What you need to know
- 2.15(iii)sex-linkedDescribes a gene carried on a sex chromosome, usually the X chromosome, so the pattern of inheritance differs between males and females. inheritance, including red–green colour blindness.
By the end, you should be able to…
- Represent X-linked alleleA different version of the same gene, found at the same locus. on the X chromosomeA long DNA molecule wound around histone proteins. After replication it is two sister chromatids joined at a centromere..
- Explain why an X-linked recessiveAn allele that is expressed in the phenotype only when no dominant allele is present (when homozygous). phenotypeThe observable characteristics of an organism, produced by its genotype interacting with the environment. is more frequent in XY individuals.
- Calculate sex-specific and overall probabilities.
Build your understanding
The allele travels with the X chromosome
In the conventional XX/XY model used in these questions, an XX parent supplies an X chromosomeA long DNA molecule wound around histone proteins. After replication it is two sister chromatids joined at a centromere. in each egg. An XY parent supplies X or Y in sperm. Most geneA sequence of DNA bases that codes for a polypeptide or a functional RNA. on X do not have a corresponding alleleA different version of the same gene, found at the same locus. at the same locusThe position of a gene on a chromosome. on Y.
Use Xᴺ for an allele associated with typical red–green colour vision and Xⁿ for a recessiveAn allele that is expressed in the phenotype only when no dominant allele is present (when homozygous). allele associated with red–green colour vision deficiency. An XⁿY individual expresses the recessive phenotypeThe observable characteristics of an organism, produced by its genotype interacting with the environment. because there is no corresponding allele on Y to mask it. Do not write a colour-vision allele on Y.
Follow the chromosomes through a cross
For XᴺXⁿ × XᴺY, the four equally likely outcomes are XᴺXᴺ, XᴺXⁿ, XᴺY and XⁿY. In the simplified model, the XX offspring have typical red–green vision; half are carriers. Half the XY offspring have the recessive phenotype.
The chance among sons is 1/2. The chance among all children is 1/4 if XX and XY conceptions are equally likely. A question’s denominator matters: “a son is affected” and “an affected son is born” can ask different probabilities.
Transmission patterns are informative
An XY father passes his X chromosome to daughters and Y to sons. Therefore, he does not pass his sex-linkedDescribes a gene carried on a sex chromosome, usually the X chromosome, so the pattern of inheritance differs between males and females. allele directly to sons. A father with the phenotype and an XᴺXᴺ mother produces XᴺXⁿ daughters and XᴺY sons under the model.
An XX individual can have an X-linked recessive phenotype if both X chromosomes carry the relevant allele. It is incorrect to claim that red–green colour vision deficiency occurs only in males. Real biology has additional complexity; use the inheritance assumptions supplied in an exam question.
Follow the process
In the XX/XY model used here: an XY parent’s X chromosome → goes into a daughter → she may carry the allele. His Y chromosome → goes into a son → that son does not receive this X-linked allele from him. A son’s X, and therefore his X-linked allele, comes from the XX parent.
The biological transmission order is not the only valid written order. You may start with “a son is affected because he inherited Xⁿ from his mother and has no second allele on Y.” Show the denominator the question asks for: among sons, or among all children.
NOT: an affected father does not pass his X-linked allele to his sons.
- XⁿY has one allele at this locus. It is not heterozygous
- Chance among sons is 1/2. Chance among all children is 1/4
- XX individuals can be affected if both X chromosomes carry Xⁿ
- Use the denominator the question names
Two traps this diagram blocks
- Do not put a colour-vision allele on the Y chromosome.
- Do not answer “1/2” and “1/4” without saying which group the denominator is.
From the diagram alone
Question
For XᴺXⁿ × XᴺY, give the probability that a son has red–green colour vision deficiency, and the probability that any child does. [3]
Show a model answer
A son receives Xⁿ or Xᴺ from his mother with equal probability, and Y from his father, so the probability a son is XⁿY is 1/2. Among all four equally likely children only one is XⁿY, so the probability for any child is 1/4, assuming sons and daughters are equally likely.
| Egg | Xᴺ sperm | Y sperm |
|---|---|---|
| Xᴺ | XᴺXᴺ: typical vision | XᴺY: typical vision |
| Xⁿ | XᴺXⁿ: carrier | XⁿY: red–green deficiency |
Make your reasoning visible
Build the explanation
X happens because Y, leading to Z. X: an affected father does not give his sons his colour-vision alleleA different version of the same gene, found at the same locus.. Y: sons receive his Y chromosomeA long DNA molecule wound around histone proteins. After replication it is two sister chromatids joined at a centromere., which has no allele at this locusThe position of a gene on a chromosome.. Z: their phenotypeThe observable characteristics of an organism, produced by its genotype interacting with the environment. depends on the X from their mother. Incomplete: “It is sex-linked so only males get it from their fathers.” Improved: “Sons receive the father’s Y, so an sex-linkedDescribes a gene carried on a sex chromosome, usually the X chromosome, so the pattern of inheritance differs between males and females. allele passes from him to daughters, not directly to sons.” The improvement names the chromosome that actually moves.
Name the denominator
State “half the sons” or “one quarter of all children” as appropriate. Show any assumed 1/2 probability of a son.
Trace paternal X carefully
A father transmits X to daughters and Y to sons in the XX/XY model.
Key explanations
- For XᴺXⁿ × XᴺY, the four equally likely genotypes are XᴺXᴺ, XᴺXⁿ, XᴺY and XⁿY.
- An XX person can show the phenotype if both X chromosomes carry Xⁿ. The condition is not restricted to males.
Avoid this misconception
The claim: An affected father passes an X-linked recessiveAn allele that is expressed in the phenotype only when no dominant allele is present (when homozygous). allele to all his sons.
The correction: His sons receive his Y chromosome. His X goes to his daughters in this model.
Why is red–green colour vision deficiency more common in boys?
The colour-vision gene is on the X chromosome. XY individuals have one X and no matching allele on Y, so a single recessive allele shows.
What the model shows
X and Y chromosomes passing from parents into gametes and a Punnett square, a pedigree of a simulated family, and a graph that can follow sons only or all children.
What to look for
Which children get the father’s X, and which get his Y. How the chance among sons differs from the chance among all children.
What it tells us
XᴺXⁿ × XᴺY gives ½ of sons affected, ¼ of all children. An affected father passes Xⁿ to all daughters and never directly to sons.
Be careful what you claim
The simplified XX/XY model from the lesson, with one gene and fictional families.
Try it in the model: add 100 children, then switch the graph to Among sons.
X-linked inheritance: colour vision
Follow X and Y chromosomes from parents into gametes and children, then compare the chance among sons with the chance among all children. All families are fictional.
What you see
- Two parents with their genotype, chromosomes and phenotype, the gametes each can make, and a Punnett square of the four equally likely offspring.
- A pedigree of the children you have simulated, a graph of the share with the phenotype as the family grows, and the expected ratios beside the observed counts.
How to use it
- Choose a cross type, then each parent’s genotype, or press a preset from the lesson.
- Press Have a child for one conception, or add 10 or 100 children at once. In X-linked mode, the graph can follow all children or sons only.
What it shows
- Genetic terms and monohybrid crosses with complete dominance and codominance (ABO), including pedigree reasoning (2.15(i)–(ii)).
- X-linked inheritance of red–green colour vision deficiency, with probabilities among sons and among all children (2.15(iii)).
Limits of the model
- One gene with full penetrance and no new mutations. Real conditions and colour vision have more complexity than this.
- The families are fictional and random. A real family’s outcomes need a clinical genetics service, not this model.
- In autosomal and ABO mode the children’s sex is chosen at random (½ each) only to draw the pedigree.
Guided tour · step 1 of 8
- The cross is XᴺXⁿ × XᴺY.
- Read the gametes and the Punnett square in the simulator.
- In Cross, just below these steps, press Have a child.
The mother’s eggs carry Xᴺ or Xⁿ. The father’s sperm carry Xᴺ or Y. Y has no colour-vision allele.
Cross
Questions
Check your understanding
Try each question before opening the suggested answer.
Practice question
Retrieval and explanation · suggested answer
For XᴺXⁿ × XᴺY, what fraction of sons are expected to have red–green colour vision deficiency?
Show suggested answer
Half: XⁿY. Sons receive either Xᴺ or Xⁿ from the mother with equal probability.
Practice question
Retrieval and explanation · suggested answer
For XⁿY × XᴺXᴺ, give the daughters’ and sons’ genotypes.
Show suggested answer
All daughters are XᴺXⁿ carriers; all sons are XᴺY. None show the recessive phenotype in this simplified cross.
Practice question
Retrieval and explanation · suggested answer
Why is XⁿY not described as heterozygous at this locus?
Show suggested answer
It has only one allele at this X-linked locus; there is no matching allele on Y. Heterozygous means two different alleles at the same locus.
Take the idea further
Apply the same specification knowledge to an unfamiliar problem. Any extra assumptions are supplied in the question.
Stretch question
An XᴺXⁿ parent and an XⁿY parent have a child. Calculate the probability of the recessive phenotype among daughters and among all children, assuming equally likely XX and XY outcomes.
Show a hint
Write all four combinations before grouping by phenotype.
Explore the reasoning
Daughters: XᴺXⁿ or XⁿXⁿ, so 1/2 affected. Sons: XᴺY or XⁿY, so 1/2 affected. Overall, two of four equally likely outcomes are affected: 1/2.
Recall this lesson
Lesson recall questions
Question 1 of 5: State what is meant by codominance.
Monohybrid inheritance and sex linkage
Practice question
Retrieval and explanation · suggested answer
State what is meant by codominance.
Lesson: 2C 2: Patterns of Inheritance
Try this from memory first.
Put your biology to work
Explore where these ideas appear in the world. Enrichment, not extra required recall.
Information should survive a colour change
2C · Sex LinkageThe National Eye Institute explains that many forms of colour vision deficiency are inherited, with common red–green forms linked to X. This makes colour choice relevant when communicating…Specification 2.15(iii)Read & use the scienceClose article
Why this matters
The National Eye Institute explains that many forms of colour vision deficiency are inherited, with common red–green forms linked to X. This makes colour choice relevant when communicating biology. For common inherited red–green forms, an X-linked recessive model helps explain why the pattern differs between people with one X chromosome and those with two. It also makes accessibility a scientific communication issue: a red and green distinction on a pedigree may disappear for some readers. Shapes, line styles and labels preserve meaning independently of hue. The application combines inheritance reasoning with experimental communication, without assuming that every form of colour vision deficiency follows the same genetic pattern.
Back to the specification
Pearson Edexcel IAL Biology · Issue 2, February 2021 · Unit 1, pp. 15–18. The application extends beyond required recall.
- 2.15(iii) understand sex linkage on the X chromosome, including red-green colour blindness in humans
Use the science
In a simplified X-linked recessive model, a carrier mother and a father without the condition have children. Explain the probability for their sons, then suggest how to label the pedigree so it works without a red/green distinction.
Explore a possible response
Each son receives Y from the father and one of the mother’s X chromosomes, so each has a 1-in-2 chance of inheriting the affected allele in this model. Use defined shading or symbols and direct genotype labels with a clear key, rather than colour alone. These probabilities apply to the specified inheritance model; not all colour vision conditions follow it.
Explore the source
- National Eye Institute: causes of colour vision deficiencyNational Eye Institute
The royal family tree meets DNA evidence
2C · Sex LinkageHaemophilia in European royal families is a famous pedigree example, but historical descriptions alone did not identify the exact molecular cause. Analysis of Romanov remains identified a likely…Specification 2.15(ii) · 2.15(iii) · 1.11Read & use the scienceClose article
Why this matters
Haemophilia in European royal families is a famous pedigree example, but historical descriptions alone did not identify the exact molecular cause. Analysis of Romanov remains identified a likely disease-causing variant in F9, an X-linked gene involved in blood clotting, supporting haemophilia B. The case shows how a family pattern and sequence evidence answer different questions. A pedigree can suggest how a condition is inherited; molecular analysis can identify a particular candidate change. Factor IX and RNA-splicing details are extension material. The useful Unit 1 connection is the route from X-linked inheritance to a changed protein and clotting function.
Back to the specification
Pearson Edexcel IAL Biology · Issue 2, February 2021 · Unit 1, pp. 15–18. The application extends beyond required recall.
- 2.15(ii) understand patterns of inheritance, including the interpretation of genetic pedigree diagrams, in the context of monohybrid inheritance
- 2.15(iii) understand sex linkage on the X chromosome, including red-green colour blindness in humans
- 1.11 understand the blood clotting process (thromboplastin release, conversion of prothrombin to thrombin and fibrinogen to fibrin) and its role in cardiovascular disease (CVD)
Use the science
In a simple X-linked recessive model, why can an affected father pass the variant to daughters but not directly to sons?
Explore a possible response
Daughters inherit his X chromosome, while sons inherit his Y. The mother’s contribution and genotype determine the full outcome for each child.
Explore the source
- Genotype analysis identifies the cause of the “royal disease”Rogaev et al., Science (2009), PubMed abstract · Abstract