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Ch. 5 - Chromosome Mapping in Eukaryotes

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Klug 12th Edition

Ch. 5 - Chromosome Mapping in Eukaryotes

Problem 28

Chapter 5, Problem 28

A number of human–mouse somatic cell hybrid clones were examined for the expression of specific human genes and the presence of human chromosomes. The results are summarized in the following table. Assign each gene to the chromosome on which it is located.

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Hello, everyone here. We have a question that is asking what the laboratory approach utilized to locate the position of a specific disease that is associated with the gene on a chromosome is called. And our answer is a positional cloning is used to locate the position of a specific disease that is associated with the gene on a chromosome, and it uses polymorphic markers. That inheritance can be traced all throughout the pedigree and the members affected by the disease. Thank you for watching. Bye.

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Related Practice

Textbook Question

Traditional gene mapping has been applied successfully to a variety of organisms including yeast, fungi, maize, and Drosophila. However, human gene mapping has only recently shared a similar spotlight. What factors have delayed the application of traditional gene-mapping techniques in humans?

Textbook Question

In a certain plant, fruit is either red or yellow, and fruit shape is either oval or long. Red and oval are the dominant traits. Two plants, both heterozygous for these traits, were testcrossed, with the following results. Determine the location of the genes relative to one another and the genotypes of the two parental plants.

Textbook Question

Two plants in a cross were each heterozygous for two gene pairs (Ab/aB) whose loci are linked and 25 mu apart. Assuming that crossing over occurs during the formation of both male and female gametes and that the A and B alleles are dominant, determine the phenotypic ratio of their offspring.

Textbook Question

A female of genotype a b c + + + produces 100 meiotic tetrads. Of these, 68 show no crossover events. Of the remaining 32, 20 show a crossover between a and b, 10 show a crossover between b and c, and 2 show a double crossover between a and b and between b and c. Of the 400 gametes produced, how many of each of the 8 different genotypes will be produced? Assuming the order a–b–c and the allele arrangement previously shown, what is the map distance between these loci?

Textbook Question

In laboratory class, a genetics student was assigned to study an unknown mutation in Drosophila that had a whitish eye. He crossed females from his true-breeding mutant stock to wild-type (brick-red-eyed) males, recovering all wild-type F₁ flies. In the F₂ generation, the following offspring were recovered in the following proportions: wild type 5/8 bright red 1/8 brown eye 1/8 white eye 1/8 The student was stumped until the instructor suggested that perhaps the whitish eye in the original stock was the result of homozygosity for a mutation causing brown eyes and a mutation causing bright red eyes, illustrating gene interaction (see Chapter 4). After much thought, the student was able to analyze the data, explain the results, and learn several things about the location of the two genes relative to one another. One key to his understanding was that crossing over occurs in Drosophila females but not in males. Based on his analysis, what did the student learn about the two genes?

Textbook Question

Drosophila melanogaster has one pair of sex chromosomes (XX or XY) and three pairs of autosomes, referred to as chromosomes II, III, and IV. A genetics student discovered a male fly with very short (sh) legs. Using this male, the student was able to establish a pure breeding stock of this mutant and found that it was recessive. She then incorporated the mutant into a stock containing the recessive gene black (b, body color located on chromosome II) and the recessive gene pink (p, eye color located on chromosome III). A female from the homozygous black, pink, short stock was then mated to a wild-type male. The F₁ males of this cross were all wild type and were then backcrossed to the homozygous b, p, sh females. The F₂ results appeared as shown in the following table. No other phenotypes were observed. Wild Pink* Black, Black, Pink, Short* Short Females 63 58 55 69 Males 59 65 51 60 *Other trait or traits are wild type. Based on these results, the student was able to assign short to a linkage group (a chromosome). Which one was it? Include your step-by-step reasoning.