Problem on Probabilty of a restriction enzyme cutting a random DNA sequence

Question

I think its a silly question to ask here. When I came to this site all I could see were the questions which asked detailed explanation behind a phenomenon and reasoning was there at first place. I am bringing quite math here!!!

First, I am a beginner.

Second, I would like to mention some problems..

1. Genomic DNA is digested with AIU I, a resrtiction enzyme which is a four base pair cutter. What is the frequency with which it will cut a DNA assuming a random distribution of bases in genome?

2. Eco RI and Rsa I restriction endonucleases require 6 bp and 4 bp sequences respectively for cleavage. In a 10 kb DNA fragment how many probable cleavage sites are present for these enzymes?

These questions were solved by my teacher using some probability results or rules. I would like to mention it here.. [you know already...!!! ;-)]

(1/4)^n *10,000 or simply (1/4)^n. By putting n = number of base pair cut by enzyme.

I get the frequency with which enzyme will cut a DNA assuming a random distribution of bases in genome by using (1/4)^n and I get probable cleavage sites are present for Eco RI and Rsa I in 10 kb DNA fragment.

But why do I do that? I mean why that formula? How to understand how it was derived? Can you explain me that?

Please suggest some good books containing concepts and numerical problems like these so that I can work on it.. [Please let me know your source]. Thank you.

Answer 1

EcoRI recognizes, binds to, and cleaves the hexamer 5'-GAATTC-3'. In a completely randomized DNA sequence, with 25 % A's, 25 % C's, 25 % G's, and 25 % T's we would expect to randomly discover the sequence GAATTC, once every 4,096 base pairs, on average. The actual distribution of EcoRI fragments (a fragment where each end contains a cleaved EcoRI site) in our random genome will vary and yield a normal distribution, or curve, with some fragments longer and some shorter. The mean value should be 4,096.

The expectation of 1 site per 4,096 bp is indeed (1/4)^6.

1/4 = 25 %

6 = the number of contiguous base pairs in the recognition site.

In a model like this it is important to keep in mind that the probability of one base occurring in the sequence is completely independent of the sequences before that base, and the sequences after that base. In other words, for each and every position the probability of getting a base that will contribute to an EcoRI site is 25 %, or 1 out of 4 = 1/4.

75% of the time, the next base will not contribute to an EcoRI site (and a quarter of the time it will).

Let's start with your 10 kb fragment, and go through from one end to the other, and mark every time a 'G' occurs. Theoretically those could all be the beginning of EcoRI sites, right? There will be approximately 2500 such 'G's. Now to match our pattern, we need an 'A' in the second position, so let's mark all those sites that contain the dinucleotide 'GA', there will be about 2500 x 0.25 = 625 of them. The next match we need is position 3, another 'A', so in a similar fashion, let's go through the positions of all those GA dinucleotides, and just mark the ones that have an 'A' at the next position. So that will be about 625 x 0.25, or approximately 156 'GAA's distrubuted randomly along the 10 kb fragment.

If we continue, essentially filtering out more and more of those 'G' sites in our initial list then we will find, on average, that there are

39 sites with GAAT

10 sites with GAATT

2 sites with GAATTC

If you don't believe me just pick a sequence and try it yourself.

10,000/4,096 ~ 2.5