Genomic dna is digested with alu 1 which is a four base pair cutter. What is the frequency with wich it will cut the dna assuming a normal distribution of bases.
(this is not a homework)
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$\begingroup$ I guess it is the probability of getting the 4 specific base pair in a row which is 1/4 *1/4*1/4*1/4. $\endgroup$– user7635691Commented Oct 28, 2017 at 3:59
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$\begingroup$ You can teach yourself to answer this type of question using this resource which I wrote some time ago: mvls.gla.ac.uk/Teaching/SLS-FTB/ren.html $\endgroup$– DavidCommented Oct 28, 2017 at 13:14
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Welcome to Biology.SE. I think this question would be better on-topic on math.SE.
Assuming the chromosome is either 4 bases long or infinitely long, then the answer is simply $\left(\frac{1}{4}\right)^4$. However, chromosomes are not infinitely long so the result should be (negligibly) lower than $\left(\frac{1}{4}\right)^4$. If chromosomes are only 3 nucleotide long, then of course, the probability will be $0$. If it is 4 bases long the probability is $\left(\frac{1}{4}\right)^4$. If it is 5 bases long the probability is $\frac{2}{4^5} = \frac{1}{2}\left(\frac{1}{4}\right)^4$.
Consider this numerical approximation, to ensure the result is correct with this simple one liner in R
n = 1e7
length(gregexpr('abcd', paste(letters[sample(1:4,n,replace=TRUE)], collapse=""), perl=TRUE)[[1]])/n
[1] 0.0039041
which is really close to $\left(\frac{1}{4}\right)^4 ≈ 0.00390625$.
So, in a genome of 3 giga bases, that represents an expected $\frac{3\cdot 10^9}{4^4} = 11,718,750$ (or ~11 millions) such locations where Alu 1 can cut.
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$\begingroup$ Remi.b what is the boundary effect. Btw thank you $\endgroup$ Commented Oct 28, 2017 at 4:04
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$\begingroup$ @user7635691 I edited the answer to add some info on that. $\endgroup$– Remi.bCommented Oct 28, 2017 at 15:21