How often would a restriction enzyme such as Not I, which has 8 nucleotides in its recognition sites 5’-GCGGCCGC-3’, cleave DNA on average? Would it cleave a species DNA more or less often if the DNA from that species were 70% A-T rich?

To determine how often a restriction enzyme like Not I would cleave DNA on average, you need to consider the frequency of its recognition site within the DNA sequence.

The recognition site for Not I is 5’-GCGGCCGC-3’, which contains eight nucleotides. This means that every time the sequence 5’-GCGGCCGC-3’ appears in the DNA, Not I has the potential to cleave it.

To estimate the occurrence of this recognition site, you need to consider the probability of each nucleotide appearing at a specific position. In a random DNA sequence, the occurrence of each nucleotide is expected to be approximately equal. Therefore, the probability of each nucleotide will be 0.25 (1/4).

Since the recognition site has eight nucleotides, the probability of finding this specific sequence in a random DNA sequence would be (0.25)^8 = 0.000244141 or approximately 0.02%.

However, it is important to note that this calculation assumes a random DNA sequence and does not take into account specific probabilities of nucleotide distributions within various species' genomes.

Regarding the DNA from a species that is 70% A-T rich, it means that 70% of the nucleotides in its genome are adenine (A) or thymine (T). This would lower the probability of finding the Not I recognition site (which is GCGGCCGC) because it contains a higher proportion of guanine (G) and cytosine (C) nucleotides.

As a result, if the DNA from a species is 70% A-T rich, it would cleave less often with the Not I restriction enzyme compared to a species with a more balanced GC content.

To determine how often a restriction enzyme such as Not I would cleave DNA on average, we need to consider the recognition site and the frequency of occurrence of that site in the DNA sequence.

The recognition site for Not I is 5’-GCGGCCGC-3’, which contains 8 nucleotides. To calculate the average frequency of occurrence, we need to know the size of the DNA sequence and the probability of each nucleotide occurring at each position.

For simplicity, let's assume a random DNA sequence. In this scenario, each nucleotide has an equal probability of occurring at each position, which is 25% (since there are four possible nucleotides - A, T, G, C).

For an 8-nucleotide recognition site, each nucleotide has a probability of 0.25 to appear at each position. So, the probability of finding a specific 8-nucleotide sequence randomly is 0.25^8, which is approximately 0.000244.

The average frequency can be calculated by dividing the length of the DNA sequence by the length of the recognition site:

Average frequency = Length of DNA sequence / Length of recognition site

Now, let's consider the second part of the question: whether Not I would cleave a species DNA more or less often if the DNA from that species were 70% A-T rich.

If the DNA from a species is 70% A-T rich, it means that 70% of the nucleotides in the DNA sequence are either A or T. As the recognition site 5’-GCGGCCGC-3’ contains G and C nucleotides, which are not A or T, the likelihood of finding this specific recognition site would decrease in an A-T rich DNA.

Therefore, if the DNA from a species is 70% A-T rich, Not I would likely cleave the DNA less often compared to a DNA sequence with a balanced nucleotide composition.

Remember that this calculation assumes a random DNA sequence and cannot account for other factors like the presence of multiple recognition sites or sequence-dependent preferences of restriction enzymes.