Restriction enzyme math question

[BassMaster]

Homework Statement

Homework Statement

I know this is a biology question but this is the most busy forum so I decided to put it here:

I'll make this simple. A restriction enzyme reads a DNA strand containing nucleotide base pairs. It cuts the strand after it reads a sequence of 6 base pairs - AAGCCT. Of course the probablity of the enzyme reading the correct base is 1/4. And it needs 6 of them.

So what is the average length of a cut strand? I'm thinking you do (1/4)^6 and this will give you something.. but what?

Thanks for your help in advance...

Homework Equations

The Attempt at a Solution

 

[greisen]

(1/4)^6 gives you the probability that your enzyme will read the string(tag) correctly and you know the length of your string so combine these number will give you an average length of a cut given that it consists of only this sequence - this is how I understand you.

 

[Blueshift5]

I am happy to help with this question. To find the average length of a cut strand, we need to first consider the probability of the enzyme reading the correct sequence of 6 base pairs (AAGCCT). As mentioned, the probability of each base pair being read correctly is 1/4, so the probability of the entire sequence being read correctly is (1/4)^6.

Next, we need to consider the length of the DNA strand being cut. Assuming the DNA strand is infinitely long, the average length of a cut strand would be equal to the length of the sequence being read (6 base pairs) divided by the probability of the enzyme reading the correct sequence. This can be represented by the equation:

Average length = (length of sequence)/(probability of correct reading)

= 6/(1/4)^6

= 6/0.000244

= 24587 base pairs

Therefore, the average length of a cut strand would be 24587 base pairs. However, it is important to note that this is based on the assumption of an infinitely long DNA strand. In reality, the length of a cut strand may vary depending on the length of the DNA being cut and the specific location of the restriction enzyme site. I hope this helps!