SPEC 3.7.1 - GENETICS, POPULATIONS, EVOLUTION & ECOSYSTEMS
Sex-Linkage, Autosomal Linkage, Multiple Alleles & Epistasis
The four inheritance patterns beyond simple Mendelian crosses — every one appears in AQA Paper 2 questions
Sex Linkage — Key Terms
Key Definition
Sex-linkage — the location of a gene on a sex chromosome (usually the X chromosome); males (XY) have only one copy of X-linked genes and are hemizygous.
Key Definition
Hemizygous — having only one copy of a gene (males for X-linked genes); neither homozygous nor heterozygous.
Carrier — a heterozygous female (X^NXⁿ) who carries a recessive X-linked allele but does not show the affected phenotype.
Why Males Are More Affected by X-Linked Conditions
Males only have one X chromosome. If their single X carries the recessive allele, there is no second copy to mask it. Females need two copies of the recessive allele (one on each X) to show the affected phenotype.
Examples: red-green colour blindness, haemophilia A — both X-linked recessive.
X-Linked Notation
AQA uses a specific notation for X-linked genes. Use it exactly — marks are awarded for correct notation.
GenotypeSexPhenotypeStatus
X^NX^NFemaleUnaffectedHomozygous dominant
X^NXⁿFemaleUnaffectedCarrier (heterozygous)
XⁿXⁿFemaleAffectedHomozygous recessive
X^NYMaleUnaffectedHemizygous dominant
XⁿYMaleAffectedHemizygous recessive
Examiner Alert
The Y chromosome carries no allele for X-linked genes. Never write Y^N or Yⁿ. Males are hemizygous — they cannot be carriers. A male either shows the condition or does not; there is no carrier state for males.
X-Linked Cross — Worked Example
Carrier female × Unaffected male (red-green colour blindness): X^BXᵇ × X^BY
where B = normal vision (dominant), b = colour blind (recessive)
X^BY
X^BX^BX^BX^BY
XᵇX^BXᵇXᵇY
GenotypePhenotypeFrequency
X^BX^BUnaffected female1/4
X^BXᵇCarrier female1/4
X^BYUnaffected male1/4
XᵇYColour-blind male1/4
Probability Statements
Probability any child is colour-blind = 1 in 4 (25%). Probability a son is colour-blind = 1 in 2 (50%). Daughters cannot be colour-blind from this cross.
Autosomal Linkage
Key Definition
Autosomal linkage — the location of two or more genes on the same autosome; linked genes do not assort independently but may be separated by crossing over.
When two genes are on the same chromosome, they tend to be inherited together. This breaks Mendel's Law of Independent Assortment and produces offspring ratios that deviate from 9:3:3:1.
Parental Types (majority)
Offspring that inherit alleles in the same combination as the parents — appear at higher frequency.
Recombinant Types (minority)
Offspring with new allele combinations produced by crossing over at prophase I — appear at lower frequency. Closer genes = fewer recombinants.
Examiner Alert
If a dihybrid cross gives ratios that deviate significantly from 9:3:3:1 (e.g. 7:1:1:7 or similar skewed ratios), the genes are linked on the same chromosome — not on different chromosomes. The chi-squared test can detect this deviation.
Multiple Alleles — ABO Blood Group
Key Definition
Multiple alleles — the existence of more than two alleles for a single gene in a population, although any one individual can only carry a maximum of two alleles.
The ABO blood group gene has three alleles: I^A, I^B, and i.
Blood GroupPossible GenotypesAllele Relationship
AI^AI^A or I^AiI^A dominant over i
BI^BI^B or I^BiI^B dominant over i
ABI^AI^BI^A and I^B codominant
Oiii recessive to both I^A and I^B
Dominance Hierarchy
I^A = I^B > i (I^A and I^B are codominant with each other; both dominant over i)
ABO — Worked Cross Example
Parent 1: Blood group A (heterozygous, I^Ai) × Parent 2: Blood group B (heterozygous, I^Bi)
I^Ai
I^BI^AI^B (AB)I^Bi (B)
iI^Ai (A)ii (O)
1 AB : 1 B : 1 A : 1 O (phenotype ratio)
Exam Application
AQA may give blood group phenotypes and ask you to deduce parental genotypes. Remember: blood group O parents (ii) can only pass on i. Blood group A or B parents could be homozygous (I^AI^A or I^BI^B) or heterozygous (I^Ai or I^Bi) — you need the offspring ratios to determine which.
Epistasis
Key Definition
Epistasis — the masking of the effect of one gene by the alleles of a different gene.
Epistasis explains why some dihybrid crosses give ratios other than 9:3:3:1. The epistatic gene modifies the expression of the hypostatic gene.
Three Types to Know
TypeWhich allele masks?Modified ratioRecessive epistasisHomozygous recessive (aa or bb) at one locus masks the other gene9:3:4Dominant epistasisDominant allele (A_) at one locus masks the other gene12:3:1Duplicate recessiveHomozygous recessive at either locus gives same phenotype9:7
Key Insight
The total always adds to 16 (the number of squares in a dihybrid Punnett grid). 9+3+4=16; 12+3+1=16; 9+7=16. If a ratio adds to 16, it is a modified dihybrid ratio caused by epistasis.
Recessive Epistasis — 9:3:4
Example: coat colour in Labrador dogs. Gene A determines whether pigment is made at all (A_ = pigment; aa = no pigment, albino). Gene B determines pigment colour (B_ = black; bb = brown). aa is epistatic to gene B — if no pigment is produced, gene B cannot be expressed.
Why Three Phenotype Classes Arise
For 9:3:4, the epistatic gene (aa) produces a different phenotype from the class it masks. This gives three distinct, visible phenotypes — not two.
Genotype classPhenotypeWhyCount/16
A_B_Black coatPigment made (A_); black colour (B_)9
A_bbBrown coatPigment made (A_); brown colour (bb)3
aaB_ + aabbAlbino (white)No pigment produced (aa epistatic — gene B irrelevant)3 + 1 = 4
9 black : 3 brown : 4 albino = 9 : 3 : 4
Key Point — Three Distinct Phenotypes
The 9:3:4 ratio has three phenotype classes, not two. The "4" class (aaB_ + aabb) is the epistatic class — aa prevents any pigment regardless of gene B. The "3" class (A_bb) is phenotypically different because pigment IS produced but it's brown, not black.
Dominant & Duplicate Recessive Epistasis
Dominant Epistasis — 12:3:1
A dominant allele at one locus masks expression of a second gene. Example: squash fruit colour — the A allele (white) masks whether the B gene produces yellow or green.GenotypePhenotypeCount/16A_B_ and A_bbWhite (A masks B entirely)9 + 3 = 12aaB_Yellow3aabbGreen112 : 3 : 1
Duplicate Recessive Epistasis — 9:7
Homozygous recessive at either gene produces the same phenotype. Both genes must have at least one dominant allele for the full phenotype to be expressed. Example: flower colour in sweet peas — both genes needed for purple pigment.GenotypePhenotypeCount/16A_B_Purple (both genes functional)9A_bb, aaB_, aabbWhite (at least one gene non-functional)3 + 3 + 1 = 79 : 7
Examiner Traps
Trap 1 — Males cannot be carriers
A male with an X-linked recessive condition is hemizygous — XⁿY. He cannot be a carrier. Carriers are only possible for females. If you write "carrier male" in an exam, you will lose marks.
Trap 2 — Daughters of affected fathers
An affected father (XⁿY) passes his Xⁿ to all daughters. All daughters of an affected father and an unaffected homozygous mother (X^NX^N) will be carriers. This is a very common exam scenario.
Trap 3 — Epistasis vs linkage
Both epistasis and autosomal linkage produce ratios that deviate from 9:3:3:1. Epistasis still has 16 total units (9:3:4, 12:3:1, 9:7). Autosomal linkage gives ratios skewed towards parental types and does NOT simplify to neat fractions of 16. The examiner tests whether you can distinguish them.
Trap 4 — Definition of epistasis
Epistasis is the masking of one gene by alleles of a different gene. Do NOT confuse with dominance (masking within the same gene locus). The examiner awards marks specifically for "different gene" in the definition.
DEEP Chain — Extended Response
Question: "Explain why a dihybrid cross between AaBb × AaBb produces a 9:7 phenotype ratio instead of the expected 9:3:3:1."
DEEP Answer Framework
D — Define
"Epistasis is the masking of the effect of one gene by the alleles of a different gene. The 9:7 ratio arises from duplicate recessive epistasis."
E — Explain the Mechanism
"Both gene A and gene B must have at least one dominant allele for the dominant phenotype to be expressed. If either gene is homozygous recessive (aa or bb), the dominant phenotype cannot be produced — regardless of the other gene's genotype."
E — Evidence / Example
"From the 16-square Punnett grid: 9 genotypes are A_B_ (dominant phenotype); the remaining 7 (A_bb, aaB_, aabb) all produce the recessive phenotype. The three expected classes from 9:3:3:1 have collapsed into one, giving 9:7."
P — Practical / Quantitative
"A chi-squared test on the observed offspring would give a value consistent with 9:7 (df = 1) rather than 9:3:3:1 (df = 3). The total always sums to 16 — this is the hallmark of a modified dihybrid ratio."
Synoptic Links
Link to Meiosis (4.04)
Sex linkage depends on the X and Y chromosomes separating at meiosis I. Crossing over at prophase I can produce recombinant genotypes in autosomal linkage. The closer genes are on a chromosome, the less likely they are to be separated by a crossover.
Link to Protein Synthesis (4.02)
Epistasis reflects gene interactions at the biochemical level. The epistatic gene often codes for an enzyme upstream in a metabolic pathway — if the upstream enzyme is absent (homozygous recessive), no substrate reaches the downstream enzyme regardless of its genotype.
Link to Chi-Squared (7.01)
Modified epistatic ratios can be tested with chi-squared. The degrees of freedom depend on how many phenotype classes the epistasis produces: 9:3:4 → df = 2; 9:7 → df = 1; 12:3:1 → df = 2.
Link to Gene Expression (8.03)
Epistasis is an example of how gene expression is context-dependent — the phenotypic effect of one gene depends on what alleles are present at another locus. This connects to the broader concept of transcription factors and gene regulation in Topic 8.
Practice Question
A woman with normal vision (X^BXᵇ) and a colour-blind man (XᵇY) have children.
(a) Draw a Punnett square and state the probability that their first son will be colour-blind. [3 marks]
(b) A cross between two pea plants (AaBb × AaBb) produces 48 offspring: 27 purple, 21 white. A student claims this is consistent with a 9:7 epistatic ratio. Evaluate this claim, stating whether you would accept or reject the null hypothesis. (χ² critical value at df = 1, p = 0.05 = 3.84) [4 marks]
D
Cross: X^BXᵇ × XᵇY. Gametes from mother: X^B, Xᵇ. Gametes from father: Xᵇ, Y.
E
Offspring: X^BXᵇ (carrier female), XᵇXᵇ (colour-blind female), X^BY (unaffected male), XᵇY (colour-blind male). Probability first son is colour-blind = 1/2 (50%).
E
Part (b): Expected from 9:7 with n = 48: purple = 9/16 × 48 = 27; white = 7/16 × 48 = 21. χ² = (27−27)²/27 + (21−21)²/21 = 0 + 0 = 0.
P
χ² (0) < critical value (3.84) → Accept null hypothesis. The difference between observed and expected is due to chance. The results are perfectly consistent with a 9:7 epistatic ratio.
Lesson Summary
Core Content — Spec 3.7.1
Sex linkage — gene on X chromosome; males hemizygous (XⁿY); females can be carriers (X^NXⁿ); Y carries no allele
Autosomal linkage — genes on same chromosome; inherited together; crossing over produces recombinants at lower frequency; deviates from 9:3:3:1
Multiple alleles — ABO blood group: I^A, I^B, i; I^A and I^B codominant; both dominant over i
Epistasis — masking of one gene by alleles of a different gene; all ratios sum to 16
Epistasis Ratios
Recessive epistasis → 9:3:4
Dominant epistasis → 12:3:1
Duplicate recessive → 9:7
All sum to 16.
X-Linked Rules
Affected males: XⁿY (hemizygous)
Carriers: X^NXⁿ (females only)
Y has NO allele — never write Y^N or Yⁿ
Sons of affected father all receive Xⁿ
Next: Sex Linkage — Key Terms