DNA can undergo reversible strand separation.

DNA strand separation (denaturation)

when ? how?

Natural:

Making RNA from DNAUnwinding by proteinsUnwinding by DNA topology

Experimental:

HeatingLow ionic strengthDisruption of hydrogen bondsurea, formamide, alkaline pH

Denaturation- renaturationDeanturation

DNA circular ?

Circular DNA & denaturation a problem, why ?

Renaturation -

Only by complementary,

Time dependent

No competition if non-related seq.

Analysis of DNA denaturation

Change in light absorption (UV) 260 nmo. 4C in Tm with every 1% GCFactors that affect Tm = Base compositionpHSalt

High temperature and neutral pH = ????

Heat denatured DNA ssDNA quick chill

dsDNA ssDNA ssDNA Alkaline pH neutral pH

Heat denatured renatured slow cooling

Renaturation: Two step reaction

Rate of renaturation

Renaturation kinetics and complexity

Rate of reassociation reaction depends on variety of sequences

Genome size, Gene number

Larger genome more genes.

Same gene appear many times ??? Different ??

What is the complexity of a genome ???

Kinetic studies

Rate of reassociation

-kC2 = dC/dt

C = Concentration of single stranded DNA at time t in seconds

K = reassociation constant

Initial concentration of DNA = C0

C = 1C0 1+ k C0 t

The rate of reassociation is controlled by C0 and tC = 1 = 1 C0 2 1+ k C0 t

50% of the reaction completed , Cot

Cot 1/2 = 1/k (k= reassociation rate constant)

DNA Renaturation Kinetics

The rate at which heat-denatured DNA sequences in solution will renature is dependent on DNA concentration, reassociation temperature, cation concentration, and viscosity (usually not a factor if the DNA is free of contaminants).

Cot = DNA conc. (mol/L) x renaturation time in sec. x a buffer factor that accounts for the effect of cations on the speed of renaturation

Greater Cot indicates slower reaction.

The same Cot value can be reached in a variety of ways For example:Nucleotide concentration = 0.050 MRenaturation time = X 344.000 secBuffer factor, 0.5 M SPB = 5.820 Cot 100.000Nucleotide concentration = 0.002 MRenaturation time = X 50,000.000 secBuffer factor, 0.12 M SPB = 1.000 Cot 100.000 The total amount of reassociation for a given Cot value does not change.

The rate at which a particular sequence will reassociate is proportional to the number of times it is found in the genome.

Given enough time, nearly all of the DNA in a heat-denatured DNA sample will reassociate.

Cot curveFraction of the DNA that is reassociated Vs log Cot

1-C/C0 Vs log cot

Rate of reassociation and DNA lengthC0t amount of the DNA in the genomeRate of reassociation is inverselyproportional to the length of reassociating DNA

Reassociation depends on the presence of complementary sequences

Eukaryotic DNA: bacterial DNA reassociation

Frequency of any sequence to be +nt in eukaryotic DNA is several hundred folds less than bacterial.

Therefore, concentration of eukaryotic DNA has to be in several hundred fold excess.

Rate of reassociation and DNA length Cot is directly related to the amount ofDNA in the genome

A fixed amount of DNA, 12 pg (C0 ) = 3000 E. coli (o.oo4 pg size) but only 4 eukarytoic genomes ( 3 pg size)

X 750 fold in DNA concentration for specific sequence

Cot indicates the total length (bp) of different sequences = complexity

C0t of the fast component = 0.25 x.0013Complexity of the fast component = ??Complexity of the genome = Rate of reassociation and Comlexity of the genomeA typical mammalian C0t curve has three plateaus.

When the same experiment is carried out using DNA purified from a complex eukaryote, such as human, then we do not see a simple sigmoidal curve. Instead we see a curve which is the sum of the reannealings of many different components.

Cot analysis

Cot analysis was first developed and utilized in the mid 1960s by Roy Britten, Eric Davidson, and associates.

It is based upon the principles of DNA renaturation kinetics.

The steps in aCot analysis

Tissue from the organism of interest is placed in an antioxidant medium and homogenized in a blender. The homogenate is filtered, plastids are preferentially lysed, and nuclei are pelleted by centrifugation.

(B) The pellet should be free of contaminating organelles (as determined by phase-contrast microscopy). DNA is isolated from purified nuclei using phenol/chloroform extractions coupled with proteinase and RNase digestions.

(C) The DNA is cut into pieces of about 450 bp by sonication or high-speed blending. (D) Fragment size is checked by agarose gel electrophoresis. Sheared DNA is precipitated, and aliquots of the DNA are dissolved in 0.03 M, 0.12 M, and 0.50 M sodium phosphate buffer (SPB) to produce solutions of known concentrations.(E) The DNA solutions are distributed into glass microcapillary tubes or glass ampoules so that each tube/ampoule contains 100 g of DNA. The ends of the tubes/ampoules are sealed.

Sonication Check the average Dissolve in different salt solutions fragment size, ppt Dispense fixed amount in glass ampules, seal

(F) One of the tubes containing a known concentration of DNA is selected.(G) The tube is placed in boiling water to denature DNA duplexes.(H) The tube then placed in a water bath set at the samples criterion (i.e., Tm - 25C). Renaturation is allowed to occur until the sample reaches a specific Cot value [Cot value = the product of the sample's nucleotide concentration (moles of nucleotides per liter), its reassociation time in seconds, and an appropriate buffer factor based upon cation concentration].Once the sample has reached the desired Cot value, the end is broken off of the tube/ampoule, and the solution is blown into a 100-fold excess of 0.03 M SPB.

0.03 M SPB+90C Tm-25C

(J) The diluted sample quickly is loaded onto a hydroxyapatite (HAP) column equilibrated with 0.03 M SPB. At this buffer concentration, all DNA binds to the HAP.(K) Once all of the solution has entered the HAP, 0.12 M SPB is added causing single-stranded DNA (ssDNA) to elute. Eluant containing ssDNA is collected in a graduated polypropylene tube.(L) After the ssDNA had been collected, 0.50 M SPB is added to the column to elute double- stranded DNA (dsDNA).

(M) The volumes of the ssDNA eluant and the dsDNA eluant are determined.(N) Exactly 0.9 ml of the centrifuged ssDNA eluant is mixed with 0.1 ml of aqueous 10 N KOH to denature any DNA duplexes. A sample of the dsDNA eluant is likewise denatured.(O) The A260 values (adjusted for light scatter at 320 nm) of the ssDNA/KOH mixture and the dsDNA/KOH mixture are determined. For a particular Cot value, the percentage of ssDNA (% ssDNA) is calculated as follows: [(Vss x Ass) x 100] [(Vss x Ass) + (Vds x Ads)] = % ssDNAwhere Vss = total volume of single-strand fraction, Vds = total volume of double-strand fraction, Ass = A260 (adjusted for light scatter) for the KOH-denatured single-strand fraction, and Ads = A260 (adjusted for light scatter) for the KOH denatured double-strand fraction.

P) The logarithms of Cot values ranging from essentially no renaturation to nearly complete renaturation are plotted against corresponding % ssDNA values to yield Cot points. A graph of Cot points ranging from little or no reassociation until reassociation approaches completion is called a Cot curve. Because a DNA sequence reassociates at a rate that is directly proportional to the number of times it occurs in the genome, sequences that occur more than once in a genome (repetitive DNA) reassociate at lower Cot values than sequences found only once per genome (single-copy DNA). From an analysis of a Cot curve one can determine genome size, relative proportions of single-copy and repetitive sequences, the fraction of the genome occupied by each frequency component, and the mean kinetic complexity of the sequences in each frequency component.

The reassociation kinetics of this eukaryotic DNA show three types of components (indicated by the shaded areas). The arrows identify the Cot1/2 values for each component. In the calculation of the complexity, E. coli DNA was assumed to have a Cot1/2 = 4. Thus the compexity of the slow component (unique DNA with a repetition frequency of 1) = (0.45) (630) (4.2 x 106 bp)/4 = 2.98 x 108 bp. The total size of the genome = 2.98 x 108 bp/0.45 = 6.6 x 108 bp.