Ch. 8 The Cellular Basis of Reproduction and Inheritance

Taylor - Campbell Biology: Concepts & Connections 10th Edition

Taylor, Simon, Dickey, Hogan10th EditionCampbell Biology: Concepts & ConnectionsISBN: 9780136538783Not the one you use?Change textbook

Taylor, Simon, Dickey, Hogan 10th Edition

Ch. 8 The Cellular Basis of Reproduction and Inheritance

Problem 17

Chapter 8, Problem 17

Sketch a cell with three pairs of chromosomes undergoing meiosis, and show how non-disjunction can result in the production of gametes with extra or missing chromosomes.

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Start by sketching a cell with three pairs of homologous chromosomes (a total of six chromosomes). Label each pair with different colors or patterns to distinguish them (e.g., Pair 1: red, Pair 2: blue, Pair 3: green). Place the chromosomes in the nucleus of the cell in their duplicated form (sister chromatids).

Illustrate the stages of meiosis I. During prophase I, show homologous chromosomes pairing up (synapsis) and crossing over. In metaphase I, align the homologous pairs along the metaphase plate. In anaphase I, depict the separation of homologous chromosomes to opposite poles. In telophase I, show the formation of two haploid cells, each with three duplicated chromosomes.

Illustrate the stages of meiosis II. In prophase II, show the chromosomes in each haploid cell preparing for division. In metaphase II, align the chromosomes along the metaphase plate. In anaphase II, depict the separation of sister chromatids to opposite poles. In telophase II, show the formation of four haploid gametes, each with three single chromosomes.

To demonstrate nondisjunction, modify the anaphase stage in either meiosis I or meiosis II. For meiosis I nondisjunction, show one homologous pair failing to separate, resulting in one cell with an extra chromosome and one cell missing a chromosome. For meiosis II nondisjunction, show one pair of sister chromatids failing to separate, resulting in one gamete with an extra chromosome, one gamete missing a chromosome, and two normal gametes.

Conclude by labeling the resulting gametes to indicate their chromosomal composition. Highlight how nondisjunction can lead to conditions such as trisomy (e.g., Down syndrome) or monosomy (e.g., Turner syndrome) if these gametes are involved in fertilization.

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Key Concepts

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

Meiosis

Meiosis is a specialized type of cell division that reduces the chromosome number by half, resulting in four genetically diverse gametes. It consists of two sequential divisions: meiosis I, where homologous chromosomes are separated, and meiosis II, where sister chromatids are separated. This process is crucial for sexual reproduction, ensuring genetic variation through recombination and independent assortment.

Nondisjunction

Nondisjunction is the failure of homologous chromosomes or sister chromatids to separate properly during meiosis. This can occur in either meiosis I or meiosis II, leading to gametes with an abnormal number of chromosomes. If such gametes participate in fertilization, they can result in aneuploidy, where the resulting zygote has extra or missing chromosomes, potentially leading to genetic disorders.

Aneuploidy

Aneuploidy refers to an abnormal number of chromosomes in a cell, which can arise from nondisjunction during meiosis. Common forms of aneuploidy include trisomy, where there is an extra chromosome (e.g., Down syndrome), and monosomy, where a chromosome is missing. The presence of aneuploidy can significantly impact development and lead to various health issues, making it a critical concept in genetics.

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Nondisjunction

Related Practice

Textbook Question

Briefly describe how three different processes that occur during a sexual life cycle increase the genetic diversity of offspring.

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Textbook Question

Discuss the factors that control the division of eukaryotic cells grown in the laboratory. Cancer cells are easier to grow in the lab than other cells.

Why do you suppose this is?

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Textbook Question

Compare cytokinesis in plant and animal cells.

In what ways are the two processes similar?

In what ways are they different?

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Textbook Question

Suppose you read in the newspaper that a genetic engineering laboratory has developed a procedure for fusing two gametes from the same person (two eggs or two sperm) to form a zygote. The article mentions that an early step in the procedure prevents crossing over from occurring during the formation of the gametes in the donor's body. The researchers are in the process of determining the genetic makeup of one of their new zygotes. Which of the following predictions do you think they would make? Justify your choice, and explain why you rejected each of the other choices.

a. The zygote would have 46 chromosomes, all of which came from the gamete donor (its one parent), so the zygote would be genetically identical to the gamete donor.

b. The zygote could be genetically identical to the gamete donor, but it is much more likely that it would have an unpredictable mixture of chromosomes from the gamete donor's parents.

c. The zygote would not be genetically identical to the gamete donor, but it would be genetically identical to one of the donor's parents.

d. The zygote would not be genetically identical to the gamete donor, but it would be genetically identical to one of the donor's grandparents.

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Textbook Question

Bacteria are able to divide on a faster schedule than eukaryotic cells. Some bacteria can divide every 20 minutes, while the minimum time required by eukaryotic cells in a rapidly developing embryo is about once per hour, and most cells divide much less often than that. State at least two testable hypotheses explaining why bacteria can divide at a faster rate than eukaryotic cells.

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Textbook Question

Red blood cells, which carry oxygen to body tissues, live for only about 120 days. Replacement cells are produced by cell division in bone marrow.

How many cell divisions must occur each second in your bone marrow just to replace red blood cells? Here is some information to use in calculating your answer: There are about 5 million red blood cells per cubic millimeter (mm³) of blood. An average adult has about 5 L (5,000 cm³) of blood. (Hint: What is the total number of red blood cells in the body?

What fraction of them must be replaced each day if all are replaced in 120 days?

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