in invitro phage packging, y must we need 2 mutated lysogen-that cant develop head and tail?
354760
a phage can package its dna in itself.\nbut the genes it packages, also constitute the lytic cycle genes which destroy the cell.\nthe phage head recognises viral dna by the cohesive site genes and so engulfs the dna flanked by this site.\nwhen we use seperate mutated bacteria to produce different parts seperately, we have the benefit of replacing the entire viral dna by our gene of intrest and the flanking it with cos sites so that the phage head takes it up as its own dna.\nthis will eliminate the genes that cause the lytic cycle and we will have successful transformation of the bacteria, without killing it.
913135
They latch onto one of your cells. Suck out the DNA. Insert their own DNA and a new virus cell is born.
974197
I asked that question, and there is no known mechanism. There is however a mechanism that keeps different species from being able to interbreed.
353721
ERRRRRR evolution!!! 2 points
623852
Well, it could be both, but I would say that a viroid is more correct. \n\nAll proteins are single stranded. But the term "single stranded" is usually applied to nucleic acids, ´cause they can form double-stranded helices. \n\nSo, bearing that in mind, I´d choose viroid.
846739
It can be either (a) or (b) or (d). This depends on how the strong the cell is or how conducive is that cell for that virus. (e) is not possible because virus and bacteria are two different organisms. Virus is much more smaller than bacteria.
517165
Viruses are not free-living...they depend on the host cell machinery to keep them "alive" and allow them to replicate.  Therefore, the only way to cultivate viruses in the laboratory is to grow them in tissue/cell cultures.  You can then allow the tissue/cell culture to incubate and isolate the viruses at a later time.\n\nI don't know what viruses in particular you're asking about, but if you are trying to grow bacteriophage, you can plate the phage suspension on a lawn of bacteria and allow plaques (areas of clearing where the phage have killed the bacteria) to form and then cultivate the viruses from there.
866554
There are more than two, depending on specifically what you are doing.\n\nGenerally transformation is of bacteria. The two major methods for doing this are plasmid transformation and phage transformation. (see link below)\n\nThere are additional methods used on eukaryotic cells.\n The Shotgun approach coats minuscule particles of gold with the DNA you want transferred and shoots them into the cells. Insertion is random.\n\n Some plants are amenable to agrobacterium mediated transformation. This is much more precise than the shotgun approach. (see link below)
116048
same number but one is just mutated
918238
You must take 2 grams of L-lysine capsules every 4 hours.  also take a gram of pure vitamin C 3 times daily.  \n\nLysine will disrupt the virus' inner mechanism and suppress it.  Lysine is a non-toxic amino acid sold in all health food stores.\nvitamin C will rev up your immune system to defeat the virus.  \n\nGo buy them now - spend some money - andhelp yourself.\n\nResults guaranteed if you obey my instructions.\n\nSo what are U waiting for ??????????????????????????
410584
Are the genes linked? I don't have my book with me here.\n\nIf not, mate a male with one of them. Take the male progeny from that mating and cross with the other mutant. Select hermaphrodites from the period of egg laying when the sex ratio is 1:1 (say the second 8-hour period after they're together). Allow these hermaphrodites to self and select double mutants from the progeny. I think this should do it, but you'd better draw it out.\n\nIf they're linked I think it will just be harder to find double mutant progeny in the final stage, again draw it out to check.
496547
Why? because you need a way to integrate your target sequence and vectors such as specialized virii or  plasmids are preferred because they do the whole integration thing without much inconvenience.    You can't just push your target dna into the cell as the dna will likely be digested rather than integrated.
1126696
it's true because phosphatase prevents vectors from closing in on themselves. This doesn't inhibit cloning but ehances it and allows a greater chance for the insert getting in the plasmid and not just the vector re-closing in on itself.\n\nIf you have any other questions about plasmids, cloning in bacteria, etc. feel free to contact me. I do this in lab almost daily.
793861
Im sorry i have to say no.why do you need 2 heads to mess something up when one head can do it by itself'
966045
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
54260
Bacteriophage lierally means "bacteria-eating" it describes a virus that infects bacteria.  You might want to look up lytic and lysogenic cycles, they will explain the "life cycle" of a virus, but im not sure you can really put a time on it, it depends on the type of virus and the type of bacteria...
859339
Well, if you consider the phages can be as small as 20 nm, any pore larger than that will allow them through.  Remember an nm is a billionth of a meter.\n\nThat's tiny.