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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 19

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.
Table showing progeny counts for two heterozygous plants testcrossed, with phenotypes red/long, yellow/oval, red/oval, and yellow/long.
Determine the location of the genes relative to one another and the genotypes of the two parental plants.

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Step 1: Identify the parental genotypes and phenotypes. Both plants are heterozygous for fruit color (Red is dominant to Yellow) and fruit shape (Oval is dominant to Long). So, the parental genotypes are likely RrOo, where R = red, r = yellow, O = oval, and o = long.
Step 2: Analyze the progeny phenotypes and their counts from the testcross. The testcross involves crossing the heterozygous plants (RrOo) with homozygous recessive plants (rroo). The progeny phenotypes are red/long, yellow/oval, red/oval, and yellow/long with their respective counts for Plant A and Plant B.
Step 3: Determine which phenotypes represent parental (non-recombinant) and recombinant types. The most frequent phenotypes correspond to parental types, while the least frequent correspond to recombinants. For Plant A, red/long and yellow/oval are most frequent, so these are parental; red/oval and yellow/long are recombinant. For Plant B, red/oval and yellow/long are most frequent, so these are parental; red/long and yellow/oval are recombinant.
Step 4: Calculate the recombination frequency (RF) for each plant using the formula: \(\text{RF} = \frac{\text{Number of recombinant progeny}}{\text{Total progeny}} \times 100\). For Plant A, sum the recombinant progeny counts (red/oval + yellow/long) and divide by total progeny (100). Repeat for Plant B.
Step 5: Interpret the recombination frequency to determine gene linkage and relative gene location. A recombination frequency less than 50% indicates that the genes are linked and close on the same chromosome. The parental genotypes can be deduced based on which allele combinations are inherited together (coupling or repulsion phase).

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Key Concepts

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

Testcross and Its Purpose

A testcross involves crossing an individual with a homozygous recessive individual to determine the genotype of the former. It reveals whether the tested individual is heterozygous or homozygous dominant for specific traits by analyzing the phenotypes of the offspring. This method is essential for mapping gene linkage and recombination.
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Gene Linkage and Recombination

Genes located close together on the same chromosome tend to be inherited together, a phenomenon called linkage. Recombination occurs when crossing over during meiosis exchanges genetic material between homologous chromosomes, producing new allele combinations. The frequency of recombinant offspring helps determine the physical distance between genes.
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Dominant and Recessive Traits and Phenotypic Ratios

Dominant traits mask the expression of recessive traits in heterozygotes. In this problem, red and oval are dominant, while yellow and long are recessive. Analyzing the phenotypic ratios of progeny from heterozygous parents helps infer parental genotypes and gene linkage, especially when deviations from expected Mendelian ratios occur.
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Related Practice
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

A female of genotype

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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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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Textbook Question

An organism of the genotype AaBbCc was testcrossed to a triply recessive organism (aabbcc). The genotypes of the progeny are presented in the following table.

If these three genes were all assorting independently, how many genotypic and phenotypic classes would result in the offspring, and in what proportion, assuming simple dominance and recessiveness in each gene pair?

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