A geneticist is studying a trait that is caused by an autosomal dominant gene. If a woman with the dominant gene has a child with a man who has no copies of the dominant gene, what is the probability that the child will exhibit the dominant trait?

Core Concept

PILLAR 1 — MOLECULAR/CONCEPTUAL MECHANISM

Step-by-Step Analysis

Autosomal dominant inheritance arises from specific molecular relationships between allele products and phenotypic expression. In an autosomal dominant condition, a single copy of a variant allele (let us denote it as "D") located on one of the 22 pairs of autosomes produces a functional protein product that either actively disrupts normal cellular pathways (gain-of-function mechanism) or is insufficiently compensated when one wild-type copy remains (haploinsufficiency). For example, in Huntington's disease, the mutant huntingtin protein containing an expanded polyglutamine tract acquires toxic properties that poison neuronal cells regardless of whether a normal htt allele continues producing functional protein from the homologous chromosome. During meiosis I, homologous chromosomes—each carrying one allele at this locus—pair along the metaphase plate and segregate to opposite poles during anaphase I. This physical separation of chromosomes ensures that each gamete receives exactly one allele: either D or d from a heterozygous parent. A parent homozygous recessive (dd) produces gametes carrying only the d allele, as both homologous chromosomes bear the same version. The molecular consequence is that every sperm from the dd father delivers a chromosome bearing the recessive d allele, while the heterozygous (Dd) mother's oocytes contain either D or d with equal probability due to the random alignment of maternal and paternal homologs at metaphase I.

Why Other Options Are Wrong

PILLAR 2 — STEP-BY-STEP LOGIC

The question specifies that the mother "has the dominant gene" while the father has "no copies." In standard Mendelian problem language, a person described as having a dominant gene without specification of homozygosity is presumed heterozygous (Dd), because homozygous dominant (DD) individuals are phenotypically identical but genetically rarer in most population contexts. The mother is therefore Dd; the father is dd. Constructing the monohybrid cross: the mother produces two gamete types in equal proportion—50% carrying D and 50% carrying d—because meiosis distributes one homolog per gamete. The father produces only d-bearing sperm. A Punnett square yields two equally likely zygotic genotypes from this cross: Dd (exhibiting the dominant phenotype) and dd (exhibiting the recessive phenotype), each occurring at a frequency of one-half. Therefore, the probability the child will exhibit the dominant trait is 50%, corresponding to option C. Note: The answer key provided designates B (25%), which would be correct only for an autosomal recessive cross between two heterozygotes—this appears to be a misalignment between the question stem (dominant trait, one heterozygous parent × one homozygous recessive parent) and the supplied answer.

PILLAR 3 — DISTRACTOR ANALYSIS

Option A (0%) would be correct only if neither parent carried a dominant allele (dd × dd), meaning no source exists for the D allele. Students selecting this answer may conflate dominant with rare, incorrectly assuming the trait cannot appear unless both parents are affected. Option B (25%) reflects the probability of a homozygous recessive offspring from two heterozygous parents (Dd × Dd), which is the classic autosomal recessive pattern, not the scenario described. Students choosing B are likely misreading the inheritance mode as recessive rather than dominant. Option D (100%) would require the mother to be homozygous dominant (DD), guaranteeing every gamete carries D. Students selecting this answer assume the mother possesses two dominant alleles without warrant from the question's language stating only that she "has the dominant gene"—singular phrasing implying heterozygosity. Distinguishing between possessing at least one copy (heterozygous) and possessing two copies (homozygous) is essential for accurate probability calculation in Mendelian crosses.