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Ch. 7 - Linkage and Chromosome Mapping in Eukaryotes
Klug - Essentials of Genetics 10th Edition
Klug10th EditionEssentials of GeneticsISBN: 9780135588789Not the one you use?Change textbook
Chapter 7, Problem 18

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.
Recommended video:
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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

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

Drosophila females homozygous for the third chromosomal genes pink and ebony (the same genes from Problem 16) were crossed with males homozygous for the second chromosomal gene dumpy. Because these genes are recessive, all offspring were wild type (normal). F1 females were testcrossed to triply recessive males. If we assume that the two linked genes, pink and ebony, are 20 mu apart, predict the results of this cross. If the reciprocal cross were made (F1 males—where no crossing over occurs—with triply recessive females), how would the results vary, if at all?

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Textbook Question

In Drosophila, two mutations, Stubble (Sb) and curled (cu), are linked on chromosome III. Stubble is a dominant gene that is lethal in a homozygous state, and curled is a recessive gene. If a female of the genotype

is to be mated to detect recombinants among her offspring, what male genotype would you choose as a mate?

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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

In Drosophila, Dichaete (D) is a mutation on chromosome III with a dominant effect on wing shape. It is lethal when homozygous. The genes ebony body (e) and pink eye (p) are recessive mutations on chromosome III. Flies from a Dichaete stock were crossed to homozygous ebony, pink flies, and the F₁ progeny, with a Dichaete phenotype, were backcrossed to the ebony, pink homozygotes. Using the results of this backcross shown in the table.

Diagram this cross, showing the genotypes of the parents and offspring of both crosses.

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Textbook Question

In Drosophila, Dichaete (D) is a mutation on chromosome III with a dominant effect on wing shape. It is lethal when homozygous. The genes ebony body (e) and pink eye (p) are recessive mutations on chromosome III. Flies from a Dichaete stock were crossed to homozygous ebony, pink flies, and the F1 progeny, with a Dichaete phenotype, were backcrossed to the ebony, pink homozygotes. Using the results of this backcross shown in the table,

What is the sequence and interlocus distance between these three genes?

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