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Ch. 5 - Chromosome Mapping in Eukaryotes
Chapter 5, Problem 18

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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Identify the genotypes of the female Drosophila. Since Stubble (Sb) is dominant and lethal in homozygous form, the female must be heterozygous for Stubble (Sb/sb). Curled (cu) is recessive, so the female must be heterozygous (cu/+) to express the wild-type phenotype.
To detect recombinants, choose a male with a genotype that will allow the expression of both recessive phenotypes in the offspring. This means the male should be homozygous recessive for both traits (sb/sb and cu/cu).
By mating the female (Sb/sb, cu/+) with a male (sb/sb, cu/cu), any recombinant offspring will express the recessive phenotypes, making them easily identifiable.
Consider the possible gametes produced by the female: Sb cu, Sb +, sb cu, and sb +. The male will produce only one type of gamete: sb cu.
Analyze the offspring phenotypes. Recombinants will be those that do not match the parental phenotypes, specifically those that express the recessive curled phenotype without the dominant Stubble phenotype.

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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. In this case, the Stubble (Sb) and curled (cu) mutations are linked on chromosome III, which means that they do not assort independently. Understanding linkage is crucial for predicting offspring genotypes and phenotypes in genetic crosses.
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Chi Square and Linkage

Dominant and Recessive Alleles

In genetics, alleles can be classified as dominant or recessive based on their expression in the phenotype. The Stubble (Sb) mutation is a dominant allele, meaning that only one copy is needed for the trait to be expressed, while the curled (cu) mutation is recessive, requiring two copies for expression. This distinction is essential for determining the expected phenotypes of the offspring from the mating.
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Variations on Dominance

Homozygosity and Lethality

Homozygosity occurs when an individual has two identical alleles for a particular gene. In this scenario, the Stubble (Sb) mutation is lethal in a homozygous state, meaning that individuals with two copies of this allele will not survive. This information is critical when selecting a male genotype for mating, as it influences the viability of the offspring and the potential for observing recombinants.
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Other Gene Interactions
Related Practice
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, Phenotype Number Dichaete 401 ebony, pink 389 Dichaete, ebony 84 pink 96 Dichaete, pink 2 ebony 3 Dichaete, ebony, pink 12 wild type 13 What is the sequence and interlocus distance between these three genes?
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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, Phenotype Number Dichaete 401 ebony, pink 389 Dichaete, ebony 84 pink 96 Dichaete, pink 2 ebony 3 Dichaete, ebony, pink 12 wild type 13 Diagram this cross, showing the genotypes of the parents and offspring of both crosses.
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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
If the cross described in Problem 18 were made, and if Sb and cu are 8.2 map units apart on chromosome III, and if 1000 offspring were recovered, what would be the outcome of the cross, assuming that equal numbers of males and females were observed?
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Textbook Question
Are mitotic recombinations and sister chromatid exchanges effective in producing genetic variability in an individual? in the offspring of individuals?
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Textbook Question
What possible conclusions can be drawn from the observations that in male Drosophila, no crossing over occurs, and that during meiosis, synaptonemal complexes are not seen in males but are observed in females where crossing over occurs?
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