Question

Given a DNA molecule with 10,000bp, in theory how many times the “GTCCGGAC” sequence is expected? And how about “ATWAT” sequence? (Where W= A or T)

Answer 1

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.

#1. Average restriction fragment size for an enzyme recognizing 8 bp resection site

Probability of finding an 8bp restriction site = (¼)⁸ = 1/ 65536

That is, there is one 8bp long restriction site per 65536 bp.

# No. of 8 bp long restriction sites =

Size of DNA molecules / Average length of restriction fragment

                                    = 10000 bp / 65536 bp = 0.15 = 0 (nearest, lower whole number)

Therefore, the number of 8 bp long restriction sites = 0   

#2. Since W may be either A or T, the probability of finding a W =

                                                                                    Probability of A + probability of T

                                                                                    = ¼ + ¼ = ½

# Now, probability of finding ATWAT = ¼ x ¼ x ½ x ¼ x ¼ = 1 / 512

No. of ATWAT restriction sites =

Size of DNA molecules / Average length of restriction fragment

                                    = 10000 bp / 512 bp = 19.53 = 0 (nearest, lower whole number)

Therefore, the number of restriction sites = 19