Two DNA replication complexes move around the E. coli chromosomes. Approximately how many bases are copied by?
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This question is not really answerable.  The E. coli chromosome is circular so it depends on how much time passes.  Also you need to know how many bases are in the chromosome.  ( 1 mega base?)   So hypothetically if an E.coli is able to replicate itself in 20 minutes at log phase growth and there is 2 replication complexes, in 20 minutes you would have approximately 4 million base pairs.
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DNA duplicates during inter phase of cell division to prepare the cell for mitosis
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its a double helix.\n\n  \n 1. DNA is made up of subunits which scientists called nucleotides.\n   2. Each nucleotide is made up of a sugar, a phosphate and a base.\n   3. There are 4 different bases in a DNA molecule:\n\n          adenine (a purine) \n          cytosine (a pyrimidine) \n          guanine (a purine) \n          thymine (a pyrimidine) \n\n   4. The number of purine bases equals the number of pyrimidine bases\n   5. The number of adenine bases equals the number of thymine bases\n   6. The number of guanine bases equals the number of cytosine bases\n   7. The basic structure of the DNA molecule is helical, with the bases being stacked on top of each other
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DNA is a double helix molecule.  The 2 strands are aniparellel (one runs 3' to 5' and the other 5' to 3').  The DNA says in the nucleus of the cell.  The way it replicates is that in uncoils itself and separated from the other strand.  Because of the 4 different nucleotides it is simple to replicate.  The nucleotides are shown as A,C,T, and G.  G goes with C, and T goes with A.  After the stands have separated the nucleotides that are floating about attach themselves to the proper match.  They attach themsleves using an enzyme called DNA polymerase. Now you have 2 complete DNA strands.
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DOUBLE-STRANDED HELIX
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There is a simple approximation that the average molecular mass for a base pair is 660. That means that a mole of a base bair would be 660 g or if you prefer 6.023*10^23 base pairs would be 660*10^9 ng.\n\nTherefore 1 ng would be (6.023*10^23)/(660*10^9) =9.13*10^11 base pairs. I am using base pairs since DNA is double stranded. (If you wanted to take into account the nucleotides of both strands then you just mutliply by 2 (18.26*10^11))\n\nFor calculating the mass of the DNA, if bp is the number of basepairs of the genome of interest then its approximate molecular weight is 660*bp and the mass is \nmass=mole*MW = mole*660*bp\n\nFrom here on it depends on how many moles of DNA you have. If you want to find out the weight of a single molecule it is MW/6.023*10^23= 660*bp/6.023*10^23 g= 660*bp/6.023*10^14 ng
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In DNA deoxyribose-phosphate backbone and nitrogenous bases.  The bases of DNA are adenine, cytosine, guanine and thymine.  In RNA ribose-phosphate backbone and nitrogenous bases are adenine, cytosine, guanine and uracil.  A deoxyribose-phosphate backbone is essentially a pentonse sugar (5 carbons) linked to a phosphate group in a 5' to 3' direction.  The bases are connected to the 1' end of the ribose groups. Pairing occurs by hydrogen bonding between A+T, A+U, and G+C between two strands.
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The four nucleotide bases, the building blocks of every piece of DNA, are represented by the letters A, C, G, and T, which stand for their chemical names: adenine, cytosine, guanine, and thymine. The A on one strand always pairs with the T on the other, whereas C always pairs with G. The two strands are said to be complementary to each other. \n\nTo copy DNA, polymerase requires two other components: a supply of the four nucleotide bases and something called a primer. DNA polymerases, whether from humans, bacteria, or viruses, cannot copy a chain of DNA without a short sequence of nucleotides to "prime" the process, or get it started. So the cell has another enzyme called a primase that actually makes the first few nucleotides of the copy. This stretch of DNA is called a primer. Once the primer is made, the polymerase can take over making the rest of the new chain. \n\nA PCR vial contains all the necessary components for DNA duplication: a piece of DNA, large quantities of the four nucleotides, large quantities of the primer sequence, and DNA polymerase. The polymerase is the Taq polymerase, named for Thermus aquaticus, from which it was isolated. \n\nThe three parts of the polymerase chain reaction are carried out in the same vial, but at different temperatures. The first part of the process separates the two DNA chains in the double helix. This is done simply by heating the vial to 90-75 degrees centigrade (about 165 degrees Fahrenheit) for 30 seconds. \n\nBut the primers cannot bind to the DNA strands at such a high temperature, so the vial is cooled to 55 degrees C. At this temperature, the primers bind or "anneal" to the ends of the DNA strands. This takes about 20 seconds. \n\nThe final step of the reaction is to make a complete copy of the templates. Since the Taq polymerase works best at around 75 degrees C (the temperature of the hot springs where the bacterium was discovered), the temperature of the vial is raised. \n\nThe Taq polymerase begins adding nucleotides to the primer and eventually makes a complementary copy of the template. If the template contains an A nucleotide, the enzyme adds on a T nucleotide to the primer. If the template contains a G, it adds a C to the new chain, and so on to the end of the DNA strand. This completes one PCR cycle.  \n\nIf Instead of G if A gets added to C, then its called as mismatch...or mutation which DNA come over by repair mechanism...\n\nhope this helps!!!
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In G1, there are 46 chromosomes, EACH CONSISTING OF A SINGLE CHROMATID.  You will have two copies of chromosome one (one from mom, the other from dad), two copies of chromosome two, etc.  These pairs are called homologues.  Consequently there are 23 homologous chromosomes in G1 and, again, there is one chromatid per chromosome in this phase.  This gives a total of 46 chromosomes.  You should know that each chromosome is decondensed (not coiled up) at this time. \n\nS phase is the synthesis or replication phase of the cell cycle.  After S phase is complete, the definition of a chromosome changes.  It is now a "replicated chromosome" that consists of 2 chromatids (some texts describe these as "sister chromatids").  Again, the total number of chromosomes is 46; however, because of the definition change of the S-phase chromosome, each chromosome how includes twice as much DNA due to replication.  After S phase there are two chromatids per chromosome.  These chromosomes are decondensed.\n\nIn G2, there are 46 chromosomes:  2 chromatids per chromosome, decondensed, with the chromosomes existing in homologous pairs.\n\nIn prophase of mitosis, the chromosomes condense to facilitate easy separation into the daughter cells.  Here there are still 2 chromatids per chromosome.\n\nIn metaphase, there are STILL 2 chromatids per chromosome, all smartly arranged at the "equator" of the cell, the metaphase plate.\n\nIn anaphase, the replicated chromosomes (i.e., the "sister chromatids") are pulled apart.  At the point that they are separated, the definition of a chromosome changes AGAIN, BACK to a single chromatid.  Each daughter cell will receive 46 of these single chromatids and, therefore, will have 46 chromosomes in 23 homologous pairs.  The chromosomes are still condensed at this point.\n\nIn telophase, the chromosome situation is exactly the same.  23 homologous PAIRS of chromosomes (46 single chromatids).  The difference here is that the chromosomes are now decondensing.\n\nSummary:  the number of chromatids per chromosome\n\nG1-1\nafter S- 2\nG2-2\nprophase-2\nmetaphase-2\nlate anaphase-1\ntelophase-1
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1. Gene transformation\n2. Junk DNA\n3. Major Histocompatibility Complex - MHC\n4. X-linked\n\nBy the way okazaki fragments are bits of DNA formed during the replication process.
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I'm not certain if this is what you mean, but prokaryotic  organisms (i.e. bacteria) have plasmids more often than do eukaryotic organisms. Not all bacterial chromosomes are plasmids, and not all plasmids are bacterial chromosomes.
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This may seem like a silly suggestion, but try representing the elements mentioned in your textbook (which I'm sure has some illustrations) with pieces of paper, matches, beans or something like that, so you can understand the events and the outcome. \nNow, if you _really_ want an explanation, here it comes (I've ommited the names of the phases, they'll be in your textbook).\n\nMeiosis includes 2 cell divisions with different characteristics. \nWe start with a diploid (2 n) cell, in which DNA duplication has occurred so that each DNA strand is double. These strands do not yet appear as chromosomes (condensed bodies); in fact, the chromatin strands have to be "relaxed" in order to duplicate. \n\n***Meiosis I: (in which the nº of chromosomes decreases to half)\n1. The chromatin (DNA strands) starts to condense and appears as small dense dots (the chromosomes) within the cell nucleus. These chromosomes will each have 2 "arms" (they look like an X or a V) because the DNA in each has been duplicated.\n2. Now all these chromosomes align in homologous pairs, with the aid of a transparent structure formed by tiny intracellular fibers (the spindle) that has formed and will help them to move in orderly fashion within the nucleus.\n3. While they keep condensing, each homologous chromosome contacts its "brother", and they exchange some fragments of their DNA ("crossing over"). Usually only one "arm" of each  chromosome participates in crossing over. \n4. In the meantime, the nuclear membrane is disintegrating...\n5. The pairs of homologues, still united, align at the middle (equator) of the spindle... \n6... and then each pair of homologues is divided, one chromosome of each pair goes toward each "pole" of the spindle. \n7. New nuclear membranes form around the 2 groups of chromosomes. Each of these groups will only have ONE of each pair of homologues. Essentially the information in each new nucleus is the same, BUT has some differences because, as you recall, there has been crossing over i.e. partial exchange of DNA between homologues. \n8. The original cell with the 2 nuclei divides and forms 2 cells, with half the nº of chromosomes of the original cell (haploid or n). Each of the chromosomes still has a duplicate DNA strand.\n\n... the cell takes a short break (you can take a break too!)... but the chromatin does not duplicate this time (it's already done that)\n\n***Meiosis II: (in which each double DNA strand becomes single again, this phase is similar to a normal mitosis).\n1. The chromatin condenses again, so that it appears as chromosomes.\n2. This time the spindle aligns the center of each chromosome at its equator.\n3. The central part of the chromosome is duplicated and the sister "arms" (chromatids) separate, each migrating to opposite poles.\n4. Now you have two groups of chromosomes, each group still has the haploid (n) nº of chromosomes (the same amount with which they started this 2nd meiosis) but now, each chromosome has a single DNA strand. \nA new nuclear membrane forms around each of these groups of chromosomes.\n5. The rest of the cell divides.\n\nSo, to sum up: you started with ONE diploid cell. You end with FOUR haploid cells. In animals, this process only happens during the production of gametes. \nIf we're talking about the formation of female gametes, then only ONE of the 4 resulting cells will become an ovule (and that only after more maturation), the other 3 cells form "polar bodies".\nIn contrast, when it's spermatozoids being produced, all FOUR resulting cells will become spermatozoids. \nThat's one of the reasons sperm is cheaper than ovules! There's just more of them in the entire animal kingdom.\n\nNow, what is the purpose of this extremely complicated (the above is a very simplified version) process? The result will be cells that have half the chromosomes of the normal body cells. These cells
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You get a continuous and discontinuous strand because DNA polymerase only makes DNA in 1 direction (5` to 3`).  The continuous strand is in that orientation anyway, so you get a nice long replication.  The opposit strand is oriented 3` to 5` relative to the polymerase, so the replication has to be broken down into multiple, short, 5` to 3` segments so that replication can occur within the same polymerase macromolecule.\n\nI had to mention  DNA polymerase, it's really impossible to explain this without saying something about how it works :P I did leave out RNA primers though!  If anyone else is curious, the RNA primers are the site of the initiation of replication of each little Okazaki fragment.
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The answer is 6.3 billion (the population of the Earth) minus any identical twins. \n\nHumans have 3 billion base pairs of DNA, most of which are identical. To get to a number of possible differences, one would have to add up each set of one difference, two differences, etc. Here is an example with a smaller DNA sequence:\n\nGCT\nIn the first position, one can have three other nucleotides, C,A, or T. Ditto with the second position, and third position. But on top of this we have to count the sequences that differ at any two positions, and at all three positions.\n\nMathematically, this is 4*4*4, so just for three base pairs we have 64 possibilities.\n\nTherefore with 3 billion positions the answer is 4 raised to the 3 billionth power; 4^3000000000. This number is impossibly huge to comprehend. By comparison, a googol is 10^100, itself larger than the number of particles in the Universe. \n\nAnd this only counts point mutations. There are nearly infinite number of insertion and/or deletion mutations that can occur on top of the point mutations.
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9\n\n3 bases for each amino acid (applies to both DNA and RNA).
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http://www.microbesonline.org/operons/gnc205922.html
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In the monomer of DNA, the carbon number of deoxyribose sugar component (circular structured sugar with 5 Carbons) is marked as the first C binding to the Nitrogen base (A, T, G or C), and the 5th C binding to the phosphate group (PO4). \n\nThe 3' and 5' is called basing on the 3rd and 5th carbon in the deoxyribose molecules. DNA polymerization is like this: and individual monomer's phosphate group (which is bound to the 5th C) is added to the 3rd C of the previous monomer in the polymerized chain. The polymerization only occurs in this direction, no reverse direction (no addition of 3rd C of an individual monomer to the phosphate group bound to a monomer in a polymerized chain. That's why they say the DNA polymerization is in 5' to 3' direction.\n\nYou can go to this link to see a picture of the monomer structure and the polymerization: http://cmgm.stanford.edu/biochem/biochem201/Slides/DNA%20Structure/004%20DNA%20Covalent%20Structure.JPG