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AP Biology Unit 5 Practice Questions: Heredity

10 original exam-style questions on Heredity. Answer each one to see the explanation — no account needed.

Question 1 of 10 · Chi-Square Analysis in Genetics

A genetics student performed a series of crosses to investigate the inheritance of two traits in pea plants: seed color (yellow, Y dominant over green, y) and seed texture (round, R dominant over wrinkled, r). The student crossed two true-breeding strains to obtain an F1 dihybrid, then performed a testcross with a homozygous recessive plant. The progeny counts and chi-square analysis are shown below.

Testcross Progeny: F1 (YyRr) × yyrr
PhenotypeObserved (O)Expected (E)(O−E)2/E(O - E)^2 / E
Yellow, Round2182001.62
Yellow, Wrinkled1862000.98
Green, Round1922000.32
Green, Wrinkled2042000.08
**Total**800800χ2=3.00\chi^2 = 3.00

Critical value for chi-square at p=0.05p = 0.05 with 3 degrees of freedom: χcrit2=7.815\chi^2_{crit} = 7.815

Yellow, round peas appeared 218 times but were expected 200 times. Which of the following provides the strongest justification for concluding that this excess is due to chance rather than genetic linkage between the two genes?
  1. A. The 218 observed offspring represent less than 5% deviation from the expected 200, which is always considered biologically insignificant regardless of sample size
  2. B. The four offspring classes each deviate by fewer than 20 individuals, and small absolute deviations confirm independent assortment in all cases
  3. C. The yellow-round class exceeds its expectation, but the chi-square test cannot detect linkage when the sample size exceeds 100 individuals
  4. D. The total chi-square value of 3.00 falls below the critical value of 7.815 at p=0.05p = 0.05, so deviations from the expected 1:1:1:1 ratio are not statistically significant
Show answer and explanation

Correct answer: D

The chi-square test formalizes whether observed deviations exceed what chance alone would produce; because χ2=3.00<χcrit2=7.815\chi^2 = 3.00 < \chi^2_{crit} = 7.815, we fail to reject the null hypothesis of independent assortment, and the excess in yellow-round offspring is attributed to sampling variation. Choice A is incorrect because percent deviation alone is not the proper statistical criterion — the chi-square test accounts for sample size, which a simple percentage comparison ignores.

Question 2 of 10 · Pedigree Analysis

The pedigree below shows the inheritance of a condition called hypophosphatemia (vitamin D-resistant rickets) across three generations of a family. Affected individuals are shown in filled symbols.

Generation II-1I-2Generation IIII-1II-2II-3II-4Generation IIIIII-1III-2III-3III-4
Based on the pedigree, which pattern of inheritance is most consistent with the transmission of hypophosphatemia?
  1. A. Autosomal recessive, because affected individuals skip generations.
  2. B. X-linked recessive, because only females are affected across all generations.
  3. C. Autosomal dominant, because the trait appears in every generation in both sexes.
  4. D. X-linked dominant, because affected females pass the trait to daughters but not sons.
Show answer and explanation

Correct answer: D

Hypophosphatemia is actually X-linked dominant: affected female I-2 passes the trait to daughter II-1 but not to sons; II-1 passes it to daughter III-1 but not III-2. Only females are affected, which is consistent with an X-linked dominant allele that is lethal in hemizygous males (or simply not expressed in males here). Autosomal dominant would be expected to affect males as well.

Question 3 of 10 · Meiosis: Crossing Over and Genetic Recombination

During meiosis, homologous chromosomes pair and exchange segments in a process called crossing over. Which of the following correctly identifies when crossing over occurs and what structure facilitates it?
  1. A. Crossing over occurs during meiosis II prophase and is facilitated by the synaptonemal complex that holds sister chromatids together
  2. B. Crossing over occurs during meiosis I prophase (prophase I) and is facilitated by the synaptonemal complex that holds homologous chromosomes together
  3. C. Crossing over occurs during interphase just before meiosis begins and requires only DNA polymerase to copy segments between chromosomes
  4. D. Crossing over occurs during meiosis I metaphase when homologs are aligned at the metaphase plate and can exchange segments under tension
Show answer and explanation

Correct answer: B

During prophase I of meiosis, homologous chromosomes undergo synapsis — the synaptonemal complex forms between them, holding homologs in close apposition and facilitating reciprocal exchange of chromatid segments (chiasmata formation) by the RecA homolog Dmc1/Rad51. Choice A is incorrect because crossing over involves homologs (not sister chromatids) and occurs in meiosis I, not meiosis II.

Question 4 of 10 · Independent Assortment and Gamete Diversity

In meiosis, independent assortment of homologous chromosome pairs at meiosis I metaphase contributes to genetic diversity in offspring. For an organism with a haploid number of n=4n = 4, how many unique chromosome combinations can be produced in gametes through independent assortment alone (ignoring crossing over)?
  1. A. 44 combinations, because each chromosome pair can only align in one of two orientations at the metaphase plate
  2. B. 88 combinations, because n=4n = 4 haploid chromosomes each have two possible orientations giving 4×2=84 \times 2 = 8
  3. C. 1616 combinations, because 2n=24=162^n = 2^4 = 16 unique gamete types are produced by independent assortment of 4 chromosome pairs
  4. D. 256256 combinations, because 444^4 represents the orientation choices for 4 pairs of chromosomes
Show answer and explanation

Correct answer: C

For an organism with nn pairs of homologous chromosomes, independent assortment generates 2n2^n possible gamete chromosome combinations; with n=4n = 4 pairs, 24=162^4 = 16 combinations. Choice B confuses nn (number of pairs) with nn (the number of individual chromosomes), incorrectly applying n×2n \times 2 instead of 2n2^n.

Question 5 of 10 · Molecular Genetics and Heredity

Researchers used gel electrophoresis to analyze DNA fragments from four individuals. Each individual was tested for a restriction fragment length polymorphism (RFLP) associated with sickle cell anemia. The restriction enzyme cuts the normal allele (HbAH^bA) at a specific site but does not cut the sickle allele (HbSH^bS). Fragments were separated and stained.

Gel Electrophoresis — RFLP Analysis— (negative pole, top) → fragments migrate toward positive pole (bottom) +LadderIndiv. 1Indiv. 2Indiv. 3Indiv. 41.3 kb1.1 kb0.8 kb0.5 kbNote: Normal allele cut into 1.1 kb + 0.5 kb fragments; sickle allele uncut = 1.3 kb band
Based on the gel results, which individual is most likely a carrier (heterozygous) for sickle cell anemia?
  1. A. Individual 1, because it shows only the cut fragments at 1.1 kb and 0.5 kb.
  2. B. Individual 2, because it shows only the uncut 1.3 kb fragment.
  3. C. Individual 3, because it shows both the 1.3 kb uncut band and the 1.1 kb + 0.5 kb cut fragments.
  4. D. Individual 4, because it shows bands at 1.1 kb and 0.5 kb, indicating homozygous normal.
Show answer and explanation

Correct answer: C

A carrier has one normal allele (cut into 1.1 kb and 0.5 kb) and one sickle allele (uncut, 1.3 kb), producing all three bands. Individual 3 displays exactly this pattern. Individual 1 and 4 are homozygous normal (only cut fragments), and Individual 2 is homozygous sickle (only uncut band).

Question 6 of 10 · X-Linked Inheritance: Carrier Analysis

In humans, color blindness is caused by an X-linked recessive allele (XbX^b). A woman with normal vision whose father was color blind marries a man with normal vision. What is the probability that their first son will be color blind?
  1. A. 0%, because the father has normal vision and does not contribute the recessive allele
  2. B. 25%, because only half of sons can inherit the recessive allele from the carrier mother, and sons have a 50% probability overall
  3. C. 50%, because the mother is a confirmed carrier and all sons inherit their single X chromosome from their mother
  4. D. 100%, because the color blind grandfather guarantees all maternal X chromosomes carry the recessive allele
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Correct answer: C

The woman's father was color blind (XbYX^b Y), so she must have inherited XbX^b from him, making her an obligate carrier (XBXbX^B X^b). Each son receives his X chromosome from the mother — with probability 12\frac{1}{2} of receiving XbX^b — and his Y from the father, giving a 50% probability of color blindness. Choice D is incorrect because the woman is heterozygous (XBXbX^B X^b), not homozygous recessive.

Question 7 of 10 · Pedigree Analysis

A geneticist studies a rare autosomal recessive disorder in a family. The pedigree below summarizes three generations. Shaded individuals are affected; circles represent females, squares represent males.

I 1 2 3 4 II 1 2 3 III 1 2 3
Based on the pedigree, which genotypes are most likely correct for individuals I-1 and II-3? (Use 'A' for dominant allele and 'a' for recessive allele.)
  1. A. I-1 is aa; II-3 is Aa
  2. B. I-1 is AA; II-3 is Aa
  3. C. I-1 is Aa; II-3 is AA
  4. D. I-1 is Aa; II-3 is Aa
Show answer and explanation

Correct answer: D

Because II-2 is affected (aa), both of his parents (I-1 and I-2) must each carry at least one 'a' allele, making I-1 Aa. II-3 is unaffected yet has an affected child (III-2, genotype aa), so she must be a carrier: Aa. Option C is wrong because an AA individual cannot produce an aa offspring when mated with an aa partner.

Question 8 of 10 · Incomplete Dominance and Non-Mendelian Inheritance

In snapdragons, flower color exhibits incomplete dominance: CRCRC^R C^R plants have red flowers, CWCWC^W C^W plants have white flowers, and CRCWC^R C^W plants have pink flowers. If two pink-flowered plants are crossed, which of the following correctly describes the expected offspring ratio?
  1. A. All offspring will be pink, because the CRC^R and CWC^W alleles blend permanently in the offspring
  2. B. 3 red : 1 white, because CRC^R is still dominant and the recessive white phenotype appears only in homozygous offspring
  3. C. 1 red : 2 pink : 1 white, because heterozygotes produce an intermediate phenotype and alleles segregate normally
  4. D. 1 red : 1 pink : 1 white : 1 colorless, because four phenotypic classes result from combining three alleles
Show answer and explanation

Correct answer: C

Pink × pink (CRCW×CRCWC^R C^W \times C^R C^W) yields 14CRCR\frac{1}{4} C^R C^R (red) : 24CRCW\frac{2}{4} C^R C^W (pink) : 14CWCW\frac{1}{4} C^W C^W (white) — a 1:2:1 phenotypic ratio characteristic of incomplete dominance where alleles do not blend but each contributes additively to phenotype. Choice B applies complete dominance ratios incorrectly to an incomplete dominance scenario.

Question 9 of 10 · Meiosis I: Prophase I and Crossing Over

Which of the following correctly identifies a key difference between prophase I of meiosis and prophase of mitosis?
  1. A. In prophase I, the nuclear envelope remains intact throughout the entire phase, whereas in mitotic prophase the nuclear envelope breaks down early
  2. B. In prophase I, homologous chromosomes undergo synapsis to form bivalents and crossing over may occur, whereas in mitotic prophase homologs do not pair
  3. C. In prophase I, chromosomes are unreplicated so only one chromatid is present per chromosome, whereas mitotic prophase begins with replicated chromosomes
  4. D. In prophase I, the spindle apparatus forms in the cytoplasm but does not contact chromosomes until metaphase II, whereas in mitosis spindle fibers attach in prophase
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Correct answer: B

Prophase I is distinguished by synapsis — homologous chromosomes pair along their entire lengths within the synaptonemal complex — and by crossing over between non-sister chromatids, which generates recombinant chromosomes; these events do not occur in mitotic prophase, where homologs remain unpaired. Choice A is incorrect because the nuclear envelope breaks down in both prophase I and mitotic prophase.

Question 10 of 10 · Incomplete Dominance

In snapdragons, red flower color (R) and white flower color (r) show incomplete dominance. When red-flowered plants are crossed with white-flowered plants, the F1 generation is entirely pink. If F1 plants are crossed with each other, what is the expected F2 phenotype ratio?
  1. A. 3 red : 1 white, following standard Mendelian dominance
  2. B. 1 red : 1 pink : 1 white : 1 other color
  3. C. 1 red : 2 pink : 1 white, reflecting blended intermediate phenotypes
  4. D. All pink, because blending is permanent once it occurs
Show answer and explanation

Correct answer: C

With incomplete dominance, Rr (pink) × Rr (pink) produces RR (red) : 2 Rr (pink) : rr (white) = 1:2:1 phenotype ratio. The genotype ratio is also 1:2:1 because each phenotype is distinct from the others. The 3:1 ratio applies only to complete dominance where Rr and RR look identical.

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