"Assuming a random sequence of DNA, calculate the probability of cleavage by the restriction enzyme HindIII which has the recognition sequence: A/AGCTT. If the E. coli genome is 46 kb in length, how many HindIII site would you predict to find in this genome

45 times

Explanation:

There are 4 nucleotides. The probability of finding one particular nucleotide (e.g. A) is 1/4.

The probability of finding a sequence of 2 nucleotides (e.g. AG) is 1/4 x 1/4 = 1/16.

The probability of finding a sequence of 3 nucleotides (e.g. AGC) is 1/4 x 1/4 x 1/4 = 1/64.

The pattern here is that the probability can be calculated by the probability of finding any particular nucleotide to the power of the length of the sequence. For a 2 nucleotide sequence, the probability is 1/(4)². For a 3 nucleotide sequence, the probability is 1/(4)³, and so on.

This is a 5 nucleotide recognition sequence, so the probability is 1/(4)⁵ = 1/1024.

If the E coli genome is 46 kb, that means there are 46,000 nucleotides. We expect to find this sequence once every 1024 bases:

46,000 ÷ 1024 = 44.92

Therefore, we expect to find this sequence approximately 45 times in the E coli genome.

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