Its a passage based question, but here is just the question:
Assuming DNA has an equal proportion of all bases, how many fragments would a restriction enzyme that recognized 4 base pairs produce from a DNA fragment that is 10,000 base pairs long?
Also the question is taken from BR bio, section 10, passage 10, question 59
Ok, so the restriction enzyme recognizes 4 base pairs. You do 4^4 because there are 4 possibilities (4 different bases in any given nucleic acid polymer) that can be at any of the 4 positions of the sequence recognized by the enzyme.
So in DNA you have A, T, C, and G as bases and in a 4 bp sequence you have positions 1, 2, 3, 4.
Say the enzyme recognized a 6 bp sequence, you would do 4^6 because there are still only 4 bases (ATCG) but now you have 6 possible positions.
The number you get from the 4^x calculation is the theoretical frequency of the recognizable palindromic sequence. The longer the sequence is, the rarer it will be found. This is why restriction enzymes that recognize smaller sequences cleave more fragments than restriction enzymes that recognize longer sequences.
Anyway, back on track. This problem tells you the restriction enzyme (RE) recognizes a 4 bp sequence. So you know that the RE will only cleave when it finds the exact 4bp sequence it is looking for. If I told you the sequence was, for example, CCGG but the DNA I show to the RE is only AT base pairs, there would be no fragments cleaved because a CCGG sequence would not exist at any point along the strand. However, the problem tells you to assume equal amounts of all bases, so you have to do a little more work.
