Study Guide And Solutions Manual For Genetic Analysis: An Integrated Approach
3rd Edition
ISBN: 9780134832258
Author: Mark F. Sanders, John L. Bowman, Peter Mirabito
Publisher: PEARSON
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Chapter 18, Problem 15P
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Here are schematic diagrams of mutant Drosophila larvae.
The left side of each pair shows a wild-type larva, with gray boxes showing the sections that are missing in the mutant larva. Which type of gene is defective in each larva: a gap gene, a pair-rule gene, or a segment-polarity gene?
A maternal effect gene in Drosophila, called torso, is found as a recessive allele that prevents the correct development of anterior- and posterior-most structures. A wild-type (homozygous) male is crossed to a female of unknown genotype. This mating produces 100% larva that are missing their anterior- and posterior-most structures and therefore die during early development. What is the genotype and phenotype of the female fly in this cross? What are the genotypes and phenotypes of the female fly’s parents? Show COMPLETE cross.
A maternal effect gene in Drosophila, called torso, is found as a recessive allele that prevents the correct development of anterior- and posterior-most structures. A wild-type (homozygous) male iscrossed to a female of unknown genotype. This mating produces 100% larva that are missing theiranterior- and posterior-most structures and therefore die during early development.
What is the genotype and phenotype of the female fly in this cross? What are the genotypes andphenotypes of the female fly’s parents? Show COMPLETE cross.
Maternal genotype:Paternal genotype:F1 genotype:F1 phenotype:
Chapter 18 Solutions
Study Guide And Solutions Manual For Genetic Analysis: An Integrated Approach
Ch. 18 - 18.1 Explain why many developmental genes encode...Ch. 18 - Bird beaks develop from an embryonic group of...Ch. 18 - 18.3 How is positional information provided along...Ch. 18 - Early development in Drosophila is atypical in...Ch. 18 - 18.5 Consider the evenskipped regulatory sequences...Ch. 18 - What is the difference between a parasegment and...Ch. 18 - Why do loss-of-function mutations in Hox genes...Ch. 18 - 18.8 Compare and contrast the specification of...Ch. 18 - Prob. 9PCh. 18 - Ablation of the anchor cell in wild type C....
Ch. 18 - 18.11 In gain-of-function and. elegans mutants,...Ch. 18 - Prob. 12PCh. 18 - Prob. 13PCh. 18 - 18.14 Given that maternal Bicoid activates the...Ch. 18 - What phenotypes do you expect in flies homozygous...Ch. 18 - The pair rule gene fushitarazu is expressed in...Ch. 18 - 18.17 In contrast to Drosophila, some insects...Ch. 18 - Prob. 18PCh. 18 - 18.19 You are traveling in the Netherlands and...Ch. 18 - 19.20 A powerful approach to identifying genes of...Ch. 18 - The Hoxd 913 genes are thought to specify digit...Ch. 18 - Three-spined stickleback fish live in lakes formed...Ch. 18 - The flowering jungle plant Lacandoniaschismatica,...Ch. 18 - 18.24 Homeotic genes are thought to regulate each...Ch. 18 - Prob. 25PCh. 18 - Basidiomycota is a monophyletic group of fungi...Ch. 18 - Prob. 27PCh. 18 - In C. elegans there are two sexes: hermaphrodite...Ch. 18 - In Drosophila, recessive mutations in the...
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- Describe the following mutations in a fruit fly: Dumpy: Curly: Apterous: Ebony: Vestigial: Antennapedia: White eye: Bar eye: Vermillion eye: Lobe eye: Sepia eye:arrow_forwardA number of genes that control expression of Hox genes in Drosophila have been identified. One of these homozygous mutants is extra sex combs, where some of the head and all of the thorax and abdominal segments develop as the last abdominal segment. In other words, all affected segments develop as posterior segments. What does this phenotype tell you about which set of Hox genes is controlled by the extra sex combs gene?arrow_forwardWhat is the genetic distance between the eye colour locus (w) and the bristle locus (sn)? answer in Mu (Map units) and to 2 decimal places? The following table summarises the results of the 2022-2023 Drosophila three-point cross involving the loci white eyes (w), miniature wings (m) and singed bristles (sn). The data is also available in the Excel file 'Drosophila Counts 2022-2023'. We strongly recommend working in Excel during this exercise.Phenotype Count+ + + 584w m sn 324w + + 227+ m sn 150+ m + 134w + sn 196 + + sn 134 w m + 92arrow_forward
- Honey bee workers from a colony in North Carolina are visiting two food sites, A and B, at 6 am as shown below. Site A is 1500 m from the hive; site B is 300 m from the hive. Which of the following statements is true regarding the genetic basis of the bees’ ability to track the passage of time and compensate for the movement of the sun? a. The bees’ clock mechanism is based on the Amfor gene, which causes a build up of AMFOR protein in the corpora pedunculata that activates the Gp9 and UU genes, whose products degrade AMFOR in a 24 hour cycle. b. The bees’ clock mechanism is based on the per gene in the optic lobes, whose activity has a 24 hour cycle regulated by the dbt and tim genes. c. The bees’ clock mechanism is based on alternating expression of two alleles for the for gene in the subesophagael ganglion: the R allele which is active at night and the s allele which is active during the day. d. The bees’ clock mechanism is based on levels of eclosion hormone released from…arrow_forwardWhat changes, if any, would you predict would occur in the pigmentation of Drosophila melanogaster with increased global warming? What type of genetic changes would you expect to see? Be as specific as you can.arrow_forwardWhat role do homeotic genes play in the development of fruit flies?arrow_forward
- To understand the genetic basis of locomotion in the diploid nematode Caenorhabditis elegans, recessive mutations were obtained, all making the worm “wiggle” ineffectually instead of moving with its usual smooth gliding motion. These mutations presumably affect the nervous or muscle systems. Twelve homozygous mutants were intercrossed, and the F1 hybrids were examined to see if they wiggled. The results were as follows, where a plus sign means that the F1 hybrid was wild type (gliding) and “w” means that the hybrid wiggled.a. Explain what this experiment was designed to test. b. Use this reasoning to assign genotypes to all 12 mutants. c. Explain why the phenotype of the F1 hybrids between mutants 1 and 2 differed from that of the hybrids between mutants 1 and 5arrow_forwardLid (Little imaginal discs) is a gene located on chromosome 2L and FRT40A refers to an FRT sequence inserted at the 40A (cytological location). Tubulin (Tub) refers to a promoter that can drive target gene expression in all the cell types. Gal80 is an inhibitor of Gal4 that blocks Gal4’s ability to activate transcription. Determine whether the following statements are true or false for flies with the following genotype: y w HsFLP; lid, FRT40A / P{Tub-Gal80}, FRT40A; P{Tub-Gal4} P{UAS-GFP} / + A. The fly will express Gal4, which will activate GFP in all the somatic cells without heat shock. B. After heat shock, FLP recombinase will be induced, which will induce mitotic recombination at FRT40A. However, lid mutant clone will be not induced because it is located closer to the centromere than FRT40A. C. After heat shock, FLP recombinase will be induced, which will induce mitotic recombination at FRT40A, resulting in the generation of homozygous lid mutant clones and twin spots that are…arrow_forwardThere is a gene in the fruit fly (Drosophila) called antennepedia. It controls the formation of which structures? What happens when it isarrow_forward
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