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Ch. 5 - Genetic Linkage and Mapping in Eukaryotes
Sanders - Genetic Analysis: An Integrated Approach 3rd Edition
Sanders3rd EditionGenetic Analysis: An Integrated ApproachISBN: 9780135564172Not the one you use?Change textbook
Chapter 5, Problem 30d

A Drosophila experiment examining potential genetic linkage of X-linked genes studies a recessive eye mutant (echinus), a recessive wing-vein mutation (crossveinless), and a recessive bristle mutation (scute). The wild-type phenotypes are dominant. Trihybrid wild-type females (all have the same genotype) are crossed to hemizygous males displaying the three recessive phenotypes. Among the 20,765 progeny produced from these crosses are the phenotypes and numbers listed in the table. Any phenotype not given is wild type.
Table displaying phenotypes and numbers from a Drosophila genetic linkage experiment, totaling 20,765 progeny.
Use chi-square analysis to demonstrate that the data in this experiment are not the result of independent assortment.

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Step 1: Understand the problem and identify the hypothesis. The null hypothesis (H₀) in this case is that the genes are assorting independently. To test this, we will use a chi-square analysis to compare the observed phenotypic frequencies with the expected frequencies under independent assortment.
Step 2: Calculate the expected frequencies for each phenotype under the assumption of independent assortment. First, determine the total number of progeny (20,765) and the proportions expected for each phenotype based on Mendelian inheritance. For three genes, independent assortment predicts a 1:1:1:1 ratio for each combination of phenotypes. Use this ratio to calculate the expected frequency for each phenotype.
Step 3: Apply the chi-square formula for each phenotype. The formula is: χ² = Σ((Oᵢ - Eᵢ)² / Eᵢ), where Oᵢ is the observed frequency and Eᵢ is the expected frequency for each phenotype. Substitute the observed and expected values for each phenotype into the formula.
Step 4: Sum the chi-square values for all phenotypes to obtain the total chi-square statistic. This value will represent the overall deviation of the observed data from the expected data under the null hypothesis.
Step 5: Compare the calculated chi-square statistic to the critical value from a chi-square distribution table. Use the degrees of freedom (df), which is calculated as the number of phenotypic categories minus 1. If the chi-square statistic exceeds the critical value at a chosen significance level (e.g., 0.05), reject the null hypothesis and conclude that the genes are not assorting independently.

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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 the same chromosome to be inherited together during meiosis. This phenomenon contrasts with independent assortment, where genes segregate independently. In the context of the Drosophila experiment, analyzing the inheritance patterns of the X-linked genes can reveal whether they are linked or assorting independently based on the observed progeny ratios.
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Chi Square and Linkage

Chi-Square Analysis

Chi-square analysis is a statistical method used to determine if there is a significant difference between observed and expected frequencies in categorical data. In genetics, it helps assess whether the distribution of phenotypes in progeny aligns with expected ratios under the assumption of independent assortment. A significant chi-square value indicates that the observed data deviates from what would be expected if the genes assorted independently, suggesting linkage.
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Phenotypic Ratios

Phenotypic ratios represent the relative frequencies of different phenotypes observed in the offspring of a genetic cross. In this experiment, the ratios of the various phenotypes, including recessive mutants and wild types, provide insight into the genetic relationships between the traits. By comparing these ratios to expected ratios under independent assortment, researchers can infer whether the genes are linked or assorting independently.
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Related Practice
Textbook Question

A Drosophila experiment examining potential genetic linkage of X-linked genes studies a recessive eye mutant (echinus), a recessive wing-vein mutation (crossveinless), and a recessive bristle mutation (scute). The wild-type phenotypes are dominant. Trihybrid wild-type females (all have the same genotype) are crossed to hemizygous males displaying the three recessive phenotypes. Among the 20,765 progeny produced from these crosses are the phenotypes and numbers listed in the table. Any phenotype not given is wild type.

Determine the gene order and identify the alleles on the homologous X chromosomes in the trihybrid females.

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Textbook Question

A Drosophila experiment examining potential genetic linkage of X-linked genes studies a recessive eye mutant (echinus), a recessive wing-vein mutation (crossveinless), and a recessive bristle mutation (scute). The wild-type phenotypes are dominant. Trihybrid wild-type females (all have the same genotype) are crossed to hemizygous males displaying the three recessive phenotypes. Among the 20,765 progeny produced from these crosses are the phenotypes and numbers listed in the table. Any phenotype not given is wild type.

Calculate the recombination frequencies between each of the gene pairs.

500
views
Textbook Question

A Drosophila experiment examining potential genetic linkage of X-linked genes studies a recessive eye mutant (echinus), a recessive wing-vein mutation (crossveinless), and a recessive bristle mutation (scute). The wild-type phenotypes are dominant. Trihybrid wild-type females (all have the same genotype) are crossed to hemizygous males displaying the three recessive phenotypes. Among the 20,765 progeny produced from these crosses are the phenotypes and numbers listed in the table. Any phenotype not given is wild type.

Compare the recombination frequencies and speculate about the source of any apparent discrepancies in the recombination data.

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Textbook Question

A genetic study of an early onset form of heart disease identifies 10 families containing members with the condition. No clear dominant or recessive pattern of inheritance is evident, but an analysis of SNP markers for five families detects a strong association with a marker on chromosome 12, and genetic linkage analysis for the marker produces a lod score of 2.2.


What do the association and lod score results suggest about this genetic marker?

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Textbook Question

A genetic study of an early onset form of heart disease identifies 10 families containing members with the condition. No clear dominant or recessive pattern of inheritance is evident, but an analysis of SNP markers for five families detects a strong association with a marker on chromosome 12, and genetic linkage analysis for the marker produces a lod score of 2.2.


What next step do you recommend for this genetic analysis?

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Textbook Question

In experiments published in 1918 that sought to verify and expand the genetic linkage and recombination theory proposed by Morgan, Thomas Bregger studied potential genetic linkage in corn (Zea mays) for genes controlling kernel color (colored is dominant to colorless) and starch content (starchy is dominant to waxy). Bregger performed two crosses. In Cross 1, pure-breeding colored, starchy-kernel plants (C1 Wx/C1 Wx) were crossed to plants pure-breeding for colorless, waxy kernels (c1 wx/c1 wx). The F₁ of this cross were test-crossed to colorless, waxy plants. The test-cross progeny were as follows:

In Cross 2, plants pure-breeding for colored, waxy kernels (C1 wx/C1 wx) and colorless, starchy kernels (c1 Wx/c1 Wx) were mated, and their F₁ were test-crossed to colorless, waxy plants. The test-cross progeny were as follows:

For each set of test-cross progeny, determine whether genetic linkage or independent assortment is more strongly supported by the data. Explain the rationale for your answer.

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