what type of data about a protein these two technique collect\nwhich aspect are they specific for? i.e In what situation it's better to use AMF and what suituation to use CD\nwhy in studying fibril assembly, ususlly both are used? how do the data provided by them complementary each other\nand what is their adv. over the crystallization method\nthere are many Q plx ans one by one thx quote reference if possible plx
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AFM collect surface atomic data\nCD collect absorbance which reveal the secondary structure\nCD 2nd structure and low resolution\nAFM higher resolution and study the whole structure
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put simply its complete bullshit... lol, solidification is just the slowing of atoms/molecules, sonce light ALWAYS (dont even go tehre with the whole medium thing, its always) travels at teh same speed... like i said before solidified light is bs. :-)
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http://cwx.prenhall.com/bookbind/pubbooks/hillchem3/medialib/media_portfolio/text_images/CH08/FG08_17.JPG
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do you mean interactions BETWEEN nucleic acids and proteins?  If so there's a lot of ways you can study that - you can do a footprint analysis - digest the nucleic acid in the presence of the protein and see what piece gets protected to see where the protein binds if you know the identity of the two interacting molecules.  If you don't know their identity you can try a gel-shift assay where DNA will move at different rates on a gel in the presence and absence of various proteins.  If you know the protein but not the DNA you can try an affinity column with the protein immobilized and then try to determine which DNA sequences bind.  You can also try to determine the crystal structure of a co-crystal if you have a lot of the two materials.  good luck!
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In SDS-PAGE you have the formation of a complex between SDS and proteins so that you have anegatively charged SDS molecule for approximately every two amino acids in the protein. This means that the protein-SDS complex will have a huge negative charge that will mask any charge of the protein itself and force the protein-SDS complex to move towards the positive electrode.\n\nI've never heard of the other thing you are mentioning. Could you expand on that? Hydrophobicity of what? DNA has a large negative charge due to its phosphate backbone and that's the reason it migrates during electrophoresis.
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try going here:\nhttp://www2.ncsu.edu/ncsu/pams/science_house/learn/CountertopChem/exp9.html\n\n\nAlso, you may find this interesting: http://biology.clc.uc.edu/fankhauser/Cheese/ROOTBEER_Jn0.htm
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Structure and Function of Macromolecules\n\n \nOrganic molecules that weigh more than 100,000 Dalton's are referred to as macromolecules.\n \nThese macromolecules are constructed of smaller units called polymers. These polymers are subdivided into their basic units called monomers.\n \nPolymers account for the molecular uniqueness of organisms. Twenty amino acids are responsible for all forms of life. These amino acids form every conceivable protein known to man.\n \nMaking and breaking of polymers:\n \nDehydration synthesis: is an anabolic process by which two molecules are chemically bonded through the use of enzymes and a loss of water. Example: glucose + glucose = maltose + water.\n \nHydrolysis: is a catabolic process by which the bond between monomers are broken by the enzyme and the addition of water. Example: Sucrose + water = glucose + fructose.\n \nCarbohydrates: include sugars and their polymers. They include monosaccharides disaccharides, and polysaccharides. The monosaccharide is a monomer, the disaccharide is a polymer,and the polysaccharides are macromolecules.\n \nMonosaccharides: The basic formula (CH2O)\nExamples: triose sugars, 3 carbons, glyceraldehyde and dihydroxyacetone; pentose sugars. 5 carbons. ribose, deoxyribose, and ribulose; hexose sugars, 6 carbons, glucose, galactose, and fructose.\n \nDisaccharides: These are double sugars with the formula C12H22O11. Notice that one molecule of water is missing from the formula. The covalent bond holding the two monomers together is called a 1-4 or 1-2 glycoside linkage. Examples: sucrose = glucose + fructose. maltose = glucose + glucose, and lactose = glucose + galactose.\n\n \nPolysaccharides: The basic formula is ( C6H10O5)n . These are macromolecules capable of acting as structural or storage molecules.\n \nStorage Polysaccharides: Starch is a plant storage polysaccharide that is composed entirely of glucose joined by @1-4 glycoside linkages. amylose is the simplest form of starch. amylopectin is more complex and is branched. Glycogen is an animal starch stored in the liver and muscles of vertebrates. It is more highly branched than amylopectin.\n \nStructural Polysaccharides: Cellulose and chitin are examples of structural polysaccharides.\nCellulose is the most abundant organic compound on earth. It is made of glucose, like starch, but they differ in the type of 1-4 linkage. Instead of an @ linkage as in starch cellulose contains a B 1-4 linkage.\nThis causes the polysaccharide to take on a step arrangement and not a linear one like in starch.\nEnzymes find it difficult to brake the B 1-4 linkage.\n \nLipids: A group of polymers that have one characteristic in common, they do not mix with water. They are hydrophobic. Some important groups are fats, phospholipids, and steroids.\n \nFats: are large molecules composed of 2 types of monomers, glycerol ( an alcohol containing 3 carbons) and 3 fatty acid molecules. The bond connecting the glycerol and fatty acids in the fat molecule is called an ester bond. There are two types of fatty acids: saturated and unsaturated. The saturated fatty acids do not contain any double bonds between the carbons, while the unsaturated fatty acids contain one or more double bonds between the carbons. These double bonds cut down on the number of hydrogen atoms that can be attached to the carbon in the molecule. This causes the molecule to bend or kink at each of the double bond sites.\n  \nCharacteristics of Fats:\n \nSaturatedUnsaturated\n1. solid at room temperature\n\n1. liquid at room temperature\n\n2. found mostly in animals\n\n2. found mostly in plants\n\n3. no double bonds between carbons\n\n3. double bonds found between carbons\n\n \n\n \nFunction of fats: acts as insulation in higher vertebrates, serves as an energy storage source 1g.=9 Kcal of energy, and shock absorber for internal organs.\n \nPhospholipids: structurally related to fats but contain 2 fatty acids and one molecule of phosphate. These molecules are fo
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Go to "protein" on wikipedia ( put, protein, in address bar ) then scroll down to "synthesis", for the process. It is too  long a story for my typing skills. The wiki article is not too technical.
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From Gene to Protein\nScientists that worked on Protein Synthesis:\n1. Archibald Garrod: First proposed the relationship between genes and proteins.\n2. George Beadle and Edward Tatum: working with Neurospora , formulated the one-gene-one-protein hypothesis. This was later changed to one-gene-one-polypeptide. Since some proteins are produced from more than one gene.\n\nThe Problem: Since there are 20 different amino acids; and these amino acids are used to make proteins, how does the information get from the DNA to the final protein product? In a DNA molecule the only variable part is the 4 different bases. How can 4 bases, or combinations there of , code for the 20 different amino acids?\n\n1. If 1 base coded for 1 amino acid = 41 (4) amino acids coded for.\n\n2. If 2 bases coded for 1 amino acid = 42 (16) amino acids coded for.\n\n3. If 3 bases coded for 1 amino acid = 43 ( 64) amino acids coded for.\n\nNotice the large gap between 2 and 3. It was concluded that each coding part of DNA (codon) consisted of 3 bases. Since 64 is larger than the 20 amino acids, it stands to reason that some amino acids are coded for by different codons. Why is this important?\n\nTranscription: DNA is too large to leave the nucleus. If the program it contains must leave the nucleus, an intermediate must be produced. This intermediate is called m-RNA. m-RNA is formed in the following way, RNA polymerase binds to, a region of the DNA called, the promoter. Here is found the start codon TAC which codes for the amino acid methionine. Once this has been established the m-RNA sequence is produced growing from the 5' - 3' region. This occurs at the rate of 60 nucleotides / sec. The terminator sequence will keep the process from going on indefinitely. The eukaryotic m-RNA must be modified before going to the ribosome, while the prokaryotic does not.\n\nTranslation: t-RNA : a specialized form of RNA used to carry a specific amino acid to the ribosome and place it in the proper position. t-RNA contains 80 nucleotides in the form of a clover leaf. The t-RNA must attach the correct amino acid to itself. The process is as follows:\n\n1. The specific amino acid reacts with a specific enzyme and ATP to release 2 P and combine the AMP and the amino acid.\n\n2. The t-RNA will replace the AMP and bond with the amino acid.\n\nRibosomes: Before translation ribosomes consist of 2 separate sub-units. As translation proceeds the ribosome parts come together to allow protein synthesis to begin.\n\nThere are 3 steps to protein synthesis:\n\n1. Initiation: m-RNA binds to the small subunit of the ribosome. There is 1 binding site for the m-RNA, and 2 binding sites for the t-RNA. The P site holds for the t-RNA carrying the growing polypeptide and the A site for the t-RNA carrying the next amino acid. AUG ( start codon ) is in place and the appropriate t-RNA delivers the amino acid Methionine. The large sub unit then joins and creates a functional ribosome.\n\n2. Elongation: The chain moves from the 5' - 3' direction caused by the use of GTP and several protein-enzyme complexes. As the polypeptide chain is produced the t-RNA move from site A to site P, and then are released into the cytoplasm. The polypeptide chain continues to grow, and is held to the ribosome by the t-RNA at the P site.\n\n3. Termination: The termination codons: UAA,UAG, and UGA stop the process by adding water to the end of the chain instead of an amino acid. In order for the protein to become functional, it must be folded into its appropriate shape,and/ or have some amino acids removed, and/or have some of the polypeptides modified by adding sugars or phosphate groups to them.\n\nThe Wobble Phenomenon: There are only 40 different types of t-RNA and 64 codons. This means that some of the t-RNA can pair up with several different codons. This can occur because the third base of a t-RNA molecule can form a hydrogen bond with more than one kind of base. U in the third position can bond with A
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is this your school homework,  are there suppossed to be other questions before these because my computer shows them starting from No. 4.\nnway,\n4.  filter the solution while still hot because as the soltion cools crystallization occurs, and you will nloose your product on the fiolter paper.\n5. crystallization can be induced by reduction in temperature i.e put in an ice bath or you can use a technique called seeding where you put a few crystals of the expected substance in the solution, the other method is to increase the concentration of the solution by evaporating it, to saturation point, ensure you do not evaporate it dry.\n6.i do not know what you started with so i can not tell.\n7.slow cooling will definately give bigger crystalls.\n8.sodium chloride reacts with silver nitrate to form a white precipitate, actually silver nitrate solution is used to analyse NaCl in industrial setting.\n\nope you will find these information sufficient.
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okay I'm gonna try this. :-)\n\nImmunohistochemistry is a way to detect certain proteins that are in your body's tissues.\n\nIn kidney diseases, there may be abnormal proteins in the kidney that you'll want to look for.\n\nSo what happens is a scientific technician will take an extremely thin slice of kidney tissue, and treat it with "antibodies" to the abnormal (bad) protein.  \n\n[An antibody is something that will stick to a certain protein, they can be made to be very selective.  So there is a specially designed antibody that will find this bad protein and stick to it.]\n\nIn addition, this antibody has a "signal" stuck to it.  Normally it's some kind of fluorescent (shiny, hehe) dye.  After treating the sample with antibodies and getting them stuck to the abnormal proteins in the tissue, the technician will then perform a reaction that will activate the dye.  Then, the sample is put under a microscope, and the picture is analyzed for presence of this dye.  If the kidney is diseased and has the abnormal "bad" protein, there will be colored dye spots all over the picture.\n\nSo in short... kidney diseases can show abnormal (bad) proteins, scientists can make things ("antibodies") that stick to these bad proteins, and then these antibodies can be "activated" to light up the dye and reveal where the bad proteins (which they are sticking to!) are.\n\nIn theory, a normal kidney tissue sample would not show any fluorescence/lighting if they were looking for a "bad" protein.  Of course, "bad" protein is a huge generalization, and I'm just trying to simplify it into layman's terms. \n\nContact me if you have other questions, hope this helps!\n\n-IJ
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You're really going to have to do your own research on this one. \n\nYou've picked a very broad topic and it will be easier for you to understand if you can turn up some basic research materials on bioreactors/fermenters with diagrams.\n\n"with diagrams" is the key there. These things are easier to understand if you visualize them.
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This technique was considered a good one since it took all possible ways of aligning the sequences into account and the melting temperature would be a good average. However it is not considered the best approach these days since sequences can be computationally aligned. There is hardly any other approach used currently, however the sequences used for the comparisons are the major source of contention. Not all sequences evolve at the same rate. Some are too critical and changes can be lost if the result causes loss of function of the gene product/action. Finding the corresponding genes in more distant organisms can also be difficult. Some approaches have considered using non-coding sequences since these are not believed to be affected by evolution and all mutations in them would be retained faithfully. If one assumes a constant rate of background mutation, these mutations would indicate the age of the sequence lineage. However, these too are difficult for cross genera comparisons.
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Try doing a Google search under the Hall or the Hall-Heroult process.    That or check out a General or an Inorganic Chemistry book from the library.  That's a good launch point.
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Circular dichroism (CD), is the differential absorption of left- and right-handed circularly polarized light. It is a form of spectroscopy used to determine the optical isomerism and secondary structure of molecules.\n\nAt a given wavelength,\n\nΔA = (AL − AR) \nwhere ΔA is the difference between absorbance of left circularly polarized (LCP) and right circularly polarized (RCP) light (this is what is usually measured). It can also be expressed as:\n\nΔA = (εL − εR) C l \nwhere εL and εR are the molar extinction coefficients for RCP and LCP light, where C is the molar concentration and l is path length.\n\nThen\n\nΔε = (εL − εR) \nis the molar circular dichroism. This is what is usually meant by the circular dichroism of the substance.\n\nAlthough ΔA is usually measured, for historical reasons most measurements are reported in degrees of ellipticity.\n\nThe molar ellipticity is:\n\n[θ] = 3298Δε. \nIn general, this phenomenon will be exhibited in absorption bands of any optically active molecule. As a consequence, circular dichroism is exhibited by biological molecules, because of the dextrorotary (e.g. some sugars) and levorotary (e.g. some amino acids) molecules they contain. Noteworthy as well is that secondary structure will also impart a distinct CD to their respective molecules. Therefore, the alpha helix of proteins and the double helix of nucleic acids have CD spectral signatures representative of their structures.\n\nCD is closely related to the optical rotary dispersion (ORD) technique, and is generally considered to be more advanced. CD is measured in or near the absorption bands of the molecule of interest, while ORD can be measured far from these bands. In principle these two spectral measurements can be interconverted through an integral transform, if all the absorptions are included in the measurements.\n\nThe ultraviolet CD spectrum of proteins can predict important characteristics of their secondary structure. CD spectra can be readily used to estimate the fraction of a molecule that is in the alpha-helix conformation, the beta-sheet conformation, the beta-turn conformation, or some other (random) conformation. These fractional assignments place important constraints on the possible secondary conformations that the protein can be in. CD cannot, in general, say where the alpha helices that are detected are located within the molecule or even completely predict how many there are. Despite this, CD is a valuable tool, especially for showing changes in conformation. It can, for instance, be used to study how the secondary structure of a molecule changes as a function of temperature or of the concentration of denaturing agents. In this way it can reveal important thermodynamic information about the molecule that cannot otherwise be easily obtained. Anyone attempting to study a protein will find CD a valuable tool for verifying that the protein is in its native conformation before undertaking extensive and/or expensive experiments with it. Also, there are a number of other uses for CD spectroscopy in protein chemistry not related to alpha-helix fraction estimation. s\n\nCD spectroscopy is usually used to study proteins in solution, and thus it complements methods that study the solid state. This is also a limitation, in that many proteins are embedded in membranes in their native state, and solutions containing membrane structures are often strongly scattering. CD is sometimes measured in thin films.\nCD has also been studied in carbohydrates, but with limited success due to the experimental difficulties associated with measurement of CD spectra in the vacuum ultraviolet (VUV) region of the spectrum (100-200 nm), where the corresponding CD bands of unsubstituted carbohydrates lie. Substituted carbohydrates with bands above the VUV region have been successfully measured.\n\nMeasurement of CD is also complicated by the fact that typical aqueous buffer systems often absorb in the range where structur
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mix it...let NaCl dissolve in water...then let the mmixture evaporate or perhaps bring the mixture into a boil 'til you can see the formation of crystals...   try it... i've encountered such experiment in my chemistry class... im not quite certain of the procedures though...it's two years ago...cant remember it well...^_^
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You are on the right track, read both books objectively and sincerely, I believe you will no doubt arrive at the truth. Kindly let me know the results of your findings at the end of your research, best of luck with your download hunt. Kind regards.