Genetic Analysis: An Integrated Approach (3rd Edition)
3rd Edition
ISBN: 9780134605173
Author: Mark F. Sanders, John L. Bowman
Publisher: PEARSON
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Chapter 18, Problem 16P
The pair
a. Given the hierarchy of gene action during Drosophila embryogenesis, what might you predict to be the result of these experiments?
b. Based on your prediction, can you predict the
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Discuss the morphological differences between the parasegments and segments of Drosophila. Discuss the evidence, providing specific examples, that suggests the parasegments of the embryo are the subdivisions for the organization of gene expression.
A mutation occurs in the Drosophila doublesex gene that prevents Tra from binding to the dsx RNA transcript. What would be the consequences of this mutation for Dsx protein expression in males? In females?
Researchers have exploited Minute mutations in orderto study the phenotypes associated with recessive lethal mutations (l−) that decrease the rate of cell divisionand thus make only very tiny homozygous mutant clones that are difficult to analyze. Many differentstrains of Drosophila carry dominant loss-of-functionMinute (M) mutations in a variety of genes encodingribosomal protein subunits. The M genes are haploinsufficient; flies with only one wild-type M+ gene copyhave a slower pace of cell division, and thus prolongeddevelopment and subtle morphological abnormalities.To circumvent the tiny clone problem, researchersgenerate GFP-marked homozygous l−/ l− clones thatare also M+/ M+, in flies that are l−/ l+ and M−/ M+.The loss of the Minute mutation only in cells withinthe clone gives the l−/ l− cells a growth advantageover their neighbors, enabling the mutant clone togrow large enough to study. Diagram chromosomesthat could be used to generate such clones
Chapter 18 Solutions
Genetic Analysis: An Integrated Approach (3rd Edition)
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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- The extracellular protein factor Decapentaplegic(Dpp) is critical for proper wing development in Drosoph-ila (Figure Q21–3A). It is normally expressed in a narrowstripe in the middle of the wing, along the anterior–pos-terior boundary. Flies that are defective for Dpp formstunted “wings” (Figure Q21–3B). If an additional copyof the gene is placed under control of a promoter that isactive in the anterior part of the wing, or in the posterior part of the wing, a large mass of wing tissue composed ofnormal-looking cells is produced at the site of Dpp expres-sion (Figure Q21–3C and D). Does Dpp stimulate cell divi-sion, cell growth, or both? How can you tell?arrow_forwardWhat would be the most likely effect of inhibiting the translation of hunchback mRNA throughout a Drosophila embryo?arrow_forwardWhich of the following applies to the female sex determination in Drosophila melanogaster fruit flies? (Mark all that apply) When the X:A ratio is 1.0, the SXL gene is transcribed during the early embryonic stages. The SXL protein inhibits the expression of female differentiaion genes The SXL protein promotes the correct splicing of the Tra pre-mRNA so that functional Tra protein is made. The presence of the SXL protein promotes the correct splicing of the SXL pre-mRNA during later embryonic stages to ensure to continuous production of SXL protein. The Tra protein will stimulate the expression of male-specific genes.arrow_forward
- 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 Prated with Ants-Wingless MW standards || | | mutanttype ODW-2 inhibits transcription of WL-1 O DW-2 polyubiquitinates WL-1 null null phenotype Model Ac O DW-2 phosphorylates WL-1 and prevents it from entering the nucleus O DW-2 cleaves WL-1 proteolytically Wings Mode of regulation DW-2 Wings Wingsarrow_forwardAurora AAurora A is a protein that acts as a kinase (transfers phosphates to molecules). Many types of cancer cells, including breast cancer cells, have higher than normal levels of this protein.Expressions of Aurora A genes in normal breast tissues (n = 10), normal tissues adjacent to tumors (n = 12) and breast tumors (n = 14).Scientists studying the production of Aurora A protein in normal frog cells observed that the amount of this protein in the cells changed throughout the cell cycle.Scientists tested chemicals that block Aurora 2 to see if they could be used as anti-cancer drugs. They found that some of the candidate drugs did slow the growth of cancer cells in cell culture in the lab. But when they tested these drugs in cancer patients to see if the drugs could slow the growth of solid tumors, they found that the benefit to patients was small when compared to the development of severe side effects such as anemia (low red blood cell count) and leukopenia (low white blood cell…arrow_forwardThe floral homeotic genes of Arabidopsis belong to the MADS-box gene family, while in Drosophila, homeotic genes belong to the homeobox gene family. In both Arabidopsis and Drosophila, members of the Polycomb gene family control expression of these divergent homeotic genes. How do Polycomb genes control expression of two very different sets of homeotic genes?arrow_forward
- In mammalian females, the majority of one X chromosome's genes are shut down in a process called X chromosome condensed and afterwards called a The inactive X is Because the shutdown is random in the embryonic cells formed by the time inactivation occurs, all mature mammalian females are a for the expression of X chromosome genes. The inactivation of one X in the female is thought to . The shutdown of the X chromosome is accomplished by the balance gene expression between males and females, a theory called gene. is the master gene for male sex determination in mammals and is found on the human Y-chromosome.arrow_forwardWhat would be the most likely result of injecting bicoid mRNA into the posterior end of a Drosophila embryo and inhibiting the translation of nanos mRNA?arrow_forwardExplain the Drosophila genetics tool for gene expression known as the UAS-GAL4 systemarrow_forward
- Suppose a researcher has three different Drosophila strains that have mutations in the bicoid gene called bicoid-A, bicoid-B, and bicoid-C; the wild type is designated bicoid +. To study these mutations, phenotypically normal female flies that are homozygous for the given bicoid mutation were obtained, and their oocytes were analyzed using a Northern blot to determine the size and/or amount of the bicoid mRNA and in situ hybridization to determine the bicoid mRNA location within the oocyte. A wild-type strain was also analyzed as a control. In both cases, the probe was complementary to the bicoid mRNA and the results are shown below. (Anterior is on the left; posterior is on the right.) Northern blot 1 2 - 3 4 In situ hybridization Wild type Lane 1. Wild type (bicoid*) Lane 2. bicoid-A Lane 3. bicoid-B Lane 4. bicoid-C bicoid-B bicoid-A bicoid-C Which mutation is likely to cause the embryo to develop two "anterior" ends? bicoid-B Obicoid-A bicoid-Carrow_forwardThe Wnt planar polarity signaling pathway nor-mally ensures that each wing cell in Drosophila has a sin-gle hair. Overexpression of the Frizzled gene from a heat-shock promoter (hs-Fz) causes multiple hairs to grow frommany cells (Figure Q15–5A). This phenotype is suppressedif hs-Fz is combined with a heterozygous deletion (DshΔ)of the Dishevelled gene (Figure Q15–5B). Do these resultsallow you to order the action of Frizzled and Dishevelledin the signaling pathway? If so, what is the order? Explainyour reasoningarrow_forwardIn a mutant mouse, it was discovered that a gene "W" is moved from its original chromosomal locus to the telomere region, where chromatin is packed into heterochromatin. Given what we have learned about eukaryotic gene regulation, can you speculate how the expression of gene W in the new mouse line would be changed as compare to the normal mouse? elevated in the mutant mouse than in the wild-type mouse. cannot be determined. O reduced in the mutant mouse than in the wild-type mouse. change in the mutant mouse as compared to the wild-type mouse.arrow_forward
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Embryology | Fertilization, Cleavage, Blastulation; Author: Ninja Nerd;https://www.youtube.com/watch?v=8-KF0rnhKTU;License: Standard YouTube License, CC-BY