Biology • Year 12 • Module 5 • Lesson 14
Mendelian Patterns — Autosomal Inheritance, Sex Linkage, Punnett Squares
Apply Punnett-square reasoning and pedigree logic to real-style scenarios — autosomal recessive carrier crosses, X-linked haemophilia, and inferring inheritance patterns from family data.
1. Carrier cross — cystic fibrosis
Cystic fibrosis (CF) is caused by an autosomal recessive allele. Let F = unaffected (dominant) and f = CF (recessive). Two carrier parents (Ff) are expecting a child. 7 marks
1.1 Complete the Punnett square below by filling each offspring genotype box. 4 marks
| F | f | |
|---|---|---|
| F | ||
| f |
1.2 State the probability that the child is (a) affected, (b) a phenotypically unaffected carrier, (c) homozygous unaffected. 3 marks
2. X-linked recessive — haemophilia in a carrier-mother / unaffected-father cross
Haemophilia is X-linked recessive. Let XH = unaffected allele and Xh = affected allele. A carrier mother (XHXh) has children with an unaffected father (XHY). 8 marks
2.1 Complete the Punnett square (label each offspring with full genotype, e.g. XHY). 4 marks
| XH (from father) | Y (from father) | |
|---|---|---|
| XH (from mother) |
||
| Xh (from mother) |
2.2 State the probability that a child of this cross is:
(a) an affected son
(b) an unaffected son
(c) a carrier daughter
(d) an affected daughter
Express each as a percentage of all children. 4 marks
3. Pedigree interpretation — albinism family
The pedigree below tracks albinism (an autosomal recessive trait) across three generations. Squares = males, circles = females, filled = affected, unfilled = unaffected. 7 marks
Albinism pedigree — autosomal recessive trait. Let A = unaffected (dominant), a = albinism (recessive).
3.1 State the genotype of II-1 and explain how you know. 2 marks
3.2 What can you deduce about the genotypes of I-1 and I-2? Justify briefly. 2 marks
3.3 III-1 is affected but his parents (II-2 and II-3) are unaffected. State the most likely genotypes of his parents and explain. 2 marks
3.4 Give one piece of evidence in this pedigree that is more consistent with autosomal recessive inheritance than with X-linked recessive inheritance. 1 mark
4. Inheritance pattern from family data
Three large families were surveyed for a particular condition. Family sizes are similar. Researchers recorded how many sons and daughters were affected in each family. 6 marks
| Family | Sons | Affected sons | Daughters | Affected daughters | Parents (phenotype) |
|---|---|---|---|---|---|
| F1 | 6 | 3 | 5 | 0 | Mother unaffected; father unaffected |
| F2 | 4 | 2 | 6 | 0 | Mother unaffected; father unaffected |
| F3 | 5 | 2 | 4 | 0 | Mother unaffected; father unaffected |
4.1 Identify the pattern that distinguishes affected sons from affected daughters in this dataset. 2 marks
4.2 Justify, with reference to lesson content, why this pattern is more consistent with X-linked recessive inheritance than with autosomal recessive inheritance. 3 marks
4.3 Predict the most likely genotype of the mothers, using XH/Xh notation. 1 mark
5. Diagram critique — student's Punnett square
A Year 12 student has drawn the Punnett square below to answer the prompt "Two heterozygous parents (Aa × Aa) — predict offspring." There are three biological errors. Identify each and write the correction. 6 marks (2 per error: 1 identify, 1 correct)
5.1 Error 1: What is wrong?
Correction:
5.2 Error 2: What is wrong?
Correction:
5.3 Error 3: What is wrong?
Correction:
Aa × Aa Punnett figure in Card 2.
Q1 — Cystic fibrosis carrier cross (7 marks)
1.1 Completed Punnett square (4 marks, 1 per cell):
top-left FF • top-right Ff • bottom-left Ff • bottom-right ff.
1.2 (3 marks):
- (a) Affected (
ff) = 1/4 = 25%. [1] - (b) Phenotypically unaffected carrier (
Ff) = 2/4 = 50%. [1] - (c) Homozygous unaffected (
FF) = 1/4 = 25%. [1]
Q2 — Haemophilia carrier-mother / unaffected-father cross (8 marks)
2.1 Completed Punnett square (4 marks, 1 per cell):
top-left XHXH (unaffected daughter) • top-right XHY (unaffected son) • bottom-left XHXh (carrier daughter) • bottom-right XhY (affected son).
2.2 (4 marks, 1 each):
- (a) Affected son =
XhY= 25% of all children (= 50% of sons). - (b) Unaffected son =
XHY= 25% of all children (= 50% of sons). - (c) Carrier daughter =
XHXh= 25% of all children (= 50% of daughters). - (d) Affected daughter = 0% — the father provides his only X chromosome (
XH), so no daughter can be homozygousXhXh.
Q3 — Albinism pedigree (7 marks)
3.1 (2 marks). II-1 is affected, so her genotype must be aa (homozygous recessive). [1] She inherited one a from each parent, which means both parents must each carry at least one a. [1]
3.2 (2 marks). I-1 and I-2 are both unaffected but produced an affected daughter, so each must carry the a allele — they are both heterozygous carriers, genotype Aa. [1] If either were AA, no aa child would be possible. [1]
3.3 (2 marks). III-1 is aa, so each parent (II-2 and II-3) must each carry at least one a. Both are unaffected, so they are heterozygous carriers, Aa × Aa. [1] II-2 is the son of two carriers (I-1 and I-2), so being Aa is consistent; II-3 married in but must also be a carrier for the affected child to occur. [1]
3.4 (1 mark). An affected female (II-1) is present. X-linked recessive traits very rarely produce affected females (they would need to inherit Xh from both parents, including an affected father, which is not shown). Autosomal recessive inheritance produces affected males and females at roughly equal probability. [1] Accept also: father I-1 is unaffected but has an affected daughter — under X-linked recessive an affected daughter would require an affected father.
Q4 — Inheritance pattern from family data (6 marks)
4.1 (2 marks). Across all three families, only sons are affected; no daughters are affected (0/15 daughters vs 7/15 sons). [1] Affected sons appear in roughly half of the male children. [1]
4.2 (3 marks). X-linked recessive traits appear more frequently in males because males have only one X chromosome, so a single recessive allele on the X is enough to express the trait. [1] If the trait were autosomal recessive, affected sons and daughters would be expected in roughly equal numbers — daughters could be aa just as easily as sons. [1] The complete absence of affected daughters across three families, combined with both parents being unaffected (mothers must be carriers), is the classic signature of X-linked recessive inheritance — and is also consistent with no father-to-son transmission of the X-linked allele. [1]
4.3 (1 mark). Mothers are most likely carriers: XHXh. [1]
Q5 — Diagram critique (6 marks)
5.1 Error 1 (gamete labelling). The student labelled the top row "A, A" and the side column "a, a" — that would mean one parent only produces A gametes (homozygous dominant AA) and the other only a gametes (homozygous recessive aa), which is not an Aa × Aa cross. [1] Correction: both parents are heterozygous Aa, so each parent produces both A and a gametes — the top row should read A, a and the side column should also read A, a. [1]
5.2 Error 2 (all cells Aa). Filling all four inner cells with Aa is impossible in a real heterozygous cross — it would require every gamete combination to produce a heterozygote, which only happens when one parent is AA and the other aa. [1] Correction: with A, a on both axes the correct inner cells are AA, Aa, Aa, aa — i.e. genotype ratio 1 : 2 : 1. [1]
5.3 Error 3 (certainty language). Saying "4 out of every 4 children will definitely be heterozygous" confuses probability with certainty and assumes a guaranteed family outcome. [1] Correction: a Punnett square predicts the probability of each genotype per fertilisation event, not a fixed family ratio. The correct wording is "each child has a 50% probability of being heterozygous (Aa), 25% homozygous dominant (AA) and 25% homozygous recessive (aa)". [1]