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

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. 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.

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. 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 ($X^H X^H$).

• 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.

Worked example: haemophilia

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

Mother $X^H X^h$ × Father $X^h Y$.

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.

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 the question.

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.

Examples 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 squared (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.

Try this

Q1. Duchenne muscular dystrophy is X-linked recessive. A non-carrier woman marries an affected man. Predict the genotypes and phenotypes of all children, distinguishing daughters from sons. [3 marks]

• Cue. Father passes his X with the allele to all daughters (all carriers) and his Y to all sons (all unaffected). No affected children in this cross.

Q2. In a population of 10 000 Australian-born males, approximately 800 are red-green colour blind. Calculate the expected frequency of affected females and explain why it is much lower. [3 marks]

• Cue. q = 0.08 in males; female frequency is q squared = 0.0064, so about 64 in 10 000. Females need both X chromosomes to carry the allele.

Q3. A pedigree shows an unaffected man and a phenotypically unaffected woman with two affected sons and one unaffected daughter. (a) Deduce the most likely inheritance pattern. (b) State the mother's genotype using $X^A$/$X^a$ notation. (c) Calculate the probability that the daughter is a carrier. [1+1+2 marks]

• Cue. (a) X-linked recessive: affected sons from unaffected mother. (b) $X^A X^a$ (carrier). (c) Given she is unaffected, the conditional probability is 2/3 since the three unaffected daughter genotypes from a carrier mother and unaffected father are XX, XX, Xx in ratio 1:1, so 1 in 2... refine: 1/2 carrier among all daughters, conditional on unaffected the answer is 1/2.