How does DNA cleaved by a restriction enzyme that produces blunt ends get incorporated in a vector?
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Ligation (joining of DNA ends) happens when phosphate group at the end of DNA strand is joined by ligase enzyme to hydroxy group on another end of the same or other DNA strand, by phosphodiester bond.\n\nSticky ends help this process by fixing DNA ends in place and in correct configuration, but are not necessary. In fact, when there is no phosphate group at both strands of DNA, even sticky ends cannot be joined permanently.\n\nBlunt-end ligation is less efficient than sticky end ligation, but still goes on. DNA ends simply collide in the solution and get joined by ligase that forms phosphodiester bonds. That's why "water exclusion" polymers that increase effective concentration of DNA ends (and thus frequency of collisions between them) are used to help blunt ligation.
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To be honest with you i don't quite understand your question, it is often that restriction enzymes are used to create genomic library for which you look into the overlapping sequence and then be able to determine which fragment is next to which. If that is what you have done, you should have used the fragmented DNA sequence to determine the orginall DNA sequence that you digested it with the enzyme, to confirm the sequence you ahve found is correct, you can use PCR, you got the hypothetical sequence, thus by making primers from the start and the end of the sequence (doesn't matter which way round) you should have amplify the same sequence which you started with, you got the amplified DNA (same as the orginal DNA because you used primers that complemented the very start and the very end of the sequence which you found out after the restruction enzyme mapping) run it on gel electrophoresis with a DNA size marker that will include the size of DNA you have sequenced, if the amplified DNA size is the same as the sequence you have found from restriction mapping, that means your primers did complemented from the start and the end, that means the sequence you thought is actually the correct sequnece.\nI am sorry, i don't think i am making this clear enough, but anyway, that's the best i can think of.
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agarose forms a matrix the DNA or RNA fragments have to pass through when the current is applied.  the longer the nucleic acid fragments the harder it is to pass though the gel so the fragments seperate by size.
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Easiest method involves the pooling of the source (best source is wheat germ due to ease) next grind the germ then add some detergent solution- this will lyse the cell wall (the grinding breaks down the plant wall) next is to add some form of alcohol (isopropanol or high % ethanol is best (91% Isopropanol is available at drug stores)) Add the alcohol and the DNA will float to the top with the solution- spool on a glass rod.
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as someone already mentioned, there is one process called alternative splicing. a single gene, a single DNA sequence, can be transformed into several proteins. RNAm is formed in the process of transcription, but it cannot leave the cell nucleus without being processed by several enzymes, which remove certain bits of RNA (introns) and leave others (exons). These exons that remain are to exit the nucleus and be translated into proteins.\nThe thing is, these enzymes that "process" DNA can do it in several different ways. For the same sequence, there can be an enzyme that recognises certain bits of DNA as introns and another enzyme that recognises different bits of the SAME sequence as introns. Thus being, the two molecules of processed DNA to exit the nucleus will be different and, consequently, yield different proteins.\nthis multiplicity of codification is an evolutionary advantadge of eukarya, since it allows a lot more information to be contained in the same sequence of DNA, thus occupying less space within the cell; this might have been one of the key events that allowed the development of more complex cells, from prokaryotes to eukaryotes.\n\njust to complement the given information: these several proteins that can be coded in the same gene, usually belong to the same protein family or else have a similar function, e.g. in the same process)
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C)gel electrophoresis. \nIn gene splicing ther's cutting of DNA At specific regions. DNA Cloning involves the magnification or production of identical copies of DNA
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The DNA has to be wrapped up very tightly to fit all of it into the nucleus.  This is a complex process involving chromatin (DNA and associated proteins).  For example, histones are groups of protein molecules that the DNA strand wraps around to help compact its shape.  For a gene to be expressed (transcribed) a section of the the DNA has to become 'unwrapped' and 'exposed' to the transcription machinery.  Methylation and Acetylation of histones play significant roles in this process.\n\nAny good, recent, biochemistry textbook will explain these issues.  If you go to pubmed, they have an online book section.  Click on any of the biochemistry books and enter into the search field "DNA, chromatin", or "DNA, histones", or "gene expression, chromatin" and you will find lots of info. and probably pictures too.\n\nBest wishes and good luck.
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Certain ligases which cut DNA at known sequence points are used to excise the gene of interest from a particular strand of DNA. This excised (or cut-out) gene is then incorporated into the DNA of a bacterium.  The bacterium, in the process of replication, will add the gene into its own plasmid and produce the protein. Alternatively, I believe viral vectors are also used to introduce genes into a particular host "factory" cell that then produces the required protein. Not totally informed on that part, though.
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Nearly Complete Genomic Libraries of Higher Organisms Can Be Prepared by λ Cloning\n\nWith the availability of λ phage cloning vectors, preparation of genomic libraries for higher organisms, including humans, is feasible. A genomic library is a set of λ (or plasmid) clones that collectively contain every DNA sequence in the genome of a particular organism. Figure 7-12 summarizes the general procedure for constructing a λ genomic library. The λ DNA first is treated with a restriction enzyme to produce fragments called λ vector arms, which have sticky ends and together contain all the genes necessary for lytic growth. This step frees the nonessential region in the middle of the λ genome; this region is separated from the λ arms and discarded. Genomic DNA then is extracted from a cell type that contains all the genetic information of the organism under study. Sperm cells or cells of an early embryo often are used as sources of mammalian DNA. The extracted DNA then is cleaved by a restriction enzyme to produce ≈20-kb fragments with sticky ends complementary to the sticky ends on the λ vector arms being used.\n\nThe λ arms and the collection of genomic DNA fragments are mixed in about equal amounts. The complementary sticky ends on the fragments and λ arms hybridize and then are joined covalently by DNA ligase. Each of the resulting recombinant DNA molecules contains a foreign DNA fragment located between the two arms of the λ vector DNA. The ligated recombinant DNAs then are packaged into λ virions in vitro as described above. Only DNA molecules of the correct size can be packaged to produce fully infectious recombinant λ virions.\n\nFinally, the recombinant λ virions are plated on a lawn of E. coli cells to generate a large number of recombinant λ plaques. Since each plaque arises from a single recombinant virion, all the progeny λ phages that develop are genetically identical and constitute a clone carrying a particular genomic DNA insert. The different plaques correspond to distinct phage clones, each carrying a different DNA insert, and collectively they constitute a λ genomic library. Multiple λ vectors have been constructed containing different restriction sites, so that restriction fragments generated by a variety of restriction enzymes can be cloned in λ vectors.\n\nIn constructing a library of genomic DNA, the DNA commonly is cleaved with a restriction enzyme that recognizes a 4-bp restriction site (e.g., Sau3A, as shown in Figure 7-12). A specific 4-bp sequence will occur on average once every 44 = 256 base pairs. Complete digestion of the human haploid genome, which contains ≈3 × 109 base pairs, would yield somewhat more than 107 nonoverlapping different fragments. However, to increase the probability that all regions of the genome are successfully cloned and represented in the λ genomic library, the genomic DNA usually is only partially digested to yield overlapping restriction fragments of ≈20 kb (Figure 7-13). In a large λ library constructed from such overlapping restriction fragments, a specific sequence of genomic DNA (e.g., the gene encoding β-globin) often is contained in or extends over several “overlapping” clones. By a technique called chromosome walking, described in the next chapter, overlapping genomic clones can be ordered.\n\nThe size of a genomic library for a given organism depends on the amount of DNA in that organism's haploid genome. If the human genome of about 3 × 109 base pairs is cleaved into 20-kb fragments for insertion into a λ vector, then roughly 1.5 × 105 different recombinant λ phage virions would be required to constitute a complete library. Because the restriction fragments of human DNA are incorporated into phages randomly, about 106 recombinant phages are necessary to assure that each region of human DNA has a 90 – 95 percent chance of being included.\n\nEach plaque produced by a recombinant bacteriophage λ contains large numbers of recombi
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The function of DnaB helicase is to couple ATP hydrolysis with the unwinding of the DNA double helix at the replication fork, creating two antiparallel strands of ssDNA. DNA polymerase and DnaB helicase are complexed via the tao subunit of DNA polymerase which serves to stabilize the hexameric structure of the helicase. This now immobilized protein complex allows for DNA to pass through it. There are other helicases involved with replication but this is the only one found to be essential (knockout mutations are fatal) and the only one that can stimulate primase. If you need more info, please formulate a somewhat more coherent and specific question.
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I'm not sure how you intend to destroy one strand and not the other while the are rapped around each other, but the two strands can be easily separated using naturally occuring rep enzymes.\n\nIf the appropriate enzymes and nucleotides are present, the missing strand will be replaced with a new, identical strand.  In the absence of the appropriate components, the single-stranded DNA will continue to exist for a while, but will slowly degrade as it is less stable than the double-stranded form.  Also, the DNA will likely fold up on itself as there are almost certainly numerous regions which are complimentary to each other.\n\n__________\nDo you mean identical strands or complementary strands?  If they're identical, they're not two halves of a double-stranded molecule, so what happens to one will have no relation to what happens to the other.  If you mean complementary, you could stand to be a bit clearer on what you expect the laser to do to the strand that is irradiated.
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Experimentally you can do primer extension, which will show the +1 site, from there you can identify the RBS and -10 and -35 sequences.\n\nYou can also do mobility shift assays with RNA polymerase to show that a certain piece of DNA contains the promoter, then do DNAse protection assays to show where the polymerase is actually binding.
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Here's a great animation that helps explain it:  http://www.johnkyrk.com/DNAreplication.html
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The lipid parts have a cation binding site to bind with the negative charges of the DNA.  The final complex is called a liposome.  \n\nIt then carries the DNA into cell and into the nucleus.  You still have to use a bunch of reagents in the process.  This is just from my research background.  The specifics are long.  You can email me for them
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Control of Gene Expression: The lac operon consists of one regulatory gene (the i gene) and three structural genes (z, y, and a). The i gene codes for the repressor of the lac operon. The z gene codes for b-galactosidase (b-gal), which is primarily responsible for the hydrolysis of the disaccharide, lactose into its monomeric units, galactose and glucose. The y gene codes for permease, which increases permeability of the cell to b-galactosides. The a gene encodes a transacetylase. During normal growth on a glucose-based medium, the lac repressor is bound to the operator region of the lac operon, preventing transcription. However, in the presence of an inducer of the lac operon, the repressor protein binds the inducer and is rendered incapable of interacting with the operator region of the operon. RNA polymerase is thus able to bind at the promoter region, and transcription of the operon ensues. The lac operon is repressed, even in the presence of lactose, if glucose is also present. This repression is maintained until the glucose supply is exhausted. The repression of the lac operon under these conditions is termed catabolite repression and is a result of the low levels of cAMP that result from an adequate glucose supply. The repression of the lac operon is relieved in the presence of glucose if excess cAMP is added. As the level of glucose in the medium falls, the level of cAMP increases. Simultaneously there is an increase in inducer binding to the lac repressor. The net result is an increase in transcription from the operon. The ability of cAMP to activate expression from the lac operon results from an interaction of cAMP with a protein termed CRP (for cAMP receptor protein). The protein is also called CAP (for catabolite activator protein). The cAMP-CRP complex binds to a region of the lac operon just upstream of the region bound by RNA polymerase and that somewhat overlaps that of the repressor binding site of the operator region. The binding of the cAMP-CRP complex to the lac operon stimulates RNA polymerase activity 20-to-50-fold.
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dna ligation is the joining of linear dna fragments together using covalent bonds. more specifically, dna ligation involves creating a phosphodiester bond between the 3' hydroxyl end of one nucleotide and the 5' phosphate of another.
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DNA's most amazing characteristic is that it will bind to itself in a very specific manner, forming a helix. Figure 1 illustrates two DNA molecules binding. The variable groups of the polymer interact with eachother to mediate this binding. Adenine interacts with Thymine and Cytosine interacts with Guanine. No other bonds are energetically favorable, for example Cytosine cannot bond with itself. The bonds that these variable groups form are called are called base pairs and are illustrated in Figure 2. Thus, by knowing the sequence of bases of one strand of DNA we immediately know the sequence of the DNA strand which will bind to it, this strand is called the reverse complement or just the complementary strand. In Figure 1 the right hand strand is the reverse complement of the left hand side and vice-versa.\n\n\nFrom G. S. Sent, Molecular Biology of Bacterial Viruses. Copyright 1963 by W. H. Freeman and Company\n\nFigure Two:Illustration of DNA basepairing. Thermodynamically the interactions Adenine, Guanine, Cytosine, and Thymine are driven by both the hydrophobic stacking interactions hydrogen bond interactions. The specificity of the base pairing is driven mainly by hydrogen bonding and stearic interference.\nThis relationship to the complementary strand has many implications that make life as we know it possible, and allows all sorts of molecular biology techniques. In one of the greatest understatements of all time Watson and Crick said it best in their original paper on the structure of DNA:\n\n"It has not escaped our notice that the specific pairing we have postulated immediately suggests a possible copying mechanism for the genetic material.".