In Drosophila, loss
How will you determine where and when the mouse genes are expressed?
How will you create loss
How will you determine whether the mouse genes have redundant functions?
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GENETIC ANALYSIS: INTEGRATED - ACCESS
- 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_forwardSuppose 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_forwardIn the sea urchin, early development may occur even in the presence of actinomycin D, which inhibits RNA synthesis. However, if actinomycin D is present early in development but is removed a few hours later, all development stops. In fact, if actinomycin D is present only between the sixth and eleventh hours of development, events that normally occur at the fifteenth hour are arrested. What conclusions can be drawn concerning the role of gene transcription between hours 6 and 15?arrow_forward
- Explain why loss-of-function hedgehog and smoothened mutations yield the same phenotype in flies, but a loss-of- function patched mutation yields the opposite phenotype.arrow_forwardThe restriction digests of the cloned Drosophila gene can provide direct visible evidence of a mutation, as these samples come from a clone of the gene. In order to similarly detect a mutation in the copy of the endogenous gene within the Drosophila genome, a mechanism for specifically detecting restriction fragments from that gene among the complex set of fragments generated in a restriction digest of the entire Drosophila genome. Remember that the Drosophila melanogaster genome consists of ~123,000 kb. For a 10.2 kb Drosophila gene, what fraction of the genome does this gene constitute?arrow_forwardIn Drosophila, both fushi tarazu (ftz) and engrailed (eng) genes encode homeobox transcription factors and are capable of eliciting the expression of other genes. Both genes work at about the same time during development and in the same region to specify cell fate in body segments. To discover if ftz regulates the expression of engrailed;if engrailed regulates ftz; or if both are regulated by another gene, you perform a mutant analysis. In ftz embryos (ftz/ ftz) engrailed protein is absent; in engrailed embryos (eng/eng) ftz expression is normal. What does this tell you about the regulation of these two genes—does the engrailed gene regulate ftz, or does the ftz gene regulate engrailed?arrow_forward
- 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?arrow_forwardThere are similarities and differences during regulation of gene expression in both prokaryotes and eukaryotes. Promoters, transcription factors and RNA polymerase are essential elements in transcription but their properties and function may differ.a) Predict the outcome or consequences of mRNA transcription by RNA polymerase II in eukaryote without the presence of transcription factors (TF).arrow_forwardThe 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_forward
- 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 clonesarrow_forwardIn the module, you have learned about P-element mediated transgenesis in Drosophila and the concept of using transgenes to rescue mutant phenotypes. In the figure below, you will see a wild type fly with its natural eye colour and three mutants with their eye colours changed to vermillion, white and rosy, respectively. A schematic of P-element mediated transgenesis (as shown in the lectures) is also included in the figure. Please inspect the schematic carefully and choose which of the following statements is true: I. Injection of the white experimental transgene into the vermillion mutant embryo will not change the vermillion mutant phenotype II. Injection of the white experimental transgene in the rosy mutant embryo will change rosy eye colour to red (wild type) III. Injection of the white experimental transgene in the white mutant embryo will not change the white mutant phenotype IV. Injection of the white experimental transgene in the rosy mutant…arrow_forwardExplain one experimental strategy for determining the functional role of the mouse HoxD-3 gene.arrow_forward
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