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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2:44 minutes

Problem 31b

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 a woman who is wild type have a son with Jacob syndrome (XYY) who has hemophilia.

Verified step by step guidance

Understand the genetic basis of the traits: Both color blindness and hemophilia are X-linked recessive traits, meaning the genes responsible for these conditions are located on the X chromosome. Males (XY) inherit their X chromosome from their mother and their Y chromosome from their father, while females (XX) inherit one X chromosome from each parent.

Analyze the parents' genotypes: The man is color blind, so his X chromosome must carry the allele for color blindness (X^c). The woman is wild type, meaning her X chromosomes do not carry the alleles for color blindness or hemophilia (X⁺).

Determine the son's genotype: The son has Jacob syndrome (XYY), meaning he has two Y chromosomes and one X chromosome. He also has hemophilia, so his X chromosome must carry the allele for hemophilia (X^h).

Identify the source of nondisjunction: Nondisjunction is the failure of chromosomes to separate properly during meiosis. Since the son has two Y chromosomes, nondisjunction must have occurred in the father during spermatogenesis. This resulted in a sperm cell with both Y chromosomes.

Determine the stage of nondisjunction: If nondisjunction occurred during the first meiotic division, the two Y chromosomes would fail to separate, leading to a sperm cell with two Y chromosomes. If nondisjunction occurred during the second meiotic division, a single Y chromosome would fail to separate, but this would not result in a sperm cell with two Y chromosomes. Therefore, nondisjunction must have occurred during the first meiotic division in the father.

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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 both X-linked recessive traits, meaning that males (who have one X and one Y chromosome) are more likely to express these traits if they inherit the affected X chromosome. Females, having two X chromosomes, can be carriers without expressing the traits if they have one normal X chromosome.

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 the question, identifying where nondisjunction occurs is crucial for understanding the genetic makeup of the offspring, particularly in relation to the Jacob syndrome (XYY) condition.

Meiotic Division

Meiotic division consists of two rounds of cell division (meiosis I and meiosis II) that result in four genetically diverse gametes. The first meiotic division separates homologous chromosomes, while the second separates sister chromatids. The timing of nondisjunction—whether it occurs in the first or second meiotic division—affects the resulting gametes and can lead to conditions like Jacob syndrome, which involves an extra Y chromosome.

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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 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.

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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?