Predict an X-linked cross from a carrier mother and unaffected father
Use the right denominator: half the sons but one quarter of all offspring express the trait.
Section B16.3 · Monohybrid inheritance
Exam scope and assessment notes
Mapped to the named 0973 section using teaching from the corresponding single-science course. A 0973-specific review of every Core and Supplement objective is still in progress. Written practice is self-checked; it does not demonstrate supervised laboratory competence or award an official Cambridge grade. This qualification reports two grades on the 9–1 scale; confirm the entry route for your zone and exam series.
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 gamete types can the XNY father contribute here?
What you will learn
- Construct the XNXn crossed with XNY square
- Separate daughter son and all-offspring phenotype probabilities
Original teaching model. Pedigrees and records are fictional; allele and probability assumptions are stated.
| Feature | Meaning |
|---|---|
| Mother and father | XNXn × XNY |
| Daughters | XNXN or XNXn; no model trait, half carriers |
| Sons | XNY or XnY; no trait:trait = 1:1 |
| All offspring | No trait:trait = 3:1; trait probability 1/4 |
Construct the XNXn crossed with XNY square
The supplied usual XX/XY model gives the mother genotype XNXn and the father XNY. The mother’s gametes can carry XN or Xn; the father’s can carry XN or Y. Combine one contribution from each in a two-by-two square. A Y contribution does not carry a second N or n allele at the tracked locus.
The four equal positions are XNXN, XNY, XNXn and XnY. The two XX daughter possibilities are XNXN, not carrying n, and XNXn, an unaffected carrier under the model. The two XY son possibilities are XNY without the feature and XnY expressing it. Genetic carrier status is not an additional visible affected phenotype in this classroom rule.
Separate daughter son and all-offspring phenotype probabilities
Among daughters, the model predicts no expression and half carry n. Among sons, expression and no expression each have probability one half, a 1:1 ratio. Across all offspring with the stipulated equal sex probabilities, one of four positions expresses the feature: one quarter, compared with three quarters not expressing it. Overall no trait:trait is therefore 3:1.
State the denominator before quoting a percentage. Fifty per cent of sons expressing the feature is compatible with twenty-five per cent of all offspring expressing it. For eighty expected offspring, the model gives twenty in each of the four genotype categories and twenty expressing sons overall. Actual offspring counts are not guaranteed to match the expected pattern exactly.
Sources for this lesson
Pause and explain
What is the trait probability among all offspring in the equal-sex model?
Worked example
For XNXn × XNY, distinguish the chance of the model trait among sons from its chance among all offspring.
Show the worked solution
- Combine maternal XN/Xn with paternal XN/Y to obtain XNXN, XNY, XNXn and XnY in four equal-probability positions.
- Only XnY expresses the trait. It is one of the two son positions (1/2 of sons) and one of all four positions (1/4 of all offspring).
Answer XNXn × XNY gives XNXN, XNY, XNXn and XnY. Trait among sons = 1/2; trait among all offspring = 1/4 under the equal-probability model.
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.
Fill the square with maternal XN/Xn rows and paternal XN/Y columns and assign each expression category.
Give me a hint
Combine one maternal X with one paternal contribution.
Compare my reasoning
- List XNXN, XNY, XNXn and XnY.
- Apply the model expression rule.
XNXN: unaffected daughter; XNY: unaffected son; XNXn: unaffected carrier daughter; XnY: expressing son.
Look for these in your work
- Correct four genotype combinations
- Correct expression and carrier assignments
For 80 expected offspring under this model, give expressing sons, total sons and the trait percentage among sons versus all offspring.
Give me a hint
Twenty in each of four equal genotype positions gives two different denominators.
Compare my reasoning
- Calculate twenty expressing sons and forty total sons.
- Calculate 20/40 and 20/80 as percentages.
Expected expressing sons 20 and total sons 40. Trait among sons is 50%; trait among all offspring is 25%.
Look for these in your work
- Correct expected counts
- Correct fifty and twenty-five per cent denominators
Explain why an unaffected daughter can carry n and why this does not alter the 3:1 all-offspring no-trait:trait ratio under the rule.
Give me a hint
Carrier genotype and expressed phenotype are separate classifications.
Compare my reasoning
- Identify XNXn as carrying n with N present.
- Apply the supplied N-dominant rule.
- Count the one expressing XnY position.
XNXn carries n but does not express the feature because N is present under the model. Three positions do not express it and one XnY position does, giving no trait:trait = 3:1 across all offspring.
Look for these in your work
- Separates carrier from expression
- Correct overall labelled ratio
Common mistakes
- Calling unaffected carrier daughters affected in this rule
- Using the sons-only half as an all-offspring half
- Giving the father’s Y an N allele
What can you explain now?
What is the trait probability among all offspring in the equal-sex model?
Compare with the explanation
One quarter
Only XnY among the four equal positions expresses the trait.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Without your notes, construct the xnxn crossed with xny square and separate daughter son and all-offspring phenotype 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.