QUESTION 140 (6 marks)
The pedigree shows a known fully penetrant X-linked dominant trait. Allele causes the trait; does not. There are no new mutations. Individual I-2 is unaffected.
a) State the genotypes of I-1, I-2 and II-1. [3 marks]
b) II-1 has an unaffected male partner. Assuming equal probabilities of male and female offspring, calculate the probability that their next child is an affected daughter. [2 marks]
c) Explain why an affected father does not transmit his X-linked allele to his sons. [1 mark]
Practice marking scheme
Answer
I-1: XᴰY; I-2: XᵈXᵈ. II-1: XᴰXᵈ. With an unaffected male partner, II-1 has a 1/4 probability of an affected daughter and 1/2 probability of an affected child of either sex.
Working
An affected father transmits Xᴰ to every daughter and Y to every son. The unaffected mother supplies Xᵈ. II-1 is therefore heterozygous. Her cross with XᵈY gives XᴰXᵈ, XᵈXᵈ, XᴰY and XᵈY with equal probabilities, assuming equal sex probabilities.
Marking criteria
- States I-1 XᴰY. [1 mark]
- States I-2 XᵈXᵈ. [1 mark]
- States II-1 XᴰXᵈ. [1 mark]
- Uses probability 1/2 of a daughter and 1/2 maternal D transmission. [1 mark]
- Obtains 1/4 for an affected daughter. [1 mark]
- States that a son receives his father’s Y rather than his X. [1 mark]
Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.
View the QCAA syllabusCompare your working with the guide above.