So we've now been through all of the first stages of initiation, also counting choice initiation, spreading an establishment of X inactivation, and we've reached the maintenance phase. This phase that goes on and it's the most stable phase for decades at a time. So what do we know about maintenance? Well, we know the imprinted X inactivation, which occurs in the placenta and random X inactivation that occurs in the embryo and the adult have slightly different requirements for maintenance. They both involve this progressive layering of redundant of epigenetic marks that are mitotically heritable. This is a hallmark of both of these. However, the exact epigenetic marks on each of them vary slightly. So, I'd like to go through and discuss two of the factors Dnmt1 and Smchd1, which I'll say is Smchd1 and they're involvement in the maintenance of X inactivation. In both of these cases, we know that if you have an inactive state already established and removed Dnmt1 or Smchd1, you get reactivation of that previously silenced inactive X chromosome. And this really tells us that they must be involved in that maintenance phase of X inactivation, since when removed, we don't maintain activation any longer. So Dnmt1 is the DNA methyltransferase that we spoke about when we first thought about DNA methylation back in Week 1. So it's a maintenance methyltransferase because it recognises a hemimethylated DNA substrate. That means that of the two strands, only one is methylated. And this hemimethylated form exists after cell division, because the daughter strand hasn't yet been methylated. And so, we know this DNA methyltransferase is involved in maintaining DNA methylation states after cell division. So we know that DNA methylation is one of the final steps involved in X chromosome inactivation. It happens relatively late in terms of the timing of X inactivation in embryonic stem cell differentiation or in vivo. And so the studies on the role of Dnmt1 have come about from a many and various studies, but what I'd like to talk about are the ones that have used embryos with particular genetics. So, in this case, they're looking at embryos that are null for Dnmt1, so, in other words, they've got two knockout alleles, both the copies of Dnmt1 are knocked out. And so, these are Dnmt1 null or Dnmt1 knockout embryos, and I've looked at these in comparison to Dnmt1 wild-type embryos that have their normal complement of Dnmt1 levels. And they've looked at both males and females. But in this case, of course, it's the females that are important. So, in addition to this Dnmt1 null versus wild-type comparison, they're looking at the expression of the X chromosome using a transgene located on the X chromosome. So, in this case, it's a lacZ transgene. And lacZ is a bacterial gene, that when it's exposed to a particular substrate, can produce a blue pigment. And so, this gives a very easy way to be able to detect whether or not lacZ is being expressed or not because you can test whether or not you can make blue pigment or not in the particular embryo or placental tissue. So in this case what they did was they created the embryos that I'll tell you about with their particular genetics. And they looked and they sectioned the embryo to look for the proportion of the cells that were blue when exposed to this substrate. And they looked in the placenta to look at the proportion of cells that were blue. And because they were looking at the embryo and the placenta, they were looking at random and imprinted X inactivation in turn. So, what they looked at, at embryonic day 8.5, so this is remembering that mouse X inactivation random X inactivation and is initiated at E5.5, so a several days after that. They're checking to see, in both the embryo and in the placenta the level of expression of lacZ by the proportion of blue cells. So in each case, the paternal X chromosome, here, written as Xp, has the lacZ transgene on it, whereas, the maternal X chromosome is wild-type. This means that the paternal X chromosome having lacZ will mean that in the imprinted X inactivation that occurs in the placenta, lacZ will be silenced or should be silenced. And so this means that you shouldn't find any expression of lacZ and no blue pigment in the placenta. So indeed, if you look at the wild-type, the dead empty one, wild-type embryos, you find that there are no blue cells in the placenta. And so at just as expected, because of imprinted X inactivation. Whereas in the embryo, because of random X inactivation, you would expect that half of the time the chromosome, the paternal X chromosome, carrying that lacZ transgene would be inactive, and you would get a cell that was not blue. And half of the time it would be active, and so you would get the cell that was blue. And so in this case you see 50% of the cells are blue in the embryo due to random X inactivation. So if you then look at very the same genetics but now include a Dnmt1 null in the Dnmt1 null embryo, you see the same thing. There's no difference at this stage. We still see normal random X inactivation, 50% blue cells, and in the placenta, no blue cells because of imprinted X inactivation. However, they also looked at a stage later, one day later in embryonic times. So, embryonic day 9.5, and now they found a difference. So at this stage the wild-type embryos still showed 50% blue cells in the embryo and no blue cells in the placenta, just like you'd expect. But the Dmnt1 null embryos found that in the placenta you still had no blue cells. So normal imprinted X inactivation. But in the embryo now and only the embryo, you found 100% blue cells. So this indicates that random X inactivation failed by E9.5, but imprinted X inactivation was normal. And so, since they knew that at E8.5, everything was normal but by E9.5 we had reactivation. This means that it must be a failure in the maintenance of the X inactivation because you can detect there was a previously silent X chromosome. So it wouldn't be an error in the initiation of X inactivation. Otherwise, you would never detect that silent X chromosome in the first place. But rather, a problem with the maintenance of X inactivation, since you then resulted in a reactivation of an X chromosome, which had previously been observed to be silent at E8.5. So, what we can summarise this result, we know that in Dnmt1 null female embryos, they have normal imprinted X inactivation in the placenta. But they have reactivation of random X inactivation in the embryo. And, therefore, we can conclude that Dnmt1 is required for random X inactivation, but is not required for imprinted X inactivation. And so, there is a differentiation between these two types of X inactivation and their requirements for Dnmt1. So, it's interesting in this light to think that we actually know imprinted X inactivation is not nearly as stable as random X inactivation. Perhaps this makes sense, because imprinted X inactivation only needs to occur for the life of the placenta, which is just for the period of gestation, in comparison to random X inactivation, which way have to go on for decades and decades up to 100 years. And so, we know DNA methyltransferase is one, is this maintenance methyltransferase, which helps ensure mitotic heritability of DNA methylation. So potentially, imprinted X inactivation, which is more labile, which doesn't involve Dnmt1, is perhaps more labile for exactly that reason because it doesn't involve DNA methylation. In the next lecture, we'll consider the same sort of crosses and embryos, but this time considering Smchd1.