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REASSOCI - REASSOCIATION KINETICS Dr P J Laipis Department...

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Unformatted text preview: REASSOCIATION KINETICS Dr. P. J. Laipis Department of Biochemistry, University of Florida Original References: Britten and Kohne, Science 161:529 (1968). Britten et al. Methods in Enzymology 32:363 (1974). Objectives 1. Know the general techniques and the basic principles on which the analysis of reassociation kinetics depends. 2. Recognize the kinds of information Cot analysis can provide about DNA and RNA molecules, and the kinds it cannot provide. 3. Know what Cot analysis says about the structure of eukaryotic chromosomes. INTRODUCTION Cot analysis, or the analysis of the rates of reassociation between separated strands of either DNA or RNA was an extremely important technique in determining the information content of nucleic acids. Ideally, once one knew the physical structure of the genome of a virus or of man, the next important piece of data would be to know the sequence of bases in that DNA or RNA genome. Unfortunately, direct sequencing techniques were tedious, and in the case of long DNA molecules, effectively impossible. However, by analyzing the kinetics of reassociation between separated strands of DNA molecules, or between DNA and RNA molecules in solution, it was possible to obtain estimates of 1) the size (or molecular complexity) of the various nucleic acid molecules that are present, 2) the relative proportion of each size class of molecules in the population, and 3) the actual number of molecules (repetition rate) in each size class. In the next two sections, the theory behind Cot analysis, and some of the actual techniques used will be discussed. The last section briefly discusses a number of examples of the types of information that can be obtained. THEORY The separated complementary strands of nucleic acids will, under appropriate conditions, reassociate to form base—paired duplex molecules. The rate at which this reassociation occurs depends on four factors: 1) the concentration of cations, which decrease the intermolecular ionic repulsion between the negatively charged strands of nucleic acid molecules (below 0.01 M Na ion, the reassociation reaction is effectively blocked), 2) the temperature of incubation, which must be high enough to destabilize intrastrand secondary structure, but below the Tm, or melting temperature of the native double—strand molecule, 3) the size of the DNA fragments (viscosity affects the rates of movements of molecules in solution), and 4) the concentration of molecules, which determines the number of intermolecular collisions which will occur in a unit of time. Experimentally, if the cation concentration, temperature, and size of the nucleic acid molecules are held constant, then the amount of reassociation will depend only on the time allowed and the concentration (measured in moles of nucleotides/liter) of molecules in the reaction. The rate of disappearance of single strands is given by the expression: dC in which C is the concentration of single—stranded fragments in moles/liter, t is the time in seconds, and k is a second order rate constant, depending on cation concentration, temperature, fragment size, and the sequence complexity of the nucleic acid. This last point is extremely important in the understanding of this process, and yet it is fairly obvious. For reassociation to occur, each single—strand fragment of DNA, for example, must find its complementary fragment. Let us consider two viruses, one of which has a genome 5,000 nucleotides long, and the other 250,000 nucleotides long, and further note that these genomes are unique; that is, there are no repeating sequences in either genome (this observation is true for viruses and prokaryotic organisms). Thus, if the initial concentration of each genome in moles/L of nucleotides is the same, say 106 nucleotides, and each genome is sheared into small, identical size fragments, say 500 nucleotides long, to reduce viscosity effects on hybridization, then in the reaction tube containing the small virus genome, there will be about 10 different sheared fragments produced from each strand, and each fragment will be present 200 times in the total reaction mixture (ie. 106 nucleotides total/5 x 103 nucleotides/genome = 200 genomes, each of which gives 10 unique frag— ments). Similarly, for the large genome, there are 500 unique fragments, but each is present only 4 times in the total reaction. It will be 50 times easier for a single—strand fragment in the small genome to find its complement, than for the same size fragment in the large virus genome to do the same, and consequently, the rate of reassociation will be 50 times faster, since all other conditions are constant and identical. This second order rate equation can be rearranged to: =49 =kdt C2 and integrated from t=o and C=CO to yield: l_i=kt c c O This can be rearranged to give equation (2): 1 £1::_______ (2). C l~+kC t 0 0 At t=O, all the DNA is single—stranded, so C=Co and Co equals the total DNA present in the reaction. Equation (2) further shows that C/Co, the fraction of DNA remaining single—stranded, is a function of Cot, the product of the initial concentration and the time allowed for reaction. This expression is conveniently plotted as shown in Figure 1, to give a “Cot” curve. to ’“i Ideal reassociation thne cause Qfi Halfzcocfion Hnalflqm o, [MUM mum 35.! 1 Hum I {In} '4 ‘,~J.,|lg.u ': 0.01 0.1 I 10 100 Concentration x time (C 0t) Figure 1. Cot curve of an ideal second—order reassociation reaction. Note that 80% of the reaction occurs over a 2—log interval of Cot. (From R. Britten and D. Kohne, Science 161, 529, 1968) Finally, Cot% is the Cot value at which C/Co = %, ie. the reassociation is half finished. By plugging into equation (2), Cot% = l/k, and so it is easy to determine k and thus the sequence complexity of the molecules under experiment, (k, the rate constant, is inversely proportional to the complexity) from a plot of reassociation kinetics as shown in Figure 1. This experimental determination of Cot% then leads to an estimate of the sequence complexity of the DNA molecules under examination. Alternatively, if the complexity is known, the concentration of a sequence can be determined. These alternatives are discussed below. EXPERIMENTAL PRACTICE Experimentally, DNA or RNA is sheared into fragments of about 300—600 nucleotides in size, generally by some type of controlled, high—speed blending. These small fragments are then denatured by boiling briefly and quick—cooled to prevent any renaturation. The dissociated fragments are then incubated in a controlled temperature bath at standard salt concentrations (.18 N cation), and at intervals, samples are removed and the percent of nucleic acid remaining single—stranded Various techniques can be used to determine the amount of DNA or RNA which is single—stranded. Various nucleases are available which digest either single—stranded DNA or RNA, but will not attack double—— stranded DNA or RNA, or DNA:RNA hybrid molecules. It is convenient to start with radioactively labeled nucleic acids, and determine the amount of radioactivity which is either digested or remains resistant as a function of time. Optical methods can also be used. A second method involves the use of hydroxyapatite (hydrated calcium phosphate) columns or some other cationic resin. Under appropriate salt conditions, double—stranded DNA or RNA, or DNA:RNA hybrids will be retained on hydroxyapatite or the resins, while the single—stranded nucleic acids pass through. Higher salt solutions are used to wash the bound, double—strand molecules from the column and collected separately. Detection again is by radioactivity or optical methods. Finally, since single—stranded DNA absorbs more UV light than double— stranded DNA (the hyperchromic effect), the relative proportions of double and single stranded DNA can be monitored by observing the decrease in absorbance of 260 nm light as the reaction proceeds. The use of single—stranded, highly radioactive RNA or DNA "probes" has become important, since it allows the quantitation of sequences. EXPERIMENTAL RESULTS I. C°t% is proportional to the complexity of the DNA population being studied. In Figure 2 (adapted from Britten and Kohne, Science 161:529, 1968) the reassociation kinetics of a number of nucleic acid molecules from different sources, each done separately, are plotted together. Nucleotide pairs in genome, determined independently 1* lo ‘0’ )u’ i 10‘ :05? 13‘ ha’ 10' ' 'U G) H .‘3 U 0 m (0 N O “a: I: O I: 0 N I— [L a» 4 4 - to M) w" m" 10'2 0.: I to too woo 3' C°t(mom xsecflned Figure 2. Reassociation of nucleic acids, sheared to BOO—nucleotide fragments, from various sources It is obvious that the Cot%’s for each nucleic acid are different, and that they are proportional to the complexity of the nucleic acid in nucleotide pairs. If the reaction has been carried out under standard conditions, then Cot% x 5 x 105 L x nuc. pairs/mole*sec equals the complexity, in nucleotide pairs, of the DNA molecule. For all prokaryotic organisms, the complexity of their genome is equal to the molecular weight. From Figure 2, the Cot% of E. coli DNA is 10 mole sec/L, which when multiplied by the above constant, yields 5 x 106 nucleotide pairs as the complexity of the E. coli genome, which is in excellent agreement with the molecular weight determined by other techniques. (By electron microscopic measurements, E. coli DNA has a molecular weight of 2.5 x 109 daltons, and thus contains about 5 x 106 base pairs, since one base pair = 600 daltons.) II. The Cot curves of eukaryotic organisms are unfortunately not simple. Figure 3 shows an idealized Cot curve from a eukaryotic organism such as the mouse. Clearly, the mouse genome contains sequences of different complexities. 100 80 Z Remaining Single— 60 Stranded 40 20 0 10-3 104- 10-1 1 10 102 103 104 105 Cot (mole sec/liter) There are three separate classes of DNA sequences present in the DNA of this hypothetical mouse. By noticing the C°t%’s and the plateaus between species and by remembering that the hybridization reaction is 80% complete within two logs of concentration, we can immediately determine that 10% of the DNA renatures very rapidly, with an apparent (%t% of 104, 20% of the DNA renatures more slowly, with an apparent Cot of 10*, and that 70% of the mouse DNA has a Cot% of about 7 x 103 The observed Cot% values must be corrected for the differences in concentration. Since the Cot% of the fastest renaturing component I is 10*, but it is only 10% of the total mixture, it has a “real" Cot% of 104, because if it were present alone, there would be ten times more DNA, and thus it would hybridize ten times faster. Similarly, the Cot% “pure” of the other two species are really 2 x 10‘2 (.2 x 104) and 5 x lO‘3 (.7 X 7 X 10*), ie, in a mixture, the observed Cot% x the fraction that the DNA species is of the total gives the Cot% of a pure preparation of that DNA species. Once the C°t% “pure" is known for a DNA species, then it is easy to calculate the molecular complexity of that DNA molecule. And, if we know the total molecular weight of the genome (2 X 1012 daltons or 3.3 X 109 base pairs for the mouse), it is easy to determine the number of copies of each sequence in the genome. So, for the mouse whose Cot curve is presented above, we calculate: Calculated # of Apparent Fraction of t% Complexity copies Cot% Total Genome in in genome basepairs 10‘3 10% (3.3 X 108 10'4 50 base pairs) 10‘1 20% (6.6 x 108 2 x 10'2 104 ~7 x 104 base pairs) I II III 7 X 103 70% (2.3 x 109 5 X 103 2.5 x 109 ~1 base pairs) That is, the mouse, and in general, all eukaryotic organisms have several classes of DNA sequences in their genomes. A fraction of the DNA is made up of highly repetitive sequences; short, 4—500 nucleotide long sequences that are repeated many, many times in the genome. Another fraction of the genome is comprised of DNA sequences that are moderately repetitious; longer, 103 — 104 base pairs in length and repeated several thousand to hundreds of thousands of times. Finally, a major fraction of the genome is made up of sequences present only once in the entire genome, the "unique" DNA of the organism. Note that this technique says nothing about the arrangement of these sequences in the DNA, nor does it say all the short, highly repetitive, or moderately repetitive fragments are identical in sequence —— just that the sequences are the same length. CURRENT USAGE Information derived by these methods led to an understanding of chromosome structure. Nowadays, these techniques are rarely used for this purpose. However, calculations of hybridization kinetics are widely used in various blotting methods, in quantitative measurement of mRNA levels, and in various PCR procedures. In many of these processes, one of the two strands of the DNA or RNA is not present in equal concentration and quantitative information is derived by measuring the amount of probe “driven” into a hybrid. What is determined is not the sequence complexity, which is known, but the time and concentration of DNA or RNA necessary for the hybridization to go to completion. We will expand on this in later material. ...
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