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 19

Chapter 5, Problem 19

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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Determine the recombination frequency between the Sb and cu genes. Since they are 8.2 map units apart, the recombination frequency is 8.2%, or 0.082. This means that 8.2% of the offspring will be recombinant, while the remaining 91.8% will be parental types.

Calculate the expected number of recombinant offspring. Multiply the total number of offspring (1000) by the recombination frequency (0.082). This will give the total number of recombinant offspring.

Divide the recombinant offspring into two equal groups, as there are two types of recombinant gametes (one with Sb and the other with cu). Each recombinant type will represent half of the total recombinant offspring.

Calculate the expected number of parental offspring. Subtract the number of recombinant offspring from the total number of offspring (1000). The remaining offspring will be parental types. Divide this number into two equal groups, as there are two types of parental gametes (one with Sb and cu together, and the other with neither Sb nor cu).

Summarize the expected outcome of the cross by listing the number of offspring for each genotype or phenotype category (two parental types and two recombinant types). Ensure that the total adds up to 1000.

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Key Concepts

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

Genetic Mapping

Genetic mapping is a method used to determine the location of genes on a chromosome and the distance between them. The distance is often measured in map units, where 1 map unit corresponds to a 1% chance of recombination occurring between two genes during meiosis. In this case, the distance of 8.2 map units between the Sb and cu genes indicates a relatively low likelihood of recombination, which affects the expected offspring ratios.

Recombination Frequency

Recombination frequency refers to the proportion of offspring that exhibit a combination of traits different from those of the parents due to crossing over during meiosis. It is calculated based on the number of recombinant offspring divided by the total number of offspring. In this scenario, with a recombination frequency of 8.2%, we can predict the expected ratios of parental and recombinant phenotypes among the 1000 offspring.

Punnett Square and Expected Ratios

A Punnett square is a diagram used to predict the genetic makeup of offspring from a cross between two individuals. By applying the recombination frequency to the Punnett square, we can calculate the expected phenotypic ratios of the offspring. Given the genetic distance and the total number of offspring, we can derive the expected numbers of each phenotype resulting from the cross.

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Chi Square Analysis

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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

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

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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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?

957

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