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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 29

A female of genotype
Diagram showing gene loci a, b, and c with plus alleles aligned, illustrating crossover event data and mapping calculations.
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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Step 1: Identify the parental and recombinant types based on the crossover data. The parental genotype is + + + for loci a, b, and c. From the 100 tetrads, 68 show no crossover (parental), 20 show a single crossover between a and b, 10 show a single crossover between b and c, and 2 show a double crossover between a and b and between b and c.
Step 2: Calculate the number of gametes for each genotype. Each tetrad produces 4 gametes, so multiply the number of tetrads in each category by 4 to get the total gametes. For example, no crossover tetrads produce 68 × 4 = 272 parental gametes.
Step 3: Determine the genotypes of gametes produced by each crossover type. Single crossovers between a and b will produce recombinant gametes differing at loci a and b, but not c. Single crossovers between b and c will produce recombinant gametes differing at loci b and c, but not a. Double crossovers will produce gametes recombinant at both intervals.
Step 4: Assign the number of gametes to each of the 8 possible genotypes based on the crossover events. Remember that each tetrad produces 4 gametes, and the crossover events determine which alleles are combined in each gamete.
Step 5: Calculate the map distances between loci using the recombination frequencies. Use the formula for map distance in centiMorgans (cM): \(\text{Map distance} = \frac{\text{Number of recombinant tetrads}}{\text{Total tetrads}} \times 100\). Calculate separately for intervals a–b and b–c, considering single and double crossovers appropriately.

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Key Concepts

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

Genetic Recombination and Crossover Events

Genetic recombination occurs during meiosis when homologous chromosomes exchange segments, resulting in crossover events. These crossovers create new allele combinations in gametes, increasing genetic diversity. Understanding single and double crossovers is essential for predicting gamete genotypes and calculating recombination frequencies.
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Recombination after Double Strand Breaks

Genetic Mapping and Map Distance

Genetic mapping estimates the physical distance between genes based on recombination frequencies. Map distance is measured in centiMorgans (cM), where 1% recombination equals 1 cM. Calculating map distances involves analyzing crossover data to determine how often genes are separated during meiosis, reflecting their relative positions on a chromosome.
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Mapping Genes

Gamete Genotype Prediction from Tetrad Analysis

Tetrad analysis examines the four products of a single meiosis to determine crossover patterns and resulting gamete genotypes. By categorizing tetrads into parental, single crossover, and double crossover types, one can predict the frequency of each genotype among gametes. This approach helps quantify genetic linkage and recombination.
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Gamete Genotypes
Related Practice
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.

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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.

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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

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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