In view of the above, reassociation of any particular DNA can be described by the rate constant K (in units, nucleotide moles liter-1
) or in the form of its reciprocal i.e. Cot1/2
(in nucleotides moles x sec/liter). It is obvious that reassociation depends on concentration (Co) and time of incubation (t), i.e. Cot
as described above. A higher Cot1/2
means slower reaction and lower Cot1/2
meant faster reaction. This parameter Cot1/2
varies for different organisms and also for different DNA fractions from the same organism.
of a reaction will be directly proportional to the DNA content, if there is no repetition of sequences within a single genome. For instance, if the bacterial genome has 0.004 pg DNA (4.2 x 106
bp) and if Co is 12 pg per unit volume, it will have 3000 copies of each DNA sequence. Contrary to this if a eukaryotic genome is 4 pg, then there will be only 3 copies in 12 pg i.e. concentration is 1000 times less than in bacterial genome, provided there is no repetition of sequences in the genome of 4 pg. In view of this, Cot1/2
will indicate the length of all the different sequences (represented only once) in a genome, which will be less than the length of total DNA in a genome when there is repetition. This will be described as kinetic complexity of the genome, and can be determined by knowing Cot1/2
and comparing it with Cot1/2
of a genome with known complexity (e.g. E. coli,
which has a genome = .004 pg DNA = 4.2 x 106
bp with Cot1/2
= 9). The following relationship is used for calculating the kinetic complexity.
Figure 28.5 shows reassociation curves of several DNAs, which are all kinetically pure, meaning that DNA is homogeneous from the point of view of complexity. In eukaryotes, the genome contains more than one such pure components as shown in the comparison made in Figure 28.6. In this figure calf DNA has been shown to have two components, each with its characteristic Cot1/2
However, more than two such components can also be found as shown for wheat in Figure 28.7. Proportion of each component is determined by using the formula 1 - C/C0
(C = concentration at t1/2
for the corresponding component). From this proportion, chemical complexity of the component can be determined, if chemical complexity of the genome is known. For instance, if genome size is 12 x 108
bp and the component represents 25% of the genome, then the chemical complexity
of this component is 3 x 108
bp. If kinetic complexity is known from the equation given earlier
, then repetition frequency (f)
of repetitive DNA component can be determined using the following formula :
The results of an exercise to determine the size of three different DNA components and their Cot1/2 values along with kinetic complexity and the repetition frequencies are shown in Figure 28.8.
|Fig. 28.7. Reassociation pattern of wheat genomic DNA showing more then two homogeneous components (fast, intermediate and slow reassociation)
|Fig. 28.8. Reassociation kinetics of eukaryotic DNA, showing calculation of complexity and repetition frequency.