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.
Ch, 15 - The Chromosomal Basis of Inheritance
Urry11th EditionCampbell BiologyISBN: 9789357423311
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Table of contents
- Ch. 1 - Evolution, the Themes of Biology, and Scientific Inquiry8
- Ch. 2 - The Chemical Context of Life8
- Ch. 3 - Water and Life6
- Ch. 4 - Carbon and the Molecular Diversity of Life9
- Ch. 5 - The Structure and Function of Large Biological Molecules9
- Ch. 6 - A Tour of the Cell6
- Ch. 7 - Membrane Structure and Function6
- Ch. 8 - An Introduction to Metabolism6
- Ch. 9 - Cellular Respiration and Fermentation10
- Ch. 10 - Photosynthesis7
- Ch. 11 - Cell Communication7
- Ch. 12 - The Cell Cycle10
- Ch. 13 - Meiosis and Sexual Life Cycles10
- Ch. 14 - Mendel and the Gene Idea14
- Ch, 15 - The Chromosomal Basis of Inheritance7
- Ch. 16 - The Molecular Basis of Inheritance9
- Ch. 17 - Gene Expression: From Gene to Protein9
- Ch. 18 - Regulation of Gene Expression10
- Ch. 19 - Viruses6
- Ch. 20 - DNA Tools and Biotechnology8
- Ch. 21 - Genomes and Their Evolution8
- Ch. 22 - Descent with Modification: A Darwininan View of Life6
- Ch. 23 - The Evolution of Populations5
- Ch. 24 - The Origin of Species6
- Ch. 25 - The History of Life on Earth6
- Ch. 26 - Phylogeny and the Tree of Life7
- Ch. 27 - Bacteria and Archaea8
- Ch. 28 - Protists7
- Ch. 29 - Plant Diversity I: How Plants Colonized Land7
- Ch. 30 - Plant Diversity II: The Evolution of Seed Plants7
- Ch. 31 - Fungi5
- Ch. 32 - An Overview of Animal Diversity4
- Ch. 33 - An introduction to Invertebrates6
- Ch. 34 - The Origin and Evolution of Vertebrates7
- Ch. 35 - Vascular Plant Structure, Growth, and Development8
- Ch. 36 - Resource Acquisition and Transport in Vascular Plants8
- Ch. 37 - Soil and Plant Nutrition10
- Ch. 38 - Angiosperm Reproduction and Biotechnology8
- Ch. 39 - Plant Responses to Internal and External Signals8
- Ch. 40 - Basic Principles of Animal Form and Function8
- Ch. 41 - Animal Nutrition7
- Ch. 42 - Circulation and Gas Exchange7
- Ch. 43 - The Immune System6
- Ch. 44 - Osmoregulation and Excretion6
- Ch. 45 - Hormones and the Endocrine System8
- Ch. 46 - Animal Reproduction8
- Ch. 47 - Animal Development8
- Ch. 48 - Neurons, Synapses, and Signaling6
- Ch. 49 - Nervous Systems8
- Ch. 50 - Sensory and Motor Mechanisms5
- Ch. 51 Animal Behavior6
- Ch. 52 - An Introduction to Ecology and the Biosphere7
- Ch. 53 - Population Ecology10
- Ch. 54 - Community Ecology9
- Ch. 55 - Ecosystems and Restoration Ecology8
- Ch. 56 - Conservation Biology and Global Change6
Chapter 15, Problem 9
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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Understand that recombination frequency is a measure of genetic linkage and is used to estimate the distance between genes on a chromosome. It is expressed as a percentage, where 1% recombination frequency is equivalent to 1 map unit (centimorgan).
Identify the recombination frequencies given in the problem: 14% with the vestigial wing locus and 26% with the brown eye locus. These frequencies suggest the relative distances between gene a and the other two loci on the chromosome.
Visualize the chromosome as a linear map where each locus is positioned based on recombination frequencies. The vestigial wing locus and the brown eye locus are reference points, and gene a is somewhere between them.
Calculate the approximate position of gene a by considering the recombination frequencies as distances. If gene a is 14 map units from the vestigial wing locus and 26 map units from the brown eye locus, it suggests that gene a is closer to the vestigial wing locus than to the brown eye locus.
Place gene a on the chromosome map by estimating its position relative to the other loci. Since gene a is 14 map units from the vestigial wing locus and 26 map units from the brown eye locus, it is likely located between these two loci, closer to the vestigial wing locus.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Genetic Recombination
Genetic recombination refers to the process by which two DNA molecules exchange genetic material, resulting in new combinations of alleles. In Drosophila, recombination occurs during meiosis, and the frequency of recombination between two loci can indicate their relative positions on a chromosome. Higher recombination frequencies suggest greater physical distances between genes.
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Linkage Maps
Linkage maps are diagrams that show the relative positions of genes on a chromosome based on recombination frequencies. These maps are constructed by analyzing the recombination data from genetic crosses, allowing researchers to estimate the distance between genes. In this context, the recombination frequencies between gene a and other loci help determine its approximate location on the chromosome.
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Chromosomal Mapping
Chromosomal mapping involves determining the physical location of genes on a chromosome. By using recombination frequencies, scientists can infer the order and distance between genes. In the given question, the recombination frequencies of gene a with the vestigial wing and brown eye loci provide clues to its position, suggesting it is closer to the vestigial wing locus than the brown eye locus.
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Related Practice
Textbook Question
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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 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?
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