Ch, 15 - The Chromosomal Basis of Inheritance

Campbell - Campbell Biology 11th Edition

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Urry 11th Edition

Ch, 15 - The Chromosomal Basis of Inheritance

Problem 6

Chapter 15, Problem 6

A wild-type fruit fly (heterozygous for gray body color and red eyes) is mated with a black fruit fly with purple eyes. The offspring are wild-type, 721; black purple, 751; gray purple, 49; black red, 45. What is the recombination frequency between these genes for body color and eye color? Using information from problem 3, what fruit flies (genotypes and phenotypes) would you mate to determine the order of the body color, wing size, and eye color genes on the chromosome?

Verified step by step guidance

Identify the phenotypes and genotypes of the parent fruit flies. The wild-type fruit fly is heterozygous for gray body color (Gg) and red eyes (Rr), while the black fruit fly with purple eyes is homozygous recessive (ggpp).

Determine the phenotypes of the offspring and their respective numbers. The offspring are: wild-type (gray body, red eyes) - 721, black body, purple eyes - 751, gray body, purple eyes - 49, black body, red eyes - 45.

Calculate the total number of offspring by adding all the given offspring numbers together. Total offspring = 721 + 751 + 49 + 45.

Identify recombinant phenotypes, which are the least frequent phenotypes resulting from crossing over. In this case, gray body with purple eyes (49) and black body with red eyes (45) are recombinant.

Calculate the recombination frequency using the formula: Recombination frequency = (Number of recombinant offspring / Total number of offspring) * 100. Plug in the values to find the percentage.

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Key Concepts

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

Recombination Frequency

Recombination frequency is a measure of the likelihood that two genes located on the same chromosome will be separated during meiosis due to crossing over. It is calculated by dividing the number of recombinant offspring by the total number of offspring, then multiplying by 100 to express it as a percentage. This frequency helps in mapping the relative positions of genes on a chromosome.

Genotype and Phenotype

Genotype refers to the genetic constitution of an organism, specifically the alleles present for a particular trait, while phenotype is the observable expression of those traits. In the context of fruit flies, genotypes can include combinations of alleles for body color and eye color, and phenotypes are the physical characteristics, such as gray body color or red eyes. Understanding these concepts is crucial for predicting offspring traits in genetic crosses.

Gene Mapping

Gene mapping is the process of determining the relative positions of genes on a chromosome and the distances between them. This is often achieved through genetic crosses and analyzing offspring ratios to infer the order of genes based on recombination frequencies. By understanding the arrangement of genes, researchers can predict inheritance patterns and identify genetic linkage.

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Related Practice

Textbook Question

Using the information from problem 4, scientists do a further testcross using a heterozygote for height and nose morphology. The offspring are tall upturned snout, 40; dwarf upturned snout, 9; dwarf downturned snout, 42; tall downturned snout, 9. Calculate the recombination frequency from these data, and then use your answer from problem 4 to determine the correct order of the three linked genes.

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Textbook Question

A wild-type fruit fly (heterozygous for gray body color and normal wings) is mated with a black fly with vestigial wings. The offspring have the following phenotypic distribution: wild-type, 778; black vestigial, 785; black normal, 158; gray vestigial, 162. What is the recombination frequency between these genes for body color and wing size? Is this consistent with the results of the experiment in Figure 15.9?

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A planet is inhabited by creatures that reproduce with the same hereditary patterns seen in humans. Three phenotypic characters are height (T=tall,t=dwart), head appendages (A=antennae,a=no antennae), and nose morphology (S=upturned snout,s=downturned snout). Since the creatures are not 'intelligent,' Earth scientists are able to do some controlled breeding experiments using various heterozygotes in testcrosses. For tall heterozygotes with antennae, the offspring are tall antennae, 46; dwarf antennae, 7; dwarf no antennae, 42; tall no antennae, 5. For heterozygotes with antennae and an upturned snout, the offspring are antennae upturned snout, 47; antennae downturned snout, 2; no antennae downturned snout, 48; no antennae upturned snout, 3. Calculate the recombination frequencies for both experiments.

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Textbook Question

You design Drosophila crosses to provide recombination data for gene a, which is located on the chromosome shown in Figure 15.12. Gene a has recombination frequencies of 14% with the vestigial wing locus and 26% with the brown eye locus. Approximately where is a located along the chromosome?

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