Biology 9700/42 — October/November 2016

Background Concept

Humans have 23 pairs of chromosomes: 22 pairs of autosomes and 1 pair of sex chromosomes. The sex chromosomes are X and Y. Females are XX, so they carry two copies of every gene on the X chromosome; males are XY, so they carry only one copy of X-linked genes. The Y chromosome is much smaller and carries very few genes, so most sex-linked traits in A-level Biology are described as X-linked.

Because males have only one X chromosome, any recessive allele on it is expressed in the phenotype — there is no second allele to mask it. This is why X-linked recessive conditions (such as haemophilia, red–green colour blindness, and the form of DI described in this question) appear far more often in males than in females.

Understanding the Question

The question is a one-mark 'explain the term' item. The mark scheme accepts 'allele/gene carried on the X chromosome' (or 'sex chromosome' is allowed). A complete sentence is expected.

Approach

State the location of the allele clearly. It is not enough to say 'on a chromosome' — you must name the sex chromosome, ideally X.

Step-by-Step Reasoning

• Identify the locus of the gene: a sex chromosome (X is preferred).
• Combine into a single sentence: 'Sex linkage means the allele is located on the X chromosome.'

Key Takeaways

• Sex linkage = gene/allele on a sex chromosome (X in most A-level contexts).
• This explains the unequal sex ratios seen in X-linked recessive conditions.

Common Mistakes

• Writing 'a gene on a chromosome' — too vague; the mark scheme requires the sex chromosome.
• Saying 'on the Y chromosome' — almost always wrong for A-level questions, because the named sex-linked conditions (haemophilia, colour blindness, DI, Duchenne muscular dystrophy) are X-linked.

Things to Be Careful About

• If the question were about holandric (Y-linked) inheritance, you would specify the Y chromosome. The Cambridge syllabus treats sex linkage as X-linked unless stated otherwise.

Background Concept

Inheritance of a sex-linked (X-linked) recessive allele follows special rules:

• A female has two X chromosomes, so she can be homozygous normal ($\text{X}^A\text{X}^A$), a carrier/heterozygous ($\text{X}^A\text{X}^a$, phenotypically normal), or homozygous affected ($\text{X}^a\text{X}^a$, with DI).
• A male has one X and one Y, so he has only one allele for any X-linked gene. He is either normal ($\text{X}^A\text{Y}$) or affected ($\text{X}^a\text{Y}$); there are no male carriers.
• An affected father passes his $\text{X}^a$ to every daughter, making all daughters at least carriers.
• An affected son must have received $\text{X}^a$ from his mother (his Y comes from his father), so the mother must be a carrier or herself affected.

In this question, the parents are not affected, but they have an affected son. The mother must therefore be a carrier, and the father is normal.

Understanding the Question

The question gives you the four-line scaffold (symbols, parental genotypes, gametes, offspring genotypes, offspring phenotypes) and asks for a complete genetic diagram. Four marks are available: one for correct symbols, one for correct parental genotypes and gametes, one for the offspring genotypes, and one for correctly assigning the phenotypes in the right order.

The clue that the inheritance is X-linked recessive is in part (b): the faulty receptor allele is sex-linked, and the affected child's sex is implied to be male (because an affected daughter would need an affected father, who is not present). So the child with DI is a son with genotype $\text{X}^a\text{Y}$.

Approach

1. Define the symbols with the recessive allele in lower case (a) attached to the X chromosome, and the dominant normal allele as A.
2. Use the fact that neither parent is affected but they have an affected son to deduce that the mother is $\text{X}^A\text{X}^a$ and the father is $\text{X}^A\text{Y}$.
3. List all four gamete types.
4. Combine them in a Punnett square.
5. Convert each offspring genotype into a phenotype, and list them in the order the Punnett square is read.

Step-by-Step Reasoning

• Symbols (1 mark). The mark scheme requires the dominant normal allele as A and the recessive DI allele as a, written as superscripts on the X (e.g. $\text{X}^A$, $\text{X}^a$). Do not use a slash — superscripts attached to the X chromosome are the conventional CIE notation.
• Parental genotypes and gametes (1 mark). Mother $\text{X}^A\text{X}^a$ produces eggs carrying $\text{X}^A$ or $\text{X}^a$ (in equal numbers). Father $\text{X}^A\text{Y}$ produces sperm carrying $\text{X}^A$ or Y.
• Punnett square and offspring genotypes (1 mark). The four combinations of these gametes give the four offspring genotypes shown in the square: $\text{X}^A\text{X}^A$, $\text{X}^A\text{X}^a$, $\text{X}^A\text{Y}$, $\text{X}^a\text{Y}$.
• Phenotypes in the correct order (1 mark). Reading the square in the same order as the genotypes, the phenotypes are: female normal, female normal (carrier), male normal, male with DI. The mark scheme insists the phenotypes appear in the same order as the offspring genotypes, not grouped by sex.

Key Takeaways

• When two unaffected parents have an affected son for an X-linked recessive condition, the mother must be a carrier ($\text{X}^A\text{X}^a$) and the father is normal ($\text{X}^A\text{Y}$).
• X-linked recessive conditions appear predominantly in males; an affected female would require an affected father (and at least a carrier mother).
• Always write the Punnett square in a fixed order so that the offspring phenotypes can be listed in the same order.

Common Mistakes

• Writing the mother as $\text{X}^A\text{X}^A$ (homozygous normal) — this cannot produce an affected son, so it fails the question.
• Putting the allele on the autosome (e.g. 'aa' parents) — the question states the allele is sex-linked.
• Using $\text{X}^{\text{A}}$ notation with the allele on its own (e.g. 'X with A') rather than as a superscript — CIE convention is $\text{X}^A$.
• Forgetting that the carrier daughter ($\text{X}^A\text{X}^a$) is phenotypically normal, not affected.
• Listing the phenotypes out of order or grouping all normals together — the mark scheme requires the phenotype directly under or next to its corresponding genotype.

Things to Be Careful About

• CIE accepts either superscript notation ($\text{X}^A$) or written-out form ('X carrying A / X with A') — the superscript is cleaner.
• The Y chromosome carries no allele for this gene; never write $\text{Y}^a$ or $\text{X}^A\text{Y}^a$.
• Probability: this cross gives a 1 in 4 chance of an affected child overall, but every affected child must be male. Mentioning this is good practice, though it is not required for the four marks.