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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 6b

In Drosophila, the map positions of genes are given in map units numbering from one end of a chromosome to the other. The X chromosome of Drosophila is 66 m.u. long. The X-linked gene for body color—with two alleles, y⁺ for gray body and y for yellow body—resides at one end of the chromosome at map position 0.0. A nearby locus for eye color, with alleles w⁺ for red eye and w for white eye, is located at map position 1.5. A third X-linked gene, controlling bristle form, with f⁺ for normal bristles and f for forked bristles, is located at map position 56.7. At each locus the wild-type allele is dominant over the mutant allele.


Do you expect any of these gene pair(s) to assort independently? Explain your reasoning.

Verified step by step guidance
1
Understand the concept of genetic linkage and independent assortment: Genes that are located close to each other on the same chromosome are said to be linked and tend to be inherited together. Independent assortment occurs when genes are located far apart on the same chromosome or on different chromosomes, allowing them to segregate independently during meiosis.
Review the map positions of the genes provided: The body color gene (y⁺/y) is at position 0.0, the eye color gene (w⁺/w) is at position 1.5, and the bristle form gene (f⁺/f) is at position 56.7. These positions are measured in map units (m.u.), where 1 m.u. corresponds to a 1% recombination frequency.
Determine the distances between the genes: Calculate the genetic distances between each pair of genes. For example, the distance between the body color gene and the eye color gene is |1.5 - 0.0| = 1.5 m.u., and the distance between the body color gene and the bristle form gene is |56.7 - 0.0| = 56.7 m.u. Similarly, calculate the distance between the eye color gene and the bristle form gene.
Evaluate the likelihood of independent assortment: Genes that are more than 50 m.u. apart are expected to assort independently because the recombination frequency approaches 50%, which is equivalent to random assortment. Compare the calculated distances to this threshold to determine whether any gene pairs are likely to assort independently.
Conclude based on the distances: Based on the distances calculated, determine which gene pairs are close enough to be considered linked (less than 50 m.u.) and which are far enough apart to assort independently (greater than or equal to 50 m.u.). Provide reasoning for your conclusion, referencing the principles of genetic linkage and recombination.

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Key Concepts

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

Gene Linkage

Gene linkage refers to the tendency of genes located close to each other on the same chromosome to be inherited together during meiosis. This occurs because linked genes do not assort independently, as they are physically connected on the chromosome. In the case of Drosophila, the proximity of the body color gene (0.0 m.u.) and the eye color gene (1.5 m.u.) suggests they are likely linked, while the bristle form gene (56.7 m.u.) is farther away, indicating a potential for independent assortment.
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Map Units and Recombination Frequency

Map units, or centimorgans (cM), are a measure of genetic distance based on the frequency of recombination between two loci during meiosis. One map unit corresponds to a 1% chance of recombination occurring between two genes. The closer two genes are on a chromosome, the lower the likelihood of recombination, which affects their inheritance patterns. In this scenario, the small distance between the body color and eye color genes suggests they are less likely to assort independently.
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Dominance and Alleles

Dominance refers to the relationship between alleles, where the presence of a dominant allele masks the expression of a recessive allele. In the context of the Drosophila genes mentioned, the wild-type alleles (y⁺, w⁺, f⁺) are dominant over their respective mutant alleles (y, w, f). Understanding dominance is crucial for predicting phenotypic outcomes in offspring, especially when considering how linked genes may influence traits together rather than independently.
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Related Practice
Textbook Question

In tomato plants, purple leaf color is controlled by a dominant allele A, and green leaf by a recessive allele a. At another locus, hairy leaf H is dominant to hairless leaf h. The genes for leaf color and leaf texture are separated by 16 m.u. on chromosome 5. On chromosome 4, a gene controlling leaf shape has two alleles: a dominant allele C that produces cut-leaf shape and a recessive allele c that produces potato-shaped leaves.

The cross of a purple, hairy, cut plant heterozygous at each gene to a green, hairless potato plant produces the following progeny:

 

Give the genotypes of parental and progeny plants in this experiment.

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Textbook Question

In tomato plants, purple leaf color is controlled by a dominant allele A, and green leaf by a recessive allele a. At another locus, hairy leaf H is dominant to hairless leaf h. The genes for leaf color and leaf texture are separated by 16 m.u. on chromosome 5. On chromosome 4, a gene controlling leaf shape has two alleles: a dominant allele C that produces cut-leaf shape and a recessive allele c that produces potato-shaped leaves.

Fully explain the number and frequency of each phenotype class.

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Textbook Question

In Drosophila, the map positions of genes are given in map units numbering from one end of a chromosome to the other. The X chromosome of Drosophila is 66 m.u. long. The X-linked gene for body color—with two alleles, y⁺ for gray body and y for yellow body—resides at one end of the chromosome at map position 0.0. A nearby locus for eye color, with alleles w⁺ for red eye and w for white eye, is located at map position 1.5. A third X-linked gene, controlling bristle form, with f⁺ for normal bristles and f for forked bristles, is located at map position 56.7. At each locus the wild-type allele is dominant over the mutant allele.


In a cross involving these three X-linked genes, do you expect any gene pair(s) to show genetic linkage? Explain your reasoning.

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Textbook Question

In Drosophila, the map positions of genes are given in map units numbering from one end of a chromosome to the other. The X chromosome of Drosophila is 66 m.u. long. The X-linked gene for body color—with two alleles, y⁺ for gray body and y for yellow body—resides at one end of the chromosome at map position 0.0. A nearby locus for eye color, with alleles w⁺ for red eye and w for white eye, is located at map position 1.5. A third X-linked gene, controlling bristle form, with f⁺ for normal bristles and f for forked bristles, is located at map position 56.7. At each locus the wild-type allele is dominant over the mutant allele.


A wild-type female fruit fly with the genotype y⁺w⁺f/ywf⁺ is crossed to a male fruit fly that has yellow body, white eye, and forked bristles. Predict the frequency of each progeny phenotype class produced by this mating.

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Textbook Question

In Drosophila, the map positions of genes are given in map units numbering from one end of a chromosome to the other. The X chromosome of Drosophila is 66 m.u. long. The X-linked gene for body color—with two alleles, y⁺ for gray body and y for yellow body—resides at one end of the chromosome at map position 0.0. A nearby locus for eye color, with alleles w⁺ for red eye and w for white eye, is located at map position 1.5. A third X-linked gene, controlling bristle form, with f⁺ for normal bristles and f for forked bristles, is located at map position 56.7. At each locus the wild-type allele is dominant over the mutant allele.


Explain how each of the predicted progeny classes is produced.

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Textbook Question

Genes A, B, and C are linked on a chromosome and found in the order A–B–C. Genes A and B recombine with a frequency of 8%, and genes B and C recombine at a frequency of 24%. For the cross a⁺b⁺c/abc⁺ × abc/abc, predict the frequency of progeny genotypes. Assume interference is zero.

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