Chemical complexity vs sequence (kinetic) complexity
(C = concentration of single stranded DNA at time t)
(K = reassociation rate constt.)
By integrating the above equation (consult a book on Integral Calculus, if necessary), between the limits of C0 at t = 0 and C at time t1/2 (when reaction is half complete), we get the following expression.
Fig. 28.5. Reassociation patterns of double stranded DNA from various sources. All curves are typical S-shaped indicating homogeneity, but they are displaced to different values to Cot due to genome size.
Fig. 28.6. Reassociation patterns of bacterial and calf DNA. Note that the reaction on calf DNA takes place in two stages, one early representing repetitive DNA and the other representing single copy DNA. Midpoints of these two stages are indicated by dotted vertical lines, separated by a factor of 1000,000 (for details see text), (redrawn from Sci. Amer. 222:. April, 1970).
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.
The Cot1/2 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 value.
then repetition frequency (f) of repetitive DNA component can be determined using the following formula :
Fig. 28.5. Reassociation patterns of double stranded DNA from various sources. All curves are typical S-shaped indicating homogeneity, but they are displaced to different values to Cot due to genome size.
Fig. 28.6. Reassociation patterns of bacterial and calf DNA. Note that the reaction on calf DNA takes place in two stages, one early representing repetitive DNA and the other representing single copy DNA. Midpoints of these two stages are indicated by dotted vertical lines, separated by a factor of 1000,000 (for details see text), (redrawn from Sci. Amer. 222:. April, 1970).
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.