Ch. 5 - Chromosome Mapping in Eukaryotes

Klug - Concepts of Genetics 12th Edition

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Klug 12th Edition

Ch. 5 - Chromosome Mapping in Eukaryotes

Problem 16b

Chapter 5, Problem 16b

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?

Verified step by step guidance

Step 1: Identify the parental and recombinant phenotypes from the table. The parental phenotypes are the most frequent classes, which are 'Dichaete' (401) and 'ebony, pink' (389). The recombinant phenotypes are the less frequent classes, such as 'Dichaete, ebony' (84), 'pink' (96), 'Dichaete, pink' (2), 'ebony' (3), 'Dichaete, ebony, pink' (12), and 'wild type' (13).

Step 2: Determine the gene order by analyzing double crossover classes. The double crossover phenotypes are the rarest classes, which are 'Dichaete, pink' (2) and 'ebony' (3). These indicate which gene is in the middle because double crossovers affect the middle gene. Compare these to the parental types to infer the gene order.

Step 3: Calculate the recombination frequencies between each pair of genes. Use the formula for recombination frequency (RF): \(\text{RF} = \frac{\text{Number of recombinant offspring}}{\text{Total number of offspring}} \times 100\) Calculate RF for each gene pair by summing the appropriate recombinant classes that involve crossovers between those genes.

Step 4: Assign the gene order based on the double crossover analysis and recombination frequencies. The gene in the middle will be the one involved in the double crossover classes. The two outer genes will have recombination frequencies that add up approximately to the total distance between the outermost genes.

Step 5: Calculate the interlocus distances (map units) between the genes using the recombination frequencies. The distances between genes are expressed in centiMorgans (cM), where 1% recombination frequency equals 1 cM. Use the recombination frequencies calculated to find the distances between each pair of genes.

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Key Concepts

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

Genetic Linkage and Recombination

Genetic linkage refers to genes located close together on the same chromosome that tend to be inherited together. Recombination occurs during meiosis when crossing over between homologous chromosomes can separate linked genes, producing new allele combinations. The frequency of recombination between two genes is proportional to their physical distance on the chromosome, measured in map units or centimorgans (cM).

Three-Point Test Cross Mapping

A three-point test cross involves crossing an individual heterozygous for three genes with a homozygous recessive individual to analyze recombination frequencies. By examining the phenotypes and their numbers in the progeny, one can determine the gene order and calculate the distances between genes. Double crossover events are key to identifying gene order and refining map distances.

Interpreting Phenotypic Classes in Linkage Analysis

Phenotypic classes in a backcross reflect parental and recombinant genotypes. The most frequent classes represent parental (non-recombinant) types, while less frequent classes indicate single or double crossovers. Identifying these classes allows calculation of recombination frequencies and helps deduce gene order and map distances.

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Textbook Question

Another cross in Drosophila involved the recessive, X-linked genes yellow (y), white (w), and cut (ct). A yellow-bodied, white-eyed female with normal wings was crossed to a male whose eyes and body were normal but whose wings were cut. The F₁ females were wild type for all three traits, while the F₁ males expressed the yellow-body and white-eye traits. The cross was carried to an F₂ progeny, and only male offspring were tallied. On the basis of the data shown here, a genetic map was constructed.

Were any double-crossover offspring expected?

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Textbook Question

Another cross in Drosophila involved the recessive, X-linked genes yellow (y), white (w), and cut (ct). A yellow-bodied, white-eyed female with normal wings was crossed to a male whose eyes and body were normal but whose wings were cut. The F₁ females were wild type for all three traits, while the F₁ males expressed the yellow-body and white-eye traits. The cross was carried to an F₂ progeny, and only male offspring were tallied. On the basis of the data shown here, a genetic map was constructed.

Could the F₂ female offspring be used to construct the map? Why or why not?

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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

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

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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