[MUSIC] Then comes a section which is a little bit awkward. They discuss the fact that most negative mutants are recessive to the wild type. That is in neurospora, all the strains that cannot make leucine, if you made them as a leu plus strain the diploid is leu plus. There is no negative dominant at this stage of the game. Of course, this was done for. It has not been done for coli because diploid E coli were not existing at the time. We will see them next week, partial deployed. But they assume that because most of the mutagens are recessive, in neurospora most of the nutritional mutagens are also recessive, expect it to be recessive in E coli. They say that this is a reasonable assumption So then they explain how they make the mating. Basically, they grow the cult the cells together for a number of hours. And then they watch the cells and they plate them on minimum [INAUDIBLE]. So then they immediately describe the fact that the reversion can be a problem if you have a single. For instance, the strain 58 or strain 679 could revert to B plus or to T plus. But, if you ask 58-161 to revert to B plus, it will two and plus will revert to both plus. You will not detect it. So, they start with B minus, N minus, which is strain, this one. I will change the color. This strain. And they mix it with a strain that is P- T- which is here, 679-183. So they mix this one and this one, originally derived from 58 and 679. Themselves originally derived from K12. And they select for Y type recombinance, i.e. B+, M+, P+, T+. These are called prototroph because they don't require anything but glucose and salt to grow. And they isolate them at a frequency, in this experiment, of 10 to the minus 7. If they cultivate separately 58-161 and 679-183 and ask for phototroph, they don't have any. So, the controls are less than 10 to the 10. 10 to the 10, this is a number of cells they've tested and here they have 10 to the -7. So the differential in numbers is at least 1000 between the recombinant and the. Of course, in this experiment, they don't know whether a t minus p minus strain acquire DNA or genes, or information from the other one, or whether the B minus N minus acquired information from the other. They have no idea about direction. The only thing they know, is that if you have a million people another way you get a mutant, a recombinant. How does it work? They have no idea. Now they say there are lots of different ways by which you could have such a result. There are lots of different ways. One way would be to have a cell, one of the cell, and the other cell is glued to the first cell so that they help each other. The blind and the paralyzed man. The blind is carrying the paralyzed man, the paralyzed man is seeing where they go. This is a kind of syntrophy, or it's been called symbiosis. There are lots of different ways. That's one possibility. When they look at the cells under the microscope, the cells that glow without any addition. On the microscope, they only see regular E coli. They don't see doublets. So this, no. Now, there's another possibility. It's that they have a cell which is a fusion between a red and a green a zygote. This is a zygote hypothesis. Well, if the zygote hypothesis is correct, the simplest idea is that the zygote would be twice as big as the single cell. While I exaggerate on my drawing but, this is what they would expect, they don't see this, they only see the normal sights one element. Second element, if they start with a green cell, that is T1 sensitive. In the red cell, that is T1 resistant. Of course one of them is B minus. N minus is otherwise P minus, T minus. And you do the same experiment. And then you select the. The ones that are +++. And you pick them one by one. And you grow them. If the cells are a mixture, a zygote. You would expect they would have the receptor on the green surface and like the receptor on the red surface they would be killed by the viruses. They should be all killed by the viruses. But what they find is that, a fraction approximately 20% or 30% are T1 sensitive, and 70% are T1 resistant. But the cells from one cell, one colony, one event, one recombination, are either sensitive or resistant not both. And they're all the same, that's a very strong argument to say that they are not no mixing of this kind or this kind. They also did an experiment against the deployed hypothesis. Okay, you can think of the diploid as being larger than the haploid. That's normal. But if this cell, the recombinant, was something like this, this would have two chromosome, so I shouldn't draw inside this cell. This cell should have two chromosomes, like you and me, and all of us. In most of our cells, except the gonads, we have two chromosomes, one from daddy, one from mommy. And this would be a simple case of complementation. Now they take this cell, that R prototroph recombinant. They treat them with UV and they try to isolate auxotroph mutant. And they isolate one auxotroph which requires nicotimic acid. Nic-, unable to synthesize nicotine, nicotinic acid. Now, if this cell was diploid, the chances that you would mutate both nic genes from both chromosomes are very, very low. The fact that you can isolate auxotroph suggests that the recombinant is a haploid recombinant. Okay, in order to test for transformation, because that's and you see here, it's claimed that Avery was not taken seriously. That his experiments were not appreciated by the community. In fact, they spent about half the paper trying to determine whether they have transformation or non-transformation. And they show that it's unlikely to be transformation because you need contact of the cells, you cannot do it with a culture medium. You can not take a supernatum from the cells, or an extract from the cell, and obtain recombinance the way Avery was able to do with. Okay, and finally. Finally, they go through an extensive analysis of their recombination experiment. This analysis involves in this case, triple mutant. One is a y10 and one is a y24. You see that the first cell is biotin minus phenalyne cistine minus, it's one of the parent and the other one is theanine minus, leucine minus, B1 minus. And on top of this one is resistant to T1 and the other is sensitive. Those are the parent. And then they isolate some recombinants that are the prototrophs. So these are the prototroph. They have 86 prototroph that grow on minimum medium. Some of them are resistant to T1, some are sensitive. In the same experiment they also plated an aliquad on a medium containing nothing but vitamin B1. This medium ,here on line four there's no biotene is by B minus or B plus, but they isolate 37 colonies on a plate with vitamin B1 and they count how many are plus and how many are minus. And 36 are plus out of 37. 97% of the colony that are prodigal for the other marker are plus for B1. This is very high. But in some cases it's very low. If you take, for instance, here the next line, in this case they will isolate cells in the presence of turning. Width, so that t minus and t plus are congruent. Isolate 31 colony, and out of the 31 colony, two are t plus and 29 are t minus. So the percentage vary according to the. So this is their best evidence for recombination. They don't have, they have very limited quantitative analysis because this is the best quantitative analysis they have.
