I was reading this little 'article' about homologous recombination and knockout mouse. According to this article you first remove embryonic stem cells from a gray-fur blastocyst, then insert the construct and place them back into the blastocyst. Further the text states:

A knockout mouse has had both alleles of a particular gene replaced with an inactive allele. This is usually accomplished by using homologous recombination to replace one allele followed by two or more generations of selective breeding until an breeding pair are isolated that have both alleles of the targeted gene inactivated or knocked out.

However what I don't get here is that we have cells from an embryo, and thus have cells with both alleles. Hence, the regions surrounding these alleles will be homologous, so why would this step result in heterozygous (and thus require selective breeding afterwards) instead of homozygous mouse direcly?


1 Answer 1


It is a simple matter of probability. If the probability of replacing one allele is p then the probability of two alleles getting replaced would be p2 (note that the latter will be smaller because probabilities are less than one).

If you have a selectable marker such as GFP then you'll end up selecting both heterozygotes and homozygotes but the frequency of the former will be much higher especially when you consider the fact that the probability of homologous recombination mediated knock-out/in is very low in mammalian cells (10-7 – 10-5). Therefore, inbreeding is an easier way to obtain homozygotes than screening the ESCs. Moreover, selective breeding is inevitable because in the first round will just produce chimeras (not all ESCs are replaced with the modified ones).

Technologies like CRISPR are now being used to accelerate the rate of gene replacement.

  • $\begingroup$ Thankyou, I have one additional question. Why would you not just replace all ESCs? Or is this simply not technically possible? $\endgroup$
    – KingBoomie
    Commented Apr 30, 2018 at 15:43
  • $\begingroup$ @RickBeeloo It would be very challenging. Almost impossible. $\endgroup$
    Commented May 1, 2018 at 9:32

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