In this video we'll look at
the evolution and growth of types.
The types can be anything -
it can be genotypes in the population
or molecules in a test tube.
At first, let's look at the growth
of a single type.
So, x represents how many
of this type we have -
how many molecules,
what's the concentration or anything.
w is the growth rate of this type,
and if we go from one time step
to the next,
x prime is... the concentration of
the number of molecules
in the next time step.
If we want to look t time steps
into the future, xt,
we simply take w to the power of t,
so xt is equal to w to the power of t
times x zero.
Let's look what happens when we do that.
So I'm taking w to be 1.1
and I start with x0 equals 1.
We see the growth of this single type.
This type increases from 1
to 10 and further.
And now, let's see what happens
when we have a growth rate
lower than one.
So, let's take a growth rate of 0.9.
Everything else is the same.
Now we see that the type - of course -
decreases in frequency.
It starts at 1 and then
exponentially decreases towards 0.
This decrease below one is called
the "extinction threshold."
When a type - when a molecule -
is below the extinction threshold,
it simply will disappear.
Most molecules that we have around us
have this property,
and only molecules that really
replicate themselves as high-fidelity
can go above this extinction threshold.
The extinction threshold is separate
from the error catastrophe.
The error catastrophe
is what we'll talk about here.
The error catastrophe doesn't happen
when a type goes extinct,
but instead, when it cannot be maintained
versus other times.
We can represent more than one type
as a vector in a population,
so let's look at two types.
x is now a vector.
We have 1 of type 1; 1 of type 2.
And, when x is a vector,
we can simply represent the gross
as a diagonal matrix.
W is the growth matrix
and here I'm taking 1.1 and 0.95.
We go from one time step to the next
simply by taking x prime equal to W,
the matrix times x, matrix multiplication.
If we multiply W times x,
you see we get 1.1 and 0.95.
So, type 1 grew by 10 percent;
type 2 decreased by 5 percent.
If we want to apply W twice,
we apply W times W times x0,
and you see that we put W on the left.
So, we add more and more Ws on the left,
and this is because I chose
to take x to be a column vector.
If I had taken x as a row vector,
I could have put x on the left
and then we would have added Ws
on the right.
We can take a look at x at time t
by simply taking the matrix power of W
to the power of t times x0.
Let's do that.
So, we start again with one one.
We take the W matrix that we had before,
and here you see the growth
of the two types.
Type 1 really increases exponentially
in frequency,
type 2 decreases because its growth rate
is lower than 1 and goes to 0.
So, type one took over the population.
You might think that now
this describes any population.
If we add mutation,
then we would just have type 1
as the wild type,
as the type that is present.
And, all other types just
are represented as small mutants.
It turns out that this is not exactly
a right description
of a species in most cases.
In most cases,
a species will not simply be represented
by one type that is the optimal
and all other types that appear
at really low rates.
Let's plot this plot that we had here,
and log as a log plot.
Now we can see
that type 1 increases exponentially,
which in a log plot is a straight line,
and type 2 decreases exponentially.
Let's add mutations.
So now, I add a mutation rate
of 10 percent.
The... mutations are simply represented
by stochastic matrix M,
where - in this case -
type 1 maintains itself
with a probability of 1 minus mu
and turns to type 2
with probability mu.
Type 2 doesn't mutate - it stays as type 2
and doesn't mutate to type 1 at first.
So, we take this M,
we start again from the same x0
as we had before.
And now, let's see what happens
when we mutate.
So, since type 1 mutates to type 2,
now you see there is less of type 1
and more of type 2.
Now we want to represent the dynamics
with both mutation and growth.
We can simply do this by multiplying
the matrices W times M.
It's also possible to multiply M times W
in a different order,
and this simply says which is done first -
gross or mutation.
Since gross follows mutation,
follows gross, follows mutation -
it doesn't really matter
in which order you do it.
I simply chose W times M
because the algebra is easier.
If we take W times M the matrix
and diagonalize it,
it's easier to look at the matrix power.
So, if W times M is equal to
V times D times V minus one,
where D is a diagonal matrix
and V is the eigenvector matrix
of all the eigenvectors
of the matrix W times M,
it's easy to calculate W times M
to the power of T
because it's simply
V times D to the t times V to the minus 1.
D to the t is dominated by the largest
eigenvalue just like we saw before -
that type 1 grew at a higher rate
than all the other types - than type 2.
When D is diagonal,
it will be dominated
by the largest positive eigenvalue.
Let's look at this.
So, now I'm taking a mutation rate
of one percent
and the same M matrix
as we had before.
And, let's look at the growth.
So now, with mutation,
you see that type 1 increases
exponentially just like it did before.
Type 2 at first decreases,
just like it did...
also like it did before,
but then it starts increasing -
this is a bit weird.
Let's continue the graph a bit further.
So, instead of doing just 50 steps,
let's do 150 steps.
What you can see is
type 2 decreases first,
but then starts to also increase
exponentially at the same rate as type 1.
Let's look at the eigenvector
that corresponds to this growth.
So, you see that the matrix W times M
has two eigenvectors
and with two corresponding eigenvalues.
One eigenvector is zero one
and the other one is an eigenvector
where both type 1 and type 2 are present,
which has a growth rate of 1.089.
So it has a growth rate
of higher than 1,
whereas the other one has a growth rate
of lower than 1.
Therefore, this is the dominant
eigenvector.
And, this distribution where both type 1
and type 2 are present
is called a "quasispecies."
So, we see that instead of
just having one type in a population
with very few... mutants,
both types are present
and grow exponentially.
Let's change the mutation rate.
Instead of having a mutation rate
of 1 percent,
let's change it to 8 percent.
Now you see both types grow
exponentially,
but type 2 - the green type -
is now present at a higher frequency
than type 1.
So, the quasispecies is now mainly
composed of type 2 with very few type 1.
Let's see what it looks like
in the eigenvalue.
Again, you see there's two eigenvectors
with two corresponding eigenvalues -
the largest eigenvalue is this 1.01.
And, it has both type 1 and type 2
where type 2 is present
at a higher frequency.
If we increase the mutation rate
even further -
let's go to...
0.2...
Now, you see.. that both types
don't increase - they decrease.
Type 1 decreases exponentially
at a faster rate
and type two decreases exponentially
probably with a rate of 0.95.
If we now look at the eigenvalues,
you see again there's two eigenvalues
and two eigenvectors.
The larger... eigenvalue this time
is the 0.95.
It's corresponding eigenvector is this one
that has only type 2 present
and not type 1.
The other eigenvector
has a negative component of type 2,
so it can't actually exist.
This crossing where type 1 is lost
and type 2 is the only one
that is present in the population
at a high enough error rate
is what we call the "error catastrophe."
Let's look at this a bit further.
So now, I will want to plot
instead of going manually
through different values of S.
Let's just go through
a large sequence of S
and plot what happens with them.
So, here's what we get.
On the x-axis...
actually, I'm on modified mu,
so the x-axis represents mu -
the mutation rate.
And, the y-axis...
is the largest eigenvalue.
So this is the growth rate
of the quasispecies - of the types -
that grow to the highest
of the various rates.
So you see when S is very low
we have a growth rate very close to 1.1,
and the type that has both type 1
and type 2 increases exponentially.
This eigenvector, as mu increases,
goes lower and lower,
until at this point it actually crosses
and becomes lower
than the other eigenvalue -
the eigenvalue with a growth rate
of 0.95.
At this point, type 1 will disappear
from the population
and only type 2 will be present.
So, this transition is what we call
the "error catastrophe" -
the mutation rate is too high.
In the mutation matrix that I had before,
I only let type 1 mutate.
So, type 1 mutated into type 2
but type 2 did not mutate back to type 1.
This is where the error catastrophe
is strongest.
If we allowed back mutation -
so, here I'm now working with a matrix
where type 2 can mutate to type 1
with a low rate of mu
divided by a hundred.
Let's see what happens then.
I plot... oops!
I plot exactly the same...
I plot exactly the same as I did before,
and now you see
that the two eigenvalues don't cross.
This eigenvalue decreases
and then becomes almost 0.95,
and this eigenvalue stays at 0.95
and then decreases almost linearly,
but they never cross.
The reason is that type 1 is still present
in the population -
it's simply... generated by mutations
from type 2.
So... there is not a full crossing,
and therefore,
it's not a real phase transition.
Let's derive this error catastrophe
when it happens.
So, we have now M
without the back mutation
one minus mu mu,
and W with a growth rate
of w for type 1.
We multiply the two matrices
and this is the result
of the multiplication.
So, we simply take 1 minus 2 mu
times W,
mu times 0 and so on.
This is the result.
Now, in order to calculate an eigenvector,
we simply multiply this by a vector,
1 alpha,
and we want to calculate when
multiplying this vector times this matrix
will give us a vector
that has exactly the same ratios.
This will be an eigenvector of this -
of the matrix W times M.
In order... for it to stay
at the same ratio,
the ratio between this and this
has to be like the ratio
between 1 and alpha.
Therefore, mu plus alpha divided by 1
minus mu times w
has to be equal to alpha.
We can simply do a bit of algebra
and then we get that alpha
has to be equal to mu
divided by w times 1 minus mu
minus 1.
So, in order for alpha to be positive
in order for there to be an eigenvalue
where both types can be represented,
or both type 1 and type 2 are present,
this needs to be positive.
For this to be positive,
the denominator needs to be positive,
which means that w times 1 minus mu
has to be bigger than 1.
Another way to write it
is to write w as 1 plus s
where s is the growth rate beyond 1.
And then, we need to have 1 plus s
times 1 minus mu is bigger than 1,
which means that for small s
and small mu,
s has to be approximately
bigger than mu.
So, the error threshold happens
when the growth rate is bigger than
the mutation rate
or when the mutation rate is low enough.
Let's switch now to a representation
of the error catastrophe
as it was originally introduced
by Manfred Eigen and Peter Schuster
where they looked at...
instead of just looking at two types,
they looked at the genome of length L,
which means that they had four
to the L types.
And, I assume for simplicity
that there is one optimal sequence:
A-A-G-C and so on.
But, I would like to be able to talk about
just a binary sequence
instead of four bases.
So, let's look at the binary genome
in which 1 means that the sequence
is identical to the optimum
and zero means that it's different
from the optimum.
Let's look at the per site mutation rate
instead of a global rate.
So, nu is now the mutation rate per site.
And, just for simplicity,
I assume it's the same
to go from optimum to non-optimum
and back.
So, now we look just at ones and zeros,
and... we assume
that there's just one mutation
per time step.
So, nu is low enough in L
versus L
so that there's only one... mutation
per time step.
We'll call i the number of zeros,
and we'll simply look at the types
as the number of zeros
that... we have in the sequence.
So, i represents number of zeros;
L minus i the number of ones.
Now we can calculate the probability
to go from i zeros to i plus 1.
In order for that to happen,
a mutation needs to happen,
so it's L times nu.
And then, the mutation needs to hit a 1
and mutate it into a 0.
So, it's L minus i times L...
times L nu,
which gives us simply
L minus i times nu,
and similar for decreasing one...
the number of zeros.
Let's see what it does.
So, we have a nu of 1 percent
and a length of L,
L equals 10.
And now, we need to build
the mutation matrix.
We assume that we can only mutate
one up or one down,
and I simply take the expression
that we just calculated.
This gives us the following matrix.
You can see that the matrix
is different from 0
only along the diagonal
and one step away
because we only allow one mutation.
Now we can also build
the... growth rate matrix.
It's diagonal and only has one entry
that's different from 1.
So, I take a growth rate of 1.3,
which is for type 1.
Now, we can look at
the maximal eigenvalue -
the eigenvector with the maximum
eigenvalue for this matrix W times M,
and we see that type 1 is represented
at the highest frequency
and other types also represented.
So, this would be the quasispecies
in this case.
The population would be represented
by the optimal type -
types with one mutation,
two mutations and so on.
What happens now
when we increase the mutation rate?
Sorry, when we decrease
the selection pressure.
Let's go now to...
a benefit of 0.08,
so a growth rate of 1.08.
You can see that now
the optimal type is represented
with quite a low frequency
and all the other types
are represented quite highly.
In order to get a better overview
of what happens,
let's look at various ss.
So, we have plotted four different ss.
At sequence 0.3 - like we had before -
you see that type 1
has the highest representation,
one mutation away a bit lower and so on.
s equal to 0.2 -
it spreads out a bit more.
s equal to 0.1 - it's very far spread.
At 0.09, you see that type 1 is already
at a fairly low frequency.
And, the highest type is the type
that you would get mostly
if you did just a random coin flip
with five ones and five zeros.
And then, as s goes to 0.07,
you see that type 1
is almost totally non-present,
and only this type is represented.
We can also represent it as a graph.
In a graph,
here on the x-axis, is s.
s goes from 0 to 0.8,
and the y-axis is the frequency
of the various types.
Black represents the optimal type -
this is the type with one mutation,
two mutations,
three mutations and so on.
You see that when the benefit is 0.8 -
so growth rate is 1.8 -
type 1 is represented
at the highest frequency.
As s goes down,
the representation of type 1 decreases
and of the other types increases,
until we reach this threshold
where type 1 almost disappears
from the population
and the highest representative type
is the type that has five mutations.
This is simply at this point...
we simply look at coin flips.
So, then the selection...
doesn't affect the population anymore.
Let's calculate
where this transition happens.
So, we said that the chance to go
from i to i plus 1
is L minus i times nu.
Now, we want to ask -
can we maintain the zero type?
So, when i is equal to 0.
We want to maintain zero type
versus going from 0 to 1.
Therefore, we look at just
a transition rate from 0 to 1.
That rate is L times nu.
An imitation will increase
the number of... zeros.
So, what is now the error catastrophe -
the global mutation rate -
not precise,
but for the whole genome now
to go from 0 to 1,
is L times nu.
And now, we will therefore have
the same error threshold as before.
s needs to be bigger than mu,
which is L times nu.
So, the error threshold will happen
when s is bigger than L
times the per site mutation rate.
This is a very nice expression.
Eigen and Schuster used it to calculate
an error threshold for
how good can a molecule replicate.
So, if a molecule has a replication of...
with a fidelity of one percent,
it cannot code for this fidelity
of one percent
with a molecule that's larger than
100 bases.
If it has one per mille,
it can't code it with a molecule
larger than a thousand bases.
This shows that in order to...
you need to cross
the threshold of fidelity
with a short enough molecule.
This expression can also represent
what it means to be neutral
versus non-neutral.
How big does the difference have to be
between one type and another
for evolution to be able to maintain
this difference?
In order to look at this,
let's locate... the population
that has not just a difference
between the best type and all the others,
but instead there's a...
a stairs between one type and the next.
So, the highest type has zero mutations,
then the next highest growth rate
is one mutation,
two mutations and so on,
with some kind of...
randomly chosen growth rates.
The difference - this threshold
between type 0 and type 1
or between type 1 and type 2 -
I've called Delta s,
the change in growth rate.
And now, we need to be able to maintain
this difference versus the mutation.
If we look at the change in growth rate
from i to i plus 1 if that is Delta s i,
we want to ask -
when can this be maintained
versus the mutation?
And, what needs to happen is that
Delta s i needs to be bigger
than L minus i,
the chance to go from i to i minus 1...
from i to i plus 1 times nu.
So, we can take a random example.
Here I simply choose random ss
with random...
steps,
so the x-axis in black
represents the Delta s i,
the step from...
the change in growth rate
from i to i plus 1.
And, the red line represents
L minus i times nu.
This is the effect of the mutations.
In order to be able to maintain
a certain number of mutations,
the black line has to be
above the red line.
So, this means that this population
can maintain this point
versus the effect of mutation,
whereas it cannot maintain these points.
So, the difference between this and this,
as far as this population is concerned,
is neutral.
This point can be...
can be taken...can be maintained,
but this point again cannot be maintained.
So, we can see that the mutation rate
defines what is seen as neutral
versus non-neutral.
And, the distribution in a population
has to be above a narrow threshold,
which can be crossed -
as you see here - several times.
Thank you.