if there is a restriction enzyme that cuts a piece of DNA at 2 recognition sites,how many fragments on the gel
958167
Theoratically 3 fragments, imagine a strain line, the first cut will produce 2 fragments, a second cut in one of the fragment will produce the 3rd one. there are other factors you need to consider such as the resolution range of your gel, if the cut produces fragments vary in size a lot you might not be able to detect all the fragments. also if the site where is double nick is produced is at the very end of the orginal piece, you might not be able to see the fragment that is very short in length such as 2 or 3 bp long etc....
93891
The important thing is to keep the helical structure clearly visible. Don't worry about the side hydrogens or details of the molecules.\n\nKeep in mind that DNA is a double-stranded helix, and the strands don't revolve around each other identically. Which means there are two sizes of 'grooves': the small grooves are a bit over half the width of the large ones.\n\nIf you use styrofoam balls for the nucleotides and join them with drinking straws (a handy way to do this project), take care that the angle subtended by three of the balls be around 35 degrees.\n\nAbout 10 balls should complete a cycle, which means ball number 11 must be above ball number 1 in a straight line.\n\nDo a strand first, so that it looks fairly twisted yet linear. Add the other strand afterwards. A length of about 20 nucleotides per strand (= 40 in total = 40 small styrofoam balls) should be enough.\n\nGo to a crafts store and buy a bag of balls (they come in 50, 100, etc.) and a bag of 100 drinking straws. Color the balls blue for cytosine, red for thymine, green for guanine and orange for adenine. Pair orange with red and blue with green when building the second strand.\n\nHope this helps.
1038957
The quickest absorption will be of the 2-glucose "polymer."  That way, every cut by the enzyme yields two glucoses.  With any other chain length, each cut would yield one glucose until the last cut, which would yield 2.\n\nAnything wrong with amalyse?
472133
I dont know if this will help but here is a web site I found!\n\nhttp://www.biochemj.org/bj/119/0017/1190017.pdf
277411
What a wonderfully esoteric question!  I've been able to find one site (or a doc on a site) that says,\n\n"The effective spring constant of the DNA is 0.073 pN/nm .."\nhttp://picasso.ucsf.edu/~mollyd/ProblemSet2.doc\n\nBut, I don't know if this will meet your needs.  Here's my search terms.  Maybe you can browse these hits and see if something is more appropriate.\n\nhttp://www.google.com/search?q=dna+%22spring+constant%22+pN/nm+&hl=en&lr=&rls=GGLJ,GGLJ:2006-08,GGLJ:en&start=30&sa=N
891338
In terms of how many restriction endonucleases there are "in the world", the answer could theoretically be in the millions.  Different strains of the same bacterial species can make different RE, so you can imagine the potential for variety in nature.\n\nAs for commercially available: new england biolabs sells about 230 different enzymes.\n\nAs for what's commonly used, it depends on the lab and the target organism.  The lab I got my PhD in, we worked with a low gc bacteria, and we used very different enzymes than we use in my current lab, working with a high gc bacteria.  In either case, we used 20-30 different enzymes.
342631
Before a cell can divide, it must duplicate all its DNA. In eukaryotes, this occurs during S phase of the cell cycle. \n\nA portion of the double helix is unwound by a helicase. \nA molecule of a DNA polymerase binds to one strand of the DNA and begins moving along it in the 3' to 5' direction, using it as a template for assembling a leading strand of nucleotides and reforming a double helix. In eukaryotes, this molecule is called DNA polymerase delta (δ). \nBecause DNA synthesis can only occur 5' to 3', a molecule of a second type of DNA polymerase (epsilon, ε, in eukaryotes) binds to the other template strand as the double helix opens. This molecule must synthesize discontinuous segments of polynucleotides (called Okazaki fragments). Another enzyme, DNA ligase I then stitches these together into the lagging strand. \n\nWhen the replication process is complete, two DNA molecules — identical to each other and identical to the original — have been produced. Each strand of the original molecule has \nremained intact as it served as the template for the synthesis of \na complementary strand.\nThis mode of replication is described as semi-conservative: one-half of each new molecule of DNA is old; one-half new
505328
Transformation\nNucleotides\nBase
93644
I use a 1 KB DNA ladder for electrophoresis and it comes with a paper to actually tell you which band represents which size fragment. Maybe you could look on the tech company's website. If you do not need to know the actual fragment size of each band, maybe you could just compare to make sure the band you need is there in the first place.
244250
You have to identify two sequences that are >6kb apart from one another that flank the sequence you are attempting to amplify. Better yet, you can get you a well understood cloning vector (plasmid) that already has primers develped for it that flank the polyrestriction site region (ie. puc19 [see link below], find the needed restriction enzyme sites that flank your desired 6kb sequence and clone the DNA into the vector. Be sure to check your sequence for any internal restriction sites. This may require you to do a partial digestion and recovery.
44968
An Okazaki fragment is a relatively short fragment of DNA created on the lagging strand during DNA replication.
374804
Consider a regular climbing ladder.  If the ladder was sawed in half by sawing each rung crosscut thru the middle than each half of the ladder would represent one  DNA strand.  Each new organism receives a single strand from each of it's parents.  The two strands join to make the new organism's double stranded DNA.  The rail or sides of the ladder would represent the sugar and phosphate link backbone of the DNA.  The sawed rung would represent the nucleic acid bases.  The bases are classified as either purines or pyrimidines.  The two purines are Adenine (A) and Guanine (G) which are similiar to the compound Purine.  The two pyrimidine bases are Thymine (T) and Cytosine (C) which are structured similiar to the compound Pyrimidine.  In RNA Thymidine is replaced by Uracil.  The collective term for these structures is "nitrogenous base".  Each base contains either/or hydrogen (h), nitrogen (n) and oxygen (o) atoms located on the structure where the "cut" would exhist on the sawed ladder rungs.  When these bases move close together there is an attraction between these atoms and a bridge forms called a hydrogen bond  or h-bond.Thymine has an o and a h.  Adenine has a h which is attracted to T's o and an n which is attracted to T' h. A and T always h-bond or pair.  Cytosine has an h,n,and o.  Guanine has two h's and an o.  one h on guanine is attracted to the o on cytosine, the other to the n.  The o is attracted to the h on cytosine.  Therefore C always pairs with G and is held together by three h-bonds, where A and T are paired by two h-bonds.  In total the h-bonds strongly bind the two DNA single strands together but one at a time they can be broken to seperate the strands much like a zipper opening link by link.  Since C always pairs with G their proportion is is always the same.  The same is true for A and T.  Each Dna single strand has what is called a 3' (three prime hydroxyl) and a 5' (five prime phosphate) end.  When the two single strands pair to form thd Double strand they Form with the 3' end of one strand joining at the 5' end of the other strand.  This is known as antiparallel since they are running in opposite directions.
111490
Ahh...thats easy...After putting the DNA in the holes, and turning on the electric field, DNAs move to the (+) pole because they have a (-) charge....Small DNAs move faster and they separate so the result is seen on the gel...
614020
Here it is! With no hustle!\n\n\nFor pure DNA sample at 260nm wavelength measurements,\n\nOD of 1 = 50 micrograms/mL\n\nFor pure RNA, at the same wavelength\n\nOD of 1 = 40 micrograms/mL\n\nPeace!
902893
your question doesnt make sense\ni know theres such thing as a DNase, that chews up DNA, but a replication digestive enzyme is like an oxymoron, its two different processes
589633
1)there are 5 to 9 nucleotides on 5' mRNA named Shine Delgarno which are the match for 16SrRNA on 3' in the small subunit of ribosome.so 16SRNA conduct ribosome to AUG by binding to shine....and helps to initiate the protein synthesis.\n\n2)there are special sites on 16SrRNA which have a direct reaction with antisense on tRNA\n\n3)it binds to IF3(initiation factor3)
616933
DNA synthesis only goes in 5'-3' direction.  this is fine on a 3'-5' strand, since it follows the direction of the DNA double helix opening and replication is continous. The problem is with the 5'-3' strand, the lagging strand, since its complimentary strand goes 3'-5' direction.  so a primase makes a short stretch of RNA on the DNA fragment and produces okazaki fragments on this 5'-3' strand and synthesizes new DNA in a direction opposite of the replication fork.  As DNA unwinds, more and more okazaki fragments are formed, exonuclease remove the RNA primers and ligase joins the fragments into a continous strand