What are wrong denominators?

"50% of sons are affected" is different from "25% of all children are affected sons." Read what the question asks you to count.

What is read the command word?

"Predict" wants offspring genotypes and phenotypes; "determine" or "explain how a cross could show" wants you to compare outcomes under two hypotheses (sex-linked vs autosomal).

What are separate sons from daughters?

State the son outcome and the daughter outcome separately - the whole point of sex-linkage is that they differ.

What is quote the right denominator?

Decide whether the question wants a fraction of sons, a fraction of daughters, or a fraction of all children, and answer in those terms.

§-Syllabus dot point

NSWBiologySyllabus dot point

Inquiry Question 4: How can the genetic similarities and differences within and between species be compared?

Investigate the inheritance patterns including but not limited to: sex-linkage, codominance, incomplete dominance, multiple alleles

A focused answer to the HSC Biology Module 5 dot point on sex-linked (X-linked) inheritance. Why X-linked recessive disorders affect males more than females, the standard worked Punnett squares for carrier mothers, named examples (haemophilia, colour blindness, Duchenne muscular dystrophy), the criss-cross pedigree pattern, and worked HSC past exam questions.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-28

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
In context

What this dot point is asking

NESA wants you to explain sex-linked (X-linked) inheritance patterns and use Punnett squares to predict offspring probabilities for X-linked traits. The common command words here are "predict", "explain", "construct" (a Punnett square) and "determine" (whether a trait is sex-linked or autosomal). This dot point sits within a broader cluster of non-Mendelian inheritance patterns (sex-linkage, codominance, incomplete dominance, multiple alleles). This page focuses on sex-linkage.

Marks are earned for correct notation (allele written as a superscript on the X), a correctly worked Punnett square, the correct genotype and phenotype ratios, and a clear link between the male-only-has-one-X biology and the observed pattern.

The answer

Why sex-linked matters

In mammals (including humans), biological sex is determined by the sex chromosomes. Females are XX; males are XY. Most genes on the X chromosome have no equivalent on the much shorter Y chromosome.

This produces an asymmetry. Females have two copies of each X-linked gene. Males have only one copy (they are hemizygous). So a recessive allele on the X chromosome shows up immediately in a male (he has no second X to mask it), but only in females who are homozygous for the recessive allele.

The result: X-linked recessive disorders are far more common in males than in females. Classic examples include haemophilia, colour blindness, and Duchenne muscular dystrophy.

Notation

Use superscripts on the X chromosome to show the allele.

$X^H$ = dominant (normal/unaffected) allele.
$X^h$ = recessive (affected) allele.
Y = no allele (Y is irrelevant for X-linked traits).

Female genotypes can be $X^H X^H$ (unaffected, homozygous), $X^H X^h$ (unaffected carrier), or $X^h X^h$ (affected).
Male genotypes can be $X^H Y$ (unaffected) or $X^h Y$ (affected).

Why no male carriers

A male only has one X chromosome. He either has the recessive allele (and is affected) or he doesn't (and is unaffected). There is no carrier state for males in X-linked recessive inheritance.

Standard carrier-mother cross

Carrier mother ( $X^H X^h$ ) × unaffected father ( $X^H Y$ ).

Daughters: 50% unaffected non-carrier, 50% unaffected carrier. No affected daughters.
Sons: 50% unaffected, 50% affected.

The two key statistics from this cross.

50% of sons are affected.
50% of daughters are carriers (none are affected).

Affected father, unaffected mother

Affected father ( $X^h Y$ ) × unaffected non-carrier mother ( $X^H X^H$ ).

	$X^h$	Y
$X^H$	$X^H X^h$	$X^H Y$
$X^H$	$X^H X^h$	$X^H Y$

All daughters are carriers ( $X^H X^h$ ).
All sons are unaffected ( $X^H Y$ ).

This is why an affected man cannot pass the X-linked allele to his sons (he passes Y to sons, not his X). All his daughters become carriers, however.

The criss-cross pattern

Because an affected father gives his $X^h$ to every daughter (making her a carrier) and gives Y to every son, the allele appears to "skip" the second generation in the male line and re-emerge in grandsons. The classic pedigree signature is an affected grandfather, unaffected carrier daughters, and affected grandsons - the allele crossing from the male line, through a carrier female, back into the male line.

In a pedigree, a filled square (affected male) at the top and another filled square two rows down, linked through an unaffected mother, is the giveaway of X-linked recessive inheritance.

Worked example: haemophilia

A carrier mother and an affected father have children. Predict offspring outcomes.

Mother $X^H X^h$ × Father $X^h Y$ .

	$X^h$	Y
$X^H$	$X^H X^h$ (carrier daughter)	$X^H Y$ (unaffected son)
$X^h$	$X^h X^h$ (affected daughter)	$X^h Y$ (affected son)

Daughters: 50% carrier, 50% affected.
Sons: 50% unaffected, 50% affected.

This is the only standard cross that produces affected daughters in X-linked recessive inheritance.

Model exam answers

Worked example 1 - predicting a carrier-mother cross (3 marks)

Question. A woman who is a carrier for haemophilia (X-linked recessive) has children with an unaffected man. Use a Punnett square to predict the proportion of their sons and daughters expected to be affected or carriers.

Model answer. Let $X^H$ be the unaffected allele and $X^h$ the haemophilia allele. The mother is $X^H X^h$ and the father is $X^H Y$ .

	$X^H$	$X^h$
$X^H$	$X^H X^H$	$X^H X^h$
Y	$X^H Y$	$X^h Y$

Daughters ( $X^H X^H$ or $X^H X^h$ ) are all unaffected, with $\tfrac{1}{2}$ being carriers. Sons ( $X^H Y$ or $X^h Y$ ) are $\tfrac{1}{2}$ unaffected and $\tfrac{1}{2}$ affected. So 50% of sons are expected to have haemophilia and 50% of daughters are expected to be carriers, with no affected daughters.

Marker's note. One mark for correct genotypes and a correctly set-out Punnett square, one mark for the correct daughter outcome (none affected, half carriers), one mark for the correct son outcome (half affected). Bare "H/h" notation without the X loses the notation mark.

Worked example 2 - determining sex-linked vs autosomal (4 marks)

Question: A recessive coat colour appears more often in male mice than female mice. Explain how a single cross could confirm whether the trait is X-linked recessive rather than autosomal recessive.
Model answer: Cross an affected female with an unaffected male. If the trait is X-linked recessive, the affected female is $X^a X^a$ and the male is $X^A Y$ : all sons inherit her $X^a$ and the Y, so all sons are affected ( $X^a Y$ ), while all daughters inherit her $X^a$ and his $X^A$ , so all daughters are unaffected carriers ( $X^A X^a$ ). If instead the trait is autosomal recessive, the affected female is $aa$ and an unaffected male is $AA$ (all offspring unaffected) or $Aa$ (half affected, equally in both sexes). The key signal is the SEX difference: if every son is affected and no daughter is affected, the trait is X-linked recessive; if affected offspring appear equally in both sexes, it is autosomal.
Marker's note: Top band requires the decisive cross named, both predicted outcomes (X-linked and autosomal) modelled with correct notation, and an explicit decision rule that links the sex pattern of the offspring to the conclusion. Vague answers that only describe one hypothesis cap at 2 marks.

Worked example 3 - reading a pedigree (3 marks)

Question: A pedigree shows an unaffected couple with two affected sons and one unaffected daughter for an X-linked recessive disorder. State the genotype of each parent and the probability that the unaffected daughter is a carrier.
Model answer: The father is unaffected, so he is $X^H Y$ . The mother is unaffected but has affected sons, who must have received $X^h$ from her, so she is the carrier $X^H X^h$ . From $X^H X^h \times X^H Y$ , all daughters are unaffected ( $X^H X^H$ or $X^H X^h$ in a $1:1$ ratio). Because every daughter of this cross is unaffected, knowing she is unaffected gives no extra information, so the probability she is a carrier is $\tfrac{1}{2}$ .
Marker's note: One mark for each parental genotype with correct reasoning, one mark for the $\tfrac{1}{2}$ carrier probability. A common error is to quote $\tfrac{2}{3}$ (the conditional probability used for autosomal recessive carriers when an affected genotype is possible) - here all daughters are unaffected, so the conditioning changes nothing and the answer is $\tfrac{1}{2}$ .

Common sex-linkage traps

Wrong notation: Always write the allele as a superscript on the X. Writing just "H" or "h" without the X is a 1-mark deduction because it hides the sex-linkage.
Forgetting the Y has no allele: The Y chromosome does not carry the X-linked gene. Males are NEVER carriers of X-linked traits.
Wrong denominators: "50% of sons are affected" is different from "25% of all children are affected sons." Read what the question asks you to count.
Confusing X-linked dominant and X-linked recessive: Most exam questions focus on X-linked recessive. X-linked dominant is rare; if a question says dominant, the pattern flips.
Misusing the 2/3 carrier rule: The $\tfrac{2}{3}$ conditional probability belongs to autosomal recessive carriers when an affected genotype is possible. For an X-linked cross where all daughters come out unaffected, the conditional carrier probability stays $\tfrac{1}{2}$ .

In context

Example 1. Red-green colour blindness in Australian schoolboys. Roughly 8 percent of Australian males of European ancestry are red-green colour blind, compared with about 0.5 percent of females. The trait is X-linked recessive. A male only needs one copy of the affected allele on his single X chromosome to express colour blindness; a female needs the affected allele on both X chromosomes. Mathematically, if the allele frequency in the population is $q = 0.08$ , the male incidence is $q$ (8 percent) while the female incidence is $q^2$ (0.64 percent). NSW Education's screening for trade apprenticeships in electrical work uses Ishihara plates because miswiring an electrical panel by colour error is a safety hazard.

Example 2. Haemophilia in Queen Victoria's descendants. Queen Victoria was a carrier of the X-linked recessive allele for haemophilia B. She passed the allele to several daughters, who became carriers, and to her son Leopold, who was affected and died at 30. Through royal intermarriage across Europe, the allele entered the Spanish, German and Russian royal families. The most famous affected descendant was Tsarevich Alexei of Russia. Modern genetic counsellors at Westmead Hospital use exactly the same Punnett-square logic applied to Victoria's pedigree when advising families with an X-linked condition that any son of a carrier has a 50 percent chance of being affected.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2020 HSC4 marksExplain how a cross between individuals I and II could be used to determine whether the inheritance of colour in the fish is sex-linked or autosomal. [Individual I is an orange male, individual II is a yellow (recessive) female; yellow has been shown to be recessive.]

Show worked answer →

Top marks (4) require explaining the possible outcomes of the cross AND relating differences in those outcomes to justify the type of inheritance, communicated succinctly with appropriate terms/formats (e.g. Punnett squares).

Sample answer (marking guidelines):

If sex-linked: I and II would be X^A Y and X^a X^a respectively. The cross would produce all male offspring yellow and all female offspring orange.
If autosomal: II would be aa, while I would be AA or Aa. If I is AA all offspring are orange; if I is Aa, then 50% of offspring would be yellow and 50% orange, with colours distributed equally between male and female offspring.
Therefore, the absence of any orange male offspring from this cross would confirm the inheritance is sex-linked.

Markers noted poor use of Punnett-square format and limited understanding of the difference between autosomal and sex-linked inheritance.

Source: NESA 2020 HSC Biology examination and marking guidelines.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marksDefine the terms "carrier" and "sex-linked" as they apply to X-linked recessive inheritance.

Show worked solution →

Carrier (1 mark). A carrier is an individual who has one copy of a recessive allele but does not express the trait because a dominant allele masks it. For an X-linked recessive trait, a carrier is a heterozygous female ( $X^H X^h$ ) who is phenotypically unaffected but can pass the allele to her offspring.

Sex-linked (1 mark). A sex-linked trait is controlled by a gene located on a sex chromosome (almost always the X chromosome in HSC examples), so its inheritance pattern differs between males and females.

foundation3 marksDuchenne muscular dystrophy is X-linked recessive. A non-carrier woman has children with an affected man. Predict the genotypes and phenotypes of all children, distinguishing daughters from sons.

Show worked solution →

Parental genotypes (1 mark): Mother $X^H X^H$ (unaffected non-carrier); father $X^h Y$ (affected).
Daughters (1 mark): Every daughter inherits her father's only X (the $X^h$ ) and one of her mother's $X^H$ alleles, so all daughters are $X^H X^h$ , that is, unaffected carriers.
Sons (1 mark): Every son inherits his father's Y and one of his mother's $X^H$ alleles, so all sons are $X^H Y$ , that is, unaffected.

No children are affected in this cross; the key teaching point is that an affected father passes the allele to all daughters but to none of his sons.

foundation3 marksA carrier mother (

X^H X^h

) has children with an unaffected father (

X^H Y

) for an X-linked recessive condition. State the proportion of sons affected, the proportion of daughters affected, and the proportion of daughters who are carriers. Show your reasoning with a Punnett square.

Show worked solution →

Punnett square (1 mark): Mother gametes $X^H$ and $X^h$ ; father gametes $X^H$ and Y. Offspring: $X^H X^H$ , $X^H X^h$ , $X^H Y$ , $X^h Y$ .
Sons (1 mark): Sons are $X^H Y$ or $X^h Y$ in equal proportion, so $\tfrac{1}{2}$ (50%) of sons are affected.
Daughters (1 mark): Daughters are $X^H X^H$ or $X^H X^h$ in equal proportion. No daughter is affected (0%), and $\tfrac{1}{2}$ (50%) of daughters are carriers.

core4 marksRed-green colour blindness is X-linked recessive. In a population, the frequency of the affected allele is

q = 0.08

. Calculate the expected proportion of affected males and the expected proportion of affected females, and explain why the two values differ so greatly.

Show worked solution →

Affected males (1 mark): A male is affected if his single X carries the recessive allele, so the male incidence equals the allele frequency: $q = 0.08$ , that is, 8%.
Affected females (1 mark): A female is affected only if BOTH X chromosomes carry the recessive allele, so the female incidence equals $q^2 = 0.08^2 = 0.0064$ , that is, 0.64%.
Explanation (2 marks): Award 1 mark for stating that males are hemizygous (only one X), so a single recessive allele is enough to make a male affected, whereas a female needs two. Award the second mark for linking this to the maths: requiring two independent copies multiplies the probability ( $q \times q = q^2$ ), which for a small $q$ is far smaller than $q$ itself, so affected females are much rarer than affected males.

core4 marksA phenotypically unaffected couple have a son with haemophilia (X-linked recessive) and a phenotypically unaffected daughter. (a) Deduce the genotype of each parent. (b) Calculate the probability that the daughter is a carrier. Justify your answer.

Show worked solution →

Part (a) - parents (2 marks). The father is unaffected, so he is $X^H Y$ (1 mark). The mother is unaffected but has an affected son, who must have inherited his $X^h$ from her, so she must carry one $X^h$ ; being unaffected she is the heterozygous carrier $X^H X^h$ (1 mark).

Part (b) - daughter (2 marks). From the cross $X^H X^h \times X^H Y$ , daughters are $X^H X^H$ or $X^H X^h$ in a $1:1$ ratio, and BOTH are unaffected (1 mark). The daughter is known to be unaffected, but all daughters of this cross are unaffected, so this gives no extra information; the conditional probability that she is a carrier is therefore $\tfrac{1}{2}$ , that is, 50% (1 mark).

core5 marksA breeder crosses a carrier female cat (

X^O X^o

) with a male carrying the recessive allele (

X^o Y

) for an X-linked recessive coat condition. Construct a Punnett square, give the genotypic and phenotypic ratios of the offspring, and state which is the only standard X-linked recessive cross that produces affected daughters.

Show worked solution →

Punnett square (2 marks): Female gametes $X^O$ and $X^o$ ; male gametes $X^o$ and Y. Offspring: $X^O X^o$ , $X^o X^o$ , $X^O Y$ , $X^o Y$ (1 mark for correct gametes, 1 mark for correct cells).
Genotypic ratio (1 mark): $1\ X^O X^o : 1\ X^o X^o : 1\ X^O Y : 1\ X^o Y$ .
Phenotypic ratio (1 mark): Daughters: $\tfrac{1}{2}$ carrier (unaffected), $\tfrac{1}{2}$ affected. Sons: $\tfrac{1}{2}$ unaffected, $\tfrac{1}{2}$ affected. Overall $\tfrac{1}{4}$ of offspring are affected daughters and $\tfrac{1}{4}$ are affected sons.
Affected daughters (1 mark): A carrier mother crossed with an AFFECTED father ( $X^o Y$ ) is the only standard X-linked recessive cross that can produce affected daughters, because a daughter needs a recessive allele from BOTH parents and only an affected father supplies an $X^o$ to every daughter.

exam6 marksA geneticist is studying a coat-colour trait in a strain of mice and suspects it is X-linked recessive rather than autosomal recessive. Design a single test cross that would distinguish the two hypotheses, predict the offspring outcomes under each hypothesis, and explain how the results would let you decide. Use appropriate notation and Punnett squares.

Show worked solution →

The decisive cross (1 mark): Cross an affected female with an unaffected (wild-type) male. This reciprocal cross is decisive because the outcome differs sharply depending on whether the gene is on the X chromosome or an autosome.
If X-linked recessive (2 marks): Affected female is $X^a X^a$ ; unaffected male is $X^A Y$ . Every son inherits his mother's $X^a$ and his father's Y, so ALL sons are $X^a Y$ (affected). Every daughter inherits her mother's $X^a$ and her father's $X^A$ , so ALL daughters are $X^A X^a$ (unaffected carriers). Prediction: all sons affected, all daughters unaffected - a clear sex difference (criss-cross inheritance).
If autosomal recessive (2 marks): Affected female is $aa$ ; unaffected male is $AA$ (or $Aa$ ). If the male is $AA$ , all offspring are $Aa$ and unaffected, with no difference between sexes. If the male is $Aa$ , half the offspring are affected, again with no difference between the sexes.
Decision rule (1 mark): If the trait appears in ALL the sons but NONE of the daughters (a sex-linked difference), the gene is X-linked recessive. If the affected and unaffected offspring are distributed equally between the sexes, it is autosomal. Award full marks only where the candidate states the test cross, gives both predictions with correct notation, and links the observed sex pattern to a decision.

exam7 marksHaemophilia is an X-linked recessive disorder. Evaluate the claim that "an X-linked recessive disorder can never pass directly from an affected father to his son." In your answer, use Punnett squares to model the relevant crosses, address the role of the carrier state, and discuss the limits of the claim.

Show worked solution →

Restating and supporting the claim (2 marks): A father passes his Y chromosome (not his X) to every son, so an affected father ( $X^h Y$ ) gives each son a Y and the son's single X comes from the mother. Modelling $X^h Y \times X^H X^H$ : sons are $X^H Y$ (all unaffected) and daughters are $X^H X^h$ (all carriers). The father's affected allele therefore reaches his daughters, never his sons - supporting the claim for father-to-son transmission of the allele itself.
The carrier state and how a son CAN be affected (2 marks): A son becomes affected only by inheriting $X^h$ from his MOTHER. If the mother is a carrier ( $X^H X^h$ ), modelling $X^H X^h \times X^H Y$ gives sons $X^H Y$ and $X^h Y$ in a $1:1$ ratio, so half the sons are affected. The allele a father passes to his daughters can re-emerge in his GRANDSONS, because those carrier daughters can pass $X^h$ to their own sons (the criss-cross pattern).
Evaluating the limits (2 marks): The claim is true for the direct father-to-son step of the SAME allele, because of Y-only transmission to sons. It is misleading if read as "the disorder cannot run in the male line," because it reappears in grandsons via carrier daughters. The claim also assumes standard X-linkage; it would not hold for an autosomal or Y-linked trait, or in the rare case of a son who is affected by a new mutation.
Communication (1 mark): Awarded for a sustained, logically sequenced judgement that uses correct $X^H$ / $X^h$ /Y notation and at least one correct Punnett square to support each part of the evaluation.