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Classical papers in molecular genetics

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University of Geneva

Classical papers in molecular genetics

97 个评分

You have all heard about the DNA double helix and genes. Many of you know that mutations occur randomly, that the DNA sequence is read by successive groups of three bases (the codons), that many genes encode enzymes, and that gene expression can be regulated. These concepts were proposed on the basis of astute genetic experiments, as well as often on biochemical results. The original articles were these concepts appeared are however not frequently part of the normal curriculum of biologists, biochemists and medical students. This course proposes to read study and discuss a small selection of these classical papers, and to put these landmarks in their historical context. Most of the authors displayed interesting personal histories and many of their contributions go beyond not only the papers we will read but probably all their scientific papers. Our understanding of the scientific process, of the philosophy underlying the process of scientific discovery, and on the integration of new concepts is not only important for the history of science but also for the mental development of creative science.

从本节课中

Session 11

The major start point in the modern history of lysogeny, written by Lwoff and Guttman, is not available in English. We will start this session with the study by Bertani of the Li strain of E. coli that carries 3 lysogenic prophages. Each prophage can be induced independently of the other two, and the relative rate of spontaneous induction depends on the physiological state of the cells. Bertani devised a method for detecting single bursts and could calculate that the rate of induction is about 1 per 50000 cells per generation, a rate much lower that that observed by Lwoff with the B. megatherium prophage.After a brief outline of the regulation of phage lambda, we will discuss the lethal phenotype of a thermo-inducible prophage that cannot excise from the chromosome. Suppressors of this phenotype belong to two main classes: mutations in genes O or P prevent phage replication while x mutations prevent the expression of the early operon that includes O and P. Upon shift to the high temperature, the repressor becomes inactivated. If the cells are incubated at the high temperature for several generations and returned to the permissive temperature where the repressor can be active, the x mutants do slowly recover immunity while the O and P mutants remain permanently non immune. We now know that x mutations inactivate an early operon that contains O and P. These mutations have been called “physiological deletions”. The phenotype of the x mutants is recessive, providing the first evidence for cro, a new gene in the x-O-P operon. Mutants that do not make Cro have been isolated. Unlike wild type phage that preferentially turn on the lytic program, cro mutants are unable to form plaques because lysogeny is the preferential program. Cro also represses the other early operon since its expression is increased with both x and cro mutants

与讲师见面

Dominique Belin
Professor
Department of Pathology and Immunology University of Geneva Medical School

[MUSIC]

This little note in the [FOREIGN].

Is not really the story of Crowe, but

it's necessary to understand the story of Crowe.

The paper started with one observation.

We knew from a long time, that if you use lysogen you get a lytic cycle.

So that was not a surprise.

At that time, the Pasteur lab had identified a mutation

in c1857, in c1 called 857.

Which is a temperature-sensitive mutation.

At 30 degrees, the repressor binds.

At 40 degrees it doesn't bind.

So you can keep a c1857 lysogen at 30,

which you heat it to 40 degrees, you get a cycle, a lytic cycle.

So far, easy.

Now, in doing some other experiments people were actually trying to induce.

A lysogen that carries a c1857 mutation and an N-mutation.

And remember, N is the protein which is required

to go from this very early to this early part of the cycle.

So if you don't have N, you don't express the left operand.

And you don't express, you don't excise the DNA from the chromosome.

But you still express about 10% of O and P, which is on the right.

And O and P are replication genes.

And so, basically what happens is that the chromosome becomes a jamming chromosome,

because replication comes from the origin, and replication comes from the prophate.

And all of that is a mess in the cell duct.

That was the observation.

The prediction was that, if you

try to isolate survivors from this strain, you should get mutants in O and

P, in O or in P which were the two known replication genes.

Surprisingly [FOREIGN] identify X mutants and

the X mutants people did not know where

they were the X mutants were neither O nor

P and S had not been used as a letter.

W had been used, X had not been used so they called it X.

So how do you study these X mutants?

You can show that the prophage is present and

carries the suppressor mutation,

the mutation that helps you to survive.

You infect these cells with lambda [FOREIGN] at 30 degree,

you isolate the, you get a phage lysate, you

infect non-lysogenic cells at 30 degree, and you isolate lysogen at 30 degree.

Only the prophage can lysogenize.

Lamlarvae cannot lysogenize.

And you can show that these cells, these lysogen now survive at 42 degrees.

They're immune at 30, so they carry the prophage and they survive.

So they carry the suppressant.

The ex-mutation could be mapped.

And the map to the immunity region.

How do you know that?

Well, let's take a mutant in O or in P.

You infect this lysogen and with a land at 434 phage, and

you look for things that can grow under 434 lysogen.

To grow in a 434 lisogen, you need to be N+, and you need to be OP+.

So you ask for recombination in these regions.

And of course you are turbulent so

[INAUDIBLE] for the tubicles you have C1857.

So you can ask the same thing with the X mutants.

What happens with the X mutants?

With the X mutants, in order for the phage to recombine, you want to have this.

This makes no difference, because it's the same DNA.

It's the same DNA because X is under the immunity region.

Now if the phage has X It doesn't express the O and

P genes, so it doesn't grow.

You don't get phage which grow on a su0434 lysogen.

In this case, you get the phage.

In this case, you don't get the phage.

Because an X is phenotypically O minus and P minus.

So they characterize this X mutant.

These are the new lysogen that they made out of their original strains.

This was a parent, and this is one X mutant.

This is one O mutant.

This is one P mutant.

As an example.

And we can look at what's going on with this.

Four different lysogens.

So this is a lysogen.

This is a lysogen, and this is a lysogen.

If you infect with the lambda clear at 30 degree no growth

because the repressor is active.

The bacteria are immune.

At 40 degree growth because the repressor is inactive.

This is standard.

Lamda [FOREIGN] should grow whether or not there is a repressor.

It grows at all temperatures.

So this is expected.

This is expected.

Now, you ask or we ask just before whether you can recombine with 434.

The O or P mutants.

Can we combine this one?

The X mutant cannot recombine.

Neither with 434, nor with 21.

It's under the immunity region.

And then they ask a very important question.

Can the X mutant make O and P?

Can they compliment an O minus, or a P minus?

And in fact, the X mutant cannot compliment

neither an O minus Nor a P minus.

So they don't make either protein.

In contrast, the O mutant can complement a P.

And the P mutant can complement an O.

This is exactly what we expected.

So X Mutants are under the immunity region,

and they are phenotypically O- P-.

They were considered and are promoter mutants.

Now that was the status of the letter.

And then they ask the following thing.

This strain cannot grow at 40 degree

because it's killed by replication then.

But this guy should, they can grow at 42.

What happens when you put them back at 30 degrees?

They both have the repressor, the C1857 repressor so

they should make a more stable protein.

Stable at 30 degree, it should have repression.

Will they have repression?

And this is what started the next paper.

So this was published two years later in KNAS, and

the abstract of the paper is very simple, a new gene in lambda is described.

This gene called cro, for control of repressor in

other genes, prevents expression of immunity and

regulates the expression of the genes to the left.

They've identified the cro gene and they identify mutants of

the cro gene which have been isolated and characterized.

So I put you under the abstract, this map you've seen.

This map is the one map that says that

very early you make N and cro.

Later, slightly later, you make all the left operon and

the right operon, including N and cro.

This is a more detailed map from the paper that shows you some

of the genes that we'll discuss.

XS is a gene necessary to come out of the chromasome.

XO is an enzymes and exonucleus and we discussed C1.

We discussed that's a repressor.

Cro we discussed.

X mutants are a promoter mutants and the rest we can ignore.

And o and p we discussed.

Okay.

So the first figure of the paper is an experiment

that asks whether you can synthesize

repressor when you go back to 30 degrees.

So you grow the cells first at 41 degrees or 40 degrees overnight,

and then you dilute them and keep them at 30 degrees.

And each of these time points you infect with the lambda clear and

you ask what fraction of the cells are immune and

what fraction of the cells are non-immune.

A non-immune cell will make a plaque, an immune cell will make no plaque.

So, if you start with the circles,

the circles are c1857 and minus x minus.

This is promoter mutant.

If you have the promoter mutant,

the cells will recover immunity over a period of four hours.

And it takes time, because it takes at least one hour to go one log,

two hours to go two log, about four hours to go three logs.

So immunity is recovered with X mutant.

With O mutants, you don't recover immunity.

The difference between the X and the O.

Now, you have to wait several hours to recover immunity but you can recover it.

Now they ask, we have to find this X.

Who is dominant?

X plus or X minus.

We have two phenotypes.

We have the phenotype of the triangle and we have the phenotype of the circle.

Two phenotypes, you can ask which one is dominant.

You put one on an F prime and the other one on the chromosome and then you act.

In both cases you get exactly the same result.

If you are one x plus, x plus is dominant.

X mutants is recessive.

The double heterozygotes do not recover immunity.

From this they concluded that there is

a product diffusion acting in trance that

is made from X and prevents [INAUDIBLE].

And this is [INAUDIBLE].

Now, this product, they wanted to ask,

weather the product of this cro gene is immunity specific or not?

In other words, can the cro from lambda work on the DNA from 434?

Can the cro of 434 work on the DNA of lambda?

So in this case, they will look for

recovery of immunity, again.

It's the same experiment, recovery of immunity, but

with cells that have one or two prophage, with different immunities.

So, the first line is the very easy one.

The string has immunity 434 and is x.

Will the string recover immunity against 434?

Yes, it is immune.

It recovers immunity.

Why?

Because it's an x, and the x doesn't need cro.

Then you have two prophages, one which is the same as before.

This is a typo in the original paper.

And this one is an o minus.

So you have the x, 434x, and lambda.

The 434 recovers immunity.

But the lambda one does not recover immunity.

It stays non-immune.

It remains non-immune.

It’s incapable of synthesizing promoter because it

can't express the cro gene and the experiment can be done.

Vice versa, when in this case the vice versa experiment

is immunity against 434 is never regained in

an O-mutant, and is regained in the next mutant.

And I put you on the bottom where the mutations are,

so it's immunity specific.

Then the author decided to

isolate cro minus mutant.

In order to isolate cro minus mutant, they started with a strain that is lysogenic.

Lysogenic for c1 857, N, and O.

This strain will survive at 42 degree, but if you put it back at 30 it

does not recover unity because cro prevents synthesis of the repressor.

So what you can do is you can plate at 30 degrees and look for

cells that have recovered immunity, that are resistant to lambda clear, cI.

These will be either mutant in x or mutant in cro.

You can show that the mutants in x lose p-expression and the mutants in cro don't.

So you get chromines.

So once you want to isolate this phage, what you do is a cross again with 434.

I think I forgot here to put the O-mutant.

This is the original phage.

The phage has now a fourth mutation, the cro mutation.

And you isolate it as a C1857 cro mutant.

It has a temperature sensitive repressor, and no active cro.

And the plating properties of this cell is shown here.

It cannot plate at 30 degree.

Why?

Very low plaque formation at 30 degree.

It plates totally normally at 42 degree.

Why is that?

Because this trade does not make the repressor for your repressor synthesis,

so it makes enough repressor so that all the impacted cell become lysogenic.

You never see a plaque.

A lysogene bacteria is not making the plaque,

it's lost In this load of other colonies.

Now, if you provide cro,

then you can lysogenize,, you can plate at 30 degrees.

This is expressing cro, and the cro is a lambda cro has to be the lambda cro.

This is the lambda clear expressed.

And this allows plaque formation at 30 degree.

So the cro mutant don't plate at 30, they plate at 40.

So you have to realize that now you have cI mutants,

cro mutant, x mutant, virulent mutants.

Now one of the property at the time when people were doing these experiments,

the only enzyme you could assay was exonuclease.

And exonuclease is an enzyme that can be assayed because

it's very easy It's made from the left operator.

There is no problem and you can quantitate it.

But in order to get xo you have to be N+.

So they had to put these prophage into this train, which is su+.

So that now the end mutations are suppressed.

The old mutations are suppressed.

The x-mutation is not and the cro-mutation are not.

And so now you can measure expression of the left operand over time.

If the phage is a normal phage.

Like a phage.

It will start making exonuclease at around 15 minute and

reach a plateau at around 35, 40 minutes.

This is a wild type.

The cro mutant or the x mutants because it's the same.

An x mutant doesn't make cro and a cro mutant doesn't make cro.

All of these make a lot more exo.

Even at 20 minutes and at 40 minutes, look at the amount of exo.

It's about two and a half, four times more.

They do not stop exosynthesis because

cro is necessary to shut off the host.

And, in fact, what I showed you here is exactly what's going on.

If you have cro you will go from here to here.

This requires cro in order not to transcribe these parts.

If cro is not made, you keep making this.

It's complicated, but if you think of it,

it's not any more complicated than any little auto running machine.

It knows when to start, it knows when to stop, it knows when to change program and

change gear, this is like an automatic vehicle.

Lambda is such a beautiful little automatic vehicle,

that is has it's circuitry all organized within a few thousands of base papers.

A few thousands of nucleotides and four regulatory genes are enough.

Five regulatory genes are enough to make such a complex decision.

It's a very short time spent on lambda, but

I didn't want you to have the impression that

the only phage that was worked on was D4.

Lambda also was a very important phage.

Unfortunately a lot of papers on Lambda are not easy to read.

Either they are too short, notes like the [FOREIGN], or they're way too long.

So I've chosen some.

This one may have been the hardest of the series,

but when you reread it, it's a jewel.

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