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Ch. 10 - Eukaryotic Chromosome Abnormalities and Molecular Organization
Sanders - Genetic Analysis: An Integrated Approach 3rd Edition
Sanders3rd EditionGenetic Analysis: An Integrated ApproachISBN: 9780135564172Not the one you use?Change textbook
All textbooksSanders 3rd EditionCh. 10 - Eukaryotic Chromosome Abnormalities and Molecular OrganizationProblem 31d
Chapter 10, Problem 31d

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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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 (Xcb) and hemophilia (Xh). His genotype is XcbhY. 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 (Xh), while the other two X chromosomes are normal (X+). Her genotype is X+X+Xh.
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 (Xcbh) or a Y chromosome, so the nondisjunction must have occurred in the mother, who contributed two X chromosomes (X+ and Xh).
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 Xh). If it occurred in the second meiotic division, the mother would produce gametes with two identical X chromosomes (e.g., X+X+ or XhXh). Since the daughter has one Xh 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.
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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.
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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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Related Practice
Textbook Question

A healthy couple with a history of three previous spontaneous abortions has just had a child with cri-du-chat syndrome, a disorder caused by a terminal deletion of chromosome 5. Their physician orders karyotype analysis of both parents and of the child. The karyotype results for chromosomes 5 and 12 are shown here. Are the chromosomes in the child consistent with those expected in a case of cri-du-chat syndrome? Explain your reasoning.

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.

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

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

Textbook Question

A healthy couple with a history of three previous spontaneous abortions has just had a child with cri-du-chat syndrome, a disorder caused by a terminal deletion of chromosome 5. Their physician orders karyotype analysis of both parents and of the child. The karyotype results for chromosomes 5 and 12 are shown here. Which parent has an abnormal karyotype? How can you tell? What is the nature of the abnormality?

Textbook Question

A healthy couple with a history of three previous spontaneous abortions has just had a child with cri-du-chat syndrome, a disorder caused by a terminal deletion of chromosome 5. Their physician orders karyotype analysis of both parents and of the child. The karyotype results for chromosomes 5 and 12 are shown here. Why does this parent have a normal phenotype?