How is localization of Gap gene expression achieved in the Drosophil.pdf

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How is localization of Gap gene expression achieved in the Drosophila embryo? (what is the mechanism?) Solution Drosophila has segmented body pattern, along the longitudinal axis. Gap gene is involved in development of the segmented embryos. Specialized transcription factors (mostly zinc fingers) are coded by gap genes which directly control the expression of set of genes (the pair rule genes) involved in segmentation. There is a cascade of transcription factors, which establish the segemented body pattern of drosophila. This starts during early development stage, wherein, the maternal gene products gets localized at the polar ends of the egg. Zygotic segmentation genes has the information for positioning of transcription factors. This regulatory hierarchy is led by gap gene. Along the longitudinal axis, expression of gap genes occurs in preblastoderm. Its discrete expression pattern defines specific and overlapping sets of segment primordia. The transcription factors encoded by gap genes form concentration gradients which are overlapping in nature, and the protein gradients thus formed provide cues to form repeated patterns. Thus the pattern of segments from blastoderm embryo are copied. Therefore, basically it is the maternal transcription factors which regulate the region-specific expression of zygotic transcription factors. The transcription factors encoded by gap genes regulate the expression of pair-rule genes and homeotic genes. Thus, it is the localized gene expression of gap gene which ultimately leads to the formation of the periodic pattern; mechanism known as the pair-rule gene expression. Studies of gap gene expression has demonstrated that from drosophila blastoderm, organisms with similar phenotype can be developed, despite of varitions occurring in the genotype, the phenomenon known as canalization..

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Discuss the relationship between gap rule gene mutations, pair-rule gene mutations, and segment polarity gene mutations in Drosophila melanogaster. Solution Segmentation in Drosophila is controlled by gene expression cascade that occur in a sequential order. The maternal effect genes form the anterior and posterior poles in the embryo. The spatial pattern of segments is controlled by transcription regulation of gap genes. The gap genes express the overlapping segments of anterior and posterior poles. The gap genes regulate next set of genes called pair rule genes. These genes are expressed in 7 stripes corresponding to each segment. Pair rule genes express the transcription factors regulating the segment polarity genes. These genes are expressed as 14 segmentally repeated stripes. The segment polarity genes express regulatory proteins and transcription factors that maintain interaction between cells and finally, homeotic genes control the character of each segment. The missing segments producing gaps are known by Krüppel mutants resulting in missing of T1- A5 segments. Hunchback mutants are missing head segments and knirps mutants lack posterior segments. Mutations in pair-rule genes- result in half the number of parasegments.The fushi- tarazu mutants lack odd-numbered parasegments and even-skipped mutants lack even-numbered parasegments. Mutations in segment polarity genes- result in duplication of the anterior or posterior half within each segment. The gooseberry and engrailed mutants have the posterior part of each segment replaced by the anterior part of the adjacent segment..

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we are interested in comparing the mean costs of electrical energy produced by nuclear and coal powered electrical power plants. Pilot studies have indicated the variation in the costs associated with the two types of plants are sn= $.224 and sc= $ .20.How many plants of each type should be sampled in order to estimate the difference mean costs with a 95% confidence interval to within $.01? Solution sn = 0.224 and sc = 0.20 Let sample sizes be n1 and n2 SE = 0.224/rtn1 + 0.20/rtn2 Margin of error = z alpha/2 (SE) For 95% z = 1.96 Hence 1.96(0.224/rtn1 + 0.20/rtn2) = 0.01 As there is one equation and two variables assume n1 = n2 Then we have 0.424/rtn1 = 0.01/1.96 = 0.0051 Or rt n1 = 8.313 n1 =69.111 or n >70 for each sample..

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What are the genetic differences between a processed pseudogene and a mutated pseudogene and how do their typical locations relative to their functional paralogous genes differ? Solution Pseudogenes are the genes that have lost some part or entire of its original ability, of their real gene relative and thus they have a altered or no gene expression activity. Most commonly, pseudogenes are formed as a result of accumulation of multiple mutations in a single gene, whose gene product is not essential for survival. Pseudogenes may not be fully functional and may perform regulatory functions. Thus a pseudogene has its association with a functional gene, and both of them share a common ancestor. Processed pseudogenes are the result of retrotransposition, i.e, a part of the mRNA or hnRNA transcript from a gene spontaneously reverses back (transcribed again) into DNA and gets inserted into chromosomal DNA.Thus they are processed once and then reverses back to result in a pseudogene. As a result, random copies of genes are produced in vitro. When such pseudogenes are inserted into the genome, they usually contain a poly-A tail,they are without introns (introns are spliced out) and also lack the upstream promoters. Thus after transposition event, they become non-functional pseudogenes. Such pseudogenes are randomly inserted into the genome. On the other hand, mutated pseudogenes arise after multiple mutation events which prevent a gene from normal transcription and make them less functional or non-functional; like gene duplication event caused by homologous recombination (repititive sequences on misaligned chromosomes become non-functional) or one allele of gene having deleterious mutation resulting in loss of function, non-sense mutations, unequal cross overs and sister chromatid exchanges etc. In such pseudogenes, the location of genes remain same as their functional paralogous genes..

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Assume you found a new worm species while working at Friday Harbor Marine Lab last summer. It has some cells at the vegetal pole that form right after cleavage, then move into the embryo. If you cut off the cells, the animals that developed were sterile. Using your knowledge gained from lecture, what can you tell me about what these cells might be doing? What experiments might you do to test your hypothesis? What are enhancers? How do they function to promote gene expression, especially since they may be some distance from the transcribed gene?? Do enhancers interact with promoters and basal transcription factors? Solution Lower invertebrates (including worms) develop via autonomous specification. In this type of specification, fate of cells is decided in early blastula stage. If in early stages, any cell is lost, then the organ/cell part/body part destined to be formed from that lost cell will not be formed. Such pattern of cell division and cleavage in early stages is known as mosaic division. In the above case, the new worm species isolated is a lower invertebrate. And the lost cells, are those which were responsible for formation of gonads. Now, because of autonomous specification, loss of cells, responsible for development of gonads leads to formation of sterile worms. To test the hypothesis, you can take a control animal (worm) without the \"cut off cells” and a test animal with cells cut out. Enhancers are short regions in DNA, which bind to transcription factors. Transcription factors in turn bind to promoter. So, with the help of enhancer binding of promoter, RNA polymerase and transcription factors is speeded up. Overall the process of transcription is enhanced. Enhancers can be located anywhere in the DNA. They may far away from the gene; or may be present in some other gene; or can be present in the exon itself. They can be reversed also. Enhancer of one gene can affect the transcription of another gene. There overall function is to recruit RNA polymerase to the promoter. This can be done by chromatin modeling in several ways (For detailed description of this part, please post separately). No, enhancers do not interact directly with promoters and BTFs. They interact with other transcription factors which in turn interact with a \'Transcription activator protein\'; which in turn, binds to promoter..

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