Problem-solving templates

Problem-solving strategies applied to reconfigurable manufacturing systems entail developing (a) domain independent solution(s) templates at the macro level, (b) capability models for application specific domain dependent problems at the micro level, and (c) coordination models to integrate models developed in... 

Self-organized materials with high surface area and pore size 3-25 nm was produced used templating and coassembly. The highly porous nature of the ordered combined with low adsorption and emission in the visible spectrum, facile diffusion makes them good candidate for optical and chemical sensor and provide new avenues for encapsulation/ immobilization processes and solve the problems mentioned above. 

In protozoa this problem is solved by the addition of preexisting oligodeoxynucleotide blocks to the 3 ends of DNA. These blocks are composed of tandemly repeated units of (T2G4)n or (T4G4)n, where n is approximately 50. The enzyme that adds these polymers requires a primer but not a template. These oligonucleotide block polymers are called telomers. Another DNA polymerase,... 

The ends of a linear chromosome are not readily replicated by cellular DNA polymerases. DNA replication requires a template and primer, and beyond the end of a linear DNA molecule no template is available for the pairing of an RNA primer. Without a special mechanism for replicating the ends, chromosomes would be shortened somewhat in each cell generation. The enzyme telomerase solves this problem by adding telomeres to chromosome ends. 

For the very low density varieties of the cases shown in Figure 2 and, more particularly, Figure 4 (curve 7), for which initiation is slow compared to both termination (release from end of template) and polymerization, a simpler treatment, in which the interference of one ribosome with another is totally neglected, should suffice. In this case an equation of the form of Eq. (1), herein only applied to the problem of DNA synthesis, should be valid, but Eqs. (2) and (3) should be modified to account for repetitive initiation at site 1 and continuing release from site K, respectively Eqs. (4) and (6) will not apply. In the even more restricted (but perhaps biochemically relevant) case in which, in addition to neglecting ribosome interference, one may also neglect the back reaction (kb x 0), one may solve this system of equations (Eq. (1), plus Eqs. (2) and (3) modified as described) very easily by taking Laplace transforms.13 This is the only case with repetitive initiation for which we have been able to find solutions for the transient, as well as steady state, behavior. 

The T-odd bacteriophages Tl, T3, T5, and T7 are medium-sized phage with linear duplex DNA genomes. Replication of linear DNA in these and in many other genomes presents a problem. Even if the RNA primer segment is made at the very 3 end of the template strand, there will be a gap in the final replicated strand when the primer is digested out. Since there is no known enzyme that will add to the 3 end of a chain, this gap will remain unfilled. The problem is solved by terminal redundancy, the presence of a common 260-nucleotide... 

Since eukaryotic chromosomes are linear, the ends of these chromosomes require a special solution to ensure complete replication. This can be seen in figure 26.26. At the very end of a linear duplex a primer is necessary to initiate DNA replication. After RNA primer removal there is bound to be a gap at the 5 end of the newly synthesized DNA chains. Since DNA synthesis always requires a primer the usual way of filling this gap is not going to solve the problem. This dilemma is overcome by a special structure at the ends (telomeres) of eukaryotic chromosomes and a special type of reverse transcriptase (telomerase) that synthesizes telomeric DNA. In many eukaryotes the telomeres contain short sequences (frequently hexamers) that are tan-demly repeated many times. Telomerase contains an RNA that binds to the 3 ends and also serves as a template for the extension of these ends. Prior to replication, the 3 ends of the chromosome are extended with additional tandemly repeated hexamers. The 3 ends are extended sufficiently so that there is room to accommodate an RNA primer. In this way there is no net loss of DNA from the 5 ends as a result of replication. After replication the 3 end is somewhat... 

The nature of the iminic nitrogen substituent influences the cycloaddition pathway (4 + 2 versus 3 + 2) followed in the reactions of a-nitrosoalkenes with alkyl/aryl-substituted acyclic imines.4 The problem of rotamer control in Lewis acid-catalysed 3 + 2- and 4 + 2-cycloaddition reactions of a./S-disubstituted acryloylimides was solved by the use of N-H imide templates.5... 

The following text will give you a walkthrough of the use of this spreadsheet and the McCabe-Thiele equilibrium curve. MTPlot72 is applicable as a template in solving any liquid-liquid problem. 

One cannot deny the fact that the origin of unlimited heredity is an unsolved problem. Perhaps compartmentation will help solve this problem as well, so that long templates could self-replicate within vesicles without enzymatic aid. 

Currently the dideoxy method probably is the method of choice for sequencing DNA by primed synthesis methods and is generally applicable to any DNA that can be obtained in single-stranded form. In the last two years two important advances have been made which have essentially solved the problem of preparing the DNA template in single-stranded form and these are described in the following Section and in Chapter 4. The first method, which... 

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