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I have seen that the formula to calculate the number of times a given sequence of nucleotides occur in a target genome is derived from that to calculate the expected frequency of restriction sites:

a = (g/2)^G+C × ((1-g)/2)^A+T,

where:

a = probability
g = G+C content of the target genome
C+G = number of G and C in the stretch
A+T = number of A and T in the stretch.

An example was given for the mitochondrial genome and the restriction site for EcoRI (GAATTC). I calculated, based on the example:

a = (0.44/2)^2 × (1-0.44)^4 = 0.0005

given that the length m of the mitochndrial genome is 16 000 bp, the number of occurrences is:

n = am = 0.0005 × 16 000 = 4.92 (against 4.80 reported in the paper).

I then tried to calculate the occurrence of a primer based targeting E. coli: GTGTCCATTTATACGGACATCCATG. The GC content of E. coli is 50.8%, thus:

a = (0.58/2)^11 × (0.42)^14 = 1.22×10^-6 * 3.24×10^-10 = 3.95×10^-16

and the number of occurrences is:

n = 3.95×10^-16 × 16*10^6 = 6.32×10-9

Looks to me, that the primer should not occur at all in the E. coli genome.

Are the formula and its application correct? Also, I don't like the fact that one needs to know the CG content beforehand (although, one can assume 50% content) and an exponent at the 11th power; is there an alternative formula?

Thank you.

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Strictly speaking the question is from the realm of probability theory. If the probability of a nucleotide $X$ in the genome is $p_X$ ($\sum_x p_X=1$), then the probability to encounter a particular sequence, where nucleotide $X$ occurs $n_X$ times, is $$ \prod_X p_X^{n_X} = p_A^{n_A}p_C^{n_C}p_G^{n_G}p_T^{n_T}. $$ The formula given in the question is a particular case for $p_C=p_G=g/2$, $p_A=p_T=(1-g)/2$, $$ \left(\frac{g}{2}\right)^{n_C+n_G}\left(\frac{1-g}{2}\right)^{n_A + n_T} $$ which is obviously always the case for DNA, where the numbers of Cs and Gs (or As and Ts) are always the same, but the more general formula could serve for RNA or proteins.

What one should keep in mind is that the formula above is for a sequence with a particular order of the nucleotides! If the order is immaterial, and only the content of the sequence matters, the formula has to be multiplied by a binomial factor, which is effectively summing over all the possible arrangements of the nucleotides. Thus, for the case of DNA we obtain $$ {n_A + n_C + n_G + n_T \choose n_C + n_G} \left(\frac{g}{2}\right)^{n_C+n_G}\left(\frac{1-g}{2}\right)^{n_A + n_T} $$ You can check yourself that the difference due to the binomial factor is striking!

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  • $\begingroup$ Thank you. Would you have a reference paper for this formula? $\endgroup$
    – Gigiux
    Dec 1 '20 at 13:39
  • $\begingroup$ @Gigiux I can't give a reference for specific use in biology. Generally this is known as the binomial distribution: en.wikipedia.org/wiki/Binomial_distribution $\endgroup$ Dec 1 '20 at 13:43
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Are the formula and its application correct?

Your formula is correct, and Vadim gives a good discussion of the underlying theory. However, concerning your application...

Looks to me, that the primer should not occur at all in the E. coli genome.

This is the point of primers -- because of their length, they are unlikely to occur at random in a given DNA sequence. So, starting with a known DNA sequence, you can create a primer that matches that sequence and have good confidence that you are not getting off-target priming, which may give you false-positive amplification in PCR. This is precisely what the authors do in your linked paper.1 Using the known sequences for Rfb and SLT-1 genes in E. coli O157:H7, the authors make primers for the specific detection of those genes in drinking water, and, by extension, the presence of the organism that harbors those genes.


  1. Bonetta S, Borelli E, Bonetta S, Conio O, Palumbo F, Carraro E. Development of a PCR protocol for the detection of Escherichia coli O157:H7 and Salmonella spp. in surface water. Environ Monit Assess. 2011 Jun;177(1-4):493-503.
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  • $\begingroup$ I tried to calculate the values given your binomial formula. The number of EcoRI sites in E. coli should be: a = (0.44/2)^1 * (0.44/2)^1 * ((1-.44)/2)^2 * (0.56/2)^2 = = 0.22^2 * 0.28^2 = 0.0484 * 0.0784 = 0.0038 which is ten times the value given in the paper. $\endgroup$
    – Gigiux
    Dec 2 '20 at 7:31
  • $\begingroup$ The expected number of restriction sites is: n = 0.0038 * 1.6*10^7 = 60 712.96. This is hundredfold more than the expected value of 645 reported by this paper (mmbr.asm.org/content/mmbr/62/3/985.full.pdf). In the paper, they calculated the number of expected 6-mer in a genome as long as that of E. coli (1113) but they did nor provide the calculation. $\endgroup$
    – Gigiux
    Dec 2 '20 at 7:32
  • $\begingroup$ As for the primer, the probability of findig it in the E. coli genome is: a = 0.25^5 * 0.25^6 * 0.25^8 * 0.25^6 = = 8.8*10^-16 n = 8.8*10^-16 * 1.6*10^7 = = 1.4*10^-08. Again, there should be -- as acvill pointed out -- unique. $\endgroup$
    – Gigiux
    Dec 2 '20 at 7:32
  • $\begingroup$ I did not understand how the binomial factor is defined. It should be: B = n! / x! (n-x)! where n is the number of trials and x the number of success. Thus it should look something like: B = 16 000! / (6!*15 994!) B' = 1.6*10^7! / (25!*1.6*10^7!) which are very big numbers. I understood this calculation was an easy one. Where am I getting it wrong? Thank you $\endgroup$
    – Gigiux
    Dec 2 '20 at 7:32

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