Question

Suppose you have a single intact E. coli chromosome (4.6 x 106 bp), how many fragments would you expect to obtain with TseI? Please show your calculations

Answer 1

The recognition for the restriction enzyme TseI is 5'-GCWGC-3'. Now we are given the information that the template DNA is 4.6 x 10⁶ base pairs long. Since we are not given the sequence, we have to assume that each position on the DNA sequence can be an A, T, G ,or a C, and that each of these are equally probable. This means that the probability of a nucleotide in the first position of any given sequence being A is 1/4. This is the same for the other nucleotides. So the probability of a stretch of DNA in the template sequence being the same as the recognition sequence for the enzyme TseI is the probability of occurrence of the sequence GCWGC, where W is either and A or a T.

Now probability of the first site being G = 1/4

Probability of second site being C = 1/4

Probability of third site being an A or a T = P(A) + P(T) = 1/4 + 1/4 = 1/2

Probability of the fourth site being G = 1/4, and lastly

Probably of the fifth site being C = 1/4.

So the overall probability of occurrence of the sequence GCWGC is the product the individual probabilities of each site being filled by the correct nucleotide, i.e., 1/4 x 1/4 x 1/2 x 1/4 x 1/4 = 1/512 .

The total number of expected restriction sites on the E. coli chromosome is the total number of base pairs multiplied with the probability of finding a sequence of nucleotides that correspond to the TseI restriction sequence.

So total number of cuts = 4600000 x 1/512 = 4600000/512 = 8984 restriction sites. Since the E. coli chromosome is circular, this many cuts make the same number of fragments, i.e. 8984 fragments. (If it was linear, this many cuts would lead to 8985 fragments).

I hope this helps :)