Ans. DNA consists of 4 types of bases (A, C, T, G). A base at each position has the same probability of (1/4) at all positions- that is, probability of finding a “A” in [AAGCTT] at position 1 (first nucleotide) is (1/4); the same is also true for position 2 ; or any other base at any position. It is because the identity of a base at one position does not affect the identity of another base at different position.
So, the probability of finding 1 (any one) base out of total 4 type of base = 1/4.
The probability of finding a restriction site in a DNA molecule = (¼)n,
where n is the number of nucleotides in the restriction site.
Now,
#1. The probability of finding HaeII (n =6) restriction site = (1/4)6 = 1/ 4096 i.e. there is 1 restriction site every 4096 base pairs for dsDNA molecule.
Therefore, average size of fragments = 4096 base pairs
#2. No. of restriction fragments = Chromosome size / size of restriction fragment
= (2.13 x 106 bp) / 4096 bp
= 520
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