B33: Sex-Linked Inheritance
How some characteristics are linked to sex chromosomes
HIGHER ONLY: This topic is only examined on the Higher tier paper. You do not need this if you are taking the Foundation paper.
What Are Sex-Linked Genes?
Sex-linked genes are genes located on the sex chromosomes (X or Y). Most sex-linked genes are carried on the X chromosome because the Y chromosome is smaller and carries fewer genes.
| Feature | X Chromosome | Y Chromosome |
|---|---|---|
| Size | Larger | Smaller |
| Number of genes | Many (approximately 800) | Few (approximately 50–60) |
| Sex-linked genes | Carries many sex-linked genes | Carries very few sex-linked genes |
Why Males Are More Likely to Have X-Linked Disorders
Males have only one X chromosome (XY), so if they inherit a recessive allele on their X chromosome, there is no second X chromosome with a dominant allele to mask it. Females have two X chromosomes (XX), so a dominant allele on the other X can mask a recessive allele.
- A male with one recessive allele on his X chromosome will express the disorder
- A female with one recessive allele on one X chromosome is a carrier — the dominant allele on her other X chromosome masks it
- A female needs two recessive alleles (one on each X) to have the disorder, which is much less likely
X-Linked Disorders
Colour Blindness
Colour blindness (red-green colour blindness) is caused by a recessive allele on the X chromosome. We write: XB = normal vision, Xb = colour blindness allele.
Haemophilia
Haemophilia is a disorder where blood does not clot properly. It is caused by a recessive allele on the X chromosome. We write: XH = normal blood clotting, Xh = haemophilia allele.
| Genotype | Sex | Phenotype |
|---|---|---|
| XBXB | Female | Normal vision |
| XBXb | Female | Normal vision (carrier) |
| XbXb | Female | Colour blind |
| XBY | Male | Normal vision |
| XbY | Male | Colour blind |
Worked Examples with Punnett Squares
Worked Example 1: Carrier Female × Normal Male (Colour Blindness)
Mother: XBXb (carrier). Father: XBY (normal vision).
| XB | Y | |
|---|---|---|
| XB | XBXB | XBY |
| Xb | XBXb | XbY |
Offspring:
- XBXB — female, normal vision (25%)
- XBXb — female, carrier, normal vision (25%)
- XBY — male, normal vision (25%)
- XbY — male, colour blind (25%)
None of the daughters are colour blind, but 50% of daughters are carriers. 50% of sons are colour blind.
Worked Example 2: Carrier Female × Colour Blind Male (Colour Blindness)
Mother: XBXb (carrier). Father: XbY (colour blind).
| Xb | Y | |
|---|---|---|
| XB | XBXb | XBY |
| Xb | XbXb | XbY |
Offspring:
- XBXb — female, carrier (25%)
- XbXb — female, colour blind (25%)
- XBY — male, normal vision (25%)
- XbY — male, colour blind (25%)
50% of daughters are colour blind. 50% of sons are colour blind. This is the only cross where a female can be colour blind — she must inherit two recessive alleles.
Worked Example 3: Haemophilia — Carrier Female × Normal Male
Mother: XHXh (carrier). Father: XHY (normal clotting).
| XH | Y | |
|---|---|---|
| XH | XHXH | XHY |
| Xh | XHXh | XhY |
Offspring:
- 25% — female, normal (XHXH)
- 25% — female, carrier (XHXh)
- 25% — male, normal (XHY)
- 25% — male, haemophiliac (XhY)
50% of sons have haemophilia. 50% of daughters are carriers.
Worked Example 4: Colour Blind Male × Homozygous Normal Female
Father: XbY (colour blind). Mother: XBXB (homozygous normal).
| Xb | Y | |
|---|---|---|
| XB | XBXb | XBY |
| XB | XBXb | XBY |
All offspring have normal vision. All daughters are carriers (XBXb). All sons have normal vision (XBY). This shows that a colour blind father cannot pass the allele to his sons — he passes his Y chromosome to sons, not his X. But he can pass the allele to his daughters through his X chromosome.
Comparison: Inheritance Patterns
| Cross | Daughters | Sons |
|---|---|---|
| Carrier female × Normal male | 50% carrier, 50% normal | 50% affected, 50% normal |
| Carrier female × Affected male | 50% carrier, 50% affected | 50% affected, 50% normal |
| Normal female × Affected male | 100% carriers | 100% normal |
| Affected female × Normal male | 100% carriers | 100% affected |
Practice Questions
Q1: Higher Explain why males are more likely than females to have X-linked recessive disorders.
Q2: Higher A carrier female (XHXh) and a normal male (XHY) have children. What is the probability that a son will have haemophilia? Show your working with a Punnett square.
Q3: Higher Can a colour blind father and a homozygous normal mother (XBXB) have a colour blind son? Explain your answer.
Q4: Higher A woman with normal vision whose father was colour blind marries a man with normal vision. What is the probability their first child will be a colour blind male?
Answers
- Males have only one X chromosome (XY), so if they inherit a recessive allele on that X chromosome, there is no second X with a dominant allele to mask it — the recessive allele is always expressed. Females have two X chromosomes (XX), so even if one X carries the recessive allele, the other X is likely to carry a dominant allele that masks it. Females need two recessive alleles to express the disorder, which is much less likely.
- Punnett square: XHXh × XHY → XHXH, XHXh, XHY, XhY. Sons are XHY (normal) and XhY (haemophiliac). Probability of a son having haemophilia = 50%.
- No. The father is XbY and passes his Y chromosome to his sons (not his X). The mother is XBXB and can only pass on XB to her children. All sons will be XBY (normal vision). However, all daughters will be XBXb (carriers).
- The woman’s father was colour blind (XbY), so she must have inherited Xb from him. Since she has normal vision, she must be XBXb (carrier). Her husband is XBY. Cross: XBXb × XBY → XBXB, XBXb, XBY, XbY. Probability of a colour blind male (XbY) = 25% (1 in 4 of all children).
Exam Tips
- When writing sex-linked alleles, always include the chromosome: XB not just B. This shows the examiner you understand the allele is on the X chromosome. Males are written as XBY or XbY
- A colour blind father cannot pass colour blindness to his sons — he gives them his Y chromosome, not his X. But he can pass the allele to his daughters through his X chromosome
- When a question says a woman’s father had an X-linked disorder, you can deduce her genotype — she must carry the allele because she inherited one of her X chromosomes from him
- Always state phenotypes as well as genotypes in your answer for full marks
🔢 Maths Skills
Mathematical Skills
Sex-linked Punnett squares: when the gene is on the X chromosome, include X and Y in the square. For XBXb × XBY, each outcome has a probability of 1/4 = 25%. The probability of a SON being colour blind = 1/2 = 50% (only look at the male offspring row). The probability of ANY child being colour blind = 1/4 = 25%.
Conditional probability: "probability of an affected son" is different from "probability of a son being affected". In the first, the denominator is all children; in the second, only male children.
⚠️ Common Misconceptions
Watch Out!
Students often think males can be carriers of X-linked conditions. Wrong: Males can be carriers of X-linked disorders Correct: Males have only one X chromosome, so a single recessive allele on that X is always expressed — they either have the condition or they don't
Students often think sex-linked means only one sex is affected. Wrong: Sex-linked conditions only affect one sex Correct: Both sexes can be affected, but males are much more likely because they have only one X chromosome
✍️ 6-Mark Question
Extended Answer
6 marks: Explain why colour blindness is more common in males than in females. Use a Punnett square to support your answer.
Colour blindness is caused by a recessive allele on the X chromosome (Xb). Males have only one X chromosome (XY), so if they inherit Xb from their mother, they will express the condition — there is no second X with a dominant allele to mask it. Females have two X chromosomes (XX), so they need two copies of the recessive allele (XbXb) to be colour blind, which is much less likely. A carrier female (XBXb) has normal vision because the dominant XB masks Xb. Punnett square for XBXb × XBY: 25% XBXB (normal female), 25% XBXb (carrier female), 25% XBY (normal male), 25% XbY (colour blind male). 50% of sons are affected but 0% of daughters are colour blind.
Mark scheme: 1 mark for identifying X-linked recessive, 1 mark for males having one X, 1 mark for Punnett square, 1 mark for phenotype outcomes, 1 mark for females needing two copies, 1 mark for concluding explanation
📊 AO3: Analyse & Evaluate
Analysis and Evaluation
A pedigree shows: a woman with normal vision has a colour blind son and a normal son. Her father was colour blind. Determine the woman's genotype and the probability that her next child will be a colour blind female. What further information would you need to determine whether her daughter is a carrier?
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🎬 Video Resources
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