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Ch. 5 - Chromosome Mapping in Eukaryotes
Klug - Concepts of Genetics  12th Edition
Klug12th EditionConcepts of Genetics ISBN: 9780135564776Not the one you use?Change textbook
Chapter 5, Problem 31a

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.
Table showing F2 phenotypic counts of Drosophila by sex for wild, pink, black short, and black pink short traits.
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.

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Step 1: Understand the genetic background and the traits involved. The traits are black body color (b) on chromosome II, pink eye color (p) on chromosome III, and short legs (sh) of unknown chromosomal location. The mutant alleles are recessive, and the female used in the backcross is homozygous recessive for all three traits (b, p, sh). The male parent in the backcross is wild type for all traits.
Step 2: Analyze the F2 phenotypic classes and their counts. The phenotypes are Wild type, Pink only, Black and Short together, and Black, Pink, and Short together. Since the female parent is homozygous recessive for all three traits, the phenotypes in the progeny reflect the alleles contributed by the male parent and the recombination events between the genes.
Step 3: Determine linkage by comparing expected independent assortment ratios to observed data. If short legs (sh) were on a different chromosome from black (b) or pink (p), the phenotypes involving sh would assort independently, and the phenotypic ratios would reflect independent assortment. However, if sh is linked to either b or p, the phenotypes involving those traits will show non-independent assortment, with certain combinations appearing more frequently.
Step 4: Compare the phenotypic counts of 'Black, Short' and 'Black, Pink, Short' classes. If short legs (sh) is linked to black (b) on chromosome II, these two classes should be more frequent and appear together more often than expected by chance. Conversely, if sh is linked to pink (p) on chromosome III, the 'Pink' and 'Black, Pink, Short' classes would show linkage patterns.
Step 5: Conclude the linkage group of short legs (sh) by identifying which chromosome's recessive trait co-segregates with sh. The chromosome whose recessive trait appears together with sh more frequently than expected indicates linkage. Therefore, based on the phenotypic distribution, assign short legs (sh) to the chromosome that shows this pattern of linkage.

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Key Concepts

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

Linkage and Chromosome Mapping

Linkage refers to genes located close together on the same chromosome that tend to be inherited together. By analyzing offspring phenotypes and their frequencies, geneticists can determine whether genes are linked and assign them to specific chromosomes. This concept is essential for interpreting the inheritance patterns in the F2 generation and identifying the chromosome carrying the short (sh) gene.
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Chi Square and Linkage

Recessive Traits and Pure Breeding

A recessive trait only appears phenotypically when an organism has two copies of the recessive allele (homozygous recessive). Pure breeding stocks are homozygous for the trait, ensuring consistent expression in offspring. Understanding this helps explain the initial crosses and the phenotypic ratios observed in the progeny.
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Sex Chromosomes vs. Autosomes in Drosophila

Drosophila melanogaster has one pair of sex chromosomes (XX or XY) and three pairs of autosomes (II, III, IV). Knowing the chromosomal location of genes (e.g., black on chromosome II, pink on III) allows geneticists to deduce linkage groups by comparing expected and observed phenotypic ratios, especially when incorporating a new gene like short.
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Related Practice
Textbook Question

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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Textbook Question

A female of genotype

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?

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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. 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?

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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.

The student repeated the experiment, making the reciprocal cross, with F₁ females backcrossed to homozygous b, p, sh males. She observed that 85 percent of the offspring fell into the given classes, but that 15 percent of the offspring were equally divided among b + p, b + +, + sh p, and + sh + phenotypic males and females. How can these results be explained, and what information can be derived from the data?

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Textbook Question

In Drosophila, a female fly is heterozygous for three mutations, Bar eyes (B), miniature wings (m), and ebony body (e). Note that Bar is a dominant mutation. The fly is crossed to a male with normal eyes, miniature wings, and ebony body. The results of the cross are as follows.

111 miniature; 101 Bar, ebony

29 wild type; 31 Bar, miniature, ebony

117 Bar; 35 ebony

26 Bar, miniature; 115 miniature, ebony

Interpret the results of this cross. If you conclude that linkage is involved between any of the genes, determine the map distance(s) between them.

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Textbook Question

The gene controlling the Xg blood group alleles (Xg⁺ and Xg⁻) and the gene controlling a newly described form of inherited recessive muscle weakness called episodic muscle weakness (EMWX) (Ryan et al., 1999) are closely linked on the X chromosome in humans at position Xp22.3 (the tip of the short arm). A male with EMWX who is Xg⁻ marries a woman who is Xg⁺ and they have eight daughters and one son, all of whom are normal for muscle function, the male being Xg⁺ and all the daughters being heterozygous at both the EMWX and Xg loci. Following is a table that lists three of the daughters with the phenotypes of their husbands and children. Create a pedigree that represents all data stated above and in the following table.


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