"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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answer: after carbon dioxide enters the leaf through stomata it moves into the mesophyll cells where photosynthesis occurs and glucose is constructed. respiration occurs when glucose (sugar produced during photosynthesis) combines with oxygen to produce useable cellular energy.

explanation: