How can I find the DNA sequence for Vitreoscilla Hemglobin gene? Protocol for how to do a PCR reaction?
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gene:\nhttp://www.ncbi.nlm.nih.gov/entrez/viewer.fcgi?db=nucleotide&val=46561731\nhttp://www.ncbi.nlm.nih.gov/entrez/viewer.fcgi?db=nucleotide&val=10444342\nhttp://www.ncbi.nlm.nih.gov/entrez/viewer.fcgi?db=nucleotide&val=311024\nhttp://www.ncbi.nlm.nih.gov/entrez/viewer.fcgi?db=nucleotide&val=48496\nhttp://www.ncbi.nlm.nih.gov/entrez/viewer.fcgi?db=nucleotide&val=155319\nhttp://www.ncbi.nlm.nih.gov/entrez/viewer.fcgi?db=nucleotide&val=155317\n\nPCR:\nhttp://www.protocol-online.org/prot/Molecular_Biology/PCR/Standard_PCR/index.html
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K.C.'s answer works if you have a selective media that will work. Also (or in addition to that) you may want to plate out colonies and then PCR out a unique sequence or two.\n\nThis wouldn't be definitive but if you design your primers right it may work, but also may be very time consuming.\n\nAlso you may want to try restriction digests to remap your gene. This would work well, if you have a good map.\n\nGood luck, what a bummer
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Here's one from Biodidac (great site for illustrations):\nhttp://biodidac.bio.uottawa.ca/Thumbnails/showimage.cfm?File_name=Aves065b&File_type=gif
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http://www.api-pt.com/pdfs/2005Amicro.pdf\nhttp://www.life.umd.edu/classroom/bsci424/PathogenDescriptions/Enterobacteriaceae.htm\nhttp://www.indstate.edu/thcme/micro/pdf/chapter13.pdf\nhttp://www.austincc.edu/microbugz/images/03pvulgaris.jpg
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human don't have plasmid. but if you're asking how to test a bacterium for human growth hormone (hgh) recombination into its plasmid, you can perform is PCR test for the HGH mRNA. that should be the most direct proof.\nor you can do a DNA sequencing, but that takes longer.
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You can find practical information from Qiagen website if you use their kits which are quite popular these days.  \n\nIt depends on which methods you use for isolation.  The alkali lysis (see link below) was commonly used to rupture bacteria.  Plasmid DNA (smaller in size and more soluble) is separated from baterial genomic DNA (larger and less soluble).  The plasmid DNA is then separate from proteins by phenol extraction where the proteins are removed to the phenol phase while plasmid DNA stays in the aquous phase.  The plasmid DNA is then precipated with ethanol to complete the purification.\n\nQiagen's kit and others overcome the use of phenol (which is toxic and smelly) by using a solid phase that bind DNA to seperate the contaminating proteins.  The DNA is later eluted from the solid phase to complete the purification.\n\nHere are additional links:\nhttp://biotech.biology.arizona.edu/labs/DNA_isolation_plasmid.html\nThe concept of alkali lysis method of plasmid DNA isolation is shown here:\nhttp://a32.lehman.cuny.edu/molbio_course/plasmid_protocol.htm
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There have been hundreds of thousands of such experiments done. Moreover you can easily do them yourself using some nutriemt agar and an incubator with a few hundred dollars to pay someone to do the genetic workup.\n\ntalk.origins is your one stop shop for these types of quetsions. The site is highly reputable, well refernces, cites peer reveiewed science journals and most importantly lists literally hundreds of experimenst that show exactly what you describe.\n\nI've attached a couple link to pages that list a dozenor so such mutation studies, but the site has many hundreds of others.
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http://faculty.clintoncc.suny.edu/faculty/Michael.Gregory/files/Bio%20101/Bio%20101%20Lectures/DNA/dna.htm
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well, its a little more complicated than the above person mentions.\n\nAside from finding the right person to do so the process is listed below:\n1)  Restriction Enzymes break apart several parts of your DNA\n2) All the DNA sequences are put into a microfilament inside a gel.\n3) The gel is attached to an anode and a cathode at one end.\n4) the heavier and more electronexgative pieces move to the anode.\n5) this gives you a unique fingerprinting of your dna. The end result is a bunch of lines on gel which is unique to yourself.\n\nIf you want to get the sequences of your DNA figured out the project is much more complicated and would take many many years. However, the human genome project did it. And since every person has a 1% difference in their alleles (that is right ONLY ONE PERCENT!) it is very similar. Moreover out of 1000 nucleotide bases in a sequence of a polypeptide or DNA, 999 are the same as any other given person. So essentially the human genome project maps your genome very very closely.
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Here's a video of a neutrophil chasing a bacterium.  It's video number 4.\n\nhttp://www.hopkinsmedicine.org/cellbio/devreotes/videos.htm\n\nIf you can't save it or download it, email me, and I'll see what I can do.
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there are databases with which you can do homology searches. best to know what genes you are looking for though as genome wide homology may be difficult to search for\ntry the ensemble database:\nhttp://www.ensembl.org/index.html\notherwise a broad literature search might be better suited. odds are someone else has already done a similar comparison.\n\nfound this old review\n\nhttp://www.sciencedirect.com/science?_ob=MImg&_imagekey=B6VS0-45765F9-2J-1&_cdi=6248&_user=911953&_orig=search&_coverDate=06%2F30%2F1997&_qd=1&_sk=999929996&view=c&wchp=dGLbVtz-zSkWA&md5=03f894feac6d92afb553850237f34bcf&ie=/sdarticle.pdf\n\nmight be of some help. if the link doesnt work, can be found using a Pubmed search.
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This article answers your question.\n\nhttp://www.hhmi.org/cgi-bin/askascientist/highlight.pl?kw=&file=answers%2Fgeneral%2Fans_043.html
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Complementation Testing\nOccasionally, multiple mutations of a single wild type phenotype are observed. The appropriate genetic question to ask is whether any of the mutations are in a single gene, or whether each mutations represents one of the several genes necessary for a phenotype to be expressed. The simplest test to distinguish between the two possibilities is the complementation test. The test is simple to perform --- two mutants are crossed, and the F1 is analyzed. If th e F1 expresses the wild type phenotype, we conclude each mutation is in one of two possible genes necessary for the wild type phenotype. When it is shown that shown genetically that two (or more) genes control a phenotype, the genes are said to form a complementation group. Alternatively, if the F1 does not express the wild type phenotype, but rather a mutant phenotype, we conclude that both mutations occur in the same gene.\nThese two results can be explained by considering the importance of genes to phenotypic function. If two separate genes are involved, each mutant will have a lesion in one gene while maintaining a wild type copy of the second gene. When the F1 is produc ed, it will expresses the mutant allele of gene A and the wild type allele of gene B (each contributed by one of the mutant parents). The F1 will also express the wild type allele for gene A and the mutant allele for gene B (contributed by the other muta nt parent). Because the F1 is expressing both of the necessary wild type alleles, the wild type phenotype is observed. \n\nConversely, if the mutations are in the same gene, each homolog will express a mutant version of the gene in the F1. Without a normal functioning gene product in the individual, a mutant phenotype occurs. \n\nEye color in Drosphila is a good model to demonstrate the complementation test. A wide array of spontaneous mutations have been studied. These experiments demonstrated that five genes (white, ruby, vermillion, garnet and carnation) controlling ey e color reside within 60 cM of each other on the X chromosome. The dominant wild type allele for each gene produces the deep red eyes. The mutant alleles produce a different color. If mutants from any of these five genes are crossed, the F1 would expre ss deep red eye color (wild type phenotype). \n\nFive different alleles (buff, coral, apricot, white and cherry) are also known to exist for the white gene, each representing a mutation at a different position in the gene. If mutant flies for any of the five white alleles are crossed, the F1 offspring would have a mutant eye color. Therefore, two genes involved in the expression of a phenotype complement each other. But complementation bet ween two alleles of the same gene does not occur
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Wow, this answer is gonna be a long one...   There are numerous studies of the charting of the positions of genes on a DNA molecule or chromosome and the distance between genes of Escherisia Coli. It basically comes down to good old conjugation and transduction.  Having sequenced the E. coli K-12 genome, scienctists are creating a knockout of every gene, a clone of every ORF, and an extensive gene expression data set under a variety of physiological conditions. These clones and data sets are made available freely to the community.  Scienctists have also sequenced six additional enterobacterial strains which are pathogens closely related to E. coli K-12, and many plasmids. The comparative genomics aim of this effort is to characterize the gene pool of horizontally transferred elements and virulence determinants known as the pathosphere.\n\nI put a reference on it so be sure to check it out yourself. \n\nA mutator gene, mutD5, whose phenotype is conditional, has been identified in Escherichia coli. By P1 transduction it has been shown to lie at about 5.7 min on the chromosome, being co-transduced with proA and argF. In rich medium, streptomycin- and nalidixic acid-resistant mutation frequencies are 50 to 100 times higher than those in minimal medium. In minimal medium, the mutD5-induced mutation frequencies are still 50 to 100 times above co-isogenic wild-type (mut+) levels. Similar results were obtained with all markers tested. Mutant frequencies can be raised by thymidine in the medium at concentrations as low as 0.04 μM, or by the endogenous generation of thymidine from thymine plus a deoxyribosyl donor. Deoxyadenosine, various ribonucleosides, thymine, and 2-deoxyribose do not stimulate mutation. None of these effects are related to growth rate, since growth rate and mutation rate can be decoupled completely.\n\nThe gntP gene, located between the fim and uxu loci in Escherichia coli K-12, has been cloned and characterized. Nucleotide sequencing of a region encompassing the gntP gene revealed an open reading frame of 447 codons with significant homology to the Bacillus subtilis gluconate permease. Northern (RNA) blotting indicated that the gntP gene was monocistronic and was transcribed as an mRNA with an apparent molecular size of 1.54 kb. The transcriptional start point was determined by primer extension analysis. The gntP gene was found to be under catabolite repression and was not induced by gluconate. Also, expression seemed to be stringently controlled. Several observations indicated that the GntP protein is an inner membrane protein; it contains characteristic membrane-spanning regions and was isolated predominantly from the inner-membrane fraction of fractionated host cells. A topology analysis predicted a protein with 14 membrane-spanning segments. The inability of a mutant strain to grow on gluconate minimal medium could be relieved by introduction of a plasmid encoding the gntP gene. Finally, the kinetics of GntP-mediated gluconate uptake were investigated, indicating an apparent Km for gluconate of 25 microM.
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Without using molecular methods in a bio lab, you would have to rely on phenotypic testing - ie. the expression of the gene.\n\nin microbiology:\ntest the presence of the gene product and compare them to what is expected\n\nMolecular methods would be more reliable and accurate in detecting a mutation. If you know the sequence of the normal gene you can amplify it by PCR and visualise it by agarose gel elecrophoresis - in which case, the mutated gene would give an altered product if any at all. Or you could use real-time-PCR with a labelled probe complementary to the gene sequence. During the PCR cycle, the probe will bind loosely to the mutated gene but strongly to the natural gene. So during the high temperature stages, the mix with the mutated gene will lose the signal from the probe before the mix with the natural gene.\n\nThere are other methods, but I'm not quite sure what you're dealing with...
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40.\nhttp://www.research.att.com/~njas/sequences/b000602.txt
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For ssDNA, a dimer (a two base sequence) is the answer...this would be the smallest DNA sequence possible from a PCR reaction. The organic extraction part is not the limiting factor until the number of bases in the sequence becomes very large. Extraction regime should be desalting, followed by Ethanol precipitation, and then either Gel electrophoresis or Liquid Chromatography.