Pearson eText Genetic Analysis: An Integrated Approach -- Instant Access (Pearson+)
3rd Edition
ISBN: 9780135564172
Author: Mark Sanders, John Bowman
Publisher: PEARSON+
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Chapter 18, Problem 17P
In contrast to Drosophila, some insects (e.g., centipedes) have legs on almost every segment posterior to the head. Based on your knowledge of Drosophila, propose a genetic explanation for this
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Chapter 18 Solutions
Pearson eText Genetic Analysis: An Integrated Approach -- Instant Access (Pearson+)
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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- What is the signaling pathway that mediates the organizing activity of the A/P organizer in the Drosophila wing disc? Describe two experiments that suggest this pathway functions to organize pattern and promote growth along the anterior/posterior axis of wing imaginal discs.arrow_forwardThe homeotic mutation Antennapedia causes mutant Drosophila to have legs in place of antennae and is a dominant gain-of-function mutation. List all the properties of such mutations. How does the Antennapedia gene change antennae into legs?arrow_forwardBriefly describe how the bicoid and nanos genes help to determine the anterior–posterior axis of the fruit fly.arrow_forward
- provide one example from the Drosophila AP toolkit for "Toolkit genes can be classified according to the phenotypes caused by their mutation. Similar mutant phenotypes often reflect genes that function in a single developmental pathway. Distinct pathways exist for the generation of body axes, for example, and for the formation and identity of fields."arrow_forwardPax6 is a gene that is involved in mammalian eye development. Mutations in this gene result in a lack of iris in humans, and mice that are homozygous for the mutation lack eyes completely. Mutations in the eyeless gene in the fruitfly Drosophila result in reduced eyes. Expressing Pax6 from mammals in other tissues of Drosophila during its embryonic development can result in an ectopic eye (e.g. an eye expressed in a different tissue type). An example of an ectopic eye on the antenna of Drosophila is shown below. Part A: The eye of humans, mouse, and fly are an example of the following (check all that apply) A. Structural analog B. Structural homolog C. Developmental homolog D. Genetic homolog Part B: Darwin's hypothesis of natural selection has two components: Species are not static but change through time Species are related, and not independent. In what ways does the human, mouse, and fly eye support or contradict one or both of the two components of…arrow_forwardFollowing a mutagenesis experiment to identify novel genes affecting the circadian clock in Drosophila melanogaster you discover several mutants. You start considering two of those mutants that you call c and d. The two homozygous c/c and d/d are arrhythmic (arrhythmic is the definition of their phenotype), whereas the two heterozygous C/c and D/d are rhythmic (rhythmic is the definition of their phenotype) with a 24h period. You make two true-breeding stocks: stock 3 homozygous for c and stock 4 homozygous for d. You cross them in both directions and in both cases you observe complementation with no difference between males and females. Then you take the progeny of one cross, for instance the F1 of Females 3 x Males 4, and you perform a Testcross. Out of 1000 flies resulting from the lestcross only 125 are rhythmic. Using the number of flies expected for the Parental and the Recombinant genotypes and the formula that defines recombination frequency, what is the distance between loci C…arrow_forward
- Following a mutagenesis experiment to identify novel genes affecting the circadian clock in Drosophila melanogaster you discover several mutants. You start considering two of those mutants that you call c and d. The two homozygous c/c and d/d are arrhythmic (arrhythmic is the definition of their phenotype), whereas the two heterozygous C/c and D/d are rhythmic (rhythmic is the definition of their phenotype) with a 24h period. You make two true-breeding stocks: stock 3 homozygous for c and stock 4 homozygous for d. You cross them in both directions and in both cases you observe complementation with no difference between males and females. Then you take the progeny of one cross, for instance the F1 of Females 3 x Males 4, and you perform a Testcross. Out of 1000 flies resulting from the Testcross only 125 are rhythmic. Considering that out of 1000 flies resulting from the Testcross only 125 are rhythmic how many flies in total are approximatively expected for the Parental genotypes?…arrow_forwardFollowing a mutagenesis experiment to identify novel genes affecting the circadian clock in Drosophila melanogaster you discover several mutants. You start considering two of those mutants that you call c and d. The two homozygous c/c and d/d are arrhythmic (arrhythmic is the definition of their phenotype), whereas the two heterozygous C/c and D/d are rhythmic (rhythmic is the definition of their phenotype) with a 24h period. You make two true-breeding stocks: stock 3 homozygous for c and stock 4 homozygous for d. You cross them in both directions and in both cases you observe complementation with no difference between males and females. Then you take the progeny of one cross, for instance the F1 of Females 3 x Males 4, and you perform a Testcross. Out of 1000 flies resulting from the Testcross only 125 are rhythmic. What is the genotype of Males 4? Select only one answer. 01. C/C D/D O 2. C/C D/d O3. C/C d/d O 4. C/C D/D O 5. c/c D/d O 6. C/cd/d 07. c/c D/D O 8. C/C D/d O 9. c/cd/d Karrow_forwardFollowing a mutagenesis experiment to identify novel genes affecting the circadian clock in Drosophila melanogaster you discover several mutants. You start considering two of those mutants that you call C and d. The two homozygous C/C and d/d are arrhythmic (arrhythmic is the definition of their phenotype), whereas the two heterozygous C/C and D/d are rhythmic (rhythmic is the definition of their phenotype) with a 24h period. You make two true-breeding stocks: stock 3 homozygous for C and stock 4 homozygous for d. You cross them in both directions and in both cases you observe complementation with no difference between males and females. Then you take the progeny of one cross, for instance the F1 of Females 3 x Males 4, and you perform a Testcross. Out of 1000 flies resulting from the Testcross only 125 are rhythmic. Using the number of flies expected for the Parental and the Recombinant genotypes and the formula that defines recombination frequency, what is the distance between loci C…arrow_forward
- Following a mutagenesis experiment to identify novel genes affecting the circadian clock in Drosophila melanogaster you discover several mutants. You start considering two of those mutants that you call c and d. The two homozygous c/c and d/d are arrhythmic (arrhythmic is the definition of their phenotype), whereas the two heterozygous C/c and D/d are rhythmic (rhythmic is the definition of their phenotype) with a 24h period. You make two true-breeding stocks: stock 3 homozygous for c and stock 4 homozygous for d. You cross them in both directions and in both cases you observe complementation with no difference between males and females. Then you take the progeny of one cross, for instance the F1 of Females 3 x Males 4, and you perform a Testcross. Out of 1000 flies resulting from the Testcross only 125 are rhythmic. What is the genotype of the F1 (progeny of Females 3 x Males 4). Select only one answer. 1. C/c D/d 2. C/C d/d 3. C/C D/D 4. c/c D/d…arrow_forwardFollowing a mutagenesis experiment to identify novel genes affecting the circadian clock in Drosophila melanogaster you discover several mutants. You start considering two of those mutants that you call c and d. The two homozygous c/c and d/d are arrhythmic (arrhythmic is the definition of their phenotype), whereas the two heterozygous C/c and D/d are rhythmic (rhythmic is the definition of their phenotype) with a 24h period. You make two true-breeding stocks: stock 3 homozygous for c and stock 4 homozygous for d. You cross them in both directions and in both cases you observe complementation with no difference between males and females. Then you take the progeny of one cross, for instance the F1 of Females 3 x Males 4, and you perform a Testcross. Out of 1000 flies resulting from the Testcross only 125 are rhythmic. Which is the genotype of the rhythmic flies resulting from the testcross? Select only one answer 1. C/C d/d 2. C/c D/d 3. C/c d/d 4. c/c…arrow_forwardFollowing a mutagenesis experiment to identify novel genes affecting the circadian clock in Drosophila melanogaster you discover several mutants. You start considering two of those mutants that you call c and d. The two homozygous c/c and d/d are arrhythmic (arrhythmic is the definition of their phenotype), whereas the two heterozygous C/c and D/d are rhythmic (rhythmic is the definition of their phenotype) with a 24h period. You make two true-breeding stocks: stock 3 homozygous for c and stock 4 homozygous for d. You cross them in both directions and in both cases you observe complementation with no difference between males and females. Then you take the progeny of one cross, for instance the F1 of Females 3 x Males 4, and you perform a Testcross. Out of 1000 flies resulting from the Testcross only 125 are rhythmic. Comparing the number of flies expected for the Parental and the Recombinant genotypes, what can be concluded about the C and D loci? Select only one answer. 1. The…arrow_forward
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