Concept

Sex linkage

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Statement

Inheritance of genes carried on the sex chromosomes.

Why it matters

law-of-segregation and independent-assortment describe how alleles behave when genes sit on different, freely-assorting chromosomes; sex linkage is the first major exception the introductory course meets, because the X and Y chromosomes are not a matched, interchangeable pair. A gene carried on the X chromosome is transmitted, and expressed, on a fundamentally different schedule in the two sexes, and recognising that schedule is what lets a pedigree be read correctly — distinguishing an X-linked recessive trait from an ordinary autosomal recessive trait long before any molecular test is available.

The pattern also matters clinically: human conditions including red-green colour blindness, haemophilia A, and Duchenne muscular dystrophy are all X-linked recessive, and their characteristic excess of affected males, with unaffected carrier mothers, is a direct, everyday consequence of the mechanism proved below.

Hypotheses

Sex is determined by a heteromorphic chromosome pair (XX in one sex, XY in the other, as in mammals and Drosophila), and the two sex chromosomes segregate at meiosis exactly as law-of-segregation requires of any homologous pair.The X and Y differ substantially in size and gene content — the human Y carries only a few dozen genes against several hundred to a thousand-plus on the X — so an individual with one X and one Y is not simply "heterozygous" at every X-linked locus; that individual has only one copy of most X-linked genes to begin with. A male (XY) is hemizygous for X-linked genes: he carries exactly one allele at each such locus, not two.With no second X-linked allele present to mask it, any allele a male carries on his single X, dominant or recessive, is expressed in his phenotype. A female, carrying two X chromosomes, can be heterozygous and have a recessive allele masked by a dominant one, exactly as at an ordinary autosomal locus. The X and Y share small pseudoautosomal regions that recombine at meiosis like ordinary autosomal segments.Genes located in these regions are physically on the sex chromosomes but are not sex-linked in the inheritance sense developed here — they behave, transmission-wise, like autosomal genes, because both sexes end up with two functional copies. True sex linkage applies only to the much larger non-recombining portion of the X (and, separately and much more rarely, to genes unique to the Y).

Proof

\[ \text{At meiosis, a male transmits his X to every daughter and his Y to every son; a female transmits one of her two X's to every offspring, regardless of its sex.} \]

This follows directly from standard meiotic segregation (law-of-segregation) applied to the sex-chromosome pair itself: a son necessarily receives the father's Y (the only way to become XY) and the mother's contributed X; a daughter necessarily receives the father's X and the mother's contributed X. A

\[ \text{A recessive X-linked allele is phenotypically expressed in every male that carries it (hemizygosity), but only in females homozygous for it.} \]

Directly from the Hypotheses: a male has no second X-linked allele available to mask a recessive one, while a female's second X can carry the dominant allele and mask it, exactly as in ordinary autosomal dominance. A heterozygous female is an unaffected carrier. A

\[ \text{Consequently, an X-linked recessive trait appears far more often in males than in females of the same population.} \]

Any male carrying a single copy of the recessive allele is affected (Step 2); a female requires two independently-inherited copies, a rarer event whenever the allele's frequency in the population is low, exactly as Hardy–Weinberg predicts for any recessive allele at low frequency — but here compounded further because males need only one copy, not two. A

\[ \text{Reciprocal crosses give different results for an X-linked gene, unlike an autosomal gene.} \]

Cross a homozygous recessive female (\(X^aX^a\)) to a wild-type male (\(X^AY\)): all sons are \(X^aY\) (affected, inheriting the mother's only available X), all daughters are \(X^AX^a\) (unaffected carriers, inheriting the father's dominant \(X^A\)) — a "criss-cross" pattern. The reverse cross (wild-type female × affected male) gives a different outcome entirely (Worked examples). This reciprocal-cross asymmetry, absent for any autosomal gene where the two reciprocal crosses are equivalent, is the diagnostic signature of X-linkage. A

\[ \text{A male never transmits an X-linked allele to a son.} \]

By Step 1, a father contributes only his Y chromosome to a son; his X-linked genotype is therefore passed exclusively to his daughters, every one of whom is an obligate recipient of it. This "no father-to-son transmission" rule is a second, independent diagnostic test usable directly on a pedigree. A

Result

\[ X^{A}X^{a}\ (\text{carrier}) \times X^{A}Y \ \Rightarrow\ \tfrac14 X^AX^A,\ \tfrac14 X^AX^a,\ \tfrac14 X^AY,\ \tfrac14 X^aY \]

Reading. A carrier mother crossed to an unaffected father produces daughters that are all phenotypically unaffected (half carriers) but sons that are affected with probability one half — a 1:1 split among sons that has no autosomal-recessive analogue at this allele frequency, and that immediately signals X-linkage on inspection of a pedigree.

Scope. Applies to genes on the non-recombining portion of the X in any XX/XY system (Hypotheses); does not apply to pseudoautosomal genes, and reverses in identity (not in mechanism) for ZW systems such as birds and butterflies, where the female is the heterogametic sex.

Corollaries & converses

An X-linked dominant allele shows the opposite skew: every daughter of an affected father is affected (she must receive his single, dominant-bearing X), while no son of an affected father is affected (Step 5) — a distinct, recognisable pedigree pattern from the recessive case worked above.
Y-linked (holandric) genes, transmitted father-to-son exclusively and never to daughters, are the limiting case of Step 1 applied to the Y rather than the X; because the human Y carries very few genes outside sex determination itself, clinically significant holandric traits are rare.
linkage-recombination extends this same logic to genes that lie near each other on an ordinary autosome: sex-linkage is, in effect, the special case of complete linkage between a gene and the sex-determining chromosome itself, rather than partial linkage between two autosomal loci.

Fails without

Drop male hemizygosity (Hypotheses): if males instead carried two functional X-linked alleles like females, a recessive allele would be masked in males exactly as it is in females, erasing both the male-excess pattern of Step 3 and the reciprocal-cross asymmetry of Step 4 — the entire diagnostic signature of sex linkage depends specifically on males having only one copy to express.
Ignore the pseudoautosomal-region caveat (Hypotheses, Tier 3): a gene located in a pseudoautosomal region recombines between X and Y at meiosis and effectively behaves as autosomal, so both sexes end up with two recombining copies; treating such a gene's inheritance data as if it must show the criss-cross pattern of Step 4 gives a false negative for linkage and can lead to misclassifying a genuinely sex-chromosome-borne gene as autosomal.

Common errors

Describing a male's genotype at an X-linked locus with two allele symbols (e.g. "\(X^AX^a\)"), as though he were heterozygous; a male is hemizygous and carries exactly one allele at that locus (Hypotheses).
Assuming any trait more common in males than females must be Y-linked; the overwhelming majority of clinically relevant sex-linked human conditions (colour blindness, haemophilia, Duchenne muscular dystrophy) are X-linked recessive, not Y-linked (Discussion).
Assuming that because a father is affected by an X-linked recessive condition, his sons must also be affected; by Step 5, a father's X-linked genotype is never transmitted to sons at all, only to daughters (who become obligate carriers, not necessarily affected, unless the mother also contributes a recessive allele).
Treating an X-linked recessive pedigree and an ordinary autosomal recessive pedigree as diagnostically identical; the reciprocal-cross asymmetry of Step 4 and the male-excess pattern of Step 3 are absent from autosomal inheritance and are exactly what distinguishes the two on inspection.

Discussion

Thomas Hunt Morgan's 1910 discovery of the white-eye mutation in Drosophila melanogaster was the first direct experimental demonstration that a specific gene resides on a specific, identifiable chromosome, and it did so precisely by exhibiting the reciprocal-cross asymmetry proved in Step 4: crossing a white-eyed male to red-eyed females gave all red-eyed offspring in the F1, but the reciprocal cross (white-eyed females to red-eyed males) gave red-eyed daughters and white-eyed sons, tracking the X chromosome exactly. The result provided crucial early support for the chromosomal theory of inheritance that Walter Sutton and Theodor Boveri had proposed a few years earlier on cytological grounds alone.

In female mammals, only one of the two X chromosomes in each cell is transcriptionally active; the other is largely silenced by X-inactivation, a process established early in embryonic development and then clonally inherited by all descendant cells (discovered by Mary Lyon, hence "Lyonization"). Because the choice of which X to inactivate is essentially random cell-by-cell, a heterozygous female is a genetic mosaic, with different patches of tissue expressing one X-linked allele or the other — visible directly in phenomena such as the patchy coat colour of tortoiseshell cats, which are always female for exactly this reason.

Common misconception: that "sex-linked" is synonymous with "X-linked recessive." Sex linkage covers X-linked dominant and Y-linked (holandric) inheritance equally, each with its own distinct pedigree signature (Corollaries); X-linked recessive simply happens to be the pattern behind the most familiar human examples.

Worked examples

\[ \text{Morgan's cross: white-eyed female } (X^wX^w) \times \text{red-eyed male } (X^WY) \]

All daughters are \(X^WX^w\) (red-eyed carriers, receiving \(X^W\) from their father); all sons are \(X^wY\) (white-eyed, receiving their only X from their white-eyed mother). Every son is affected and every daughter is an unaffected carrier — the criss-cross pattern of Step 4, running from mother to son. A

\[ \text{Reciprocal cross: red-eyed female } (X^WX^W) \times \text{white-eyed male } (X^wY) \]

All daughters are \(X^WX^w\) (red-eyed carriers, as before) but all sons are now \(X^WY\) (red-eyed, receiving their only X from their homozygous red-eyed mother) — in sharp contrast to Worked Example 1, where every son was affected. The two reciprocal crosses give opposite results for the sons, exactly the asymmetry Step 4 predicts and that no autosomal gene would ever show. A

\[ \text{Cross 1 (mother affected): all sons affected} \qquad \text{Cross 2 (father affected): no sons affected} \]

Reading. Simply swapping which parent carries the recessive allele flips the outcome for the sons completely, while the daughters' outcome (all unaffected carriers) stays the same in both crosses — a signature no autosomal cross can reproduce.

Scope. The identical logic applies to any X-linked recessive trait, including the human examples named in Discussion; pedigree analysis routinely uses exactly this reciprocal asymmetry to distinguish X-linked from autosomal recessive inheritance.

Further example. A geneticist is studying a rare X-linked recessive disorder in a large, randomly mating population. The frequency of the disease-causing allele (\(a\)) in the population is known to be \(q = 0.01\). Assuming the population is in Hardy-Weinberg equilibrium for this locus, calculate the expected frequency of affected males and affected females. Additionally, calculate the probability that a female who is a carrier (\(X^AX^a\)) and a male who is unaffected (\(X^AY\)) will have a son with the disorder.

\[ p = 1 - q = 1 - 0.01 = 0.99 \]

In a two-allele system, the sum of allele frequencies must equal 1. Since \(q\) is the frequency of the recessive allele \(a\), \(p\) is the frequency of the dominant allele \(A\). A

\[ f(\text{affected males}) = q = 0.01 \]

Males are hemizygous for X-linked genes (genotype \(X^AY\) or \(X^aY\)). They inherit their single X chromosome from their mother. Therefore, the frequency of affected males is directly equal to the frequency of the recessive allele \(a\) in the population. A

\[ f(\text{affected females}) = q^2 = (0.01)^2 = 0.0001 \]

Females are diploid for X-linked genes. To be affected, a female must be homozygous recessive (\(X^aX^a\)). Under Hardy-Weinberg equilibrium, the frequency of the homozygous recessive genotype is the square of the recessive allele frequency. A

\[ P(\text{son affected} | X^AX^a \times X^AY) = 0.5 \]

The cross is between a carrier mother (\(X^AX^a\)) and an unaffected father (\(X^AY\)). Sons receive the Y chromosome from the father and one X chromosome from the mother. The mother passes \(X^A\) or \(X^a\) with equal probability (0.5). Thus, 50% of sons will be \(X^AY\) (unaffected) and 50% will be \(X^aY\) (affected). A

\[ \text{Affected males: } 0.01 \quad \text{Affected females: } 0.0001 \quad P(\text{son affected}) = 0.5 \]

Reading. This example highlights the key difference in expression between sexes for X-linked recessive traits. Males are affected at a frequency 100 times higher than females (\(q\) vs \(q^2\)) when \(q\) is small. In a specific mating of a carrier female and normal male, there is a 50% risk for sons, while daughters will all be phenotypically normal (though 50% will be carriers).

Scope. This calculation holds for any X-linked recessive trait in a population meeting Hardy-Weinberg assumptions (random mating, no selection, no mutation, large population size). It does not apply to X-linked dominant traits or to populations with significant sex-biased selection or non-random mating.

Problems

A colour-blind man and a woman with normal vision, whose father was colour-blind, have children. Give the expected phenotypes and genotypes of their daughters and sons, assuming colour blindness is X-linked recessive.
Solution
The mother's colour-blind father means she is an obligate carrier, \(X^CX^c\). The father is \(X^cY\). Cross \(X^CX^c \times X^cY\): daughters are \(\tfrac12\,X^CX^c\) (carriers, normal vision) and \(\tfrac12\,X^cX^c\) (colour-blind, since they receive \(X^c\) from both parents); sons are \(\tfrac12\,X^CY\) (normal) and \(\tfrac12\,X^cY\) (colour-blind).
A pedigree shows an X-linked recessive trait appearing in a daughter whose mother is unaffected. What must be true of the mother's and father's genotypes?
Solution
An affected daughter is \(X^aX^a\), so she must receive one \(X^a\) from each parent: the unaffected mother must be a carrier (\(X^AX^a\)) and the father must himself be affected (\(X^aY\), hemizygous, per Step 2), since he has only one X-linked allele to contribute and it must be \(X^a\) for the daughter to be homozygous recessive.
Explain, using Step 5, why an X-linked recessive trait can never be transmitted directly from an affected father to an affected son without the mother also contributing the recessive allele.
Solution
By Step 5 (a direct consequence of meiotic segregation of the sex chromosomes, Step 1), a father transmits his Y chromosome, not his X, to every son; his X-linked genotype is therefore invisible to his sons entirely. A son's X-linked phenotype depends only on the single X he receives from his mother — so for a son to be affected, the mother must supply an \(X^a\) allele (as either a carrier or affected herself), regardless of the father's own genotype or phenotype.
A woman with normal vision (genotype \(X^CX^c\)) marries a man with normal vision (genotype \(X^CY\)). They have a son. What is the probability that the son is colour-blind? If they have a daughter, what is the probability that she is a carrier?
Solution
The cross is \(X^CX^c \times X^CY\). For sons: The father contributes \(Y\). The mother contributes either \(X^C\) or \(X^c\) with equal probability (1/2). Genotypes of sons: \(1/2\, X^CY\) (normal) and \(1/2\, X^cY\) (colour-blind). Probability son is colour-blind = \(1/2\). For daughters: The father contributes \(X^C\). The mother contributes either \(X^C\) or \(X^c\) with equal probability (1/2). Genotypes of daughters: \(1/2\, X^CX^C\) (normal, non-carrier) and \(1/2\, X^CX^c\) (normal, carrier). Probability daughter is a carrier = \(1/2\).
In a population, the frequency of an X-linked recessive allele \(a\) is \(q = 0.02\). Assuming Hardy-Weinberg equilibrium, calculate the expected frequency of affected males and affected females in the population.
Solution
Let \(p\) be the frequency of the dominant allele \(A\) and \(q\) be the frequency of the recessive allele \(a\). Given \(q = 0.02\), then \(p = 1 - q = 1 - 0.02 = 0.98\). Males are hemizygous (\(X^AY\) or \(X^aY\)). The frequency of affected males is equal to the frequency of the allele \(a\): \[ f(\text{affected males}) = q = 0.02 \] Females are diploid (\(X^AX^A\), \(X^AX^a\), \(X^aX^a\)). The frequency of affected females is the frequency of the homozygous recessive genotype: \[ f(\text{affected females}) = q^2 \] \[ q^2 = (0.02)^2 = 0.0004 \] Thus, the expected frequency of affected males is 0.02 (or 2%) and the expected frequency of affected females is 0.0004 (or 0.04%).