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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 31a

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?

Verified step by step guidance
1
Understand the problem: The question involves interpreting the results of a genetic study, specifically the association of a SNP marker with a condition and the significance of a LOD score. The goal is to determine what these results suggest about the genetic marker on chromosome 12.
Step 1: Recall the concept of association. A strong association between a SNP marker and a condition suggests that the marker is located near a gene or genetic region that may influence the condition. This is because markers that are physically close to causal genes tend to be inherited together due to linkage disequilibrium.
Step 2: Understand the LOD score. The LOD (logarithm of the odds) score is a statistical measure used in genetic linkage analysis to evaluate the likelihood that a particular marker is linked to a trait. A LOD score of 3 or higher is generally considered strong evidence of linkage, while a score below 2 is considered weak evidence. A score of 2.2 falls in a gray area, suggesting moderate evidence of linkage.
Step 3: Interpret the results. The strong association with the SNP marker on chromosome 12 suggests that this region may contain a gene or genetic element contributing to the early onset heart disease. However, the LOD score of 2.2 indicates that the evidence for linkage is not definitive and further studies are needed to confirm the relationship.
Step 4: Suggest next steps. To strengthen the evidence, researchers could increase the sample size by studying more families, perform fine-mapping to narrow down the region of interest, or use functional studies to identify candidate genes in the region and test their role in the condition.

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Key Concepts

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

SNP Markers

Single Nucleotide Polymorphisms (SNPs) are variations at a single position in a DNA sequence among individuals. They are the most common type of genetic variation and can be used as markers in genetic studies to identify associations with diseases. In this context, the strong association of a SNP marker on chromosome 12 with early onset heart disease suggests that this marker may be linked to the genetic factors contributing to the condition.
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Lod Score

The lod score (logarithm of the odds) is a statistical measure used to evaluate the likelihood that two loci are linked, compared to the likelihood that they are not linked. A lod score of 2.2 indicates a significant likelihood of linkage, suggesting that the genetic marker on chromosome 12 is likely associated with the heart disease in the studied families. Generally, a lod score above 3 is considered strong evidence for linkage.
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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 phenomenon is important in understanding the inheritance patterns of traits and diseases. In this case, the genetic linkage analysis showing a lod score of 2.2 implies that the marker on chromosome 12 may be located near a gene that influences the early onset of heart disease, supporting the hypothesis of a genetic basis for the condition.
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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.

Calculate the recombination frequencies between each of the gene pairs.

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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.

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

424
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.

Use chi-square analysis to demonstrate that the data in this experiment are not the result of independent assortment.

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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.

403
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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:

Calculate the recombination frequency for each of the progeny groups.

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