediate phenotype in heterozygotes and a 1:2:1 phenotypic ratio in F₂ crosses.
Question 5 · Medium
A researcher is studying two genes, A and B, in fruit flies. When the genes are on the same chromosome, they tend to be inherited together more often than expected under independent assortment. A cross yields 42% AB, 42% ab, 8% Ab, and 8% aB gametes. Which of the following best explains these results?
- A
  The genes assort independently because crossing over occurs between them at a frequency of exactly 50%
  Why not A: 50% recombination frequency produces equal parental and recombinant classes — that is the definition of independent assortment; the data show unequal frequencies.
- B
  The genes are linked and 16% recombination frequency indicates they are 16 cM apartCorrect
- C
  The genes are on different chromosomes and crossing over between them produces the recombinant gametes
  Why not C: Genes on different chromosomes assort independently (50% recombination); the observed 16% recombinant frequency indicates linkage.
- D
  Gene B is epistatic to gene A, suppressing its expression in recombinant classes
  Why not D: Epistasis affects phenotypic ratios, not gamete frequencies; this question is about recombination, not epistasis.
Explanation
Genetic linkage occurs when two genes are located on the same chromosome and tend to be inherited together. The recombination frequency (map distance) = (recombinant gametes) / (total gametes) × 100. Recombinant gametes are Ab and aB: 8% + 8% = 16%. Therefore, genes A and B are 16 cM (centimorgans) apart. Parental types (AB and ab) are more frequent because crossing over occurs between the loci only 16% of the time. If the genes were unlinked, each gamete class would be ~25%.

Key takeaway
Linked genes show recombination frequencies below 50%; the recombination frequency (in percent) equals the map distance in centimorgans.
Question 6 · Medium
A woman is heterozygous for a balanced chromosomal translocation between chromosomes 14 and 21. During meiosis, if the translocated chromosome 21 and the normal chromosome 21 both segregate to the same cell, which of the following describes the result in the offspring if that cell is fertilized by a normal sperm?
- A
  The offspring has monosomy 21 and is unaffected
  Why not A: Monosomy 21 would result from receiving fewer copies of chromosome 21 material, and monosomy for autosomes is typically lethal, not asymptomatic.
- B
  The offspring has an extra copy of chromosome 21 material (trisomy 21, Down syndrome)Correct
- C
  The offspring is a carrier of the translocation with no extra chromosome 21 material
  Why not C: A carrier receives the translocation chromosome but only one copy of chromosome 21 material; this scenario describes both normal 21 and translocated 21 going to the same cell.
- D
  The offspring has a deletion of chromosome 14 material
  Why not D: The question specifies the segregation of chromosome 21 copies; whether chromosome 14 material is affected depends on the specific translocation, but the key outcome here is trisomy 21 material.
Explanation
In a Robertsonian translocation between chromosomes 14 and 21, the long arm of chromosome 21 is attached to chromosome 14. The carrier has: the translocated chromosome (14q;21q), normal chromosome 14, and one normal chromosome 21 (45 chromosomes total). During meiosis, if the translocated chromosome (carrying 21q) and the normal chromosome 21 co-segregate into the same gamete, that gamete has two copies of chromosome 21 material. Fertilization by a normal sperm adds a third copy, resulting in trisomy 21 (Down syndrome). This is the mechanism of familial (translocation-based) Down syndrome, which is distinct from the more common nondisjunction form.

Key takeaway
Translocation carriers can produce gametes with extra chromosomal material, leading to offspring with trisomy even without traditional nondisjunction.
Question 7 · Medium
In a cross between two individuals who both carry recessive alleles for cystic fibrosis (Cc × Cc), what is the probability that their first three children are ALL unaffected?
- A
  $(\frac{1}{4})^{3} = \frac{1}{64}$
  Why not A: This is the probability of all three being affected (cc), not unaffected.
- B
  $(\frac{3}{4})^{3} = \frac{27}{64}$ Correct
- C
  $\frac{3}{4}$
  Why not C: Three-quarters is the probability that any single child is unaffected; for three independent children, probabilities must be multiplied.
- D
  $(\frac{1}{2})^{3} = \frac{1}{8}$
  Why not D: One-half is the carrier frequency, not the unaffected probability; unaffected includes both CC and Cc genotypes.
Explanation
From a Cc × Cc cross, the probability that any given child is unaffected (CC or Cc) = 3/4, and affected (cc) = 1/4. Because each child is an independent event, the probability that all three are unaffected is (3/4)³ = 27/64 ≈ 0.42. This uses the product rule, which states that the probability of two or more independent events all occurring equals the product of their individual probabilities.

Key takeaway
For multiple independent inheritance events, multiply individual probabilities (product rule); probability all three unaffected = (3/4)³ = 27/64.
Question 8 · Medium
A scientist is studying a gene located on the X chromosome. She crosses a wild-type female (X^+ X^+) with a mutant male (X^m Y). She then crosses the F₁ females with wild-type males. What phenotypic ratio would be expected in F₂ males?
- A
  All F₂ males are wild-type
  Why not A: F₁ females are carriers (X^+ X^m); half of their sons receive X^m, making them mutant.
- B
  1/2 wild-type : 1/2 mutant malesCorrect
- C
  All F₂ males are mutant
  Why not C: F₁ carrier females pass X^+ to half their sons and X^m to the other half.
- D
  3/4 wild-type : 1/4 mutant males
  Why not D: This ratio applies to autosomal dominant/recessive crosses; males are hemizygous for X-linked alleles.
Explanation
F₁ females are all carriers: X^+X^m (received X^+ from mother and X^m from father). When crossed with wild-type males (X^+Y): F₂ males can receive either X^+ (wild-type) or X^m (mutant) from the carrier mother, each with probability 1/2. Because males are hemizygous (XY), the mutant allele is always expressed if inherited. Therefore, F₂ males are 1/2 wild-type (X^+Y) and 1/2 mutant (X^mY). This classic reciprocal cross pattern reveals X-linked inheritance.

Key takeaway
X-linked recessive traits skip a generation in males when transmitted through a carrier female (criss-cross inheritance).
Question 9 · Medium
In chickens, feather color is controlled by two genes. Allele E (dominant) produces black feathers; allele e (homozygous recessive, ee) produces red feathers when at least one C allele is present. However, any bird homozygous recessive at the C locus (cc) is white regardless of the E locus. This is an example of recessive epistasis. When two white birds both with genotype EEcc are crossed with birds of genotype EECc, what fraction of offspring are expected to be white?
- A
  $\frac{3}{4}$
  Why not A: Three-quarters is the proportion of C_ genotypes (non-white if other conditions met), not the white proportion.
- B
  $\frac{1}{2}$ Correct
- C
  $\frac{1}{4}$
  Why not C: One-quarter is the proportion of cc from a Cc × Cc cross; here one parent is cc (all gametes carry c) so the cc proportion is higher.
- D
  $\frac{1}{8}$
  Why not D: One-eighth would arise only if both C loci and a third locus were segregating; here only the C locus is relevant for white color.
Explanation
The EE genotype is fixed in both parents (EEcc × EECc), so all offspring are EE — the E locus does not vary. Focus on the C locus: cc × Cc gives offspring: 1/2 Cc (colored) and 1/2 cc (white). All cc offspring are white because the C locus epistasis rule applies: cc → white regardless of E genotype. Therefore, exactly 1/2 of all offspring are white (genotype EEcc). This cross illustrates recessive epistasis, where homozygosity at one locus masks expression at another.

Key takeaway
In recessive epistasis, the epistatic locus (cc) masks color expression; cross cc × Cc gives 1/2 white (cc) offspring.
Question 10 · Hard
Genomic imprinting results in differential expression of a gene depending on whether it was inherited from the mother or the father. The Igf2 gene in mice is only expressed from the paternal chromosome. If an Igf2 deletion is inherited from the father, the offspring will lack Igf2 expression entirely. What would happen if the same deletion were inherited from the mother?
- A
  The offspring would also lack Igf2 expression because the deletion removes the coding sequence regardless of parent of origin
  Why not A: Imprinting means expression depends on parent of origin; the maternal allele is silenced regardless of its sequence, so deletion of a silenced allele has no additional effect on expression.
- B
  The offspring would have normal Igf2 expression because the paternal allele (which is expressed) is still intactCorrect
- C
  The offspring would overexpress Igf2 because losing maternal repression amplifies paternal expression
  Why not C: The maternal allele is already imprinted (silenced), not actively repressing the paternal allele; its deletion does not change paternal expression level.
- D
  The offspring would show the deletion phenotype in maternal tissues only
  Why not D: Imprinting is gene-specific and applies in all tissues, not selectively to maternal-derived cells.
Explanation
Genomic imprinting involves epigenetic silencing of one parental allele through mechanisms such as DNA methylation. For Igf2, the maternal allele is imprinted (epigenetically silenced), while the paternal allele is expressed. If the deletion is inherited from the mother, it affects only the already-silenced maternal copy — the expressed paternal allele is intact and unaffected. Therefore, the offspring has normal Igf2 expression. Only paternal deletion eliminates expression. This asymmetry in phenotypic consequence based on parent of origin is a hallmark of imprinted genes.

Key takeaway
For imprinted genes, the phenotypic effect of a mutation depends on parent of origin; a mutation in the already-silenced allele has no expression consequence.
Question 11 · Hard
A woman with blood type A (genotype I^A i) and a man with blood type B (genotype I^B i) have four children. Based on ABO blood type genetics, which of the following combinations of blood types is possible among their four children?
- A
  All four children have blood type AB
  Why not A: Only I^A I^B offspring have type AB; the cross can also produce type A (I^A i), type B (I^B i), and type O (ii) children.
- B
  Children with blood types A, B, AB, and O are all possibleCorrect
- C
  Only blood types A and B are possible because both parents have dominant alleles
  Why not C: Both parents carry the recessive i allele; ii offspring have blood type O.
- D
  Only blood types AB and O are possible in a 1:1 ratio
  Why not D: AB (I^A I^B) and O (ii) each have a 1/4 chance, but A (I^A i) and B (I^B i) are also equally likely at 1/4 each.
Explanation
ABO blood type is controlled by three alleles (I^A, I^B, i) at a single locus. I^A and I^B are codominant; i is recessive to both. Cross: I^A i × I^B i yields four equally probable genotypes: I^A I^B (blood type AB, 1/4), I^A i (blood type A, 1/4), I^B i (blood type B, 1/4), and ii (blood type O, 1/4). All four ABO blood types are possible in the offspring. The expected phenotypic ratio is 1 AB : 1 A : 1 B : 1 O.

Key takeaway
When both parents are heterozygous for ABO (I^A i × I^B i), all four blood types are possible in equal proportions.
Question 12 · Hard
Two genes, R and S, are located 20 cM apart on chromosome 3. An individual with genotype RS/rs (in coupling) mates with an rs/rs individual. A student predicts the offspring will show a 1:1 ratio of parental to recombinant phenotypes. Is the student correct, and what is the actual expected ratio of all four phenotypic classes?
- A
  Correct; 50% parental (RS and rs) and 50% recombinant (Rs and rS)
  Why not A: 50% recombinant frequency describes unlinked genes; 20 cM linkage means only 20% recombinant gametes.
- B
  Incorrect; expected ratio is 40% RS : 40% rs : 10% Rs : 10% rSCorrect
- C
  Incorrect; all offspring are parental (RS or rs) because genes on the same chromosome never recombine
  Why not C: Crossing over does occur between linked genes, just less frequently than 50%; 20 cM means 20% recombination.
- D
  Incorrect; expected ratio is 25% RS : 25% rs : 25% Rs : 25% rS because genes >10 cM apart assort randomly
  Why not D: Equal frequencies occur only when genes are unlinked (>50 cM or on different chromosomes); 20 cM linkage still shows preferential parental combinations.
Explanation
For linked genes 20 cM apart, crossing over occurs between them in 20% of meioses. The gametes from the RS/rs parent are: parental types RS (40%) and rs (40%), recombinant types Rs (10%) and rS (10%). In a testcross (RS/rs × rs/rs), offspring phenotype directly reflects gamete type from the heterozygous parent. Therefore: 40% RS (R+S+), 40% rs (r-s-), 10% Rs (R+s-), 10% rS (r-S+). The total recombinant frequency = 20%, matching the map distance. The student's prediction was wrong because 50% recombinant frequency only applies to unlinked genes.

Key takeaway
Map distance in cM equals the percent recombinant offspring; 20 cM linkage yields 40:40:10:10 parental-to-recombinant gamete frequency.