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In this video we'll look at
the evolution and growth of types.
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The types can be anything -
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it can be genotypes in the population
or molecules in a test tube.
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At first, let's look at the growth
of a single type.
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So, x represents how many
of this type we have -
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how many molecules,
what's the concentration or anything.
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w is the growth rate of this type,
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and if we go from one time step
to the next,
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x prime is... the concentration of
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the number of molecules
in the next time step.
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If we want to look t time steps
into the future, xt,
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we simply take w to the power of t,
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so xt is equal to w to the power of t
times x zero.
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Let's look what happens when we do that.
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So I'm taking w to be 1.1
and I start with x0 equals 1.
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We see the growth of this single type.
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This type increases from 1
to 10 and further.
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And now, let's see what happens
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when we have a growth rate
lower than one.
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So, let's take a growth rate of 0.9.
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Everything else is the same.
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Now we see that the type - of course -
decreases in frequency.
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It starts at 1 and then
exponentially decreases towards 0.
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This decrease below one is called
the "extinction threshold."
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When a type - when a molecule -
is below the extinction threshold,
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it simply will disappear.
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Most molecules that we have around us
have this property,
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and only molecules that really
replicate themselves as high-fidelity
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can go above this extinction threshold.
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The extinction threshold is separate
from the error catastrophe.
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The error catastrophe
is what we'll talk about here.
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The error catastrophe doesn't happen
when a type goes extinct,
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but instead, when it cannot be maintained
versus other times.
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We can represent more than one type
as a vector in a population,
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so let's look at two types.
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x is now a vector.
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We have 1 of type 1; 1 of type 2.
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And, when x is a vector,
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we can simply represent the gross
as a diagonal matrix.
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W is the growth matrix
and here I'm taking 1.1 and 0.95.
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We go from one time step to the next
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simply by taking x prime equal to W,
the matrix times x, matrix multiplication.
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If we multiply W times x,
you see we get 1.1 and 0.95.
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So, type 1 grew by 10 percent;
type 2 decreased by 5 percent.
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If we want to apply W twice,
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we apply W times W times x0,
and you see that we put W on the left.
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So, we add more and more Ws on the left,
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and this is because I chose
to take x to be a column vector.
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If I had taken x as a row vector,
I could have put x on the left
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and then we would have added Ws
on the right.
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We can take a look at x at time t
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by simply taking the matrix power of W
to the power of t times x0.
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Let's do that.
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So, we start again with one one.
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We take the W matrix that we had before,
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and here you see the growth
of the two types.
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Type 1 really increases exponentially
in frequency,
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type 2 decreases because its growth rate
is lower than 1 and goes to 0.
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So, type one took over the population.
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You might think that now
this describes any population.
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If we add mutation,
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then we would just have type 1
as the wild type,
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as the type that is present.
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And, all other types just
are represented as small mutants.
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It turns out that this is not exactly
a right description
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of a species in most cases.
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In most cases,
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a species will not simply be represented
by one type that is the optimal
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and all other types that appear
at really low rates.
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Let's plot this plot that we had here,
and log as a log plot.
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Now we can see
that type 1 increases exponentially,
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which in a log plot is a straight line,
and type 2 decreases exponentially.
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Let's add mutations.
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So now, I add a mutation rate
of 10 percent.
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The... mutations are simply represented
by stochastic matrix M,
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where - in this case -
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type 1 maintains itself
with a probability of 1 minus mu
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and turns to type 2
with probability mu.
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Type 2 doesn't mutate - it stays as type 2
and doesn't mutate to type 1 at first.
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So, we take this M,
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we start again from the same x0
as we had before.
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And now, let's see what happens
when we mutate.
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So, since type 1 mutates to type 2,
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now you see there is less of type 1
and more of type 2.
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Now we want to represent the dynamics
with both mutation and growth.
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We can simply do this by multiplying
the matrices W times M.
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It's also possible to multiply M times W
in a different order,
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and this simply says which is done first -
gross or mutation.
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Since gross follows mutation,
follows gross, follows mutation -
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it doesn't really matter
in which order you do it.
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I simply chose W times M
because the algebra is easier.
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If we take W times M the matrix
and diagonalize it,
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it's easier to look at the matrix power.
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So, if W times M is equal to
V times D times V minus one,
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where D is a diagonal matrix
and V is the eigenvector matrix
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of all the eigenvectors
of the matrix W times M,
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it's easy to calculate W times M
to the power of T
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because it's simply
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V times D to the t times V to the minus 1.
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D to the t is dominated by the largest
eigenvalue just like we saw before -
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that type 1 grew at a higher rate
than all the other types - than type 2.
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When D is diagonal,
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it will be dominated
by the largest positive eigenvalue.
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Let's look at this.
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So, now I'm taking a mutation rate
of one percent
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and the same M matrix
as we had before.
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And, let's look at the growth.
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So now, with mutation,
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you see that type 1 increases
exponentially just like it did before.
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Type 2 at first decreases,
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just like it did...
also like it did before,
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but then it starts increasing -
this is a bit weird.
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Let's continue the graph a bit further.
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So, instead of doing just 50 steps,
let's do 150 steps.
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What you can see is
type 2 decreases first,
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but then starts to also increase
exponentially at the same rate as type 1.
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Let's look at the eigenvector
that corresponds to this growth.
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So, you see that the matrix W times M
has two eigenvectors
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and with two corresponding eigenvalues.
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One eigenvector is zero one
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and the other one is an eigenvector
where both type 1 and type 2 are present,
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which has a growth rate of 1.089.
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So it has a growth rate
of higher than 1,
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whereas the other one has a growth rate
of lower than 1.
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Therefore, this is the dominant
eigenvector.
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And, this distribution where both type 1
and type 2 are present
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is called a "quasispecies."
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So, we see that instead of
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just having one type in a population
with very few... mutants,
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both types are present
and grow exponentially.
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Let's change the mutation rate.
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Instead of having a mutation rate
of 1 percent,
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let's change it to 8 percent.
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Now you see both types grow
exponentially,
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but type 2 - the green type -
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is now present at a higher frequency
than type 1.
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So, the quasispecies is now mainly
composed of type 2 with very few type 1.
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Let's see what it looks like
in the eigenvalue.
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Again, you see there's two eigenvectors
with two corresponding eigenvalues -
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the largest eigenvalue is this 1.01.
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And, it has both type 1 and type 2
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where type 2 is present
at a higher frequency.
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If we increase the mutation rate
even further -
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let's go to...
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0.2...
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Now, you see.. that both types
don't increase - they decrease.
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Type 1 decreases exponentially
at a faster rate
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and type two decreases exponentially
probably with a rate of 0.95.
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If we now look at the eigenvalues,
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you see again there's two eigenvalues
and two eigenvectors.
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The larger... eigenvalue this time
is the 0.95.
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It's corresponding eigenvector is this one
that has only type 2 present
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and not type 1.
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The other eigenvector
has a negative component of type 2,
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so it can't actually exist.
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This crossing where type 1 is lost
and type 2 is the only one
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that is present in the population
at a high enough error rate
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is what we call the "error catastrophe."
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Let's look at this a bit further.
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So now, I will want to plot
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instead of going manually
through different values of S.
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Let's just go through
a large sequence of S
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and plot what happens with them.
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So, here's what we get.
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On the x-axis...
actually, I'm on modified mu,
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so the x-axis represents mu -
the mutation rate.
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And, the y-axis...
is the largest eigenvalue.
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So this is the growth rate
of the quasispecies - of the types -
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that grow to the highest
of the various rates.
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So you see when S is very low
we have a growth rate very close to 1.1,
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and the type that has both type 1
and type 2 increases exponentially.
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This eigenvector, as mu increases,
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goes lower and lower,
until at this point it actually crosses
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and becomes lower
than the other eigenvalue -
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the eigenvalue with a growth rate
of 0.95.
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At this point, type 1 will disappear
from the population
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and only type 2 will be present.
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So, this transition is what we call
the "error catastrophe" -
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the mutation rate is too high.
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In the mutation matrix that I had before,
I only let type 1 mutate.
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So, type 1 mutated into type 2
but type 2 did not mutate back to type 1.
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This is where the error catastrophe
is strongest.
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If we allowed back mutation -
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so, here I'm now working with a matrix
where type 2 can mutate to type 1
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with a low rate of mu
divided by a hundred.
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Let's see what happens then.
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I plot... oops!
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I plot exactly the same...
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I plot exactly the same as I did before,
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and now you see
that the two eigenvalues don't cross.
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This eigenvalue decreases
and then becomes almost 0.95,
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and this eigenvalue stays at 0.95
and then decreases almost linearly,
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but they never cross.
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The reason is that type 1 is still present
in the population -
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it's simply... generated by mutations
from type 2.
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So... there is not a full crossing,
and therefore,
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it's not a real phase transition.
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Let's derive this error catastrophe
when it happens.
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So, we have now M
without the back mutation
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one minus mu mu,
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and W with a growth rate
of w for type 1.
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We multiply the two matrices
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and this is the result
of the multiplication.
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So, we simply take 1 minus 2 mu
times W,
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mu times 0 and so on.
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This is the result.
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Now, in order to calculate an eigenvector,
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we simply multiply this by a vector,
1 alpha,
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and we want to calculate when
multiplying this vector times this matrix
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will give us a vector
that has exactly the same ratios.
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This will be an eigenvector of this -
of the matrix W times M.
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In order... for it to stay
at the same ratio,
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the ratio between this and this
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has to be like the ratio
between 1 and alpha.
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Therefore, mu plus alpha divided by 1
minus mu times w
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has to be equal to alpha.
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We can simply do a bit of algebra
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and then we get that alpha
has to be equal to mu
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divided by w times 1 minus mu
minus 1.
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So, in order for alpha to be positive
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in order for there to be an eigenvalue
where both types can be represented,
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or both type 1 and type 2 are present,
this needs to be positive.
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For this to be positive,
the denominator needs to be positive,
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which means that w times 1 minus mu
has to be bigger than 1.
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Another way to write it
is to write w as 1 plus s
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where s is the growth rate beyond 1.
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And then, we need to have 1 plus s
times 1 minus mu is bigger than 1,
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which means that for small s
and small mu,
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s has to be approximately
bigger than mu.
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So, the error threshold happens
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when the growth rate is bigger than
the mutation rate
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or when the mutation rate is low enough.
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Let's switch now to a representation
of the error catastrophe
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as it was originally introduced
by Manfred Eigen and Peter Schuster
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where they looked at...
instead of just looking at two types,
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they looked at the genome of length L,
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which means that they had four
to the L types.
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And, I assume for simplicity
that there is one optimal sequence:
239
00:14:17,000 --> 00:14:19,540
A-A-G-C and so on.
240
00:14:20,560 --> 00:14:22,570
But, I would like to be able to talk about
241
00:14:22,720 --> 00:14:25,620
just a binary sequence
instead of four bases.
242
00:14:25,820 --> 00:14:27,900
So, let's look at the binary genome
243
00:14:28,080 --> 00:14:31,240
in which 1 means that the sequence
is identical to the optimum
244
00:14:31,380 --> 00:14:33,900
and zero means that it's different
from the optimum.
245
00:14:34,520 --> 00:14:38,160
Let's look at the per site mutation rate
instead of a global rate.
246
00:14:38,340 --> 00:14:41,560
So, nu is now the mutation rate per site.
247
00:14:41,820 --> 00:14:44,400
And, just for simplicity,
I assume it's the same
248
00:14:44,560 --> 00:14:47,040
to go from optimum to non-optimum
and back.
249
00:14:47,740 --> 00:14:50,240
So, now we look just at ones and zeros,
250
00:14:50,380 --> 00:14:52,100
and... we assume
251
00:14:52,260 --> 00:14:54,430
that there's just one mutation
per time step.
252
00:14:54,430 --> 00:14:56,780
So, nu is low enough in L
versus L
253
00:14:56,940 --> 00:15:00,260
so that there's only one... mutation
per time step.
254
00:15:00,640 --> 00:15:02,980
We'll call i the number of zeros,
255
00:15:03,140 --> 00:15:04,760
and we'll simply look at the types
256
00:15:04,920 --> 00:15:07,980
as the number of zeros
that... we have in the sequence.
257
00:15:08,900 --> 00:15:13,480
So, i represents number of zeros;
L minus i the number of ones.
258
00:15:13,760 --> 00:15:18,680
Now we can calculate the probability
to go from i zeros to i plus 1.
259
00:15:18,840 --> 00:15:21,460
In order for that to happen,
a mutation needs to happen,
260
00:15:21,460 --> 00:15:23,260
so it's L times nu.
261
00:15:23,440 --> 00:15:27,700
And then, the mutation needs to hit a 1
and mutate it into a 0.
262
00:15:27,880 --> 00:15:31,440
So, it's L minus i times L...
times L nu,
263
00:15:31,600 --> 00:15:33,880
which gives us simply
L minus i times nu,
264
00:15:34,020 --> 00:15:38,280
and similar for decreasing one...
the number of zeros.
265
00:15:40,020 --> 00:15:41,660
Let's see what it does.
266
00:15:41,840 --> 00:15:44,900
So, we have a nu of 1 percent
267
00:15:45,060 --> 00:15:47,940
and a length of L,
L equals 10.
268
00:15:49,840 --> 00:15:52,180
And now, we need to build
the mutation matrix.
269
00:15:52,360 --> 00:15:55,600
We assume that we can only mutate
one up or one down,
270
00:15:55,780 --> 00:15:59,360
and I simply take the expression
that we just calculated.
271
00:16:00,410 --> 00:16:02,080
This gives us the following matrix.
272
00:16:02,080 --> 00:16:04,460
You can see that the matrix
is different from 0
273
00:16:04,640 --> 00:16:07,540
only along the diagonal
and one step away
274
00:16:07,720 --> 00:16:09,680
because we only allow one mutation.
275
00:16:10,520 --> 00:16:15,220
Now we can also build
the... growth rate matrix.
276
00:16:15,380 --> 00:16:18,840
It's diagonal and only has one entry
that's different from 1.
277
00:16:19,140 --> 00:16:23,180
So, I take a growth rate of 1.3,
which is for type 1.
278
00:16:25,400 --> 00:16:28,920
Now, we can look at
the maximal eigenvalue -
279
00:16:29,100 --> 00:16:34,740
the eigenvector with the maximum
eigenvalue for this matrix W times M,
280
00:16:35,300 --> 00:16:39,500
and we see that type 1 is represented
at the highest frequency
281
00:16:39,680 --> 00:16:41,780
and other types also represented.
282
00:16:41,960 --> 00:16:44,300
So, this would be the quasispecies
in this case.
283
00:16:44,500 --> 00:16:47,360
The population would be represented
by the optimal type -
284
00:16:47,520 --> 00:16:50,540
types with one mutation,
two mutations and so on.
285
00:16:50,860 --> 00:16:53,920
What happens now
when we increase the mutation rate?
286
00:16:55,120 --> 00:16:59,540
Sorry, when we decrease
the selection pressure.
287
00:17:01,740 --> 00:17:03,200
Let's go now to...
288
00:17:05,960 --> 00:17:10,900
a benefit of 0.08,
so a growth rate of 1.08.
289
00:17:13,240 --> 00:17:15,099
You can see that now
290
00:17:15,280 --> 00:17:19,040
the optimal type is represented
with quite a low frequency
291
00:17:19,200 --> 00:17:22,220
and all the other types
are represented quite highly.
292
00:17:22,619 --> 00:17:25,240
In order to get a better overview
of what happens,
293
00:17:25,400 --> 00:17:28,079
let's look at various ss.
294
00:17:28,260 --> 00:17:30,500
So, we have plotted four different ss.
295
00:17:30,660 --> 00:17:32,530
At sequence 0.3 - like we had before -
296
00:17:32,530 --> 00:17:36,020
you see that type 1
has the highest representation,
297
00:17:36,200 --> 00:17:38,640
one mutation away a bit lower and so on.
298
00:17:38,800 --> 00:17:41,760
s equal to 0.2 -
it spreads out a bit more.
299
00:17:41,940 --> 00:17:45,680
s equal to 0.1 - it's very far spread.
300
00:17:45,860 --> 00:17:50,440
At 0.09, you see that type 1 is already
at a fairly low frequency.
301
00:17:50,620 --> 00:17:53,490
And, the highest type is the type
that you would get mostly
302
00:17:53,490 --> 00:17:59,580
if you did just a random coin flip
with five ones and five zeros.
303
00:18:00,520 --> 00:18:03,980
And then, as s goes to 0.07,
you see that type 1
304
00:18:04,160 --> 00:18:09,240
is almost totally non-present,
and only this type is represented.
305
00:18:10,260 --> 00:18:11,980
We can also represent it as a graph.
306
00:18:12,140 --> 00:18:14,880
In a graph,
here on the x-axis, is s.
307
00:18:15,040 --> 00:18:17,200
s goes from 0 to 0.8,
308
00:18:17,440 --> 00:18:20,920
and the y-axis is the frequency
of the various types.
309
00:18:22,120 --> 00:18:24,200
Black represents the optimal type -
310
00:18:24,680 --> 00:18:27,060
this is the type with one mutation,
two mutations,
311
00:18:27,220 --> 00:18:28,560
three mutations and so on.
312
00:18:28,720 --> 00:18:33,280
You see that when the benefit is 0.8 -
so growth rate is 1.8 -
313
00:18:33,600 --> 00:18:37,060
type 1 is represented
at the highest frequency.
314
00:18:37,240 --> 00:18:38,860
As s goes down,
315
00:18:39,060 --> 00:18:43,840
the representation of type 1 decreases
and of the other types increases,
316
00:18:44,020 --> 00:18:46,280
until we reach this threshold
317
00:18:46,460 --> 00:18:50,260
where type 1 almost disappears
from the population
318
00:18:50,440 --> 00:18:55,260
and the highest representative type
is the type that has five mutations.
319
00:18:55,420 --> 00:18:59,760
This is simply at this point...
we simply look at coin flips.
320
00:18:59,920 --> 00:19:05,300
So, then the selection...
doesn't affect the population anymore.
321
00:19:06,520 --> 00:19:09,020
Let's calculate
where this transition happens.
322
00:19:10,540 --> 00:19:13,620
So, we said that the chance to go
from i to i plus 1
323
00:19:13,760 --> 00:19:15,620
is L minus i times nu.
324
00:19:16,360 --> 00:19:19,420
Now, we want to ask -
can we maintain the zero type?
325
00:19:19,600 --> 00:19:21,360
So, when i is equal to 0.
326
00:19:21,520 --> 00:19:24,860
We want to maintain zero type
versus going from 0 to 1.
327
00:19:25,060 --> 00:19:28,100
Therefore, we look at just
a transition rate from 0 to 1.
328
00:19:28,280 --> 00:19:30,460
That rate is L times nu.
329
00:19:30,660 --> 00:19:36,140
An imitation will increase
the number of... zeros.
330
00:19:38,110 --> 00:19:41,200
So, what is now the error catastrophe -
the global mutation rate -
331
00:19:41,200 --> 00:19:43,240
not precise,
but for the whole genome now
332
00:19:43,400 --> 00:19:46,460
to go from 0 to 1,
is L times nu.
333
00:19:46,700 --> 00:19:50,440
And now, we will therefore have
the same error threshold as before.
334
00:19:50,600 --> 00:19:54,340
s needs to be bigger than mu,
which is L times nu.
335
00:19:54,940 --> 00:19:56,610
So, the error threshold will happen
336
00:19:56,610 --> 00:19:59,940
when s is bigger than L
times the per site mutation rate.
337
00:20:00,320 --> 00:20:02,260
This is a very nice expression.
338
00:20:03,260 --> 00:20:08,440
Eigen and Schuster used it to calculate
an error threshold for
339
00:20:08,880 --> 00:20:11,740
how good can a molecule replicate.
340
00:20:11,920 --> 00:20:16,780
So, if a molecule has a replication of...
with a fidelity of one percent,
341
00:20:17,140 --> 00:20:19,440
it cannot code for this fidelity
of one percent
342
00:20:19,580 --> 00:20:22,820
with a molecule that's larger than
100 bases.
343
00:20:23,260 --> 00:20:24,500
If it has one per mille,
344
00:20:24,500 --> 00:20:28,580
it can't code it with a molecule
larger than a thousand bases.
345
00:20:30,260 --> 00:20:32,640
This shows that in order to...
you need to cross
346
00:20:32,640 --> 00:20:36,040
the threshold of fidelity
with a short enough molecule.
347
00:20:37,440 --> 00:20:39,300
This expression can also represent
348
00:20:39,480 --> 00:20:42,560
what it means to be neutral
versus non-neutral.
349
00:20:42,740 --> 00:20:46,100
How big does the difference have to be
between one type and another
350
00:20:46,740 --> 00:20:49,270
for evolution to be able to maintain
this difference?
351
00:20:50,120 --> 00:20:51,310
In order to look at this,
352
00:20:51,310 --> 00:20:54,520
let's locate... the population
that has not just a difference
353
00:20:54,680 --> 00:20:56,760
between the best type and all the others,
354
00:20:56,940 --> 00:20:59,780
but instead there's a...
355
00:21:00,620 --> 00:21:03,680
a stairs between one type and the next.
356
00:21:05,020 --> 00:21:08,660
So, the highest type has zero mutations,
357
00:21:08,820 --> 00:21:11,400
then the next highest growth rate
is one mutation,
358
00:21:11,560 --> 00:21:14,720
two mutations and so on,
with some kind of...
359
00:21:16,920 --> 00:21:18,900
randomly chosen growth rates.
360
00:21:19,700 --> 00:21:23,040
The difference - this threshold
between type 0 and type 1
361
00:21:23,200 --> 00:21:24,680
or between type 1 and type 2 -
362
00:21:24,680 --> 00:21:28,340
I've called Delta s,
the change in growth rate.
363
00:21:29,000 --> 00:21:34,620
And now, we need to be able to maintain
this difference versus the mutation.
364
00:21:35,800 --> 00:21:41,920
If we look at the change in growth rate
from i to i plus 1 if that is Delta s i,
365
00:21:42,880 --> 00:21:43,960
we want to ask -
366
00:21:44,140 --> 00:21:47,420
when can this be maintained
versus the mutation?
367
00:21:47,620 --> 00:21:49,720
And, what needs to happen is that
368
00:21:49,900 --> 00:21:53,720
Delta s i needs to be bigger
than L minus i,
369
00:21:53,900 --> 00:21:59,920
the chance to go from i to i minus 1...
from i to i plus 1 times nu.
370
00:22:03,180 --> 00:22:05,300
So, we can take a random example.
371
00:22:05,600 --> 00:22:10,240
Here I simply choose random ss
with random...
372
00:22:11,600 --> 00:22:12,820
steps,
373
00:22:13,000 --> 00:22:16,720
so the x-axis in black
represents the Delta s i,
374
00:22:16,880 --> 00:22:18,400
the step from...
375
00:22:19,180 --> 00:22:22,700
the change in growth rate
from i to i plus 1.
376
00:22:22,880 --> 00:22:26,540
And, the red line represents
L minus i times nu.
377
00:22:26,700 --> 00:22:28,880
This is the effect of the mutations.
378
00:22:29,060 --> 00:22:33,060
In order to be able to maintain
a certain number of mutations,
379
00:22:33,240 --> 00:22:35,330
the black line has to be
above the red line.
380
00:22:35,330 --> 00:22:39,980
So, this means that this population
can maintain this point
381
00:22:41,020 --> 00:22:42,760
versus the effect of mutation,
382
00:22:42,920 --> 00:22:45,340
whereas it cannot maintain these points.
383
00:22:45,960 --> 00:22:49,800
So, the difference between this and this,
as far as this population is concerned,
384
00:22:49,960 --> 00:22:51,260
is neutral.
385
00:22:51,440 --> 00:22:53,200
This point can be...
386
00:22:55,560 --> 00:22:57,360
can be taken...can be maintained,
387
00:22:57,540 --> 00:22:59,590
but this point again cannot be maintained.
388
00:22:59,590 --> 00:23:02,480
So, we can see that the mutation rate
389
00:23:02,640 --> 00:23:06,100
defines what is seen as neutral
versus non-neutral.
390
00:23:06,260 --> 00:23:10,840
And, the distribution in a population
has to be above a narrow threshold,
391
00:23:11,080 --> 00:23:13,740
which can be crossed -
as you see here - several times.
392
00:23:14,660 --> 00:23:15,780
Thank you.