You have been analyzing wing development mutants in Drosophila and have collected thegenetic/Western data below. Your epistasis analysis (not shown), suggested your current modelthat DW-2 is upstream of WL-1. What is a plausible mechanism for how DW-2 regulates WL-1?mutant namewingless-1doublewing-2Prated withAnts-WinglessMWstandards|| | |mutanttypeODW-2 inhibits transcription of WL-1O DW-2 polyubiquitinates WL-1nullnullphenotypeModel AcO DW-2 phosphorylates WL-1 and prevents it from entering the nucleusO DW-2 cleaves WL-1 proteolyticallyWingsMode of regulationDW-2WingsWings

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Description: You have been analyzing wing development mutants in Drosophila and have collected thegenetic/Western data below. Your epistasis analysis (not shown), suggested your current modelthat DW-2 is upstream of WL-1. What is a plausible mechanism for how DW

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You have been analyzing wing development mutants in Drosophila and have collected the genetic/Western data below. Your epistasis analysis (not shown), suggested your current model that DW-2 is upstream of WL-1. What is a plausible mechanism for how DW-2 regulates WL-1? mutant name wingless-1 doublewing-2 Probed with Ant-Wingless MW standards - mutanttype O DW-2 polyubiquitinates WL-1 null ODW-2 inhibits transcription of WL-1 null - phenotype O DW-2 phosphorylates WL-1 and prevents it from entering the nucleus DW-2 cleaves WL-1 proteolytically Model Ac Mode of regulation Wings DW-2 T Wings

You have been analyzing wing development mutants in Drosophila and have collected the genetic/Western data below. Your epistasis analysis (not shown), suggested your current model that DW-2 is upstream of WL-1. What is a plausible mechanism for how DW-2 regulates WL-1? mutant name wingless-1 doublewing-2 Probed with Ant-Wingless MW standards - mutanttype O DW-2 polyubiquitinates WL-1 null ODW-2 inhibits transcription of WL-1 null - phenotype O DW-2 phosphorylates WL-1 and prevents it from entering the nucleus DW-2 cleaves WL-1 proteolytically Model Ac Mode of regulation Wings DW-2 T Wings

Biology (MindTap Course List)

11th Edition

ISBN:9781337392938

Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. Berg

Publisher:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. Berg

Chapter17: Developmental Genetics

Section: Chapter Questions

Problem 3TYU: The anteriorposterior axis of a Drosophila embryo is first established by certain (a) homeotic genes...

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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…
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 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…
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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. Using the number of flies expected for the Parental and the Recombinant genotypes and the formula that defines recombination frequency, what is the recombination…
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