[MUSIC] Okay, now we are reaching the beautiful experiment. Before we do that, I just want to make sure that we all are at the same level of understanding. What the results are compatible with amber being UAG and ochre being UAA. However, however, if the amber codon was GGG, and the ochre code was GGA, all the results that we've seen so far would be compatible with this possibility. As long as there are codons in this case, AA G or A, that are also nonsense. So what, if you read the paper carefully, you are driven to UAG and UAA. Because they're very smart in their writing. But they let you accept, they let you think freely. And if you want, you can actually think of many other possibility that would be acceptable and compatible with the data. But, and that's where come the last experiment. The last experiment will demonstrate that the amber and ochre have a U and an A. And if the amber has a U, an A, and a G, then necessarily, the ochre would have UAA. What this doesn't prove is that this order is correct. The order could be different. Could be GAU, AAU. The order we don't know yet. We're getting at the composition. But what they will do in the last experiment is to show you that the amber codon has both a U and an A. How can you demonstrate that the amber codon has both a U and an A? Well we have to go back to this drawing. If the amber codon has both a U and an A, that means that the wild type codons that can give rise to the amber codon must have a G and a C, or a C. But, If the base in the message is an a, that means that the sense strand has been modified and codes for an A. This is UAG. Which means that what is the code on that gives right to UAG? It's UGG. In this case, the U is on the antisense strand. So the UAG is derived from this C, which can only be seen after replication. Because this is not yet a UAG, because when this molecule of DNA infects the cell, only one strand is read, before replication. Before replication, one strand is copied, the other strand is ignored. So the mutation on the other strand is ignored until replication, in this case. In this case, the mutation is immediately apparent because the modification of the base is on the DNA strand that codes for the message. So this is immediate, and this is only after replication. So what they did is actually to see whether they could demonstrate that. So this experiment is a bit complicated. Let's get the set up. The set up is they will isolate mutants after mutagenesis with hydroxylamine. So you take phage, you trade with hydroxylamine. And then you will infect either E Coli B, on which everybody can grow all the R2 mutants, or E Coli K SU0 SU0 on which only the R+ can grow. And they lyse the cells after infection. So it's a large experiment with many phages. There are liters of infections and they get literally liters of phage. And they use a lot of mutagenesis so that they really push the damage as much as they can so that they lose a lot of phages, but they don't care. And they get about 1% R. Then they characterize the R. R1, R plaques on B grows on K lambda. So it's just picking from one plate to the next. Number of R1 about the same in the two sets, about 2000, so the set are comparable. Then they identify the R2, the lead key they ignore, and then they have the amber and the ochre. And they have a set of 319 amber, and 121 ochre, and 83 ochre in set B, 121, 82 in set K. Those are all identified by picking on two different strains, nothing more complicated than that, it's a lot of phoric practically, but it's very simple. An amber will grow on a SU amber, an ochre will grow on an SU ochre. Simple. An amber will also grow on an SU ochre. But the ochre will not grow on an SU amber. So it's very simple. So they have a collection of mutants at different places. They don't know what they are. So they map the mutant by the Benzer system, by either deletion mapping in svelt segments, subsegments, sub-subsegments. And finally they look where the mutants are. Because they know all the original mutants, they try to get how many times did they recover In their set, what was originally called HB 118? And the amber HB 118 is recovered 27 times in the set B, 15 times in the set A. HB 129, 14, 25, 12, 16. EM84, 29, 21, 28, 12. All of this, and probably N19 too, are recovered at the same frequency roughly in the two sets. The same frequency in the two sets. That is, if you go once through K, you don't lose them. Why don't we lose them? Because these mutants, when they infect K, are not expressed immediately. So they can replicate and give rights to progeny and we'll recover the page. So the one which do replicate are the one that start from a CAG codon, because the C is on the mRNA strand. The one that I immediately expressed will be dead. They will not survive one cycle of an infection through K lambda, so they would be UGG. So if we go back to this, we have seen that there are mutants like HB118 HB129, S199 that are recovered with the same frequency in the both cells. Not the same number, but the same frequency. And you have mutants like N97, 44 to 1, 31 to 2, 44 to 3, 21 to 1. Now these guys are UGG codons because they're immediately lost. Because the C on the base is red like a U, and so you have an A in the codon. So these are the UGG codons. And the blue are the CAG codons. Just by this very simple experiment of going though one infection where you kill some of the amber, but not the others. And of course you do recover them at some frequency, but at much lower frequency. So, this is very impressive. What is even more impressive is that the number of UAG, UGG and CAG codons they identified is the total number of UGG and CAG codons in the sequence. They picked most of them, all of them in this case. Now that's quite impressive. Now, how do we deal with a set of ochre? Well the set of ochre, is, there are not enough mutants. But there are, if I call them the blue mutants, they are blue mutants. Like N55 is a blue mutant, 360 is a blue mutant, X20 is a blue mutant for sure, N7 is a blue mutant. In fact, they recover four out of five of the mutants. There's one other, three and five. This is one and this is one. This is sites that we know are ochre, CAA ochre. These are only CAA. Why? Because the other transition we cannot see. The other transition is UGA. And UGA is a stop codon by itself. So, you cannot see the other transition, you can only see one, induced by hydroxylamine in this gene. So, it's in the R2BC strand they fear they got four out of five of the codons, and here they got six out of nine of the codons. And here they got all of them. All the CAG and UGG codon were found. Now, you may ask, well, what about, let me take another color. What about H E122? How do you interpret H E122? There is one isolating set B0 and set K. Well it turns out that I sequenced H E122 for totally different reasons. And HE122 is a U. The Y-type codon is UAU, which cannot be induced by hydroxylamine to UAG, because it has no Gs. So this is a transversion, and this occurs. It happens. Cosmic rays, whatever, it happens. Errors of DNA replication, all of this happens. But it's at low level. It's a background. This is a background mutation. The same thing for C2O4. Now at the end of this, they had some idea about the composition. Must have a U, must have an A, and must have a G or C. And there the biochemistry comes back because they identify the amino acid by pure biochemistry analysis of protein sequences. And they identify revertance of amber mutants, they didn't look at revertance of ochre, they only look at the revertance of amber. So, the revertants of amber. And there were a lot. They used the head protein of phage T4, revertants of an amber, or actually of several ambers. And they got, reverted to, induced by 2-aminopurine, they got CAG and UGG, the two transitions, glutamine and tryptophan. From the fully revertance, they also got glutamine and tryptophan. And they even analyzed a series of sites. They analyzed 115 revertants, 62 had a tryptophan. And at least one of the other 50 plus had a glutamine. So all of that was perfectly consistent. So, so far they had this. Now, they also identify revertants, not the Brenner group but the Garon group with Fourier, identified revertants that had the tyrosine, that had the leucine, that had the lysine or an arginine, later found to be lysine, the serine, and the glutamine. So far, so good. All of that chemistry fits the genetics. And now, all of this was put together by Nirenberg who used synthetic oligonucleotides to decipher the code. And when he did that, one of the first he identified was UGG tryptophan, same as this. Then he identified UUG as leucine, same as this. He identified CAG as glutamine, again, and CAG as glutamine. And the others, lysine and serine and glutamate were identified. And the tyrosine, the other ones were identified later. So going through a pure genetic approach to identify, to predict the codon structure, and then biochemistry to fit that with the protein sequences, is quite marvelous. At the end of the paper, they discuss both suppression and nonsense. Their idea is that there are tRNAs that are necessary for. Inserting amino acids, charge deioning. So they proposed that there are tRNA that have no amino acids and are used for termination. That's perfectly logical, and wrong. Because now we know that it's a set of proteins that terminate translation. But, they couldn't possibly know that. The funny thing about it, if you want, is that these proteins have been crystallized, and their structure in space looks quite a lot like a tRNA. So, the notion of that it's a special tRNA is still reasonably bad. And then they discussed suppression. And suppression they say can happen for many reasons. You can have change in the activating enzyme that linked amino acid to the tRNA. You can have a change in the nonsense reading tRNA, if there's such a nonsense reading tRNA. Or you can have a change in a tRNA that allows reading an expansion of the code. And in fact, that's what's the case in all these suppressors. It's an expansion of the coding capacity of the tRNA. And this, of course, they could not predict. Today we know that E Coli has more than 80 genes for tRNAs. 20 are the minimal, 61 is more than enough, and in fact there are 80 genes. And a lot of these genes can be converted into suppressor tRNA. So they discussed ambiguity of the code. They don't know anything about the ochre, because they don't have the biochemistry for the ochre. And they are careful in the discussion to separate what they are looking at, which is specific suppression of one codon from generalized errors in protein synthesis. And if you remember, in the Watson and Crick DNA structure of the DNA session, we'll also discuss a paper by Speier, who actually proved that the DNA polymerase is the main component of fidelity. And the reason he started these experiments was because he'd heard that the ribosome can affect translation fidelity. So it's a very nice circle that these ideas do in time. Of course they are presented to you in different sessions because we have to keep some structure. But a lot of that was actually going on at the same time.