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Not all genes follow Mendel's law of independent assortment. Genes located on the sex chromosomes show sex-linked inheritance patterns, and genes located close together on the same autosome show autosomal linkage. Both depart from the classic 9:3:3:1 dihybrid ratio and are important topics in the Edexcel A-Level Biology specification.
Humans have 22 pairs of autosomes and one pair of sex chromosomes:
The Y chromosome is much smaller than the X chromosome and carries relatively few genes. The X chromosome carries many genes that have no corresponding allele on the Y chromosome.
| Feature | X chromosome | Y chromosome |
|---|---|---|
| Size | Large (~800 genes) | Small (~50 genes) |
| Inherited from | Mother (always) or Father (to daughters) | Father (to sons only) |
| Key gene | Many — including genes for colour vision, blood clotting | SRY (sex-determining region Y) |
| Pseudoautosomal region | Present — small homologous region at tips | Present — allows pairing with X during meiosis |
A sex-linked gene is one that is located on a sex chromosome. In practice, nearly all sex-linked genes discussed at A-Level are located on the X chromosome (X-linked), because the Y chromosome carries very few genes.
Alleles are written as superscripts on the sex chromosome:
Haemophilia A is an X-linked recessive condition caused by a deficiency in clotting factor VIII.
Cross: Carrier female (X^H X^h) × Normal male (X^H Y)
Punnett Square:
| X^H | X^h | |
|---|---|---|
| X^H | X^H X^H (normal female) | X^H X^h (carrier female) |
| Y | X^H Y (normal male) | X^h Y (affected male) |
Offspring:
Red-green colour blindness is X-linked recessive. Let X^B = normal vision, X^b = colour blind.
Cross: Carrier female (X^B X^b) × Colour-blind male (X^b Y)
| X^B | X^b | |
|---|---|---|
| X^b | X^B X^b (carrier female) | X^b X^b (colour-blind female) |
| Y | X^B Y (normal male) | X^b Y (colour-blind male) |
Offspring:
Duchenne muscular dystrophy (DMD) is X-linked recessive. The dystrophin gene on the X chromosome is mutated. Let X^D = normal, X^d = DMD.
Cross: Carrier female (X^D X^d) × Normal male (X^D Y)
| X^D | X^d | |
|---|---|---|
| X^D | X^D X^D (normal female) | X^D X^d (carrier female) |
| Y | X^D Y (normal male) | X^d Y (affected male) |
Offspring: 50% of sons will be affected. No daughters will be affected (though 50% will be carriers). This pattern — affected males born to unaffected parents — is the hallmark of X-linked recessive inheritance.
Queen Victoria was a carrier of haemophilia (X^H X^h). Through her descendants, the allele spread to the royal families of Spain, Germany, and Russia. Her son Leopold had haemophilia (X^h Y), and several grandsons were affected. This is a real historical example of X-linked recessive inheritance tracked through a pedigree over multiple generations.
Look for these clues:
Two genes are linked when they are located on the same chromosome. Linked genes tend to be inherited together because they travel on the same chromosome during meiosis. They do not assort independently.
If genes A/a and B/b are on different chromosomes, a dihybrid test cross (AaBb × aabb) gives a 1:1:1:1 ratio.
If genes A/a and B/b are linked on the same chromosome, the test cross deviates from 1:1:1:1. The parental combinations (e.g. AB and ab) are much more frequent than recombinant combinations (e.g. Ab and aB).
Linked genes are not permanently stuck together. Crossing over during prophase I of meiosis can separate linked alleles, producing recombinant gametes.
| Gamete type | Description | Frequency |
|---|---|---|
| Parental | Same allele combination as in the parent | High (majority) |
| Recombinant | New allele combination due to crossing over | Low (minority) |
The recombination frequency depends on the distance between the two genes:
In Drosophila, body colour and wing length are linked on the same chromosome:
A test cross is performed: BbVv × bbvv
If the genes were unlinked, we would expect a 1:1:1:1 ratio.
Observed offspring:
Total = 900. Recombinants = 100/900 = 11.1%
This tells us the genes are linked with a recombination frequency of 11.1%, corresponding to a map distance of 11.1 centimorgans (cM).
The recombination frequency between two linked genes is proportional to the physical distance between them. This principle is used to construct genetic maps (also called linkage maps).
Exam tip: Recombination frequency can never exceed 50%, because at that point the genes behave as though they are on different chromosomes (assorted independently). This is because a maximum of one crossover typically occurs in any region.
| Feature | Independent assortment | Autosomal linkage |
|---|---|---|
| Gene location | Different chromosomes | Same chromosome |
| Test cross ratio | 1:1:1:1 | Deviation from 1:1:1:1 |
| Parental types | 50% of offspring | >50% of offspring |
| Recombinant types | 50% of offspring | <50% of offspring |
| Effect of crossing over | Not relevant | Produces recombinants |
| Condition | Gene location | Type | Prevalence |
|---|---|---|---|
| Haemophilia A | X chromosome | X-linked recessive | ~1 in 5,000 males |
| Red-green colour blindness | X chromosome | X-linked recessive | ~8% of males, 0.5% of females |
| Duchenne muscular dystrophy | X chromosome | X-linked recessive | ~1 in 3,500 males |
| Haemophilia B (Christmas disease) | X chromosome | X-linked recessive | ~1 in 25,000 males |
| G6PD deficiency | X chromosome | X-linked recessive | ~400 million affected worldwide |
Sex-linked genes are located on the sex chromosomes (usually the X). Males are hemizygous, so X-linked recessive conditions are far more common in males. Autosomal linkage occurs when genes are on the same chromosome — they tend to be inherited together but can be separated by crossing over. The recombination frequency measures the degree of linkage and is used to map gene positions. Both sex linkage and autosomal linkage cause deviations from the expected Mendelian ratios.
Sex linkage and autosomal linkage are the two principal extensions of Mendelian inheritance that break independent assortment. Sex linkage describes inheritance of genes on the sex chromosomes — overwhelmingly the X chromosome at A-Level, because the human Y carries far fewer protein-coding genes. Males are hemizygous at X-linked loci, so a single recessive X-linked allele is fully expressed, producing the characteristic male-biased prevalence of X-linked recessive disorders (haemophilia, red–green colour blindness, DMD), the criss-cross grandfather → carrier daughter → grandson pattern, and the absence of male-to-male transmission. Autosomal linkage describes the contrasting case in which two loci on the same autosome fail to assort independently — a direct violation of Mendel's second law, resolved at meiosis by crossing over. Recombination frequency (RF) quantifies the linkage, with 1% RF = 1 centimorgan (cM) and a 50% ceiling at which loci appear unlinked.
This material sits in Edexcel 9BI0 Topic 8 (Grey Matter — Coordination, Response and Gene Technology) under the inheritance strand, where candidates apply the rules of inheritance to sex-linked and autosomally linked loci and account for departures from the expected Mendelian ratios. Required content covers sex determination in humans (XX female, XY male; SRY on Y as the testis-determining gene); X-linked recessive inheritance with haemophilia A, red–green colour blindness and Duchenne muscular dystrophy as the standard worked examples; the superscript notation (X^H, X^h, Y) and construction of sex-linked Punnett squares; the hemizygosity of males at X-linked loci and the resulting male-biased prevalence of X-linked recessive disorders; carrier-female × normal-male, normal-female × affected-male and carrier-female × affected-male as the canonical pedigree crosses; autosomal linkage as a violation of independent assortment; the test-cross deviation from the 1 : 1 : 1 : 1 dihybrid ratio as the diagnostic for linkage; recombination frequency computed as (recombinants / total offspring) × 100, with 1% RF = 1 cM and a 50% maximum; and the role of crossing over at prophase I of meiosis in generating the recombinant minority. Synoptic links run backwards to lesson 3 (meiosis), lesson 4 (monohybrid) and lesson 5 (dihybrid), sideways to lesson 6 (codominance) and forwards to lesson 8 (chi-squared), lesson 9 (natural selection) and lesson 10 (Hardy–Weinberg). Refer to the official Pearson Edexcel 9BI0 specification document for exact wording.
Question (8 marks):
(a) A woman whose father had haemophilia A marries an unaffected man. Using X^H and X^h, deduce the woman's genotype with reasoning, construct the Punnett square, and state the probabilities of (i) an affected son, (ii) an affected daughter and (iii) a carrier daughter. (4)
(b) A Drosophila test cross of doubly heterozygous AaBb × doubly homozygous recessive aabb gives 1,000 progeny: 412 AaBb : 408 aabb : 92 Aabb : 88 aaBb. Calculate the recombination frequency, state the map distance in centimorgans, and explain whether the loci are linked. (3)
(c) Explain, in terms of events at prophase I of meiosis, how the recombinant minority class arises and why loci > 50 cM apart appear unlinked. (1)
Solution with mark scheme:
(a) M1 (AO2.1) — deduce maternal genotype. Her father had haemophilia (genotype X^h Y) and must transmit his single X^h to every daughter. The woman therefore inherits X^h from her father and X^H from her unaffected mother (assumed X^H X^H), making her genotype X^H X^h — an obligate carrier.
A1 (AO2.1) — Punnett square for X^H X^h × X^H Y.
| X^H | X^h | |
|---|---|---|
| X^H | X^H X^H (normal female) | X^H X^h (carrier female) |
| Y | X^H Y (normal male) | X^h Y (affected male) |
M1 + A1 (AO2.7) — phenotype probabilities. Each cell has probability 1/4. P(affected son) = 1/4 overall (1/2 of sons); P(affected daughter) = 0 because every daughter inherits X^H from her father, so no daughter can be X^h X^h; P(carrier daughter) = 1/4 overall (1/2 of daughters). The hallmark X-linked recessive signature — affected sons born to phenotypically unaffected parents, no affected daughters — emerges directly.
(b) M1 (AO2.2) — identify recombinants. The AaBb parent's gametes pair with the aabb parent's invariant ab gametes, so each progeny phenotype reads off the gamete from AaBb. The majority classes (AaBb = 412, aabb = 408; total 820) are parental; the minority classes (Aabb = 92, aaBb = 88; total 180) are recombinant.
A1 (AO2.2) — compute RF and map distance. RF = (180 / 1,000) × 100 = 18%. Using 1% RF = 1 cM, the map distance is 18 cM.
A1 (AO3.1) — conclude. The progeny depart sharply from the unlinked 1 : 1 : 1 : 1 expectation (250 : 250 : 250 : 250). Parental classes greatly exceed recombinant classes, so the loci are linked at 18 cM. A χ² test against 1 : 1 : 1 : 1 (lesson 8) would reject the null of independent assortment at p < 0.01.
(c) M1 (AO1.2 + AO2.1) — crossing over and the 50 cM ceiling. Recombinant gametes arise when non-sister chromatids of homologous chromosomes exchange segments at chiasmata during prophase I of meiosis, so a parental AB chromatid becomes Ab (or aB). RF saturates at 50% because multiple (even) crossovers cancel at the gamete level — half the gametes are recombinant and half parental, indistinguishable from independent assortment. Loci > 50 cM apart therefore appear unlinked even when they share a chromosome, motivating multi-point linkage maps stitched from short reliably linked intervals.
Total: 8 marks (M3 A5).
Question (6 marks): A heterozygous-carrier woman (X^B X^b) has children with a colour-blind man (X^b Y). (i) Construct the Punnett square and state expected proportions of normal-vision sons, colour-blind sons, carrier daughters and colour-blind daughters; (ii) explain why colour blindness is far more common in males than females; and (iii) using the criss-cross pattern, explain how an X-linked recessive allele passes from an affected grandfather to an affected grandson without either parent being affected.
Mark scheme decomposition by AO:
| Mark | AO | Earned by |
|---|---|---|
| 1 | AO1.1 | Constructing a 2 × 2 Punnett square from X^B X^b × X^b Y giving genotypes X^B X^b : X^b X^b : X^B Y : X^b Y in the ratio 1 : 1 : 1 : 1 |
| 2 | AO2.1 | Mapping to phenotypes: carrier daughter : colour-blind daughter : normal-vision son : colour-blind son = 1 : 1 : 1 : 1 — notably, half the daughters are colour-blind because the affected father contributes X^b and the carrier mother contributes X^b at probability 1/2 |
| 3 | AO1.2 | Identifying that males are hemizygous at X-linked loci — a single X^b allele is fully expressed because the Y carries no functional homologue, so male prevalence ≈ allele frequency q while female prevalence ≈ q² (under random mating) |
| 4 | AO2.7 | Quantifying the bias: with q ≈ 0.08, P(affected male) ≈ 8% vs. P(affected female) ≈ 0.64% — the observed ~10× male excess |
| 5 | AO3.1 | Tracing the criss-cross: affected grandfather (X^b Y) transmits X^b to every daughter (obligate carrier X^B X^b); each carrier daughter has 1/2 chance of transmitting X^b to each son; an affected grandson inherits via his mother despite neither parent being affected |
| 6 | AO3.2 | Concluding that the trait skips a generation through carrier females — the diagnostic pedigree signature of X-linked recessive inheritance, absent the sex bias of autosomal recessive |
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