Biology • Year 12 • Module 5 • Lesson 14

Mendelian Patterns — Autosomal Inheritance, Sex Linkage, Punnett Squares

Build HSC band 5–6 extended-response technique on inheritance — Punnett-square reasoning, pedigree analysis, and the precise language of probability for autosomal and X-linked traits.

Master · Extended Response

1. Extended response — determine and justify an inheritance pattern (Band 5–6)

7 marks Band 5–6

Stimulus. A four-generation pedigree shows a particular condition with the following features:

The trait appears in every generation.
Both males and females are affected, in roughly equal numbers.
Every affected individual has at least one affected parent.
Unaffected × unaffected crosses do not produce affected children.
There is one example of an affected father passing the trait to his son.

Q1. Determine and justify the most likely mode of inheritance (autosomal dominant, autosomal recessive, X-linked dominant or X-linked recessive). In your response you must:

State the conclusion explicitly.
Refer to each of the five features above.
Explain why the other three modes of inheritance are excluded.
Use a sample Punnett square (in words is fine) to show a consistent affected × unaffected cross.

Stuck? Plan first: state mode → match all five clues to that mode → exclude each of the other three → finish with a Punnett-based probability statement.

2. Stimulus-based extended response — haemophilia in the British royal family (Band 5–6)

8 marks Band 5–6

Stimulus. Haemophilia B is caused by a recessive allele on the X chromosome that disables a clotting factor. Queen Victoria of the United Kingdom (1819–1901) is believed to have been a carrier for haemophilia B. Several of her sons were affected by the condition, and several of her daughters were obligate carriers, spreading the affected allele into the Russian, Spanish and German royal families through marriage. Affected males rarely survived to reproductive age in the nineteenth century. Modern molecular testing has confirmed the haemophilia B allele in surviving genetic samples from her descendants. (After Rogaev et al. (2009), Science 326: 817.)

Q2. Analyse and evaluate, using lesson content, why this pattern of inheritance is consistent with X-linked recessive transmission, and assess the implications for the spread of the affected allele across multiple royal lineages.

In your answer:

Define X-linked recessive inheritance using chromosome logic (X and Y).
Use a Punnett square (or equivalent reasoning) for a carrier-mother × unaffected-father cross to predict the probability of affected sons, unaffected sons, carrier daughters and unaffected daughters.
Explain why Queen Victoria's daughters became carriers who spread the allele further while the affected sons did not transmit it.
Evaluate why the affected allele continued to appear in subsequent generations across multiple lineages.

Stuck? Use Card 3's haemophilia worked paragraph as your spine, attach the Card 4 "no father-to-son X transmission" rule as your exclusion argument, then evaluate the historical pattern as the synthesis.

3. Evaluate this claim (Band 5–6)

6 marks Band 5–6

"A Punnett square proves how children in a family will turn out. If two carriers have four children and the Punnett square predicts a 3:1 phenotype ratio, then three of the four children must be unaffected and exactly one must be affected. Also, because dominant alleles are more powerful than recessive alleles, the dominant phenotype is always more common in the population."

Q3. Evaluate this claim. Identify which parts are correct, which are wrong, and reformulate it into a biologically defensible statement using precise language of probability and the lesson's framing of dominance.

Stuck? Revisit lesson § Card 1 (probability vs certainty), Misconceptions box (dominance ≠ frequency), and the Trap callout in Card 2.

Answers — Do not peek before attempting

Q1 — Sample Band 6 response (7 marks), annotated

The most likely mode of inheritance is autosomal dominant. [1 — conclusion]

The trait appears in every generation, and every affected individual has at least one affected parent. This is the signature of a dominant pattern: heterozygous individuals express the trait, so the allele cannot "hide" in unaffected carriers between generations. [1 — dominant clues] Both males and females are affected in roughly equal numbers, and there is at least one case of father-to-son transmission. Both observations argue against any X-linked mode: father-to-son transmission of an X-linked allele is impossible because fathers pass Y, not X, to sons. Autosomal inheritance puts the gene on a chromosome that both sexes inherit two copies of, so affected males and females occur at similar frequency. [1 — autosomal clues + father-to-son rules out X-linked]

The three other modes are excluded as follows:

Autosomal recessive — would typically skip generations (carriers don't show the trait) and unaffected × unaffected matings can produce affected children. Neither pattern is present here. [1 — excludes AR]
X-linked recessive — would predict more affected males than females, possibly unaffected female carriers, and no father-to-son transmission. The roughly equal sex ratio of affected individuals and the observed father-to-son transmission both contradict this. [1 — excludes XLR]
X-linked dominant — an affected father would pass X^A only to daughters (he gives Y to sons), so no sons of an affected father would be affected. The stimulus has a father-to-son transmission, ruling this out. [1 — excludes XLD]

An affected heterozygote (Aa) crossed with an unaffected homozygous recessive (aa) gives a Punnett-square outcome of 50% Aa (affected) : 50% aa (unaffected) per child. Each child therefore has a 50% probability of being affected — but with three children, any 0:3 to 3:0 ratio of affected to unaffected is possible because fertilisation events are independent. [1 — Punnett + probability language]

Q2 — Sample Band 6 response (8 marks), annotated

Haemophilia B is X-linked recessive, meaning the affected allele is located on the X chromosome and is expressed only when no dominant X^H allele is present to mask it. Because males have only one X chromosome (genotype XY), a single recessive allele on that X is enough to express the trait; females (XX) need two copies. [1 — XLR definition with chromosome logic]

Queen Victoria was a phenotypically unaffected carrier with genotype X^HX^h. Her unaffected husband was X^HY. [1 — correct genotypes with allele notation] The Punnett square for X^HX^h × X^HY gives four offspring genotypes: X^HX^H, X^HX^h, X^HY, X^hY. [1 — Punnett square correct] Per child the probabilities are 25% unaffected daughter, 25% carrier daughter, 25% unaffected son and 25% affected son (equivalently 50% of sons affected, 50% of daughters carriers). These are per-fertilisation probabilities, not fixed family outcomes. [1 — probabilities stated correctly]

Victoria's carrier daughters were phenotypically unaffected, reached reproductive age and married into other royal lineages. Each carrier daughter had a 50% probability per child of passing X^h through her own meiotically formed gametes, so she could produce both carrier daughters and affected sons in her new family without ever showing the condition herself — the allele effectively travelled invisibly through female carriers. [1 — carrier-daughter transmission]

Her affected sons, by contrast, were genetic dead-ends for this allele in the 19th century. Most did not survive to reproduce, but even those who did would pass their Y chromosome (not X^h) to any sons, and would pass X^h only to daughters — automatically making any daughters obligate carriers, not transmitting the trait son-to-son. [1 — affected-son transmission limits]

The complete absence of father-to-son X transmission is what locks the inheritance pattern as X-linked rather than autosomal recessive. If the trait were autosomal recessive, affected sons could pass the affected allele directly to other sons; the strict father-Y-to-son rule from the lesson's Card 4 eliminates that possibility here. [1 — applies no-father-to-son rule]

Overall, the inheritance pattern in the royal families is fully consistent with X-linked recessive transmission: carrier daughters silently re-seeded the allele in each generation, while affected sons produced visible cases without spreading the allele further. This explains how a single carrier (Victoria) could seed haemophilia B into three distinct European royal lineages over multiple generations, despite the recessive nature of the allele. [1 — integrated evaluative judgement]

Q3 — Sample Band 6 response (6 marks)

The claim is almost entirely flawed. [1 — judgement]

What is wrong:

"A Punnett square proves how children will turn out." A Punnett square is a probability model, not a deterministic prediction. It states the chance of each genotype per fertilisation event, and each fertilisation event is independent of the others. [1 — probability not certainty]
"Three of the four children must be unaffected and exactly one must be affected." A 3:1 phenotype ratio is the expected ratio per child, not a guaranteed family outcome. With four children, any combination from 0:4 to 4:0 affected:unaffected is possible — the 3:1 ratio only emerges reliably across very many fertilisation events. [1 — 3:1 is per-child probability]
"Dominant alleles are more powerful than recessive alleles." Dominance describes whether an allele is expressed in a heterozygote — it is a relationship about phenotype expression, not about strength, health, or fitness. [1 — dominance is not "power"]
"The dominant phenotype is always more common in the population." Frequency depends on allele frequency in a population (governed by mutation, selection, drift, gene flow) — not on dominance. The lesson gives Huntington's disease as an example of a rare dominant allele; many common phenotypes are produced by recessive alleles. [1 — dominance ≠ frequency]

Defensible reformulation: "A Punnett square models the per-child probability of each offspring genotype given the parents' genotypes — for example, two carriers of an autosomal recessive condition (Aa × Aa) produce a 1:2:1 genotype ratio and 3:1 phenotype ratio per fertilisation event, not a guaranteed family outcome. Dominance refers only to expression in a heterozygote; whether an allele is dominant or recessive says nothing about its frequency in a population." [1 — biologically defensible reformulation]