Question

Not all restriction enzymes necessarily cleave DNA at rotationally symmetric sites. For example, the recognition site of the restriction enzyme Bst XI is 5' CCANNNNN^NTGG, where the symbol ^ represents the cleavage site and N means any nucleotide. Make the simplifying assumptions that equal amouts of all four nucleotides exist in the human genome, and that its base sequence is random. How frequently would you expect to find Bst XI recognition sites in the human genome

Answer 1

Answer:

The correct answer is - Bst XI will cleave the DNA on average once in 4096 base pairs: one out 4096 DNA fragments will have compatible ends

Explanation:

In the question the sequence of the Bst XI is given:

5' CCANNNNN'NTGG 3' , here ' sign is the site of restriction

It is also given that N = any nucleotide and equal amounts of nucleotide present in the human genome.

Probability of anyone nucleotide = 1/4 (G = A = T = C = 1/4 )

Nucleotide is equal so we can neglect it as its probability is 1.

Number of bases = 6

so, the probability of occurrence of the site with 6 different bases

= (1/4)^6

= 1/4096

Thus, It means the enzyme Bst XI will cut DNA at once in every 4096 bases.