Color blindness is an X-linked recessive trait. A color-blind male and a female with normal vision, whose father was color-blind, have a child. What is the probability that their first-born child will be a color-blind male?

Core Concept

PILLAR 1 — MOLECULAR/CONCEPTUAL MECHANISM

Step-by-Step Analysis

X-linked recessive inheritance arises from the asymmetric distribution of sex-determining chromosomes between human males and females. Males possess one X chromosome and one Y chromosome (XY), while females carry two X chromosomes (XX). The X chromosome is a large, gene-dense structure carrying hundreds of loci—including the OPN1LW and OPN1MW genes encoding long-wavelength (red) and medium-wavelength (green) cone opsin photopigments in the retina. The Y chromosome, by contrast, is gene-poor and lacks homologous copies of these X-linked loci. This chromosomal asymmetry creates hemizygosity in males: any allele present on a male's single X chromosome is phenotypically expressed because no second X chromosome exists to mask it. In females, diploid homologous pairing at these cone opsin loci permits dominant–recessive interactions. The wild-type OPN1LW/OPN1MW alleles produce functional cone opsins that embed in outer segment disc membranes via transmembrane α-helices, binding 11-cis-retinal as a chromophore cofactor. Mutant alleles—often arising from missense mutations altering critical nucleotide positions in the opsin coding sequence—produce misfolded, nonfunctional proteins that fail trafficking quality-control checkpoints in the endoplasmic reticulum of cone photoreceptor cells. A male inheriting a mutant X-linked opsin allele expresses complete red-green color deficiency because every retinal cone cell contains only that defective allele's product. A heterozygous female, possessing one functional and one mutant allele, retains sufficient functional opsin proteins for normal trichromatic vision through X-inactivation mosaicism, wherein roughly half her cone cells express the wild-type allele from one X and half express the mutant allele from her other X chromosome.

Why Other Options Are Wrong

PILLAR 2 — STEP-BY-STEP LOGIC

To solve this cross, we must deduce parental genotypes from the phenotypic information given. The father is color-blind, meaning his single X chromosome carries the mutant recessive allele (denoted Xᶜ Y). The mother has normal vision but her father was color-blind. Because a father passes his sole X chromosome to every daughter, this woman necessarily inherited her father's Xᶜ allele. Since she exhibits normal vision, her second X chromosome (inherited from her mother) must carry the dominant wild-type allele (X⁺). Her genotype is therefore X⁺Xᶜ—a heterozygous carrier.

We now perform a Punnett square analysis for the cross Xᶜ Y (father) × X⁺Xᶜ (mother). The father produces sperm carrying either Xᶜ or Y, each with probability 1/2. The mother produces ova carrying either X⁺ or Xᶜ, each with probability 1/2. The four equally probable offspring genotypes are: X⁺Xᶜ (carrier daughter, normal vision), XᶜXᶜ (color-blind daughter), X⁺Y (normal-vision son), and XᶜY (color-blind son). Each outcome has probability 1/4. The question asks specifically for the probability that the first-born child is a color-blind male—genotype XᶜY. This equals 1/4, or 25%.

PILLAR 3 — DISTRACTOR ANALYSIS

Option A (0) traps students who reason that because the father contributes his Xᶜ to all daughters and his Y to all sons, and because the question asks about a male child, the father's Xᶜ is irrelevant. These students fail to recognize that the mother—being a carrier—can contribute her paternally inherited Xᶜ allele to a son, who would then be hemizygous XᶜY and thus color-blind. This reflects a fundamental misunderstanding of carrier transmission in X-linked inheritance.

Option C (1/2) likely ensnares students who correctly identify that among male offspring, the probability of color blindness is 1/2 (XᶜY versus X⁺Y), but who fail to multiply by the additional 1/2 probability that the child is male in the first place. This error conflates conditional probability (given the child is male) with the joint probability (male AND color-blind) the question actually demands.

Option D (1/4) is the correct answer, representing the joint probability that the child is both male (1/2) and color-blind given male (1/2), yielding 1/2 × 1/2 = 1/4.

Option E (1) would represent certainty that every child is a color-blind male—an impossibility given that daughters (who inherit the mother's X⁺ on half of gamete fusions) and normal-vision sons (X⁺Y) are readily produced. This distractor exploits wholesale confusion about X-linked segregation patterns and probability calculations in Mendelian heredity.