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If I want to make sure that a ZFN cuts at only a certain position in a genome, how many subunits would I need to ensure this?

If the human genome is roughly 3 billion base pairs, we need a sequence that is 16 base pairs because 4^16 exceeds 3 billion (if my thoughts are correct). But then I don't quite understand how the subunits of the ZFN come into this.

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  • $\begingroup$ How unique and complex is your cut site? $\endgroup$
    – WYSIWYG
    May 18 '17 at 12:43
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    $\begingroup$ Hello! I dont have a specific sequence, it is more a general question of how many you would need to really specify a cutting. Can you help? $\endgroup$
    – O.kth
    May 18 '17 at 12:57
  • $\begingroup$ This sounds like a homework question where you need to find the length of a sequence that has a probability of occurring once in the genome. $\endgroup$
    – canadianer
    May 18 '17 at 15:58
  • $\begingroup$ @canadianer Exactly, so if the human genome is roughly 3 billion base pairs we need a sequence that is 16 base pairs because 4^16 exceeds 3 billion if my thoughts are correct. But then I don't quite understand how the subunits of the ZFN come into this.. $\endgroup$
    – O.kth
    May 18 '17 at 16:25
  • $\begingroup$ If you want to be a bit cheeky then the answer is that it does not matter anymore as everyone doing real science have moved on to Crispr/Cas9. en.m.wikipedia.org/wiki/CRISPR $\endgroup$ May 18 '17 at 22:30
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The human diploid genome is ~6.5 billion bp, so you need to find the length of a sequence that has a probability of occurring only once (assuming the sequence is completely random):

$4^n=6.5\cdot10^9$

$n\cdot log(4)=log(6.5)+9$

$n=\frac{log(6.5)+9}{log(4)}$

$n\approx16.30$

According to this article, each finger of a ZFN recognizes 3 bp, which means you'd need $\frac{18}{3}=6$ fingers to probabilistically recognize a unique site. Interestingly, the article also says:

The requirement for dimerization is a great advantage for this reason: because a monomer is not active, cleavage does not occur at single binding sites. The cleavage reagent is assembled only at the target if the fingers have adequate specificity, and the combined requirement for binding two proteins brings the overall specificity into a very useful range; e.g., two three-finger proteins specify the location of 18 bp, which is sufficient, in principle, to pick out a single target, even in a complex genome.

enter image description here

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  • $\begingroup$ Thank you once again! You have been very helpful many times! :) $\endgroup$
    – O.kth
    May 18 '17 at 17:23
  • $\begingroup$ Stupid question: isn't it sufficient to consider the haploid genome because one can approximate the sequence of the second set with the first set of chromosomes? (@O.kth: if you are happy with the answer, please up-vote and accept, thanks!). $\endgroup$ May 18 '17 at 17:30
  • $\begingroup$ @O.kth You're welcome. If this (or other) posts have answered your question, please mark them as accepted by clicking the checkmark next to the answer (and up-vote). It seems I've actually answered several of your questions and you haven't accepted any of them :( $\endgroup$
    – canadianer
    May 18 '17 at 17:32
  • $\begingroup$ @AlexDeLarge Not stupid at all. I assumed for the calculation that the entire genome was random, which is of course not true (especially when considering homologies between the chromosomes). A more accurate calculation could certainly be made but, fortunately for this question, the answer will be the same. $\endgroup$
    – canadianer
    May 18 '17 at 17:35
  • $\begingroup$ @canadianer sorry for that, I often forget but I think I have accepted and upvoted all of the good answers I've gotten now! :) $\endgroup$
    – O.kth
    May 18 '17 at 20:04

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