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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1:27 minutes

Problem 31d

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.

Verified step by step guidance

Step 1: Understand the genetic basis of the traits. Both color blindness and hemophilia are X-linked recessive traits, meaning they are carried on the X chromosome. A female has two X chromosomes (XX), while a male has one X and one Y chromosome (XY). A female must inherit two copies of the recessive allele to express the trait, while a male needs only one copy since he has only one X chromosome.

Step 2: Analyze the genotypes of the parents. The man is color blind and has hemophilia, so his X chromosome must carry both the alleles for color blindness (X^cb) and hemophilia (X^h). His genotype is X^cbhY. The woman is wild type, meaning she has normal vision and normal blood clotting. Her genotype is X⁺X⁺, where X⁺ represents the normal allele.

Step 3: Determine the genotype of the daughter. The daughter has triple X syndrome (XXX), meaning she has three X chromosomes. She has hemophilia but normal color vision. This indicates that one of her X chromosomes carries the hemophilia allele (X^h), while the other two X chromosomes are normal (X⁺). Her genotype is X⁺X⁺X^h.

Step 4: Identify the source of nondisjunction. Nondisjunction is the failure of chromosomes to separate properly during meiosis. Since the daughter has three X chromosomes, nondisjunction must have occurred in one of the parents. The father can only contribute one X chromosome (X^cbh) or a Y chromosome, so the nondisjunction must have occurred in the mother, who contributed two X chromosomes (X⁺ and X^h).

Step 5: Determine the stage of nondisjunction. If nondisjunction occurred in the first meiotic division, the mother would produce gametes with two different X chromosomes (e.g., X⁺ and X^h). If it occurred in the second meiotic division, the mother would produce gametes with two identical X chromosomes (e.g., X⁺X⁺ or X^hX^h). Since the daughter has one X^h and two X⁺, nondisjunction likely occurred in the first meiotic division in the mother.

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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 (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 the question, nondisjunction could explain the presence of an extra X chromosome in the daughter with triple X syndrome.

Meiotic Division

Meiotic division consists of two rounds of cell division (meiosis I and meiosis II) that result in four genetically diverse gametes. In meiosis I, homologous chromosomes are separated, while in meiosis II, sister chromatids are separated. Understanding which meiotic division nondisjunction occurs in is crucial for determining the genetic makeup of the offspring, as it affects the distribution of chromosomes in the resulting gametes.

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Textbook Question

A boy with Klinefelter syndrome (47,XXY) is born to a mother who is phenotypically normal and a father who has the X-linked skin condition called anhidrotic ectodermal dysplasia. The mother's skin is completely normal with no signs of the skin abnormality. In contrast, her son has patches of normal skin and patches of abnormal skin. Which parent contributed the abnormal gamete?

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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 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 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