Predict an X-linked cross from a carrier mother and expressing father
A supplied carrier-by-expressing cross can give the feature in both daughters and sons.
Section 17.4 · Monohybrid inheritance
Exam scope and assessment notes
Extended only (Supplement) · Biology 0970, 2026–2028. Use the stated specimen features and evidence. Diagrams are schematic. Written practice is self-checked; human subject-teacher review, supervised practical competence and cumulative assessment remain pending. Cambridge states that IGCSE 0970 has the same subject content as IGCSE 0610, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
Before you begin
- Recall chromosomes, genes and alleles, meiosis, haploid gametes and fertilisation restoring two chromosome sets.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which paternal gamete types fit XnY?
What you will learn
- Combine XNXn and XnY gamete contributions
- Calculate expressing daughters sons and all-offspring probabilities
Original teaching model. Pedigrees and records are fictional; allele and probability assumptions are stated.
| Feature | Meaning |
|---|---|
| Parents | XNXn × XnY |
| Daughters | XNXn or XnXn; no trait:trait = 1:1 |
| Sons | XNY or XnY; no trait:trait = 1:1 |
| All offspring | Two expressing positions out of four; overall ratio 1:1 |
Combine XNXn and XnY gamete contributions
The supplied mother is XNXn and the father XnY. Maternal gametes can carry XN or Xn; paternal gametes can carry Xn or Y. Combining the four equal positions gives XNXn, XNY, XnXn and XnY. Normalising allele order makes the first mixed female genotype XNXn rather than treating its reverse order as a different outcome.
Among daughters, XNXn is an unaffected carrier and XnXn expresses the model feature. Among sons, XNY is unaffected and XnY expresses it. Both the daughter group and the son group therefore have no trait:trait = 1:1 under the supplied expression rule. Female expression is possible; this cross illustrates why more common in males does not mean male-only.
Calculate expressing daughters sons and all-offspring probabilities
Across all offspring, two of the four equal positions express the feature and two do not, giving overall no trait:trait = 1:1. If the expected total is sixty, each genotype category has expected count fifteen. Expressing daughters fifteen plus expressing sons fifteen gives thirty expressing offspring overall, half the total. The allele key determines expression, not the shape of a pedigree symbol alone.
A prediction still depends on equal gamete transmission, independent combination, the usual XX/XY model and the stated expression rule. An unaffected mother’s phenotype by itself does not establish the supplied carrier genotype, and a finite sample is not required to have equal sex or phenotype counts. The cross is a teaching calculation rather than evidence about an actual learner or family.
Pause and explain
What is the expressing daughter genotype in this cross?
Worked example
Predict genotype categories and trait-expression ratios for XNXn × XnY.
Show the worked solution
- Combine XN/Xn maternal gametes with Xn/Y paternal gametes: XNXn, XNY, XnXn and XnY.
- Within daughters, one position is unaffected carrier and one expresses; within sons one is unaffected and one expresses. Each sex group and all offspring therefore have no trait:trait = 1:1.
Answer XNXn × XnY predicts XNXn, XNY, XnXn and XnY equally; each sex group and all offspring have no trait:trait = 1:1.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
List the four square positions from maternal XN/Xn and paternal Xn/Y, assigning daughter or son and expression category.
Give me a hint
Combine one maternal X with one paternal chromosome.
Compare my reasoning
- List XNXn and XNY from the XN row.
- List XnXn and XnY from the Xn row.
- Apply the expression key.
XNXn: unaffected carrier daughter; XNY: unaffected son; XnXn: expressing daughter; XnY: expressing son.
Look for these in your work
- Correct four genotype combinations
- Correct sex and expression categories
For 60 expected offspring under this model, give expressing-daughter, expressing-son and total expressing counts.
Give me a hint
Each of the four positions has probability one quarter.
Compare my reasoning
- Calculate 60 × 1/4 for each expressing genotype.
- Add the two expressing groups.
Expected expressing daughters 15, expressing sons 15, total expressing 30; overall trait probability is 1/2.
Look for these in your work
- Correct fifteen and fifteen
- Correct total thirty and half probability
Compare the overall phenotype prediction here with XNXn × XNY and explain which changed parental contribution matters.
Give me a hint
The father’s X contribution changes from N to n.
Compare my reasoning
- Identify paternal Xn rather than XN.
- Recognise the new XnXn expressing-daughter possibility.
- Compare overall expressing positions.
Here overall no trait:trait is 1:1 rather than 3:1 for XNXn × XNY. The expressing father supplies Xn to daughters, making XnXn possible alongside XnY.
Look for these in your work
- Correct two overall ratios
- Explains changed paternal X contribution
Common mistakes
- Claiming an XX individual cannot express the X-linked recessive feature
- Counting carrier status as affected expression
- Assuming an unaffected maternal phenotype alone proves XNXn
What can you explain now?
What is the expressing daughter genotype in this cross?
Compare with the explanation
XnXn
It receives n on both X contributions under the rule.
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Make it stick
Without your notes, combine xnxn and xny gamete contributions and calculate expressing daughters sons and all-offspring probabilities. Use the supplied allele key and state what the evidence can establish.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.