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
- Genetic Cloning
  9m
- Methods for Analyzing DNA
  9m
19. Cancer Genetics29m
- Overview of Cancer
  21m
- Cancer Mutations
  7m
20. Quantitative Genetics1h 26m
21. Population Genetics50m
- Hardy Weinberg
  19m
- Allelic Frequency Changes
  31m
22. Evolutionary Genetics29m

6. Chromosomal Variation

Chromosomal Mutations: Aneuploidy

Genetics

6. Chromosomal Variation

Chromosomal Mutations: Aneuploidy

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1:39 minutes

Problem 31c

Textbook Question

For the following crosses, determine as accurately as possible the genotypes of each parent, the parent in whom nondisjunction occurs, and whether nondisjunction takes place in the first or second meiotic division. Both color blindness and hemophilia, a blood-clotting disorder, are X-linked recessive traits. In each case, assume the parents have normal karyotypes.
A color-blind man and a woman who is wild type have a daughter with Turner syndrome (XO) who has normal color vision and blood clotting.

Verified step by step guidance

Step 1: Understand the genetic basis of the traits and the chromosomal abnormality. Color blindness and hemophilia are X-linked recessive traits, meaning they are carried on the X chromosome. Turner syndrome (XO) results from a missing X chromosome, leading to a 45,X karyotype.

Step 2: Analyze the genotypes of the parents. The man is color-blind, so his genotype is X^cY, where X^c represents the X chromosome carrying the color-blind allele. The woman is wild type, meaning she has two normal X chromosomes (X⁺X⁺).

Step 3: Determine the possible gametes produced by each parent. The man can produce gametes with either X^c or Y. The woman can produce gametes with X⁺.

Step 4: Consider the nondisjunction event. Since the daughter has Turner syndrome (XO) and normal color vision, she must have inherited one X chromosome from her mother (X⁺) and no sex chromosome from her father. This indicates nondisjunction occurred in the father during meiosis.

Step 5: Identify the stage of nondisjunction. If nondisjunction occurred in the first meiotic division, the father would produce gametes with both X^c and Y or no sex chromosome at all. If it occurred in the second meiotic division, the father would produce gametes with either X^c, Y, or no sex chromosome. Since the daughter inherited no sex chromosome from the father, nondisjunction likely occurred in the second meiotic division.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

X-linked Inheritance

X-linked inheritance refers to the pattern of inheritance for genes located on the X chromosome. In this case, color blindness and hemophilia are X-linked recessive traits, meaning that males (XY) are more likely to express these traits since they have only one X chromosome. Females (XX) can be carriers if they have one affected X chromosome but typically do not express the trait unless both X chromosomes are affected.

Nondisjunction

Nondisjunction is the failure of homologous chromosomes or sister chromatids to separate properly during cell division, leading to gametes with an abnormal number of chromosomes. This can occur during either the first or second meiotic division. In the context of Turner syndrome (XO), nondisjunction likely occurred during the formation of the egg or sperm, resulting in a missing X chromosome in the daughter.

Turner Syndrome

Turner syndrome is a chromosomal disorder characterized by the presence of a single X chromosome (XO) in females, leading to various developmental and physical features. Individuals with Turner syndrome typically have normal intelligence but may experience short stature, infertility, and other health issues. In this scenario, the daughter with Turner syndrome has normal color vision and blood clotting, indicating that she inherited a normal X chromosome from her mother.

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Related Practice

Textbook Question

Most cases of Turner syndrome are attributed to nondisjunction of one or more of the sex chromosomes during gametogenesis, from either the male or female parent. However, some females possess a rare form of Turner syndrome in which some of the cells of the body (somatic cells) lack an X chromosome, while other cells have the normal two X chromosomes. Often detected in blood and/or skin cells, such individuals with mosaic Turner syndrome may exhibit relatively mild symptoms. An individual may be specified as 45,X(20)/46,XX(80) if, for example, 20 percent of the cells examined were X monosomic. How might mitotic events cause such mosaicism, and what parameter(s) would likely determine the percentages and distributions of X0 cells?

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Textbook Question

A 3-year-old child exhibited some early indication of Turner syndrome, which results from a 45,X chromosome composition. Karyotypic analysis demonstrated two cell types: 46,XX (normal) and 45,X. Propose a mechanism for the origin of this mosaicism.

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Textbook Question

A normal female is discovered with 45 chromosomes, one of which exhibits a Robertsonian translocation containing most of chromosomes 15 and 21. Discuss the possible outcomes in her offspring when her husband contains a normal karyotype.

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Textbook Question

For the following crosses, determine as accurately as possible the genotypes of each parent, the parent in whom nondisjunction occurs, and whether nondisjunction takes place in the first or second meiotic division. Both color blindness and hemophilia, a blood-clotting disorder, are X-linked recessive traits. In each case, assume the parents have normal karyotypes.

A man who is color blind and has hemophilia and a woman who is wild type have a daughter with triple X syndrome (XXX) who has hemophilia and normal color vision.

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Textbook Question

A man who is color blind and a woman who is wild type have a son with Jacob syndrome (XYY) who has hemophilia.

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Textbook Question

A man and a woman who each has the wild-type phenotype have a son with Klinefelter syndrome (XXY) who has hemophilia.

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Multiple Choice

Nondisjunction during meiosis can result in which of the following chromosomal abnormalities?