Sex Linkage and Autosomal Linkage

Mendel's law of independent assortment holds true only when the two genes in question are on different chromosomes. Two important departures from this idealised pattern arise when genes are linked physically on the same chromosome. Sex linkage refers to genes located on the X (or, rarely, the Y) sex chromosome; their distinctive inheritance pattern reflects the difference in sex-chromosome composition between males (XY) and females (XX). Autosomal linkage refers to genes located on the same non-sex chromosome; linked genes do not assort independently but are inherited together more often than would be expected by chance, with the degree of linkage measured by recombination frequency. Together these phenomena explain why pedigrees of conditions like haemophilia, colour blindness and Duchenne muscular dystrophy show their characteristic patterns, and why Thomas Morgan's work on fruit-fly genetics in the early twentieth century transformed our understanding of the gene.

Spec mapping: This lesson sits in AQA 7402 Section 3.7.1 — Inheritance, extending the Mendelian framework to include linkage. The relevant content covers sex determination in mammals, X-linked inheritance patterns, the distinction between sex-linkage and autosomal-linkage, recombination frequency, and the use of pedigrees to identify inheritance patterns. (Refer to the official AQA specification document for exact wording.)

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Sex Determination in Humans

• Humans have 23 pairs of chromosomes: 22 pairs of autosomes and 1 pair of sex chromosomes.
• Females have two X chromosomes (XX); males have one X and one Y chromosome (XY).
• The SRY gene on the Y chromosome triggers male development by initiating the production of testes-determining factor (TDF).
• During meiosis, a female produces gametes that all contain one X chromosome, while a male produces gametes containing either an X or a Y chromosome.
• The sex of offspring is determined by whether the sperm carries an X or a Y chromosome.

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X-Linked (Sex-Linked) Inheritance

Key Definition: Sex-linked genes are genes located on the X chromosome (or, rarely, the Y chromosome). Because the X chromosome is much larger and carries many more genes than the Y chromosome, "sex-linkage" at A-Level essentially means X-linkage. Y-linked inheritance exists (e.g. SRY, AZF genes) but is much less significant in disease.

Key Features of X-Linked Inheritance

• Males are hemizygous for X-linked genes — they have only one copy, so any allele on the X chromosome is expressed regardless of dominance.
• Females can be homozygous dominant, heterozygous (carriers), or homozygous recessive.
• X-linked recessive conditions are more common in males because males need only one copy of the recessive allele to be affected. The frequency ratio (male:female) approximates 1/q, where q is the allele frequency — for very rare alleles, conditions are essentially male-restricted.
• An affected father cannot pass an X-linked condition to his sons (sons inherit the Y chromosome from their father).
• An affected father always passes the X-linked allele to his daughters, making them at least carriers.
• A carrier mother has a 50% chance of passing the recessive allele to each child, regardless of sex.

Examples of X-linked recessive conditions

Condition	Affected gene	Phenotype
Haemophilia A	F8 (clotting factor VIII)	Impaired blood clotting; bleeding into joints and tissues
Haemophilia B	F9 (clotting factor IX)	As haemophilia A but rarer
Red-green colour blindness	OPN1LW / OPN1MW	Inability to distinguish red and green hues
Duchenne muscular dystrophy	DMD (dystrophin)	Progressive muscle wasting; early childhood onset
G6PD deficiency	G6PD	Haemolytic anaemia after certain triggers (broad beans, drugs)
X-linked agammaglobulinaemia	BTK	Severe immune deficiency due to failure of B-cell maturation

The pedigree pattern in all these conditions is characteristic: affected males inheriting from unaffected carrier mothers; affected fathers transmitting through their unaffected carrier daughters to half their grandsons; the trait apparently "skipping" generations through carrier females.

Writing Sex-Linked Genotypes

Alleles are written as superscripts on the X chromosome symbol:

• X^H = normal allele (e.g., for blood clotting)
• X^h = recessive allele (e.g., for haemophilia)
• The Y chromosome is written simply as Y (no corresponding allele)

Genotype	Sex	Phenotype
X^H X^H	Female	Normal
X^H X^h	Female	Carrier (unaffected)
X^h X^h	Female	Affected (haemophilia)
X^H Y	Male	Normal
X^h Y	Male	Affected (haemophilia)

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Worked Example 1 — Haemophilia

Haemophilia A is an X-linked recessive condition in which the blood fails to clot properly due to a deficiency in clotting factor VIII.

Cross: Carrier female × Normal male Parental genotypes: X^H X^h × X^H Y

Gametes: X^H or X^h (mother); X^H or Y (father)

	X^H	Y
X^H	X^H X^H (normal female)	X^H Y (normal male)
X^h	X^H X^h (carrier female)	X^h Y (affected male)

Offspring:

• 1/4 normal female
• 1/4 carrier female
• 1/4 normal male
• 1/4 haemophiliac male

Key observations:

• 50% of sons are affected.
• No daughters are affected, but 50% of daughters are carriers.
• The condition appears to "skip generations" — an unaffected carrier mother can produce affected sons.

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Worked Example 2 — Red-Green Colour Blindness

Red-green colour blindness is another X-linked recessive condition.

Cross: Colour-blind female × Normal male Parental genotypes: X^b X^b × X^B Y

	X^B	Y
X^b	X^B X^b (carrier female)	X^b Y (colour-blind male)
X^b	X^B X^b (carrier female)	X^b Y (colour-blind male)

Offspring:

• All daughters are carriers (X^B X^b) — phenotypically normal.
• All sons are colour-blind (X^b Y).

Exam Tip: In sex-linked crosses, always write the sex chromosomes explicitly (e.g., X^H X^h, not just Hh). Examiners expect to see the X and Y chromosome notation. Failing to show this loses marks.

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Worked Example 4 — Calculating Carrier Risk in a Pedigree

Sex-linkage questions frequently move beyond a single cross to ask for the probability that an individual is a carrier, or that a future child is affected. These items reward disciplined use of the multiplication and addition rules of probability applied to the sex-chromosome logic already established.

Consider a woman, Anna, whose maternal grandfather was affected by haemophilia A but whose father and brothers are all unaffected. Anna wants to know the probability that a son of hers would be affected.

Step 1 — establish the grandmother's genotype. Anna's affected maternal grandfather was X^h Y. Every daughter of an affected father inherits his X^h, so Anna's mother received X^h from her father. Anna's maternal grandmother was phenotypically normal; assuming she was not herself a carrier, she contributed X^H. Anna's mother is therefore an obligate carrier: X^H X^h.

Step 2 — find the probability that Anna is a carrier. Anna's mother (X^H X^h) crosses with Anna's unaffected father (X^H Y). Anna, being an unaffected daughter, is either X^H X^H or X^H X^h with equal probability, because half of a carrier mother's daughters are carriers. The fact that Anna's father and brothers are unaffected gives no additional information about Anna's own genotype — her brothers' phenotypes are independent outcomes of the same cross. So P(Anna is a carrier) = 1/2.

Step 3 — find the probability that a son of Anna's is affected, given she is a carrier. If Anna is X^H X^h, then in a cross with an unaffected partner (X^H Y) half of her sons receive X^h and are affected. So P(affected son | Anna is a carrier) = 1/2.

Step 4 — combine using the multiplication rule. The two events must both occur, so P(Anna is a carrier AND has an affected son) = 1/2 × 1/2 = 1/4. If the question instead asks for the probability that a given child is an affected boy, multiply by a further 1/2 for the child being male: 1/2 × 1/2 × 1/2 = 1/8.

This structure — establish obligate-carrier genotypes from the pedigree, assign a carrier probability to the individual of interest, then chain the conditional cross probabilities with the multiplication rule — is the template for every X-linked risk calculation. A common mark-losing error is to forget the initial 1/2 for the individual's own carrier status and quote the conditional 1/2 as if the carrier state were certain.

By the end of this lesson you should be able to: explain sex determination in mammals and the meaning of hemizygosity; predict the outcomes of X-linked crosses using correct X^allele / Y notation; recognise X-linked recessive inheritance in a pedigree; calculate carrier and affected-offspring probabilities using the multiplication rule; distinguish sex-linkage from autosomal linkage; and calculate recombination frequency to map linked genes.

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Recognising X-Linked Inheritance in Pedigrees

Features that suggest X-linked recessive inheritance in a pedigree:

• The condition is more common in males than females.
• Affected males often have unaffected mothers who are carriers.
• The condition is never passed directly from father to son (because fathers give their Y chromosome to sons).
• All daughters of an affected male will be at least carriers.
• If a female is affected, her father must also be affected and her mother must be at least a carrier.

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Autosomal Linkage

Key Definition: Autosomal linkage occurs when two or more genes are located on the same autosome (non-sex chromosome). Linked genes do not assort independently during meiosis and tend to be inherited together. The strength of linkage is inversely related to the physical distance between the loci on the chromosome.

Consequences of Linkage

When two genes are linked, the predictions of Mendel's second law break down in a quantifiable way:

• A dihybrid heterozygote (AaBb, with A and B on the same chromosome) does not produce four types of gamete in equal 1:1:1:1 proportions.
• Instead, parental-type gametes (with the original allele combinations inherited from the parents) are far more common than recombinant-type gametes.
• Recombinant gametes are produced only when crossing over occurs between the two gene loci during prophase I of meiosis.
• The closer two genes are on the chromosome, the less likely a chiasma will form between them, and the smaller the recombination frequency.

The expected 9:3:3:1 ratio of a dihybrid cross between two double heterozygotes is replaced by a different ratio reflecting the over-representation of parental phenotypes. The actual numerical pattern depends on the recombination frequency and the configuration (coupling vs repulsion) of the parental chromosomes.

Coupling and Repulsion

The arrangement of alleles on homologous chromosomes affects which gamete types are "parental":

• Coupling (cis): Both dominant alleles are on the same chromosome (AB / ab). Parental gametes: AB and ab. Recombinant gametes: Ab and aB. F₁ offspring resemble the dominant phenotype with the typical parental combination.
• Repulsion (trans): One dominant and one recessive allele are on each chromosome (Ab / aB). Parental gametes: Ab and aB. Recombinant gametes: AB and ab. F₁ offspring of a test cross show the heterozygous parental phenotype combinations more frequently than the homozygous recombinant ones.

The distinction matters because in a test cross of an AaBb individual with an aabb partner, the offspring phenotypes directly reflect the gamete frequencies of the dihybrid parent. Comparing the observed phenotypic proportions with the 1:1:1:1 expected from independent assortment reveals the linkage configuration.

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Recombination Frequency and Gene Mapping

Recombination frequency = (Number of recombinant offspring / Total number of offspring) × 100%

The recombination frequency captures, in a single number, how strongly two genes are linked. It is most easily measured from a test cross of a dihybrid individual against a double recessive partner, where the offspring phenotypes directly reflect the gametes of the dihybrid parent.

• A recombination frequency of 50% indicates the genes are on different chromosomes (or so far apart on the same chromosome that they effectively assort independently).
• A recombination frequency significantly less than 50% indicates the genes are linked.
• A recombination frequency of 0% would indicate complete linkage — no crossing over ever occurs between the two loci. This is rare; most loci show some recombination.
• The lower the recombination frequency, the closer together the genes are on the chromosome.

Worked example — calculating recombination frequency

A test cross between a dihybrid female (AaBb, with A and B on the same chromosome in coupling) and a homozygous recessive male (aabb) produces 1000 offspring with the following phenotypes:

• 410 AB (parental)
• 425 ab (parental)
• 75 Ab (recombinant)
• 90 aB (recombinant)

Recombinant total: 75 + 90 = 165 Total offspring: 1000 Recombination frequency: (165 / 1000) × 100% = 16.5%

Therefore the two genes are 16.5 centiMorgans apart on the chromosome. The parental gametes (AB and ab) confirm the coupling configuration; if the configuration had been repulsion (Ab / aB), the parental gametes would have been Ab and aB and the recombinants would have been AB and ab.

Constructing Genetic Maps

Recombination frequencies can be used to construct genetic maps (linkage maps) showing the relative positions of genes on a chromosome.