Final answer:
A restriction enzyme that recognizes a 6-base pair sequence in a DNA with six different bases present at equal frequency is expected to cleave every 1 in 46,656 base pairs, assuming a random distribution of the bases.
Step-by-step explanation:
The student is inquiring about the frequency at which a restriction enzyme with a 6-base recognition site would be expected to cleave DNA if the DNA has six different bases present (A, G, C, T, X, and Y) at random and at equal frequency. With six different bases and equal frequency of each base, there are a total of 66 (46,656) different possible combinations for any given 6-base sequence.
Considering that a restriction enzyme recognizes a specific 6-base pair sequence, and assuming that there is no particular bias in the sequence distribution of DNA, the enzyme would be expected to cleave every 1 in 46,656 base pairs, on average, under the assumption of random distribution of these six bases. However, this is a simplified calculation that doesn't take into account potential sequence biases or other biological complexities.
In reality, DNA-protein interactions, as well as chromatin structure, can influence the accessibility and cleavage frequency of restriction enzymes. For example, certain DNA sequences may be more tightly bound by histones, making them less accessible to enzymes, while others might be in open conformation and more easily cleaved. Additionally, the true distribution of bases and sequences in an organism's genome is not perfectly random, which can affect the cleavage frequency.