Not all restriction enzymes necessarily cleave DNA at rotationally symmetric sites. For example, the recognition site of the restriction enzyme Bst XI is 5' CCA^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

The frequency of Bst XI recognition sites in the human genome is once every 65536 bases in the human genome.

The frequency of cutting in a random DNA sequence for a given restriction enzyme, F = once per every 4^n nucleotide bases.

Given the recognition site of the restriction enzyme Bst XI as 5' CCA^NTGG,

Therefore, the number of nucleotide bases, n in the recognition sequence of Bst XI is eight nucleotide bases i.e. CCA

Assuming that equal amounts of all four nucleotides exist in the human genome, and that its base sequence is random:

Using the formula,  frequency = 4^n

for Bst XI, n = 8

Frequency = once per 4⁸ nucleotide bases

Frequency = once per 65536 nucleotide bases

Therefore, the frequency of Bst XI recognition sites in the human genome is once every 65536 bases in the human genome.

Learn more about restriction enzymes at: link

The frequency to be expected in order to find Bst XI identification sites in the human genome would be:

- Bst XI will cleave the DNA on average once in 4096 base pairs: one out of 4096 DNA fragments will have compatible ends.

1. Number of bases = 6so, 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.

Learn more about "bases" here:

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