Sex linkage

Concept

Sex linkage

T-016Home BU-103Threads information · evolution

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.

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.