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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
All textbooksKlug 12th EditionCh. 5 - Chromosome Mapping in EukaryotesProblem 32
Chapter 5, Problem 32

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.

Verified step by step guidance
1
Step 1: Identify the parental genotypes and phenotypes. The female fly is heterozygous for three mutations: Bar eyes (B), miniature wings (m), and ebony body (e). The male fly has normal eyes (b+), miniature wings (m), and ebony body (e). Note that Bar (B) is dominant, while miniature (m) and ebony (e) are recessive.
Step 2: Analyze the offspring phenotypes and group them into parental and recombinant classes. Parental phenotypes will match the original parental combinations, while recombinant phenotypes will result from crossing over between linked genes. For example, compare the phenotypes of the offspring to the parental genotypes to identify which are parental and which are recombinant.
Step 3: Determine if linkage is involved. If the genes are unlinked, the offspring phenotypes will follow Mendel's law of independent assortment, resulting in equal proportions of all combinations. If linkage is involved, the parental phenotypes will appear more frequently than the recombinant phenotypes. Use the offspring counts to assess this.
Step 4: Calculate the recombination frequency for each pair of genes. Recombination frequency is calculated as: \( \text{Recombination Frequency} = \frac{\text{Number of Recombinants}}{\text{Total Offspring}} \times 100 \). Identify the recombinant phenotypes for each gene pair (e.g., Bar and miniature, Bar and ebony, miniature and ebony) and calculate the recombination frequency for each pair.
Step 5: Use the recombination frequencies to determine the map distances between the genes. Recombination frequency (in percentage) is approximately equal to the map distance in centimorgans (cM). Arrange the genes in a linear order based on the calculated distances to construct a genetic map.

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Key Concepts

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

Genetic Linkage

Genetic linkage refers to the tendency of genes located close to each other on a chromosome to be inherited together during meiosis. This occurs because linked genes do not assort independently, leading to a higher frequency of parental combinations in offspring. Understanding linkage is crucial for interpreting genetic crosses, as it can affect the expected ratios of phenotypes.
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Chi Square and Linkage

Recombination Frequency

Recombination frequency is the measure of how often recombination occurs between two genes during meiosis, expressed as a percentage. It is used to estimate the genetic distance between genes on a chromosome, with 1% recombination frequency corresponding to 1 map unit (centimorgan). This concept is essential for determining the linkage and relative positions of genes based on offspring phenotypes.
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Recombination after Single Strand Breaks

Phenotypic Ratios

Phenotypic ratios represent the relative frequencies of different phenotypes observed in the offspring of a genetic cross. In this context, analyzing the ratios of the offspring's traits helps identify whether the genes are linked or assorting independently. By comparing the observed ratios to expected Mendelian ratios, one can infer the genetic relationships and potential linkage between the traits.
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Mutations and Phenotypes
Related Practice
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. No other phenotypes were observed.

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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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

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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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) 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.

For each of the offspring, indicate whether or not a crossover was required to produce the phenotypes that are given.

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Textbook Question

Because of the relatively high frequency of meiotic errors that lead to developmental abnormalities in humans, many research efforts have focused on identifying correlations between error frequency and chromosome morphology and behavior. Tease et al. (2002) studied human fetal oocytes of chromosomes 21, 18, and 13 using an immunocytological approach that allowed a direct estimate of the frequency and position of meiotic recombination. Below is a summary of information that compares recombination frequency with the frequency of trisomy for chromosomes 21, 18, and 13.

What conclusions can be drawn from these data in terms of recombination and nondisjunction frequencies? How might recombination frequencies influence trisomic frequencies?

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