In Drosophila subobscura, the presence of a recessive gene called grandchildless (gs) causes the offspring of homozygous females, but not those of homozygous males, to be sterile. Can you offer an explanation as to why females and not males are affected by the mutant gene?
Table of contents
- 1. Introduction to Genetics51m
- 2. Mendel's Laws of Inheritance3h 37m
- 3. Extensions to Mendelian Inheritance2h 41m
- 4. Genetic Mapping and Linkage2h 28m
- 5. Genetics of Bacteria and Viruses1h 21m
- 6. Chromosomal Variation1h 48m
- 7. DNA and Chromosome Structure56m
- 8. DNA Replication1h 10m
- 9. Mitosis and Meiosis1h 34m
- 10. Transcription1h 0m
- 11. Translation58m
- 12. Gene Regulation in Prokaryotes1h 19m
- 13. Gene Regulation in Eukaryotes44m
- 14. Genetic Control of Development44m
- 15. Genomes and Genomics1h 50m
- 16. Transposable Elements47m
- 17. Mutation, Repair, and Recombination1h 6m
- 18. Molecular Genetic Tools19m
- 19. Cancer Genetics29m
- 20. Quantitative Genetics1h 26m
- 21. Population Genetics50m
- 22. Evolutionary Genetics29m
2. Mendel's Laws of Inheritance
Sex-Linked Genes
Problem 12d
Textbook Question
A woman's father has ornithine transcarbamylase deficiency (OTD), an X-linked recessive disorder producing mental deterioration if not properly treated. The woman's mother is homozygous for the wild-type allele.
Identify a male with whom the woman could produce a daughter with OTD.

1
Understand the inheritance pattern: Ornithine transcarbamylase deficiency (OTD) is an X-linked recessive disorder. This means the gene responsible for the disorder is located on the X chromosome, and males (XY) are more likely to express the disorder because they have only one X chromosome.
Analyze the woman's genotype: Since the woman's father has OTD, he must have passed his affected X chromosome to her. Her mother is homozygous for the wild-type allele, so the woman is heterozygous (X^OT/X^+), where X^OT represents the affected allele and X^+ represents the wild-type allele.
Determine the genotype of a male partner: To produce a daughter with OTD, the daughter must inherit the affected X chromosome (X^OT) from her mother and another affected X chromosome (X^OT) from her father. This means the male partner must have OTD and carry the affected X chromosome (X^OT).
Explain the male partner's genotype: Since males are XY, a male with OTD would have the genotype X^OT/Y. This ensures that he can pass the affected X chromosome (X^OT) to his daughters.
Conclude the pairing: The woman (X^OT/X^+) must have children with a male who has OTD (X^OT/Y) to produce a daughter with the genotype X^OT/X^OT, which would result in the daughter having OTD.

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
X-Linked Recessive Inheritance
X-linked recessive inheritance refers to genetic conditions that are associated with genes located on the X chromosome. Males, having one X and one Y chromosome, are more likely to express these disorders because they have only one copy of the X chromosome. In contrast, females have two X chromosomes, so they can be carriers without showing symptoms if they have one normal allele.
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X-Inactivation
Genotype of the Woman
The woman in the scenario has a father with ornithine transcarbamylase deficiency, meaning he has the affected genotype (X^dY). Since her mother is homozygous for the wild-type allele (X^WX^W), the woman must be a carrier (X^dX^W), possessing one affected X chromosome from her father and one normal X chromosome from her mother.
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Gamete Genotypes
Possible Male Genotype
To produce a daughter with ornithine transcarbamylase deficiency, the male partner must provide an X chromosome carrying the affected allele (X^d). Since males have one X and one Y chromosome, a male with the genotype X^dY would be necessary. This male would pass on the X^d chromosome to any daughters, resulting in a 50% chance of them inheriting the disorder if the woman is a carrier.
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Gamete Genotypes
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